f

Most Natural Numbers Are Uncomputable

```I prove most natural numbers are so large
even an "infinitely" fast computer can not

First, I will describe a simple set theory
problem and then I show how this problem
is related to the computability of natural
numbers.

I want to create a family of sets.
The first member of this family is
the set of all natural numbers, N.
Each member after the first is created
by removing the least element of
the previous set.

A0 = {0,1,2,3,...}
A1 = {1,2,3,4,...}
A2 = (2,3,4,5,...}
....

Is the empty set a member of this
family of sets?

Assume the empty set is a member.
Then there exists another member
with exactly one element. Let this
set be {z}. z must be the largest
natural number.

Since there is no largest natural number,
the empty set can not be a member of this
family of sets.

This proves it is impossible to
remove an infinite number of elements
from a set if we only remove one element
at a time.

Assume I have an infinitely fast computer,
like a Zeno Machine (ZM).
http://en.wikipedia.org/wiki/Zeno_machine

A ZM is a type of Accelerate Turing Machine.
A ZM performs the first operation in 1/2
second, the 2nd operation in 1/4 second,
and the nth operation in 1/(2^n) second.
A ZM can perform an infinite number of
operations in one second.
(The time required to perform these
operations is irrelevant for this proof.)

Like a Turing Machine, a ZM reads and writes
one character at a time from a tape.

Assume I have an infinitely long blank tape.
(Later, I will show that I only need to assume
the tape is finite, but unbounded.)

I want to test if this tape is actually infinitely long.

Assume the ZM will generate an error
if it runs out of tape.
Running out of tape would prove the
input tape was finite.

I also want to be sure that every finite
position on the tape is blank.
To do so, I create a simple three state
Turing Machine.

State 1:
On any input other than blank generate an
error and halt.
On blank input move tape head one posiiton
to the right and switch to state 2.

State 2:
On any input other than blank generate an
error and halt.
On blank input move tape head one posiiton
to the left and switch to state 3.

State 3:
On any input other than blank generate an
error and halt.
On blank input write a '1', move tape head
one posiiton to the right, and switch to state 1.

This program can never overwrite the last
blank space on the tape. Of course,
if the tape is infinitely long, there
will never be a "last" blank position.

Now, I run my program on the ZM for an
hour or two. I want to be sure the ZM
has performed an "infinite" number of
operations.

If the input tape is finite, the ZM
will halt with an error. Assume the
ZM hasn't halted after an hour.

What is on the tape after we halt the ZM?

Consider my set theory problem.
The first set in the family, N,
represents the positions of blank
spaces on my input tape when I
start the ZM.

After each '1' written by the ZM,
the next set in the family represents
the remaining blank positions on the
tape.

There must still be an infinite number
of blank spaces on the input tape.
If the ZM had overwritten every position
on the tape, the empty set would be a
member of my family of sets.

Since each blank is at a finite position,
there must exist a smallest blank position.

Even though the ZM is "infinitely"
fast it has only written a finite
number of 1's. (less than the smallest
blank position).

This proves nearly all natural numbers
are too large to be read or written by
any sequential computer.

Notice that I still don't know if the
input tape was infinitely long. There
are an infinite number of finite tapes
too long to be read by the ZM.

According to Wikipedia:
http://en.wikipedia.org/wiki/Decidable_language

"A recursive language is a formal language for which there exists a
Turing machine which will, when presented with any input string, halt
and accept if the string is in the language, and halt and reject
otherwise. The Turing machine always halts; it is known as a decider
and is said to decide the recursive language."

The proof above shows the language of unary
representations of natural numbers is not
recursive.

A TM can certainly recognize "small" natural
numbers, but "large" natural numbers
are indistinguishable from an infinite string.

Even an infinitely fast Turing Machine can not
decide if an input tape contains a unary
representation of a natural number.

Normally, we assume a TM can read any finite
string if we wait "long enough".

My proof shows this assumption is false.
Even if we have an infinitely fast computer
and wait forever, it is impossible for
any sequential computer to read most
finite strings.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/15/2009 5:05:13 AM
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```On 2009-07-15, RussellE <reasterly@gmail.com> wrote:
> Is the empty set a member of this
> family of sets?

Since every set in your family is indexed by a natural number, no.

> This proves it is impossible to remove an infinite number of
> elements from a set if we only remove one element at a time.

Non sequitur.  All it proves is that removing any finite number of
elements from an infinite set leaves an infinite set.

> Assume I have an infinitely fast computer, like a Zeno Machine (ZM).

A dubious assumption to begin with.  Most logically coherent models of
"Zeno machine" have conditions on their "final" state dependent upon
whether the sequence of computation reached an eventually static state
or not.  Without any such conditions, the machine is ill-specified.

> State 1:
[etc]

Yet another example of an ill-specified Zeno machine.  As the rest of

- Tim
```
 0
tim669 (185)
7/15/2009 6:17:26 AM
```On Tue, 14 Jul 2009, RussellE wrote:

> I want to create a family of sets.
> The first member of this family is
> the set of all natural numbers, N.

> Each member after the first is created
> by removing the least element of
> the previous set.
>
> A0 = {0,1,2,3,...}
> A1 = {1,2,3,4,...}
> A2 = (2,3,4,5,...}
> ...
>
> Is the empty set a member of this family of sets?
>
Never.  Exercise.  For all n in N, n + 1 in A_n.

> Assume the empty set is a member.

which anything coherent can be proven.

> Then there exists another member
> with exactly one element. Let this
> set be {z}. z must be the largest
> natural number.
>
> Since there is no largest natural number,
> the empty set can not be a member of this
> family of sets.
>
> This proves it is impossible to
> remove an infinite number of elements
> from a set if we only remove one element
> at a time.
>
> Assume I have an infinitely fast computer,
> like a Zeno Machine (ZM).
> http://en.wikipedia.org/wiki/Zeno_machine
>
> A ZM is a type of Accelerate Turing Machine.
> A ZM performs the first operation in 1/2
> second, the 2nd operation in 1/4 second,
> and the nth operation in 1/(2^n) second.
> A ZM can perform an infinite number of
> operations in one second.
> (The time required to perform these
> operations is irrelevant for this proof.)
>
> Like a Turing Machine, a ZM reads and writes
> one character at a time from a tape.
>
> Assume I have an infinitely long blank tape.
> (Later, I will show that I only need to assume
> the tape is finite, but unbounded.)
>
> I want to test if this tape is actually infinitely long.
>
> Assume the ZM will generate an error
> if it runs out of tape.
> Running out of tape would prove the
> input tape was finite.
>
> I also want to be sure that every finite
> position on the tape is blank.
> To do so, I create a simple three state
> Turing Machine.
>
> State 1:
> On any input other than blank generate an
> error and halt.
> On blank input move tape head one posiiton
> to the right and switch to state 2.
>
> State 2:
> On any input other than blank generate an
> error and halt.
> On blank input move tape head one posiiton
> to the left and switch to state 3.
>
> State 3:
> On any input other than blank generate an
> error and halt.
> On blank input write a '1', move tape head
> one posiiton to the right, and switch to state 1.
>
> This program can never overwrite the last
> blank space on the tape. Of course,
> if the tape is infinitely long, there
> will never be a "last" blank position.
>
> Now, I run my program on the ZM for an
> hour or two. I want to be sure the ZM
> has performed an "infinite" number of
> operations.
>
> If the input tape is finite, the ZM
> will halt with an error. Assume the
> ZM hasn't halted after an hour.
>
> What is on the tape after we halt the ZM?
>
> Consider my set theory problem.
> The first set in the family, N,
> represents the positions of blank
> spaces on my input tape when I
> start the ZM.
>
> After each '1' written by the ZM,
> the next set in the family represents
> the remaining blank positions on the
> tape.
>
> There must still be an infinite number
> of blank spaces on the input tape.
> If the ZM had overwritten every position
> on the tape, the empty set would be a
> member of my family of sets.
>
> Since each blank is at a finite position,
> there must exist a smallest blank position.
>
> Even though the ZM is "infinitely"
> fast it has only written a finite
> number of 1's. (less than the smallest
> blank position).
>
> This proves nearly all natural numbers
> are too large to be read or written by
> any sequential computer.
>
> Notice that I still don't know if the
> input tape was infinitely long. There
> are an infinite number of finite tapes
> too long to be read by the ZM.
>
> According to Wikipedia:
> http://en.wikipedia.org/wiki/Decidable_language
>
> "A recursive language is a formal language for which there exists a
> Turing machine which will, when presented with any input string, halt
> and accept if the string is in the language, and halt and reject
> otherwise. The Turing machine always halts; it is known as a decider
> and is said to decide the recursive language."
>
> The proof above shows the language of unary
> representations of natural numbers is not
> recursive.
>
> A TM can certainly recognize "small" natural
> numbers, but "large" natural numbers
> are indistinguishable from an infinite string.
>
> Even an infinitely fast Turing Machine can not
> decide if an input tape contains a unary
> representation of a natural number.
>
> Normally, we assume a TM can read any finite
> string if we wait "long enough".
>
> My proof shows this assumption is false.
> Even if we have an infinitely fast computer
> and wait forever, it is impossible for
> any sequential computer to read most
> finite strings.
>
>
> Russell
> - 2 many 2 count
>
```
 0
marsh6245 (32)
7/15/2009 7:24:46 AM
```On Jul 14, 11:17=A0pm, Tim Little <t...@little-possums.net> wrote:
> On 2009-07-15, RussellE <reaste...@gmail.com> wrote:
>
> > Is the empty set a member of this
> > family of sets?
>
> Since every set in your family is indexed by a natural number, no.
>
> > This proves it is impossible to remove an infinite number of
> > elements from a set if we only remove one element at a time.
>
> Non sequitur. =A0All it proves is that removing any finite number of
> elements from an infinite set leaves an infinite set.
>
> > Assume I have an infinitely fast computer, like a Zeno Machine (ZM).
>
> A dubious assumption to begin with. =A0Most logically coherent models of
> "Zeno machine" have conditions on their "final" state dependent upon
> whether the sequence of computation reached an eventually static state
> or not. =A0Without any such conditions, the machine is ill-specified.

The final state of the ZM is well defined.
The output tape will contain a finite string of 1's
followed by a blank. No other final state is possible.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/15/2009 9:02:02 AM
```Zeno machines are not Turing machines, so you haven't said anything

But your post seems to be saying that if you count off (remove) all
the naturals from the naturals you are left with the empty set. N - N
= {}. Groundbreaking!

```
 0
hobkesh (5)
7/16/2009 2:31:23 AM
```Musatov wrote:
ab wrote:
> Zeno machines are not Turing machines, so you haven't said anything
>
> But your post seems to be saying that if you count off (remove) all
> the naturals from the naturals you are left with the empty set. N - N
> = {}. Groundbreaking!

Not so fast trivial injector!

The sequence (xs) converges h-bounded effectively to some x, hence x
is o(2n)- bounded computable.Proof. For any natural number n.

--
Musatov
```
 0
marty.musatov (1143)
7/16/2009 3:30:20 AM
```On Jul 15, 7:31=A0pm, ab <hobk...@gmail.com> wrote:

> Zeno machines are not Turing machines, so you haven't said anything

Zeno machines are Accelerated Turing Machines.
You are correct that Zeno machine differs from a TM.
Even so, I show the Zeno machine is no more powerful than
a TM when it comes to recognizing natural numbers.
Neither can recognize large natural numbers.

> But your post seems to be saying that if you count off (remove) all
> the naturals from the naturals you are left with the empty set. N - N
> =3D {}. Groundbreaking!

I show it is impossible to remove all the naturals
from the set of all naturals if you are removing
each natural one at a time.
If you could there would be a last natural.

From this it follows that it is impossible for a
TM to read all the positions on an infinite tape
if it reads the tape one position at a time.
If a TM can't read an infinite tape then
there exists finite tapes it can't read either.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/16/2009 4:32:04 AM
```RussellE <reasterly@gmail.com> writes:

> On Jul 15, 7:31 pm, ab <hobk...@gmail.com> wrote:
>
>> Zeno machines are not Turing machines, so you haven't said anything
>
> Zeno machines are Accelerated Turing Machines.
> You are correct that Zeno machine differs from a TM.
> Even so, I show the Zeno machine is no more powerful than
> a TM when it comes to recognizing natural numbers.
> Neither can recognize large natural numbers.

You don't define this notion of recognising natural numbers.  A TM can
be written to recognise any given natural number so you can't mean
that.  If you mean subsets of N, then that would appear to be false
since Zeno machines can recognise some sets that TM's can't (the
classic example being the set of encodings of halting TMs).

>> But your post seems to be saying that if you count off (remove) all
>> the naturals from the naturals you are left with the empty set. N - N
>> = {}. Groundbreaking!
>
> I show it is impossible to remove all the naturals
> from the set of all naturals if you are removing
> each natural one at a time.
> If you could there would be a last natural.
>
> From this it follows that it is impossible for a
> TM to read all the positions on an infinite tape
> if it reads the tape one position at a time.

This is either trivially true or trivially false depending on how you
define your terms.  You are using a paradox inherent in the definition
of a Zeno machine to talk about "all the positions" as if is was well
defined for TMs.

> If a TM can't read an infinite tape then
> there exists finite tapes it can't read either.

How does that follow?  Most the terms are open to more than one
definition, so I am having to guess what you mean by them but I can't
see anything in your argument that supports this step no matter how I
define "read a tape", "infinite tape" and "finite tape".

--
Ben.
```
 0
ben.usenet (6790)
7/16/2009 2:31:43 PM
```"RussellE" <reasterly@gmail.com> wrote in message
On Jul 15, 7:31 pm, ab <hobk...@gmail.com> wrote:

> Zeno machines are not Turing machines, so you haven't said anything

Zeno machines are Accelerated Turing Machines.

***************************
There is another old paradox that applies to machines of this type.

Imagine at t =1 we add 100 numbered balls 1 to 100 to an urn, but remove
ball number 1.  At t = 1/2 we add balls 101-200 but remove ball  number 2.
At t=1/4 we add balls 201-300, remove ball 3, etc.

How many balls are there at t=0 ?

Well, none. Because clearly ball number n was removed at step n for all n.

But equally clearly, at every step you are adding 99 balls in total, so how
can it be zero?

```
 0
webbfamily (36)
7/16/2009 2:38:55 PM
```On Jul 16, 7:31=A0am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:
> > On Jul 15, 7:31=A0pm, ab <hobk...@gmail.com> wrote:

> >> Zeno machines are not Turing machines, so you haven't said anything
>
> > Zeno machines are Accelerated Turing Machines.
> > You are correct that Zeno machine differs from a TM.
> > Even so, I show the Zeno machine is no more powerful than
> > a TM when it comes to recognizing natural numbers.
> > Neither can recognize large natural numbers.
>
> You don't define this notion of recognising natural numbers. =A0A TM can
> be written to recognise any given natural number so you can't mean
> that.

That is exactly what I mean. There exists finite strings too large
to be read by any Turing Machine.

I used the language of unary representations of naturals
in my example. A unary representation is a finite string
of 1's followed by a blank. The exists unary representations
of natural numbers too large to be read by any TM.

> =A0If you mean subsets of N, then that would appear to be false
> since Zeno machines can recognise some sets that TM's can't (the
> classic example being the set of encodings of halting TMs).

I helped derive that proof in this newsgroup several years ago.
Unfortunately, the Halting proof is wrong. It assumes a Zeno Machine
can perform an infinite number of operations. This proof shows
a ZM can only perform a finite number of operations.

> >> But your post seems to be saying that if you count off (remove) all
> >> the naturals from the naturals you are left with the empty set. N - N
> >> =3D {}. Groundbreaking!
>
> > I show it is impossible to remove all the naturals
> > from the set of all naturals if you are removing
> > each natural one at a time.
> > If you could there would be a last natural.
>
> > From this it follows that it is impossible for a
> > TM to read all the positions on an infinite tape
> > if it reads the tape one position at a time.
>
> This is either trivially true or trivially false depending on how you
> define your terms. =A0You are using a paradox inherent in the definition
> of a Zeno machine to talk about "all the positions" as if is was well
> defined for TMs.

I start by assuming a ZM can perform an infinite number of
operations. Then I show the ZM can never perform more
that a finite number of read or writes.

My proof is little more than a proof that all natural numbers
are finite. Each operation performed by the ZM (or any TM)
can be assigned a natural number. There is a step 1, a step 2,
etc. There can never be a step infinity.

If the ZM halts or is halted, I can show it has performed
a finite number of operations.

The only way the ZM can halt on its own while running
my program would be if the input tape was finite length.
Assuming the input tape is infinite, the ZM will never halt
unless I halt it. This is true despite the fact I assumed
the ZM could perform an infinite number of operations
in one second.

> > If a TM can't read an infinite tape then
> > there exists finite tapes it can't read either.
>
> How does that follow? =A0Most the terms are open to more than one
> definition, so I am having to guess what you mean by them but I can't
> see anything in your argument that supports this step no matter how I
> define "read a tape", "infinite tape" and "finite tape".

I show that even a ZM can only read or write a finite number
of positions. Assume I run the ZM for an hour and then halt it.
The tape will contain a finite string of 1's followed by a blank.
Obviously, there exists finite strings of 1's longer than
the finite string written by the ZM.

The ZM writes a 1 on every position it reads, so the string
of 1's also tells us how many positions were read by the ZM.

No TM can write a string longer than the one written by the ZM.
I am already assuming the ZM can perform an infinite number
of operations. Assuming a TM can write an uncountable number
of 1's wouldn't change the proof.

I think of the string written by the ZM as a "limit" to the
length of strings any TM can read or write.
Since there exists finite strings the ZM can't read,
there must be finite strings too large for any TM to read.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/16/2009 10:33:44 PM
```On Jul 16, 3:33=A0pm, RussellE <reaste...@gmail.com> wrote:

> There exists finite strings too large
> to be read by any Turing Machine.

You've not proven any such thing.

> My proof is little more than a proof that all natural numbers
> are finite.

We already know that all natural numbers are finite.

MoeBlee
```
 0
jazzmobe (307)
7/16/2009 11:09:51 PM
```On Jul 16, 4:09=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 16, 3:33=A0pm, RussellE <reaste...@gmail.com> wrote:

>
> We already know that all natural numbers are finite.
>

Then it shouldn't be hard to see why any sequential set
of operations that terminates must be finite.
I can assign a natural number to each operation and
there is a "last" operation when the TM halts or is halted.

This must be true even if we assume the TM can perform
an infinite number of operations in finite time.

The standard definition of a TM assumes the input tape is finite.
Somehow, a human has to decide if the input is finite.
I don't know how any human could do this, but I can
prove a TM can not decide for all strings if the
string is finite or infinite.

If a TM can't distinguish all finite strings from infinite strings
there must exist finite strings the TM can't recognise.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/17/2009 12:25:35 AM
```On Jul 16, 8:25=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 16, 4:09=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 16, 3:33=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > We already know that all natural numbers are finite.
>
> Then it shouldn't be hard to see why any sequential set
> of operations that terminates must be finite.
> I can assign a natural number to each operation and
> there is a "last" operation when the TM halts or is halted.
>
> This must be true even if we assume the TM can perform
> an infinite number of operations in finite time.
>
> The standard definition of a TM assumes the input tape is finite.
> Somehow, a human has to decide if the input is finite.
> I don't know how any human could do this, but I can
> prove a TM can not decide for all strings if the
> string is finite or infinite.
>
> If a TM can't distinguish all finite strings from infinite strings
> there must exist finite strings the TM can't recognise.
>
> Russell
> - 2 many 2 count

Hi, Russell:

You are being very silly.

Of course a Turing Machine cannot properly
recognize that an infinite string/input is
finite.

This does not mean a Turing Machine cannot
recognize that a finite string/input is
finite.

One typical arrangement is that the input
consisting of 0's and 1's (or other input
symbols in consecutive order) is terminated
by a blank (but feel free to posit any other
distinguished symbol as EOT =3D end of tape).

regards, chip
```
 0
hardmath (81)
7/17/2009 12:36:48 AM
```On Jul 16, 5:25=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 16, 4:09=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 16, 3:33=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > We already know that all natural numbers are finite.
>
> Then it shouldn't be hard to see why any sequential set
> of operations that terminates must be finite.
> I can assign a natural number to each operation and
> there is a "last" operation when the TM halts or is halted.

Okay, as far as it goes.

> This must be true even if we assume the TM can perform
> an infinite number of operations in finite time.

Don't worry about that. TMs do not perform an infinite number of
operations in a finite number of steps (by the way, notice 'steps', or
'primitive computations', etc. rather than 'time').

> The standard definition of a TM assumes the input tape is finite.

Any given input is finite. There is no finite upper bound though on
the length of the input.

> Somehow, a human has to decide if the input is finite.

This does not depend on human insight. Turing machines may be
described as purely mathematical objects.

> I don't know how any human could do this, but I can
> prove a TM can not decide for all strings if the
> string is finite or infinite.

So what? Who claims that a Turing machine can make such a decision?

> If a TM can't distinguish all finite strings from infinite strings
> there must exist finite strings the TM can't recognise.

That's definitely The Non Sequitur Of The Week.

Anyway, your best bet would be to make arguments that don't turn on
vagueness held in such terms as "recognize", "time", etc. Better just
to prove actual theorems about Turing machines as they are
mathematically defined. I mean, do you even KNOW a mathematical
definition of 'Turing machine'?

MoeBlee

```
 0
jazzmobe (307)
7/17/2009 12:48:08 AM
```On Jul 16, 5:36=A0pm, Chip Eastham <hardm...@gmail.com> wrote:
> On Jul 16, 8:25=A0pm, RussellE <reaste...@gmail.com> wrote:

> > If a TM can't distinguish all finite strings from infinite strings
> > there must exist finite strings the TM can't recognise.

> This does not mean a Turing Machine cannot
> recognize that a finite string/input is
> finite.

Yes it does.

By definition, the input for a TM must be finite.
That is why I am using a Zeno Machine.
I start by assuming a ZM can read an infinitely long tape.

I will write a program to recognise unary representations
of natural numbers. A unary representation is a string
of 1's followed by a blank position.

If the ZM reads a character other than '1' or blank
it halts and rejects the string.

If the ZM reads a '1' it writes a '0' and moves the
head position one position to the right.

If the ZM reads a blank it halts and accepts the string.

I have a tape that contains a unary representation of
a natural number or it contains an infinitely long string of 1's.

I assume the ZM performs the first read in 1/2 second,
the next in 1/4 second, etc. The ZM performs an
infinite number of operations in 1 second.

Can the ZM determine if the tape contains a unary
representation of a natural number?

I start the ZM. If the ZM halts in less than 1 second
it will correctly tell us if the input was accepted or rejected.

Assume the ZM is still running after 1 second.
Remember, the ZM can't halt as long as it is reading 1's.
Can I now assume the input contains an infinite
string of 1's?

No, I can not. If I stop the ZM now, I will find the
input tape contains a finite string of 0's followed
by a '1'. This proves the ZM has not read every
position on the tape and I can not assume there
isn't a blank in one of the unread positions.

The ZM can not recognise all finite strings of '1'.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/17/2009 2:14:42 AM
```RussellE <reasterly@gmail.com> writes:

> On Jul 16, 7:31 am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > On Jul 15, 7:31 pm, ab <hobk...@gmail.com> wrote:
>
>> >> Zeno machines are not Turing machines, so you haven't said anything
>>
>> > Zeno machines are Accelerated Turing Machines.
>> > You are correct that Zeno machine differs from a TM.
>> > Even so, I show the Zeno machine is no more powerful than
>> > a TM when it comes to recognizing natural numbers.
>> > Neither can recognize large natural numbers.
>>
>> You don't define this notion of recognising natural numbers.  A TM can
>> be written to recognise any given natural number so you can't mean
>> that.
>
> That is exactly what I mean. There exists finite strings too large
> to be read by any Turing Machine.

At what point does the obvious machine that starts at the beginning
and halts at the end go wrong?

> I used the language of unary representations of naturals
> in my example. A unary representation is a finite string
> of 1's followed by a blank. The exists unary representations
> of natural numbers too large to be read by any TM.

See above.

>>  If you mean subsets of N, then that would appear to be false
>> since Zeno machines can recognise some sets that TM's can't (the
>> classic example being the set of encodings of halting TMs).
>
> I helped derive that proof in this newsgroup several years ago.
> Unfortunately, the Halting proof is wrong. It assumes a Zeno Machine
> can perform an infinite number of operations. This proof shows
> a ZM can only perform a finite number of operations.

Then a TM can emulate a ZM so why did you bother with at all?

<big snip -- sorry just could follow it>

>> > If a TM can't read an infinite tape then
>> > there exists finite tapes it can't read either.
>>
>> How does that follow?  Most the terms are open to more than one
>> definition, so I am having to guess what you mean by them but I can't
>> see anything in your argument that supports this step no matter how I
>> define "read a tape", "infinite tape" and "finite tape".
>
> I show that even a ZM can only read or write a finite number
> of positions. Assume I run the ZM for an hour and then halt it.
> The tape will contain a finite string of 1's followed by a blank.
> Obviously, there exists finite strings of 1's longer than
> the finite string written by the ZM.
>
> The ZM writes a 1 on every position it reads, so the string
> of 1's also tells us how many positions were read by the ZM.
>
> No TM can write a string longer than the one written by the ZM.
> I am already assuming the ZM can perform an infinite number
> of operations.

I thought you now know that to be false (not sure if this is by
definition or proof, but you said it was false above).

> Assuming a TM can write an uncountable number
> of 1's wouldn't change the proof.
>
> I think of the string written by the ZM as a "limit" to the
> length of strings any TM can read or write.
> Since there exists finite strings the ZM can't read,
> there must be finite strings too large for any TM to read.

You claim that a ZM can only read a finite part of the tape, but you
don't show that this finite part is bounded for all ZMs.  Why do you
call it a "limit"?

--
Ben.
```
 0
ben.usenet (6790)
7/17/2009 3:08:10 AM
```On Jul 16, 8:08=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

> > No TM can write a string longer than the one written by the ZM.
> > I am already assuming the ZM can perform an infinite number
> > of operations.

> I thought you now know that to be false (not sure if this is by
> definition or proof, but you said it was false above).

You are correct. I prove a TM or ZM can only read or write
a finite number of positions. So why do I use a ZM?

The standard definition of TM assumes the input is finite.
This simply avoids the entire issue of infinite strings.
It assumes a human has somehow decided in advance
that the input string is finite.

I am using ZM's to eliminate this assumption.
I start by assuming a ZM can read an infinite string.

Another common assumption is that a TM can
read any finite string if we just wait "long enough",
or we wait until the TM has performed "enough"
operations.

I prove this assumption to be false.
No matter how long we wait, or how many
operations the TM performs, there will be unread
finite positions on the tape.

This is true even if the TM is "infinitely" fast and
can perform an "infinite" number of operations.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/17/2009 3:56:22 AM
```On Jul 16, 8:08=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

> You claim that a ZM can only read a finite part of the tape, but you
> don't show that this finite part is bounded for all ZMs. =A0Why do you
> call it a "limit"?

I can show the string must be finite for any ZM.
You are correct that the string produced by a particular ZM
may not be a limit for all ZM's.

But it seems reasonable the string produced by any ZM
would be a limit for any TM that is not a ZM.

It would be very strange if a TM could write a longer
string than a ZM, since we are assuming the ZM is
more powerful than a TM.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/17/2009 4:05:37 AM
```On 2009-07-16 10:38:55 -0400, "Peter Webb"
<webbfamily@DIESPAMDIEoptusnet.com.au> said:

>
> "RussellE" <reasterly@gmail.com> wrote in message
> On Jul 15, 7:31 pm, ab <hobk...@gmail.com> wrote:
>
>> Zeno machines are not Turing machines, so you haven't said anything
>
> Zeno machines are Accelerated Turing Machines.
>
> ***************************
> There is another old paradox that applies to machines of this type.
>
> Imagine at t =1 we add 100 numbered balls 1 to 100 to an urn, but
> remove ball number 1.  At t = 1/2 we add balls 101-200 but remove ball
> number 2. At t=1/4 we add balls 201-300, remove ball 3, etc.
>
> How many balls are there at t=0 ?
>
> Well, none. Because clearly ball number n was removed at step n for all n.
>
> But equally clearly, at every step you are adding 99 balls in total, so
> how can it be zero?

Can we conclude from this that "naïve" Zeno machines of the kind
RussellE is describing are unsound as logical constructs? Or is this a
rhetorical paradox, rather than a formal one?

I ask, because that looks a lot like a sketch for a proof that A - B ≠
0 for sets A, B of cardinality ℵ₀ and A ≠ B, to me...

Free-associating,
-o

```
 0
angrybaldguy (338)
7/17/2009 5:44:52 AM
```RussellE <reasterly@gmail.com> writes:

> On Jul 16, 8:08 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>> > No TM can write a string longer than the one written by the ZM.
>> > I am already assuming the ZM can perform an infinite number
>> > of operations.
>
>> I thought you now know that to be false (not sure if this is by
>> definition or proof, but you said it was false above).
>
> You are correct. I prove a TM or ZM can only read or write
> a finite number of positions. So why do I use a ZM?

Why is there still no definition of what you mean.  For example, is
there a natural number N such that there is no TM which halts after
visiting >= N positions on its tape?  Maybe you mean that all TMs that
halt will have visited a finite number of tape positions?

I don't think the can get any further without a better definition of
what you are proving.

> The standard definition of TM assumes the input is finite.
> This simply avoids the entire issue of infinite strings.
> It assumes a human has somehow decided in advance
> that the input string is finite.

That was indeed what was decided.

> I am using ZM's to eliminate this assumption.
> I start by assuming a ZM can read an infinite string.
>
> Another common assumption is that a TM can
> read any finite string if we just wait "long enough",
> or we wait until the TM has performed "enough"
> operations.
>
> I prove this assumption to be false.

I missed it in all the ill-defined words.

> No matter how long we wait, or how many
> operations the TM performs, there will be unread
> finite positions on the tape.

That is so obviously true (and well-know) I am not sure you need to
prove it again.  Is that the final conclusion?

> This is true even if the TM is "infinitely" fast and
> can perform an "infinite" number of operations.

Since you don't define these terms no one can really comment.  If you
can say what you mean by infinitely fast and infinite operations,
there would be something to debate.

--
Ben.
```
 0
ben.usenet (6790)
7/17/2009 1:47:53 PM
```RussellE <reasterly@gmail.com> writes:

> On Jul 16, 8:08 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>> You claim that a ZM can only read a finite part of the tape, but you
>> don't show that this finite part is bounded for all ZMs.  Why do you
>> call it a "limit"?
>
> I can show the string must be finite for any ZM.
> You are correct that the string produced by a particular ZM
> may not be a limit for all ZM's.
>
> But it seems reasonable the string produced by any ZM
> would be a limit for any TM that is not a ZM.

Things that "seem reasonable" is not a firm foundation when dealing
with paradoxical notions like Zeno machines.

If you define what sort of ZM you are talking about (for example you
are very vague about halting states, saying things like "if I halt the
machine...") then the claim could be examined by someone other than
yourself.

> It would be very strange if a TM could write a longer
> string than a ZM, since we are assuming the ZM is
> more powerful than a TM.

Not at all.  If, as you claim, you have defined them in such a way
that a particular ZM halts after examining only a finite number of
tape positions (I think this is a common definition, but I am not sure
about that) then of course there exist TMs that can examine more
positions.

--
Ben.
```
 0
ben.usenet (6790)
7/17/2009 1:56:36 PM
```On Jul 16, 8:56=A0pm, RussellE <reaste...@gmail.com> wrote:

> The standard definition of TM assumes the input is finite.
> This simply avoids the entire issue of infinite strings.
> It assumes a human has somehow decided in advance
> that the input string is finite.

No, it doesn't. We have a mathematical definition of 'Turing machine'
and of 'computation of Turing machine on a finite string'. There is no
inconsistency in that. Whatever personal sense you have of some
illegitimacy in Turing machines or set theory or whatever, it is not
coming out from you as an actual mathematical argument.

> Another common assumption is that a TM can
> read any finite string if we just wait "long enough",
> or we wait until the TM has performed "enough"
> operations.
>
> I prove this assumption to be false.

It's not merely an assumption. It just follows from the definitions.
Though, it may help not to be distracted by the vagaries of such as
"long enough" or "wait", etc.

> No matter how long we wait, or how many
> operations the TM performs, there will be unread
> finite positions on the tape.

That is not at issue. (Where the "tape" is infinite, the input is on
some finite portion of that tape.)

> This is true even if the TM is "infinitely" fast and
> can perform an "infinite" number of operations.

No such scenario is required to comprehend that a Turing machine takes
finite strings, and "recognizes" every entry in a finite string.

MoeBlee

```
 0
jazzmobe (307)
7/17/2009 5:23:02 PM
```I have a simpler version of the proof that doesn't
require Zeno machines or infinite tapes.

Assume I have a TM and a finite, unbounded tape.
Finite and unbounded means the tape is finite,
but I can add a finite number of positions to the
end of the tape if I need it to be longer.

A language is a set of strings. A language is recursive
if there exists a TM that can always decide
if the string on a tape is one of the strings
in the language. The TM must be able to halt
and accept the input if the input string is in the
language or it must halt and reject the input
if the string is not in the language.

I define the language of unary representations of
natural numbers as any finite string of 1's.

I will assume the only symbols on a tape are 1's.
This means any finite tape is a member of this
language. All a TM has to do to decide this
language is read to the end of the tape, halt,
and accept the input.

This tape is finite, but unbounded,
so I can add positions to the end of
the tape if I need to.

Each time the TM moves the tape head
one position to the right, I will add a '1'
to the end of the tape.

Clearly, this tape is finite and
the string on the tape is a member
of the language of unary representations.
Yet, no TM can halt and accept this input.
This proves the language of unary
representations is not recursive.

Russell
- 2 many 2 count
..
```
 0
reasterly (337)
7/17/2009 7:00:47 PM
```On Jul 17, 12:00=A0pm, RussellE <reaste...@gmail.com> wrote:
> I have a simpler version of the proof that doesn't
> require Zeno machines or infinite tapes.
>
> Assume I have a TM and a finite, unbounded tape.
> Finite and unbounded means the tape is finite,
> but I can add a finite number of positions to the
> end of the tape if I need it to be longer.
>
> A language is a set of strings. A language is recursive
> if there exists a TM that can always decide
> if the string on a tape is one of the strings
> in the language. The TM must be able to halt
> and accept the input if the input string is in the
> language or it must halt and reject the input
> if the string is not in the language.
>
> I define the language of unary representations of
> natural numbers as any finite string of 1's.
>
> I will assume the only symbols on a tape are 1's.
> This means any finite tape is a member of this
> language. All a TM has to do to decide this
> language is read to the end of the tape, halt,
> and accept the input.
>
> This tape is finite, but unbounded,
> so I can add positions to the end of
> the tape if I need to.
>
> Each time the TM moves the tape head
> one position to the right, I will add a '1'
> to the end of the tape.
>
> Clearly, this tape is finite and
> the string on the tape is a member
> of the language of unary representations.
> Yet, no TM can halt and accept this input.
> This proves the language of unary
> representations is not recursive.
>
> Russell
> - 2 many 2 count
> .

To create an 'infinite' guide in parallel line we use a Tape Measure
as a tool.

1. Start from an 'Endpoint' to create a finite guide line with a guide
point at the origin.

Finite n-tape automata over possibly infinite alphabets seems a good a
place as any to start.

What do you think?

--
Martin Musatov
```
 0
marty.musatov (1143)
7/17/2009 7:22:29 PM
```On 2009-07-17 15:00:47 -0400, RussellE <reasterly@gmail.com> said:

> Each time the TM moves the tape head
> one position to the right, I will add a '1'
> to the end of the tape.

You've stepped outside of the Turing machine formalism here. You don't
get to "edit" the tape between steps. A Turing machine is a function f
from a state vector to a new state vector, and a Turing machine's
execution is described by an unbroken sequence of applications of that
function to some initial state S0, possibly ending at some halting
state Sn:

{s0,
s1 = f s0,
s2 = f s1
= f.f s0,
s3 = f s2
= f.f s1
= f.f.f s0,
s4 = f s3
= f.f s2
= f.f.f s1
= f.f.f.f s0,
....}

By "adding a 1" to the tape, you're describing the sequence

{s0,
s1 = f.G s0,
s2 = f.G s1
= f.G.f.G s0,
s3 = f.G s2
= f.G.f.G s1
= f.G.f.G.f.G s0,
...}

where G is computable by a Turing machine that adds a cell to a finite
string to the end of the input[0].

This is actually a legal construction for a Turing machine, but not for
the one you described in prose. A Turing machine that passed through
every state in the second trace would not halt given any finite tape
composed of 1s (ncluding the empty tape), but this proves nothing about
machines that do halt on every finite input tape.

Confusing Turing machines with some physical model that can be
manipulated and toyed with using intuition is a depressingly common
misunderstanding. Don't do it. They're purely mathematical constructs;
the "machine" and "tape" and "step" nomenclature is a convenient
metaphor, not a literal description.

-o

[0] Proving the existence of a Turing machine G is left as an exercise.

```
 0
angrybaldguy (338)
7/17/2009 9:31:39 PM
```On Jul 17, 2:31=A0pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-17 15:00:47 -0400, RussellE <reaste...@gmail.com> said:
>
> > Each time the TM moves the tape head
> > one position to the right, I will add a '1'
> > to the end of the tape.
>
> You've stepped outside of the Turing machine formalism here. You don't
> get to "edit" the tape between steps. A Turing machine is a function f
> from a state vector to a new state vector, and a Turing machine's
> execution is described by an unbroken sequence of applications of that
> function to some initial state S0, possibly ending at some halting
> state Sn:
>
> {s0,
> =A0s1 =3D f s0,
> =A0s2 =3D f s1
> =A0 =A0 =3D f.f s0,
> =A0s3 =3D f s2
> =A0 =A0 =3D f.f s1
> =A0 =A0 =3D f.f.f s0,
> =A0s4 =3D f s3
> =A0 =A0 =3D f.f s2
> =A0 =A0 =3D f.f.f s1
> =A0 =A0 =3D f.f.f.f s0,
> ...}
>
> By "adding a 1" to the tape, you're describing the sequence
>
> {s0,
> =A0s1 =3D f.G s0,
> =A0s2 =3D f.G s1
> =A0 =A0 =3D f.G.f.G s0,
> =A0s3 =3D f.G s2
> =A0 =A0 =3D f.G.f.G s1
> =A0 =A0 =3D f.G.f.G.f.G s0,
> =A0...}
>
> where G is computable by a Turing machine that adds a cell to a finite
> string to the end of the input[0].
>
> This is actually a legal construction for a Turing machine, but not for
> the one you described in prose. A Turing machine that passed through
> every state in the second trace would not halt given any finite tape
> composed of 1s (ncluding the empty tape), but this proves nothing about
> machines that do halt on every finite input tape.
>
> Confusing Turing machines with some physical model that can be
> manipulated and toyed with using intuition is a depressingly common
> misunderstanding. Don't do it. They're purely mathematical constructs;
> the "machine" and "tape" and "step" nomenclature is a convenient
> metaphor, not a literal description.
>
> -o
>
> [0] Proving the existence of a Turing machine G is left as an exercise.

What if we determine a deterministic Turing machine M to be the
partial function f such that .....?

--
Musatov
```
 0
marty.musatov (1143)
7/17/2009 10:01:15 PM
```RussellE <reasterly@gmail.com> writes:

> I have a simpler version of the proof that doesn't
> require Zeno machines or infinite tapes.
>
> Assume I have a TM and a finite, unbounded tape.
> Finite and unbounded means the tape is finite,
> but I can add a finite number of positions to the
> end of the tape if I need it to be longer.
>
> A language is a set of strings. A language is recursive
> if there exists a TM that can always decide
> if the string on a tape is one of the strings
> in the language. The TM must be able to halt
> and accept the input if the input string is in the
> language or it must halt and reject the input
> if the string is not in the language.

You could avoid all this wording by referencing the standard
definitions.  You would also avoid ambiguity by doing so.  For example
the above does not address the issue of blanks which I assume are on
all but a finite number of tape cells.

> I define the language of unary representations of
> natural numbers as any finite string of 1's.
>
> I will assume the only symbols on a tape are 1's.
> This means any finite tape is a member of this
> language. All a TM has to do to decide this
> language is read to the end of the tape, halt,
> and accept the input.

No need.  The TM can halt and accept immediately.  You have defined
things in such a way that there is nothing to decide anymore.

> This tape is finite, but unbounded,
> so I can add positions to the end of
> the tape if I need to.
>
> Each time the TM moves the tape head
> one position to the right, I will add a '1'
> to the end of the tape.

This is not a TM.  You need to name it and define it.  Then you can
argue about what sets it may or may not decide.  I'll call it an RTM
below.

> Clearly, this tape is finite and
> the string on the tape is a member
> of the language of unary representations.
> Yet, no TM can halt and accept this input.
> This proves the language of unary
> representations is not recursive.

It is not clear at all.  This is not a TM but a new abstraction.  It
is not even clear if the "input" is indeed a member of the set.  Some
definitions would save you a lot of time.

If you decide that whatever this growing input represents is indeed a
member of the set then a one-state RTM will do (as above) but if you
define it to not in the set then a two-state machine does the job
(halt and accept only on blank).

--
Ben.
```
 0
ben.usenet (6790)
7/17/2009 10:23:37 PM
```On Jul 17, 2:31=A0pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-17 15:00:47 -0400, RussellE <reaste...@gmail.com> said:

> > Each time the TM moves the tape head
> > one position to the right, I will add a '1'
> > to the end of the tape.
>
> You've stepped outside of the Turing machine formalism here. You don't
> get to "edit" the tape between steps.

I am not editing the tape. The tape isn't long enough to
hold the unimaginably huge, but finite, string of 1's I want to put on
it.
I am allowed to add a finite number of positions to the tape.

The TM has not read the part of the tape I describe,
so I have not changed the current state of the TM in any way.

I am not assuming the tape changes.
I am describing a finite and unbounded input tape.

> Confusing Turing machines with some physical model that can be
> manipulated and toyed with using intuition is a depressingly common
> misunderstanding. Don't do it. They're purely mathematical constructs;
> the "machine" and "tape" and "step" nomenclature is a convenient
> metaphor, not a literal description.

Few people would disagree with me if I stated there exist
finite strings no physical computer can read.

Even a computer that performs an operation every 10^(-44)
second (Planck Time http://en.wikipedia.org/wiki/Planck_time)
and ran for 30 billion years would perform less than
10^60 operations.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/18/2009 12:10:02 AM
```"I guess it is true, what they say!" he exclaimed. "A single well
placed question is worth a thousand conjectures."

On Jul 17, 3:23=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:
> > I have a simpler version of the proof that doesn't
> > require Zeno machines or infinite tapes.
>
> > Assume I have a TM and a finite, unbounded tape.
> > Finite and unbounded means the tape is finite,
> > but I can add a finite number of positions to the
> > end of the tape if I need it to be longer.
>
> > A language is a set of strings. A language is recursive
> > if there exists a TM that can always decide
> > if the string on a tape is one of the strings
> > in the language. The TM must be able to halt
> > and accept the input if the input string is in the
> > language or it must halt and reject the input
> > if the string is not in the language.
>
> You could avoid all this wording by referencing the standard
> definitions. =A0You would also avoid ambiguity by doing so. =A0For exampl=
e
> the above does not address the issue of blanks which I assume are on
> all but a finite number of tape cells.
>
> > I define the language of unary representations of
> > natural numbers as any finite string of 1's.
>
> > I will assume the only symbols on a tape are 1's.
> > This means any finite tape is a member of this
> > language. All a TM has to do to decide this
> > language is read to the end of the tape, halt,
> > and accept the input.
>
> No need. =A0The TM can halt and accept immediately. =A0You have defined
> things in such a way that there is nothing to decide anymore.
>
> > This tape is finite, but unbounded,
> > so I can add positions to the end of
> > the tape if I need to.
>
> > Each time the TM moves the tape head
> > one position to the right, I will add a '1'
> > to the end of the tape.
>
> This is not a TM. =A0You need to name it and define it. =A0

Fine. I name it and define it a Martinian Machine or MM. It starts
with a finite tape of all 1's. This tape is finite, but unbounded, so
I may add positions to the end of the tape if I need to.

Here is a rough schematic of the MM.

---- >         ---- >
TAPE  head =3D> TAPE  write 1 (slack)
< ----        < ----

Then you can
> argue about what sets it may or may not decide. =A0I'll call it an MM
> below.
>
> > Clearly, this tape is finite and
> > the string on the tape is a member
> > of the language of unary representations.
> > Yet, no TM can halt and accept this input.
> > This proves the language of unary
> > representations is not recursive.
>
> It is not clear at all. =A0This is not a TM but a new abstraction. =A0It
> is not even clear if the "input" is indeed a member of the set. =A0Some
> definitions would save you a lot of time.
>
> If you decide that whatever this growing input represents is indeed a
> member of the set then a one-state MM will do (as above) but if you
> define it to not in the set then a two-state machine does the job
> (halt and accept only on blank).
>
> --
> Ben.- Hide quoted text -
>
> - Show quoted text -

MM
```
 0
marty.musatov (1143)
7/18/2009 12:22:21 AM
```On 2009-07-17 20:10:02 -0400, RussellE <reasterly@gmail.com> said:

> On Jul 17, 2:31�pm, Owen Jacobson <angrybald...@gmail.com> wrote:
>> On 2009-07-17 15:00:47 -0400, RussellE <reaste...@gmail.com> said:
>
>>> Each time the TM moves the tape head
>>> one position to the right, I will add a '1'
>>> to the end of the tape.
>>
>> You've stepped outside of the Turing machine formalism here. You don't
>> get to "edit" the tape between steps.
>
> I am not editing the tape. The tape isn't long enough to
> hold the unimaginably huge, but finite, string of 1's I want to put on
> it.

You've stepped outside of the Turing machine formalism here, too. The
tape is, from the definition of a Turing machine, long enough to hold
the entire initial state, and any state reached during comptuation. The
only constraint is that those states are, individually, finite.

The formalism can loosely be described by an *infinitely* long tape, of
which only finitely many cells are non-blank at any step, without
changing the capabilities of the model at all, so "the tape isn't long
enough to hold my finite number" is not true.

> I am allowed to add a finite number of positions to the tape.

You are allowed to add nothing at all, except to the initial state. The
tape's contents at each step are solely determined by the preceeding
state and tape contents, and the machine's transition function.

> The TM has not read the part of the tape I describe,
> so I have not changed the current state of the TM in any way.

*sigh* Yes, you have. The state of a TM includes the entire tape, not

Seriously, have you read *any* of the literature on this? Or did you
skim Wikipedia and then set out to tear down the mathematical
establishment? Even
<http://plato.stanford.edu/entries/turing-machine/#FormalDefinition> is
a decent start.

If, after reading that link, you still can't see why the entire tape is
part of the state of a machine, consider:

Machine X and machine Y have the same transition function, initial
state, and set of states, and are run for an identical number of steps
prior to the following situation. When the machine encounters a zero in

At a given step, machine X's is in state 1 and its tape contains ... 0
0 [1] 1 0 0 ..., with the head's location denoted by []. After two
steps, machine X's tape becomes ... 0 0 0 0 [0] 0 ..., with the tape's
head two cells to the right of its original position; the machine halts.

At the same step, machine Y's is also in state 1, and its tape contains
.... 0 0 [1] 1 1 0 ... . If only the tape cell under the head is part of
the machine's state, then, since the machine's behaviour is the same as
machine X's for a given state, it must also halt with a zero under the
head after two more steps. However, after two steps, the tape contains
.... 0 0 0 0 [1] 0 ... .

Since Y arrived at a different configuration from X after the same
number of steps, Y's state differed at every step. The only difference
between those two machines at every step is the tape contents, which at
least some of the time only differed at cells *not* under the head. So
those cells also contribute to machine Y's state.

> I am not assuming the tape changes.
> I am describing a finite and unbounded input tape.

Once again: you're describing two different machines: one (machine A)
which recognizes the language of finite strings of 1s and one (machine
B) which never halts for any input. That machine B (where the tape's
non-blank portion grows over time) never halts for a given input has no
bearing on whether machine A ever halts for the same input.

>> Confusing Turing machines with some physical model that can be
>> manipulated and toyed with using intuition is a depressingly common
>> misunderstanding. Don't do it. They're purely mathematical constructs;
>> the "machine" and "tape" and "step" nomenclature is a convenient
>> metaphor, not a literal description.
>
> Few people would disagree with me if I stated there exist
> finite strings no physical computer can read.

Irrelevant to what is and is not computable. Computability is a purely
mathematical construct.

> Even a computer that performs an operation every 10^(-44)
> second (Planck Time http://en.wikipedia.org/wiki/Planck_time)
> and ran for 30 billion years would perform less than
> 10^60 operations.

Irrelevant to what is and is not computable. Computability is a purely
mathematical construct. Any Turing machine that halts on a given input
does so after a finite number of steps - but it can be a very large
finite number.

You might be thinking about tractability. Many computable problems are
intractable: a brute-force answer to a million-node travelling salesman
problem is computable (the algorithm isn't even hard to write), but
intractable.

Stating that most natural numbers are intractable to compute is not
news, either...

-o

```
 0
angrybaldguy (338)
7/18/2009 2:12:28 AM
```On Jul 17, 7:12=A0pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-17 20:10:02 -0400, RussellE <reaste...@gmail.com> said:

> The formalism can loosely be described by an *infinitely* long tape, of
> which only finitely many cells are non-blank at any step, without
> changing the capabilities of the model at all, so "the tape isn't long
> enough to hold my finite number" is not true.

I am trying to prove there is a uncomputable natural number.
I can't tell you exactly how big the tape is in advance.
If I could, the number wouldn't be uncomputable.
The best I can hope for is to show such a number exists.

Once again from Wikipedia: http://en.wikipedia.org/wiki/Turing_recognizable

A recursively enumerable language is a formal language for which there
exists a
Turing machine (or other computable function) that will halt and
accept when
presented with any string in the language as input but may either halt
and reject
or loop forever when presented with a string not in the language.

I realize this is not the most formal definition, but it is enough

I want to prove the language of unary representations of natural
numbers
is not recursively enumerable.

A unary representation is a continous string of 1's followed by a
blank.

I invite you to write a TM to recognize this language.
You may use whatever formalism you like.
But, I get to provide the input tape.
I guarantee this tape has a finite number of non-blank positions.

Now, I want to talk about uncomputable natural numbers.
How many operations does your TM have to perform before
you decide your TM will "loop forever"?

The very definition of Turing recognizable assumes we can
tell when a TM loops forever.

can you prove my tape does not contain a unary
representation of a natural number?
If not, the language of unary representations is not
recusively enumerable.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/18/2009 4:03:03 AM
```RussellE wrote:
....
> The very definition of Turing recognizable assumes we can
> tell when a TM loops forever.

No, it does not. You may be confusing "recognizable" and "decidable".

Patricia
```
 0
pats (3556)
7/18/2009 4:19:48 AM
```On 2009-07-18 00:03:03 -0400, RussellE <reasterly@gmail.com> said:

> On Jul 17, 7:12�pm, Owen Jacobson <angrybald...@gmail.com> wrote:
>> On 2009-07-17 20:10:02 -0400, RussellE <reaste...@gmail.com> said:
>
>> The formalism can loosely be described by an *infinitely* long tape, of
>> which only finitely many cells are non-blank at any step, without
>> changing the capabilities of the model at all, so "the tape isn't long
>> enough to hold my finite number" is not true.
>
> I am trying to prove there is a uncomputable natural number.

Since every natural number is representable as a finite set, and every
finite set is countable, you are trying to prove a contradiction.
<http://mathworld.wolfram.com/CountableSet.html> is probably a good
place to start.

> I can't tell you exactly how big the tape is in advance.

The tape is, by definition, large enough to contain any finite number

> If I could, the number wouldn't be uncomputable.

Well, shit, then your "proof" is broken. You probably should've thought
of that before you started waving technical terms around.

> The best I can hope for is to show such a number exists.
>
> Once again from Wikipedia: http://en.wikipedia.org/wiki/Turing_recognizable
>
> A recursively enumerable language is a formal language for which there
> exists a
> Turing machine (or other computable function) that will halt and
> accept when
> presented with any string in the language as input but may either halt
> and reject
> or loop forever when presented with a string not in the language.
>
> I realize this is not the most formal definition, but it is enough
> for me to gripe about.

You can gripe about it all you want, but if you want to present a
worthwhile argument, you'll need a formal definition. No exceptions.

> I want to prove the language of unary representations of natural
> numbers is not recursively enumerable.

I'll come back to this, because it's hilarious.

> A unary representation is a continous string of 1's followed by a
> blank.

More conventionally, a unary representation is any member of the
language {1}*, where * is the Kleene star
(http://planetmath.org/encyclopedia/KleeneStar.html). We'll take the
set {1} U {b} as the symbol set for our Turing machine, using b as the
'blank' symbol.

For the purposes of this exercise, let's exclude the empty string,
making our language {1}* - {""}.

> I invite you to write a TM to recognize this language.
> You may use whatever formalism you like.
> But, I get to provide the input tape.
> I guarantee this tape has a finite number of non-blank positions.

Given a symbol set Q = {b, 1}.
Given a state set S = {J, A}, an initial state s0 = J (for reJect, not
to be confused with the direction 'R') and an accept state sA = A.
Given the directions D = {L, R} for 'move one cell left' and 'move one
cell right'.

Then, we have a transition function f : Q x S -> Q x S x D which is
defined for all non-halting configurations in Q x S and undefined for
all halting configurations.

f(1, J) = (1, A, R)
f(1, A) = (1, A, R)
f(b, J) undefined
f(b, A) undefined

If the machine is state 1 and halts, then it accepts the input.

An example run:

[...[1]1 1 b ...], s = J: f(...) = (1, A, R)
[... 1[1]1 b ...], s = A: f(...) = (1, A, R)
[... 1 1[1]b ...], s = A: f(...) = (1, A, R)
[... 1 1 1[b]...], s = A: f(...) undefined, halt.

The machine has halted in state A, so the input was accepted. It should
be fairly obvious that, no matter how many 1s appear in order on the
tape, for that tape there is some finite natural number of steps that
leads to the machine halting in the accept state.

An example non-accepting run:

[... [b] ...], s = J: f(...) undefined, halt.

The machine has halted, but, as it's not in an accepting state, the
input was rejected.

If we wanted to accept {1}* instead of {1}* - {""}, then change the
machine such that both states are accepting states. On encountering a
blank at the first step, the machine would halt as before, but we would
interpret that as a second accepting state.

> Now, I want to talk about uncomputable natural numbers.
> How many operations does your TM have to perform before
> you decide your TM will "loop forever"?

None. The TM described above always halts, for all input tapes
containing only finitely many non-blank symbols.

> The very definition of Turing recognizable assumes we can
> tell when a TM loops forever.

If it halts, it will do so after finitely many steps. So, for any
machine and input, there is some finite number of steps after which you
can conclusively say that the machine will not halt.

It's even possible to decide, given all three of a Turing machine M, an
input tape t, and a natural number n, whether the M will halt on t
after n steps using only a Turing machine.

> If your TM loops forever when reading my input tape, can you prove my
> tape does not contain a unary representation of a natural number?

Let's posit a machine that *does* not-halt when given an initial state
that doesn't contain a unary representation of a natural number at the
initial head position. Then, yes, if that machine doesn't halt, then we
can conclude the input tape is not in the language {1}* - {""}.

This is the contrapositive
(http://mathworld.wolfram.com/ModusTollens.html) of "if the input tape
is in the language {1}* - {""}, then the machine halts in an accept
state." No further work is required.

> If not, the language of unary representations is not
> recusively enumerable.

A good many things are true if you accept contradictory premises, yes.

Given your propensity for cutting out anything resembling formal
arguments before replying to posts, though, I don't think you care.

-o

Oh, hell, that's right. I said I'd come back to this:

> I want to prove the language of unary representations of natural
> numbers is not recursively enumerable.

Given + as the string concatenation operator
(http://planetmath.org/encyclopedia/Length4.html):

f(0) = ""
f(n) = "1" + f(n - 1)

is a primitive recursive function that enumerates the language {1}*. So
{1}* is recursively enumerable.

This isn't a proof, exactly, but it is a pretty compelling argument.
The real proof is the Turing machine construction, above, which agrees
with this argument.

```
 0
angrybaldguy (338)
7/18/2009 6:37:27 AM
```On 2009-07-18 00:19:48 -0400, Patricia Shanahan <pats@acm.org> said:

> RussellE wrote:
> ...
>> The very definition of Turing recognizable assumes we can
>> tell when a TM loops forever.
>
> No, it does not. You may be confusing "recognizable" and "decidable".
>
> Patricia

That's a good point. The TM I described in
<200907180237278930-angrybaldguy@gmailcom> both recognizes and decides
the given language. I probably should've stuck to just recognizing the
language, but the resulting TM is actually more complicated (I think
you need three states rather than two).

For the record:

A langauge is recognizable if and only if there exists a Turing machine
that halts and accepts given an input tape that *is* in the language.

A language is decidable if and only if there exists a Turing machine
that halts and accepts if the input tape is in the language, and halts
and rejects otherwise.

-o

```
 0
angrybaldguy (338)
7/18/2009 6:42:05 AM
```On Jul 17, 11:37=A0pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-18 00:03:03 -0400, RussellE <reaste...@gmail.com> said:

> > A unary representation is a continous string of 1's followed by a
> > blank.
>
> More conventionally, a unary representation is any member of the
> language {1}*, where * is the Kleene star
> (http://planetmath.org/encyclopedia/KleeneStar.html). We'll take the
> set {1} U {b} as the symbol set for our Turing machine, using b as the
> 'blank' symbol.

OK

> For the purposes of this exercise, let's exclude the empty string,
> making our language {1}* - {""}.

I like zero, but sure, we can exclude the empty string.

> > I invite you to write a TM to recognize this language.
> > You may use whatever formalism you like.
> > But, I get to provide the input tape.
> > I guarantee this tape has a finite number of non-blank positions.
>
> Given a symbol set Q =3D {b, 1}

I would prefer a symbol set that allows for non-empty finite strings
that aren'.t unary representations.
How about a symbol set Q =3D {b, 0, 1}?

> Given a state set S =3D {J, A}, an initial state s0 =3D J (for reJect, no=
t
> to be confused with the direction 'R') and an accept state sA =3D A.
> Given the directions D =3D {L, R} for 'move one cell left' and 'move one
> cell right'.
>
> Then, we have a transition function f : Q x S -> Q x S x D which is
> defined for all non-halting configurations in Q x S and undefined for
> all halting configurations.
>
> f(1, J) =3D (1, A, R)
> f(1, A) =3D (1, A, R)
> f(b, J) undefined
> f(b, A) undefined
>
> If the machine is state 1 and halts, then it accepts the input.
>
> An example run:
>
> [...[1]1 1 b ...], s =3D J: f(...) =3D (1, A, R)
> [... 1[1]1 b ...], s =3D A: f(...) =3D (1, A, R)
> [... 1 1[1]b ...], s =3D A: f(...) =3D (1, A, R)
> [... 1 1 1[b]...], s =3D A: f(...) undefined, halt.

So, this TM moves right if it reads a '1' and halts

> The machine has halted in state A, so the input was accepted. It should
> be fairly obvious that, no matter how many 1s appear in order on the
> tape, for that tape there is some finite natural number of steps that
> leads to the machine halting in the accept state.

Why do you think this is obvious?
It is an assumption, at best, and may simply be false.

> An example non-accepting run:
>
> [... [b] ...], s =3D J: f(...) undefined, halt.
>
> The machine has halted, but, as it's not in an accepting state, the
> input was rejected.
>
> If we wanted to accept {1}* instead of {1}* - {""}, then change the
> machine such that both states are accepting states. On encountering a
> blank at the first step, the machine would halt as before, but we would
> interpret that as a second accepting state.
>
> > Now, I want to talk about uncomputable natural numbers.
> > How many operations does your TM have to perform before
> > you decide your TM will "loop forever"?
>
> None. The TM described above always halts, for all input tapes
> containing only finitely many non-blank symbols.

Would you give a proof please?
Preferably without assuming a TM can read any finite string.
What happens if the input is an infinitely long string of 1's?

> > The very definition of Turing recognizable assumes we can
> > tell when a TM loops forever.
>
> If it halts, it will do so after finitely many steps.

Yes.

> So, for any
> machine and input, there is some finite number of steps after which you
> can conclusively say that the machine will not halt.

How many steps would that be?
Are you expecting me to tell you in advance how many
non-blank positions are on the tape?

> It's even possible to decide, given all three of a Turing machine M, an
> input tape t, and a natural number n, whether the M will halt on t
> after n steps using only a Turing machine.
>
> > If your TM loops forever when reading my input tape, can you prove my
> > tape does not contain a unary representation of a natural number?
>
> Let's posit a machine that *does* not-halt when given an initial state
> that doesn't contain a unary representation of a natural number at the
> initial head position. Then, yes, if that machine doesn't halt, then we
> can conclude the input tape is not in the language {1}* - {""}.

How would you prove that?
Did the TM read every finite position on the tape?
If the TM hasn't read every finite position on the tape,
how do you know the tape isn't a unary representation?
Is the TM still running? If so, it can't have read
every finite position on the tape.

I am confident the tape I gave you will cause your TM
to "loop forever".

> This is the contrapositive
> (http://mathworld.wolfram.com/ModusTollens.html) of "if the input tape
> is in the language {1}* - {""}, then the machine halts in an accept
> state." No further work is required.
>
> > If not, the language of unary representations is not
> > recusively enumerable.
>
> A good many things are true if you accept contradictory premises, yes.
>
> Given your propensity for cutting out anything resembling formal
> arguments before replying to posts, though, I don't think you care.
>
> -o
>
> Oh, hell, that's right. I said I'd come back to this:
>
> > I want to prove the language of unary representations of natural
> > numbers is not recursively enumerable.
>
> Given + as the string concatenation operator
> (http://planetmath.org/encyclopedia/Length4.html):
>
> f(0) =3D ""
> f(n) =3D "1" + f(n - 1)
>
> is a primitive recursive function that enumerates the language {1}*. So
> {1}* is recursively enumerable.
>
> This isn't a proof, exactly, but it is a pretty compelling argument.
> The real proof is the Turing machine construction, above, which agrees
> with this argument.

Yes, there are a number of ways to define recursively enumerable.
I prefer the one I quoted from Wikipedia because of the "loop forever"
part.
I am still curious how one knows a TM loops forever.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/18/2009 8:56:25 AM
```On Jul 17, 11:37=A0pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-18 00:03:03 -0400, RussellE <reaste...@gmail.com> said:
>
> > I want to prove the language of unary representations of natural
> > numbers is not recursively enumerable.
>
> I'll come back to this, because it's hilarious.

You came back to it, but what of the hilarity?

I myself want to prove that almost all natural numbers are
really really big. I mean just freakin' huge.

Proof: consider some natural number n. The number 10,000 * n
is totally way bigger than that. And you can even repeat this
process like, fifty times or more. I know, right? Ginormous.

Marshall
```
 0
7/18/2009 2:10:38 PM
```On 2009-07-18 04:56:25 -0400, RussellE <reasterly@gmail.com> said:

> On Jul 17, 11:37�pm, Owen Jacobson <angrybald...@gmail.com> wrote:
>> On 2009-07-18 00:03:03 -0400, RussellE <reaste...@gmail.com> said:
>
>>> A unary representation is a continous string of 1's followed by a
>>> blank.
>>
>> More conventionally, a unary representation is any member of the
>> language {1}*, where * is the Kleene star
>> (http://planetmath.org/encyclopedia/KleeneStar.html). We'll take the
>> set {1} U {b} as the symbol set for our Turing machine, using b as the
>> 'blank' symbol.
>
> OK
>
>> For the purposes of this exercise, let's exclude the empty string,
>> making our language {1}* - {""}.
>
> I like zero, but sure, we can exclude the empty string.
>
>>> I invite you to write a TM to recognize this language.
>>> You may use whatever formalism you like.
>>> But, I get to provide the input tape.
>>> I guarantee this tape has a finite number of non-blank positions.
>>
>> Given a symbol set Q = {b, 1}
>
> I would prefer a symbol set that allows for non-empty finite strings
> that aren'.t unary representations.
> How about a symbol set Q = {b, 0, 1}?

You should've said so up front.

This isn't difficult; we can add one more state and the appropriate
transitions so that the machine, on encountering a 0, transitions to a
reject state other than J and then halts. Construction is left as an
exercise.

>> Given a state set S = {J, A}, an initial state s0 = J (for reJect, not
>> to be confused with the direction 'R') and an accept state sA = A.
>> Given the directions D = {L, R} for 'move one cell left' and 'move one
>> cell right'.
>>
>> Then, we have a transition function f : Q x S -> Q x S x D which is
>> defined for all non-halting configurations in Q x S and undefined for
>> all halting configurations.
>>
>> f(1, J) = (1, A, R)
>> f(1, A) = (1, A, R)
>> f(b, J) undefined
>> f(b, A) undefined
>>
>> If the machine is state 1 and halts, then it accepts the input.
>>
>> An example run:
>>
>> [...[1]1 1 b ...], s = J: f(...) = (1, A, R)
>> [... 1[1]1 b ...], s = A: f(...) = (1, A, R)
>> [... 1 1[1]b ...], s = A: f(...) = (1, A, R)
>> [... 1 1 1[b]...], s = A: f(...) undefined, halt.
>
> So, this TM moves right if it reads a '1' and halts

It also switches to its accept state if it reads a '1', and starts in a
reject state.

>> The machine has halted in state A, so the input was accepted. It should
>> be fairly obvious that, no matter how many 1s appear in order on the
>> tape, for that tape there is some finite natural number of steps that
>> leads to the machine halting in the accept state.
>
> Why do you think this is obvious?
> It is an assumption, at best, and may simply be false.

For every input tape t composed of symbols from Q = {b, 1}, if t
contains n '1's and n is a natural number, the above machine will halt
after n transitions.

Every tape containing n '1's (where n is a natural number) has a blank
at the position to the right of the n'th '1' (counting from the
leftmost). At every transition, either the machine has encountered a 1
and moves to the right, bringing it closer to this blank, or it has
encountered a blank, in which case it halts.

If the n'th '1', counting from the left, did not have a blank after it,
the tape would contain more than n '1's.

If, for *every* natural number m, t contains more than m '1's, then t
is not a tape with a finite number of '1's.

>> An example non-accepting run:
>>
>> [... [b] ...], s = J: f(...) undefined, halt.
>>
>> The machine has halted, but, as it's not in an accepting state, the
>> input was rejected.
>>
>> If we wanted to accept {1}* instead of {1}* - {""}, then change the
>> machine such that both states are accepting states. On encountering a
>> blank at the first step, the machine would halt as before, but we would
>> interpret that as a second accepting state.
>>
>>> Now, I want to talk about uncomputable natural numbers.
>>> How many operations does your TM have to perform before
>>> you decide your TM will "loop forever"?
>>
>> None. The TM described above always halts, for all input tapes
>> containing only finitely many non-blank symbols.
>
> Would you give a proof please?
> Preferably without assuming a TM can read any finite string.
> What happens if the input is an infinitely long string of 1's?

See above.

You already ruled out an input tape with infinitely many '1's on it,
but for the sake of completeness, that is the only tape on which the
above machine will not halt. Since at each step it has an infinite
number of '1's left to examine, then it does not halt. Since the
machine does not halt, that tape is not in the language.

You see? We don't need to run the machine at all to demonstrate that it
does not accept that tape; it's enough to prove that, if run, it would
not halt in an accepting state.

>>> The very definition of Turing recognizable assumes we can
>>> tell when a TM loops forever.
>>
>> If it halts, it will do so after finitely many steps.
>
> Yes.
>
>> So, for any
>> machine and input, there is some finite number of steps after which you
>> can conclusively say that the machine will not halt.
>
> How many steps would that be?

A function of the input tape and the machine. You seem to be looking
for some constant; there isn't one.

(This function is not computable, but does exist; its range is the
natural numbers plus omega.)

> Are you expecting me to tell you in advance how many
> non-blank positions are on the tape?

Yes; it's part of the Turing machine formalism that the input tape is
completely specified. It is enough to say that 'there are n non-blank
positions on the tape, where n is a natural number', though: it's
enough that for every value of n, the machine halts after n steps as
shown above.

>> It's even possible to decide, given all three of a Turing machine M, an
>> input tape t, and a natural number n, whether the M will halt on t
>> after n steps using only a Turing machine.
>>
>>> If your TM loops forever when reading my input tape, can you prove my
>>> tape does not contain a unary representation of a natural number?
>>
>> Let's posit a machine that *does* not-halt when given an initial state
>> that doesn't contain a unary representation of a natural number at the
>> initial head position. Then, yes, if that machine doesn't halt, then we
>> can conclude the input tape is not in the language {1}* - {""}.
>
> How would you prove that?

By modus tollens from the definition of "Turing recognizable." In
Language -> Halts & Accepts, �Halts, therefore �In Language.

I think you meant "how do you prove it won't halt?" In order to do
that, we'd need a definition of the specific Turing machine you're
interested in. There's no general proof that "no Turing machine will
halt" for a tape not in some Turing recognizable language, only proofs
that a given Turing machine will or won't halt for specific inputs.

Note that the machine I actually gave *does* halt for all finite input
tapes. It's more convenient to prove that than to prove that it doesn't
halt.

> Did the TM read every finite position on the tape?
> If the TM hasn't read every finite position on the tape,
> how do you know the tape isn't a unary representation?
> Is the TM still running? If so, it can't have read
> every finite position on the tape.
>
> I am confident the tape I gave you will cause your TM
> to "loop forever".

Since the tape you gave me, in your own words, "has a finite number of
non-blank positions", and since the machine I actually constructed as
an argument halts for *all* finite numbers of non-blank positions, no,
it won't. If you want one that doesn't halt for inputs not in the
language, construct it yourself and we'll walk through proving that it
doesn't halt unless the input tape is in the language.

-o

```
 0
angrybaldguy (338)
7/18/2009 5:01:09 PM
```On Jul 18, 10:01 am, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-18 04:56:25 -0400, RussellE <reaste...@gmail.com> said:

> >> It should
> >> be fairly obvious that, no matter how many 1s appear in order on the
> >> tape, for that tape there is some finite natural number of steps that
> >> leads to the machine halting in the accept state.
>
> > Why do you think this is obvious?
> > It is an assumption, at best, and may simply be false.
>
> For every input tape t composed of symbols from Q =3D {b, 1}, if t
> contains n '1's and n is a natural number, the above machine will halt
> after n transitions.

I admit defeat then. Any formal system that assumes a TM can
read any finite string can easily prove every natural is computable.
It's like proving the set of natural numbers exists once you
assume the Axiom of Infinity. Easier probably.
Set theorist tend to make things hard for themselves.

I can prove no TM can read every finite position on a tape.
I gave this proof in the first post. The set of unread positions
is always infinite.

If a TM can't read every finite position isn't it at least
worth considering there may be finite positions it can't read?

> >>> Now, I want to talk about uncomputable natural numbers.
> >>> How many operations does your TM have to perform before
> >>> you decide your TM will "loop forever"?
>
> >> None. The TM described above always halts, for all input tapes
> >> containing only finitely many non-blank symbols.
>
> > Would you give a proof please?
> > Preferably without assuming a TM can read any finite string.
> > What happens if the input is an infinitely long string of 1's?

> You already ruled out an input tape with infinitely many '1's on it,

And you took my word for it?!?
You can't trust me. Ask anyone in this newsgroup.
It is possible I gave you a tape with an infinite string of 1's.

> but for the sake of completeness, that is the only tape on which the
> above machine will not halt. Since at each step it has an infinite
> number of '1's left to examine, then it does not halt. Since the
> machine does not halt, that tape is not in the language.
>
> You see? We don't need to run the machine at all to demonstrate that it
> does not accept that tape; it's enough to prove that, if run, it would
> not halt in an accepting state.

If we already know if the tape will be accepted or rejected,
why have a TM at all? I still haven't told you what is on the tape.
The point of the exercise is to show there exists a TM
that recognises any string that is a member of the language.

> >>> The very definition of Turing recognizable assumes we can
> >>> tell when a TM loops forever.
>
> >> If it halts, it will do so after finitely many steps.
>
> > Yes.
>
> >> So, for any
> >> machine and input, there is some finite number of steps after which yo=
u
> >> can conclusively say that the machine will not halt.
>
> > How many steps would that be?
>
> A function of the input tape and the machine. You seem to be looking
> for some constant; there isn't one.

There is no way a noncomputable number is a constant.
At least, it won't be a computable constant.

> (This function is not computable, but does exist; its range is the
> natural numbers plus omega.)
>
> > Are you expecting me to tell you in advance how many
> > non-blank positions are on the tape?
>
> Yes; it's part of the Turing machine formalism that the input tape is
> completely specified. It is enough to say that 'there are n non-blank
> positions on the tape, where n is a natural number', though: it's
> enough that for every value of n, the machine halts after n steps as
> shown above.

I guess this isn't hard to show if we assume it is true.

> >>> If your TM loops forever when reading my input tape, can you prove my
> >>> tape does not contain a unary representation of a natural number?
>
> >> Let's posit a machine that *does* not-halt when given an initial state
> >> that doesn't contain a unary representation of a natural number at the
> >> initial head position. Then, yes, if that machine doesn't halt, then w=
e
> >> can conclude the input tape is not in the language {1}* - {""}.
>
> > How would you prove that?
>
> By modus tollens from the definition of "Turing recognizable." In
> Language -> Halts & Accepts, =ACHalts, therefore =ACIn Language.
>
> I think you meant "how do you prove it won't halt?" In order to do
> that, we'd need a definition of the specific Turing machine you're
> interested in. There's no general proof that "no Turing machine will
> halt" for a tape not in some Turing recognizable language, only proofs
> that a given Turing machine will or won't halt for specific inputs.
>
> Note that the machine I actually gave *does* halt for all finite input
> tapes. It's more convenient to prove that than to prove that it doesn't
> halt.
>
> > Did the TM read every finite position on the tape?
> > If so, what was the last position your TM read?
> > If the TM hasn't read every finite position on the tape,
> > how do you know the tape isn't a unary representation?
> > Is the TM still running? If so, it can't have read
> > every finite position on the tape.
>
> > I am confident the tape I gave you will cause your TM
> > to "loop forever".
>
> Since the tape you gave me, in your own words, "has a finite number of
> non-blank positions", and since the machine I actually constructed as
> an argument halts for *all* finite numbers of non-blank positions, no,
> it won't. If you want one that doesn't halt for inputs not in the
> language, construct it yourself and we'll walk through proving that it
> doesn't halt unless the input tape is in the language.

I used a TM to construct the tape I gave you.
The fact that I used a TM is enough to prove the tape
has a finite number of non-blank positions.

This TM is very simple. The TM writes a '1' and moves the
tape head one position to the right. There is no halt state,
so this TM will "loop forever".

I have been running this TM for years on my Zeno machine
which performs an infinite number of operations every second.
There should be a pretty big string of 1's on that tape.

I am bending the rules a little. My TM is still running when
I give you the tape. The TM "loops forever". But, I assure you
there are a finite number of non-blank positions on the tape.

Yes, we can combine your recognizer TM and my tape
generator TM into a single TM, but, that isn't the issue.

I challenged you to write a TM that can recognize a
unary represetation on ANY tape I might give you.

The tape I gave you has a finite number of non-blank
positions and can be completely specified with a TM.
The tape contains a string that is a member of this
this tape. Your TM has failed to recognize a valid input.

Some people will complain the input tape is still "changing".
Not true. The input tape was completely specified
when I wrote my TM. Of course, I can't tell you how
many 1's are on the tape. The number is uncomputable.
But I can prove the tape has a finite number of 1's
followed by a blank.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/18/2009 9:32:51 PM
```RussellE wrote:
> On Jul 18, 10:01 am, Owen Jacobson <angrybald...@gmail.com> wrote:
>> On 2009-07-18 04:56:25 -0400, RussellE <reaste...@gmail.com> said:
>
>>>> It should
>>>> be fairly obvious that, no matter how many 1s appear in order on the
>>>> tape, for that tape there is some finite natural number of steps that
>>>> leads to the machine halting in the accept state.
>>> Why do you think this is obvious?
>>> It is an assumption, at best, and may simply be false.
>> For every input tape t composed of symbols from Q = {b, 1}, if t
>> contains n '1's and n is a natural number, the above machine will halt
>> after n transitions.
>
> I admit defeat then. Any formal system that assumes a TM can
> read any finite string can easily prove every natural is computable.
> It's like proving the set of natural numbers exists once you
> assume the Axiom of Infinity. Easier probably.
> Set theorist tend to make things hard for themselves.

Being able to recognize any natural number seems to me to be a desirable
property of a mathematical abstraction for reasoning about the limits of
computation. Why do you see it as a defeat? Do you think it is an
undesirable property?

>
> I can prove no TM can read every finite position on a tape.
> I gave this proof in the first post. The set of unread positions
> is always infinite.
>
> If a TM can't read every finite position isn't it at least
> worth considering there may be finite positions it can't read?

Could you rewrite this a bit more formally? I am totally confused about
quantification. Which references to "TM" are about "for all TMs" and
which are about "there exists a TM such that"?

....

> I used a TM to construct the tape I gave you.
> The fact that I used a TM is enough to prove the tape
> has a finite number of non-blank positions.
>
> This TM is very simple. The TM writes a '1' and moves the
> tape head one position to the right. There is no halt state,
> so this TM will "loop forever".
>
> I have been running this TM for years on my Zeno machine
> which performs an infinite number of operations every second.
> There should be a pretty big string of 1's on that tape.
>
> I am bending the rules a little. My TM is still running when
> I give you the tape. The TM "loops forever". But, I assure you
> there are a finite number of non-blank positions on the tape.
>
> Yes, we can combine your recognizer TM and my tape
> generator TM into a single TM, but, that isn't the issue.
>
> I challenged you to write a TM that can recognize a
> unary represetation on ANY tape I might give you.
>
> The tape I gave you has a finite number of non-blank
> positions and can be completely specified with a TM.
> The tape contains a string that is a member of this
> this tape. Your TM has failed to recognize a valid input.

The abstract machine whose tape is being changed externally while it is
running is simply not a TM. Take another look at e.g. the Wikipedia
page, with especial reference to
http://en.wikipedia.org/wiki/Turing_machine#Formal_definition. Anything
less formal than that is too informal to be of any use for anyone who is
trying to stretch the limits. You end up playing word games with
informal models that are just for casual discussion.

If you want to reason about models of parallel computation in which one
machine's tape is being modified by another machine while the first
machine is running you need to define your own abstract machine model
and demonstrate that it somehow aids in reasoning about computation.

Patricia
```
 0
pats (3556)
7/18/2009 9:53:15 PM
```On Jul 18, 2:53=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:

> The abstract machine whose tape is being changed externally while it is
> running is simply not a TM.

You can't have it both ways. You can't say the output
of a TM is completely specified by the TM and then
complain the TM is still running.

Assume I create a TM to write the digits of PI to a tape.
Is the output on this tape completely specified?

The tape with my uncomputable number is completely
specified by the TM I used to create it.
And I can prove the tape contains a finite string of 1's.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/18/2009 10:14:47 PM
```RussellE wrote:
> On Jul 18, 2:53 pm, Patricia Shanahan <p...@acm.org> wrote:
>> RussellE wrote:
>
>> The abstract machine whose tape is being changed externally while it is
>> running is simply not a TM.
>
>
> You can't have it both ways. You can't say the output
> of a TM is completely specified by the TM and then
> complain the TM is still running.
>
> Assume I create a TM to write the digits of PI to a tape.
> Is the output on this tape completely specified?
>
> The tape with my uncomputable number is completely
> specified by the TM I used to create it.
> And I can prove the tape contains a finite string of 1's.

Sorry, I should have worded that a bit more formally. By "changed
externally" I meant "changed by any means other than the single cell
write specified by the TM's table".

Patricia
```
 0
pats (3556)
7/18/2009 10:22:17 PM
```On Jul 18, 3:22=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:
> > On Jul 18, 2:53 pm, Patricia Shanahan <p...@acm.org> wrote:
> >> RussellE wrote:
>
> >> The abstract machine whose tape is being changed externally while it i=
s
> >> running is simply not a TM.
>
> > You can't have it both ways. You can't say the output
> > of a TM is completely specified by the TM and then
> > complain the TM is still running.
>
> > Assume I create a TM to write the digits of PI to a tape.
> > Is the output on this tape completely specified?
>
> > The tape with my uncomputable number is completely
> > specified by the TM I used to create it.
> > And I can prove the tape contains a finite string of 1's.
>
> Sorry, I should have worded that a bit more formally. By "changed
> externally" I meant "changed by any means other than the single cell
> write specified by the TM's table".

I haven't changed anything.

Assume I give you a tape and tell you this tape
contains the digits of PI in binary.
I also show you the TM I used to create this tape.
Is the string on this tape completely specified?

If so, then my tape with an uncomputable number
is also completely specified. I can't tell you how
long the string of 1's is, but I can prove the tape
has a finite number of 1's followed by a blank.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/18/2009 11:34:40 PM
```RussellE wrote:
> On Jul 18, 3:22 pm, Patricia Shanahan <p...@acm.org> wrote:
>> RussellE wrote:
>>> On Jul 18, 2:53 pm, Patricia Shanahan <p...@acm.org> wrote:
>>>> RussellE wrote:
>>>> The abstract machine whose tape is being changed externally while it is
>>>> running is simply not a TM.
>>> You can't have it both ways. You can't say the output
>>> of a TM is completely specified by the TM and then
>>> complain the TM is still running.
>>> Assume I create a TM to write the digits of PI to a tape.
>>> Is the output on this tape completely specified?
>>> The tape with my uncomputable number is completely
>>> specified by the TM I used to create it.
>>> And I can prove the tape contains a finite string of 1's.
>> Sorry, I should have worded that a bit more formally. By "changed
>> externally" I meant "changed by any means other than the single cell
>> write specified by the TM's table".
>
> I haven't changed anything.
>
> Assume I give you a tape and tell you this tape
> contains the digits of PI in binary.
....

The moment you tell me that, I know that it is not a Turing machine
tape. 0 and 1 both appear infinitely many times in the binary expansion
of PI. If it were a Turing machine tape, only the blank symbol would
appear infinitely many times at any step in the computation.

Patricia
```
 0
pats (3556)
7/18/2009 11:47:19 PM
```On Jul 18, 4:47=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:

> > I haven't changed anything.
>
> > Assume I give you a tape and tell you this tape
> > contains the digits of PI in binary.
>
> The moment you tell me that, I know that it is not a Turing machine
> tape. 0 and 1 both appear infinitely many times in the binary expansion
> of PI. If it were a Turing machine tape, only the blank symbol would
> appear infinitely many times at any step in the computation.

Are you saying PI is uncomputable?

Remember, the string of 1's in my uncomputable number is finite.
This is a finite string written by a TM.
Are you suggesting there exists a finite string a TM can't write?

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/19/2009 2:56:04 AM
```RussellE wrote:
> On Jul 18, 4:47 pm, Patricia Shanahan <p...@acm.org> wrote:
>> RussellE wrote:
>
>>> I haven't changed anything.
>>> Assume I give you a tape and tell you this tape
>>> contains the digits of PI in binary.
>> The moment you tell me that, I know that it is not a Turing machine
>> tape. 0 and 1 both appear infinitely many times in the binary expansion
>> of PI. If it were a Turing machine tape, only the blank symbol would
>> appear infinitely many times at any step in the computation.
>
> Are you saying PI is uncomputable?

PI is a computable number in the sense of
http://en.wikipedia.org/wiki/Computable_number#Formal_definition.
Informally, it can be calculated to any required finite number of
decimal places.

It is not computable in the sense I think you may mean. No Turing
machine can generate its exact binary expansion.

Patricia
```
 0
pats (3556)
7/19/2009 3:27:40 AM
```On Jul 18, 8:27=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:

> PI is a computable number in the sense ofhttp://en.wikipedia.org/wiki/Com=
putable_number#Formal_definition.
> Informally, it can be calculated to any required finite number of
> decimal places.

The uncomputable number can be calculated to any
required finite number of positions.
Every position is a '1'.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/19/2009 3:50:32 AM
```RussellE wrote:
> On Jul 18, 8:27 pm, Patricia Shanahan <p...@acm.org> wrote:
>> RussellE wrote:
>
>
>> PI is a computable number in the sense ofhttp://en.wikipedia.org/wiki/Computable_number#Formal_definition.
>> Informally, it can be calculated to any required finite number of
>> decimal places.
>
> The uncomputable number can be calculated to any
> required finite number of positions.
> Every position is a '1'.

Perhaps you should begin by defining *exactly* what you mean by
"uncomputable number".

There are real numbers that are not computable according to the usual
definition that are not approximated by a string of ones.

Patricia
```
 0
pats (3556)
7/19/2009 4:04:39 AM
```On Jul 18, 9:04=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:
> > On Jul 18, 8:27 pm, Patricia Shanahan <p...@acm.org> wrote:
> >> RussellE wrote:
>
> >> PI is a computable number in the sense ofhttp://en.wikipedia.org/wiki/=
Computable_number#Formal_definition.
> >> Informally, it can be calculated to any required finite number of
> >> decimal places.
>
> > The uncomputable number can be calculated to any
> > required finite number of positions.
> > Every position is a '1'.
>
> Perhaps you should begin by defining *exactly* what you mean by
> "uncomputable number".

I define a one state TM. This TM writes a '1' at the current position
and moves the tape head one to position to the right.
This TM has no halt state so it will just keep writing 1's forever.

I can prove the output of this TM is a finite string of 1's followed
by a blank. The position of the first blank is an uncomputable
natural number.

This tape is specified by every definition you have asked for.
The tape has a finite number of non-blank positions.
I can tell you what is on the tape for any "required number"
of positions. Every computable position contains a '1'.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/19/2009 6:03:38 AM
```"RussellE" <reasterly@gmail.com> wrote in message
On Jul 18, 9:04 pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:
> > On Jul 18, 8:27 pm, Patricia Shanahan <p...@acm.org> wrote:
> >> RussellE wrote:
>
> >> PI is a computable number in the sense
> >> ofhttp://en.wikipedia.org/wiki/Computable_number#Formal_definition.
> >> Informally, it can be calculated to any required finite number of
> >> decimal places.
>
> > The uncomputable number can be calculated to any
> > required finite number of positions.
> > Every position is a '1'.
>
> Perhaps you should begin by defining *exactly* what you mean by
> "uncomputable number".

I define a one state TM. This TM writes a '1' at the current position
and moves the tape head one to position to the right.
This TM has no halt state so it will just keep writing 1's forever.

I can prove the output of this TM is a finite string of 1's followed
by a blank.

*********************
No you can't. You can't prove it because its not true. If you think you can
prove it, try doing so.

```
 0
webbfamily (36)
7/19/2009 6:13:06 AM
```On Jul 18, 11:13=A0pm, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
> "RussellE" <reaste...@gmail.com> wrote in message
>
> On Jul 18, 9:04 pm, Patricia Shanahan <p...@acm.org> wrote:
>
> > RussellE wrote:
> > > On Jul 18, 8:27 pm, Patricia Shanahan <p...@acm.org> wrote:
> > >> RussellE wrote:
>
> > >> PI is a computable number in the sense
> > >> ofhttp://en.wikipedia.org/wiki/Computable_number#Formal_definition.
> > >> Informally, it can be calculated to any required finite number of
> > >> decimal places.
>
> > > The uncomputable number can be calculated to any
> > > required finite number of positions.
> > > Every position is a '1'.
>
> > Perhaps you should begin by defining *exactly* what you mean by
> > "uncomputable number".
>
> I define a one state TM. This TM writes a '1' at the current position
> and moves the tape head one to position to the right.
> This TM has no halt state so it will just keep writing 1's forever.
>
> I can prove the output of this TM is a finite string of 1's followed
> by a blank.
>
> *********************
> No you can't. You can't prove it because its not true. If you think you c=
an
> prove it, try doing so.

OK

I want to keep track of all of the blank positions on the tape.
When the TM starts the tape is all blank.
The set of blank positions is the set of all natural numbers.

A0 =3D {1,2,3,...}

Every time the TM writes a '1' I update my list.
A1 =3D {2,3,4,...}
A2 =3D {3,4,5,...}
A3 =3D {4,5,6,...}
....

For every step performed by the TM the set
of blank positions is infinite.
Since the set of blank positions is non-empty,
there is a smallest blank position.
This proves the string of 1's on the tape is finite.
The output tape is a finite string of 1's followed
by a blank. And no, I can't tell you where the
first blank is. The position of the first blank
is an uncomputable natural number.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/19/2009 7:33:12 AM
```Martin wrote:
RussellE wrote:
> On Jul 18, 11:13=A0pm, "Peter Webb"
> <webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
> > "RussellE" <reaste...@gmail.com> wrote in message
> >
..
> > On Jul 18, 9:04 pm, Patricia Shanahan <p...@acm.org> wrote:
> >
> > > RussellE wrote:
> > > > On Jul 18, 8:27 pm, Patricia Shanahan <p...@acm.org> wrote:
> > > >> RussellE wrote:
> >
> > > >> PI is a computable number in the sense
> > > >> ofhttp://en.wikipedia.org/wiki/Computable_number#Formal_definition=
..
> > > >> Informally, it can be calculated to any required finite number of
> > > >> decimal places.
> >
> > > > The uncomputable number can be calculated to any
> > > > required finite number of positions.
> > > > Every position is a '1'.
> >
> > > Perhaps you should begin by defining *exactly* what you mean by
> > > "uncomputable number".
> >
> > I define a one state TM. This TM writes a '1' at the current position
> > and moves the tape head one to position to the right.
> > This TM has no halt state so it will just keep writing 1's forever.
> >
> > I can prove the output of this TM is a finite string of 1's followed
> > by a blank.
> >
> > *********************
> > No you can't. You can't prove it because its not true. If you think you=
can
> > prove it, try doing so.
>
> OK
>
> I want to keep track of all of the blank positions on the tape.
> When the TM starts the tape is all blank.
> The set of blank positions is the set of all natural numbers.
>
> A0 =3D {1,2,3,...}
>
> Every time the TM writes a '1' I update my list.
> A1 =3D {2,3,4,...}
> A2 =3D {3,4,5,...}
> A3 =3D {4,5,6,...}
> ...
>
> For every step performed by the TM the set
> of blank positions is infinite.
> Since the set of blank positions is non-empty,
> there is a smallest blank position.
> This proves the string of 1's on the tape is finite.
> The output tape is a finite string of 1's followed
> by a blank. And no, I can't tell you where the
> first blank is. The position of the first blank
> is an uncomputable natural number.
>
>
> Russell
> - 2 many 2 count

What if the Martin Machine ran like thus:

Hint: Don't ask me what a Martin Machine is it is exactly like a
Turing Machine accept it comes only in blue and has a trademarked
name. I like blue machines to try something new.

[0] [010] =3D=3D [M < 0]
[0] [101] =3D=3D [M > 0]
[0] [111] =3D=3D [M =3D 0]
[0] [00100] =3D=3D [M < 1]
[0] [11011] =3D=3D [M > 1]
[0] [11111] =3D=3D [M =3D 1]
[0] [0001000] =3D=3D [M < 2]
[0] [1110111] =3D=3D [M > 2]
[0] [1111111] =3D=3D [M =3D 2]
[0] [000010000] =3D=3D [M < 3]
[0] [111101111] =3D=3D [M > 3]
[0] [111111111] =3D=3D [M =3D 3]
[0] [00000100000] =3D=3D [M < 4]
[0] [11111011111] =3D=3D [M > 4]
[0] [11111111111] =3D=3D [M =3D 4]
[0] [0000001000000] =3D=3D [M < 5]
[0] [1111110111111] =3D=3D [M > 5]
[0] [1111111111111] =3D=3D [M =3D 5]
[0] [000000010000000] =3D=3D [M < 6]
[0] [111111101111111] =3D=3D [M > 6]
[0] [111111111111111] =3D=3D [M =3D 6]
[0] [00000000100000000] =3D=3D [M < 7]
[0] [11111111011111111] =3D=3D [M > 7]
[0] [11111111111111111] =3D=3D [M =3D 7]
[0] [0000000001000000000] =3D=3D [M < 8]
[0] [1111111110111111111] =3D=3D [M > 8]
[0] [1111111111111111111] =3D=3D [M =3D 8]
[0] [000000000010000000000] =3D=3D [M < 9]
[0] [111111111101111111111] =3D=3D [M > 9]
[0] [111111111111111111111] =3D=3D [M =3D 9]
[0] [00000000000100000000000] =3D=3D [M < 10]
[0] [11111111111011111111111] =3D=3D [M > 10]
[0] [11111111111111111111111] =3D=3D [M =3D 10]
[0] [0000000000001000000000000] =3D=3D [M < 11]
[0] [1111111111110111111111111] =3D=3D [M > 11]
[0] [1111111111111111111111111] =3D=3D [M =3D 11]
[0] [000000000000010000000000000] =3D=3D [M < 12]
[0] [111111111111101111111111111] =3D=3D [M > 12]
[0] [111111111111111111111111111] =3D=3D [M =3D 12]
[0] [00000000000000100000000000000] =3D=3D [M < 13]
[0] [11111111111111011111111111111] =3D=3D [M > 13]
[0] [11111111111111111111111111111] =3D=3D [M =3D 13]
[0] [0000000000000001000000000000000] =3D=3D [M < 14]
[0] [1111111111111110111111111111111] =3D=3D [M > 14]
[0] [1111111111111111111111111111111] =3D=3D [M =3D 14]
[0] [000000000000000010000000000000000] =3D=3D [M < 15]
[0] [111111111111111101111111111111111] =3D=3D [M > 15]
[0] [111111111111111111111111111111111] =3D=3D [M =3D 15]
[0] [00000000000000000100000000000000000] =3D=3D [M < 16]
[0] [11111111111111111011111111111111111] =3D=3D [M > 16]
[0] [11111111111111111111111111111111111] =3D=3D [M =3D 16]
[0] [0000000000000000001000000000000000000] =3D=3D [M < 17]
[0] [1111111111111111110111111111111111111] =3D=3D [M > 17]
[0] [1111111111111111111111111111111111111] =3D=3D [M =3D 17]
[0] [000000000000000000010000000000000000000] =3D=3D [M < 18]
[0] [111111111111111111101111111111111111111] =3D=3D [M > 18]
[0] [111111111111111111111111111111111111111] =3D=3D [M =3D 18]
[0] [00000000000000000000100000000000000000000] =3D=3D [M < 19]
[0] [11111111111111111111011111111111111111111] =3D=3D [M > 19]
[0] [11111111111111111111111111111111111111111] =3D=3D [M =3D 19]
[0] [0000000000000000000001000000000000000000000] =3D=3D [M < 20]
[0] [1111111111111111111110111111111111111111111] =3D=3D [M > 20]
[0] [1111111111111111111111111111111111111111111] =3D=3D [M =3D 20]
[0] [000000000000000000000010000000000000000000000] =3D=3D [M < 21]
[0] [111111111111111111111101111111111111111111111] =3D=3D [M > 21]
[0] [111111111111111111111111111111111111111111111] =3D=3D [M =3D 21]
[0] [00000000000000000000000100000000000000000000000] =3D=3D [M < 22]
[0] [11111111111111111111111011111111111111111111111] =3D=3D [M > 22]
[0] [11111111111111111111111111111111111111111111111] =3D=3D [M =3D 22]
[0] [0000000000000000000000001000000000000000000000000] =3D=3D [M < 23]
[0] [1111111111111111111111110111111111111111111111111] =3D=3D [M > 23]
[0] [1111111111111111111111111111111111111111111111111] =3D=3D [M =3D 23]
[0] [000000000000000000000000010000000000000000000000000] =3D=3D [M < 24]
[0] [111111111111111111111111101111111111111111111111111] =3D=3D [M > 24]
[0] [111111111111111111111111111111111111111111111111111] =3D=3D [M =3D 24]
[0] [00000000000000000000000000100000000000000000000000000] =3D=3D [M <
25]
[0] [11111111111111111111111111011111111111111111111111111] =3D=3D [M >
25]
[0] [11111111111111111111111111111111111111111111111111111] =3D=3D [M =3D
25]
[0] [0000000000000000000000000001000000000000000000000000000] =3D=3D [M <
26]
[0] [1111111111111111111111111110111111111111111111111111111] =3D=3D [M >
26]
[0] [1111111111111111111111111111111111111111111111111111111] =3D=3D [M =3D
26]
[0] [000000000000000000000000000010000000000000000000000000000] =3D=3D [M
< 27]
[0] [111111111111111111111111111101111111111111111111111111111] =3D=3D [M
> 27]
[0] [111111111111111111111111111111111111111111111111111111111] =3D=3D [M
=3D 27]

Is syi set?
```
 0
scriber17 (147)
7/19/2009 1:34:18 PM
```RussellE wrote:
> On Jul 18, 9:04 pm, Patricia Shanahan <p...@acm.org> wrote:
....
>> Perhaps you should begin by defining *exactly* what you mean by
>> "uncomputable number".
>
> I define a one state TM. This TM writes a '1' at the current position
> and moves the tape head one to position to the right.
> This TM has no halt state so it will just keep writing 1's forever.
>
> I can prove the output of this TM is a finite string of 1's followed
> by a blank. The position of the first blank is an uncomputable
> natural number.
>
> This tape is specified by every definition you have asked for.
> The tape has a finite number of non-blank positions.
> I can tell you what is on the tape for any "required number"
> of positions. Every computable position contains a '1'.

This seems more like an example than a definition. If it is a typical
example, then what you would call "uncomputable number" is what I would
call "ill-defined number".

There is no such thing as "the output" for a non-halting TM.

The state of the tape after n steps, for any natural number n, is well
defined, but I don't think that is what you mean. Assuming empty initial
tape, after n steps the first blank is n steps to the right of the
initial cell.

The position of the first blank is uncomputable only in the sense that
no Turing machine can compute the square of the natural number n without
being told the value of n or data from which it can be derived.

Patricia
```
 0
pats (3556)
7/19/2009 4:23:26 PM
```On Jul 19, 9:23=A0am, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:

> >> Perhaps you should begin by defining *exactly* what you mean by
> >> "uncomputable number".
>
> > I define a one state TM. This TM writes a '1' at the current position
> > and moves the tape head one to position to the right.
> > This TM has no halt state so it will just keep writing 1's forever.
>
> > I can prove the output of this TM is a finite string of 1's followed
> > by a blank. The position of the first blank is an uncomputable
> > natural number.
>
> > This tape is specified by every definition you have asked for.
> > The tape has a finite number of non-blank positions.
> > I can tell you what is on the tape for any "required number"
> > of positions. Every computable position contains a '1'.

> This seems more like an example than a definition. If it is a typical
> example, then what you would call "uncomputable number" is what I would
> call "ill-defined number".

The unary representation of an uncomputable number is a
finite sting of 1's that can't be read or written by any TM
that halts.

What is ill defined about my number?
I even provide a TM that outputs this number.

> There is no such thing as "the output" for a non-halting TM.

Turing would have been very upset to learn this.
In his paper "ON COMPUTABLE NUMBERS, WITH AN
APPLICATION TO THE ENTSCHEIDUNGSPROBLEM"
http://www.comlab.ox.ac.uk/activities/ieg/e-library/sources/tp2-ie.pdf

Turing speaks about circular and circle free machines.
A circular machine "never writes down more than a finite number of
symbols"..
Turing then goes on to show how circle free (non-halting) machines
can be used to compute the binary representation of real numbers.

> The state of the tape after n steps, for any natural number n, is well
> defined,

This is true for any computable natural number, n.

I have proven the string from my TM must be finite.
This is a finite string of 1's that can't be read or written
by any TM that halts.

Russell
- 2 many 2 count
```
 0
7/19/2009 6:35:54 PM
```Kyle Easterly <keileeneast@gmail.com> writes:
<snip>
> I have proven the string from my TM must be finite.

This is to confuse "argue" with "prove".  Arguments about infinities
are notoriously slippery.

> This is a finite string of 1's that can't be read or written
> by any TM that halts.

Since we can both write a machine that can "read" any such string and
then halt, you must also know that that claim is false.

> Russell
> - 2 many 2 count

Posting under multiple identities is not going to help you to be
taken seriously.

--
Ben.
```
 0
ben.usenet (6790)
7/19/2009 9:42:45 PM
```On Jul 19, 2:42=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> Kyle Easterly <keileene...@gmail.com> writes:
>
> <snip>
>
> > I have proven the string from my TM must be finite.
>
> This is to confuse "argue" with "prove". =A0Arguments about infinities
> are notoriously slippery.

Yes they are. Is there some part of my argument you don't agree with?

> > This is a finite string of 1's that can't be read or written
> > by any TM that halts.
>
> Since we can both write a machine that can "read" any such string and
> then halt, you must also know that that claim is false.

I have already described such a string.

> Posting under multiple identities is not going to help you to be
> taken seriously.

My daughter checked her email while I was away from the computer.
Sorry for the confusion.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/19/2009 10:22:33 PM
```RussellE <reasterly@gmail.com> writes:

> On Jul 19, 2:42 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> Kyle Easterly <keileene...@gmail.com> writes:
>>
>> <snip>
>>
>> > I have proven the string from my TM must be finite.
>>
>> This is to confuse "argue" with "prove".  Arguments about infinities
>> are notoriously slippery.
>
> Yes they are. Is there some part of my argument you don't agree
> with?

Yes, and I think I pointed out what they were at the time but the
thread has moved on so much I no longer know what your argument is.
Please don't repeat it for my benefit.  I would appreciate some more
formality but that is up to you.

If you have a TM that writes 1s in consecutive cells and never halts
and you are claiming that some property of this TM is finite, I'd like
to see the machine (it must be very simple) and a more formal
statement of what is finite.  There is no obvious thing called "the
string from this TM".

>> > This is a finite string of 1's that can't be read or written
>> > by any TM that halts.
>>
>> Since we can both write a machine that can "read" any such string and
>> then halt, you must also know that that claim is false.
>
> I have already described such a string.

You have neither defined it nor proved any properties about it.  The
reason I am suspicious is simply that "a finite string of 1's" is
trivially "read" (I have to inset my own definition here but I think
we probably agree on what "reading a string" is) by an obvious halting
TM.  This suggests that your description would not stand up to being
formalised.  To persuade some else you have to avoid vague
descriptions and terms like "the string from my TM" (or at least you
must define them).

>> Posting under multiple identities is not going to help you to be
>> taken seriously.
>
> My daughter checked her email while I was away from the computer.
> Sorry for the confusion.

Ah, OK.  It did not really look a typical nym-shift!

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 12:51:47 AM
```On Jul 19, 5:51=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:
> > On Jul 19, 2:42=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> >> Kyle Easterly <keileene...@gmail.com> writes:
>
> >> <snip>
>
> >> > I have proven the string from my TM must be finite.
>
> >> This is to confuse "argue" with "prove". =A0Arguments about infinities
> >> are notoriously slippery.
>
> > Yes they are. Is there some part of my argument you don't agree
> > with?
>
> Yes, and I think I pointed out what they were at the time but the
> thread has moved on so much I no longer know what your argument is.
> Please don't repeat it for my benefit. =A0I would appreciate some more
> formality but that is up to you.
>
> If you have a TM that writes 1s in consecutive cells and never halts
> and you are claiming that some property of this TM is finite, I'd like
> to see the machine (it must be very simple) and a more formal
> statement of what is finite. =A0There is no obvious thing called "the
> string from this TM".

This TM has a single state.
On any input it writes a '1' at the current position and
moves the tape head one position to the right.
That is it. The TM loops forever writing 1's.

I can define this TM as a 4-tuple if you want.
I am not that familiar with that formalism.
I normally work with things like busy beavers.
Written as a BB this would be:
Any: 1R
This means on any input write a 1 and move right.
Let's call this the Big Number TM (BNTM).

> >> > This is a finite string of 1's that can't be read or written
> >> > by any TM that halts.
>
> >> Since we can both write a machine that can "read" any such string and
> >> then halt, you must also know that that claim is false.
>
> > I have already described such a string.
>
> You have neither defined it nor proved any properties about it.

I have given the proof this string must be finite several times.
Basicly, I show the TM will never run out of blank positions.
=A0
> The
> reason I am suspicious is simply that "a finite string of 1's" is
> trivially "read" (I have to inset my own definition here but I think
> we probably agree on what "reading a string" is) by an obvious halting
> TM.

Can a halting TM read the string produced by the BNTM?

Several people have complained the output of BNTM
is "ill defined".The output of BNTM is as well defined
as the output of any TM that always moves right
and never writes to a position twice.

If I described a TM that outputs the digits of Pi in binary,
few people would complain the output of the Pi TM
was ill defined.

I will give a short version of the proof this string
must be finite.

Most people seem to think the output of BNTM
is an infinite string of 1's. They reason the TM
will "eventually" write a '1' to every finite position.

I will now prove if a TM can write over every
blank position on a tape, the length of the
blank tape must be fixed and finite.

If the TM overwrites every blank position on the tape,
then the set of of blank positions will be the empty set.
I can assign a natural number to each step the TM
performs. I want to list all of the blank position after
each step:

S0 =3D {1,2,3,4,...} =3D set of blank positions at start.
S1 =3D {2,3,4,5,...) =3D set of blank positions after first write
S2 =3D {3,4,5,6,...}
....
....
....
Sz-1 =3D {z}
Sz =3D {} =3D empty set when last position overwritten

z must be a natural number.
The TM has completed its task in a finite number
of steps proving the length of the blank tape is finite.
It also proves the TM has written to a finite number
of positions.

So, even if BNTM writes to every position on the tape,
it has still only written a finite number of characters.

Of course, if BNTM has not written to
every blank position on the tape then there
are still blank positions on the tape, again
proving the output of BNTM is finite.

The output of BNTM must be a finite string of 1's
followed by a blank. And no TM that halts can read it.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/20/2009 2:22:18 AM
```
Marshall wrote:
> On Jul 17, 11:37 pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> > On 2009-07-18 00:03:03 -0400, RussellE <reaste...@gmail.com> said:
> >
> > > I want to prove the language of unary representations of natural
> > > numbers is not recursively enumerable.
> >
> > I'll come back to this, because it's hilarious.
>
> You came back to it, but what of the hilarity?
>
> I myself want to prove that almost all natural numbers are
> really really big. I mean just freakin' huge.
>
> Proof: consider some natural number n. The number 10,000 * n
> is totally way bigger than that. And you can even repeat this
> process like, fifty times or more. I know, right? Ginormous.

It is easily demonstrated that nearly all (all but a finite number
of) naturals are large.  The problem is putting a constructive bound
on the the set of small (i.e., non-large) naturals.

--
hz
```
 0
herbzet (87)
7/20/2009 2:58:19 AM
```RussellE <reasterly@gmail.com> writes:

> On Jul 19, 5:51 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > On Jul 19, 2:42 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> >> Kyle Easterly <keileene...@gmail.com> writes:
>>
>> >> <snip>
>>
>> >> > I have proven the string from my TM must be finite.
>>
>> >> This is to confuse "argue" with "prove".  Arguments about infinities
>> >> are notoriously slippery.
>>
>> > Yes they are. Is there some part of my argument you don't agree
>> > with?
>>
>> Yes, and I think I pointed out what they were at the time but the
>> thread has moved on so much I no longer know what your argument is.
>> Please don't repeat it for my benefit.  I would appreciate some more
>> formality but that is up to you.
>>
>> If you have a TM that writes 1s in consecutive cells and never halts
>> and you are claiming that some property of this TM is finite, I'd like
>> to see the machine (it must be very simple) and a more formal
>> statement of what is finite.  There is no obvious thing called "the
>> string from this TM".
>
> This TM has a single state.
> On any input it writes a '1' at the current position and
> moves the tape head one position to the right.
> That is it. The TM loops forever writing 1's.
>
> I can define this TM as a 4-tuple if you want.
> I am not that familiar with that formalism.
> I normally work with things like busy beavers.
> Written as a BB this would be:
>   Any: 1R
> This means on any input write a 1 and move right.
> Let's call this the Big Number TM (BNTM).

That's what I thought.  OK.

>> >> > This is a finite string of 1's that can't be read or written
>> >> > by any TM that halts.
>>
>> >> Since we can both write a machine that can "read" any such string and
>> >> then halt, you must also know that that claim is false.
>>
>> > I have already described such a string.
>>
>> You have neither defined it nor proved any properties about it.
>
> I have given the proof this string must be finite several times.

You have not defined it.  What is "this string".  There are a whole
bunch of strings (of 1s) on the tape at various stages in the
execution.  They are all finite and they can all be "read" by another
(halting) TM machine.

> Basicly, I show the TM will never run out of blank positions.

Of course.  At no stage can there be more than a finite number of
non-blanks.  None of these strings represent an un-computable
number.

>> The
>> reason I am suspicious is simply that "a finite string of 1's" is
>> trivially "read" (I have to inset my own definition here but I think
>> we probably agree on what "reading a string" is) by an obvious halting
>> TM.
>
> Can a halting TM read the string produced by the BNTM?

A halting TM can "read" any of the tapes produced by BNTM at any point
in its execution.  "The [single] string produced" is something in your
imagination.  You'd like it to be simultaneously finite and
un-readable in a finite number of transitions by another TM.  These
two are contradictory, which tells me that there is no such string.

If you can formalise the definition of the "the string" produced by a
TM that does not halt, we can talk about it, but the only meaning I
can think of will not help you (it is simply an infinite sequence of
1s).

> Several people have complained the output of BNTM
> is "ill defined".The output of BNTM is as well defined
> as the output of any TM that always moves right
> and never writes to a position twice.

If we model the meaning on Turing's notion of computable numbers then,
yes, there is such a string, but it has no end (just as pi has no last
digit).  That does not help you.  How do you define "the string" in
such a way that it is finite?

> If I described a TM that outputs the digits of Pi in binary,
> few people would complain the output of the Pi TM
> was ill defined.

Quite.  But if you claimed it was a finite string ("because the
machine never runs out of blanks") you would get the same opposition.

> I will give a short version of the proof this string
> must be finite.
>
> Most people seem to think the output of BNTM
> is an infinite string of 1's. They reason the TM
> will "eventually" write a '1' to every finite position.
>
> I will now prove if a TM can write over every
> blank position on a tape, the length of the
> blank tape must be fixed and finite.
>
> If the TM overwrites every blank position on the tape,
> then the set of of blank positions will be the empty set.

Note the "if".

> I can assign a natural number to each step the TM
> performs. I want to list all of the blank position after
> each step:
>
> S0 = {1,2,3,4,...} = set of blank positions at start.
> S1 = {2,3,4,5,...) = set of blank positions after first write
> S2 = {3,4,5,6,...}
> ...
> ...
> ...
> Sz-1 = {z}
> Sz = {} = empty set when last position overwritten
>
> z must be a natural number.
> The TM has completed its task in a finite number
> of steps proving the length of the blank tape is finite.
> It also proves the TM has written to a finite number
> of positions.

All you  have shown is that your assumption at the top is false.  You
have shown that assuming a TM can exhaust the tape gives you a
contradiction.  You must conclude that a TM can't exhaust the tape.

Instead you need to start by telling us how you define "the
string" that results from a non-halting TM.  Then you can reason about
it.

Here is my definition: at every step, s, of a TM's execution, the tape
can be seen a function from Z to Sigma (the TM's alphabet including
blank), i.e. a TM defines F: N x Z -> Sigma.  We define the result of
a TM to be the function R:

R(x) = y if there exists k in N s.t. F(m, x) = y for all m >= k.

This string is finite if there exists i and j in Z s.t. R(x) = ' ' for
all x <= i and x >= j.  R[BNTM] is not finite by this definition.

> So, even if BNTM writes to every position on the tape,
> it has still only written a finite number of characters.

So that assumption has to be rejected.  At every stage, a TM's tape
has a finite number of non-blank symbol.

> Of course, if BNTM has not written to
> every blank position on the tape then there
> are still blank positions on the tape, again
> proving the output of BNTM is finite.

Ah yes.  That is my view (see the more formal definition above).  If
we bend the definition of a TM so that the "result" of one can be the
"input" for another, then no TM can read BNTM's result and halt.  Note
we do need a new definition here -- a classic TM always has a finite
number of non-blanks on its input tape -- but we don't get any odd
extended kind of TM.

> The output of BNTM must be a finite string of 1's
> followed by a blank. And no TM that halts can read it.

Not by my reasoning, no.

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 3:37:32 AM
```Ben Bacarisse wrote:
> RussellE <reasterly@gmail.com> writes:
....
>> Several people have complained the output of BNTM
>> is "ill defined".The output of BNTM is as well defined
>> as the output of any TM that always moves right
>> and never writes to a position twice.
>
> If we model the meaning on Turing's notion of computable numbers then,
> yes, there is such a string, but it has no end (just as pi has no last
> digit).  That does not help you.  How do you define "the string" in
> such a way that it is finite?

One of the issues here is the distinction between the members of a
sequence and the limit of a sequence. A computable number can be the
limit of the sequence of numbers produced by a non-terminating TM
without ever actually appearing on the tape.

Patricia
```
 0
pats (3556)
7/20/2009 3:49:07 AM
```On Jul 19, 8:37=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:
> > On Jul 19, 5:51=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> >> RussellE <reaste...@gmail.com> writes:
> >> > On Jul 19, 2:42=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> >> >> Kyle Easterly <keileene...@gmail.com> writes:
>
> >> >> <snip>
>
> >> >> > I have proven the string from my TM must be finite.
>
> >> >> This is to confuse "argue" with "prove". =A0Arguments about infinit=
ies
> >> >> are notoriously slippery.
>
> >> > Yes they are. Is there some part of my argument you don't agree
> >> > with?
>
> >> Yes, and I think I pointed out what they were at the time but the
> >> thread has moved on so much I no longer know what your argument is.
> >> Please don't repeat it for my benefit. =A0I would appreciate some more
> >> formality but that is up to you.
>
> >> If you have a TM that writes 1s in consecutive cells and never halts
> >> and you are claiming that some property of this TM is finite, I'd like
> >> to see the machine (it must be very simple) and a more formal
> >> statement of what is finite. =A0There is no obvious thing called "the
> >> string from this TM".
>
> > This TM has a single state.
> > On any input it writes a '1' at the current position and
> > moves the tape head one position to the right.
> > That is it. The TM loops forever writing 1's.
>
> > I can define this TM as a 4-tuple if you want.
> > I am not that familiar with that formalism.
> > I normally work with things like busy beavers.
> > Written as a BB this would be:
> > =A0 Any: 1R
> > This means on any input write a 1 and move right.
> > Let's call this the Big Number TM (BNTM).
>
> That's what I thought. =A0OK.
>
> >> >> > This is a finite string of 1's that can't be read or written
> >> >> > by any TM that halts.
>
> >> >> Since we can both write a machine that can "read" any such string a=
nd
> >> >> then halt, you must also know that that claim is false.
>
> >> > I have already described such a string.
>
> >> You have neither defined it nor proved any properties about it.
>
> > I have given the proof this string must be finite several times.
>
> You have not defined it. =A0What is "this string". =A0There are a whole
> bunch of strings (of 1s) on the tape at various stages in the
> execution. =A0They are all finite and they can all be "read" by another
> (halting) TM machine.
>
> > Basicly, I show the TM will never run out of blank positions.
>
> Of course. =A0At no stage can there be more than a finite number of
> non-blanks. =A0None of these strings represent an un-computable
> number.
> =A0 =A0
>
> >> The
> >> reason I am suspicious is simply that "a finite string of 1's" is
> >> trivially "read" (I have to inset my own definition here but I think
> >> we probably agree on what "reading a string" is) by an obvious halting
> >> TM.
>
> > Can a halting TM read the string produced by the BNTM?
>
> A halting TM can "read" any of the tapes produced by BNTM at any point
> in its execution. =A0"The [single] string produced" is something in your
> imagination. =A0You'd like it to be simultaneously finite and
> un-readable in a finite number of transitions by another TM. =A0These
> two are contradictory, which tells me that there is no such string.
>
> If you can formalise the definition of the "the string" produced by a
> TM that does not halt, we can talk about it, but the only meaning I
> can think of will not help you (it is simply an infinite sequence of
> 1s).
>
> > Several people have complained the output of BNTM
> > is "ill defined".The output of BNTM is as well defined
> > as the output of any TM that always moves right
> > and never writes to a position twice.
>
> If we model the meaning on Turing's notion of computable numbers then,
> yes, there is such a string, but it has no end (just as pi has no last
> digit). =A0That does not help you. =A0How do you define "the string" in
> such a way that it is finite?
>
> > If I described a TM that outputs the digits of Pi in binary,
> > few people would complain the output of the Pi TM
> > was ill defined.
>
> Quite. =A0But if you claimed it was a finite string ("because the
> machine never runs out of blanks") you would get the same opposition.
>
> > I will give a short version of the proof this string
> > must be finite.
>
> > Most people seem to think the output of BNTM
> > is an infinite string of 1's. They reason the TM
> > will "eventually" write a '1' to every finite position.
>
> > I will now prove if a TM can write over every
> > blank position on a tape, the length of the
> > blank tape must be fixed and finite.
>
> > If the TM overwrites every blank position on the tape,
> > then the set of of blank positions will be the empty set.
>
> Note the "if".
>
>
>
>
>
> > I can assign a natural number to each step the TM
> > performs. I want to list all of the blank position after
> > each step:
>
> > S0 =3D {1,2,3,4,...} =3D set of blank positions at start.
> > S1 =3D {2,3,4,5,...) =3D set of blank positions after first write
> > S2 =3D {3,4,5,6,...}
> > ...
> > ...
> > ...
> > Sz-1 =3D {z}
> > Sz =3D {} =3D empty set when last position overwritten
>
> > z must be a natural number.
> > The TM has completed its task in a finite number
> > of steps proving the length of the blank tape is finite.
> > It also proves the TM has written to a finite number
> > of positions.
>
> All you =A0have shown is that your assumption at the top is false. =A0You
> have shown that assuming a TM can exhaust the tape gives you a
> contradiction. =A0You must conclude that a TM can't exhaust the tape.
>
> Instead you need to start by telling us how you define "the
> string" that results from a non-halting TM. =A0Then you can reason about
> it.
>
> Here is my definition: at every step, s, of a TM's execution, the tape
> can be seen a function from Z to Sigma (the TM's alphabet including
> blank), i.e. a TM defines F: N x Z -> Sigma. =A0We define the result of
> a TM to be the function R:
>
> =A0R(x) =3D y if there exists k in N s.t. F(m, x) =3D y for all m >=3D k.

> This string is finite if there exists i and j in Z s.t. R(x) =3D ' ' for
> all x <=3D i and x >=3D j. =A0R[BNTM] is not finite by this definition.

I am still trying to understand your system,
but I think I have to object to this part.
You seem to be saying a string is finite if and only if
we can compute a finite bound to the size of the string.

R[BNTM] is provably finite whether we can compute a
finite bound or not. This could even be the definition
of an uncomputable natural number:
A natural number that can't be bound by any
computable natural number.

> > So, even if BNTM writes to every position on the tape,
> > it has still only written a finite number of characters.
>
> So that assumption has to be rejected. =A0At every stage, a TM's tape
> has a finite number of non-blank symbol.
>
> > Of course, if BNTM has not written to
> > every blank position on the tape then there
> > are still blank positions on the tape, again
> > proving the output of BNTM is finite.
>
> Ah yes. =A0That is my view (see the more formal definition above). =A0If
> we bend the definition of a TM so that the "result" of one can be the
> "input" for another, then no TM can read BNTM's result and halt. =A0Note
> we do need a new definition here -- a classic TM always has a finite
> number of non-blanks on its input tape -- but we don't get any odd
> extended kind of TM.
>
> > The output of BNTM must be a finite string of 1's
> > followed by a blank. And no TM that halts can read it.
>
> Not by my reasoning, no.

I have shown the BNTM writes a finite string of 1's if
it writes to every position on the tape and BNTM
writes a finite string of 1's if it doesn't write to
every position on the tape.

There is no third option here.
R[BNTM] is a finite string of 1's.

True, we can't compute a finite bound on
the length of the string of 1's,
But, there is no question about what is on the tape.
It is provably not an infinite string of 1's.

Whether or not a TM halts is irrelevant to what is
on the tape it produces, assuming the TM doesn't
write to the same position more than once

If I wrote a TM to compute the digits of Pi in binary,
you wouldn't question what is on the tape
(you might, but most people wouldn't).
The fact the Pi TM doesn't halt doesn't
change what is on the tape it produces.

I'm sure lots of people questioned what was
on the Pi tape in 1939 when Turing wrote his paper
and they should continue to do so now.

As I think you alluded to above, if you accept
my argument then the output of the Pi TM
will also have a blank in an uncomputable
position.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/20/2009 10:06:33 AM
```Patricia Shanahan <pats@acm.org> writes:

> Ben Bacarisse wrote:
>> RussellE <reasterly@gmail.com> writes:
> ...
>>> Several people have complained the output of BNTM
>>> is "ill defined".The output of BNTM is as well defined
>>> as the output of any TM that always moves right
>>> and never writes to a position twice.
>>
>> If we model the meaning on Turing's notion of computable numbers then,
>> yes, there is such a string, but it has no end (just as pi has no last
>> digit).  That does not help you.  How do you define "the string" in
>> such a way that it is finite?
>
> One of the issues here is the distinction between the members of a
> sequence and the limit of a sequence. A computable number can be the
> limit of the sequence of numbers produced by a non-terminating TM
> without ever actually appearing on the tape.

Quite.  My definition was an attempt to formalise this limit.  I don't
know if the definition is really sound since I made it up as I was
going along, but it is a starting point.  The OP won't like the
definition (because it does lead to any odd consequences that I can
see) but then the only forward is for him to try to define this result
string in a way that can be reasoned about.

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 12:00:54 PM
```On Jul 19, 8:37=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

> > Of course, if BNTM has not written to
> > every blank position on the tape then there
> > are still blank positions on the tape, again
> > proving the output of BNTM is finite.
>
> Ah yes. =A0That is my view (see the more formal definition above). =A0If
> we bend the definition of a TM so that the "result" of one can be the
> "input" for another, then no TM can read BNTM's result and halt. =A0Note
> we do need a new definition here -- a classic TM always has a finite
> number of non-blanks on its input tape -- but we don't get any odd
> extended kind of TM.
>
> > The output of BNTM must be a finite string of 1's
> > followed by a blank. And no TM that halts can read it.
>
> Not by my reasoning, no.

I think I can better explain my objection to your
definition of finite.

Assume you are correct and R[BNTM] is not finite.
This means every position of R[BNTM] is a '1'.
But, I have already shown if BNTM can write to
every position on a tape then the tape itself is
of fixed and finite length.

By proving R[BNTM] is not finite you are also
proving the set of all natural numbers IS finite.

I don't think you want to do that.

I can make this more formal.

For every position n in N, BNTM will either write
to position n or it won't write to position n.

Assume BNTM writes to every position n in N.
If a TM can write to every position of a tape,
the tape itself is of fixed and finite length.

The set of all natural numbers is not finite.
Therefore, there exists a position BNTM
does not write to. This position is an
uncomputable natural number.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/20/2009 12:35:53 PM
```RussellE <reasterly@gmail.com> writes:

> On Jul 19, 8:37 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > On Jul 19, 5:51 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> >> RussellE <reaste...@gmail.com> writes:
>> >> > On Jul 19, 2:42 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> >> >> Kyle Easterly <keileene...@gmail.com> writes:
>>
>> >> >> <snip>
>>
>> >> >> > I have proven the string from my TM must be finite.
>>
>> >> >> This is to confuse "argue" with "prove".  Arguments about infinities
>> >> >> are notoriously slippery.
>>
>> >> > Yes they are. Is there some part of my argument you don't agree
>> >> > with?
>>
>> >> Yes, and I think I pointed out what they were at the time but the
>> >> thread has moved on so much I no longer know what your argument is.
>> >> Please don't repeat it for my benefit.  I would appreciate some more
>> >> formality but that is up to you.
>>
>> >> If you have a TM that writes 1s in consecutive cells and never halts
>> >> and you are claiming that some property of this TM is finite, I'd like
>> >> to see the machine (it must be very simple) and a more formal
>> >> statement of what is finite.  There is no obvious thing called "the
>> >> string from this TM".
>>
>> > This TM has a single state.
>> > On any input it writes a '1' at the current position and
>> > moves the tape head one position to the right.
>> > That is it. The TM loops forever writing 1's.
>>
>> > I can define this TM as a 4-tuple if you want.
>> > I am not that familiar with that formalism.
>> > I normally work with things like busy beavers.
>> > Written as a BB this would be:
>> >   Any: 1R
>> > This means on any input write a 1 and move right.
>> > Let's call this the Big Number TM (BNTM).
>>
>> That's what I thought.  OK.
>>
>> >> >> > This is a finite string of 1's that can't be read or written
>> >> >> > by any TM that halts.
>>
>> >> >> Since we can both write a machine that can "read" any such string and
>> >> >> then halt, you must also know that that claim is false.
>>
>> >> > I have already described such a string.
>>
>> >> You have neither defined it nor proved any properties about it.
>>
>> > I have given the proof this string must be finite several times.
>>
>> You have not defined it.  What is "this string".  There are a whole
>> bunch of strings (of 1s) on the tape at various stages in the
>> execution.  They are all finite and they can all be "read" by another
>> (halting) TM machine.
>>
>> > Basicly, I show the TM will never run out of blank positions.
>>
>> Of course.  At no stage can there be more than a finite number of
>> non-blanks.  None of these strings represent an un-computable
>> number.
>>
>>
>> >> The
>> >> reason I am suspicious is simply that "a finite string of 1's" is
>> >> trivially "read" (I have to inset my own definition here but I think
>> >> we probably agree on what "reading a string" is) by an obvious halting
>> >> TM.
>>
>> > Can a halting TM read the string produced by the BNTM?
>>
>> A halting TM can "read" any of the tapes produced by BNTM at any point
>> in its execution.  "The [single] string produced" is something in your
>> imagination.  You'd like it to be simultaneously finite and
>> un-readable in a finite number of transitions by another TM.  These
>> two are contradictory, which tells me that there is no such string.
>>
>> If you can formalise the definition of the "the string" produced by a
>> TM that does not halt, we can talk about it, but the only meaning I
>> can think of will not help you (it is simply an infinite sequence of
>> 1s).
>>
>> > Several people have complained the output of BNTM
>> > is "ill defined".The output of BNTM is as well defined
>> > as the output of any TM that always moves right
>> > and never writes to a position twice.
>>
>> If we model the meaning on Turing's notion of computable numbers then,
>> yes, there is such a string, but it has no end (just as pi has no last
>> digit).  That does not help you.  How do you define "the string" in
>> such a way that it is finite?
>>
>> > If I described a TM that outputs the digits of Pi in binary,
>> > few people would complain the output of the Pi TM
>> > was ill defined.
>>
>> Quite.  But if you claimed it was a finite string ("because the
>> machine never runs out of blanks") you would get the same opposition.
>>
>> > I will give a short version of the proof this string
>> > must be finite.
>>
>> > Most people seem to think the output of BNTM
>> > is an infinite string of 1's. They reason the TM
>> > will "eventually" write a '1' to every finite position.
>>
>> > I will now prove if a TM can write over every
>> > blank position on a tape, the length of the
>> > blank tape must be fixed and finite.
>>
>> > If the TM overwrites every blank position on the tape,
>> > then the set of of blank positions will be the empty set.
>>
>> Note the "if".
>>
>> > I can assign a natural number to each step the TM
>> > performs. I want to list all of the blank position after
>> > each step:
>>
>> > S0 = {1,2,3,4,...} = set of blank positions at start.
>> > S1 = {2,3,4,5,...) = set of blank positions after first write
>> > S2 = {3,4,5,6,...}
>> > ...
>> > ...
>> > ...
>> > Sz-1 = {z}
>> > Sz = {} = empty set when last position overwritten
>>
>> > z must be a natural number.
>> > The TM has completed its task in a finite number
>> > of steps proving the length of the blank tape is finite.
>> > It also proves the TM has written to a finite number
>> > of positions.
>>
>> All you  have shown is that your assumption at the top is false.  You
>> have shown that assuming a TM can exhaust the tape gives you a
>> contradiction.  You must conclude that a TM can't exhaust the tape.
>>
>> Instead you need to start by telling us how you define "the
>> string" that results from a non-halting TM.  Then you can reason about
>> it.
>>
>> Here is my definition: at every step, s, of a TM's execution, the tape
>> can be seen a function from Z to Sigma (the TM's alphabet including
>> blank), i.e. a TM defines F: N x Z -> Sigma.  We define the result of
>> a TM to be the function R:
>>
>>  R(x) = y if there exists k in N s.t. F(m, x) = y for all m >= k.
>
>
>> This string is finite if there exists i and j in Z s.t. R(x) = ' ' for
>> all x <= i and x >= j.  R[BNTM] is not finite by this definition.
>
> I am still trying to understand your system,
> but I think I have to object to this part.
> You seem to be saying a string is finite if and only if
> we can compute a finite bound to the size of the string.

There is no requirement to be able to compute the bound but the string
in only finite if there is a bound.  The two are not quite the same.

Anyway, the point is to give you an idea of what a definition of R
might look like.  There is no point in going on using vague language
since that is how you are able to construct a self-contradictory
definition of R in the first place.

> R[BNTM] is provably finite whether we can compute a
> finite bound or not.

No.  You have not defined R[BNTM] yet.  All you are doing is talking
and talking about them is just a parlour game.

Unless you define what *you* mean by the result of this machine you
can't prove anything about it at all.

>  This could even be the definition
> of an uncomputable natural number:
> A natural number that can't be bound by any
> computable natural number.

It could be, yes.  You'd then have to prove that such a thing exists.
I can define a Bacarisse number as an even prime > 2 but that does not
mean there are any.

>> > So, even if BNTM writes to every position on the tape,
>> > it has still only written a finite number of characters.
>>
>> So that assumption has to be rejected.  At every stage, a TM's tape
>> has a finite number of non-blank symbol.
>>
>> > Of course, if BNTM has not written to
>> > every blank position on the tape then there
>> > are still blank positions on the tape, again
>> > proving the output of BNTM is finite.
>>
>> Ah yes.  That is my view (see the more formal definition above).  If
>> we bend the definition of a TM so that the "result" of one can be the
>> "input" for another, then no TM can read BNTM's result and halt.  Note
>> we do need a new definition here -- a classic TM always has a finite
>> number of non-blanks on its input tape -- but we don't get any odd
>> extended kind of TM.
>>
>> > The output of BNTM must be a finite string of 1's
>> > followed by a blank. And no TM that halts can read it.
>>
>> Not by my reasoning, no.
>
> I have shown the BNTM writes a finite string of 1's if
> it writes to every position on the tape and BNTM
> writes a finite string of 1's if it doesn't write to
> every position on the tape.

I am going to stop now because I don't enjoy chatting how odd
infinities are.  I will gladly discuss the consequences of any
definition of R[BNTM] that you want to post, but I don't want to go
round again and again talking about the consequences of what you think
it is.

> There is no third option here.
> R[BNTM] is a finite string of 1's.
>
> True, we can't compute a finite bound on
> the length of the string of 1's,
> But, there is no question about what is on the tape.
> It is provably not an infinite string of 1's.
>
> Whether or not a TM halts is irrelevant to what is
> on the tape it produces, assuming the TM doesn't
> write to the same position more than once
>
> If I wrote a TM to compute the digits of Pi in binary,
> you wouldn't question what is on the tape
> (you might, but most people wouldn't).
> The fact the Pi TM doesn't halt doesn't
> change what is on the tape it produces.

If you take my definition of R, then R[TMpi] is the binary
representation of pi and it is not finite.  Similarly R[BNTM] is not
finite and does not correspond with the unary representation of any
natural number.  You reject my definition of R, so I will wait for
yours.

> I'm sure lots of people questioned what was
> on the Pi tape in 1939 when Turing wrote his paper
> and they should continue to do so now.
>
> As I think you alluded to above, if you accept
> my argument then the output of the Pi TM
> will also have a blank in an uncomputable
> position.

Too woolly.  Take my definition of R[TMpi] and a state table for TMpi
then I could show that there is no x >= 0 s.t. R(x) = ' '.  If that is
what your words mean then so be it, but a clearer wording is "there is
no finite position which will be left blank".  Because there is no
such x, you are enjoying the consequences of assuming a contradiction.

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 12:42:15 PM
```RussellE <reasterly@gmail.com> writes:

> On Jul 19, 8:37 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>> > Of course, if BNTM has not written to
>> > every blank position on the tape then there
>> > are still blank positions on the tape, again
>> > proving the output of BNTM is finite.
>>
>> Ah yes.  That is my view (see the more formal definition above).  If
>> we bend the definition of a TM so that the "result" of one can be the
>> "input" for another, then no TM can read BNTM's result and halt.  Note
>> we do need a new definition here -- a classic TM always has a finite
>> number of non-blanks on its input tape -- but we don't get any odd
>> extended kind of TM.
>>
>> > The output of BNTM must be a finite string of 1's
>> > followed by a blank. And no TM that halts can read it.
>>
>> Not by my reasoning, no.
>
> I think I can better explain my objection to your
> definition of finite.
>
> Assume you are correct and R[BNTM] is not finite.
> This means every position of R[BNTM] is a '1'.
> But, I have already shown if BNTM can write to
> every position on a tape then the tape itself is
> of fixed and finite length.

No, you have not.  You have talked about it.  I *can* show that
R[BNTM](x) = '1' for all x.  It follows from my definition of R and
from the properties of the machine you called BNTM.

a non-terminating TM" and you might get further.

> By proving R[BNTM] is not finite you are also
> proving the set of all natural numbers IS finite.

I didn't, but since R[BNTM](x) = '1' for all x in N, then if this does
indeed have the consequence you describe you have a paper to write.  A
small footnote is all I ask!

> I don't think you want to do that.
>
> I can make this more formal.

Ah!

> For every position n in N, BNTM will either write
> to position n or it won't write to position n.

Yes.

> Assume BNTM writes to every position n in N.

This needs to be formalised, but if you mean that there is no n for
which BNTM does not write a '1' in position n, then it is a sound
assumption.  In fact it can be proved.

> If a TM can write to every position of a tape,
> the tape itself is of fixed and finite length.

No, it shows that the sets N and "the positions the TM can write to"
are the same size (i.e. that there is a bijection between them) and
nothing more.

> The set of all natural numbers is not finite.

Therefore neither is the set of positions written to by BNTM.

> Therefore, there exists a position BNTM
> does not write to. This position is an
> uncomputable natural number.

No.

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 2:00:00 PM
```On Jul 17, 9:03=A0pm, RussellE <reaste...@gmail.com> wrote:

> I am trying to prove there is a uncomputable natural number.

Why are you wasting your time trying to do that? Why not learn some
computability theory, starting with the basics, instead?

What are computable or not are sets of tuples of natural numbers. For
any natural number n, the {n} is computable.

> Once again from Wikipedia:http://en.wikipedia.org/wiki/Turing_recognizabl=
e
>
> A recursively enumerable language is a formal language for which there
> exists a
> Turing machine (or other computable function) that will halt and
> accept when
> presented with any string in the language as input but may either halt
> and reject
> or loop forever when presented with a string not in the language.
>
> I realize this is not the most formal definition, but it is enough
> for me to gripe about.

> I want to prove the language of unary representations of natural
> numbers
> is not recursively enumerable.

Why do you want to prove such a thing?

> I invite you to write a TM to recognize this language.
> You may use whatever formalism you like.
> But, I get to provide the input tape.
> I guarantee this tape has a finite number of non-blank positions.

Such trivial machines can be found in ordinary textbooks on the
subject.

> Now, I want to talk about uncomputable natural numbers.
> How many operations does your TM have to perform before
> you decide your TM will "loop forever"?

That's an ill-premised question. It may be that we don't decide that a
TM with given input halts or not.

> The very definition of Turing recognizable assumes we can
> tell when a TM loops forever.

No it doesn't. How in the world did you ever form such an impression?
You'd do better to actually read the mathematics than to inject your
own incorrect premises.

> can you prove my tape does not contain a unary
> representation of a natural number?

No. For that you need not recursive enumerability but rather recursive
decidability (or whatever appropriate term) .

> If not, the language of unary representations is not
> recusively enumerable.

No, you're confused about recursive enumerability means. Recursive
enumerability means that if the answer is "yes" then the machine will
say "yes" in a finite number of steps. Recursive decidability (or
whatever appropriate term to be used) means that if the answer is
"yes" then the machine will say "yes" in a finite number of steps AND
if the answer is "not then the machine will say "no" in a finite
number of steps.

MoeBlee
```
 0
jazzmobe (307)
7/20/2009 2:20:05 PM
```On Jul 20, 7:00=A0am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

> > For every position n in N, BNTM will either write
> > to position n or it won't write to position n.
>
> Yes.
>
> > Assume BNTM writes to every position n in N.
>
> This needs to be formalised, but if you mean that there is no n for
> which BNTM does not write a '1' in position n, then it is a sound
> assumption. =A0In fact it can be proved.
>
> > If a TM can write to every position of a tape,
> > the tape itself is of fixed and finite length.
>
> No, it shows that the sets N and "the positions the TM can write to"
> are the same size (i.e. that there is a bijection between them) and
> nothing more.

I am asking for the positions BNTM will write to,
not the positions it can write to. You don't get to
say the TM can write to position n, but it hasn't
done so yet.

The output of a TM is provably finite. If you
claim BNTM will write to every position then
you have created a bijection between the naturals
and a finite set.

> > The set of all natural numbers is not finite.
>
> Therefore neither is the set of positions written to by BNTM.

The set of positions written to by BNTM will always be finite.

> > Therefore, there exists a position BNTM
> > does not write to. This position is an
> > uncomputable natural number.
>
> No.

You seem to be demanding I tell you which
position BNTM won't write to. I can't.
The best I can do is show such a position must exist.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/20/2009 7:28:29 PM
```Let me try to explain this better.
Consider the set of positions written to by BNTM
after each step:

S0 = {}
S1 = {1}
S2 = {1,2}
S3 = {1,2,3}
....

This family of sets does not contain
the set of all natural numbers.
This is easily proven because each
set has a largest position.

I think your argument can be stated as follows:

For every n there exists Sn which contains
n proving BNTM writes to position n.
This proves BNTM writes to every finite position.

My argument is that the set of positions
written to by BNTM must a member of the
family of sets I described above.
The set of all natural numbers is not a member
of this family of sets. BNTM can not write to every
finite position.

You are saying the limit of the positions written
is the set of all natural numbers, which is
certainly correct. I am arguing you don't
get to choose the limit. The set of
all natural numbers is not in the set
of possible outcomes.

We can never reach the limit and there
are positions on the tape BNTM will
never write to.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/20/2009 9:35:22 PM
```On Jul 16, 10:14=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 16, 5:36=A0pm, Chip Eastham <hardm...@gmail.com> wrote:
>
> > On Jul 16, 8:25=A0pm, RussellE <reaste...@gmail.com> wrote:
> > > If a TM can't distinguish all finite strings from infinite strings
> > > there must exist finite strings the TM can't recognise.
> > This does not mean a Turing Machine cannot
> > recognize that a finite string/input is
> > finite.
>
> Yes it does.
>

[snip rest of nattering]

Hi, RussellE:

You've editted my post to make it appear
I've said something quite different from
what I did by the simple expedient of
changing contexts.  What I said "does not
mean" that a Turing machine cannot
recognize finite strings/inputs is that
a Turing machine cannot recognize infinite
strings/inputs.  I then went on to sketch
how it is that a Turing machine can
recognize finite strings, which you also

I'm afraid that willfully misconstruing
another's words indicates a desire for
nothing more than to be allowed to wallow
in confusion of your own making.

```
 0
hardmath (81)
7/20/2009 9:49:34 PM
```RussellE <reasterly@gmail.com> writes:

> On Jul 20, 7:00 am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>> > For every position n in N, BNTM will either write
>> > to position n or it won't write to position n.
>>
>> Yes.
>>
>> > Assume BNTM writes to every position n in N.
>>
>> This needs to be formalised, but if you mean that there is no n for
>> which BNTM does not write a '1' in position n, then it is a sound
>> assumption.  In fact it can be proved.
>>
>> > If a TM can write to every position of a tape,
>> > the tape itself is of fixed and finite length.
>>
>> No, it shows that the sets N and "the positions the TM can write to"
>> are the same size (i.e. that there is a bijection between them) and
>> nothing more.
>
> I am asking for the positions BNTM will write to,
> not the positions it can write to. You don't get to
> say the TM can write to position n, but it hasn't
> done so yet.
>
> The output of a TM is provably finite. If you
> claim BNTM will write to every position then
> you have created a bijection between the naturals
> and a finite set.
>
>> > The set of all natural numbers is not finite.
>>
>> Therefore neither is the set of positions written to by BNTM.
>
> The set of positions written to by BNTM will always be finite.
>
>> > Therefore, there exists a position BNTM
>> > does not write to. This position is an
>> > uncomputable natural number.
>>
>> No.
>
> You seem to be demanding I tell you which
> position BNTM won't write to. I can't.
> The best I can do is show such a position must exist.

I would like you to define what you mean with enough formality that is
it possible reason about it.  In particular you need to replace my
attempt at a definition of the output of a TM with your own so that we

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 10:07:17 PM
```RussellE wrote:
> Let me try to explain this better.
> Consider the set of positions written to by BNTM
> after each step:
>
> S0 = {}
> S1 = {1}
> S2 = {1,2}
> S3 = {1,2,3}
> ...
....
> My argument is that the set of positions
> written to by BNTM must a member of the
> family of sets I described above.

Why?

First, of course, you need to define what you mean by "the set of
positions written to by BNTM". One reasonable definition would be the
union over all natural numbers n of Sn.

Having the union of the members as a member is a non-trivial property.
Some sets of sets do, some don't. For example, the union is a member of
{{1},{5},{1,5}}, but not {{1},{5}}.

In this particular case, it is obvious the union is not a member.
Consider some Si. i+1 is not a member of Si, but is a member of its
immediate successor in your list, S(i+1). For any natural number i, Si
is not equal to the union of all the Sn.

If you have some other definition of "the set of positions written to by
BNTM", you need to state it, and prove, not assume, membership in the
family of sets for that definition.

Patricia
```
 0
pats (3556)
7/20/2009 10:16:36 PM
```RussellE <reasterly@gmail.com> writes:

> Let me try to explain this better.
> Consider the set of positions written to by BNTM
> after each step:
>
> S0 = {}
> S1 = {1}
> S2 = {1,2}
> S3 = {1,2,3}
> ...
>
> This family of sets does not contain
> the set of all natural numbers.
> This is easily proven because each
> set has a largest position.
>
> I think your argument can be stated as follows:
>
> For every n there exists Sn which contains
> n proving BNTM writes to position n.
> This proves BNTM writes to every finite position.
>
> My argument is that the set of positions
> written to by BNTM must a member of the
> family of sets I described above.
> The set of all natural numbers is not a member
> of this family of sets. BNTM can not write to every
> finite position.

It is all good until that last conclusion.  It does not follow from
what goes before.  See below for another example of the same logical
leap.

I think it would help you if you tried to define what you mean by that
last sentence.  If I said, in conversation, "M can not write to every
finite position (to the right of the start)" I would mean that there
exists some n such that the machine M does not write beyond position n
for all execution steps, even if M never halts (i.e. it would be
something I most likely have to prove by induction "for all steps").
You obviously don't mean that (because it patently false for BNTM) so
what *do* you mean?

> You are saying the limit of the positions written
> is the set of all natural numbers, which is
> certainly correct. I am arguing you don't
> get to choose the limit. The set of
> all natural numbers is not in the set
> of possible outcomes.

That is normal for a limit.  No intermediate is every equal to the
limit so the set of intermediates never includes it.

> We can never reach the limit and there
> are positions on the tape BNTM will
> never write to.

Yes to the first and no to the second.  By what axiom or law do you
conclude that because the machine never writes all the positions there
are positions it can't write?  It is part of the fun of the parlour game

If you pin things down with some more formal definitions then you
either find that BNTM does, in fact, "reach the limit" (because that
is how you define the output of such a machine) or you find that there
is no cell it won't write despite "not writing to them all".  The
problem is in the language, not in the machine.

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 10:43:26 PM
```On Jul 20, 3:43=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

> > Let me try to explain this better.
> > Consider the set of positions written to by BNTM
> > after each step:
>
> > S0 =3D {}
> > S1 =3D {1}
> > S2 =3D {1,2}
> > S3 =3D {1,2,3}
> > ...
>
> > This family of sets does not contain
> > the set of all natural numbers.
> > This is easily proven because each
> > set has a largest position.
>
> > I think your argument can be stated as follows:
>
> > For every n there exists Sn which contains
> > n proving BNTM writes to position n.
> > This proves BNTM writes to every finite position.
>
> > My argument is that the set of positions
> > written to by BNTM must a member of the
> > family of sets I described above.
> > The set of all natural numbers is not a member
> > of this family of sets. BNTM can not write to every
> > finite position.
>
> It is all good until that last conclusion. =A0It does not follow from
> what goes before. =A0See below for another example of the same logical
> leap.
>
> I think it would help you if you tried to define what you mean by that
> last sentence. =A0If I said, in conversation, "M can not write to every
> finite position (to the right of the start)" I would mean that there
> exists some n such that the machine M does not write beyond position n
> for all execution steps, even if M never halts (i.e. it would be
> something I most likely have to prove by induction "for all steps").
> You obviously don't mean that (because it patently false for BNTM) so
> what *do* you mean?
>
> > You are saying the limit of the positions written
> > is the set of all natural numbers, which is
> > certainly correct. I am arguing you don't
> > get to choose the limit. The set of
> > all natural numbers is not in the set
> > of possible outcomes.
>
> That is normal for a limit. =A0No intermediate is every equal to the
> limit so the set of intermediates never includes it.
>
> > We can never reach the limit and there
> > are positions on the tape BNTM will
> > never write to.
>
> Yes to the first and no to the second. =A0By what axiom or law do you
> conclude that because the machine never writes all the positions there
> are positions it can't write? =A0It is part of the fun of the parlour gam=
e
> of talking about infinities that such things seem odd or paradoxical.
>
> If you pin things down with some more formal definitions then you
> either find that BNTM does, in fact, "reach the limit" (because that
> is how you define the output of such a machine) or you find that there
> is no cell it won't write despite "not writing to them all". =A0The
> problem is in the language, not in the machine.

I think this line line should say:
you find that there is no computable cell
it won't write despite "not writing to them all".

So, how do I prove BNTM doesn't write to them all?

Consider the set of all positions BNTM will ever write to.
Assume this set is the set of all natural numbers.

This means the set of blank positions is empty.
Assign a natural number to each step performed
by BNTM and make a list of all blank positions after
each step:

S0 =3D {1,2,3,...}
S1 =3D {2,3,4,...}
S2 =3D {3,4,5,...}
....
....
....
Sz-1 =3D {z}
Sz =3D {}

Assuming BNTM writes to every finite
position allows me to prove the set
of all natural numbers is finite.

Therefore, the set of all positions BNTM will
ever write to can not be the set of all natural numbers.
There must be at least one position not is this set.

All I have to do is prove such a position exists.
I certainly can't tell you what the position is.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/20/2009 11:45:22 PM
```RussellE <reasterly@gmail.com> writes:
<snip: the same argument again>

I accept that don't agree with my response the last time you posted
this argument, but it is simpler just to say that stand by what you
said before to leave it at that.  There is no point on going round and
round the same argument.

--
Ben.
```
 0
ben.usenet (6790)
7/20/2009 11:55:49 PM
```On Jul 20, 4:55=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

> I accept that don't agree with my response the last time you posted
> this argument, but it is simpler just to say that stand by what you
> said before to leave it at that. =A0There is no point on going round and
> round the same argument.

This has been a very long thread.

I understand you want me to rigorously define what
"string on the tape" means.

Even so, I think I prove whatever string is
produced by BNTM, this string is not an
inifinite string of 1's.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/21/2009 12:51:39 AM
```RussellE wrote:
> On Jul 20, 4:55 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>> I accept that don't agree with my response the last time you posted
>> this argument, but it is simpler just to say that stand by what you
>> said before to leave it at that.  There is no point on going round and
>> round the same argument.
>
> OK. I appreciate your responses.
> This has been a very long thread.
>
> I understand you want me to rigorously define what
> "string on the tape" means.
>
> Even so, I think I prove whatever string is
> produced by BNTM, this string is not an
> inifinite string of 1's.

You still need to define the string produced by BNTM, and prove that

To my mind, there is no such thing as "whatever string is produced by
BNTM". It produces an infinite sequence of strings.

Patricia
```
 0
pats (3556)
7/21/2009 12:56:16 AM
```On Jul 20, 5:56=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:

> To my mind, there is no such thing as "whatever string is produced by
> BNTM". It produces an infinite sequence of strings.

We can consider the infinite sequence of strings
produced by BNTM. We can represent each string
as the set of positions that contain a '1'.

Not a single one of these strings is an infinite string of 1's.

If we want to talk about strings written by a non-halting TM,
we should at least limit ourselves to strings we know a TM
can write.

The only way BNTM will ever write an infinite string of 1's
would be if there is a last natural number.

I have to question formal systems that tell me
BNTM does write an infinite string.

I realize these formal systems define infinite to
mean unbounded by any natural number.
I agree the size of the strings produced by BNTM
isn't bound by any (computable) natural number.

But, most formal systems also define infinite to
mean every finite position. I don't think these two
concepts are identical.

I propose there could be three types of strings:

1) Finite strings where the length can be bounded
by a natural number.

2) Undetermined strings where we know the string
is finite, but we can't place a bound on its length.

3) Infinite strings where the value at every position
on the tape is known.

A TM can write a type 1 string or a type 2 string,
but not a type 3 string. BNTM writes a type 2 string.

Just some ideas. It has been proven thinking
about infinity long enough will drive you crazy.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/21/2009 4:38:35 AM
```RussellE <reasterly@gmail.com> writes:

> On Jul 20, 5:56 pm, Patricia Shanahan <p...@acm.org> wrote:
>> RussellE wrote:
>
>> To my mind, there is no such thing as "whatever string is produced by
>> BNTM". It produces an infinite sequence of strings.
>
> We can consider the infinite sequence of strings
> produced by BNTM. We can represent each string
> as the set of positions that contain a '1'.
>
> Not a single one of these strings is an infinite string of 1's.
>
> If we want to talk about strings written by a non-halting TM,
> we should at least limit ourselves to strings we know a TM
> can write.
>
> The only way BNTM will ever write an infinite string of 1's
> would be if there is a last natural number.
>
> I have to question formal systems that tell me
> BNTM does write an infinite string.

The usual one does not do this so you keep to conventional thinking.
It does say that the string is unbounded.  Calling this "an infinite
string" is a useful shorthand but playing fast a loose with the term
writes more that a finite number of 1s. yet the result is infinite"
sounds odd but replace infinite with unbounded and there is no
oddity.  I don't have a problem with the shorthand because it just

> I realize these formal systems define infinite to
> mean unbounded by any natural number.
> I agree the size of the strings produced by BNTM
> isn't bound by any (computable) natural number.
>
> But, most formal systems also define infinite to
> mean every finite position. I don't think these two
> concepts are identical.

Every finite position gets written to.  How is that not the same as
the previous "unbounded" definition of infinite that you accept?

> I propose there could be three types of strings:
>
> 1) Finite strings where the length can be bounded
> by a natural number.
>
> 2) Undetermined strings where we know the string
> is finite, but we can't place a bound on its length.
>
> 3) Infinite strings where the value at every position
> on the tape is known.

In what sense is the finite length of a string of type 2 not a bound
on its length?  It is trivial to write a TM that manipulates strings
of type 1 and 2.  In fact it is easy to write a TM that computes the
bound of a type 2 string unless you have invented some odd meaning of
the finite that allows for a finite string to have no end (its end
will be the first non-blank to the right of the start position).

> A TM can write a type 1 string or a type 2 string,
> but not a type 3 string. BNTM writes a type 2 string.

No, it writes strings of type 1.  You showed that many times.  When
does it get round to writing one of type 2?

--
Ben.
```
 0
ben.usenet (6790)
7/21/2009 2:59:42 PM
```On Jul 19, 7:22=A0pm, RussellE <reaste...@gmail.com> wrote:

> Can a halting TM read the string produced by the BNTM?

That is typical of the way you get in trouble by using loose language.
There is NO "THE string produced by the BNTM'. Rather, for each finite
number of executed steps, there is a finite string of 1's to the left
of the write-heard.

> Several people have complained the output of BNTM
> is "ill defined".The output of BNTM is as well defined
> as the output of any TM that always moves right
> and never writes to a position twice.

Define "output". The BNTM never stops to produce a single output.
Rather, there is a denumerable SEQUENCE of outputs.

> If I described a TM that outputs the digits of Pi in binary,
> few people would complain the output of the Pi TM
> was ill defined.

No, again your loose language. That TM outputs the digits, ONE AT A
TIME, so that there is a denumerable SEQUENCE of outputS. PLURAL -
outputS.

> I will give a short version of the proof this string
> must be finite.

At any given execution, the string is finite. No one disputes that.
But there is a denumerable SEQUENCE of executions.

> Most people seem to think the output of BNTM
> is an infinite string of 1's.

No, there is no such things as "THE output of the BNTM". Rather, there
is a denumerable SEQUENCE of outputS. OutputS - plural.

> They reason the TM
> will "eventually" write a '1' to every finite position.

For any natural number n, your BNTM writes a string of n 1's. But it
is not the case that the BNTM ever executes an output that is an
infinite string.

> I will now prove if a TM can write over every
> blank position on a tape, the length of the
> blank tape must be fixed and finite.

At any GIVEN execution, the TM can write over only finitely many
positions. But for each position, the TM may write over that
position.

> If the TM overwrites every blank position on the tape,
> then the set of of blank positions will be the empty set.
> I can assign a natural number to each step the TM
> performs. I want to list all of the blank position after
> each step:
>
> S0 =3D {1,2,3,4,...} =3D set of blank positions at start.
> S1 =3D {2,3,4,5,...) =3D set of blank positions after first write
> S2 =3D {3,4,5,6,...}
> ...
> ...
> ...
> Sz-1 =3D {z}
> Sz =3D {} =3D empty set when last position overwritten

NO!! There is no "last position".

You are simply making the SAME MISTAKE you've been making for YEARS
(how many years now?).

Why don't you at least TRY to get rid of this fallacy you carry with
you everywhere?

> z must be a natural number.

There is no such natural number. This is the SAME MISTAKE you've been
making for YEARS.

MoeBlee
```
 0
jazzmobe (307)
7/21/2009 3:48:33 PM
```On Jul 20, 2:49=A0pm, Chip Eastham <hardm...@gmail.com> wrote:
> On Jul 16, 10:14=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > On Jul 16, 5:36=A0pm, Chip Eastham <hardm...@gmail.com> wrote:
>
> > > On Jul 16, 8:25=A0pm, RussellE <reaste...@gmail.com> wrote:
> > > > If a TM can't distinguish all finite strings from infinite strings
> > > > there must exist finite strings the TM can't recognise.
> > > This does not mean a Turing Machine cannot
> > > recognize that a finite string/input is
> > > finite.
>
> > Yes it does.
>
> [snip rest of nattering]
>
> Hi, RussellE:
>
> You've editted my post to make it appear
> I've said something quite different from
> what I did by the simple expedient of
> changing contexts. =A0What I said "does not
> mean" that a Turing machine cannot
> recognize finite strings/inputs is that
> a Turing machine cannot recognize infinite
> strings/inputs. =A0I then went on to sketch
> how it is that a Turing machine can
> recognize finite strings, which you also
>
> I'm afraid that willfully misconstruing
> another's words indicates a desire for
> nothing more than to be allowed to wallow
> in confusion of your own making.

I apologize if I mistconstrued your meaning.
I certainly didn't mean to.

I often snip big chunks when I respond.
I find long posts to be difficult to read
and my response are too long even
when I do snip the OP's post.

I am grateful you accepted my challenge
to creat a recognizer TM. The TM you
described did exactly what I asked for.
And, it has all the properties you describe
in the formal language you are using.

As you have seen, it is difficult to prove
uncomputable numbers exist in a formal
system that assumes such things can't exist.

I was bending the rules, so, in a way, I was
trying to trick you. I apologize.

I hope there are no hard feelings.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/21/2009 6:47:09 PM
```On Jul 21, 2:47=A0pm, RussellE <reaste...@gmail.com> wrote:

[snip]
> I was bending the rules, so, in a way, I was
> trying to trick you. I apologize.
>
> I hope there are no hard feelings.
>
> Russell
> - 2 many 2 count

sincerely, chip

```
 0
hardmath (81)
7/21/2009 7:30:02 PM
```On Jul 21, 7:59=A0am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

> > I have to question formal systems that tell me
> > BNTM does write an infinite string.
>
> The usual one does not do this so you keep to conventional thinking.
> It does say that the string is unbounded. =A0Calling this "an infinite
> string" is a useful shorthand but playing fast a loose with the term
> can lead to seemingly paradoxical remarks. =A0The remark "the TM never
> writes more that a finite number of 1s. yet the result is infinite"
> sounds odd but replace infinite with unbounded and there is no
> oddity. =A0I don't have a problem with the shorthand because it just
> means "unbounded" in my head.

I am essentially trying to show the difference between
unbounded and infinite. "The result is infinite" is
provably false. The result is unbounded.

> > I realize these formal systems define infinite to
> > mean unbounded by any natural number.
> > I agree the size of the strings produced by BNTM
> > isn't bound by any (computable) natural number.
>
> > But, most formal systems also define infinite to
> > mean every finite position. I don't think these two
> > concepts are identical.
>
> Every finite position gets written to. =A0How is that not the same as
> the previous "unbounded" definition of infinite that you accept?

Every finite position does NOT get written to.
If a TM can write to every position on a tape,
the tape is provably bounded and finite.

A string can be unbounded and still not be infinite.
This is the definition I am using for uncomputable.

> > I propose there could be three types of strings:
>
> > 1) Finite strings where the length can be bounded
> > by a natural number.
>
> > 2) Undetermined strings where we know the string
> > is finite, but we can't place a bound on its length.
>
> > 3) Infinite strings where the value at every position
> > on the tape is known.
>
> In what sense is the finite length of a string of type 2 not a bound
> on its length?

We don't know what the length is.
The length of the tape is unbounded.
We can only prove it must be finite.

> =A0It is trivial to write a TM that manipulates strings
> of type 1 and 2. =A0In fact it is easy to write a TM that computes the
> bound of a type 2 string unless you have invented some odd meaning of
> the finite that allows for a finite string to have no end (its end
> will be the first non-blank to the right of the start position).

Exactly. The end is the first blank.
But, we can't know where that blank is.

I am trying to formalize the difference between
unbounded and infinite.

Consider the set of all positions BNTM will ever write to.
I have shown why assuming this set is the set of all

By assuming the TM has run to completion, ie,
every position it will ever write to, we are assuming
the tape is finite.

This proves there must be at least one position
missing from the set of all positions written to
by BNTM. This is the position of the first blank.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/21/2009 7:52:20 PM
```On Jul 21, 8:48=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 19, 7:22=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > Can a halting TM read the string produced by the BNTM?
>
> That is typical of the way you get in trouble by using loose language.
> There is NO "THE string produced by the BNTM'. Rather, for each finite
> number of executed steps, there is a finite string of 1's to the left
> of the write-heard.

You wouldn't object if I talked about THE output produced
by a TM that calculates Pi.

Are you saying there is no such thing as THE tape with
the digits of Pi, only an infinite sequence of finite strings?
(I would.)

Do we agree that no matter how many steps BNTM
performs, there will be a finite strng of 1's to the left

> > Several people have complained the output of BNTM
> > is "ill defined".The output of BNTM is as well defined
> > as the output of any TM that always moves right
> > and never writes to a position twice.
>
> Define "output". The BNTM never stops to produce a single output.
> Rather, there is a denumerable SEQUENCE of outputs.

I define THE output of BNTM as the set of all positions
BNTM will ever write to. If you have a better definition,
I would appreciate if you told me what it is.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/21/2009 8:11:37 PM
```RussellE wrote:
> On Jul 21, 8:48 am, MoeBlee <jazzm...@hotmail.com> wrote:
>> On Jul 19, 7:22 pm, RussellE <reaste...@gmail.com> wrote:
>>
>>> Can a halting TM read the string produced by the BNTM?
>> That is typical of the way you get in trouble by using loose language.
>> There is NO "THE string produced by the BNTM'. Rather, for each finite
>> number of executed steps, there is a finite string of 1's to the left
>> of the write-heard.
>
> You wouldn't object if I talked about THE output produced
> by a TM that calculates Pi.

The closest any TM can get to calculating PI is producing a sequence of
string representations of numbers with finite representations, such as
terminating binary fractions, for which PI is the limit of the numbers,
as the step number tends to infinity.

> Are you saying there is no such thing as THE tape with
> the digits of Pi, only an infinite sequence of finite strings?

Correct.

> I define THE output of BNTM as the set of all positions
> BNTM will ever write to. If you have a better definition,
> I would appreciate if you told me what it is.

"THE output of the BNTM" is a strange term to use for a string that
obviously never appears on the BNTM tape. However, that strangeness
would not matter if you were to use the term consistently and not attach
any assumptions to it other than those you prove from its definition.

Personally, given the fact that the BNTM generates an infinite sequence,
I would define its output as a function from the natural numbers into
the set of finite strings over its output alphabet, such that the
natural number n is mapped to the string that is on the output tape
after exactly n steps.

Patricia
```
 0
pats (3556)
7/21/2009 8:40:23 PM
```On Jul 21, 1:11=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 21, 8:48=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 19, 7:22=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > Can a halting TM read the string produced by the BNTM?
>
> > That is typical of the way you get in trouble by using loose language.
> > There is NO "THE string produced by the BNTM'. Rather, for each finite
> > number of executed steps, there is a finite string of 1's to the left
> > of the write-heard.
>
> You wouldn't object if I talked about THE output produced
> by a TM that calculates Pi.

You need to define 'output' or 'calculates'. There may be two senses
of, say, 'output'. (1) The TM halts and the output is the resulting
tape configuration, or (2) Each step produces a tape configuration,
which is the output for that STEP.

In sense (1), your BNTM HAS NO OUTPUT, since it doesn't halt. In sense
(2), for each natural number n, there is a string of n 1's that is the
output of a step.

As to pi, a TM does NOT output a SINGLE tape configuration that has
each positioned digit of pi sequentially. Rather, either the TM leaves
a tape configuration with a digit at each STEP and continues without
halting. Or, a TM leaves a tape configuration that is a digit of pi,
halts, then the TM may leave the next digit, and halt, up to any
finite number of positions.

Again, a TM may calculate (halt) any FIINITE number of the digits of
pi. Or, a TM may just keep working, listing digits of pi, but never
halting. There is no situation in which a TM lists all digits of pi in
a finite number of steps; nor a situation in which a TM lists all
digits of pi and halts.

In technical terms, we say the TM ENUMERATES the digits of pi, and a
TM may DECIDE any PARTICULAR digit of pi. But the TM does NOT print as
some single final output the ENTIRE infinite expansion of pi.

The difference between (a) a single INFINITE output (which a TM does
NOT produce) and (b) successively putting out, for each n, a string of
n 1's, so that for any n, EVENTUALLY, the TM produces that string of n
1's.

(And such informal, descriptive words, such as 'eventually', that seem
to refer to time or other extra-mathematical notions may be dispensed
when we put this in formal mathematical terminology.)

Got it now?

> Are you saying there is no such thing as THE tape with
> the digits of Pi, only an infinite sequence of finite strings?
> (I would.)

Whatever there is on "tapes" (again 'tape' is just a figure of speech,
there are no actual tapes), there is no situation in which a TM prints
ALL the digits of pi and halts. Rather, a TM prints EACH digit of pi,
one by one, and KEEPS GOING. At no point in the operation are ALL the
digits of pi printed, but rather, at any given stage, for any given
position in the sequence of digits of pi, the machine either has
printed, or will print, the digit in that position.

Got it now?

> Do we agree that no matter how many steps BNTM
> performs, there will be a finite strng of 1's to the left

AT ANY GIVEN stage, after any finite number of steps, the BNTM leaves
a finite string of 1's. And there is NO SUCH THING as a TM completing
an infinite number of steps. Rather, either the TM halts in a finite
number of steps or the machine does not halt, in which case, for any
finite number, the machine will complete that many finite number of
steps - unbounded.

Just like if you were immortal and all you did was count natural
numbers. At any particular point, you will have counted only finitely
many natural numbers, but for any natural number, you will eventually
count it - unbounded.

Got it now?

> > > Several people have complained the output of BNTM
> > > is "ill defined".The output of BNTM is as well defined
> > > as the output of any TM that always moves right
> > > and never writes to a position twice.
>
> > Define "output". The BNTM never stops to produce a single output.
> > Rather, there is a denumerable SEQUENCE of outputs.
>
> I define THE output of BNTM as the set of all positions
> BNTM will ever write to. If you have a better definition,
> I would appreciate if you told me what it is.

I mentioned above. I'd have to check the literature, but my sense is
that a GIVEN output is either a result upon a TM HALTING or the result
at some given finite point in the non-halting operation.

Anyway, so now you should see the basic principle (aside from the
technical details): You have to distinguish between what is on the
tape at any given STEP and what is on the tape after halting occurs
(and IF halting occurs). If halting does NOT occur, then there is no
FINAL output. Yes, for any natural number (or finite sequence of 1's,
or digits of pi, whatever), that natural number will be printed to the
tape but there is no point at which the TM has printed every natural
number.

And that is different from the situation - in which we are not LIMITED
TO TM's - where we may speak of an infinite function with its infinite
range. There is a difference between the FUNCTION a TM computes and
any actual particular TM computation.

Really, PLEASE, why don't you just get a good book on recursion and
computability (that will have a chapter at least on TM's). It's been
how many years - five, ten, MORE? - that you've been floundering in
misconceptions about all of this. Just read a good book and you can
INFORM yourself and get past your continual recycling of your
misconceptions.

MoeBlee

```
 0
jazzmobe (307)
7/21/2009 8:47:29 PM
```On Jul 21, 3:52=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 21, 7:59=A0am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>
> > RussellE <reaste...@gmail.com> writes:
> > > I have to question formal systems that tell me
> > > BNTM does write an infinite string.
>
> > The usual one does not do this so you keep to conventional thinking.
> > It does say that the string is unbounded. =A0Calling this "an infinite
> > string" is a useful shorthand but playing fast a loose with the term
> > can lead to seemingly paradoxical remarks. =A0The remark "the TM never
> > writes more that a finite number of 1s. yet the result is infinite"
> > sounds odd but replace infinite with unbounded and there is no
> > oddity. =A0I don't have a problem with the shorthand because it just
> > means "unbounded" in my head.
>
> I am essentially trying to show the difference between
> unbounded and infinite. "The result is infinite" is
> provably false. The result is unbounded.
>
> > > I realize these formal systems define infinite to
> > > mean unbounded by any natural number.
> > > I agree the size of the strings produced by BNTM
> > > isn't bound by any (computable) natural number.
>
> > > But, most formal systems also define infinite to
> > > mean every finite position. I don't think these two
> > > concepts are identical.
>
> > Every finite position gets written to. =A0How is that not the same as
> > the previous "unbounded" definition of infinite that you accept?
>
> Every finite position does NOT get written to.
> If a TM can write to every position on a tape,
> the tape is provably bounded and finite.
>
> A string can be unbounded and still not be infinite.
> This is the definition I am using for uncomputable.
>
> > > I propose there could be three types of strings:
>
> > > 1) Finite strings where the length can be bounded
> > > by a natural number.
>
> > > 2) Undetermined strings where we know the string
> > > is finite, but we can't place a bound on its length.
>
> > > 3) Infinite strings where the value at every position
> > > on the tape is known.
>
> > In what sense is the finite length of a string of type 2 not a bound
> > on its length?
>
> We don't know what the length is.
> The length of the tape is unbounded.
> We can only prove it must be finite.
>
> > =A0It is trivial to write a TM that manipulates strings
> > of type 1 and 2. =A0In fact it is easy to write a TM that computes the
> > bound of a type 2 string unless you have invented some odd meaning of
> > the finite that allows for a finite string to have no end (its end
> > will be the first non-blank to the right of the start position).
>
> Exactly. The end is the first blank.
> But, we can't know where that blank is.
>
> I am trying to formalize the difference between
> unbounded and infinite.
>
> Consider the set of all positions BNTM will ever write to.
> I have shown why assuming this set is the set of all

What exactly is a BNTM?

I know what a TM (Turing Machine) is. What puzzles me in trying to
read this thread is you seem to be using the opposite of a TM. Turing
looked for machines that never stopped.
>
> By assuming the TM has run to completion, ie,
> every position it will ever write to, we are assuming
> the tape is finite.

Turing's tapes were always infinite. I'm confused and missing some
context obviously.

>
> This proves there must be at least one position
> missing from the set of all positions written to
> by BNTM. This is the position of the first blank.
>
> Russell
> - 2 many 2 count

```
 0
edprochak (546)
7/21/2009 8:55:22 PM
```On Jul 21, 12:52=A0pm, RussellE <reaste...@gmail.com> wrote:

> Every finite position does NOT get written to.
> If a TM can write to every position on a tape,
> the tape is provably bounded and finite.

NO! That is wrong. You are simply DOGMATICALLY asserting what is
incorrect.

AGAIN, at no point has the machine written to every position. But for
every position, at SOME point, the machine will write to it (given
that that machine is programmed to do that, of course).

> A string can be unbounded and still not be infinite.

A given string is either finite or infinite.

What is unbounded is not the string but rather the PROCESS of
producing longer and longer strings.

> I am trying to formalize the difference between
> unbounded and infinite.

That might be hard to do (at least in classical mathematics; you might
have better luck in certain other mathematics).  Better would be for
you to START by understanding the formalizations we have. Then you can
see how the notion of 'unbounded' can be an informal notion. (I mean
aside from the ordinary sense of a set that is not bounded per some
relation.)

> Consider the set of all positions BNTM will ever write to.
> I have shown why assuming this set is the set of all

It is not a contradiction that for any n, the nth position will be
written to, but that any given stage of computation, there are still
positions that haven't been written to.

And, again, part of the problem is that your notion of a TM and of all
this subject matter is trapped in virtually complete vagueness. If you
were to understand the actual formal definitions, then you would see
how it makes fine sense to say

"It is not a contradiction that for any n, the nth position will be
written to, but that any given stage of computation, there are still
positions that haven't been written to."

as a quite informal rendering of a rigorous formulation.

> By assuming the TM has run to completion, ie,
> every position it will ever write to, we are assuming
> the tape is finite.

NO. That is ANOTHER misconception. Rather, the correct statement is
that if the TM prints to every position on the tape and halts, then
the tape is finite.

> This proves there must be at least one position
> missing from the set of all positions written to
> by BNTM. This is the position of the first blank.

Again, for your BNTM there is NO SUCH THING as the set of all
positions written to upon HALTING, because there is no halting.

Again, at any given point in the INFINITELY ONGOING operation, there
are blank positions, but for every blank position (to the right),
there "will be" a point in the operation such that that postion will
be written to.

See it now?

MoeBlee

```
 0
jazzmobe (307)
7/21/2009 9:06:17 PM
```On Jul 21, 1:55=A0pm, Ed Prochak <edproc...@gmail.com> wrote:

> What exactly is a BNTM?

It's his term for a TM that starts by writing '1', moves right, writes

> > By assuming the TM has run to completion, ie,
> > every position it will ever write to, we are assuming
> > the tape is finite.
>
> Turing's tapes were always infinite. I'm confused and missing some
> context obviously.

Don't worry, he's very confused.

MoeBlee
```
 0
jazzmobe (307)
7/21/2009 9:15:26 PM
```On Jul 21, 1:47=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 21, 1:11=A0pm, RussellE <reaste...@gmail.com> wrote:

> The difference between (a) a single INFINITE output (which a TM does
> NOT produce) and (b) successively putting out, for each n, a string of
> n 1's, so that for any n, EVENTUALLY, the TM produces that string of n
> 1's.

Not for any n, See below.

> Just like if you were immortal and all you did was count natural
> numbers. At any particular point, you will have counted only finitely
> many natural numbers, but for any natural number, you will eventually
> count it - unbounded.

Unbounded does not mean infinite.

> Anyway, so now you should see the basic principle (aside from the
> technical details): You have to distinguish between what is on the
> tape at any given STEP and what is on the tape after halting occurs
> (and IF halting occurs). If halting does NOT occur, then there is no
> FINAL output. Yes, for any natural number (or finite sequence of 1's,
> or digits of pi, whatever), that natural number will be printed to the
> tape but there is no point at which the TM has printed every natural
> number.

I didn't ask for a FINAL output. I asked for the set of positions
BNTM will ever write to. BNTM either writes to a position or
it does not. "Eventually" has nothing to do with my definition.

You are saying "eventually" BNTM writes a 1 to every finite
position. This means the set of blank postions is empty.
You know this proof, MoeBlee. I know you don't want me
to prove there is a largest natural number again.

The set of positions written to by BNTM can not be the
set of all natural numbers. The set must be finite.
There exists finite positions no TM can ever read.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/21/2009 9:26:10 PM
```On Jul 21, 2:26=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 21, 1:47=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 21, 1:11=A0pm, RussellE <reaste...@gmail.com> wrote:
> > The difference between (a) a single INFINITE output (which a TM does
> > NOT produce) and (b) successively putting out, for each n, a string of
> > n 1's, so that for any n, EVENTUALLY, the TM produces that string of n
> > 1's.
>
> Not for any n, See below.

Yes, for any n. Nothing you wrote below refutes this.

> > Just like if you were immortal and all you did was count natural
> > numbers. At any particular point, you will have counted only finitely
> > many natural numbers, but for any natural number, you will eventually
> > count it - unbounded.
>
> Unbounded does not mean infinite.

So what? I didn't say it does.

> > Anyway, so now you should see the basic principle (aside from the
> > technical details): You have to distinguish between what is on the
> > tape at any given STEP and what is on the tape after halting occurs
> > (and IF halting occurs). If halting does NOT occur, then there is no
> > FINAL output. Yes, for any natural number (or finite sequence of 1's,
> > or digits of pi, whatever), that natural number will be printed to the
> > tape but there is no point at which the TM has printed every natural
> > number.
>
> I didn't ask for a FINAL output. I asked for the set of positions
> BNTM will ever write to. BNTM either writes to a position or
> it does not. "Eventually" has nothing to do with my definition.

Clearly, you're not LISTENING to me; not even TRYING to understand.

> You are saying "eventually" BNTM writes a 1 to every finite
> position. This means the set of blank postions is empty.

WHAT "set of blank positions"? You're just shifting, yet again, your
fallacy.

At any given stage in the ongoing operation of BNTM, there are
infinitely many rightward blank positions. But for every position, at
SOME stage (and onward) in the ongoing operation of BNTM, that
position will not be blank.

EVERY rightward position is blank at SOME stage. Moreover, they're ALL
blank at the very outset. But for every position, there is a stage at
which that position will NOT be blank and not be blank for all ongoing
stages. That does not contradict that at no stage are all the
positions written to. There IS NO final stage, no stage(omega), etc.

> You know this proof, MoeBlee. I know you don't want me
> to prove there is a largest natural number again.

What are you talking about? There is no largest natural number.

> The set of positions written to by BNTM can not be the
> set of all natural numbers. The set must be finite.
> There exists finite positions no TM can ever read.

You're just rebabbling misconception you've already said, without even
addressing the explanation I gave you.

There seems little point attempting to communicate with you.

MoeBlee

```
 0
jazzmobe (307)
7/21/2009 9:41:18 PM
```RussellE <reasterly@gmail.com> writes:

> On Jul 21, 8:48 am, MoeBlee <jazzm...@hotmail.com> wrote:
>> On Jul 19, 7:22 pm, RussellE <reaste...@gmail.com> wrote:
>>
>> > Can a halting TM read the string produced by the BNTM?
>>
>> That is typical of the way you get in trouble by using loose language.
>> There is NO "THE string produced by the BNTM'. Rather, for each finite
>> number of executed steps, there is a finite string of 1's to the left
>> of the write-heard.
>
> You wouldn't object if I talked about THE output produced
> by a TM that calculates Pi.

Most people would unless you define it.  Pi is usually shown to be
computable in a way that completely avoids defining what the output of
a non-halting TM is.  I showed you a definition that is just about
usable (I could be used to show that a single non-halting TM does
indeed calculate pi, for example) but you did not like it, probably

Please define, formally, what *you* mean by the output of a non-haling
TM.  Then we can see if the result is indeed this paradoxical finite
but unbounded, uncomputable string.

<snip>
--
Ben.
```
 0
ben.usenet (6790)
7/21/2009 10:28:36 PM
```On Jul 21, 3:28=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:
> > On Jul 21, 8:48=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> >> On Jul 19, 7:22=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> >> > Can a halting TM read the string produced by the BNTM?
>
> >> That is typical of the way you get in trouble by using loose language.
> >> There is NO "THE string produced by the BNTM'. Rather, for each finite
> >> number of executed steps, there is a finite string of 1's to the left
> >> of the write-heard.
>
> > You wouldn't object if I talked about THE output produced
> > by a TM that calculates Pi.
>
> Most people would unless you define it. =A0Pi is usually shown to be
> computable in a way that completely avoids defining what the output of
> a non-halting TM is. =A0I showed you a definition that is just about
> usable (I could be used to show that a single non-halting TM does
> indeed calculate pi, for example) but you did not like it, probably
>
> Please define, formally, what *you* mean by the output of a non-haling
> TM. =A0Then we can see if the result is indeed this paradoxical finite
> but unbounded, uncomputable string.

Someone asked what BNTM stands for.
It means Big Number Turing Machine.

I define the output of BNTM as the set of all positions
BNTM has, is, or will ever write to. Whether of not
BNTM will ever write to positions n does not depend
on BNTM halting. BNTM can write to position n or it can't.

Everyone has been claiming BNTM "can" write to every
finite position on the tape. This means the set of
positions BNTM can NEVER write to is empty.
If the set of positions BNTM can never write to is empty,
I can prove there is a largest natural number.
I have given the proof numerous times.

Either there is a largest natural number or the
set of positions BNTM can never write to is non-empty.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/21/2009 11:15:00 PM
```RussellE <reasterly@gmail.com> writes:

> On Jul 21, 3:28 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > On Jul 21, 8:48 am, MoeBlee <jazzm...@hotmail.com> wrote:
>> >> On Jul 19, 7:22 pm, RussellE <reaste...@gmail.com> wrote:
>>
>> >> > Can a halting TM read the string produced by the BNTM?
>>
>> >> That is typical of the way you get in trouble by using loose language.
>> >> There is NO "THE string produced by the BNTM'. Rather, for each finite
>> >> number of executed steps, there is a finite string of 1's to the left
>> >> of the write-heard.
>>
>> > You wouldn't object if I talked about THE output produced
>> > by a TM that calculates Pi.
>>
>> Most people would unless you define it.  Pi is usually shown to be
>> computable in a way that completely avoids defining what the output of
>> a non-halting TM is.  I showed you a definition that is just about
>> usable (I could be used to show that a single non-halting TM does
>> indeed calculate pi, for example) but you did not like it, probably
>>
>> Please define, formally, what *you* mean by the output of a non-haling
>> TM.  Then we can see if the result is indeed this paradoxical finite
>> but unbounded, uncomputable string.
>
> Someone asked what BNTM stands for.
> It means Big Number Turing Machine.
>
> I define the output of BNTM as the set of all positions
> BNTM has, is, or will ever write to. Whether of not
> BNTM will ever write to positions n does not depend
> on BNTM halting. BNTM can write to position n or it can't.

That's a good start, but that means the output is a set (presumably of
numbers, yes?) rather than a sequence or a function defining a
sequence.

Why do you think this output set is finite?  It is either equal to or
isomorphic with the set of natural numbers.

> Everyone has been claiming BNTM "can" write to every
> finite position on the tape. This means the set of
> positions BNTM can NEVER write to is empty.
> If the set of positions BNTM can never write to is empty,
> I can prove there is a largest natural number.
> I have given the proof numerous times.

But it is not a valid proof.  There is no contradiction embedded in
the idea that the set of (positive) positions the machine can never
write to is empty.

> Either there is a largest natural number or the
> set of positions BNTM can never write to is non-empty.

Not by my reasoning, but obviously you disagree.

--
Ben.
```
 0
ben.usenet (6790)
7/21/2009 11:23:51 PM
```On Jul 21, 4:23=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:
> > On Jul 21, 3:28=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> >> RussellE <reaste...@gmail.com> writes:
> >> > On Jul 21, 8:48=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> >> >> On Jul 19, 7:22=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> >> >> > Can a halting TM read the string produced by the BNTM?
>
> >> >> That is typical of the way you get in trouble by using loose langua=
ge.
> >> >> There is NO "THE string produced by the BNTM'. Rather, for each fin=
ite
> >> >> number of executed steps, there is a finite string of 1's to the le=
ft
> >> >> of the write-heard.
>
> >> > You wouldn't object if I talked about THE output produced
> >> > by a TM that calculates Pi.
>
> >> Most people would unless you define it. =A0Pi is usually shown to be
> >> computable in a way that completely avoids defining what the output of
> >> a non-halting TM is. =A0I showed you a definition that is just about
> >> usable (I could be used to show that a single non-halting TM does
> >> indeed calculate pi, for example) but you did not like it, probably
>
> >> Please define, formally, what *you* mean by the output of a non-haling
> >> TM. =A0Then we can see if the result is indeed this paradoxical finite
> >> but unbounded, uncomputable string.
>
> > Someone asked what BNTM stands for.
> > It means Big Number Turing Machine.
>
> > I define the output of BNTM as the set of all positions
> > BNTM has, is, or will ever write to. Whether of not
> > BNTM will ever write to positions n does not depend
> > on BNTM halting. BNTM can write to position n or it can't.
>
> That's a good start, but that means the output is a set (presumably of
> numbers, yes?) rather than a sequence or a function defining a
> sequence.
>
> Why do you think this output set is finite? =A0It is either equal to or
> isomorphic with the set of natural numbers.

Only if the set of all natural numbers is finite.

> > Everyone has been claiming BNTM "can" write to every
> > finite position on the tape. This means the set of
> > positions BNTM can NEVER write to is empty.
> > If the set of positions BNTM can never write to is empty,
> > I can prove there is a largest natural number.
> > I have given the proof numerous times.
>
> But it is not a valid proof. =A0There is no contradiction embedded in
> the idea that the set of (positive) positions the machine can never
> write to is empty.

Yes there is.
I have given the proof several times.
If there is some part of the proof you don't agree with,
please tell me what it is.

For each step create a list of the positions BNTM
hasn't written to. Can this list ever be empty?

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/21/2009 11:39:32 PM
```On Jul 21, 4:15=A0pm, RussellE <reaste...@gmail.com> wrote:

> > Please define, formally, what *you* mean by the output of a non-haling
> > TM. =A0Then we can see if the result is indeed this paradoxical finite
> > but unbounded, uncomputable string.
>
> Someone asked what BNTM stands for.
> It means Big Number Turing Machine.
>
> I define the output of BNTM as the set of all positions
> BNTM has, is, or will ever write to. Whether of not
> BNTM will ever write to positions n does not depend
> on BNTM halting. BNTM can write to position n or it can't.

That's not a formal definition. You were asked for a formal
definition.

But, taken informally, your notion so far is okay.

> Everyone has been claiming BNTM "can" write to every
> finite position on the tape. This means the set of
> positions BNTM can NEVER write to is empty.

If the tape is one-way, okay. Still informal, but no problems so far.

> If the set of positions BNTM can never write to is empty,
> I can prove there is a largest natural number.
> I have given the proof numerous times.

No you haven't.

The set of positions that the BNTM will not write to is empty. There
is no largest natural number. You've not proven otherwise.

> Either there is a largest natural number or the
> set of positions BNTM can never write to is non-empty.

Either you're a troll or you have a serious cognitive block.

It's fine for one to say that one does not accept that there exist
infinite sets. But it's quite another thing to mal-reason as you do.

logic (if different from the ordinary logic of classical mathematics)
from which you think you prove:

The set of positions (on a one-way tape) that the BNTM will not write
to is empty -> there is a largest natural number.

explained to you.

And remember, there IS NO last position.

MoeBlee
```
 0
jazzmobe (307)
7/21/2009 11:47:58 PM
```On Jul 21, 4:47=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 21, 4:15=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > Please define, formally, what *you* mean by the output of a non-halin=
g
> > > TM. =A0Then we can see if the result is indeed this paradoxical finit=
e
> > > but unbounded, uncomputable string.
>
> > Someone asked what BNTM stands for.
> > It means Big Number Turing Machine.
>
> > I define the output of BNTM as the set of all positions
> > BNTM has, is, or will ever write to. Whether of not
> > BNTM will ever write to positions n does not depend
> > on BNTM halting. BNTM can write to position n or it can't.
>
> That's not a formal definition. You were asked for a formal
> definition.
>
> But, taken informally, your notion so far is okay.
>
> > Everyone has been claiming BNTM "can" write to every
> > finite position on the tape. This means the set of
> > positions BNTM can NEVER write to is empty.
>
> If the tape is one-way, okay. Still informal, but no problems so far.
>
> > If the set of positions BNTM can never write to is empty,
> > I can prove there is a largest natural number.
> > I have given the proof numerous times.
>
> No you haven't.
>
> The set of positions that the BNTM will not write to is empty. There
> is no largest natural number. You've not proven otherwise.
>
> > Either there is a largest natural number or the
> > set of positions BNTM can never write to is non-empty.
>
> Either you're a troll or you have a serious cognitive block.
>
> It's fine for one to say that one does not accept that there exist
> infinite sets. But it's quite another thing to mal-reason as you do.
>
> Or, please state your axioms (or even informal principles) and your
> logic (if different from the ordinary logic of classical mathematics)
> from which you think you prove:
>
> The set of positions (on a one-way tape) that the BNTM will not write
> to is empty -> there is a largest natural number.
>
> explained to you.

> And remember, there IS NO last position.

You claim BNTM will "eventually" write to every position on the tape..
After each step performed by BNTM I want to create the
set of blank positions:

S0 =3D {1,2,3,...}  - BNTM starts with a blank tape
S1 =3D {2,3,4,...}  - BNTM writes the first '1'
S2 =3D {3,4,5,...}  - BTNM writes the 2nd '1'
S3 =3D {4,5,6,...}
....
....
....
Sz-1 =3D {z}
Sz =3D {}  - "eventually" BNTM overwrites the last blank position

There is no last position. Therefore, BNTM can NOT write
to every position on the tape. Not "eventually". Not ever.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/22/2009 1:40:23 AM
```On Jul 21, 2:41=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> You're just rebabbling misconception you've already said, without even
> addressing the explanation I gave you.

"Rebabbling" is an awesome word. I salute you! What
a useful concept for anyone who reads usenet.

Marshall

```
 0
7/22/2009 1:40:32 AM
```RussellE <reasterly@gmail.com> writes:
<snip>
> For each step create a list of the positions BNTM
> hasn't written to. Can this list ever be empty?

.... and round we go again.

Simply re-stating the same things again and again is not a good way to
convince people that are right, but maybe that is not your intention
here.  Maybe you are a new-age mathematician and all you need is to
be heard.  If so, consider yourself well and truly heard.

If you want to persuade other mathematicians, you will have to start
doing mathematics and that means formal definitions and rigorous
arguments.

--
Ben.
```
 0
ben.usenet (6790)
7/22/2009 2:13:29 AM
```On Jul 21, 7:13=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:
>
> <snip>
>
> > For each step create a list of the positions BNTM
> > hasn't written to. Can this list ever be empty?
>
> ... and round we go again.
>
> Simply re-stating the same things again and again is not a good way to
> convince people that are right, but maybe that is not your intention
> here. =A0Maybe you are a new-age mathematician and all you need is to
> be heard. =A0If so, consider yourself well and truly heard.
>
> If you want to persuade other mathematicians, you will have to start
> doing mathematics and that means formal definitions and rigorous
> arguments.

You want me to give my argument in a formal system that
assumes a TM can compute any natural number.
After I am done, you will use this assumption, in one
of its many guises, to "prove" my argument is wrong.

We both agree BNTM will never write an infinite string.
There is really no question of what is on the tape at
any step. There is a finite string of 1's.

Yet, you keep telling me that "eventually" BNTM will
write an infinite string. And you want me to believe

Do you have a problem with the way I define
THE output of BNTM as the set of positions written to?
Are you claiming this set is undefined?
Will you allow me to talk about the set of
blank positions after each step?

These are the only two sets my proof requires.
I will try to define these two sets with as much
rigor as you require.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/22/2009 2:50:08 AM
```On Jul 16, 11:56=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 16, 8:08=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>
> > RussellE <reaste...@gmail.com> writes:
> > > No TM can write a string longer than the one written by the ZM.
> > > I am already assuming the ZM can perform an infinite number
> > > of operations.
> > I thought you now know that to be false (not sure if this is by
> > definition or proof, but you said it was false above).
>
> You are correct. I prove a TM or ZM can only read or write
> a finite number of positions. So why do I use a ZM?
>
> The standard definition of TM assumes the input is finite.
> This simply avoids the entire issue of infinite strings.
> It assumes a human has somehow decided in advance
> that the input string is finite.
>
> I am using ZM's to eliminate this assumption.
> I start by assuming a ZM can read an infinite string.
>
> Another common assumption is that a TM can
> read any finite string if we just wait "long enough",
> or we wait until the TM has performed "enough"
> operations.

If I understand the Zeno Machine correctly, then you only ever have to
wait for one second, even for an infinite number of operations. So
even feeding an infinite tape to the ZM returns another inifinite
output tape in a finite time. So your conclusion fails.

>
> I prove this assumption to be false.
> No matter how long we wait, or how many
> operations the TM performs, there will be unread
> finite positions on the tape.
>
> This is true even if the TM is "infinitely" fast and
> can perform an "infinite" number of operations.

Wrong.

```
 0
edprochak (546)
7/22/2009 3:26:32 AM
```On 2009-07-20 19:45:22 -0400, RussellE <reasterly@gmail.com> said:

> On Jul 20, 3:43�pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>>> Let me try to explain this better.
>>> Consider the set of positions written to by BNTM
>>> after each step:
>>
>>> S0 = {}
>>> S1 = {1}
>>> S2 = {1,2}
>>> S3 = {1,2,3}
>>> ...
>>
>>> This family of sets does not contain
>>> the set of all natural numbers.
>>> This is easily proven because each
>>> set has a largest position.
>>
>>> I think your argument can be stated as follows:
>>
>>> For every n there exists Sn which contains
>>> n proving BNTM writes to position n.
>>> This proves BNTM writes to every finite position.
>>
>>> My argument is that the set of positions
>>> written to by BNTM must a member of the
>>> family of sets I described above.
>>> The set of all natural numbers is not a member
>>> of this family of sets. BNTM can not write to every
>>> finite position.
>>
>> It is all good until that last conclusion. �It does not follow from
>> what goes before. �See below for another example of the same logical
>> leap.
>>
>> I think it would help you if you tried to define what you mean by that
>> last sentence. �If I said, in conversation, "M can not write to every
>> finite position (to the right of the start)" I would mean that there
>> exists some n such that the machine M does not write beyond position n
>> for all execution steps, even if M never halts (i.e. it would be
>> something I most likely have to prove by induction "for all steps").
>> You obviously don't mean that (because it patently false for BNTM) so
>> what *do* you mean?
>>
>>> You are saying the limit of the positions written
>>> is the set of all natural numbers, which is
>>> certainly correct. I am arguing you don't
>>> get to choose the limit. The set of
>>> all natural numbers is not in the set
>>> of possible outcomes.
>>
>> That is normal for a limit. �No intermediate is every equal to the
>> limit so the set of intermediates never includes it.
>>
>>> We can never reach the limit and there
>>> are positions on the tape BNTM will
>>> never write to.
>>
>> Yes to the first and no to the second. �By what axiom or law do you
>> conclude that because the machine never writes all the positions there
>> are positions it can't write? �It is part of the fun of the parlour gam
> e
>> of talking about infinities that such things seem odd or paradoxical.
>>
>> If you pin things down with some more formal definitions then you
>> either find that BNTM does, in fact, "reach the limit" (because that
>> is how you define the output of such a machine) or you find that there
>> is no cell it won't write despite "not writing to them all". �The
>> problem is in the language, not in the machine.
>
> I think this line line should say:
> you find that there is no computable cell
> it won't write despite "not writing to them all".
>
> So, how do I prove BNTM doesn't write to them all?
>
> Consider the set of all positions BNTM will ever write to.
> Assume this set is the set of all natural numbers.
>
> This means the set of blank positions is empty.
> Assign a natural number to each step performed
> by BNTM and make a list of all blank positions after
> each step:
>
> S0 = {1,2,3,...}
> S1 = {2,3,4,...}
> S2 = {3,4,5,...}
> ...
> ...
> ...
> Sz-1 = {z}
> Sz = {}
>
> Assuming BNTM writes to every finite
> position allows me to prove the set
> of all natural numbers is finite.

Let S'0 = {} and S'n = S'(n - 1) U {n}, so:

S'0 = {}
S'1 = {1}
S'2 = {1, 2}
....

Let Sn = N - S'n, where N is the set of natural numbers. We arrive at
the sets Sn as above.

1. For every natural number x, there is a set S'(f(x)) that contains x.
2. There is no set x that contains every natural number.

You seem to believe these two statements are contradictory; under most
modern definitions, they are not. At least you're in good company,
though... This mistake comes up a lot.

We can take S'n to represent the tape's state for your BNTM after n
transitions: if a tape cell indexed by x is non-blank after step n,
then x is in S'n. From above, for every natural number X, there is a
tape where the cell indexed by X has been written to. But there is no
natural number X at which every cell of the tape has been written to.

From 1, we can also conclude that the union of all sets S'n is N. From
2, we can conclude that N is not one of the sets S'n.

> Therefore, the set of all positions BNTM will
> ever write to can not be the set of all natural numbers.
> There must be at least one position not is this set.

For every index x that the BNTM writes to, it does so after only f(x)
steps, where f(x) is a natural number. When x is a natural number, f(x)
= x (provable from your construction). f(x) is defined for all natural
numbers x (provable from your construction).

Assume the existence of an index y that is not a natural number, and
that is eventually written to by your TM. Then it is reached after f(y)
steps, where f(y) is a natural number. Therefore there exists some
natural number x such that f(x) = f(y).

Therefore the index written after y steps is also written to after some
natural number of steps. So there are no indexes that are written to
after ONLY a number of steps not in N.

Therefore, the set of all indexes BNTM writes to is N at most. We can
also prove that it's N at least, so it's simply N.

In summary:

You've argued that since every tape cell is written to after only a
finite number of operations, we can conclude that after some finite
number of steps your TM produces a tape where every cell has been
written. This is flatly not so, in any standard construction for Turing
machines.

-o

```
 0
angrybaldguy (338)
7/22/2009 4:14:58 AM
```On Jul 21, 5:15=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 21, 1:55=A0pm, Ed Prochak <edproc...@gmail.com> wrote:
>
> > What exactly is a BNTM?
>
> It's his term for a TM that starts by writing '1', moves right, writes
>
> > > By assuming the TM has run to completion, ie,
> > > every position it will ever write to, we are assuming
> > > the tape is finite.
>
> > Turing's tapes were always infinite. I'm confused and missing some
> > context obviously.
>
> Don't worry, he's very confused.
>
> MoeBlee

Wow. I went back and read from the start of this thread.
Yes, he is VERY confused.

Ed
```
 0
edprochak (546)
7/22/2009 4:25:41 AM
```On Jul 21, 7:13=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> RussellE <reaste...@gmail.com> writes:

>=A0Maybe you are a new-age mathematician and all you need is to
> be heard. =A0If so, consider yourself well and truly heard.

I'm really not a troll.
I will admit I am a heretic.
I don't believe the set of all natural numbers exists.

I have seen the proof that for every n,
it is "obvious" BNTM writes to position n,

Maybe I am mentally defective, but I really don't see
why it is "obvious" a TM can read and write to any
finite position.

Can someone provide a proof that for every n,
BNTM writes to position n? Preferbly without
assuming BNTM can read and write to every position.
Its easy to prove something you have already assumed to be true.

We can use Patricia's idea of an infinite sequence
of strings. Would you show me how to create a bijection
between the set of all natural numbers and the
strings written by BNTM without assuming
BNTM "obviously" writes to every finite position.

I will settle for a proof there is a bijection between
the output of BNTM and the set of all natural numbers.
I want to know which string in the infinite sequence
of strings produced by BNTM allows me to
create such a bijection.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/22/2009 5:23:29 AM
```On Jul 21, 9:14=A0pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-20 19:45:22 -0400, RussellE <reaste...@gmail.com> said:
>
>
>
>
>
> > On Jul 20, 3:43=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
> >> RussellE <reaste...@gmail.com> writes:
>
> >>> Let me try to explain this better.
> >>> Consider the set of positions written to by BNTM
> >>> after each step:
>
> >>> S0 =3D {}
> >>> S1 =3D {1}
> >>> S2 =3D {1,2}
> >>> S3 =3D {1,2,3}
> >>> ...
>
> >>> This family of sets does not contain
> >>> the set of all natural numbers.
> >>> This is easily proven because each
> >>> set has a largest position.
>
> >>> I think your argument can be stated as follows:
>
> >>> For every n there exists Sn which contains
> >>> n proving BNTM writes to position n.
> >>> This proves BNTM writes to every finite position.
>
> >>> My argument is that the set of positions
> >>> written to by BNTM must a member of the
> >>> family of sets I described above.
> >>> The set of all natural numbers is not a member
> >>> of this family of sets. BNTM can not write to every
> >>> finite position.
>
> >> It is all good until that last conclusion. =A0It does not follow from
> >> what goes before. =A0See below for another example of the same logical
> >> leap.
>
> >> I think it would help you if you tried to define what you mean by that
> >> last sentence. =A0If I said, in conversation, "M can not write to ever=
y
> >> finite position (to the right of the start)" I would mean that there
> >> exists some n such that the machine M does not write beyond position n
> >> for all execution steps, even if M never halts (i.e. it would be
> >> something I most likely have to prove by induction "for all steps").
> >> You obviously don't mean that (because it patently false for BNTM) so
> >> what *do* you mean?
>
> >>> You are saying the limit of the positions written
> >>> is the set of all natural numbers, which is
> >>> certainly correct. I am arguing you don't
> >>> get to choose the limit. The set of
> >>> all natural numbers is not in the set
> >>> of possible outcomes.
>
> >> That is normal for a limit. =A0No intermediate is every equal to the
> >> limit so the set of intermediates never includes it.
>
> >>> We can never reach the limit and there
> >>> are positions on the tape BNTM will
> >>> never write to.
>
> >> Yes to the first and no to the second. =A0By what axiom or law do you
> >> conclude that because the machine never writes all the positions there
> >> are positions it can't write? =A0It is part of the fun of the parlour =
gam
> > e
> >> of talking about infinities that such things seem odd or paradoxical.
>
> >> If you pin things down with some more formal definitions then you
> >> either find that BNTM does, in fact, "reach the limit" (because that
> >> is how you define the output of such a machine) or you find that there
> >> is no cell it won't write despite "not writing to them all". =A0The
> >> problem is in the language, not in the machine.
>
> > I think this line line should say:
> > you find that there is no computable cell
> > it won't write despite "not writing to them all".
>
> > So, how do I prove BNTM doesn't write to them all?
>
> > Consider the set of all positions BNTM will ever write to.
> > Assume this set is the set of all natural numbers.
>
> > This means the set of blank positions is empty.
> > Assign a natural number to each step performed
> > by BNTM and make a list of all blank positions after
> > each step:
>
> > S0 =3D {1,2,3,...}
> > S1 =3D {2,3,4,...}
> > S2 =3D {3,4,5,...}
> > ...
> > ...
> > ...
> > Sz-1 =3D {z}
> > Sz =3D {}
>
> > Assuming BNTM writes to every finite
> > position allows me to prove the set
> > of all natural numbers is finite.
> > This is a contradiction.
>
> Let S'0 =3D {} and S'n =3D S'(n - 1) U {n}, so:
>
> S'0 =3D {}
> S'1 =3D {1}
> S'2 =3D {1, 2}
> ...
>
> Let Sn =3D N - S'n, where N is the set of natural numbers. We arrive at
> the sets Sn as above.
>
> 1. For every natural number x, there is a set S'(f(x)) that contains x.
> 2. There is no set x that contains every natural number.
>
> You seem to believe these two statements are contradictory; under most
> modern definitions, they are not. At least you're in good company,
> though... This mistake comes up a lot.
>
> We can take S'n to represent the tape's state for your BNTM after n
> transitions: if a tape cell indexed by x is non-blank after step n,
> then x is in S'n. From above, for every natural number X, there is a
> tape where the cell indexed by X has been written to. But there is no
> natural number X at which every cell of the tape has been written to.
>
> From 1, we can also conclude that the union of all sets S'n is N.

We can?

> From
> 2, we can conclude that N is not one of the sets S'n.

Not one of the strings produced by BNTM is infinite.
Agreed.

> Therefore, the set of all indexes BNTM writes to is N at most. We can
> also prove that it's N at least, so it's simply N.

How do you prove it is the least?

You assume for every n, BNTM produces a string of length n.
Using this assumption you show the limit of the length of
strings written by BNTM is N.

A limit is not a string. A string is something written by BNTM.
And you have already shown BNTM never writes an infinite string.
The limit is not one of the strings actually written by BNTM.
By assuming BNTM can write to every finite position,
your are "proving" BNTM writes an infinite string.
Since we know this is false, there must be something
wrong with this assumption.

And you haven't said anything refuting my proof.
At what step did BNTM overwrite the "last" blank position?

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/22/2009 6:10:16 AM
```On Jul 21, 9:14=A0pm, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-07-20 19:45:22 -0400, RussellE <reaste...@gmail.com> said:

I wasn't sure what you were trying to prove at first,
but now I think see it.

> Let Sn =3D N - S'n, where N is the set of natural numbers. We arrive at
> the sets Sn as above.
>
> 1. For every natural number x, there is a set S'(f(x)) that contains x.

Since this is exactly what I am trying to disprove,
I am not sure I can agree with this. But, OK.

> 2. There is no set x that contains every natural number.
>
> You seem to believe these two statements are contradictory; under most
> modern definitions, they are not. At least you're in good company,
> though... This mistake comes up a lot.

In set theory we can assume the TM performs an infinite
number of operations in a single step.
But, these creaky old TM's can only perform one
operation at a time.

> We can take S'n to represent the tape's state for your BNTM after n
> transitions:

OK

> if a tape cell indexed by x is non-blank after step n,
> then x is in S'n. From above, for every natural number X, there is a
> tape where the cell indexed by X has been written to.

Hmmm.

> But there is no
> natural number X at which every cell of the tape has been written to.

Agreed.

> From 1, we can also conclude that the union of all sets S'n is N.

You have been talking to those set theorists again,
haven't you. You really don't want to get me started
on union and intersection. It is bad enough you have
already assumed BNTM writes every finite string of 1's

> From
> 2, we can conclude that N is not one of the sets S'n.

Agreed.

> > Therefore, the set of all positions BNTM will
> > ever write to can not be the set of all natural numbers.
> > There must be at least one position not is this set.
>
> For every index x that the BNTM writes to, it does so after only f(x)
> steps, where f(x) is a natural number. When x is a natural number, f(x)
> =3D x (provable from your construction). f(x) is defined for all natural
> numbers x (provable from your construction).

I wasn't sure what you were trying to prove here,
but now I see you are trying to show there is no
position BNTM doesn't write to.

I would call y an uncomputable number.

> Assume the existence of an index y that is not a natural number, and
> that is eventually written to by your TM.

I think it might be easier to assume BNTM never
writes to position y and then prove a contradiction.

> Then it is reached after f(y)
> steps, where f(y) is a natural number.

Aren't you assuming what you are trying to prove here?
I haven't even told you what y is. In fact, I can't. y is
uncomputable.

> Therefore there exists some
> natural number x such that f(x) =3D f(y).

Even if true, how would we know this?
All we know is that y is some huge number.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/22/2009 7:09:30 AM
```RussellE <reasterly@gmail.com> writes:

> On Jul 21, 7:13 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>
>> Maybe you are a new-age mathematician and all you need is to
>> be heard.  If so, consider yourself well and truly heard.
>
> I'm really not a troll.
> I will admit I am a heretic.
> I don't believe the set of all natural numbers exists.

It may well be possible to set up a system of logic and arithmetic in
which N does not exists -- only a sequence of ever larger finite
approximations to it.  Who knows?  I don't and, oddly enough, I am
pretty sure that you don't either.

It is almost certain that in such an impoverished system, very little
can be proved.  Induction is surely ruled out and there won't be many
theorems about numbers.  It is possible that TMs can't even be
defined.

Any dramatic conclusions will, of course, simply be a consequence of
having rejected a large chunk of what everyone else thinks of as
mathematics so you won't even be able to shock anyone with
"uncomputable naturals" or "finite but unbounded" strings and so on.
The most impressive results from this sort of meta-mathematics are
usually about how much remains the same, not how oddly different the
new system is.  Either way, it is a lifetime's work -- good luck.

I am being slightly disingenuous here.  In truth I think you are
talking rubbish.  The effect of such a change will probably be
astoundingly dull, but I have to admit that I don't really know.

> I have seen the proof that for every n,
> it is "obvious" BNTM writes to position n,

What set is n a member of?  It can't be N anymore.  Are all sets
finite in your system?  These are large rhetorical questions.  I don't
really want to know, but I think you should know some of the things
that will need to be pinned down.

> Maybe I am mentally defective, but I really don't see
> why it is "obvious" a TM can read and write to any
> finite position.

Presumably you reject proofs by induction?  If so you can't show much

> Can someone provide a proof that for every n,
> BNTM writes to position n? Preferbly without
> assuming BNTM can read and write to every position.
> Its easy to prove something you have already assumed to be true.

Who knows?  I don't even know what n is anymore.  Presumably set union
is now defined only for finite sets of finite sets so the union of all
the approximations to N is not defined.  Maybe this infinite union is
defined but in some new way?  One day you will tell us.

> We can use Patricia's idea of an infinite sequence
> of strings. Would you show me how to create a bijection
> between the set of all natural numbers and the
> strings written by BNTM without assuming
> BNTM "obviously" writes to every finite position.

Only in my world where there is N and proof by induction.

> I will settle for a proof there is a bijection between
> the output of BNTM and the set of all natural numbers.
> I want to know which string in the infinite sequence
> of strings produced by BNTM allows me to
> create such a bijection.

I'll settle for a definition of a natural number is Russel Maths.

--
Ben.
```
 0
ben.usenet (6790)
7/22/2009 11:19:54 AM
```RussellE <reasterly@gmail.com> writes:

> On Jul 21, 7:13 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>>
>> <snip>
>>
>> > For each step create a list of the positions BNTM
>> > hasn't written to. Can this list ever be empty?
>>
>> ... and round we go again.
>>
>> Simply re-stating the same things again and again is not a good way to
>> convince people that are right, but maybe that is not your intention
>> here.  Maybe you are a new-age mathematician and all you need is to
>> be heard.  If so, consider yourself well and truly heard.
>>
>> If you want to persuade other mathematicians, you will have to start
>> doing mathematics and that means formal definitions and rigorous
>> arguments.
>
> You want me to give my argument in a formal system that
> assumes a TM can compute any natural number.

Not anymore.  You don't accept that N exists so none of my maths makes
any sense to you.  Maybe when your book is published others will know
what can and can not be done in the new world of yours, but you can
hardly expect anyone to talk to you about it until you have explained
the basics.

> After I am done, you will use this assumption, in one
> of its many guises, to "prove" my argument is wrong.

yet know any theorems in your new maths.

> We both agree BNTM will never write an infinite string.
> There is really no question of what is on the tape at
> any step. There is a finite string of 1's.
>
> Yet, you keep telling me that "eventually" BNTM will
> write an infinite string. And you want me to believe
> this is not a contradiction.

Not anymore.  I now want to know things like what n+1 means and are
there an infinite number of primes?  What are the natural numbers now
and is there a bijection between them and some finite set?  Heck, what
is a set now?

> Do you have a problem with the way I define
> THE output of BNTM as the set of positions written to?

And how!  I thought we had some ground rules.  It is more than absurd
for you to even get this far in the debate without revealing that we
are talking about fundamentally different things.  Yo need to define
everything up to TMs before I even know what you are talking about!

> Are you claiming this set is undefined?
> Will you allow me to talk about the set of
> blank positions after each step?
>
> These are the only two sets my proof requires.
> I will try to define these two sets with as much
> rigor as you require.

Start with what a number is now that N does not exists.  We can go
form there though I suspect it may not be a debate with me.

--
Ben.
```
 0
ben.usenet (6790)
7/22/2009 11:31:09 AM
```"Ben Bacarisse" <ben.usenet@bsb.me.uk> wrote in message
news:0.22ff94d38b5a5a4ba2a3.20090722121954BST.871vo9m0bp.fsf@bsb.me.uk...
> RussellE <reasterly@gmail.com> writes:
>
>> On Jul 21, 7:13 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>>> RussellE <reaste...@gmail.com> writes:
>>
>>> Maybe you are a new-age mathematician and all you need is to
>>> be heard. If so, consider yourself well and truly heard.
>>
>> I'm really not a troll.
>> I will admit I am a heretic.
>> I don't believe the set of all natural numbers exists.
>
> It may well be possible to set up a system of logic and arithmetic in
> which N does not exists -- only a sequence of ever larger finite
> approximations to it.  Who knows?  I don't and, oddly enough, I am
> pretty sure that you don't either.
>

I think I do.

Simply remove the Axiom of infinity and consider only finite sets, or finite
ordinals, no N.

> It is almost certain that in such an impoverished system, very little
> can be proved.  Induction is surely ruled out and there won't be many
> theorems about numbers.  It is possible that TMs can't even be
> defined.
>

Induction works. Arithmetic works.

> Any dramatic conclusions will, of course, simply be a consequence of
> having rejected a large chunk of what everyone else thinks of as
> mathematics so you won't even be able to shock anyone with
> "uncomputable naturals" or "finite but unbounded" strings and so on.
> The most impressive results from this sort of meta-mathematics are
> usually about how much remains the same, not how oddly different the
> new system is.  Either way, it is a lifetime's work -- good luck.
>

Eliminating the axiom of infinity means you lose all the interesting bits of
set theory.

> I am being slightly disingenuous here.  In truth I think you are
> talking rubbish.  The effect of such a change will probably be
> astoundingly dull, but I have to admit that I don't really know.
>
>> I have seen the proof that for every n,
>> it is "obvious" BNTM writes to position n,
>
> What set is n a member of?  It can't be N anymore.  Are all sets
> finite in your system?  These are large rhetorical questions.  I don't
> really want to know, but I think you should know some of the things
> that will need to be pinned down.
>
>> Maybe I am mentally defective, but I really don't see
>> why it is "obvious" a TM can read and write to any
>> finite position.
>
> Presumably you reject proofs by induction?  If so you can't show much
> about numbers and you probably can't prove much about TMs either.
>
>> Can someone provide a proof that for every n,
>> BNTM writes to position n? Preferbly without
>> assuming BNTM can read and write to every position.
>> Its easy to prove something you have already assumed to be true.
>
> Who knows?  I don't even know what n is anymore.  Presumably set union
> is now defined only for finite sets of finite sets so the union of all
> the approximations to N is not defined.  Maybe this infinite union is
> defined but in some new way?  One day you will tell us.
>
>> We can use Patricia's idea of an infinite sequence
>> of strings. Would you show me how to create a bijection
>> between the set of all natural numbers and the
>> strings written by BNTM without assuming
>> BNTM "obviously" writes to every finite position.
>
> Only in my world where there is N and proof by induction.
>
>> I will settle for a proof there is a bijection between
>> the output of BNTM and the set of all natural numbers.
>> I want to know which string in the infinite sequence
>> of strings produced by BNTM allows me to
>> create such a bijection.
>
> I'll settle for a definition of a natural number is Russel Maths.
>
> --
> Ben.

```
 0
webbfamily (36)
7/22/2009 1:11:02 PM
```Ben Bacarisse wrote:
> RussellE <reasterly@gmail.com> writes:
>
>> On Jul 21, 7:13 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>>> RussellE <reaste...@gmail.com> writes:
>>>  Maybe you are a new-age mathematician and all you need is to
>>> be heard.  If so, consider yourself well and truly heard.
>> I'm really not a troll.
>> I will admit I am a heretic.
>> I don't believe the set of all natural numbers exists.
>
> It may well be possible to set up a system of logic and arithmetic in
> which N does not exists -- only a sequence of ever larger finite
> approximations to it.  Who knows?  I don't and, oddly enough, I am
> pretty sure that you don't either.
>
> It is almost certain that in such an impoverished system, very little
> can be proved.  Induction is surely ruled out and there won't be many
> theorems about numbers.  It is possible that TMs can't even be
> defined.

I agree with Ben from a mathematical point of view. However, I'm
primarily a practical programmer, so I'm going to comment on the
consequences for Turing machines as a useful abstraction for
reasoning about the limits of algorithms.

The assumptions that the TM tape is unbounded and that a non-halting TM
goes on running for ever remove some difficult issues that make
reasoning about real world computers messy.

Patricia
```
 0
pats (3556)
7/22/2009 1:51:25 PM
```"Peter Webb" <webbfamily@DIESPAMDIEoptusnet.com.au> writes:

> "Ben Bacarisse" <ben.usenet@bsb.me.uk> wrote in message
> news:0.22ff94d38b5a5a4ba2a3.20090722121954BST.871vo9m0bp.fsf@bsb.me.uk...
>> RussellE <reasterly@gmail.com> writes:
>>
>>> On Jul 21, 7:13 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>>>> RussellE <reaste...@gmail.com> writes:
>>>
>>>> Maybe you are a new-age mathematician and all you need is to
>>>> be heard. If so, consider yourself well and truly heard.
>>>
>>> I'm really not a troll.
>>> I will admit I am a heretic.
>>> I don't believe the set of all natural numbers exists.
>>
>> It may well be possible to set up a system of logic and arithmetic in
>> which N does not exists -- only a sequence of ever larger finite
>> approximations to it.  Who knows?  I don't and, oddly enough, I am
>> pretty sure that you don't either.
>>
>
> I think I do.
>
> Simply remove the Axiom of infinity and consider only finite sets, or
> finite ordinals, no N.
>
>
>> It is almost certain that in such an impoverished system, very little
>> can be proved.  Induction is surely ruled out and there won't be many
>> theorems about numbers.  It is possible that TMs can't even be
>> defined.
>>
>
> Induction works. Arithmetic works.

I thought induction was tied to this axiom, but this is not an area of
expertise of mine.

<snip>
--
Ben.
```
 0
ben.usenet (6790)
7/22/2009 2:21:20 PM
```On Jul 21, 6:40=A0pm, RussellE <reaste...@gmail.com> wrote:

> You claim BNTM will "eventually" write to every position on the tape..
> After each step performed by BNTM I want to create the
> set of blank positions:
>
> S0 =3D {1,2,3,...} =A0- BNTM starts with a blank tape
> S1 =3D {2,3,4,...} =A0- BNTM writes the first '1'
> S2 =3D {3,4,5,...} =A0- BTNM writes the 2nd '1'
> S3 =3D {4,5,6,...}
> ...
> ...
> ...
> Sz-1 =3D {z}
> Sz =3D {} =A0- "eventually" BNTM overwrites the last blank position

No. AGAIN, for the thousandth time, there is no "last blank
position".

There is no "Sz", no "stage z".

Please don't just skip over this:

We say, (1) "For every natural number n, there is a stage S such that,
at stage S, position n is written to.

We do NOT say, (2) "There is a stage S such that, for every natural
number n, at stage S position n is written to"

Those are very different. Do you see the difference? Your S(z) comes
from (2), not from (1). And (1) does NOT entail (2).

Here's an analogy:

Suppose (hypothetically, just for illustration) that I"m immortal and
all I do is count natural numbers. So, for every natural number, there
is a moment that I'll count it. But it is not true that there is a
moment such that I'll have counted every natural number.

Again, for every natural number, at some point, I'll have counted it.
But it is not true that there is some point that I'll have counted
every natural numbers.

With the BNTM, for every natural number n, there is a stage S such
that, at stage S, position n gets written to. But there is no stage S
such that for every natural number n, we have n written to.

It's the difference between "For all n, there exists an S..." and
"There exists an S such that for all n..." You can see that those are
different by another simple analogy:

At a certain party, for every boy there is a girl to dance with; but
that's different from saying that there is a girl that dances with
every boy.

When we say "For every boy there is a girl to dance with" we have NOT
committed ourselves to saying "There is a girl that dances with every
boy".

When we say "For every natural number there is a stage at which that
natural number is written" we have NOT committed ourselves to saying
"There is a stage at which every natural number is written".

(1) For every boy b, there exists a girl g such that b dances with g.

(2) There exists a girl g such that, for every boy b, we have b dances
with g.

(1) and (2) are different.

So, again, we say, "For every natural number n, there is a stage S
such that at stage S we have position n written to"; but you turn that
into "There is a stage S such that for every natural number n, we have
position n written to."

That you do that is evidenced by your use of "Sz", as if that is some
last stage where every position has been written to, as if our
statement entails that there is such a last stage. No, our statement
does NOT entail such a last stage, as I've now explained to you yet
again.

> There is no last position. Therefore, BNTM can NOT write
> to every position on the tape. Not "eventually". Not ever.

There is no last position. And there is no time (stage, whatever) at
which every position is written to. Yes, we agree. But for every
position, there is a time (stage, whatever) such that the position
will be written to (this is what is meant by "eventually"). And all
these temporal terms ('time', 'eventually', 'has been', etc.) are just
metaphors for certain formalizations, so that any PROOF regarding
these formalizations cannot rely on the metaphors but rather must come
only from the axioms, formal definitions, and logic.The metaphors can
be relied upon only to "explain" or "make a picture for the mind to
grasp". The actual mathematical proof though may not resort to the
metaphors for any actual DEDUCTION.

I've explained this as clearly as possible. If you still don't get it,
then my surmise is that you don't WANT to get it.

MoeBlee

```
 0
jazzmobe (307)
7/22/2009 3:51:42 PM
```On Jul 21, 7:50=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 21, 7:13=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>
>
>
> > RussellE <reaste...@gmail.com> writes:
>
> > <snip>
>
> > > For each step create a list of the positions BNTM
> > > hasn't written to. Can this list ever be empty?
>
> > ... and round we go again.
>
> > Simply re-stating the same things again and again is not a good way to
> > convince people that are right, but maybe that is not your intention
> > here. =A0Maybe you are a new-age mathematician and all you need is to
> > be heard. =A0If so, consider yourself well and truly heard.
>
> > If you want to persuade other mathematicians, you will have to start
> > doing mathematics and that means formal definitions and rigorous
> > arguments.
>
> You want me to give my argument in a formal system that
> assumes a TM can compute any natural number.
> After I am done, you will use this assumption, in one
> of its many guises, to "prove" my argument is wrong.
>
> We both agree BNTM will never write an infinite string.
> There is really no question of what is on the tape at
> any step. There is a finite string of 1's.
>
> Yet, you keep telling me that "eventually" BNTM will
> write an infinite string.

NO NO NO. We do NOT say that. We say that for any given position,
eventually that position will be written to. We never say there is
ever an infinite string that the TM writes.

(Note: Yes, in set theory (in which theory TMs can be formulated)
there do exist infinite strings. But they are not written by
***TMs***. A TM only writes a finite string. But a TM can keep going
to add to that finite string so that there is no upper bound on the
finite strings that TM writes.)

Again, its what I mentioned about "for all, there exists..." and
"there exists, for all...":

For every finite string of 1's, there exists a point at which the BNTM
will have written that string. But it is not the case that there
exists a point at which the BNTM will have written all the finite
strings (or even one infinite string).

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 3:59:01 PM
```On Jul 21, 10:23=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> Maybe I am mentally defective, but I really don't see
> why it is "obvious" a TM can read and write to any
> finite position.
>
> Can someone provide a proof that for every n,
> BNTM writes to position n? Preferbly without
> assuming BNTM can read and write to every position.
> Its easy to prove something you have already assumed to be true.

You are concerned with assuming the conclusion. That's a good
thing to be worried about. (I don't think it's happening here, but
it's a valid concern.) Just be sure to distinguish between assuming
the conclusion, and rejecting any conditions that leads to the
conclusion, on the basis that you don't like the conclusion.

Let's for the moment just talk about how the abstract Turing Machine
works. Let's talk specifically about the "move the head to the right"
operation. Just that.

Does it always succeed, or not? Is it guaranteed to work under all
circumstances, or are there some circumstances under which it
might fail?

The usual definition of Turing Machine says that it always succeeds.
If you are using a formalism under which it might fail, you have to
tell us that, otherwise we assume you are using the usual formalism.

Do you see that if you have a TM with a right-unbounded tape,
and you start at the left edge, and the move-right operation always
succeeds, then you can get to any position on the tape?

Do you see that if you cannot get to some position on the tape,
then that necessarily must mean that some move-right operation
has to fail?

Whether you can get to any position or not is a consequence of
the definition of TM you are using. Under the usual definition,
you can get to any position.

Marshall
```
 0
7/22/2009 4:15:23 PM
```On Jul 21, 10:23=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 21, 7:13=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>
> > RussellE <reaste...@gmail.com> writes:
> >=A0Maybe you are a new-age mathematician and all you need is to
> > be heard. =A0If so, consider yourself well and truly heard.
>
> I'm really not a troll.
> I will admit I am a heretic.
> I don't believe the set of all natural numbers exists.

Fine. But th4ere is a difference between (1) denying that there exists
a set of all the natural numbers and (2) making totally illogical
arguments about Turing machines and other illogical arguments trying
to show Z set theory is inconsistent.

These are two SEPARATE (though related) matters. (1) There does not
exist a set of all the natural numbers. (2) Z is inconsistent. The
first is a belief you have, and you're welcome to have it. But the
second is a matter of whether there is an actual formal derivation
(using only first order logic with identity applied to the Z axioms)
of a formula P and a formula ~P. The claim that there does not exist a
set of all the natural numbers does not entail that there exists a
derivation of a formula P and a formula ~P in set theory. And all your
arguments trying to show a contradiction have just been WRONG.

> I have seen the proof that for every n,
> it is "obvious" BNTM writes to position n,
>
> Maybe I am mentally defective, but I really don't see
> why it is "obvious" a TM can read and write to any
> finite position.

By induction: The TM writes to square 1. Then if the TM writes to
square n, for any natural number n, then by moving right and writing
'1', the TM writes to square n+1. So the TM writes to every square (on
a one-way tape; for a two-tape, it's also easy to show).

> Can someone provide a proof that for every n,
> BNTM writes to position n? Preferbly without
> assuming BNTM can read and write to every position.

All I assumed is that the BNTM writes to square 1 and then moves right
to move to the next square, and for every square it has written to, it
moves right and writes to the next square.

That is exactly what YOU specified the BNTM to be.

> Its easy to prove something you have already assumed to be true.

I can write the actual TM for you. The actual set of quadruplets. Oh,
but do you even know anything about the actual formal specification of
what a TM is?

> We can use Patricia's idea of an infinite sequence
> of strings. Would you show me how to create a bijection
> between the set of all natural numbers and the
> strings written by BNTM without assuming
> BNTM "obviously" writes to every finite position.
>
> I will settle for a proof there is a bijection between
> the output of BNTM and the set of all natural numbers.
> I want to know which string in the infinite sequence
> of strings produced by BNTM allows me to
> create such a bijection.

The bijection f is this:

f is function from the set of natural numbers onto the set of finite
strings of 1's produced at each step by the BNTM.

f(0) =3D the string <1>  .... i.e., f(0) is the result of the first time
the BNTM writes '1''.
Then, for all n in w, we put f(n+1) =3D the result of the BNTM moving
right and printing '1' after the BNTM has produced f(n).

That is a legal definition by recursion of a function f on the set of
natural numbers. And it is easy to prove by truly rote induction that
f is 1-1 and onto. (If you can't perform that rote induction, then you
need to get a book and STUDY it so that you can do the induction as a
simple exercise.)

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 4:17:55 PM
```On Jul 21, 11:10=A0pm, RussellE <reaste...@gmail.com> wrote:

> By assuming BNTM can write to every finite position,
> your are "proving" BNTM writes an infinite string.

Nope. Explained to you a thousand times already.

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 4:19:18 PM
```On Jul 22, 12:09=A0am, RussellE <reaste...@gmail.com> wrote:

> In set theory we can assume the TM performs an infinite
> number of operations in a single step.

NO NO NO. TOTALLY WRONG. We CANNOT assume such a thing in set theory.

You're just making arbitrarily incorrect claims.

In Z set theory, we can formulate TM's and the important theorems
about TMs. We absolutely do NOT assume in set theory that a TM can
perform infinitely many operations in a single step. Indeed, our set
theoretic formulation provides that it is NOT the case that a TM can
perform infinitely many operations in a single step.

You are TOTALLY confused. You have this BACKWARDS.

How PATHETIC that you've spent, what, ten years? more? on this subject
and you don't understand even the first thing about it, and you'll
spend another ten years, the rest of your life even, still completely
mixed up about it, as you won't interecede your own ignorance by just

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 4:26:22 PM
```> "Ben Bacarisse" <ben.use...@bsb.me.uk> wrote in message

> > It may well be possible to set up a system of logic and arithmetic in
> > which N does not exists -- only a sequence of ever larger finite
> > approximations to it. =A0

Arithmetic, okay. But analysis is another story.

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 4:29:33 PM
```On Jul 21, 10:23=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 21, 7:13=A0pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>
> > RussellE <reaste...@gmail.com> writes:
> >=A0Maybe you are a new-age mathematician and all you need is to
> > be heard. =A0If so, consider yourself well and truly heard.
>
> I'm really not a troll.
> I will admit I am a heretic.
> I don't believe the set of all natural numbers exists.
>
> I have seen the proof that for every n,
> it is "obvious" BNTM writes to position n,
>
> Maybe I am mentally defective, but I really don't see
> why it is "obvious" a TM can read and write to any
> finite position.

What would stop it from eventually writing each and every finite
position?  As others have pointed out,
You assume that the BNTM can always move right at each and every step.

Yes, it's a contradiction that the set of positions the BNTM will not
write to is empty given that the BNTM
is always only at a finite step and never finishes and thus at each
and every step, that set of positions is infinite.
EVERY axiom, including the axiom of infinity, embeds a contradiction
like this and removes it from play
to create a larger contradiction free scheme.

karl m
```
 0
malbrain (59)
7/22/2009 4:43:17 PM
```On Jul 22, 8:51=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 21, 6:40=A0pm, RussellE <reaste...@gmail.com> wrote:

> > You claim BNTM will "eventually" write to every position on the tape..
> > After each step performed by BNTM I want to create the
> > set of blank positions:
>
> > S0 =3D {1,2,3,...} =A0- BNTM starts with a blank tape
> > S1 =3D {2,3,4,...} =A0- BNTM writes the first '1'
> > S2 =3D {3,4,5,...} =A0- BTNM writes the 2nd '1'
> > S3 =3D {4,5,6,...}
> > ...
> > ...
> > ...
> > Sz-1 =3D {z}
> > Sz =3D {} =A0- "eventually" BNTM overwrites the last blank position
>
> No. AGAIN, for the thousandth time, there is no "last blank
> position".
>
> There is no "Sz", no "stage z".
>
> Please don't just skip over this:
>
> We say, (1) "For every natural number n, there is a stage S such that,
> at stage S, position n is written to.

I agree this is what you are saying.

> We do NOT say, (2) "There is a stage S such that, for every natural
> number n, at stage S position n is written to"

You are saying BNTM will "eventually" write to every position.
How is this different from saying "eventually" BNTM writes an infinite
string?

> Those are very different. Do you see the difference? Your S(z) comes
> from (2), not from (1). And (1) does NOT entail (2).
>
> Here's an analogy:
>
> Suppose (hypothetically, just for illustration) that I"m immortal and
> all I do is count natural numbers. So, for every natural number, there
> is a moment that I'll count it. But it is not true that there is a
> moment such that I'll have counted every natural number.

I like your analogy, but you are saying the immortal does
count all the natural numbers "eventually".
I am tryng to prove the universe ends and the immortal dies
long before he counts all the natural numbers.

> Again, for every natural number, at some point, I'll have counted it.
> But it is not true that there is some point that I'll have counted
> every natural numbers.

But, this is what you are saying. You are claiming BNTM
will write every finite string of 1's "eventually".

> With the BNTM, for every natural number n, there is a stage S such
> that, at stage S, position n gets written to. But there is no stage S
> such that for every natural number n, we have n written to.
>
> It's the difference between "For all n, there exists an S..." and
> "There exists an S such that for all n..." You can see that those are
> different by another simple analogy:
>
> At a certain party, for every boy there is a girl to dance with; but
> that's different from saying that there is a girl that dances with
> every boy.
>
> When we say "For every boy there is a girl to dance with" we have NOT
> committed ourselves to saying "There is a girl that dances with every
> boy".
>
> When we say "For every natural number there is a stage at which that
> natural number is written" we have NOT committed ourselves to saying
> "There is a stage at which every natural number is written".

Yes you have.

I define THE output of BNTM as the set of all positions BNTM
definition, so, I will assume I can use it.

You don't get to say BNTM hasn't written to every position yet,
but eventually it will. BNTM either writes to position n or it does
not.

If, as you and most other people have claimed, BNTM will eventually
write to every finite position, then you are claiming that yes, there
is
a stage when BNTM has written every natural number.
True, you are saying it will do so "eventually", But, you
are saying it does so.

My proof uses what are sometimes called "downward infinite
membership chains". Most set theories have have an axiom
saying downward infinite membership chains don't exist.
This axiom is call the Axiom of regularity or the Axiom of
Foundation. http://en.wikipedia.org/wiki/Axiom_of_regularity

You have to assume the Axiom of regularity is true because
you can't prove it is true from the other axioms.
I am sure MoeBlee will say set theory is consistent
whether we assume Foundation or not.
I am not going to argue with him because I know he has
studied systems without Foundation.

You can see why most set theorists prefer systems
with an Axiom of Regularity. Set theories without
Foundation have lots of strange properties and
unscrupulous people, like myself, can use
downward infinite membership chains to try to
prove "infinite" sets are really finite.

I bring this up because, as far as I know,
computation theory doesn't have an
Axiom of regularity. I can make a downward
infinite membership chain if I want to.

A TM can only perform one operation at a time.
If, as you are claiming, BNTM "eventually" writes
to every location on the tape, then I can prove
there is a "last" position on the tape with a
downward infinite membership chain that
includes the empty set.

Russell
- 2 many 2 count
```
 0
7/22/2009 6:50:10 PM
```MoeBlee wrote:

>In Z set theory, we can formulate TM's and the important theorems
>about TMs. We absolutely do NOT assume in set theory that a TM can
>perform infinitely many operations in a single step. Indeed, our set
>theoretic formulation provides that it is NOT the case that a TM can
>perform infinitely many operations in a single step.

I believe a lot of confusion about TMs is caused by the often repeated
meme that it 'contains an infinite tape'.  This is very misleading
(if not outright wrong) and it suggests that the operation of a TM
is *not* a stepwise process on finite pieces of tape.  But it is.

--
Reinier
```
 0
rp
7/22/2009 6:58:50 PM
```On Jul 22, 9:15=A0am, Marshall <marshall.spi...@gmail.com> wrote:
> On Jul 21, 10:23=A0pm, RussellE <reaste...@gmail.com> wrote:
>
>
>
> > Maybe I am mentally defective, but I really don't see
> > why it is "obvious" a TM can read and write to any
> > finite position.
>
> > Can someone provide a proof that for every n,
> > BNTM writes to position n? Preferbly without
> > assuming BNTM can read and write to every position.
> > Its easy to prove something you have already assumed to be true.
>
> You are concerned with assuming the conclusion. That's a good
> thing to be worried about. (I don't think it's happening here, but
> it's a valid concern.) Just be sure to distinguish between assuming
> the conclusion, and rejecting any conditions that leads to the
> conclusion, on the basis that you don't like the conclusion.
>
> Let's for the moment just talk about how the abstract Turing Machine
> works. Let's talk specifically about the "move the head to the right"
> operation. Just that.
>
> Does it always succeed, or not? Is it guaranteed to work under all
> circumstances, or are there some circumstances under which it
> might fail?

"Move right" always succeeds.

I am not saying there is a poisiton where the TM can't move right.
I am saying there are positions the TM will never get to
no matter how many times it moves right. I think this
is true even if we assume the TM can move right an
infinte number of times.

> The usual definition of Turing Machine says that it always succeeds.
> If you are using a formalism under which it might fail, you have to
> tell us that, otherwise we assume you are using the usual formalism.
>
> Do you see that if you have a TM with a right-unbounded tape,
> and you start at the left edge, and the move-right operation always
> succeeds, then you can get to any position on the tape?

No.

> Do you see that if you cannot get to some position on the tape,
> then that necessarily must mean that some move-right operation
> has to fail?

No. You just haven't moved right far enough and you never will.

> Whether you can get to any position or not is a consequence of
> the definition of TM you are using. Under the usual definition,
> you can get to any position.

I agree most formal systems assume a TM can read any finite
position. But, this is an assumption. I am trying to show why

Russell
- 2 many 2 count
```
 0
7/22/2009 7:12:05 PM
```On Jul 22, 11:50=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
> On Jul 22, 8:51=A0am, MoeBlee <jazzm...@hotmail.com> wrote:

> > We say, (1) "For every natural number n, there is a stage S such that,
> > at stage S, position n is written to.
>
> I agree this is what you are saying.
>
> > We do NOT say, (2) "There is a stage S such that, for every natural
> > number n, at stage S position n is written to"
>
> You are saying BNTM will "eventually" write to every position.
> How is this different from saying "eventually" BNTM writes an infinite
> string?

I've explained that already, over and over. And see rest of the post
too.

Also, I've said that such words as 'eventually' are only metaphor
anyway.

And see more comments in my post now below:

> > Those are very different. Do you see the difference? Your S(z) comes
> > from (2), not from (1). And (1) does NOT entail (2).
>
> > Here's an analogy:
>
> > Suppose (hypothetically, just for illustration) that I"m immortal and
> > all I do is count natural numbers. So, for every natural number, there
> > is a moment that I'll count it. But it is not true that there is a
> > moment such that I'll have counted every natural number.
>
> I like your analogy, but you are saying the immortal does
> count all the natural numbers "eventually".

I'm saying that when I say "counts all natural numbers eventually" I
mean "For every number, there will be a moment that number will be
counted" and I am NOT saying "there is a moment when every number has
been counted".

I don't even need the word 'eventually'.

> I am tryng to prove the universe ends and the immortal dies
> long before he counts all the natural numbers.

That's a different matter.

The assumption of TMs is that there are certain TMs that on certain
input continue to work without end. YOUr BNTM is such a TM. But again,
that is METAPHORICAL talk anyway. The actual specification of TMs is
that certain TMs don't halt. That is, there exist certain sets of
quadruples (TMs are a certain kind of FINITE set of quadruples) that
are TMs but such that they do not halt on certain input (i.e., there
generated from the finite set).

> > Again, for every natural number, at some point, I'll have counted it.
> > But it is not true that there is some point that I'll have counted
> > every natural numbers.
>
> But, this is what you are saying. You are claiming BNTM
> will write every finite string of 1's "eventually".

See above.

> > With the BNTM, for every natural number n, there is a stage S such
> > that, at stage S, position n gets written to. But there is no stage S
> > such that for every natural number n, we have n written to.
>
> > It's the difference between "For all n, there exists an S..." and
> > "There exists an S such that for all n..." You can see that those are
> > different by another simple analogy:
>
> > At a certain party, for every boy there is a girl to dance with; but
> > that's different from saying that there is a girl that dances with
> > every boy.
>
> > When we say "For every boy there is a girl to dance with" we have NOT
> > committed ourselves to saying "There is a girl that dances with every
> > boy".
>
> > When we say "For every natural number there is a stage at which that
> > natural number is written" we have NOT committed ourselves to saying
> > "There is a stage at which every natural number is written".
>
> Yes you have.

No. You merely SAY that.

> I define THE output of BNTM as the set of all positions BNTM
> definition, so, I will assume I can use it.

SUGGESTIVE by the metaphor. Especially, you must not conflate your
above definition with other sensed of 'output' that we use that CLASH

Why even risk such confusion by using 'output' in a way that clashes
with the ordinary sense?

Why not use, say, 'projected-put' of a TM?

Definition:

The 'projected-put' of a TM T on a certain input t
=3D
{u | u is a square such that there is a stage n such that u is written
to at stage n}

Note: even 'square' and 'stage' and 'tape' are metaphorical, but at
least we know how to translate them unequivocally into the rigorous
formulation if we need to.

Then, with your BNTM, the projected-put (for any input) is the entire
tape (if it is a one-way tape). And, for every natural number n, there
is a stage s such that the number of 1's written on the tape is n.

But it is NOT the case that there is a stage s such that the number of
1's written on the tape is infinite.

And it is NOT the case that there is a stage s such that very square u
has been written to.

> You don't get to say BNTM hasn't written to every position yet,
> but eventually it will. BNTM either writes to position n or it does
> not.
>
> If, as you and most other people have claimed, BNTM will eventually
> write to every finite position, then you are claiming that yes, there
> is
> a stage when BNTM has written every natural number.

No, your simply repeating the mistake I've explained to you over and
over. Eventually, I'll need to stop indulging you with my posts that
you use as yet another turn to re-state your mistake.

> True, you are saying it will do so "eventually", But, you
> are saying it does so.

No, go by the EXACT statement I made. In fact, if this 'eventually' is
what is causing you to mess this up, then DROP the word 'eventually'.

The exact statement is:

For every square u, there is a stage s such that at stage s, u is
written to.

or

For every natural number n, there is a stage s such that at stage s,
the number of 1's on the tape is n.

YOU MUST NOT TWIST that into

There is a stage s such that every square u is written to

nor into

There is a stage s such that the number of 1's on the tape is
infinite.

> My proof uses what are sometimes called "downward infinite
> membership chains". Most set theories have have an axiom
> saying downward infinite membership chains don't exist.
> This axiom is call the Axiom of regularity or the Axiom of
> Foundation.http://en.wikipedia.org/wiki/Axiom_of_regularity

No, you've not properly used the axiom of regularity.

You're used the FALLACY of just assuming that there is a final stage.
There is no such final stage.

> You have to assume the Axiom of regularity is true because
> you can't prove it is true from the other axioms.
> I am sure MoeBlee will say set theory is consistent
> whether we assume Foundation or not.

I'll say this: As far as I know, and as far as every indication I've
seen, ZF is consistent, and I've not seen any good reason to claim
that ZF is not consistent.

> I am not going to argue with him because I know he has
> studied systems without Foundation.
>
> You can see why most set theorists prefer systems
> with an Axiom of Regularity. Set theories without
> Foundation have lots of strange properties and
> unscrupulous people, like myself, can use
> downward infinite membership chains to try to
> prove "infinite" sets are really finite.

No, WRONG. We don't get the strange properties merely by NOT having
the axiom of regularity. But rather we get the strange properties by

Moreover, as far as I know, the main use of regularity has to do with
"shaping" the "world of sets" into a cumulative hierarchy. Indeed, the
axiom of regularity is, in ZFC (maybe even just in ZF?) equivalent to
saying that every set is a member of some level of the hierarchy.

> I bring this up because, as far as I know,
> computation theory doesn't have an
> Axiom of regularity. I can make a downward
> infinite membership chain if I want to.

In ZFC without the axiom of regularity, you can't prove that there
exist downward membership chains and you can't prove that there DON'T
exist downward membership chains.

If you want to use downward membership chains, then you need to
specify your exact axioms that prove the existence of such things. But
then you won't be talking about ZFC anymore. Sure, you might be able
to get a contradiction from some OTHER set of axioms you posit. So
what?

> A TM can only perform one operation at a time.
> If, as you are claiming, BNTM "eventually" writes
> to every location on the tape, then I can prove
> there is a "last" position on the tape with a
> downward infinite membership chain that
> includes the empty set.

No, you end your post by restating the falsehood you started with.

The pity is that meanwhile, as you have for over a DECADE(?) saddled
yourself with misconception, you seem bent on doing that indefinitely,
as meanwhile, you miss learning a bunch of fascinating stuff in which
TMs play an important role connected thereby to many other beautiful
concepts. And by the way, much of this stuff is material that
finitists, constructivists, and others agree upon too.

MoeBlee

```
 0
jazzmobe (307)
7/22/2009 7:41:06 PM
```On Jul 22, 11:50=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
> a
> downward infinite membership chain that
> includes the empty set.

If by 'include' you mean the empty set is a member, then what you
wrote is silliness. The empty set can't be a member of downward
infinite membership chain. Just THINK about it for a second.

MoeBlee

```
 0
jazzmobe (307)
7/22/2009 7:45:24 PM
```Kyle Easterly wrote:
> On Jul 22, 9:15 am, Marshall <marshall.spi...@gmail.com> wrote:
>> On Jul 21, 10:23 pm, RussellE <reaste...@gmail.com> wrote:
>>
>>
>>
>>> Maybe I am mentally defective, but I really don't see
>>> why it is "obvious" a TM can read and write to any
>>> finite position.
>>> Can someone provide a proof that for every n,
>>> BNTM writes to position n? Preferbly without
>>> assuming BNTM can read and write to every position.
>>> Its easy to prove something you have already assumed to be true.
>> You are concerned with assuming the conclusion. That's a good
>> thing to be worried about. (I don't think it's happening here, but
>> it's a valid concern.) Just be sure to distinguish between assuming
>> the conclusion, and rejecting any conditions that leads to the
>> conclusion, on the basis that you don't like the conclusion.
>>
>> Let's for the moment just talk about how the abstract Turing Machine
>> works. Let's talk specifically about the "move the head to the right"
>> operation. Just that.
>>
>> Does it always succeed, or not? Is it guaranteed to work under all
>> circumstances, or are there some circumstances under which it
>> might fail?
>
> "Move right" always succeeds.
>
> I am not saying there is a poisiton where the TM can't move right.
> I am saying there are positions the TM will never get to
> no matter how many times it moves right. I think this
> is true even if we assume the TM can move right an
> infinte number of times.

Do you accept mathematical induction on the natural numbers as a valid
proof technique?

Patricia
```
 0
pats (3556)
7/22/2009 7:47:46 PM
```On Jul 22, 11:58=A0am, rp@raampje.(none) (Reinier Post) wrote:
> MoeBlee wrote:
> >In Z set theory, we can formulate TM's and the important theorems
> >about TMs. We absolutely do NOT assume in set theory that a TM can
> >perform infinitely many operations in a single step. Indeed, our set
> >theoretic formulation provides that it is NOT the case that a TM can
> >perform infinitely many operations in a single step.
>
> I believe a lot of confusion about TMs is caused by the often repeated
> meme that it 'contains an infinite tape'. =A0This is very misleading
> (if not outright wrong) and it suggests that the operation of a TM
> is *not* a stepwise process on finite pieces of tape. =A0But it is.

It would be fine with me if people no longer resorted to such phrases
as "infinite tape of a TM". However, when someone is using such an
expression, it is sometimes easier for communication to go along, as
long as the expression is not taken improperly. Meanwhile, I quite
agree with you about the finititeness of every aspect of actual formal
TMs.

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 7:50:40 PM
```On Jul 22, 12:12=A0pm, Kyle Easterly <keileene...@gmail.com> wrote:

> I am not saying there is a poisiton where the TM can't move right.
> I am saying there are positions the TM will never get to
> no matter how many times it moves right. I think this
> is true even if we assume the TM can move right an
> infinte number of times.

misinformed and illogical.

> > The usual definition of Turing Machine says that it always succeeds.
> > If you are using a formalism under which it might fail, you have to
> > tell us that, otherwise we assume you are using the usual formalism.
>
> > Do you see that if you have a TM with a right-unbounded tape,
> > and you start at the left edge, and the move-right operation always
> > succeeds, then you can get to any position on the tape?
>
> No.

> > Do you see that if you cannot get to some position on the tape,
> > then that necessarily must mean that some move-right operation
> > has to fail?
>
> No. You just haven't moved right far enough and you never will.

You've never proven that.

> > Whether you can get to any position or not is a consequence of
> > the definition of TM you are using. Under the usual definition,
> > you can get to any position.
>
> I agree most formal systems assume a TM can read any finite
> position. But, this is an assumption. I am trying to show why

No, it is NOT an assumption. Rather, we have DEFINITIONS that combined
with our axioms (or, if informally, our basic principles) and logic
proves that certain Turing machines are such that on certain inputs,
for every square u, there is a stage S such that at stage S we have u
gets written to. (And, by the way, definitions given properly, such as
So the only contradiction can come from the axioms).

MoeBlee

```
 0
jazzmobe (307)
7/22/2009 7:56:50 PM
```On Jul 22, 12:47=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> Kyle Easterly wrote:

> Do you accept mathematical induction on the natural numbers as a valid
> proof technique?

Not really. I think its like saying because F is true for 2,3 and 4,
F is true for 10^1000. Maybe F is true for 10^100, but you proven
this by showing F is true for 2,3 and 4.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/22/2009 8:15:27 PM
```Am Wed, 22 Jul 2009 12:56:50 -0700 (PDT) schrieb MoeBlee:

> On Jul 22, 12:12�pm, Kyle Easterly <keileene...@gmail.com> wrote:
>>
>> I agree most formal systems assume a TM can read any finite
>> position. But, this is an assumption.

And it's indeed a WRONG assumption! After all soon or later the TM will
just break down (run out of lubricant, be worn-out, etc.). Moreover we all
know that there's an upper bound for the length of any tape in our
universe! (Note that the resources are finite in our universe!)

@MoeBlee: Still fighting against mills? :-)

Herb
```
 0
Herbert
7/22/2009 8:17:32 PM
```Am Wed, 22 Jul 2009 13:15:27 -0700 (PDT) schrieb RussellE:

>> Do you accept mathematical induction on the natural numbers as a valid
>> proof technique?
>>
> Not really. I think its like saying because F is true for 2,3 and 4,
> F is true for 10^1000. Maybe F is true for 10^100, but you proven
> this by showing F is true for 2,3 and 4.
>
> Russell
> - 2 many 2 count

Oh, you are troll! :-)

Moe LOVES trolls - he just can't help. :-)

Herb
```
 0
Herbert
7/22/2009 8:19:28 PM
```On Jul 22, 1:15=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 22, 12:47=A0pm, Patricia Shanahan <p...@acm.org> wrote:
>
> > Kyle Easterly wrote:
> > Do you accept mathematical induction on the natural numbers as a valid
> > proof technique?
>
> Not really. I think its like saying because F is true for 2,3 and 4,
> F is true for 10^1000. Maybe F is true for 10^100, but you proven
> this by showing F is true for 2,3 and 4.
>
> Russell
> - 2 many 2 count

Russell, I know you know more than this.  Induction also requires that
truth for F at n implies truth for F at n+1.

karl m
```
 0
malbrain (59)
7/22/2009 8:23:52 PM
```On Jul 22, 12:41=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 22, 11:50=A0am, Kyle Easterly <keileene...@gmail.com> wrote:

> For every square u, there is a stage s such that at stage s, u is
> written to.
>
> or
>
> For every natural number n, there is a stage s such that at stage s,
> the number of 1's on the tape is n.
>
> YOU MUST NOT TWIST that into
>
> There is a stage s such that every square u is written to
>
> nor into
>
> There is a stage s such that the number of 1's on the tape is
> infinite.

I still don't see how you can claim BNTM will write to every
finite position but BNTM never actually writes to
every finite position. Either BTNM writes to every
position or it does not.

If BNTM writes to every postion on a tape, I can
prove the tape is finite. You haven't said my proof if wrong.
Are you saying my proof is valid, but it doesn't prove
what I claim it proves?

> > My proof uses what are sometimes called "downward infinite
> > membership chains". Most set theories have have an axiom
> > saying downward infinite membership chains don't exist.
> > This axiom is call the Axiom of regularity or the Axiom of
> > Foundation.http://en.wikipedia.org/wiki/Axiom_of_regularity
>
> No, you've not properly used the axiom of regularity.

You are correct. As you have already pointed out, TM's
can be formulated is nearly any set theory, including
set theories with Foundation.

My point was there is nothing in computation theory
that prevents me from creating the sets in my proof.

Neither of us really wants to work in a set theory
without Foundation.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/22/2009 8:27:40 PM
```On Jul 22, 1:15=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 22, 12:47=A0pm, Patricia Shanahan <p...@acm.org> wrote:
>
> > Kyle Easterly wrote:
> > Do you accept mathematical induction on the natural numbers as a valid
> > proof technique?
>
> Not really. I think its like saying because F is true for 2,3 and 4,
> F is true for 10^1000. Maybe F is true for 10^100, but you proven
> this by showing F is true for 2,3 and 4.

You're completely silly. That is not at all what mathematical
induction is.

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 9:33:58 PM
```On Jul 22, 1:27=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 22, 12:41=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:

> > On Jul 22, 11:50=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
> > For every square u, there is a stage s such that at stage s, u is
> > written to.
>
> > or
>
> > For every natural number n, there is a stage s such that at stage s,
> > the number of 1's on the tape is n.
>
> > YOU MUST NOT TWIST that into
>
> > There is a stage s such that every square u is written to
>
> > nor into
>
> > There is a stage s such that the number of 1's on the tape is
> > infinite.
>
> I still don't see how you can claim BNTM will write to every
> finite position but BNTM never actually writes to
> every finite position. Either BTNM writes to every
> position or it does not.

Maybe you really are trolling me.

Look at EXACTLY what I posted:

For every position n, there is a stage S such that n is written to at
stage S. But there is no stage S such that for every position n, we
have n written to at stage S.

THAT is what you need to understand. If you STILL cannot understand
it, then I need to move on; I can't just keep repeating my perfectly
clear explanation over and over.

> If BNTM writes to every postion on a tape, I can
> prove the tape is finite. You haven't said my proof if wrong.
> Are you saying my proof is valid, but it doesn't prove
> what I claim it proves?
>

I've been EXTREMELY SPECIFIC. Go back to EXACTLY all the things I've
written about your argument. I'm not going to sit here and retype it
for you.

> > > My proof uses what are sometimes called "downward infinite
> > > membership chains". Most set theories have have an axiom
> > > saying downward infinite membership chains don't exist.
> > > This axiom is call the Axiom of regularity or the Axiom of
> > > Foundation.http://en.wikipedia.org/wiki/Axiom_of_regularity
>
> > No, you've not properly used the axiom of regularity.
>
> You are correct. As you have already pointed out, TM's
> can be formulated is nearly any set theory, including
> set theories with Foundation.
>
> My point was there is nothing in computation theory
> that prevents me from creating the sets in my proof.

There is nothing that PROVIDES you the creation of such sets, UNLESS
you add axioms that we don't use.

I posted that alrady; you just ignored it.

(Moreover, the fallacy in your argument is not even regarding
foundation.)

> Neither of us really wants to work in a set theory
> without Foundation.

Don't speak for me. I work every day in a Z-"foundation". I also work
with foundation other times for other purposes.

MoeBlee

```
 0
jazzmobe (307)
7/22/2009 9:42:52 PM
```In article <e42e074b-0ab1-47f0-8877-bc9b3dedbbd4@y4g2000prf.googlegroups.com>,
RussellE  <reasterly@gmail.com> wrote:
>I still don't see how you can claim BNTM will write to every
>finite position but BNTM never actually writes to
>every finite position. Either BTNM writes to every
>position or it does not.

If you can fool all of the people some of the time, and some of the people
all of the time, does this mean

1. for every person, there exists a time when you can fool that person,
and at every time, there exists a person you can fool;

or does it mean

2. there exists a single time at which you can fool every single person,
and there exists some person that can be fooled every single time?

If you can grasp that distinction, then try replacing "fooling a person"
with "writing on a tape position."
--
Tim Chow       tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth.  ---Galileo, Dialogues Concerning Two New Sciences
```
 0
tchow (869)
7/22/2009 9:58:41 PM
```On Jul 22, 2:58=A0pm, tc...@lsa.umich.edu wrote:
com>,
>
> RussellE =A0<reaste...@gmail.com> wrote:
> >I still don't see how you can claim BNTM will write to every
> >finite position but BNTM never actually writes to
> >every finite position. Either BTNM writes to every
> >position or it does not.
>
> If you can fool all of the people some of the time, and some of the peopl=
e
> all of the time, does this mean
>
> 1. for every person, there exists a time when you can fool that person,
> =A0 =A0and at every time, there exists a person you can fool;
>
> or does it mean
>
> 2. there exists a single time at which you can fool every single person,
> =A0 =A0and there exists some person that can be fooled every single time?
>
> If you can grasp that distinction, then try replacing "fooling a person"
> with "writing on a tape position."

Tim, I've already tried that kind of analogy with him. His post there
is even in response to my analogies of that very kind. Horse just
won't drink.

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 10:55:46 PM
```On Jul 22, 1:23=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 22, 1:15=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > On Jul 22, 12:47=A0pm, Patricia Shanahan <p...@acm.org> wrote:
>
> > > Kyle Easterly wrote:
> > > Do you accept mathematical induction on the natural numbers as a vali=
d
> > > proof technique?
>
> > Not really. I think its like saying because F is true for 2,3 and 4,
> > F is true for 10^1000. Maybe F is true for 10^100, but you proven
> > this by showing F is true for 2,3 and 4.
>
> > Russell
> > - 2 many 2 count
>
> Russell, I know you know more than this. =A0Induction also requires that
> truth for F at n implies truth for F at n+1.

You are right. Simply rejecting the Axiom of Infinity
doesn't make induction go away. We can still use induction.

Even so, I am not sure we can even define "natural" numbers.
I don't think we can even rule out the possibility there
might be a largest natural number.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/22/2009 11:36:33 PM
```On Jul 22, 4:36=A0pm, RussellE <reaste...@gmail.com> wrote:

> Simply rejecting the Axiom of Infinity
> doesn't make induction go away. We can still use induction.
>
> Even so, I am not sure we can even define "natural" numbers.
> I don't think we can even rule out the possibility there
> might be a largest natural number.

We can define 'natural number' and prove there is no largest natural
number, even without the axiom of infinity.

What POSSIBLE advantage do you derive from always remaining in the
dark when instead you could just read a book and scoop up the
background you need?

MoeBlee
```
 0
jazzmobe (307)
7/22/2009 11:50:41 PM
```RussellE wrote:
> On Jul 22, 12:47 pm, Patricia Shanahan <p...@acm.org> wrote:
>> Kyle Easterly wrote:
>
>> Do you accept mathematical induction on the natural numbers as a valid
>> proof technique?
>
> Not really. I think its like saying because F is true for 2,3 and 4,
> F is true for 10^1000. Maybe F is true for 10^100, but you proven
> this by showing F is true for 2,3 and 4.

In that case, I don't see any point in continuing to participate in this
nor useful.

Patricia
```
 0
pats (3556)
7/23/2009 12:29:02 AM
```On Jul 22, 2:42=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 22, 1:27=A0pm, RussellE <reaste...@gmail.com> wrote:

> Maybe you really are trolling me.

I will admit it is fun to argue with you,
but I am not trolling you.
There are a number of issues we don't agree on.

> Look at EXACTLY what I posted:
>
> For every position n, there is a stage S such that n is written to at
> stage S. But there is no stage S such that for every position n, we
> have n written to at stage S.

You are saying BTNM never writes an ACTUAL infinite string,
but it can write every finite string of 1's.

I take it you are saying the output of BTNM is
POTENTIALLY infinite.

Please correct me if I am wrong.

> THAT is what you need to understand. If you STILL cannot understand
> it, then I need to move on; I can't just keep repeating my perfectly
> clear explanation over and over.
>
> > If BNTM writes to every postion on a tape, I can
> > prove the tape is finite. You haven't said my proof if wrong.
> > Are you saying my proof is valid, but it doesn't prove
> > what I claim it proves?
>
>
> I've been EXTREMELY SPECIFIC. Go back to EXACTLY all the things I've
> written about your argument. I'm not going to sit here and retype it
> for you.

all of your objections. I wil try to address the ones I
do understand, like the difference between actual and
potential infinity.

It is probably best to list the things we do agree on.
Please correct me if you don't agree with this list.

I have given a number of proofs in this thread.
remove the smallest element. I show you will never
get the empty set. You haven't pointed out any
flaws with this proof, so, I will assume you agree
the proof is valid.

1) You can never get the empty set by removing
one element at a time from N.

I have also shown, if you can create the empty set
from set S by removing one element at a time,
S must be a finite set.

I am not sure if you agree with this proof.
I can't find a flaw in it and you haven't pointed
out any, so I will assume we agree on this one.

2) A set is finite if we can remove every element
from it one element at a time.

3) At every step performed by BNTM there is
a finite number of 1's on the tape.

I assume you agree with this one.
Personally, I think this is enough to prove my
argument, but you obviously disagree.
You seem to think BNTM can "potentially"
write to any position of an infinite tape.

4) We can define THE output of BNTM as
the set of all positions BNTM will ever write to.

We have had a lot of arguments on how to define
the output of BNTM. You haven't complained much

When you say the set of all positions BTNM
will ever write to is N, I take this to mean BTNM
will actually write an infinite string.

You seem to be saying BTNM doesn't actually
write an an infinite string but it potentially can.

I had hoped to eliminate this with my definition
of THE output of BNTM. I am not asking for
the set of positions BNTM can "potentially"
write to. I want the positions BNTM will
actually write to.

I don't see how you can quibble about the
difference between potential and actual
infinity here. Especially when you say things
like for every n, BTNM will write to position n
after n steps.

BNTM will write to every position on the tape
or it won't. If BNTM does write to every position
I prove the tape is finite. Which part of my
argument do you disagree with?

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/23/2009 12:56:01 AM
```On Jul 22, 5:56=C2=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 22, 2:42=C2=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 22, 1:27=C2=A0pm, RussellE <reaste...@gmail.com> wrote:
> > Maybe you really are trolling me.
>
> I will admit it is fun to argue with you,
> but I am not trolling you.
> There are a number of issues we don't agree on.
>
> > Look at EXACTLY what I posted:
>
> > For every position n, there is a stage S such that n is written to at
> > stage S. But there is no stage S such that for every position n, we
> > have n written to at stage S.
>
> You are saying BTNM never writes an ACTUAL infinite string,
> but it can write every finite string of 1's.
>
> I take it you are saying the output of BTNM is
> POTENTIALLY infinite.
>
> Please correct me if I am wrong.
>
> > THAT is what you need to understand. If you STILL cannot understand
> > it, then I need to move on; I can't just keep repeating my perfectly
> > clear explanation over and over.
>
> > > If BNTM writes to every postion on a tape, I can
> > > prove the tape is finite. You haven't said my proof if wrong.
> > > Are you saying my proof is valid, but it doesn't prove
> > > what I claim it proves?
>
> > > Please be specific.
>
> > I've been EXTREMELY SPECIFIC. Go back to EXACTLY all the things I've
> > written about your argument. I'm not going to sit here and retype it
> > for you.
>
> all of your objections. I wil try to address the ones I
> do understand, like the difference between actual and
> potential infinity.
>
> It is probably best to list the things we do agree on.
> Please correct me if you don't agree with this list.
>
> I have given a number of proofs in this thread.
> One of them is where I start with N and repeatedly
> remove the smallest element. I show you will never
> get the empty set. You haven't pointed out any
> flaws with this proof, so, I will assume you agree
> the proof is valid.
>
> 1) You can never get the empty set by removing
> one element at a time from N.
>
> I have also shown, if you can create the empty set
> from set S by removing one element at a time,
> S must be a finite set.
>
> I am not sure if you agree with this proof.
> I can't find a flaw in it and you haven't pointed
> out any, so I will assume we agree on this one.
>
> 2) A set is finite if we can remove every element
> from it one element at a time.
>
> 3) At every step performed by BNTM there is
> a finite number of 1's on the tape.
>
> I assume you agree with this one.
> Personally, I think this is enough to prove my
> argument, but you obviously disagree.
> You seem to think BNTM can "potentially"
> write to any position of an infinite tape.
>
> 4) We can define THE output of BNTM as
> the set of all positions BNTM will ever write to.
>
> We have had a lot of arguments on how to define
> the output of BNTM. You haven't complained much
>
> When you say the set of all positions BTNM
> will ever write to is N, I take this to mean BTNM
> will actually write an infinite string.
>
> You seem to be saying BTNM doesn't actually
> write an an infinite string but it potentially can.
>
> I had hoped to eliminate this with my definition
> of THE output of BNTM. I am not asking for
> the set of positions BNTM can "potentially"
> write to. I want the positions BNTM will
> actually write to.
>
> I don't see how you can quibble about the
> difference between potential and actual
> infinity here. Especially when you say things
> like for every n, BTNM will write to position n
> after n steps.
>
> BNTM will write to every position on the tape
> or it won't. If BNTM does write to every position
> I prove the tape is finite. Which part of my
> argument do you disagree with?
>
> Russell
> - 2 many 2 count

An Algorithm/Proof Hybrid
> Sum 1. say: > Truth: what becomes true in its construction to be construc=
tion it is > a fact that constitutes. We may say it is a (fact) and the pro=
of > constitutes truth. > Sum 1. I am, exist. > Sum sine regno: "I am witho=
ut a kingdom." > Sic sum ut vides. "Thus I am as you see." > Cogito, ergo s=
um. "I think, therefore I am." > "Dixit duas res ei rubori fuisse." > He sa=
id that two things had abashed him > not equal (=E2=89=A0) (=3D) equal > Ci=
vis Romanus sum: "I am a Roman citizen." > Sum: present active sum, present=
infinitive esse, perfect active > fu=C4=AB > "I am as you see." > Sum. (ar=
ithmetic) 1. A quantity obtained by addition or aggregation. > I am, exists=
> The sum of 3 and 4 is 7 > =E2=89=A0 > I am, exists > 3+4=3D7 as you see.=
" > I am, exists > 3+4=3D7 "I am as you see." > I > Sum (arithmetic) An ari=
thmetic computation, especially one posed to a > student as an exercise (no=
t necessarily limited to addition.) I am, > exists =E2=89=A0 > 2. A quantit=
y of money. > 3. A summary. > 4. A central idea or point. > 5. The utmost d=
egree. > Cum veritate sum > "I am with truth." > Sum 1. Cum veritate sum > =
I am, exists with truth > =3D paene (not comparable) =E2=89=A0 > verto: Lat=
in > (vartayati), =E2=80=9C=E2=80=98he turns=E2=80=99=E2=80=9D). > present =
active vert=C5=8D, present infinitive vertere, perfect active > vert=C4=AB,=
supine versum. > I turn, revolve. > I turn around. > I exchange. > I trans=
late. > pr=C4=ABm=C5=8D 1. at first > satis: 1. enough, filled, plenty > 1.=
adequately, sufficiently > Cum Veritate Sum paene primo satis > 1. "I am w=
ith truth (not comparable) at first to satisfy" > satis pr=C4=ABm=C5=8D ver=
to > 1. But I satisfy it sufficiently, because at first I turn, revolve. > =
pr=C4=ABm=C5=8D 1. at first > Sum venio verto > 1. "Because in essence I be=
gin in reverse" > P=3D=3DNP > Q.E.D > Martin Michael Musatov =C2=A92009 Mar=
```
 0
7/23/2009 1:00:14 AM
```On Jul 22, 2:58=C2=A0pm, tc...@lsa.umich.edu wrote:
com>,
>
> RussellE =C2=A0<reaste...@gmail.com> wrote:
> >I still don't see how you can claim BNTM will write to every
> >finite position but BNTM never actually writes to
> >every finite position. Either BTNM writes to every
> >position or it does not.
>
> If you can fool all of the people some of the time, and some of the peopl=
e
> all of the time, does this mean
>
> 1. for every person, there exists a time when you can fool that person,
> =C2=A0 =C2=A0and at every time, there exists a person you can fool;
>
> or does it mean
>
> 2. there exists a single time at which you can fool every single person,
> =C2=A0 =C2=A0and there exists some person that can be fooled every single=
time?
>
> If you can grasp that distinction, then try replacing "fooling a person"
> with "writing on a tape position."
> --
> Tim Chow =C2=A0 =C2=A0 =C2=A0 tchow-at-alum-dot-mit-dot-edu
> The range of our projectiles---even ... the artillery---however great, wi=
ll
> never exceed four of those miles of which as many thousand separate us fr=
om
> the center of the earth. =C2=A0---Galileo, Dialogues Concerning Two New S=
ciences

It means you (M.I.T.) can be fooled most of the time by a machine. But
the machine can be fooled 100% of the time if you realize it is a
MACHINE.

An Algorithm/Proof Hybrid
> Sum 1. say: > Truth: what becomes true in its construction to be construc=
tion it is > a fact that constitutes. We may say it is a (fact) and the pro=
of > constitutes truth. > Sum 1. I am, exist. > Sum sine regno: "I am witho=
ut a kingdom." > Sic sum ut vides. "Thus I am as you see." > Cogito, ergo s=
um. "I think, therefore I am." > "Dixit duas res ei rubori fuisse." > He sa=
id that two things had abashed him > not equal (=E2=89=A0) (=3D) equal > Ci=
vis Romanus sum: "I am a Roman citizen." > Sum: present active sum, present=
infinitive esse, perfect active > fu=C4=AB > "I am as you see." > Sum. (ar=
ithmetic) 1. A quantity obtained by addition or aggregation. > I am, exists=
> The sum of 3 and 4 is 7 > =E2=89=A0 > I am, exists > 3+4=3D7 as you see.=
" > I am, exists > 3+4=3D7 "I am as you see." > I > Sum (arithmetic) An ari=
thmetic computation, especially one posed to a > student as an exercise (no=
t necessarily limited to addition.) I am, > exists =E2=89=A0 > 2. A quantit=
y of money. > 3. A summary. > 4. A central idea or point. > 5. The utmost d=
egree. > Cum veritate sum > "I am with truth." > Sum 1. Cum veritate sum > =
I am, exists with truth > =3D paene (not comparable) =E2=89=A0 > verto: Lat=
in > (vartayati), =E2=80=9C=E2=80=98he turns=E2=80=99=E2=80=9D). > present =
active vert=C5=8D, present infinitive vertere, perfect active > vert=C4=AB,=
supine versum. > I turn, revolve. > I turn around. > I exchange. > I trans=
late. > pr=C4=ABm=C5=8D 1. at first > satis: 1. enough, filled, plenty > 1.=
adequately, sufficiently > Cum Veritate Sum paene primo satis > 1. "I am w=
ith truth (not comparable) at first to satisfy" > satis pr=C4=ABm=C5=8D ver=
to > 1. But I satisfy it sufficiently, because at first I turn, revolve. > =
pr=C4=ABm=C5=8D 1. at first > Sum venio verto > 1. "Because in essence I be=
gin in reverse" > P=3D=3DNP > Q.E.D > Martin Michael Musatov =C2=A92009 Mar=
```
 0
7/23/2009 1:05:12 AM
```On Jul 22, 5:29=A0pm, Patricia Shanahan <p...@acm.org> wrote:
> RussellE wrote:
> > On Jul 22, 12:47 pm, Patricia Shanahan <p...@acm.org> wrote:
> >> Kyle Easterly wrote:
>
> >> Do you accept mathematical induction on the natural numbers as a valid
> >> proof technique?
>
> > Not really. I think its like saying because F is true for 2,3 and 4,
> > F is true for 10^1000. Maybe F is true for 10^100, but you proven
> > this by showing F is true for 2,3 and 4.
>
> In that case, I don't see any point in continuing to participate in this
> thread. Your version of mathematics is likely to be neither interesting
> nor useful.
>
> Patricia

If anyone (especially a woman) starts to sound bitchy or negative but
resigned on a math forum you can bet you are on to something good! --
MMMM

```
 0
7/23/2009 1:10:28 AM
```On Wed, 22 Jul 2009 18:10:28 -0700 (PDT) scribio_vide wrote:

> If anyone (especially a woman) starts to sound bitchy or negative but
> resigned on a math forum you can bet you are on to something good!

Fuck you, asshole!
```
 0
Herbert
7/23/2009 1:14:42 AM
```On Jul 22, 5:56=A0pm, RussellE <reaste...@gmail.com> wrote:

> I take it you are saying the output of BTNM is
> POTENTIALLY infinite.
>
> Please correct me if I am wrong.

'potentially' is just another wiggle word.

I've alrady said EXACTLY what your machine does in terms of strings of
1's and postions on tape (short of stating it in actual mathematical
fomulations for TMs).

> all of your objections. I wil try to address the ones I
> do understand, like the difference between actual and
> potential infinity.

I don't use a notion of a difference between actual and potential
infinity in this. I touched on the matter in response to your use of
that notion. That notion does not quite formalize (or at least a
formalization has not been provided here) in set theory.

> It is probably best to list the things we do agree on.
> Please correct me if you don't agree with this list.
>
> I have given a number of proofs in this thread.
> One of them is where I start with N and repeatedly
> remove the smallest element. I show you will never
> get the empty set. You haven't pointed out any
> flaws with this proof, so, I will assume you agree
> the proof is valid.
>
> 1) You can never get the empty set by removing
> one element at a time from N.
>
> I have also shown, if you can create the empty set
> from set S by removing one element at a time,
> S must be a finite set.
>
> I am not sure if you agree with this proof.
> I can't find a flaw in it and you haven't pointed
> out any, so I will assume we agree on this one.

All that squeeks by as barely okay as far it goes using informal
notions such as 'removing', 'one at time', etc.

Why don't you learn the actual mathematical terminology that is
unambigous and precise?

> 2) A set is finite if we can remove every element
> from it one element at a time.

Barely passable as a mathematical statement. I let it go only because
I can GUESS what you might mean.

> 3) At every step performed by BNTM there is
> a finite number of 1's on the tape.

Correct.

> I assume you agree with this one.
> Personally, I think this is enough to prove my
> argument, but you obviously disagree.
> You seem to think BNTM can "potentially"
> write to any position of an infinite tape.

Why do you insist on putting words in my mouth when I've told you
three times already that EXACTLY what I mean is exactly as I've stated
it over and over now?

> 4) We can define THE output of BNTM as
> the set of all positions BNTM will ever write to.

new term (since 'output' already has an established meaning) such as
'project-put' (informally suggestive of "we may project that...").

> We have had a lot of arguments on how to define
> the output of BNTM. You haven't complained much

NO, I DID complain. Can't you read?

> When you say the set of all positions BTNM
> will ever write to is N, I take this to mean BTNM
> will actually write an infinite string.

STOP. ENOUGH RIGHT THERE. You're just doing it again. Simply repeating
the same mistake.

If you're not trolling me, then you really do have a severe cognitive
Please, don't even bother writing back to me.

MoeBlee
```
 0
jazzmobe (307)
7/23/2009 1:17:27 AM
```On Jul 22, 5:56=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 22, 2:42=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 22, 1:27=A0pm, RussellE <reaste...@gmail.com> wrote:
> > Maybe you really are trolling me.
>
> I will admit it is fun to argue with you,

You find it fun being an ignorant goofball.

Your post ended up just repeating yet again the same mistake you've
It would be wonderful if you don't even bother writing back to me.

MoeBlee

```
 0
jazzmobe (307)
7/23/2009 1:19:57 AM
```On 2009-07-22, Kyle Easterly <keileeneast@gmail.com> wrote:
> You are saying BNTM will "eventually" write to every position.

"Eventually writes to every position" => "For all n, there exists a
step m such that the BNTM has written to position n by step m."

> How is this different from saying "eventually" BNTM writes an infinite
> string?

"Eventually writes an infinite string" => "There exists a step m such
that for all n, the BNTM has written to position n by step m."

So the difference is the order of the quantifiers when you express
them in predicate logic.  Try learning some basic logic.

- Tim
```
 0
tim669 (185)
7/23/2009 3:49:25 AM
```On 2009-07-23, RussellE <reasterly@gmail.com> wrote:
> Please correct me if I am wrong.

You are wrong.

> I have read your posts and I honestly don't understand all of your
> objections.

I don't know about MoeBlee, but for me the fundamental one is the
difference in logic between a statement that starts: "For all X, there
exists Y such that ..." and one that starts "There exists Y such that
for all X, ...".  The ordering is very important, and you keep getting
them implicitly switched in your arguments (sometimes multiple times
in the same argument).

- Tim
```
 0
tim669 (185)
7/23/2009 3:57:10 AM
```On Jul 22, 8:49=A0pm, Tim Little <t...@little-possums.net> wrote:
> On 2009-07-22, Kyle Easterly <keileene...@gmail.com> wrote:
>
> > You are saying BNTM will "eventually" write to every position.
>
> "Eventually writes to every position" =3D> "For all n, there exists a
> step m such that the BNTM has written to position n by step m."

This is true for finite or infinite strings.

> > How is this different from saying "eventually" BNTM writes an infinite
> > string?
>
> "Eventually writes an infinite string" =3D> "There exists a step m such
> that for all n, the BNTM has written to position n by step m."

This statement is false.
Let me change it slightly so it applies to any string.

Eventually writes every position =3D> There exists a step m such
that for all n, the BNTM has written to position n by step m.

This is a provably true statement.
I am pretty sure it is equivalent to the first statement.

Russell
- 2 many 2 count

> So the difference is the order of the quantifiers when you express
> them in predicate logic. =A0Try learning some basic logic.
>
> - Tim

```
 0
reasterly (337)
7/23/2009 6:56:02 AM
```"RussellE" <reasterly@gmail.com> wrote in message
On Jul 18, 11:13 pm, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
> "RussellE" <reaste...@gmail.com> wrote in message
>
> On Jul 18, 9:04 pm, Patricia Shanahan <p...@acm.org> wrote:
>
> > RussellE wrote:
> > > On Jul 18, 8:27 pm, Patricia Shanahan <p...@acm.org> wrote:
> > >> RussellE wrote:
>
> > >> PI is a computable number in the sense
> > >> ofhttp://en.wikipedia.org/wiki/Computable_number#Formal_definition.
> > >> Informally, it can be calculated to any required finite number of
> > >> decimal places.
>
> > > The uncomputable number can be calculated to any
> > > required finite number of positions.
> > > Every position is a '1'.
>
> > Perhaps you should begin by defining *exactly* what you mean by
> > "uncomputable number".
>
> I define a one state TM. This TM writes a '1' at the current position
> and moves the tape head one to position to the right.
> This TM has no halt state so it will just keep writing 1's forever.
>
> I can prove the output of this TM is a finite string of 1's followed
> by a blank.
>
> *********************
> No you can't. You can't prove it because its not true. If you think you
> can
> prove it, try doing so.

OK

I want to keep track of all of the blank positions on the tape.
When the TM starts the tape is all blank.
The set of blank positions is the set of all natural numbers.

A0 = {1,2,3,...}

Every time the TM writes a '1' I update my list.
A1 = {2,3,4,...}
A2 = {3,4,5,...}
A3 = {4,5,6,...}
....

For every step performed by the TM the set
of blank positions is infinite.

******* At every step

Since the set of blank positions is non-empty,
there is a smallest blank position.

******* At every step

This proves the string of 1's on the tape is finite.

****** At every step

The output tape is a finite string of 1's followed
by a blank.

****** At every step

And no, I can't tell you where the
first blank is.

******** At every step, the first blank is at position n+1, where n is the
step number

The position of the first blank
is an uncomputable natural number.

********** No, its n.

Russell
- 2 many 2 count

```
 0
webbfamily (36)
7/23/2009 9:44:42 AM
```On Wed, 22 Jul 2009 08:59:01 -0700 (PDT), MoeBlee <jazzmobe@hotmail.com>
said:
> On Jul 21, 7:50�pm, RussellE <reaste...@gmail.com> wrote:
>> ...
>> We both agree BNTM will never write an infinite string.
>> There is really no question of what is on the tape at
>> any step. There is a finite string of 1's.
>>
>> Yet, you keep telling me that "eventually" BNTM will
>> write an infinite string.
>
> NO NO NO. We do NOT say that. We say that for any given position,
> eventually that position will be written to. We never say there is
> ever an infinite string that the TM writes.
> ...
> For every finite string of 1's, there exists a point at which the BNTM
> will have written that string. But it is not the case that there
> exists a point at which the BNTM will have written all the finite
> strings (or even one infinite string).

The inability to distinguish (Ax)(Ey)F from (Ey)(Ax)F seems to lie at
the heart of a number of classic crank confusions.

```
 0
cmenzel (185)
7/23/2009 12:17:17 PM
```In article <065e0a44-de6a-40f3-af78-fafc7eccb0d9@k13g2000prh.googlegroups.com>, RussellE <reasterly@gmail.com> writes:

>all of your objections. I wil try to address the ones I
>do understand, like the difference between actual and
>potential infinity.

If you understand the difference between "actual infinity" and "potential
infinity", then you can provide mathematical definitions for them.

--
Michael F. Stemper
#include <Standard_Disclaimer>
This sentence no verb.
```
 0
mstemper1 (15)
7/23/2009 1:02:21 PM
```On Thu, 23 Jul 2009 12:17:17 +0000 (UTC) Chris Menzel wrote:

> The inability to distinguish (Ax)(Ey)F from (Ey)(Ax)F seems to lie at
> the heart of a number of classic crank confusions.

Yesterday MoeBlee wrote:

"If you're not trolling me, then you really do have a severe cognitive
problem."

To which I replied:

"I guess that's true for many mathematical cranks. (For example, WM is
clearly suffering from quantifier dyslexia, besides other things.)"

Same observation. :-)

Herb
```
 0
Herbert
7/23/2009 2:58:54 PM
```On Jul 23, 5:17=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wrote:
> On Wed, 22 Jul 2009 08:59:01 -0700 (PDT), MoeBlee <jazzm...@hotmail.com>
> said:

> The inability to distinguish (Ax)(Ey)F from (Ey)(Ax)F seems to lie at
> the heart of a number of classic crank confusions

Which of these statements do you claim is false for BNTM?

1) Eventually writes to every position on the tape =3D> (Ax)(Ey) BNTM
writes to position x by step y.

2) Eventually writes to every position on the tape =3D> (Ey)(Ax) BNTM
writes to position x.by step y

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/23/2009 5:32:56 PM
```On Jul 23, 10:32=A0am, RussellE <reaste...@gmail.com> wrote:
> On Jul 23, 5:17=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wrote:
>
> > On Wed, 22 Jul 2009 08:59:01 -0700 (PDT), MoeBlee <jazzm...@hotmail.com=
>
> > said:
> > The inability to distinguish (Ax)(Ey)F from (Ey)(Ax)F seems to lie at
> > the heart of a number of classic crank confusions
>
> Which of these statements do you claim is false for BNTM?
>
> 1) Eventually writes to every position on the tape =3D> (Ax)(Ey) BNTM
> writes to position x by step y.
>
> 2) Eventually writes to every position on the tape =3D> (Ey)(Ax) BNTM
> writes to position x.by step y

The VERY point is that the expression "eventually writes to every
position on the tape" is AMBIGUOUS. So we throw it out. Instead we
specify WHICH of the two we mean. And the one we mean is:

AxEy BNTM writes to position x by step y.

One  of RussellE's forms of childish game playing is to switch back to
the ambiguous (such as 'eventually', 'potentially') after WE have
already introduced the PROGRESS of moving past the ambiguous to the
unambiguous "AxEy BNTM writes to position x by step y".

MoeBlee

```
 0
jazzmobe (307)
7/23/2009 5:41:33 PM
```On Thursday 16 July 2009 18:33, RussellE wrote:
> On Jul 16, 7:31 am, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>> RussellE <reaste...@gmail.com> writes:
>> > Zeno machines are Accelerated Turing Machines.
>
> ...
>
> I helped derive that proof in this newsgroup several years ago.
> Unfortunately, the Halting proof is wrong. It assumes a Zeno Machine
> can perform an infinite number of operations. This proof shows
> a ZM can only perform a finite number of operations.
>
> ...
>
> I start by assuming a ZM can perform an infinite number of
> operations. Then I show the ZM can never perform more
> that a finite number of read or writes.
>
> My proof is little more than a proof that all natural numbers
> are finite. Each operation performed by the ZM (or any TM)
> can be assigned a natural number. There is a step 1, a step 2,
> etc. There can never be a step infinity.

proofs and other things with "infinity" in them) work with subtly
different laws than regular numbers.

If we assume that a ZM has non-physical powers (like, being able to do
each operation faster than the previous without any bound), don't be
surprised that it can accomplish counterintuitive results.

Analogously, is 1 = .99999... (an infinite string of 9s)?  One
argument might be that .9 differs from 1 by .1, .99 differs from 1 by
..01, .999 differs by .001, and so forth.  Since any finite string of
9s differs from 1, then an infinite string of 9s must differ from 1,
too.

That reasoning is incorrect.  We can show correct reasoning as follows:
x = .9999....   [1]
Multiplying by 10, we get
10x = 9.999....   [2]
Subtracting equals from equals (Eq. 1 from Eq 2), we get
10x = 9.999....
-  x =  .9999...
---------------
9x = 9
Dividing by 9, we get
x = 1
Therefore 1 = .99999...

Analogously, a ZM *can* perform a countably infinite number of
operations in a finite amount of time.

-paul-
--
Paul E. Black (p.black@acm.org)
```
 0
p.black (279)
7/23/2009 6:25:55 PM
```On Jul 23, 10:41=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 23, 10:32=A0am, RussellE <reaste...@gmail.com> wrote:
>
> > On Jul 23, 5:17=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wrote=
:
>
> > > On Wed, 22 Jul 2009 08:59:01 -0700 (PDT), MoeBlee <jazzm...@hotmail.c=
om>
> > > said:
> > > The inability to distinguish (Ax)(Ey)F from (Ey)(Ax)F seems to lie at
> > > the heart of a number of classic crank confusions
>
> > Which of these statements do you claim is false for BNTM?
>
> > 1) Eventually writes to every position on the tape =3D> (Ax)(Ey) BNTM
> > writes to position x by step y.
>
> > 2) Eventually writes to every position on the tape =3D> (Ey)(Ax) BNTM
> > writes to position x.by step y
>
> The VERY point is that the expression "eventually writes to every
> position on the tape" is AMBIGUOUS.

I don't see how you can say "every position on the tape" is
ambiguous for a finite tape.

If you are saying "every position on the tape" is ambiguous
for infinite tapes then I agree with you.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/23/2009 6:40:20 PM
```On Jul 23, 11:40=A0am, RussellE <reaste...@gmail.com> wrote:
> On Jul 23, 10:41=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 23, 10:32=A0am, RussellE <reaste...@gmail.com> wrote:
>
> > > On Jul 23, 5:17=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wro=
te:
>
> > > > On Wed, 22 Jul 2009 08:59:01 -0700 (PDT), MoeBlee <jazzm...@hotmail=
..com>
> > > > said:
> > > > The inability to distinguish (Ax)(Ey)F from (Ey)(Ax)F seems to lie =
at
> > > > the heart of a number of classic crank confusions
>
> > > Which of these statements do you claim is false for BNTM?
>
> > > 1) Eventually writes to every position on the tape =3D> (Ax)(Ey) BNTM
> > > writes to position x by step y.
>
> > > 2) Eventually writes to every position on the tape =3D> (Ey)(Ax) BNTM
> > > writes to position x.by step y
>
> > The VERY point is that the expression "eventually writes to every
> > position on the tape" is AMBIGUOUS.
>
> I don't see how you can say "every position on the tape" is
> ambiguous for a finite tape.
>
> If you are saying "every position on the tape" is ambiguous
> for infinite tapes then I agree with you.

Stop playing childish games.

I didn't say "every position on the tape" is ambiguous. What I said is
ambiguous is "Eventually writes to every position on the tape". In
particular, part of your intellectual childishness is to exploit the
ambiguity of "eventually". And exploit is unnecessary, since you've
been told a thousand times already, the correct claim is:

AxEy BNTM writes to x at stage y.

Grow up.

Address EXACTLY "AxEy BNTM writes to x at stage y" and NOT the
confusions you create by introducing and RE-introducing more ambiguous
language.

And if you purport to prove things about Turing machines, then learn
how to do it correctly. Learn a mathematical formulation of Turing
machines and prove things within such a formulation.

MoeBlee

```
 0
jazzmobe (307)
7/23/2009 7:06:45 PM
```P.S to Jul 23, 12:06=A0pm, MoeBlee <jazzm...@hotmail.com>:

As to finite/infinite tape, it is already a given that the "tape" is
infinite. That is a given metaphor that comes with a Turing machine.
And, again, note that "tape" and such things are metaphors for
mathematical formulations. In an ordinary mathematical formulation of
Turing machines there is not even a tape, let alone a question of it
being finite or infinite. But if we are to use the metaphor of there
being a tape, then it convenient to describe it as infinite.

MoeBlee

```
 0
jazzmobe (307)
7/23/2009 7:11:06 PM
```On Jul 23, 12:06=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 23, 11:40=A0am, RussellE <reaste...@gmail.com> wrote:

> > > > Which of these statements do you claim is false for BNTM?
>
> > > > 1) Eventually writes to every position on the tape =3D> (Ax)(Ey) BN=
TM
> > > > writes to position x by step y.
>
> > > > 2) Eventually writes to every position on the tape =3D> (Ey)(Ax) BN=
TM
> > > > writes to position x.by step y
>
> > > The VERY point is that the expression "eventually writes to every
> > > position on the tape" is AMBIGUOUS.
>
> > I don't see how you can say "every position on the tape" is
> > ambiguous for a finite tape.
>
> > If you are saying "every position on the tape" is ambiguous
> > for infinite tapes then I agree with you.
>
> Stop playing childish games.
>
> I didn't say "every position on the tape" is ambiguous. What I said is
> ambiguous is "Eventually writes to every position on the tape".

Then remove the word eventually. Both statements are still true.

> In
> particular, part of your intellectual childishness is to exploit the
> ambiguity of "eventually". And exploit is unnecessary, since you've
> been told a thousand times already, the correct claim is:
>
> AxEy BNTM writes to x at stage y.

BNTM writes to every position on the tape =3D>
AxEy BNTM writes to x by stage y

I agree this is a true statement.
But so is:

BNTM writes to every position on the tape =3D>
EyAx BNTM writes to x by stage y

Like you, I am shocked by the ramifications
of my proof. But I don't see any flaws in my
argument.

Russell
- You don't have to be crazy to be a mathematician, but is does help
```
 0
reasterly (337)
7/23/2009 8:34:42 PM
```In article <7eb0a462-3781-4cf8-8b37-686f258c96cc@v15g2000prn.googlegroups.com>,
RussellE  <reasterly@gmail.com> wrote:
>BNTM writes to every position on the tape =>
>AxEy BNTM writes to x by stage y
>
>I agree this is a true statement.
>But so is:
>
>BNTM writes to every position on the tape =>
>EyAx BNTM writes to x by stage y

The trouble is, nobody else seems to understand what you even *mean* by
"BNTM writes to every position on the tape."

People understand what "AxEy BNTM writes to x by stage y" means.
People understand what "EyAx BNTM writes to x by stage y" means.

But what does "BNTM writes to every position on the tape" *mean*?  Evidently
you think it means something *distinct* from both "AxEy ..." and "EyAz ...".
(For if you thought it was synonymous with "AxEy ..." then your claim would
be false, and if you thought it was synonymous with "EyAz ..." then your
claim would be tautologous.)  What is that distinct meaning?  That is what
nobody else is clear on.
--
Tim Chow       tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth.  ---Galileo, Dialogues Concerning Two New Sciences
```
 0
tchow (869)
7/23/2009 8:51:59 PM
```On Jul 23, 1:34=A0pm, RussellE <reaste...@gmail.com> wrote:

> > you've
> > been told a thousand times already, the correct claim is:
>
> > AxEy BNTM writes to x at stage y.
>
> BNTM writes to every position on the tape =3D>
> AxEy BNTM writes to x by stage y
>
> I agree this is a true statement.
> But so is:
>
> BNTM writes to every position on the tape =3D>
> EyAx BNTM writes to x by stage y

We don't USE "BNTM writes to every position on the tape".

(IF we do say it, then we make clear that what we mean is "AxEy BNTM
writes to x by stage y" and NOTHING ELSE; specifically that we do NOT
mean "EyAx BNTM writes to x by stage y"

What we say is "AxEy BNTM writes to x by stage y".

Let P be "BNTM writes to every position on the tape".
Let Q be "AxEy BNTM writes to x by stage y".
Let R be "EyAx BNTM writes to x by stage y".

So we have (by virtue of the ambiguity of 'P'):

P -> Q
and
P-> R.

So we have both Q and R.

But no, we do NOT have P, so we do NOT have both Q and R.

What we have is just Q. ALONE. PERIOD.

STOP. Do not post anything until you post that you understand:

All we have is "AxEy BNTM writes to x by stage y". ALONE. PERIOD.

Again, stop playing games with me. Just take it that we have

"AxEy BNTM writes to x by stage y" ALONE. PERIOD.

THAT is what our theory proves and our theory does NOT prove "EyAx
BNTM writes to x by stage y".

MoeBlee

```
 0
jazzmobe (307)
7/23/2009 8:54:33 PM
```On Jul 23, 1:51=A0pm, tc...@lsa.umich.edu wrote:
..com>,
>
> RussellE =A0<reaste...@gmail.com> wrote:
> >BNTM writes to every position on the tape =3D>
> >AxEy BNTM writes to x by stage y
>
> >I agree this is a true statement.
> >But so is:
>
> >BNTM writes to every position on the tape =3D>
> >EyAx BNTM writes to x by stage y
>
> The trouble is, nobody else seems to understand what you even *mean* by
> "BNTM writes to every position on the tape."
>
> People understand what "AxEy BNTM writes to x by stage y" means.
> People understand what "EyAx BNTM writes to x by stage y" means.
>
> But what does "BNTM writes to every position on the tape" *mean*? =A0Evid=
ently
> you think it means something *distinct* from both "AxEy ..." and "EyAz ..=
..".
> (For if you thought it was synonymous with "AxEy ..." then your claim wou=
ld
> be false, and if you thought it was synonymous with "EyAz ..." then your
> claim would be tautologous.) =A0What is that distinct meaning? =A0That is=
what
> nobody else is clear on.

He means that the set of positions on the tape that the BNTM doesn't
eventually write onto is empty.

karl m
```
 0
malbrain (59)
7/23/2009 8:56:59 PM
```On Jul 23, 1:51=A0pm, tc...@lsa.umich.edu wrote:
..com>,
>
> RussellE =A0<reaste...@gmail.com> wrote:
> >BNTM writes to every position on the tape =3D>
> >AxEy BNTM writes to x by stage y
>
> >I agree this is a true statement.
> >But so is:
>
> >BNTM writes to every position on the tape =3D>
> >EyAx BNTM writes to x by stage y
>
> The trouble is, nobody else seems to understand what you even *mean* by
> "BNTM writes to every position on the tape."
>
> People understand what "AxEy BNTM writes to x by stage y" means.
> People understand what "EyAx BNTM writes to x by stage y" means.
>
> But what does "BNTM writes to every position on the tape" *mean*? =A0

Ax BNTM writes to position x

> Evidently
> you think it means something *distinct* from both "AxEy ..." and "EyAz ..=
..".
> (For if you thought it was synonymous with "AxEy ..." then your claim wou=
ld
> be false, and if you thought it was synonymous with "EyAz ..." then your
> claim would be tautologous.) =A0What is that distinct meaning? =A0That is=
what
> nobody else is clear on.

Ax BNTM writes to position x
=3D>
EyAx BNTM writes to x by stage y

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/23/2009 9:12:39 PM
```On Jul 23, 2:12=A0pm, RussellE <reaste...@gmail.com> wrote:

> Ax BNTM writes to position x
> =3D>
> EyAx BNTM writes to x by stage y

WRONG!

TOTALLY FALLACIOUS!

WRONG PERIOD.

You get an "F' on the predicate logic pop quiz.

MoeBlee
```
 0
jazzmobe (307)
7/23/2009 9:16:48 PM
```On Jul 23, 2:16=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 23, 2:12=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > Ax BNTM writes to position x
> > =3D>
> > EyAx BNTM writes to x by stage y
>
> WRONG!
>
> TOTALLY FALLACIOUS!
>
> WRONG PERIOD.
>
> You get an "F' on the predicate logic pop quiz.

P.S. I'll even do you the favor of SHOWING that you're wrong:

Ax x has a nose
->
EyAx y is a nose of x.

OBVIOUSLY WRONG.

WRONG.

MoeBlee
```
 0
jazzmobe (307)
7/23/2009 9:21:17 PM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 23, 12:06�pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 23, 11:40�am, RussellE <reaste...@gmail.com> wrote:
>
>
> > > > > Which of these statements do you claim is false for BNTM?
> >
> > > > > 1) Eventually writes to every position on the tape => (Ax)(Ey) BNTM
> > > > > writes to position x by step y.
> >
> > > > > 2) Eventually writes to every position on the tape => (Ey)(Ax) BNTM
> > > > > writes to position x.by step y
> >
> > > > The VERY point is that the expression "eventually writes to every
> > > > position on the tape" is AMBIGUOUS.
> >
> > > I don't see how you can say "every position on the tape" is
> > > ambiguous for a finite tape.
> >
> > > If you are saying "every position on the tape" is ambiguous
> > > for infinite tapes then I agree with you.
> >
> > Stop playing childish games.
> >
> > I didn't say "every position on the tape" is ambiguous. What I said is
> > ambiguous is "Eventually writes to every position on the tape".
>
> Then remove the word eventually. Both statements are still true.
>
> > In
> > particular, part of your intellectual childishness is to exploit the
> > ambiguity of "eventually". And exploit is unnecessary, since you've
> > been told a thousand times already, the correct claim is:
> >
> > AxEy BNTM writes to x at stage y.
>
> BNTM writes to every position on the tape =>
> AxEy BNTM writes to x by stage y
>
> I agree this is a true statement.
> But so is:
>
> BNTM writes to every position on the tape =>
> EyAx BNTM writes to x by stage y

Do you really think so?

From what definition of tape and TM do you conclude this?
>
> Like you, I am shocked by the ramifications
> of my proof. But I don't see any flaws in my
> argument.

I would be shocked if I has any faith in your "proof" or your "argument"
>
>
> Russell
> - You don't have to be crazy to be a mathematician, but is does help

Doesn't seem to have helped you.

--
Virgil
```
 0
virgil2 (13)
7/23/2009 9:25:10 PM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 23, 1:51�pm, tc...@lsa.umich.edu wrote:
> > In article
> >
> > RussellE �<reaste...@gmail.com> wrote:
> > >BNTM writes to every position on the tape =>
> > >AxEy BNTM writes to x by stage y
> >
> > >I agree this is a true statement.
> > >But so is:
> >
> > >BNTM writes to every position on the tape =>
> > >EyAx BNTM writes to x by stage y
> >
> > The trouble is, nobody else seems to understand what you even *mean* by
> > "BNTM writes to every position on the tape."
> >
> > People understand what "AxEy BNTM writes to x by stage y" means.
> > People understand what "EyAx BNTM writes to x by stage y" means.
> >
> > But what does "BNTM writes to every position on the tape" *mean*? �
>
> Ax BNTM writes to position x
>
> > Evidently
> > you think it means something *distinct* from both "AxEy ..." and "EyAz
> > ...".
> > (For if you thought it was synonymous with "AxEy ..." then your claim would
> > be false, and if you thought it was synonymous with "EyAz ..." then your
> > claim would be tautologous.) �What is that distinct meaning? �That is what
> > nobody else is clear on.
>
> Ax BNTM writes to position x
> =>
> EyAx BNTM writes to x by stage y

Unless Russ can come up with a natural which has no successor, the above
implication is false.

--
Virgil
```
 0
virgil2 (13)
7/23/2009 9:28:19 PM
```On Jul 23, 2:28=A0pm, Virgil <virg...@nowhere.com> wrote:
> In article

Ax BNTM writes to position x
=3D>
EyAx BNTM writes to x by stage y

> Unless Russ can come up with a natural which has no successor, the above
> implication is false.

The implication above is false if you assume BTNM
can write to every position of an infinite tape.

I am not the one claiming "Ax BNTM writes to position x".
The implication above is provably true, and I have given
the proof many times.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/23/2009 10:49:16 PM
```On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:

> I am not the one claiming "Ax BNTM writes to position x".
> The implication above is provably true, and I have given
> the proof many times.

You utterly SKIPPED what I wrote about that.

Your method of argumentation is to move the discussion BACKWARDS from
exact mathematical formulation to ambiguous English and then, even
after you've been told that the ambiguous English is not what is at
stake but rather the exact mathematical formulation, you still premise
your argument on the ambiguous English. Very childish.

MoeBlee

```
 0
jazzmobe (307)
7/23/2009 10:57:29 PM
```On Jul 23, 3:57=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > I am not the one claiming "Ax BNTM writes to position x".
> > The implication above is provably true, and I have given
> > the proof many times.
>
> You utterly SKIPPED what I wrote about that.

Ax TM writes to position x =3D> EyAx TM writes x by stage y

Moeblee, if your favorite formal system can't prove
this statement is true for any TM that halts
you need to find a new formal system.

I guess I won't be surprised if it can't.
I still find it mind boggling ZFC can't prove every
natural number is finite without AoI.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/24/2009 3:22:35 AM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 23, 2:28�pm, Virgil <virg...@nowhere.com> wrote:
> > In article
>
>
> Ax BNTM writes to position x
> =>
> EyAx BNTM writes to x by stage y
>
> > Unless Russ can come up with a natural which has no successor, the above
> > implication is false.
>
> The implication above is false if you assume BTNM
> can write to every position of an infinite tape.
>
> I am not the one claiming "Ax BNTM writes to position x".
> The implication above is provably true, and I have given
> the proof many times.

"EyAx BNTM writes to x by stage y" is false in a world in which there
are BNTMs with no last "stage", because is such a world it is the case
that 'AyEx such that BNTM passes stage y before writing x'.

--
Virgil
```
 0
virgil2 (13)
7/24/2009 4:23:23 AM
```Chicken Little:
Tim Little wrote:
> On 2009-07-15, RussellE <reasterly@gmail.com> wrote:
> > Is the empty set a member of this
> > family of sets?
>
> Since every set in your family is indexed by a natural number, no.
>
>
> > This proves it is impossible to remove an infinite number of
> > elements from a set if we only remove one element at a time.
>
> Non sequitur.  All it proves is that removing any finite number of
> elements from an infinite set leaves an infinite set.
>
>
> > Assume I have an infinitely fast computer, like a Zeno Machine (ZM).
>
> A dubious assumption to begin with.  Most logically coherent models of
> "Zeno machine" have conditions on their "final" state dependent upon
> whether the sequence of computation reached an [+] static state
> or not. (Removed the word "eventually")

Active Topics
-------------

Why do things have to age slower in time? - 5 new
-------------------------------------------------
....ET's don't send smart people in flying saucers. - Thurs, Jul 23
2009 1:16
pm

Nope, Unity Hospital

Most Natural Numbers Are Computable
----------------------------------------------
....Then remove the word [eventually]. Both statements are still
true. ...BNTM
writes to every position on the tape => AxEy BNTM writes to x by stage
y I
agree this is a true statement. But so is: BNTM writes to every
position on
the tape => EyAx BNTM writes to x by stage y Like you, I am shocked by
the
ramifications -
---------------------------------------------------------------
.... But in this case, I wanted to let the folks in the ...Nutter? Mr.
Pot,
meet Mr. Kettle. Eldon, you've spent the past 10 or 15 years attacking
ANY
betterment group with lies and libel, and you were charged \$400,000 in
court-
ordered penalties for doing so. There are rumors, not subtantiated but
not
unreasonable, that there are

New theory for ultimate fate of the universe - 5 new
----------------------------------------------------
....He doesn't recognize the difference between creative writing of
fiction and
reality. - Th

Without any such conditions, the machine is ill-specified.
>
>
> > State 1:
> [etc]
>
> Yet another example of an ill-specified Zeno machine.  As the rest of
>
>
> - Tim
And based on this [so's your face].
> [etc]
>
> Yet another example of an ill-specified Zeno machine.  As the rest of
>
>
> - Tim
+Sound==solution. Add sound to the CPU to overcome halting. Computing
data carried by sound. Never halts. Always most efficient shapes
transfer waves.
```
 0
marty.musatov (1143)
7/24/2009 12:20:46 PM
```On Jul 23, 8:22=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 23, 3:57=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > I am not the one claiming "Ax BNTM writes to position x".
> > > The implication above is provably true, and I have given
> > > the proof many times.
>
> > You utterly SKIPPED what I wrote about that.
>
> Ax TM writes to position x =3D> EyAx TM writes x by stage y
>
> Moeblee, if your favorite formal system can't prove
> this statement is true for any TM that halts
> you need to find a new formal system.
>
> I guess I won't be surprised if it can't.
> I still find it mind boggling ZFC can't prove every
> natural number is finite without AoI.

I apologize MoBlee,
I realize this is your profession.
I never imagined I would find a flaw in Peano's Axiom.
I was trying to disprove the Axiom of Infinity.

There is a silver lining.
We will need a lot of professors to rewrite the text books.

Russell
- 2 many 2 count
```
 0
7/24/2009 1:56:12 PM
```On Jul 24, 6:56=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
> On Jul 23, 8:22=A0pm, RussellE <reaste...@gmail.com> wrote:
> > On Jul 23, 3:57=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > > On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > > I am not the one claiming "Ax BNTM writes to position x".
> > > > The implication above is provably true, and I have given
> > > > the proof many times.
>
> > > You utterly SKIPPED what I wrote about that.
>
> > Ax TM writes to position x =3D> EyAx TM writes x by stage y
>
> > Moeblee, if your favorite formal system can't prove
> > this statement is true for any TM that halts
> > you need to find a new formal system.
>
> > I guess I won't be surprised if it can't.
> > I still find it mind boggling ZFC can't prove every
> > natural number is finite without AoI.
>
> I apologize MoBlee,
> I realize this is your profession.
> I never imagined I would find a flaw in Peano's Axiom.
> I was trying to disprove the Axiom of Infinity.
>
> There is a silver lining.
> We will need a lot of professors to rewrite the text books.

With the cranks, the arrogance always comes out in the end.

Marshall
```
 0
7/24/2009 2:08:13 PM
```On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 23, 2:28=A0pm, Virgil <virg...@nowhere.com> wrote:
>
> > In article
>
> Ax BNTM writes to position x
> =3D>
> EyAx BNTM writes to x by stage y
>
> > Unless Russ can come up with a natural which has no successor, the abov=
e
> > implication is false.
>
> The implication above is false if you assume BTNM
> can write to every position of an infinite tape.
>
> I am not the one claiming "Ax BNTM writes to position x".

This is easily proved by induction.  The BNTM writes to position 1
during its first step.  If it is true that the BNTM has written to
position x it is also true it will write to position x+1.  Therefore
the BNTM will write to all n in N.

karl m
```
 0
malbrain (59)
7/24/2009 5:14:12 PM
```On Jul 23, 8:22=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 23, 3:57=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > I am not the one claiming "Ax BNTM writes to position x".
> > > The implication above is provably true, and I have given
> > > the proof many times.
>
> > You utterly SKIPPED what I wrote about that.
>
> Ax TM writes to position x =3D> EyAx TM writes x by stage y

You MUST be trolling. (1) The above claim is couched different from
the one I said you skipped my explanation. (2) As to the above
ludicrously illogical claim, I had refuted it too, but now you just
repeat it as you skip what I wrote about IT.

> Moeblee, if your favorite formal system can't prove
> this statement is true for any TM that halts
> you need to find a new formal system.

Are you REALLY stupid or just trolling? (1) Neither the formal system
for ZF nor ordinary mathematical reasoning permit your inference, so
your inference has nothing to do with ZF or ordinary mathematical
reasoning about Turing machines. (2) I SHOWED you why your form of
inference is quite illogical even in ordinary discourse about things
not even necessarily mathematical.

> I guess I won't be surprised if it can't.
> I still find it mind boggling ZFC can't prove every
> natural number is finite without AoI.

You're trolling or really stupid. I've ALREADY informed you that we
don't need the axiom of infinity just to prove that every natural is
finite.

MoeBlee

```
 0
jazzmobe (307)
7/24/2009 5:46:37 PM
```On Jul 24, 6:56=A0am, Kyle Easterly <keileene...@gmail.com> wrote:

> I apologize MoBlee,
> I realize this is your profession.

What are you talking about? Neither mathematics nor posting are my
profession.

MoeBlee
```
 0
jazzmobe (307)
7/24/2009 5:47:58 PM
```Ben Bacarisse wrote:
> RussellE <reasterly@gmail.com> writes:
>
>> On Jul 21, 7:13 pm, Ben Bacarisse <ben.use...@bsb.me.uk> wrote:
>>> RussellE <reaste...@gmail.com> writes:
>>>  Maybe you are a new-age mathematician and all you need is to
>>> be heard.  If so, consider yourself well and truly heard.
>> I'm really not a troll.
>> I will admit I am a heretic.
>> I don't believe the set of all natural numbers exists.
>
> It may well be possible to set up a system of logic and arithmetic in
> which N does not exists -- only a sequence of ever larger finite
> approximations to it.  Who knows?  I don't and, oddly enough, I am
> pretty sure that you don't either.

Oddly enough, the finite syntactical part of FOL and the finitely
axiomatizable Q in it is such a logical system and arithmetic.
No existence of N is required.

>
> It is almost certain that in such an impoverished system, very little
> can be proved.

"Impoverished system"? Not so. We one can prove as many Q's theorems
one can prove!

> Induction is surely ruled out and there won't be many

Induction might be ruled out but again one can prove as many Q's
theorems as one cares to prove.

> It is possible that TMs can't even be
> defined.

If in the meta level we can define infinite as "being not finite" then
it seems to me we could define TMs.

>
> Any dramatic conclusions will, of course, simply be a consequence of
> having rejected a large chunk of what everyone else thinks of as
> mathematics so you won't even be able to shock anyone with
> "uncomputable naturals" or "finite but unbounded" strings and so on.

But I'd think we could "shock" everyone else with a "dramatic conclusion"
such as "Mathematical reasoning is relativistic" or "The proof of either
GC or cGC in Q is indeterministic"! [Where "cGC" is the FOL formula
"counter-GC" which would be something like 'There are infinite counter
examples of GC'].

> The most impressive results from this sort of meta-mathematics are
> usually about how much remains the same, not how oddly different the
> new system is.

It's actually "how oddly different the new system is": Mathematical
reasoning's being relativistic isn't something "ordinary" mathematicians/
logicians would like to hear: they seem to be "afraid" of it!

> Either way, it is a lifetime's work -- good luck.

Good luck to the collective awareness of the relativity of _mathematical_
reasoning. Unfortunately!

> The effect of such a change will probably be
> astoundingly dull, but I have to admit that I don't really know.

For what it's worth, I'd not think "relativity of mathematical reasoning"
sound "dull".

```
 0
7/24/2009 6:27:50 PM
```On Jul 24, 10:47=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 24, 6:56=A0am, Kyle Easterly <keileene...@gmail.com> wrote:

> > I apologize MoBlee,
> > I realize this is your profession.
>
> What are you talking about? Neither mathematics nor posting are my
> profession.

My mistake. I assumed you were a professor when you flunked me.
You certainly know as much about the subject as many professors.

Russell
- 2 many 2 count

```
 0
7/24/2009 9:00:17 PM
```On Jul 23, 2:12=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 23, 1:51=A0pm, tc...@lsa.umich.edu wrote:
>
>
>
>
>
ps.com>,
>
> > RussellE =A0<reaste...@gmail.com> wrote:
> > >BNTM writes to every position on the tape =3D>
> > >AxEy BNTM writes to x by stage y
>
> > >I agree this is a true statement.
> > >But so is:
>
> > >BNTM writes to every position on the tape =3D>
> > >EyAx BNTM writes to x by stage y
>
> > The trouble is, nobody else seems to understand what you even *mean* by
> > "BNTM writes to every position on the tape."
>
> > People understand what "AxEy BNTM writes to x by stage y" means.
> > People understand what "EyAx BNTM writes to x by stage y" means.
>
> > But what does "BNTM writes to every position on the tape" *mean*? =A0
>
> Ax BNTM writes to position x
>
> > Evidently
> > you think it means something *distinct* from both "AxEy ..." and "EyAz =
....".
> > (For if you thought it was synonymous with "AxEy ..." then your claim w=
ould
> > be false, and if you thought it was synonymous with "EyAz ..." then you=
r
> > claim would be tautologous.) =A0What is that distinct meaning? =A0That =
is what
> > nobody else is clear on.
>
> Ax BNTM writes to position x
> =3D>
> EyAx BNTM writes to x by stage y

Perhaps it would be clearer if you expressed this as:  EyAx BNTM will
write to x by virtue of having reached stage y. (this is satisfied for
all x by y=3D1)

karl m
```
 0
malbrain (59)
7/24/2009 9:25:59 PM
```On Jul 24, 10:14=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:

> This is easily proved by induction.

How is induction anything more than the assumption
all natural numbers are "computable"?
It is easy to prove something you assume is true.

> =A0The BNTM writes to position 1
> during its first step. =A0If it is true that the BNTM has written to
> position x it is also true it will write to position x+1.

OK

> =A0Therefore
> the BNTM will write to all n in N.

How do you get this part?
All you have shown is that if BNTM writes
to position n, BNTM writes to position n+1.
I am trying to show there is no N.

My proof is actually fairly simple.
I show BNTM can only write finite strings.
Then I show, if BNTM writes to every position
on the tape, the tape must be finite.

You can't say my proof is flawed because
of induction.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/24/2009 9:33:10 PM
```On Jul 24, 2:33=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 24, 10:14=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > On Jul 23, 3:49=A0pm, RussellE <reaste...@gmail.com> wrote:
> > This is easily proved by induction.
>
> How is induction anything more than the assumption
> all natural numbers are "computable"?
> It is easy to prove something you assume is true.
>
> > =A0The BNTM writes to position 1
> > during its first step. =A0If it is true that the BNTM has written to
> > position x it is also true it will write to position x+1.
>
> OK
>
> > =A0Therefore
> > the BNTM will write to all n in N.
>
> How do you get this part?
> All you have shown is that if BNTM writes
> to position n, BNTM writes to position n+1.

The conclusion is an application of the axiom of induction:  You
establish a base case.  You show that by assuming that X is true at n
implies X is true at n+1.  You then get to conclude that X is true for
all n in N.

karl m
```
 0
malbrain (59)
7/24/2009 9:38:48 PM
```On Jul 24, 2:33=A0pm, RussellE <reaste...@gmail.com> wrote:

> I am trying to show there is no N.
>
> My proof is actually fairly simple.
> I show BNTM can only write finite strings.
> Then I show, if BNTM writes to every position
> on the tape, the tape must be finite.

I think you mean "then I show, if BNTM writes to every position of an
infinite tape" because that is assumed for TM in general.
If the tape is infinite, and for all x BNTM writes position x, how can
you say that the tape is finite, or that the output on that tape is
finite.

At each stage x, the BNTM has written a finite string.  But BNTM will
write all x, leaving no unused positions.

It is a contradiction that BNTM leaves no unused positions, yet for
all n in N the set of unused positions is infinite.

karl m
```
 0
malbrain (59)
7/24/2009 10:26:43 PM
```On Jul 24, 3:26=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 24, 2:33=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > I am trying to show there is no N.
>
> > My proof is actually fairly simple.
> > I show BNTM can only write finite strings.
> > Then I show, if BNTM writes to every position
> > on the tape, the tape must be finite.
>
> I think you mean "then I show, if BNTM writes to every position of an
> infinite tape" because that is assumed for TM in general.
> If the tape is infinite, and for all x BNTM writes position x, how can
> you say that the tape is finite, or that the output on that tape is
> finite.
>
> At each stage x, the BNTM has written a finite string. =A0But BNTM will
> write all x, leaving no unused positions.
>
> It is a contradiction that BNTM leaves no unused positions, yet for
> all n in N the set of unused positions is infinite.

I have shown Peano's Axioms are inconsistent.

If you accept my argument (I am not saying you do)
there are position on an infinite tape no TM can write.
This means there must be "unnatural" numbers that can't
be reached by the successor function.

You can see the dilemma here. I can define a number, P,
using the sucessor function and I can prove P can't be
reached with the successor function. P both is and is not
a natural number. P and not P equals contradiction.

Our definition of natural number is fundamentally flawed.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/24/2009 11:27:57 PM
```On Jul 24, 4:27=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 24, 3:26=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 24, 2:33=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > I am trying to show there is no N.
>
> > > My proof is actually fairly simple.
> > > I show BNTM can only write finite strings.
> > > Then I show, if BNTM writes to every position
> > > on the tape, the tape must be finite.
>
> > I think you mean "then I show, if BNTM writes to every position of an
> > infinite tape" because that is assumed for TM in general.
> > If the tape is infinite, and for all x BNTM writes position x, how can
> > you say that the tape is finite, or that the output on that tape is
> > finite.
>
> > At each stage x, the BNTM has written a finite string. =A0But BNTM will
> > write all x, leaving no unused positions.
>
> > It is a contradiction that BNTM leaves no unused positions, yet for
> > all n in N the set of unused positions is infinite.
>
> Yes, it is a contradiction.
> I have shown Peano's Axioms are inconsistent.
>
> If you accept my argument (I am not saying you do)
> there are position on an infinite tape no TM can write.
> This means there must be "unnatural" numbers that can't
> be reached by the successor function.
>
> You can see the dilemma here. I can define a number, P,
> using the sucessor function and I can prove P can't be
> reached with the successor function. P both is and is not
> a natural number. P and not P equals contradiction.
>
> Our definition of natural number is fundamentally flawed.

There's another contradiction.  The natural numbers within themselves
can produce no number for the count of natural numbers, yet they are
the definition of what it means to count.

karl m
```
 0
malbrain (59)
7/24/2009 11:44:51 PM
```On Jul 24, 4:27=A0pm, RussellE <reaste...@gmail.com> wrote:

> I have shown Peano's Axioms are inconsistent.

You've shown no such thing.

> If you accept my argument (I am not saying you do)
> there are position on an infinite tape no TM can write.

You've not shown any such thing.

> You can see the dilemma here. I can define a number, P,
> using the sucessor function and I can prove P can't be
> reached with the successor function.

No, you've proven no such thing.

> P both is and is not
> a natural number. P and not P equals contradiction.

> Our definition of natural number is fundamentally flawed.

MoeBlee

```
 0
jazzmobe (307)
7/24/2009 11:45:56 PM
```On Jul 24, 4:44=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:

> There's another contradiction. =A0The natural numbers within themselves
> can produce no number for the count of natural numbers, yet they are
> the definition of what it means to count.

Let me know when you elevate that from word game to actual
mathematical demonstration.

MoeBlee
```
 0
jazzmobe (307)
7/24/2009 11:47:38 PM
```On Jul 24, 4:47=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 24, 4:44=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > There's another contradiction. =A0The natural numbers within themselves
> > can produce no number for the count of natural numbers, yet they are
> > the definition of what it means to count.
>
> Let me know when you elevate that from word game to actual
> mathematical demonstration.
>
> MoeBlee

Perhaps you will enlighten us to identify the natural number that
counts the number of natural numbers?  There isn't one.  I count using
1, 2, 3...  What do  you use?

karl m
```
 0
malbrain (59)
7/24/2009 11:52:43 PM
```On Jul 24, 4:52=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 24, 4:47=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 24, 4:44=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > There's another contradiction. =A0The natural numbers within themselv=
es
> > > can produce no number for the count of natural numbers, yet they are
> > > the definition of what it means to count.
>
> > Let me know when you elevate that from word game to actual
> > mathematical demonstration.

> Perhaps you will enlighten us to identify the natural number that
> counts the number of natural numbers? =A0There isn't one. =A0I count usin=
g
> 1, 2, 3... =A0What do =A0you use?

(1) You haven't even given a hint how you'd elevate your argument to
an actual mathematical demonstration of anything

(2) That there is no natural number that counts the set of natural
numbers is not a contradiction unless you attach the claim that
natural numbers count not just finite sets but infinite sets too. Set
theory itself doesn't make such a claim, and I'm not aware of any
commentator on set theory who ever made such a claim. Of course, the
SET of natural numbers enumerates the set of natural numbers, but who
ever claimed that any individual natural number counts anything other
than certain finite sets?

MoeBlee

```
 0
jazzmobe (307)
7/25/2009 12:09:25 AM
```> With the cranks, the arrogance always comes out in the end.

Most cranks start out arrogant.
I am arrogant, of course. You have to be in this ng.
We eat non-arrogant people for breakfast.

Just because I am a crank doesn't mean I am wrong.

Russell
- Cranks 1, Establishment 0

```
 0
reasterly (337)
7/25/2009 1:21:20 AM
```Am Fri, 24 Jul 2009 16:47:38 -0700 (PDT) schrieb MoeBlee:

> On Jul 24, 4:44�pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
>> There's another contradiction. The natural numbers within themselves
>> can produce no number for the count of natural numbers, yet they are
>> the definition of what it means to count.
>
> Let me know when you elevate that from word game to actual
> mathematical demonstration.

Oh, (after reading this) this guy just went straight into my killfile. :-)

Herb
```
 0
Herbert
7/25/2009 2:07:43 AM
```On Jul 24, 4:45=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 24, 4:27=A0pm, RussellE <reaste...@gmail.com> wrote:

> > I have shown Peano's Axioms are inconsistent.
>
> You've shown no such thing.

Actually, MoeBlee, I almost hope you are right.

Russell
- Zeno was right. Motion is impossible.
```
 0
reasterly (337)
7/25/2009 4:49:02 AM
```
RussellE wrote:
> On Jul 24, 4:45=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 24, 4:27=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > I have shown Peano's Axioms are inconsistent.
> >
> > You've shown no such thing.
>
> Actually, MoeBlee, I almost hope you are right.
>
>
> Russell
> - Zeno was right. Motion is impossible.
Nope! Still moving...

Godel =3D Zeno
```
 0
marty.musatov (1143)
7/25/2009 7:17:49 AM
```On Jul 24, 5:09=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 24, 4:52=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > On Jul 24, 4:47=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > On Jul 24, 4:44=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > There's another contradiction. =A0The natural numbers within themse=
lves
> > > > can produce no number for the count of natural numbers, yet they ar=
e
> > > > the definition of what it means to count.
>
> > > Let me know when you elevate that from word game to actual
> > > mathematical demonstration.
> > Perhaps you will enlighten us to identify the natural number that
> > counts the number of natural numbers? =A0There isn't one. =A0I count us=
ing
> > 1, 2, 3... =A0What do =A0you use?
>
> (1) You haven't even given a hint how you'd elevate your argument to
> an actual mathematical demonstration of anything
>
> (2) That there is no natural number that counts the set of natural
> numbers is not a contradiction unless you attach the claim that
> natural numbers count not just finite sets but infinite sets too. Set
> theory itself doesn't make such a claim, and I'm not aware of any
> commentator on set theory who ever made such a claim. Of course, the
> SET of natural numbers enumerates the set of natural numbers, but who
> ever claimed that any individual natural number counts anything other
> than certain finite sets?
>
> MoeBlee

I haven't studied the history of mathematics in the nineteenth century
as much as i would like. I believe that the development of set theory
was intended to resolve the contradiction between counting and being
counted.

karl m
```
 0
malbrain (59)
7/25/2009 1:35:36 PM
```On Sat, 25 Jul 2009 06:35:36 -0700 (PDT), Karl Malbrain
<malbrain@yahoo.com> said:
> ...
> I haven't studied the history of mathematics in the nineteenth century
> as much as i would like. I believe that the development of set theory
> was intended to resolve the contradiction between counting and being
> counted.

Whatever "contradiction" you have in mind, the development of set theory
had nothing whatever to do with it.  Both set theory and the theory of
transfinite numbers began their existence as tools that Cantor developed
to solve a cluster of problems in algebraic number theory whose solution
required him to iterate certain operations on (in general) infinite sets
of reals into the transfinite.  If you are indeed interested in the
actual history, have a look at Joseph Dauben's book _Georg Cantor: His
Mathematics and Philosophy of the Infinite_ (which gets a few things
wrong but is particularly good on the early developements) and also
Michael Hallett, _Cantorian Set Theory and Limitation of Size_.

```
 0
cmenzel (185)
7/25/2009 3:23:17 PM
```On Jul 25, 8:23=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wrote:
> On Sat, 25 Jul 2009 06:35:36 -0700 (PDT), Karl Malbrain
> <malbr...@yahoo.com> said:
>
> > ...
> > I haven't studied the history of mathematics in the nineteenth century
> > as much as i would like. I believe that the development of set theory
> > was intended to resolve the contradiction between counting and being
> > counted.
>
> Whatever "contradiction" you have in mind, the development of set theory
> had nothing whatever to do with it. =A0Both set theory and the theory of
> transfinite numbers began their existence as tools that Cantor developed
> to solve a cluster of problems in algebraic number theory whose solution
> required him to iterate certain operations on (in general) infinite sets
> of reals into the transfinite. =A0If you are indeed interested in the
> actual history, have a look at Joseph Dauben's book _Georg Cantor: His
> Mathematics and Philosophy of the Infinite_ (which gets a few things
> wrong but is particularly good on the early developements) and also
> Michael Hallett, _Cantorian Set Theory and Limitation of Size_.

Peano Axioms were proposed by Dedekind in 1888.
http://en.wikipedia.org/wiki/Peano_axioms

Proving (or disproving) the consistency of Peano Axioms is Hilbert's
second problem.
http://en.wikipedia.org/wiki/Hilbert%27s_second_problem

I think my proof would qualify as finitistic.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/25/2009 8:25:51 PM
```On 2009-07-23, RussellE <reasterly@gmail.com> wrote:
> Eventually writes every position => There exists a step m such
> that for all n, the BNTM has written to position n by step m.

Since the informal statement "eventually writes every position" is
somewhat ambiguous, I could accept that assignment of meaning.
However, you should be aware that it is *not* the meaning used by the
previous poster.

Such ambiguity is why I prefer to not to reason with such vague
terminology, but use more precise logic and mathematics.

> This is a provably true statement.

Actually it is provably false.  If you believe otherwise, provide a
mathematical proof for it.

Since you rejected my (and the previous poster's) assignment of
logical meaning to "eventually writes every position", how would you
express informally the proposition: "for all n, there exists a step m
such that the BNTM has written to position n by step m"?

That proposition is provably true.

- Tim
```
 0
tim669 (185)
7/26/2009 2:30:24 PM
```On 2009-07-23, RussellE <reasterly@gmail.com> wrote:
> EyAx BNTM writes to x by stage y

False.  The correct statement is

AxEy BNTM writes to x by stage y.

The fact that you cannot correctly order the quantifiers is a
sufficient basis for the flaws in your proof.

- Tim
```
 0
tim669 (185)
7/26/2009 2:34:44 PM
```On 2009-07-24, RussellE <reasterly@gmail.com> wrote:
> Yes, it is a contradiction.
> I have shown Peano's Axioms are inconsistent.

No, you have shown that you are incapable of correctly using predicate
logic.

- Tim
```
 0
tim669 (185)
7/26/2009 2:38:01 PM
```On Jul 15, 1:05=A0am, RussellE <reaste...@gmail.com> wrote:
> I prove most natural numbers are so large
> even an "infinitely" fast computer can not
> read or write these numbers.
>
> First, I will describe a simple set theory
> problem and then I show how this problem
> is related to the computability of natural
> numbers.
>
> I want to create a family of sets.
> The first member of this family is
> the set of all natural numbers, N.
> Each member after the first is created
> by removing the least element of
> the previous set.
>
> A0 =3D {0,1,2,3,...}
> A1 =3D {1,2,3,4,...}
> A2 =3D (2,3,4,5,...}
> ...
>
> Is the empty set a member of this
> family of sets?
>
> Assume the empty set is a member.
> Then there exists another member
> with exactly one element. Let this
> set be {z}. z must be the largest
> natural number.
>
> Since there is no largest natural number,
> the empty set can not be a member of this
> family of sets.
>
> This proves it is impossible to
> remove an infinite number of elements
> from a set if we only remove one element
> at a time.
>
> Assume I have an infinitely fast computer,
> like a Zeno Machine (ZM).http://en.wikipedia.org/wiki/Zeno_machine
>
> A ZM is a type of Accelerate Turing Machine.
> A ZM performs the first operation in 1/2
> second, the 2nd operation in 1/4 second,
> and the nth operation in 1/(2^n) second.
> A ZM can perform an infinite number of
> operations in one second.
> (The time required to perform these
> operations is irrelevant for this proof.)
>
> Like a Turing Machine, a ZM reads and writes
> one character at a time from a tape.
>
> Assume I have an infinitely long blank tape.
> (Later, I will show that I only need to assume
> the tape is finite, but unbounded.)
>
> I want to test if this tape is actually infinitely long.
>
> Assume the ZM will generate an error
> if it runs out of tape.
> Running out of tape would prove the
> input tape was finite.
>
> I also want to be sure that every finite
> position on the tape is blank.
> To do so, I create a simple three state
> Turing Machine.
>
> State 1:
> On any input other than blank generate an
> error and halt.
> On blank input move tape head one posiiton
> to the right and switch to state 2.
>
> State 2:
> On any input other than blank generate an
> error and halt.
> On blank input move tape head one posiiton
> to the left and switch to state 3.
>
> State 3:
> On any input other than blank generate an
> error and halt.
> On blank input write a '1', move tape head
> one posiiton to the right, and switch to state 1.
>
> This program can never overwrite the last
> blank space on the tape. Of course,
> if the tape is infinitely long, there
> will never be a "last" blank position.
>
> Now, I run my program on the ZM for an
> hour or two. I want to be sure the ZM
> has performed an "infinite" number of
> operations.
>
> If the input tape is finite, the ZM
> will halt with an error. Assume the
> ZM hasn't halted after an hour.
>
> What is on the tape after we halt the ZM?
>
> Consider my set theory problem.
> The first set in the family, N,
> represents the positions of blank
> spaces on my input tape when I
> start the ZM.
>
> After each '1' written by the ZM,
> the next set in the family represents
> the remaining blank positions on the
> tape.
>
> There must still be an infinite number
> of blank spaces on the input tape.
> If the ZM had overwritten every position
> on the tape, the empty set would be a
> member of my family of sets.
>
> Since each blank is at a finite position,
> there must exist a smallest blank position.
>
> Even though the ZM is "infinitely"
> fast it has only written a finite
> number of 1's. (less than the smallest
> blank position).
>
> This proves nearly all natural numbers
> are too large to be read or written by
> any sequential computer.
>
> Notice that I still don't know if the
> input tape was infinitely long. There
> are an infinite number of finite tapes
> too long to be read by the ZM.
>
> According to Wikipedia:http://en.wikipedia.org/wiki/Decidable_language
>
> "A recursive language is a formal language for which there exists a
> Turing machine which will, when presented with any input string, halt
> and accept if the string is in the language, and halt and reject
> otherwise. The Turing machine always halts; it is known as a decider
> and is said to decide the recursive language."
>
> The proof above shows the language of unary
> representations of natural numbers is not
> recursive.
>
> A TM can certainly recognize "small" natural
> numbers, but "large" natural numbers
> are indistinguishable from an infinite string.
>
> Even an infinitely fast Turing Machine can not
> decide if an input tape contains a unary
> representation of a natural number.
>
> Normally, we assume a TM can read any finite
> string if we wait "long enough".
>
> My proof shows this assumption is false.
> Even if we have an infinitely fast computer
> and wait forever, it is impossible for
> any sequential computer to read most
> finite strings.

Well, that's only interesting to people who actually
study university AI, rather than computers though.
Since they're also the only people who think that computers
were invented for the sole purpose of reading strings.
Rather than reading 3D Electronic Books, Laser Disk Libraries,
Atomic Clock Wristwatches, Holograms, Multiplexed Fiber Optics
Data Inputs, GPS Receivers, Flat Sceen Software Debuggers,
and mp3 and mpeg files.

>
> Russell
> - 2 many 2 count

```
 0
zzbunker (103)
7/26/2009 3:14:54 PM
```On Jul 26, 11:14=A0am, "zzbun...@netscape.net" <zzbun...@netscape.net>
wrote:
> On Jul 15, 1:05=A0am, RussellE <reaste...@gmail.com> wrote:
>
>
>
>
>
> > I prove most natural numbers are so large
> > even an "infinitely" fast computer can not
> > read or write these numbers.
>
> > First, I will describe a simple set theory
> > problem and then I show how this problem
> > is related to the computability of natural
> > numbers.
>
> > I want to create a family of sets.
> > The first member of this family is
> > the set of all natural numbers, N.
> > Each member after the first is created
> > by removing the least element of
> > the previous set.
>
> > A0 =3D {0,1,2,3,...}
> > A1 =3D {1,2,3,4,...}
> > A2 =3D (2,3,4,5,...}
> > ...
>
> > Is the empty set a member of this
> > family of sets?
>
> > Assume the empty set is a member.
> > Then there exists another member
> > with exactly one element. Let this
> > set be {z}. z must be the largest
> > natural number.
>
> > Since there is no largest natural number,
> > the empty set can not be a member of this
> > family of sets.
>
> > This proves it is impossible to
> > remove an infinite number of elements
> > from a set if we only remove one element
> > at a time.
>
> > Assume I have an infinitely fast computer,
> > like a Zeno Machine (ZM).http://en.wikipedia.org/wiki/Zeno_machine
>
> > A ZM is a type of Accelerate Turing Machine.
> > A ZM performs the first operation in 1/2
> > second, the 2nd operation in 1/4 second,
> > and the nth operation in 1/(2^n) second.
> > A ZM can perform an infinite number of
> > operations in one second.
> > (The time required to perform these
> > operations is irrelevant for this proof.)
>
> > Like a Turing Machine, a ZM reads and writes
> > one character at a time from a tape.
>
> > Assume I have an infinitely long blank tape.
> > (Later, I will show that I only need to assume
> > the tape is finite, but unbounded.)
>
> > I want to test if this tape is actually infinitely long.
>
> > Assume the ZM will generate an error
> > if it runs out of tape.
> > Running out of tape would prove the
> > input tape was finite.
>
> > I also want to be sure that every finite
> > position on the tape is blank.
> > To do so, I create a simple three state
> > Turing Machine.
>
> > State 1:
> > On any input other than blank generate an
> > error and halt.
> > On blank input move tape head one posiiton
> > to the right and switch to state 2.
>
> > State 2:
> > On any input other than blank generate an
> > error and halt.
> > On blank input move tape head one posiiton
> > to the left and switch to state 3.
>
> > State 3:
> > On any input other than blank generate an
> > error and halt.
> > On blank input write a '1', move tape head
> > one posiiton to the right, and switch to state 1.
>
> > This program can never overwrite the last
> > blank space on the tape. Of course,
> > if the tape is infinitely long, there
> > will never be a "last" blank position.
>
> > Now, I run my program on the ZM for an
> > hour or two. I want to be sure the ZM
> > has performed an "infinite" number of
> > operations.
>
> > If the input tape is finite, the ZM
> > will halt with an error. Assume the
> > ZM hasn't halted after an hour.
>
> > What is on the tape after we halt the ZM?
>
> > Consider my set theory problem.
> > The first set in the family, N,
> > represents the positions of blank
> > spaces on my input tape when I
> > start the ZM.
>
> > After each '1' written by the ZM,
> > the next set in the family represents
> > the remaining blank positions on the
> > tape.
>
> > There must still be an infinite number
> > of blank spaces on the input tape.
> > If the ZM had overwritten every position
> > on the tape, the empty set would be a
> > member of my family of sets.
>
> > Since each blank is at a finite position,
> > there must exist a smallest blank position.
>
> > Even though the ZM is "infinitely"
> > fast it has only written a finite
> > number of 1's. (less than the smallest
> > blank position).
>
> > This proves nearly all natural numbers
> > are too large to be read or written by
> > any sequential computer.
>
> > Notice that I still don't know if the
> > input tape was infinitely long. There
> > are an infinite number of finite tapes
> > too long to be read by the ZM.
>
> > According to Wikipedia:http://en.wikipedia.org/wiki/Decidable_language
>
> > "A recursive language is a formal language for which there exists a
> > Turing machine which will, when presented with any input string, halt
> > and accept if the string is in the language, and halt and reject
> > otherwise. The Turing machine always halts; it is known as a decider
> > and is said to decide the recursive language."
>
> > The proof above shows the language of unary
> > representations of natural numbers is not
> > recursive.
>
> > A TM can certainly recognize "small" natural
> > numbers, but "large" natural numbers
> > are indistinguishable from an infinite string.
>
> > Even an infinitely fast Turing Machine can not
> > decide if an input tape contains a unary
> > representation of a natural number.
>
> > Normally, we assume a TM can read any finite
> > string if we wait "long enough".
>
> > My proof shows this assumption is false.
> > Even if we have an infinitely fast computer
> > and wait forever, it is impossible for
> > any sequential computer to read most
> > finite strings.
>
> =A0 Well, that's only interesting to people who actually
> =A0 study university AI, rather than computers though.
> =A0 Since they're also the only people who think that computers
> =A0 were invented for the sole purpose of reading strings.
> =A0 Rather than reading 3D Electronic Books, Laser Disk Libraries,
> =A0 Atomic Clock Wristwatches, Holograms, Multiplexed Fiber Optics
> =A0 Data Inputs, GPS Receivers, Flat Sceen Software Debuggers,
> =A0 and mp3 and mpeg files.

Or even university AI was nvented for people who study
brooms, history, and Chinese, rather than Blue Ray, HDTV,
optical computer upsurge thinking, fiber optics, post neandertal
robotics, lasers, engineering, holograms, and non
Quantum Mechanics Wal-Mart Occupations.

>
>
>
>
>
> > Russell
> > - 2 many 2 count- Hide quoted text -
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -

```
 0
zzbunker (103)
7/26/2009 4:16:24 PM
```On Jul 26, 7:30=A0am, Tim Little <t...@little-possums.net> wrote:
> On 2009-07-23, RussellE <reaste...@gmail.com> wrote:

> > Eventually writes every position =3D> There exists a step m such
> > that for all n, the BNTM has written to position n by step m.
>
> Since the informal statement "eventually writes every position" is
> somewhat ambiguous,

I agree. But, I am not the one saying BNTM writes every
position "eventually". I am saying either BNTM writes to
every position or it does not.

> I could accept that assignment of meaning.
> However, you should be aware that it is *not* the meaning used by the
> previous poster.

Which is why the word "eventually" keeps coming up.
You are saying BNTM writes to every position of any
finite tape and therefore BNTM will "eventually" write
every finite string of 1's.

> Such ambiguity is why I prefer to not to reason with such vague
> terminology, but use more precise logic and mathematics.

Like a formal system that assumes there are such
things as natural numbers?

> > This is a provably true statement.
>
> Actually it is provably false.

It is easily proven false in any system
that assumes it is false.
Obviously, I can't prove my statement
in such a system. The best I can do
I believe I have done that.

>=A0If you believe otherwise, provide a
> mathematical proof for it.

OK.
Assume I give BNTM a tape with 100 cells.
Each time BNTM writes a '1' to the tape I add
a blank cell to the end of the tape.

Do you agree BNTM will never write to every
blank cell of this tape? If so, how can you claim
BNTM can write to every position of any finite tape?
This tape is provably finite and BNTM will never
write to every position of this tape.

> Since you rejected my (and the previous poster's) assignment of
> logical meaning to "eventually writes every position", how would you
> express informally the proposition: "for all n, there exists a step m
> such that the BNTM has written to position n by step m"?
>
> That proposition is provably true.

There is no "all". There is no "n".

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/26/2009 8:06:12 PM
```["Followup-To:" header set to comp.theory.]
On 2009-07-26, RussellE <reasterly@gmail.com> wrote:
> I am saying either BNTM writes to every position or it does not.

"Writes to every position" is still somewhat ambiguous, with two
meanings - let's call them A and B (as labelled in a previous post).
You fail to distinguish them.  So instead of saying that either A is
true or A is false, you end up saying that either A is true or B is
false.

> Which is why the word "eventually" keeps coming up.  You are saying
> BNTM writes to every position of any finite tape

Actually for every position of even an infinite tape, there exists a
step at which it is written.  Note carefully the order of the
quantifiers.

> and therefore BNTM will "eventually" write every finite string of
> 1's.

It does write every finite string of 1's.  It just does not write an
infinite string of 1's.

> Like a formal system that assumes there are such things as natural
> numbers?

No, the system is logically prior to the formal definition of natural
numbers.  Try reading, and attempting to understand, a basic
introduction to predicate logic.  In particular, note that it does not
require any set of natural numbers to exist.

> The best I can do is show a contradiction.  I believe I have done
> that.

Your belief is false.  You incorrectly use predicate logic, so your
argument is not a valid proof.

> Assume I give BNTM a tape with 100 cells.  Each time BNTM writes a
> '1' to the tape I add a blank cell to the end of the tape.
>
> Do you agree BNTM will never write to every blank cell of this tape?

The set of blank cells is changing at every step, so which blank cells
do you mean by "every"?  Every cell that is blank at any step?  Every
cell that is always blank?  Every cell that is blank at some
particular step?  Also do you mean for every such cell, it is never
written to?  Or do you mean there does not exist a time by which all
such cells are written to?  Do you even see the difference?

It's yet another of your ambiguous sentences - even more ambiguous
that usual, which is saying something.  Most of your other statements
and questions are also vague to the point of being useless.  When you
can't write what you mean, it is unlikely that anyone will find any
meaning in what you write.

- Tim
```
 0
tim669 (185)
7/27/2009 3:06:47 AM
```On 2009-07-26 23:06:47 -0400, Tim Little <tim@little-possums.net> said:

> On 2009-07-26, RussellE <reasterly@gmail.com> wrote:

>> and therefore BNTM will "eventually" write every finite string of
>> 1's.
>
> It does write every finite string of 1's.  It just does not write an
> infinite string of 1's.

It's probably worth pointing out[0] that there is no natural number n
where, at the n'th step, the BNTM has written all finite strings of 1s.
It's only in the limit, after more than finitely many steps, that it
"has written" more than finitely many finite strings of 1s, and can
therefore "have written" N.

The quotes are there because the behaviour after infinite number of
iterations cannot be used to draw random conclusions about the
behaviour after finite numbers of iterations.

One of the benefits of describing systems like Turing machines in terms
of functions on sets is that you can do away with the sort of verbal
ambiguity that comes from ad-hoc descriptions. In particular, it's very
easy to demonstrate that, for a system containing

f_0 = ""
f_n = f_(n - 1) + "1"
|f_n| = n             [read |x| as 'length of x' here -oj]

that, if n is a natural number, then no f_n contains infinitely many
1s, and that every natural number of 1s has a corresponding f_n. We
can, with effort, formally prove that

Ax Ey |f_y| > x

is true, and that

Ey Ax |f_y| > x

is false for x and y as natural numbers. In fact, we can strip this
down even further:

Ax Ey y > x

is true, and

Ey Ax y > x

is false so long as x and y are natural numbers. Similarly, if we admit
values of n in f_n that are not natural numbers, then we can prove
different, interesting things, including that

Ax |f_omega| > x

where x is a natural number.

When you get to even this minimal level of formalism, RusselE's
precision, framed in words like "eventually" and "writes" to *formal*
precision, framed in terms of structured sequences of symbols, Russel's
non-sequiter "contradiction" is shown for the broken argument it is.

Words are *not* enough.

-o

[0] To Russel and those like him, *again*, because it surely hasn't
stuck yet...

```
 0
angrybaldguy (338)
7/27/2009 6:16:49 AM
```On Jul 26, 8:06=A0pm, Tim Little <t...@little-possums.net> wrote:
> ["Followup-To:" header set to comp.theory.]
> On 2009-07-26, RussellE <reaste...@gmail.com> wrote:
>
> > I am saying either BNTM writes to every position or it does not.
>
> "Writes to every position" is still somewhat ambiguous, with two
> meanings - let's call them A and B (as labelled in a previous post).
> You fail to distinguish them. =A0So instead of saying that either A is
> true or A is false, you end up saying that either A is true or B is
> false.
>
> > Which is why the word "eventually" keeps coming up. =A0You are saying
> > BNTM writes to every position of any finite tape
>
> Actually for every position of even an infinite tape, there exists a
> step at which it is written. =A0Note carefully the order of the
> quantifiers.

You are saying:
Ax BNTM writes x =3D> AxEy BNTM writes x by stage y

I am saying:
Ax BNTM writes x =3D> EyAX BNTM writes x by stage y

I am well awre of the difference.
Both statements are true for a Turing machine.

> > and therefore BNTM will "eventually" write every finite string of
> > 1's.
>
> It does write every finite string of 1's. =A0It just does not write an
> infinite string of 1's.

You are saying BNTM writes to every position of an infinite
tape, but it doesn't write an infinite string?

> > Like a formal system that assumes there are such things as natural
> > numbers?
>
> No, the system is logically prior to the formal definition of natural
> numbers. =A0Try reading, and attempting to understand, a basic
> introduction to predicate logic. =A0In particular, note that it does not
> require any set of natural numbers to exist.
>
> > The best I can do is show a contradiction. =A0I believe I have done
> > that.
>
> Your belief is false. =A0You incorrectly use predicate logic, so your
> argument is not a valid proof.
>
> > Assume I give BNTM a tape with 100 cells. =A0Each time BNTM writes a
> > '1' to the tape I add a blank cell to the end of the tape.
>
> > Do you agree BNTM will never write to every blank cell of this tape?
>
> The set of blank cells is changing at every step,

So what.
The set of blank positions changes anytime a TM writes to a tape.

> so which blank cells
> do you mean by "every"?

Ax BTNM writes x

> =A0Every cell that is blank at any step?

No.

> =A0Every
> cell that is always blank?

Yes.

> =A0Every cell that is blank at some
> particular step?

No.

>=A0Also do you mean for every such cell, it is never
> written to?

Yes.

>=A0Or do you mean there does not exist a time by which all
> such cells are written to?

Both. I am not sure there is a difference.
Why are introducing "time" again?

>=A0Do you even see the difference?

No. Either BNTM writes to every position on the
tape or it doesn't.

> It's yet another of your ambiguous sentences - even more ambiguous
> that usual, which is saying something. =A0Most of your other statements
> and questions are also vague to the point of being useless. =A0When you
> can't write what you mean, it is unlikely that anyone will find any
> meaning in what you write.

You avoided my question.
Do you agree there will always be at least
100 blank positions on the tape?
Do you agree BNTM will never write to the last
100 positions of the tape?
Do you agree "never" means not even eventually?

A common misconception seems to be that
I am going to tell you how long my tape is.
I'm not. First of all, I can't.
The length of my finite tape with the 100 blank
positions at the end is "uncomputable".
Second, and more importantly, I don't have
to tell you how long the tape is.
All I have to do is prove the tape is finite.

I think someone gave a version of the Gentzen's
consistency proof for PA. I had a lot of objections
to this proof.

Of course, there is the obvious one that no one
who rejects PA would allow "transfinite" induction.
I know the induction is provably finite at every
step, but induction is still just the assumption
my BNTM will write to every position.

I also think Gentzen makes the same mistake
you are. The argument goes something like
assume there is a position BNTM never writes to, z.
Since z is a finite number, after some finite number of
steps BNTM will write the symbol for z, ie, a string of
z 1's. End of proof.

I haven't told you what z is. Even if
BNTM wrote a string of length z,
how would you know?

You, and many others, seem to think
we must know exactly how long the tape
is before we can even talk about it.

If you are asking for a formal proof that BNTM never
writes to position 10^^^3 - 2, I can't give you one.
(There probably is some formal finite proof because
of issues with powersets, but I'm not qualified to
present such a proof.)

Russell
- The universe is one dimension.

```
 0
reasterly (337)
7/27/2009 7:48:57 AM
```On 2009-07-27, RussellE <reasterly@gmail.com> wrote:
> You are saying:
> Ax BNTM writes x => AxEy BNTM writes x by stage y

Is that supposed to be a definition, a statement of logical
implication, a step in a proof, or what?  It seems you really can't
ever be precise.

> I am saying:
> Ax BNTM writes x => EyAX BNTM writes x by stage y

Good for you.

[ Snip more garbage based on the same tired old ambiguities and
failure to reason correctly ]

As expected, your inability to write precisely appears to be a symptom
of your inability to think precisely.  I don't think there is any

- Tim
```
 0
tim669 (185)
7/27/2009 8:36:27 AM
```Drop dead shit
```
 0
7/27/2009 9:34:53 AM
```
Vindicator2...@live.com wrote:
Elgance is beauty, beauty is...the opposite of...but some say it is...

3a97fef3/t/219d22ac02d122af/d/
a75a6b18b96c170c&amp;ct=ga&amp;cd=qYON5Hz9IIE&amp;usg=AFQjCNGKO0wT-
TiDvUCbdS84I81i-uaewQ> Musatov peer review done by Mensa: "rougise"
... on why Yanks should accept his proof and M*Martin Musatov's
proofs P~NP is resolved is a given. thejsh rap from Musatov to JSH:
"What is and . ...

2009 by - 1 message - 1 author

d58b11cf40248770&amp;ct=ga&amp;cd=qYON5Hz9IIE&amp;usg=AFQjCNEoiTjz8DBwtG0uTPORBoGL5mi_6g>
Musatov marty.musa...@gmail.com sci logic alt philosophy sci math
Martin Michael Musatov wrote: Michael Gordge wrote: On Jul 26, 3:27
pm, bofl <bigflet. ...

2009 by - 20 message - 7 author

ac9606362240c03e&amp;ct=ga&amp;cd=qYON5Hz9IIE&amp;usg=AFQjCNH60QL32F1pzHqLZgNKaLHCDrKFvQ>
==============================Martin
==============================Michael =====
=========================Musatov
==============================M. ...truth
```
 0
scriber17 (147)
7/27/2009 9:45:39 AM
```Drop dead shit
```
 0
7/27/2009 9:48:27 AM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 26, 8:06�pm, Tim Little <t...@little-possums.net> wrote:
> > ["Followup-To:" header set to comp.theory.]
> > On 2009-07-26, RussellE <reaste...@gmail.com> wrote:
> >
> > > I am saying either BNTM writes to every position or it does not.
> >
> > "Writes to every position" is still somewhat ambiguous, with two
> > meanings - let's call them A and B (as labelled in a previous post).
> > You fail to distinguish them. �So instead of saying that either A is
> > true or A is false, you end up saying that either A is true or B is
> > false.
> >
> > > Which is why the word "eventually" keeps coming up. �You are saying
> > > BNTM writes to every position of any finite tape
> >
> > Actually for every position of even an infinite tape, there exists a
> > step at which it is written. �Note carefully the order of the
> > quantifiers.
>
> You are saying:
> Ax BNTM writes x => AxEy BNTM writes x by stage y
>
> I am saying:
> Ax BNTM writes x => EyAX BNTM writes x by stage y
>
> I am well awre of the difference.
> Both statements are true for a Turing machine.

What about the machine whose command sequence consists of
1. Write x
2. Move to next position on tape.
3. Go to step 1.

--
Virgil
```
 0
virgil2 (13)
7/27/2009 7:01:03 PM
```On Jul 27, 1:36=A0am, Tim Little <t...@little-possums.net> wrote:
> On 2009-07-27, RussellE <reaste...@gmail.com> wrote:

For those of you that haven't read the preceding 200+
post, Big Number TM (BNTM) write a '1' and moves right.
BNTM never halts.

> > You are saying:
> > Ax BNTM writes x =3D> AxEy BNTM writes x by stage y
>
> Is that supposed to be a definition,

No

> a statement of logical
> implication,

Yes

> a step in a proof, or what?

Sure, why not.

> =A0It seems you really can't
> ever be precise.
>
> > I am saying:
> > Ax BNTM writes x =3D> EyAX BNTM writes x by stage y

Let's call these two statements AxEy and EyAx.
I think most people will agree both statements
are true logical implications if we limit ourselves
to finite tapes of fixed length.

But, a lot of people seem to think AxEy is true and
EyAx is flase when we are talking about infinite tapes.

I am trying to show Peono's Axioms are inconsistent.
PA says if n is a natural number then n+1 is a natural number.
All I need to do to show PA is inconsistent is prove
there is a position BNTM will never write to.

PA comes before infinity. We can't talk about infinity
until we prove there are natural numbers.
This is why I am talking about a finite tape.
Every time BNTM writes to a position I will
add one blank cell to the end of the tape.
This tape is provably finite. Will BNTM write
to every position of this finite tape?

You, and many other people are claiming BNTM
will "eventually" write to every position of ANY tape.

Let's examine your claim in more detail.
After the first step, BNTM has executed
one stage and there are 101 positions
on the tape. It is obvious, to me at least,
that AxEy is false after stage 1.
There are 100 more x's than there are y's.

It should be obvious there are 100 more x's
than y's at every step. AxEy is false at
every step. There will never be enough y's
for every x.

The point of the exercise is to prove (or disprove)
the idea that BNTM writes to every position.
You don't get to say "I assume BNTM writes
to every position therefore I prove AxEy."

Once again, you have avoided my questions.
Do you agree that "never" means not even "eventually".

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/27/2009 7:54:57 PM
```On Jul 25, 8:23=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wrote:
>Joseph Dauben's book _Georg Cantor: His
> Mathematics and Philosophy of the Infinite_ (which gets a few things
> wrong but is particularly good on the early developements)

What does Dauben get wrong?

MoeBlee

```
 0
jazzmobe (307)
7/27/2009 9:47:33 PM
```On Jul 27, 2:47=C2=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 25, 8:23=C2=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wrot=
e:
>
> >Joseph Dauben's book _Georg Cantor: His
> > Mathematics and Philosophy of the Infinite_ (which gets a few things
> > wrong but is particularly good on the early developements)
>
> What does Dauben get wrong?
>
> MoeBlee

MoeBlee et. al,

Explain to me why the case is most numbers natural are uncomputable?
This does not add up. I have shown that as long as time will allow we
may count and add and subject a conceivable reached number in time.
Now if you assert the numbers are 'uncomputable' meansing they are
computable following the variables 'un' it means quite a different
thing to many people I assert.

Even if I was a mindless machine, of which the earlier may be
contested by many in this forum, I would have still produced the
results below quite easily:

Most Natural Numbers Are Uncomputable - sci.math | Google Groups13
posts - 4 authors - Last post: 16 hours ago
From: Musatov <marty.musa...@gmail.com> .... actual history, have a
look at Joseph Dauben's book _Georg Cantor: His ...
ago

Math Forum Discussionsactual history, have a look at Joseph Dauben's
book _Georg Cantor: His ..... Read, Musatov your mom is computable and
a screwable bitch ...
mathforum.org/kb/message.jspa?messageID=3D6795772...30 - 8 hours ago

Most Natural Numbers Are Uncomputable - sci.math | Google Groups7
posts - 3 authors - Last post: yesterday
From: Musatov <marty.musa...@gmail.com> .... actual history, have a
look at Joseph Dauben's book _Georg Cantor: His ...
4eb1fd54ef79780e?...

Math Forum Discussionsactual history, have a look at Joseph Dauben's
book _Georg Cantor: His ..... Read, Musatov your mom is computable and
a screwable bitch ...
mathforum.org/kb/message.jspa?messageID=3D6795847...30 - 9 hours ago

Most Natural Numbers Are Uncomputable - sci.math | Google Groups4
posts - 2 authors - Last post: 2 days ago
From: Musatov <marty.musa...@gmail.com> .... actual history, have a
look at Joseph Dauben's book _Georg Cantor: His ...

Most Natural Numbers Are Uncomputable - sci.math | Google Groups4
posts - 3 authors - Last post: yesterday
From: Musatov <marty.musa...@gmail.com> .... actual history, have a
look at Joseph Dauben's book _Georg Cantor: His ...

Most Natural Numbers Are Uncomputable - sci.math | Google Groups8
posts - 4 authors - Last post: yesterday
From: Musatov <marty.musa...@gmail.com> .... actual history, have a
look at Joseph Dauben's book _Georg Cantor: His ...

Most Natural Numbers Are Uncomputable - sci.math | Google Groups10
posts - 5 authors - Last post: 22 hours ago
From: Musatov <marty.musa...@gmail.com> .... actual history, have a
look at Joseph Dauben's book _Georg Cantor: His ...
ago

Most Natural Numbers Are Uncomputable - sci.math | Google Groups9
posts - 4 authors - Last post: yesterday
From: Musatov <marty.musa...@gmail.com> .... actual history, have a
look at Joseph Dauben's book _Georg Cantor: His ...

7-Keto- =E2=88=86 -Steroids: Key-Molecules Owning Particular
Biological ...Dauben, W.G.; Fullerton, D.S. J. Org. Chem., 1971, 36,
3277- ..... Dushkin, M.I.; Shvarts, Y.S.; Vol'skii, N.N.; Musatov.,
M.I.; ...
www.ingentaconnect.com/content/ben/mrmc/2003/00000003/.../art00005?...

```
 0
marty.musatov (1143)
7/27/2009 9:58:37 PM
```On Jul 27, 2:58=A0pm, Musatov <marty.musa...@gmail.com> wrote:

> MoeBlee et. al,

MoeBlee
```
 0
jazzmobe (307)
7/27/2009 10:02:26 PM
```On Jul 25, 1:25=A0pm, RussellE <reaste...@gmail.com> wrote:

> I think my proof would qualify as finitistic.

Only if 'finitistic' meant 'misinformed and blatantly specious', which
it does not.

MoeBlee

```
 0
jazzmobe (307)
7/27/2009 10:42:26 PM
```On Jul 24, 6:21=A0pm, RussellE <reaste...@gmail.com> wrote:

> Just because I am a crank doesn't mean I am wrong.

Quite so. But you are a crank and you're wrong.

MoeBlee
```
 0
jazzmobe (307)
7/27/2009 10:43:08 PM
```On Jul 27, 3:43=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 24, 6:21=A0pm, RussellE <reaste...@gmail.com> wrote:

> > Just because I am a crank doesn't mean I am wrong.
>
> Quite so. But you are a crank and you're wrong.

Please provide a proof PA is consistent.
To do so, you will have to show EyAx,
(or AxEy which means you have to prove
there are many y's as x's)
which also will prove PA is inconsistent.

Does this sound familiar?
Any formal system capable of arithmetic
(which means it includes PA) that
proves its own consistency is inconsistent.

I don't see how this result is anything more
than an admission PA is inconsistent.
I don't see what we gain from weakening an
inconsistent formal theory to the point we can't
prove its (in)consistency.

Maybe you are willing to assume a system
is consistent as long as we can't prove an
inconsistency, but I am not.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/28/2009 12:06:07 AM
```On Jul 27, 5:06=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 27, 3:43=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 24, 6:21=A0pm, RussellE <reaste...@gmail.com> wrote:
> > > Just because I am a crank doesn't mean I am wrong.
>
> > Quite so. But you are a crank and you're wrong.
>
> Please provide a proof PA is consistent.

I don't need to prove PA is consistent just to note that your "proofs"
are incorrect.

Anyway, Z easily proves PA consistent, as <w 0 S + *> is a model of
the PA axioms.

Also, there are more strenuous proofs available in the literature but
well beyond the scope of a post (see Gentzen's proof, or, for example,
the proof in Shoenfield textbook just mentioned in this thread).

I don't make any claim as to any epistemological import of such
proofs, but rather that as a technical matter such proofs do exist
(aside from even the most trivial proof had from taking the
consistency of PA as an axiom).

> To do so, you will have to show EyAx,
> (or AxEy which means you have to prove
> there are many y's as x's)
> which also will prove PA is inconsistent.

You don't know what you're talking about. Please don't waste my time.

> Does this sound familiar?
> Any formal system capable of arithmetic
> (which means it includes PA) that
> proves its own consistency is inconsistent.

So?

> I don't see how this result is anything more
> than an admission PA is inconsistent.

Of course you don't see, since you know virtually NOTHING about the
subject. Grow up. Read a book.

> I don't see what we gain from weakening an
> inconsistent formal theory to the point we can't
> prove its (in)consistency.

You're completely confused and malinformed.

> Maybe you are willing to assume a system
> is consistent as long as we can't prove an
> inconsistency, but I am not.

I've written different posts about the subject. You can look them up
if you like.

Meanwhile, you are over-opinionated about a subject of which you are
virtually ignorant of its basics. And you continually repeat your
claims, in face of all refutation. Then you bring up completely self-
mal-informed speciousness such as "I don't see how this result is
anything more than an admission PA is inconsistent" as if that is some
kind of continuation of the discussion. You're obnoxious.

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 12:24:42 AM
```On Jul 27, 5:24=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 27, 5:06=A0pm, RussellE <reaste...@gmail.com> wrote:

>You're obnoxious.
>
> MoeBlee

Why, thank you! I learned from the masters.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/28/2009 12:32:46 AM
```On Mon, 27 Jul 2009 15:43:08 -0700 (PDT) MoeBlee wrote:

> On Jul 24, 6:21�pm, RussellE <reaste...@gmail.com> wrote:
>>
>> Just because I am a crank doesn't mean I am wrong.
>>
> Quite so. But you are a crank and you're wrong.

A rather common coincidence.

Herb
```
 0
Herbert
7/28/2009 12:38:17 AM
```On Jul 27, 5:38=A0pm, Herbert Newman <nomail@invalid> wrote:
> On Mon, 27 Jul 2009 15:43:08 -0700 (PDT) MoeBlee wrote:
>
> > On Jul 24, 6:21=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> >> Just because I am a crank doesn't mean I am wrong.
>
> > Quite so. But you are a crank and you're wrong.
>
> A rather common coincidence.

No one seems to want to answer my question.
Does "never" mean not even "eventually"?

Most people seem to agree BNTM "never" actually
writes to every position of an infinite tape.
Why would you believe BNTM writes to
every finite position "eventually"?

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/28/2009 12:49:12 AM
```On Jul 27, 5:32=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 27, 5:24=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 27, 5:06=A0pm, RussellE <reaste...@gmail.com> wrote:
> >You're obnoxious.

> Why, thank you! I learned from the masters.

What "masters" spout misinformation and illogic based on ignorance?

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 12:51:28 AM
```On Jul 27, 5:49=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 27, 5:38=A0pm, Herbert Newman <nomail@invalid> wrote:
>
> > On Mon, 27 Jul 2009 15:43:08 -0700 (PDT) MoeBlee wrote:
>
> > > On Jul 24, 6:21=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > >> Just because I am a crank doesn't mean I am wrong.
>
> > > Quite so. But you are a crank and you're wrong.
>
> > A rather common coincidence.
>
> No one seems to want to answer my question.
> Does "never" mean not even "eventually"?

The actual MATHEMATICAL results do not depend on, nor are affected by
the vagaries of such English words as 'eventually'. Instead, I gave
you a phrasing much closer to the actual mathematical formulation.

You're technique is just to keep moving the discussion BACKWARDS from
more precise language to less precise language then tp claim to find
some problem though it was already explained to you how such seeming
difficulties dissolve when we move from ambiguous language to more
precise language. You are intellectually puerile.

You're just not interested in UNDERSTANDING the subject matter and
such answers that contribute to an understanding.

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 12:57:33 AM
```In article
RussellE <reasterly@gmail.com> wrote:

> Most people seem to agree BNTM "never" actually
> writes to every position of an infinite tape.
> Why would you believe BNTM writes to
> every finite position "eventually"?

There is a fine difference between eventually writing to each position
and eventually writing to every position.

A TM may be capable of the former without being capable of the latter.
```
 0
Virgil7091 (95)
7/28/2009 2:31:51 AM
```On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
>
> =A0RussellE <reaste...@gmail.com> wrote:
> > Most people seem to agree BNTM "never" actually
> > writes to every position of an infinite tape.
> > Why would you believe BNTM writes to
> > every finite position "eventually"?
>
> There is a fine difference between eventually writing to each position
> and eventually writing to every position.
>
> A TM may be capable of the former without being capable of the latter.

The definition I learned for all is "each and every".  Perhaps Russell
has been failing to use all when he says every.  Can you elaborate?

karl m
```
 0
malbrain (59)
7/28/2009 3:05:59 AM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 27, 7:31�pm, Virgil <Vir...@home.esc> wrote:
> > In article
> >
> > �RussellE <reaste...@gmail.com> wrote:
> > > Most people seem to agree BNTM "never" actually
> > > writes to every position of an infinite tape.
> > > Why would you believe BNTM writes to
> > > every finite position "eventually"?
> >
> > There is a fine difference between eventually writing to each position
> > and eventually writing to every position.
> >
> > A TM may be capable of the former without being capable of the latter.
>
> The definition I learned for all is "each and every".  Perhaps Russell
> has been failing to use all when he says every.  Can you elaborate?
>
> karl m

Given the TM whose only commands are:
1. Mark the tape then  go to command 2.
2. Move to next tape position then go to command 1.
consider the following statements:

A: for each tape position, x, following the start position of the TM
there is a number of operations, y,  after which, x will be marked.

B: there is a number of operations, y, after which each position x that
gets marked is marked.

Russ seems to think A and B are equivalent.
```
 0
Virgil7091 (95)
7/28/2009 3:44:01 AM
```On Jul 27, 5:06=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> To do so, you will have to show EyAx

This proves EyAx:

http://www.imdb.com/title/tt0060665/

Very highly rated!

It is also possible to prove AxEy: it's called
universal health care, but the proof only works
in Europe and Canada; in the US it fails.

You can prove PA is consistent because PA has a model.
I can't seem to find the model at the moment; probably
the right thing to do is to search for it:

Guaranteed to be a worthwhile search.

Marshall
```
 0
7/28/2009 4:17:54 AM
```On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
> =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> > > In article
>
> > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > Most people seem to agree BNTM "never" actually
> > > > writes to every position of an infinite tape.
> > > > Why would you believe BNTM writes to
> > > > every finite position "eventually"?
>
> > > There is a fine difference between eventually writing to each positio=
n
> > > and eventually writing to every position.
>
> > > A TM may be capable of the former without being capable of the latter=
..
>
> > The definition I learned for all is "each and every". =A0Perhaps Russel=
l
> > has been failing to use all when he says every. =A0Can you elaborate?
>
> > karl m
>
> Given the TM whose only commands are:
> =A0 =A01. Mark the tape then =A0go to command 2.
> =A0 =A02. Move to next tape position then go to command 1.
> consider the following statements:
>
> A: for each tape position, x, following the start position of the TM
> there is a number of operations, y, =A0after which, x will be marked.
>
> B: there is a number of operations, y, after which each position x that
> gets marked is marked.
>
> Russ seems to think A and B are equivalent.- Hide quoted text -
>
> - Show quoted text -

You changed the tense "to be" between A and B.  B should read:

B. There is  a number of operations, y, after which each position x
will be marked.

I don't see why B is not satisfied by y =3D 1.  All positions are marked

Russell went back to the original specification of a TM, e.g. no
infinite tape, just finite unbounded, and introduced a difference
between positions to be marked and positions marked.

karl m
```
 0
malbrain (59)
7/28/2009 4:31:25 AM
```On Jul 27, 9:31=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
>
>
>
>
>
> > In article
> > =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> > > > In article
>
> > > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > > Most people seem to agree BNTM "never" actually
> > > > > writes to every position of an infinite tape.
> > > > > Why would you believe BNTM writes to
> > > > > every finite position "eventually"?
>
> > > > There is a fine difference between eventually writing to each posit=
ion
> > > > and eventually writing to every position.
>
> > > > A TM may be capable of the former without being capable of the latt=
er.
>
> > > The definition I learned for all is "each and every". =A0Perhaps Russ=
ell
> > > has been failing to use all when he says every. =A0Can you elaborate?
>
> > > karl m
>
> > Given the TM whose only commands are:
> > =A0 =A01. Mark the tape then =A0go to command 2.
> > =A0 =A02. Move to next tape position then go to command 1.
> > consider the following statements:
>
> > A: for each tape position, x, following the start position of the TM
> > there is a number of operations, y, =A0after which, x will be marked.
>
> > B: there is a number of operations, y, after which each position x that
> > gets marked is marked.
>
> > Russ seems to think A and B are equivalent.- Hide quoted text -
>
> > - Show quoted text -
>
> You changed the tense "to be" between A and B. =A0B should read:
>
> B. There is =A0a number of operations, y, after which each position x
> will be marked.

sorry, make that:

B.  There is a number of operations, y, after which EVERY position x
will be marked.

> I don't see why B is not satisfied by y =3D 1. =A0All positions are marke=
d
>
> Russell went back to the original specification of a TM, e.g. no
> infinite tape, just finite unbounded, and introduced a difference
> between positions to be marked and positions marked.
>
> karl m- Hide quoted text -
>
> - Show quoted text -

karl m
```
 0
malbrain (59)
7/28/2009 6:46:09 AM
```On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
> =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> > > In article
>
> > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > Most people seem to agree BNTM "never" actually
> > > > writes to every position of an infinite tape.
> > > > Why would you believe BNTM writes to
> > > > every finite position "eventually"?
>
> > > There is a fine difference between eventually writing to each positio=
n
> > > and eventually writing to every position.
>
> > > A TM may be capable of the former without being capable of the latter=
..
>
> > The definition I learned for all is "each and every". =A0Perhaps Russel=
l
> > has been failing to use all when he says every. =A0Can you elaborate?
>
> > karl m
>
> Given the TM whose only commands are:
> =A0 =A01. Mark the tape then =A0go to command 2.
> =A0 =A02. Move to next tape position then go to command 1.
> consider the following statements:
>
> A: for each tape position, x, following the start position of the TM
> there is a number of operations, y, =A0after which, x will be marked.

So, you are saying there is a step y when AxEy is true.

> B: there is a number of operations, y, after which each position x that
> gets marked is marked.

Both of these statements are true if BNTM writes
to every position of any tape.

Russell
- Never never means never in set theory
```
 0
reasterly (337)
7/28/2009 7:16:24 AM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 27, 8:44�pm, Virgil <Vir...@home.esc> wrote:
> > In article
> > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> >
> >
> >
> >
> > > On Jul 27, 7:31�pm, Virgil <Vir...@home.esc> wrote:
> > > > In article
> >
> > > > �RussellE <reaste...@gmail.com> wrote:
> > > > > Most people seem to agree BNTM "never" actually
> > > > > writes to every position of an infinite tape.
> > > > > Why would you believe BNTM writes to
> > > > > every finite position "eventually"?
> >
> > > > There is a fine difference between eventually writing to each position
> > > > and eventually writing to every position.
> >
> > > > A TM may be capable of the former without being capable of the latter.
> >
> > > The definition I learned for all is "each and every". �Perhaps Russell
> > > has been failing to use all when he says every. �Can you elaborate?
> >
> > > karl m
> >
> > Given the TM whose only commands are:
> > � �1. Mark the tape then �go to command 2.
> > � �2. Move to next tape position then go to command 1.
> > consider the following statements:
> >
> > A: for each tape position, x, following the start position of the TM
> > there is a number of operations, y, �after which, x will be marked.
>
> So, you are saying there is a step y when AxEy is true.
>
> > B: there is a number of operations, y, after which each position x that
> > gets marked is marked.
>
> Both of these statements are true if BNTM writes
> to every position of any tape.

Not if each operation takes some positive minimal amount of time.

In that case, there will never be a time when all positions have been
marked though for each position there will be a time when it has been
marked.
```
 0
Virgil7091 (95)
7/28/2009 7:56:25 AM
```On Jul 28, 12:56=A0am, Virgil <Vir...@home.esc> wrote:
> In article
>
>
>
>
>
> =A0RussellE <reaste...@gmail.com> wrote:
> > On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> > > In article
> > > =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> > > > > In article
> > > > > <728f4d92-4497-44d6-a19a-e88eae655...@m3g2000pri.googlegroups.com=
>,
>
> > > > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > > > Most people seem to agree BNTM "never" actually
> > > > > > writes to every position of an infinite tape.
> > > > > > Why would you believe BNTM writes to
> > > > > > every finite position "eventually"?
>
> > > > > There is a fine difference between eventually writing to each pos=
ition
> > > > > and eventually writing to every position.
>
> > > > > A TM may be capable of the former without being capable of the la=
tter.
>
> > > > The definition I learned for all is "each and every". =A0Perhaps Ru=
ssell
> > > > has been failing to use all when he says every. =A0Can you elaborat=
e?
>
> > > > karl m
>
> > > Given the TM whose only commands are:
> > > =A0 =A01. Mark the tape then =A0go to command 2.
> > > =A0 =A02. Move to next tape position then go to command 1.
> > > consider the following statements:
>
> > > A: for each tape position, x, following the start position of the TM
> > > there is a number of operations, y, =A0after which, x will be marked.
>
> > So, you are saying there is a step y when AxEy is true.
>
> > > B: there is a number of operations, y, after which each position x th=
at
> > > gets marked is marked.
>
> > Both of these statements are true if BNTM writes
> > to every position of any tape.
>
> Not if each operation takes some positive minimal amount of time.
>
> In that case, there will never be a time when all positions have been
> marked though for each position there will be a time when it has been
> marked.- Hide quoted text -
>
> - Show quoted text -

With induction I can prove that the BNTM writes to all positions of
the tape.  "All positions" implies both each and every position, as
expressed by A and B.

karl m
```
 0
malbrain (59)
7/28/2009 8:27:36 AM
```On Jul 28, 12:16=A0am, RussellE <reaste...@gmail.com> wrote:
> On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:

> > A: for each tape position, x, following the start position of the TM
> > there is a number of operations, y, =A0after which, x will be marked.
>
> So, you are saying there is a step y when AxEy is true.

You don't even know what you're saying. 'y' is free in the first and
bound in the second.

> > B: there is a number of operations, y, after which each position x that
> > gets marked is marked.
>
> Both of these statements are true if BNTM writes
> to every position of any tape.

You're doing it again. Reverting back to ambiguous language after a
more precise formulation has been given. That you CONTINUE to do that,
even after it has been pointed out to you over and over, is childishly
dishonest.

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 4:30:23 PM
```On Jul 28, 9:30=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 12:16=A0am, RussellE <reaste...@gmail.com> wrote:
>
> > On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> > > A: for each tape position, x, following the start position of the TM
> > > there is a number of operations, y, =A0after which, x will be marked.
>
> > So, you are saying there is a step y when AxEy is true.
>
> You don't even know what you're saying. 'y' is free in the first and
> bound in the second.
>
> > > B: there is a number of operations, y, after which each position x th=
at
> > > gets marked is marked.
>
> > Both of these statements are true if BNTM writes
> > to every position of any tape.
>
> You're doing it again. Reverting back to ambiguous language after a
> more precise formulation has been given. That you CONTINUE to do that,
> even after it has been pointed out to you over and over, is childishly
> dishonest.
>
> MoeBlee

Perhaps I'm mis-interpreting something, but I thought that Virgil
explained that your expressions A and B referred to an expression of
"each" position on the tape, and "every" position on the tape
respectively, as the two necessary components to expressing "all"
positions on the tape.  In other words, ALL is defined as EACH and
EVERY.

karl m
```
 0
malbrain (59)
7/28/2009 4:41:13 PM
```On Jul 28, 9:41=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:

> Perhaps I'm mis-interpreting something, but I thought that Virgil
> explained that your expressions A and B referred to an expression of
> "each" position on the tape, and "every" position on the tape
> respectively, as the two necessary components to expressing "all"
> positions on the tape. =A0In other words, ALL is defined as EACH and
> EVERY.

The only things needes in this regard are the universal ('A') and
existential ('E') quantifiers, without getting bogged down in vagaries
of English words:

We prove:

AnEs n is written at step s.

We do NOT prove:

EsAn n is written at step s.

Indeed, we prove:

~EsAn n is written at step s.

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 5:06:47 PM
```On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 9:41=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > explained that your expressions A and B referred to an expression of
> > "each" position on the tape, and "every" position on the tape
> > respectively, as the two necessary components to expressing "all"
> > positions on the tape. =A0In other words, ALL is defined as EACH and
> > EVERY.
>
> The only things needes in this regard are the universal ('A') and
> existential ('E') quantifiers, without getting bogged down in vagaries
> of English words:
>
> We prove:
>
> AnEs n is written at step s.
>
> We do NOT prove:
>
> EsAn n is written at step s.
>
> Indeed, we prove:
>
> ~EsAn n is written at step s.
>
> MoeBlee

I conclude that you have proof that there is no step s where EVERY
position of the tape "is written" (all at once), and that there is
always a step s where EACH position of the tape "is written" (one at a
time).

This contradicts the proof by induction that BNTM writes ALL positions
of the tape, or the contrapositive, that there are no positions that
BTMN doesn't write.

karl m
```
 0
malbrain (59)
7/28/2009 6:05:43 PM
```On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 9:41=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > explained that your expressions A and B referred to an expression of
> > "each" position on the tape, and "every" position on the tape
> > respectively, as the two necessary components to expressing "all"
> > positions on the tape. =A0In other words, ALL is defined as EACH and
> > EVERY.
>
> The only things needes in this regard are the universal ('A') and
> existential ('E') quantifiers, without getting bogged down in vagaries
> of English words:
>
> We prove:
>
> AnEs n is written at step s.

No you don't. You can't.
If you do, you also prove PA is inconsistent.
You can never show there are at least as
many steps as positions.

You assume Ax BNTM writes x and
from this you imply AxEy writes x by step y.

> We do NOT prove:
>
> EsAn n is written at step s.

And then you inexplicably declare EsAn is false
with no shred of proof other than it would mean
PA is inconsistent.

A) Ax BNTM writex x =3D> AxEy BNTM writes x by y
B) Ax BNTM writes x =3D> EyAx BNTM writes x by y

Both of these inferences are true for any tape and any TM.

Russell
- 2 many 2 count

```
 0
7/28/2009 6:06:50 PM
```On Jul 28, 11:05=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:

> I conclude that you have proof that there is no step s where EVERY
> position of the tape "is written" (all at once), and that there is
> always a step s where EACH position of the tape "is written" (one at a
> time).

No, I didn't say I agree to go BACKWARDS from the precision of AxEy to
whatever vagaries can be exploited from such terms as "all at once",
etc.

What I said is exact, and nothing more:

AnEs n is written at step s
~EsAn n is written at step s

(where 'n is written at step s' informally stands for a certain exact
mathematical formula regarding Turing machines).

MoeBlee
```
 0
jazzmobe (307)
7/28/2009 6:14:32 PM
```On Jul 28, 11:06=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
> On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:

> > We prove:
>
> > AnEs n is written at step s.
>
> No you don't. You can't.

Sure we do; we've shown you the informal version. Then a formal
version is as easy as putting the matter all in terms of exact
mathematics of Turing machines.

> If you do, you also prove PA is inconsistent.

You just keep SAYING that.

> You can never show there are at least as
> many steps as positions.

You just keep SAYING such things.

> You assume Ax BNTM writes x and
> from this you imply AxEy writes x by step y.

No, that's not my argument.

> > We do NOT prove:
>
> > EsAn n is written at step s.
>
> And then you inexplicably declare EsAn is false
> with no shred of proof other than it would mean
> PA is inconsistent.

No, I NEVER SAID it is false because otherwise it would mean PA is
inconsistent. PLEASE DO NOT PUT WORDS IN MY MOUTH.

The proof of

~EsAn n is written at step s.

is easy to formulate in the mathematics of Turing machines (but I
didn't claim that I had given such a proof here).

> A) Ax BNTM writex x =3D> AxEy BNTM writes x by y
> B) Ax BNTM writes x =3D> EyAx BNTM writes x by y
>
> Both of these inferences are true for any tape and any TM.

You just keep SAYING that.

B is a fallacious inference.

compulsion.

MoeBlee
```
 0
jazzmobe (307)
7/28/2009 6:21:54 PM
```On Jul 28, 11:21=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 11:06=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
>
>
> > A) Ax BNTM writex x =3D> AxEy BNTM writes x by y
> > B) Ax BNTM writes x =3D> EyAx BNTM writes x by y
>
> > Both of these inferences are true for any tape and any TM.
>
> You just keep SAYING that.
>
> B is a fallacious inference.

How? I read "for ALL x a Turing Machine writes x" to mean that for
EACH x the TM writes position x, and for EVERY x the TM writes
position x.

This is just basic logic and basic language.

karl m
```
 0
malbrain (59)
7/28/2009 6:30:09 PM
```On Jul 28, 11:21=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 11:06=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
>
> > On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > We prove:
>
> > > AnEs n is written at step s.
>
> > No you don't. You can't.
>
> Sure we do; we've shown you the informal version. Then a formal
> version is as easy as putting the matter all in terms of exact
> mathematics of Turing machines.

The proof I saw uses induction.
Induction is nothing more than the assumption
Ax BNTM writes x.

> > If you do, you also prove PA is inconsistent.
>
> You just keep SAYING that.

And I will keep saying it.
PA is provably inconsistent.

> > You can never show there are at least as
> > many steps as positions.
>
> You just keep SAYING such things.
>
> > You assume Ax BNTM writes x and
> > from this you imply AxEy writes x by step y.
>
> No, that's not my argument.

> > > We do NOT prove:
>
> > > EsAn n is written at step s.
>
> > And then you inexplicably declare EsAn is false
> > with no shred of proof other than it would mean
> > PA is inconsistent.
>
> No, I NEVER SAID it is false because otherwise it would mean PA is
> inconsistent. PLEASE DO NOT PUT WORDS IN MY MOUTH.

Sorry.

> The proof of
>
> ~EsAn n is written at step s.
>
> is easy to formulate in the mathematics of Turing machines (but I
> didn't claim that I had given such a proof here).
>
> > A) Ax BNTM writex x =3D> AxEy BNTM writes x by y
> > B) Ax BNTM writes x =3D> EyAx BNTM writes x by y
>
> > Both of these inferences are true for any tape and any TM.
>
> You just keep SAYING that.

I have given the proof many times.
You know, the one about the set of blank positions.

> B is a fallacious inference.

So is A, if we are talking about infinite tapes.

> compulsion.

You are probably right.
They locked Cantor up, too.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/28/2009 6:51:34 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 12:56�am, Virgil <Vir...@home.esc> wrote:
> > In article
> >
> >
> >
> >
> >
> > �RussellE <reaste...@gmail.com> wrote:
> > > On Jul 27, 8:44�pm, Virgil <Vir...@home.esc> wrote:
> > > > In article
> > > > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > > > On Jul 27, 7:31�pm, Virgil <Vir...@home.esc> wrote:
> > > > > > In article
> > > > > > <728f4d92-4497-44d6-a19a-e88eae655...@m3g2000pri.googlegroups.com>,
> >
> > > > > > �RussellE <reaste...@gmail.com> wrote:
> > > > > > > Most people seem to agree BNTM "never" actually
> > > > > > > writes to every position of an infinite tape.
> > > > > > > Why would you believe BNTM writes to
> > > > > > > every finite position "eventually"?
> >
> > > > > > There is a fine difference between eventually writing to each
> > > > > > position
> > > > > > and eventually writing to every position.
> >
> > > > > > A TM may be capable of the former without being capable of the
> > > > > > latter.
> >
> > > > > The definition I learned for all is "each and every". �Perhaps
> > > > > Russell
> > > > > has been failing to use all when he says every. �Can you elaborate?
> >
> > > > > karl m
> >
> > > > Given the TM whose only commands are:
> > > > � �1. Mark the tape then �go to command 2.
> > > > � �2. Move to next tape position then go to command 1.
> > > > consider the following statements:
> >
> > > > A: for each tape position, x, following the start position of the TM
> > > > there is a number of operations, y, �after which, x will be marked.
> >
> > > So, you are saying there is a step y when AxEy is true.
> >
> > > > B: there is a number of operations, y, after which each position x that
> > > > gets marked is marked.
> >
> > > Both of these statements are true if BNTM writes
> > > to every position of any tape.
> >
> > Not if each operation takes some positive minimal amount of time.
> >
> > In that case, there will never be a time when all positions have been
> > marked though for each position there will be a time when it has been
> > marked.- Hide quoted text -
> >
> > - Show quoted text -
>
> With induction I can prove that the BNTM writes to all positions of
> the tape.  "All positions" implies both each and every position, as
> expressed by A and B.

What I said, and what is still true is:
If each step of my TM takes some fixed positive amount of time of there
will never be a time when all positions have been marked. But for each
position, there will be a time when it is marked.
```
 0
Virgil7091 (95)
7/28/2009 7:11:13 PM
```On Jul 28, 12:11=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
> =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 28, 12:56=A0am, Virgil <Vir...@home.esc> wrote:
> > > In article
>
> > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> > > > > In article
> > > > > <d6c9f7e2-1e80-4f00-aa81-920525b8b...@y10g2000prf.googlegroups.co=
m>,
> > > > > =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > > > On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> > > > > > > In article
> > > > > > > <728f4d92-4497-44d6-a19a-e88eae655...@m3g2000pri.googlegroups=
..com>,
>
> > > > > > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > > > > > Most people seem to agree BNTM "never" actually
> > > > > > > > writes to every position of an infinite tape.
> > > > > > > > Why would you believe BNTM writes to
> > > > > > > > every finite position "eventually"?
>
> > > > > > > There is a fine difference between eventually writing to each
> > > > > > > position
> > > > > > > and eventually writing to every position.
>
> > > > > > > A TM may be capable of the former without being capable of th=
e
> > > > > > > latter.
>
> > > > > > The definition I learned for all is "each and every". =A0Perhap=
s
> > > > > > Russell
> > > > > > has been failing to use all when he says every. =A0Can you elab=
orate?
>
> > > > > > karl m
>
> > > > > Given the TM whose only commands are:
> > > > > =A0 =A01. Mark the tape then =A0go to command 2.
> > > > > =A0 =A02. Move to next tape position then go to command 1.
> > > > > consider the following statements:
>
> > > > > A: for each tape position, x, following the start position of the=
TM
> > > > > there is a number of operations, y, =A0after which, x will be mar=
ked.
>
> > > > So, you are saying there is a step y when AxEy is true.
>
> > > > > B: there is a number of operations, y, after which each position =
x that
> > > > > gets marked is marked.
>
> > > > Both of these statements are true if BNTM writes
> > > > to every position of any tape.
>
> > > Not if each operation takes some positive minimal amount of time.
>
> > > In that case, there will never be a time when all positions have been
> > > marked though for each position there will be a time when it has been
> > > marked.- Hide quoted text -
>
> > > - Show quoted text -
>
> > With induction I can prove that the BNTM writes to all positions of
> > the tape. =A0"All positions" implies both each and every position, as
> > expressed by A and B.
>
> What I said, and what is still true is:
> If each step of my TM takes some fixed positive amount of time of there
> will never be a time when all positions have been marked. But for each
> position, there will be a time when it is marked.- Hide quoted text -
>
> - Show quoted text -

We have a contradiction between your definition of the machine and the
mathematics of induction that establishes that for ALL x the machine
writes to position x.

karl m

```
 0
malbrain (59)
7/28/2009 7:19:03 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 9:30�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 28, 12:16�am, RussellE <reaste...@gmail.com> wrote:
> >
> > > On Jul 27, 8:44�pm, Virgil <Vir...@home.esc> wrote:
> > > > A: for each tape position, x, following the start position of the TM
> > > > there is a number of operations, y, �after which, x will be marked.
> >
> > > So, you are saying there is a step y when AxEy is true.
> >
> > You don't even know what you're saying. 'y' is free in the first and
> > bound in the second.
> >
> > > > B: there is a number of operations, y, after which each position x that
> > > > gets marked is marked.
> >
> > > Both of these statements are true if BNTM writes
> > > to every position of any tape.
> >
> > You're doing it again. Reverting back to ambiguous language after a
> > more precise formulation has been given. That you CONTINUE to do that,
> > even after it has been pointed out to you over and over, is childishly
> > dishonest.
> >
> > MoeBlee
>
> Perhaps I'm mis-interpreting something, but I thought that Virgil
> explained that your expressions A and B referred to an expression of
> "each" position on the tape, and "every" position on the tape
> respectively, as the two necessary components to expressing "all"
> positions on the tape.  In other words, ALL is defined as EACH and
> EVERY.
>
> karl m

Those A and B statements are mine, not Moblee's. If you read them
carefully enough you will see that A allows separate steps for separate
marks but that B requires a single finite step finish all but finitely
many marks, after which, no further steps are needed.
```
 0
Virgil7091 (95)
7/28/2009 7:23:51 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 10:06�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 28, 9:41�am, Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > explained that your expressions A and B referred to an expression of
> > > "each" position on the tape, and "every" position on the tape
> > > respectively, as the two necessary components to expressing "all"
> > > positions on the tape. �In other words, ALL is defined as EACH and
> > > EVERY.
> >
> > The only things needes in this regard are the universal ('A') and
> > existential ('E') quantifiers, without getting bogged down in vagaries
> > of English words:
> >
> > We prove:
> >
> > AnEs n is written at step s.
> >
> > We do NOT prove:
> >
> > EsAn n is written at step s.
> >
> > Indeed, we prove:
> >
> > ~EsAn n is written at step s.
> >
> > MoeBlee
>
> I conclude that you have proof that there is no step s where EVERY
> position of the tape "is written" (all at once), and that there is
> always a step s where EACH position of the tape "is written" (one at a
> time).
>
> This contradicts the proof by induction that BNTM writes ALL positions
> of the tape, or the contrapositive, that there are no positions that
> BTMN doesn't write.

On the contrary, unless there is a last position on the tape, which
induction denies, you are wrong.
>
> karl m
```
 0
Virgil7091 (95)
7/28/2009 7:26:50 PM
```In article
Kyle Easterly <keileeneast@gmail.com> wrote:

> On Jul 28, 10:06�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 28, 9:41�am, Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > explained that your expressions A and B referred to an expression of
> > > "each" position on the tape, and "every" position on the tape
> > > respectively, as the two necessary components to expressing "all"
> > > positions on the tape. �In other words, ALL is defined as EACH and
> > > EVERY.
> >
> > The only things needes in this regard are the universal ('A') and
> > existential ('E') quantifiers, without getting bogged down in vagaries
> > of English words:
> >
> > We prove:
> >
> > AnEs n is written at step s.
>
> No you don't. You can't.
> If you do, you also prove PA is inconsistent.
> You can never show there are at least as
> many steps as positions.
>
> You assume Ax BNTM writes x and
> from this you imply AxEy writes x by step y.
>
> > We do NOT prove:
> >
> > EsAn n is written at step s.
>
> And then you inexplicably declare EsAn is false
> with no shred of proof other than it would mean
> PA is inconsistent.
>
> A) Ax BNTM writex x => AxEy BNTM writes x by y
> B) Ax BNTM writes x => EyAx BNTM writes x by y
>
> Both of these inferences are true for any tape and any TM.

Not if we have a set of naturals, N which contains 0 and for each
member, n, contains a member, n+1, greater than n, and is a subset of
any other set with those properties.

In such a set, N, it is quite true that:  Am in N, E n in N, n > m

but quite false that: En in N, Am in N, n > m.

And in ZF, for example, we have such an N.
```
 0
Virgil7091 (95)
7/28/2009 7:34:29 PM
```On Jul 28, 12:26=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
> =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > On Jul 28, 9:41=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > > explained that your expressions A and B referred to an expression o=
f
> > > > "each" position on the tape, and "every" position on the tape
> > > > respectively, as the two necessary components to expressing "all"
> > > > positions on the tape. =A0In other words, ALL is defined as EACH an=
d
> > > > EVERY.
>
> > > The only things needes in this regard are the universal ('A') and
> > > existential ('E') quantifiers, without getting bogged down in vagarie=
s
> > > of English words:
>
> > > We prove:
>
> > > AnEs n is written at step s.
>
> > > We do NOT prove:
>
> > > EsAn n is written at step s.
>
> > > Indeed, we prove:
>
> > > ~EsAn n is written at step s.
>
> > > MoeBlee
>
> > I conclude that you have proof that there is no step s where EVERY
> > position of the tape "is written" (all at once), and that there is
> > always a step s where EACH position of the tape "is written" (one at a
> > time).
>
> > This contradicts the proof by induction that BNTM writes ALL positions
> > of the tape, or the contrapositive, that there are no positions that
> > BTMN doesn't write.
>
> On the contrary, unless there is a last position on the tape, which
> induction denies, you are wrong.

I believe the proof is straight-forward.

The TM writes to position 1 (by virtue of having started).  If you
assume the TM has written position x, it is also true that it has
written position x+1 (by the programming of the machine).  By the
axiom of induction, it is true that the TM has written ALL x.  Not
just EACH x but EVERY x.

karl m
```
 0
malbrain (59)
7/28/2009 7:36:01 PM
```On Jul 28, 11:30=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 28, 11:21=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 11:06=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
>
> > > A) Ax BNTM writex x =3D> AxEy BNTM writes x by y
> > > B) Ax BNTM writes x =3D> EyAx BNTM writes x by y
>
> > > Both of these inferences are true for any tape and any TM.
>
> > You just keep SAYING that.
>
> > B is a fallacious inference.
>
> How?

It is a trivial exercise in predicate logic. I already gave
counterexamples. Please refer to them from my previous posts if you

I read "for ALL x a Turing Machine writes x" to mean that for
> EACH x the TM writes position x, and for EVERY x the TM writes
> position x.
>
> This is just basic logic and basic language.

You don't even need all this "all and each" stuff.

"Ax BNTM writes x" is adequate.

But it is NOT basic logic that "EyAx BNTM writes x by y" then follows.

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 7:41:28 PM
```On Jul 28, 12:36=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 28, 12:26=A0pm, Virgil <Vir...@home.esc> wrote:
>
>
>
>
>
> > In article
> > =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > On Jul 28, 9:41=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > > > explained that your expressions A and B referred to an expression=
of
> > > > > "each" position on the tape, and "every" position on the tape
> > > > > respectively, as the two necessary components to expressing "all"
> > > > > positions on the tape. =A0In other words, ALL is defined as EACH =
and
> > > > > EVERY.
>
> > > > The only things needes in this regard are the universal ('A') and
> > > > existential ('E') quantifiers, without getting bogged down in vagar=
ies
> > > > of English words:
>
> > > > We prove:
>
> > > > AnEs n is written at step s.
>
> > > > We do NOT prove:
>
> > > > EsAn n is written at step s.
>
> > > > Indeed, we prove:
>
> > > > ~EsAn n is written at step s.
>
> > > > MoeBlee
>
> > > I conclude that you have proof that there is no step s where EVERY
> > > position of the tape "is written" (all at once), and that there is
> > > always a step s where EACH position of the tape "is written" (one at =
a
> > > time).
>
> > > This contradicts the proof by induction that BNTM writes ALL position=
s
> > > of the tape, or the contrapositive, that there are no positions that
> > > BTMN doesn't write.
>
> > On the contrary, unless there is a last position on the tape, which
> > induction denies, you are wrong.
>
> I believe the proof is straight-forward.
>
> The TM writes to position 1 (by virtue of having started). =A0If you
> assume the TM has written position x, it is also true that it has
> written position x+1 (by the programming of the machine). =A0By the
> axiom of induction, it is true that the TM has written ALL x. =A0Not
> just EACH x but EVERY x.
>
> karl m- Hide quoted text -
>
> - Show quoted text -

Sorry, now I'm mixing tenses.  Please replace "has written" by
"writes" in the above proof.

karl m
```
 0
malbrain (59)
7/28/2009 7:42:14 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 11:21�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 28, 11:06�am, Kyle Easterly <keileene...@gmail.com> wrote:
> >
> >
> > > A) Ax BNTM writex x => AxEy BNTM writes x by y
> > > B) Ax BNTM writes x => EyAx BNTM writes x by y
> >
> > > Both of these inferences are true for any tape and any TM.

Since both the x's and the y's biject naturally with the members of N,
which member of N corresponds to the y for which you claim all x's have
been written?
> >
> > You just keep SAYING that.
> >
> > B is a fallacious inference.
>
> How? I read "for ALL x a Turing Machine writes x" to mean that for
> EACH x the TM writes position x, and for EVERY x the TM writes
> position x.
>
> This is just basic logic and basic language.
>
> karl m

But is not at all equivalent to what we have written and which you
object to without any logical basis.

The issue is to WHETHER the TM writes to each position but  on which
step it writes to which position.

As the process of writing has no last step, there is no step at which it
is complete.

To claim that something which is specially designed never to be complete
must be so anyway is incredibly dense of you.
```
 0
Virgil7091 (95)
7/28/2009 7:45:50 PM
```On Jul 28, 11:51=A0am, RussellE <reaste...@gmail.com> wrote:
> On Jul 28, 11:21=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 11:06=A0am, Kyle Easterly <keileene...@gmail.com> wrote:
>
> > > On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > We prove:
>
> > > > AnEs n is written at step s.
>
> > > No you don't. You can't.
>
> > Sure we do; we've shown you the informal version. Then a formal
> > version is as easy as putting the matter all in terms of exact
> > mathematics of Turing machines.
>
> The proof I saw uses induction.
> Induction is nothing more than the assumption
> Ax BNTM writes x.

Certain proofs use induction. Induction is not "the assumption Ax BNTM
writes to x".

> > > If you do, you also prove PA is inconsistent.
>
> > You just keep SAYING that.
>
> And I will keep saying it.

I don't doubt your ability to repeat yourself mindlessly.

> > > You can never show there are at least as
> > > many steps as positions.
>
> > You just keep SAYING such things.
>
> > > You assume Ax BNTM writes x and
> > > from this you imply AxEy writes x by step y.
>
> > No, that's not my argument.
>

And you've not shown anything incorrect in such use of induction.

> > > A) Ax BNTM writex x =3D> AxEy BNTM writes x by y
> > > B) Ax BNTM writes x =3D> EyAx BNTM writes x by y
>
> > > Both of these inferences are true for any tape and any TM.
>
> > You just keep SAYING that.
>
> I have given the proof many times.
> You know, the one about the set of blank positions.

And you've been shown the fallacies in your "proofs".

There is no sense in just keep coming back to repeat yourself.

> > B is a fallacious inference.
>
> So is A, if we are talking about infinite tapes.

There are no tapes involved in any actual proof in this regard,
infinite or otherwise. The notion of a 'tape' is metaphorical. I've
told you this several times. You just keep ignoring.

> > compulsion.
>
> You are probably right.
> They locked Cantor up, too.

They "locked him up"? Are you sure? Anyway, not on account of any
problems such as you display.

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 7:50:58 PM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 28, 11:21�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > On Jul 28, 11:06�am, Kyle Easterly <keileene...@gmail.com> wrote:
> >
> > > On Jul 28, 10:06�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > We prove:
> >
> > > > AnEs n is written at step s.
> >
> > > No you don't. You can't.
> >
> > Sure we do; we've shown you the informal version. Then a formal
> > version is as easy as putting the matter all in terms of exact
> > mathematics of Turing machines.
>
> The proof I saw uses induction.
> Induction is nothing more than the assumption
> Ax BNTM writes x.

The issue is not whether  a particular x gets written, but "when" it
gets written. If each step takes a a fixed minimum time, no matter how
often repeated, then there is never a time when all the x's have been
written but for each x there is a time at which it is written.

> And I will keep saying it.
> PA is provably inconsistent.

It is your internal logic which is inconsistent.
PA has been tested  and passed by far better minds than yours

>
> You are probably right.
> They locked Cantor up, too.

Nowdays, they let nuts like you stay free.
```
 0
Virgil7091 (95)
7/28/2009 7:52:51 PM
```On Jul 28, 12:36=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 28, 12:26=A0pm, Virgil <Vir...@home.esc> wrote:
>
>
>
> > In article
> > =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > On Jul 28, 10:06=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > On Jul 28, 9:41=A0am, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > > > explained that your expressions A and B referred to an expression=
of
> > > > > "each" position on the tape, and "every" position on the tape
> > > > > respectively, as the two necessary components to expressing "all"
> > > > > positions on the tape. =A0In other words, ALL is defined as EACH =
and
> > > > > EVERY.
>
> > > > The only things needes in this regard are the universal ('A') and
> > > > existential ('E') quantifiers, without getting bogged down in vagar=
ies
> > > > of English words:
>
> > > > We prove:
>
> > > > AnEs n is written at step s.
>
> > > > We do NOT prove:
>
> > > > EsAn n is written at step s.
>
> > > > Indeed, we prove:
>
> > > > ~EsAn n is written at step s.
>
> > > > MoeBlee
>
> > > I conclude that you have proof that there is no step s where EVERY
> > > position of the tape "is written" (all at once), and that there is
> > > always a step s where EACH position of the tape "is written" (one at =
a
> > > time).
>
> > > This contradicts the proof by induction that BNTM writes ALL position=
s
> > > of the tape, or the contrapositive, that there are no positions that
> > > BTMN doesn't write.
>
> > On the contrary, unless there is a last position on the tape, which
> > induction denies, you are wrong.
>
> I believe the proof is straight-forward.
>
> The TM writes to position 1 (by virtue of having started). =A0If you
> assume the TM has written position x, it is also true that it has
> written position x+1 (by the programming of the machine). =A0By the
> axiom of induction, it is true that the TM has written ALL x. =A0Not
> just EACH x but EVERY x.

Fine.

What that means is AxEy x is written at step y.

It does NOT mean EyAx x is written at step y.

If you cannot see the difference and that the second is fallacious,
then please get a good book on predicate calculus that explains the
difference between 'A' and 'E' and why they are cannot in general be
reversed in order (or just look at my informal counterexamples earlier

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 7:55:02 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> You changed the tense "to be" between A and B.  B should read:
>
> B. There is  a number of operations, y, after which each position x
> will be marked.
>
> I don't see why B is not satisfied by y = 1.

Among other reasons, positions 1 and  2 cannot both be marked after only
```
 0
Virgil7091 (95)
7/28/2009 8:03:58 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> > B. There is �a number of operations, y, after which each position x
> > will be marked.
>
> sorry, make that:
>
> B.  There is a number of operations, y, after which EVERY position x
> will be marked.

Not if there are more than y positions to be marked.
```
 0
Virgil7091 (95)
7/28/2009 8:06:02 PM
```
On Mon, 27 Jul 2009, MoeBlee wrote:

> On Jul 25, 8:23=A0am, Chris Menzel <cmen...@remove-this.tamu.edu> wrote:
> >Joseph Dauben's book _Georg Cantor: His
> > Mathematics and Philosophy of the Infinite_ (which gets a few things
> > wrong but is particularly good on the early developements)
>
> What does Dauben get wrong?
>
> MoeBlee

Will somebody finally tell me which numbers are most natural?
:-)=3D

Cheers, ZVK(Slavek)
```
 0
kovarik (7)
7/28/2009 8:08:35 PM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 27, 8:44�pm, Virgil <Vir...@home.esc> wrote:
> > In article
> > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> >
> >
> >
> >
> > > On Jul 27, 7:31�pm, Virgil <Vir...@home.esc> wrote:
> > > > In article
> >
> > > > �RussellE <reaste...@gmail.com> wrote:
> > > > > Most people seem to agree BNTM "never" actually
> > > > > writes to every position of an infinite tape.
> > > > > Why would you believe BNTM writes to
> > > > > every finite position "eventually"?
> >
> > > > There is a fine difference between eventually writing to each position
> > > > and eventually writing to every position.
> >
> > > > A TM may be capable of the former without being capable of the latter.
> >
> > > The definition I learned for all is "each and every". �Perhaps Russell
> > > has been failing to use all when he says every. �Can you elaborate?
> >
> > > karl m
> >
> > Given the TM whose only commands are:
> > � �1. Mark the tape then �go to command 2.
> > � �2. Move to next tape position then go to command 1.
> > consider the following statements:
> >
> > A: for each tape position, x, following the start position of the TM
> > there is a number of operations, y, �after which, x will be marked.
>
> So, you are saying there is a step y when AxEy is true.

No I'm not. I am saying that once given an x, THEN there is some step y
at (and after) which x will be marked. But  I am not saying anything
about what may transpire when that x is not given up front.
>
> > B: there is a number of operations, y, after which each position x that
> > gets marked is marked.
>
> Both of these statements are true if BNTM writes
> to every position of any tape.

Only if there is a last position on the tape. Which we do not assume. In
fact we assume the contrary, that there is no last position.
```
 0
Virgil7091 (95)
7/28/2009 8:10:22 PM
```On Jul 28, 12:23=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
> =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 28, 9:30=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > On Jul 28, 12:16=A0am, RussellE <reaste...@gmail.com> wrote:
>
> > > > On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> > > > > A: for each tape position, x, following the start position of the=
TM
> > > > > there is a number of operations, y, =A0after which, x will be mar=
ked.
>
> > > > So, you are saying there is a step y when AxEy is true.
>
> > > You don't even know what you're saying. 'y' is free in the first and
> > > bound in the second.
>
> > > > > B: there is a number of operations, y, after which each position =
x that
> > > > > gets marked is marked.
>
> > > > Both of these statements are true if BNTM writes
> > > > to every position of any tape.
>
> > > You're doing it again. Reverting back to ambiguous language after a
> > > more precise formulation has been given. That you CONTINUE to do that=
,
> > > even after it has been pointed out to you over and over, is childishl=
y
> > > dishonest.
>
> > > MoeBlee
>
> > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > explained that your expressions A and B referred to an expression of
> > "each" position on the tape, and "every" position on the tape
> > respectively, as the two necessary components to expressing "all"
> > positions on the tape. =A0In other words, ALL is defined as EACH and
> > EVERY.
>
> > karl m
>
> Those A and B statements are mine, not Moblee's. If you read them
> carefully enough you will see that A allows separate steps for separate
> marks but that B requires a single finite step finish all but finitely
> many marks, after which, no further steps are needed.

Yes, thank you.

when referring to them exactly as you explain above.

However, if as you introduced this post, A and B are meant to
distinguish EACH x from EVERY x, to constitute ALL x, I get a
contradiction with a proof by induction that both A and B are true.

There is no concept of a "last element" in applying the axiom of
induction to a premise and its basis, nor any concept of "time."

karl m

```
 0
malbrain (59)
7/28/2009 8:30:41 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 12:11�pm, Virgil <Vir...@home.esc> wrote:
> > In article
> > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> >
> >
> >
> >
> > > On Jul 28, 12:56�am, Virgil <Vir...@home.esc> wrote:
> > > > In article
> >
> > > > �RussellE <reaste...@gmail.com> wrote:
> > > > > On Jul 27, 8:44�pm, Virgil <Vir...@home.esc> wrote:
> > > > > > In article
> > > > > > <d6c9f7e2-1e80-4f00-aa81-920525b8b...@y10g2000prf.googlegroups.com>,
> > > > > > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > > > > > On Jul 27, 7:31�pm, Virgil <Vir...@home.esc> wrote:
> > > > > > > > In article
> > > > > > > > <728f4d92-4497-44d6-a19a-e88eae655...@m3g2000pri.googlegroups.co
> > > > > > > > m>,
> >
> > > > > > > > �RussellE <reaste...@gmail.com> wrote:
> > > > > > > > > Most people seem to agree BNTM "never" actually
> > > > > > > > > writes to every position of an infinite tape.
> > > > > > > > > Why would you believe BNTM writes to
> > > > > > > > > every finite position "eventually"?
> >
> > > > > > > > There is a fine difference between eventually writing to each
> > > > > > > > position
> > > > > > > > and eventually writing to every position.
> >
> > > > > > > > A TM may be capable of the former without being capable of the
> > > > > > > > latter.
> >
> > > > > > > The definition I learned for all is "each and every". �Perhaps
> > > > > > > Russell
> > > > > > > has been failing to use all when he says every. �Can you
> > > > > > > elaborate?
> >
> > > > > > > karl m
> >
> > > > > > Given the TM whose only commands are:
> > > > > > � �1. Mark the tape then �go to command 2.
> > > > > > � �2. Move to next tape position then go to command 1.
> > > > > > consider the following statements:
> >
> > > > > > A: for each tape position, x, following the start position of the
> > > > > > TM
> > > > > > there is a number of operations, y, �after which, x will be marked.
> >
> > > > > So, you are saying there is a step y when AxEy is true.
> >
> > > > > > B: there is a number of operations, y, after which each position x
> > > > > > that
> > > > > > gets marked is marked.
> >
> > > > > Both of these statements are true if BNTM writes
> > > > > to every position of any tape.
> >
> > > > Not if each operation takes some positive minimal amount of time.
> >
> > > > In that case, there will never be a time when all positions have been
> > > > marked though for each position there will be a time when it has been
> > > > marked.- Hide quoted text -
> >
> > > > - Show quoted text -
> >
> > > With induction I can prove that the BNTM writes to all positions of
> > > the tape. �"All positions" implies both each and every position, as
> > > expressed by A and B.
> >
> > What I said, and what is still true is:
> > If each step of my TM takes some fixed positive amount of time of there
> > will never be a time when all positions have been marked. But for each
> > position, there will be a time when it is marked.- Hide quoted text -
> >
> > - Show quoted text -
>
> We have a contradiction between your definition of the machine and the
> mathematics of induction that establishes that for ALL x the machine
> writes to position x.

writes to all positions does not mean that it writes to all of them
simultaneously. In each cycle of steps, it writes to only one more, so
there are at all times some that it has not yet written to.

Unless you define your machine so that it can execute an infinite number
of steps in a finite time interval. But even then, at any particular
step, it will have left infinitely many positions still to be written.
```
 0
Virgil7091 (95)
7/28/2009 8:37:51 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 12:26�pm, Virgil <Vir...@home.esc> wrote:
> > In article
> > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> >
> >
> >
> >
> > > On Jul 28, 10:06�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > On Jul 28, 9:41�am, Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > > > explained that your expressions A and B referred to an expression of
> > > > > "each" position on the tape, and "every" position on the tape
> > > > > respectively, as the two necessary components to expressing "all"
> > > > > positions on the tape. �In other words, ALL is defined as EACH and
> > > > > EVERY.
> >
> > > > The only things needes in this regard are the universal ('A') and
> > > > existential ('E') quantifiers, without getting bogged down in vagaries
> > > > of English words:
> >
> > > > We prove:
> >
> > > > AnEs n is written at step s.
> >
> > > > We do NOT prove:
> >
> > > > EsAn n is written at step s.
> >
> > > > Indeed, we prove:
> >
> > > > ~EsAn n is written at step s.
> >
> > > > MoeBlee
> >
> > > I conclude that you have proof that there is no step s where EVERY
> > > position of the tape "is written" (all at once), and that there is
> > > always a step s where EACH position of the tape "is written" (one at a
> > > time).
> >
> > > This contradicts the proof by induction that BNTM writes ALL positions
> > > of the tape, or the contrapositive, that there are no positions that
> > > BTMN doesn't write.
> >
> > On the contrary, unless there is a last position on the tape, which
> > induction denies, you are wrong.
>
> I believe the proof is straight-forward.
>
> The TM writes to position 1 (by virtue of having started).  If you
> assume the TM has written position x, it is also true that it has
> written position x+1 (by the programming of the machine).

Not unless the machine has stopped.  And it never stops.
As long as it is still running, there are unwritten positions.

>   By the
> axiom of induction, it is true that the TM has written ALL x.

By induction it is true that the TM WILL write to all x, but it cannot
have done so at any time  before the end of time.

Unless it is allowed to perform infinitely many steps in a finite time
interval.

Which is not standard for TMs.
```
 0
Virgil7091 (95)
7/28/2009 8:43:29 PM
```On Jul 28, 1:30=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> I get a
> contradiction with a proof by induction that both A and B are true.

From the mathematical formulation of 'Turing machine' in Z set theory,
using only first order logic with identity, you have not derived any

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 8:44:11 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 12:36�pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> > On Jul 28, 12:26�pm, Virgil <Vir...@home.esc> wrote:
> >
> >
> >
> >
> >
> > > In article
> > > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > > On Jul 28, 10:06�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > > On Jul 28, 9:41�am, Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > > > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > > > > explained that your expressions A and B referred to an expression of
> > > > > > "each" position on the tape, and "every" position on the tape
> > > > > > respectively, as the two necessary components to expressing "all"
> > > > > > positions on the tape. �In other words, ALL is defined as EACH and
> > > > > > EVERY.
> >
> > > > > The only things needes in this regard are the universal ('A') and
> > > > > existential ('E') quantifiers, without getting bogged down in vagaries
> > > > > of English words:
> >
> > > > > We prove:
> >
> > > > > AnEs n is written at step s.
> >
> > > > > We do NOT prove:
> >
> > > > > EsAn n is written at step s.
> >
> > > > > Indeed, we prove:
> >
> > > > > ~EsAn n is written at step s.
> >
> > > > > MoeBlee
> >
> > > > I conclude that you have proof that there is no step s where EVERY
> > > > position of the tape "is written" (all at once), and that there is
> > > > always a step s where EACH position of the tape "is written" (one at a
> > > > time).
> >
> > > > This contradicts the proof by induction that BNTM writes ALL positions
> > > > of the tape, or the contrapositive, that there are no positions that
> > > > BTMN doesn't write.
> >
> > > On the contrary, unless there is a last position on the tape, which
> > > induction denies, you are wrong.
> >
> > I believe the proof is straight-forward.
> >
> > The TM writes to position 1 (by virtue of having started). �If you
> > assume the TM has written position x, it is also true that it has
> > written position x+1 (by the programming of the machine). �By the
> > axiom of induction, it is true that the TM has written ALL x. �Not
> > just EACH x but EVERY x.
> >
> > karl m- Hide quoted text -
> >
> > - Show quoted text -
>
> Sorry, now I'm mixing tenses.  Please replace "has written" by
> "writes" in the above proof.

You assume that the TM in question has run to its end and has nothing
left to do, which would have require it to perform infinitely many
operations in finite time.
I, on the other hand, assume that any such TM will perform no more than
finitely many steps in any finite time interval, so that if such a TM
has ever been started, it must still be in operation, and there must
still be marks that it has yet to make.
```
 0
Virgil7091 (95)
7/28/2009 8:49:16 PM
```On Jul 28, 12:11=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
> =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 28, 12:56=A0am, Virgil <Vir...@home.esc> wrote:
> > > In article
>
> > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> > > > > In article
> > > > > <d6c9f7e2-1e80-4f00-aa81-920525b8b...@y10g2000prf.googlegroups.co=
m>,
> > > > > =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > > > On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> > > > > > > In article
> > > > > > > <728f4d92-4497-44d6-a19a-e88eae655...@m3g2000pri.googlegroups=
..com>,
>
> > > > > > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > > > > > Most people seem to agree BNTM "never" actually
> > > > > > > > writes to every position of an infinite tape.
> > > > > > > > Why would you believe BNTM writes to
> > > > > > > > every finite position "eventually"?
>
> > > > > > > There is a fine difference between eventually writing to each
> > > > > > > position
> > > > > > > and eventually writing to every position.
>
> > > > > > > A TM may be capable of the former without being capable of th=
e
> > > > > > > latter.
>
> > > > > > The definition I learned for all is "each and every". =A0Perhap=
s
> > > > > > Russell
> > > > > > has been failing to use all when he says every. =A0Can you elab=
orate?
>
> > > > > > karl m
>
> > > > > Given the TM whose only commands are:
> > > > > =A0 =A01. Mark the tape then =A0go to command 2.
> > > > > =A0 =A02. Move to next tape position then go to command 1.
> > > > > consider the following statements:
>
> > > > > A: for each tape position, x, following the start position of the=
TM
> > > > > there is a number of operations, y, =A0after which, x will be mar=
ked.
>
> > > > So, you are saying there is a step y when AxEy is true.
>
> > > > > B: there is a number of operations, y, after which each position =
x that
> > > > > gets marked is marked.
>
> > > > Both of these statements are true if BNTM writes
> > > > to every position of any tape.
>
> > > Not if each operation takes some positive minimal amount of time.
>
> > > In that case, there will never be a time when all positions have been
> > > marked though for each position there will be a time when it has been
> > > marked.- Hide quoted text -
>
> > > - Show quoted text -
>
> > With induction I can prove that the BNTM writes to all positions of
> > the tape. =A0"All positions" implies both each and every position, as
> > expressed by A and B.
>
> What I said, and what is still true is:
> If each step of my TM takes some fixed positive amount of time of there
> will never be a time when all positions have been marked.

Why do you keep introducing time?
I never assumed each operation takes a "fixed finite positive of
time".
In my original post I assumed a ZM can perform an infinite
number of operation in one second.

You are saying there is "never" a step y when EyAx is true
and never means not even eventually.

> But for each
> position, there will be a time when it is marked

There is "never" a step y when AxEy is true,
but there will be such a step y "eventually".

Either there is such a step y or there isn't.

Russell
- Never never means never in set theory

```
 0
reasterly (337)
7/28/2009 8:56:04 PM
```In article
Karl Malbrain <malbrain@yahoo.com> wrote:

> On Jul 28, 12:23�pm, Virgil <Vir...@home.esc> wrote:
> > In article
> > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> >
> >
> >
> >
> > > On Jul 28, 9:30�am, MoeBlee <jazzm...@hotmail.com> wrote:
> > > > On Jul 28, 12:16�am, RussellE <reaste...@gmail.com> wrote:
> >
> > > > > On Jul 27, 8:44�pm, Virgil <Vir...@home.esc> wrote:
> > > > > > A: for each tape position, x, following the start position of the
> > > > > > TM
> > > > > > there is a number of operations, y, �after which, x will be marked.
> >
> > > > > So, you are saying there is a step y when AxEy is true.
> >
> > > > You don't even know what you're saying. 'y' is free in the first and
> > > > bound in the second.
> >
> > > > > > B: there is a number of operations, y, after which each position x
> > > > > > that
> > > > > > gets marked is marked.
> >
> > > > > Both of these statements are true if BNTM writes
> > > > > to every position of any tape.
> >
> > > > You're doing it again. Reverting back to ambiguous language after a
> > > > more precise formulation has been given. That you CONTINUE to do that,
> > > > even after it has been pointed out to you over and over, is childishly
> > > > dishonest.
> >
> > > > MoeBlee
> >
> > > Perhaps I'm mis-interpreting something, but I thought that Virgil
> > > explained that your expressions A and B referred to an expression of
> > > "each" position on the tape, and "every" position on the tape
> > > respectively, as the two necessary components to expressing "all"
> > > positions on the tape. �In other words, ALL is defined as EACH and
> > > EVERY.
> >
> > > karl m
> >
> > Those A and B statements are mine, not Moblee's. If you read them
> > carefully enough you will see that A allows separate steps for separate
> > marks but that B requires a single finite step finish all but finitely
> > many marks, after which, no further steps are needed.
>
> Yes, thank you.
>
>  I read your A and B carefully, and introduced parenthetical remarks
> when referring to them exactly as you explain above.
>
> However, if as you introduced this post, A and B are meant to
> distinguish EACH x from EVERY x, to constitute ALL x, I get a
> contradiction with a proof by induction that both A and B are true.
>
> There is no concept of a "last element" in applying the axiom of
> induction to a premise and its basis, nor any concept of "time."

There is a concept of sequence, which serves the same purpose. And a TM
that proceeds through a sequence term by term, as the TM in question
must do, cannot reach any term prior to having accessed all prior terms.

But "B: EyAx by step y position x is marked", requires taking terms of
the sequence out of order, and marking all terms of the sequence before
actually getting to them.
```
 0
Virgil7091 (95)
7/28/2009 8:56:42 PM
```In article
RussellE <reasterly@gmail.com> wrote:

> On Jul 28, 12:11�pm, Virgil <Vir...@home.esc> wrote:
> > In article
> > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> >
> >
> >
> >
> > > On Jul 28, 12:56�am, Virgil <Vir...@home.esc> wrote:
> > > > In article
> >
> > > > �RussellE <reaste...@gmail.com> wrote:
> > > > > On Jul 27, 8:44�pm, Virgil <Vir...@home.esc> wrote:
> > > > > > In article
> > > > > > <d6c9f7e2-1e80-4f00-aa81-920525b8b...@y10g2000prf.googlegroups.com>,
> > > > > > �Karl Malbrain <malbr...@yahoo.com> wrote:
> >
> > > > > > > On Jul 27, 7:31�pm, Virgil <Vir...@home.esc> wrote:
> > > > > > > > In article
> > > > > > > > <728f4d92-4497-44d6-a19a-e88eae655...@m3g2000pri.googlegroups.co
> > > > > > > > m>,
> >
> > > > > > > > �RussellE <reaste...@gmail.com> wrote:
> > > > > > > > > Most people seem to agree BNTM "never" actually
> > > > > > > > > writes to every position of an infinite tape.
> > > > > > > > > Why would you believe BNTM writes to
> > > > > > > > > every finite position "eventually"?
> >
> > > > > > > > There is a fine difference between eventually writing to each
> > > > > > > > position
> > > > > > > > and eventually writing to every position.
> >
> > > > > > > > A TM may be capable of the former without being capable of the
> > > > > > > > latter.
> >
> > > > > > > The definition I learned for all is "each and every". �Perhaps
> > > > > > > Russell
> > > > > > > has been failing to use all when he says every. �Can you
> > > > > > > elaborate?
> >
> > > > > > > karl m
> >
> > > > > > Given the TM whose only commands are:
> > > > > > � �1. Mark the tape then �go to command 2.
> > > > > > � �2. Move to next tape position then go to command 1.
> > > > > > consider the following statements:
> >
> > > > > > A: for each tape position, x, following the start position of the
> > > > > > TM
> > > > > > there is a number of operations, y, �after which, x will be marked.
> >
> > > > > So, you are saying there is a step y when AxEy is true.
> >
> > > > > > B: there is a number of operations, y, after which each position x
> > > > > > that
> > > > > > gets marked is marked.
> >
> > > > > Both of these statements are true if BNTM writes
> > > > > to every position of any tape.
> >
> > > > Not if each operation takes some positive minimal amount of time.
> >
> > > > In that case, there will never be a time when all positions have been
> > > > marked though for each position there will be a time when it has been
> > > > marked.- Hide quoted text -
> >
> > > > - Show quoted text -
> >
> > > With induction I can prove that the BNTM writes to all positions of
> > > the tape. �"All positions" implies both each and every position, as
> > > expressed by A and B.
> >
> > What I said, and what is still true is:
> > If each step of my TM takes some fixed positive amount of time of there
> > will never be a time when all positions have been marked.
>
> Why do you keep introducing time?
> I never assumed each operation takes a "fixed finite positive of
> time".
> In my original post I assumed a ZM can perform an infinite
> number of operation in one second.
>
> You are saying there is "never" a step y when EyAx is true
> and never means not even eventually.
>
> > But for each
> > position, there will be a time when it is marked
>
> There is "never" a step y when AxEy is true,
> but there will be such a step y "eventually".
>
> Either there is such a step y or there isn't.

Is there any STEP y for which there is no STEP y+1?
Not according to my definition of a TM, however fast it goes.

If you don't like time, then number the cycles of the TM and the marks
it makes with successive members of N.

"There is a natural, m , such that for all naturals , n, m >= n."
But in the ZF worldm for every m in N there is an n > m.
E.g.,  For each m in N,  m+1 > m.
```
 0
Virgil7091 (95)
7/28/2009 9:07:45 PM
```On Jul 28, 1:10=A0pm, Virgil <Vir...@home.esc> wrote:
> In article

> > So, you are saying there is a step y when AxEy is true.
>
> No I'm not.

So, you agree there is never such a step y.

> I am saying that once given an x,

What makes you think I am givnig you an x?
I have already explained x is "uncomputable".
You are the one claiming AxEy is true.
You have to prove this statement is true for all x.
I only have to prove it is false for some finite x.
I don't have to tell you what x is.
I only have to prove x is finite.

> THEN there is some step y
> at (and after) which x will be marked. But =A0I am not saying anything
> about what may transpire when that x is not given up front.
>
>
>
> > > B: there is a number of operations, y, after which each position x th=
at
> > > gets marked is marked.
>
> > Both of these statements are true if BNTM writes
> > to every position of any tape.
>
> Only if there is a last position on the tape. Which we do not assume. In
> fact we assume the contrary, that there is no last position

Yes, you ASSUME there is no last position.
If BNTM writes to every position of a tape, we can
prove there is a last position written by BNTM.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/28/2009 9:17:10 PM
```On Jul 28, 12:50=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 11:51=A0am, RussellE <reaste...@gmail.com> wrote:

> > They locked Cantor up, too.
>
> They "locked him up"? Are you sure?

Cantor suffered from depression.
He was placed in a sanitorium several times.
http://en.wikipedia.org/wiki/Georg_Cantor

> Anyway, not on account of any
> problems such as you display.

On at least one occassion, Cantor went into
a depression because of public attacks on his
theories.

Cantor also believed his theories were given
to him by God. He even met with the Pope.

I am sure some people are conviced my ideas
were given to me by Satan.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/28/2009 9:36:42 PM
```On Jul 28, 2:36=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 28, 12:50=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 11:51=A0am, RussellE <reaste...@gmail.com> wrote:
> > > They locked Cantor up, too.
>
> > They "locked him up"? Are you sure?
>
> Cantor suffered from depression.
> He was placed in a sanitorium several times.

Being placed in a sanitorium for depression is not in itself being
"locked up". (It's possible Cantor was locked up.)

> > Anyway, not on account of any
> > problems such as you display.
>
> On at least one occassion, Cantor went into
> a depression because of public attacks on his
> theories.

I didn't know you go into depressions on account of attacks on the
nonsense you post. Anyway, I wasn't even counting that as among your
cognitive problems.

> Cantor also believed his theories were given
> to him by God. He even met with the Pope.

Cantor met with the Pope? Cantor wrote at least one letter to the
Pope. What is your source that Cantor MET with the Pope?

Also, Cantor believed that his mathematics were in accord with certain
theology and even that such mathematics expressed certain things
regarding theology. But did Cantor say that his work was "given to him
by God" in some sense of divine revelation beyond what ordinarily a
pious Christian of his time would say about God's role? (Maybe he
did.)

Anyway, that Cantor saw mathematics in certain religious ways is well
known, and well discussed both in Dauben's and Hallett's books.
However, none of the material that Cantor developed and that has been
incorporated into formal set theory requires any theological
convictions.

> I am sure some people are conviced my ideas
> were given to me by Satan.

Why are you sure of such a thing? Has anyone ever even suggested such
a thing to you?

MoeBlee
```
 0
jazzmobe (307)
7/28/2009 9:56:10 PM
```On Jul 27, 8:44=A0pm, Virgil <Vir...@home.esc> wrote:
> In article
> =A0Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 27, 7:31=A0pm, Virgil <Vir...@home.esc> wrote:
> > > In article
>
> > > =A0RussellE <reaste...@gmail.com> wrote:
> > > > Most people seem to agree BNTM "never" actually
> > > > writes to every position of an infinite tape.
> > > > Why would you believe BNTM writes to
> > > > every finite position "eventually"?
>
> > > There is a fine difference between eventually writing to each positio=
n
> > > and eventually writing to every position.
>
> > > A TM may be capable of the former without being capable of the latter=
..
>
> > The definition I learned for all is "each and every". =A0Perhaps Russel=
l
> > has been failing to use all when he says every. =A0Can you elaborate?
>
> > karl m
>
> Given the TM whose only commands are:
> =A0 =A01. Mark the tape then =A0go to command 2.
> =A0 =A02. Move to next tape position then go to command 1.
> consider the following statements:
>
> A: for each tape position, x, following the start position of the TM
> there is a number of operations, y, =A0after which, x will be marked.
>
> B: there is a number of operations, y, after which each position x that
> gets marked is marked.

Can you elaborate further on the "subtle difference" that allows the
TM to write to each position, yet disallows the TM to write to every
position, under the induction result that the TM writes to all
positions?

karl m
```
 0
malbrain (59)
7/28/2009 11:01:48 PM
```On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:

> Can you elaborate further

AxEy x is written at step y

and

EyAx x is written at step y

You TRULY do not understand the difference between those?

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 11:05:45 PM
```On Jul 28, 4:05=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > Can you elaborate further
>
> AxEy x is written at step y
>
> and
>
> EyAx x is written at step y
>
> You TRULY do not understand the difference between those?
>
> MoeBlee

I know the difference between them in general.  The first one says
each x is written at some step, the second one says all are never
written at any step.

I don't see the connection to a difference between each and every as
developed in "for all x, TM writes x".

karl m
```
 0
malbrain (59)
7/28/2009 11:19:04 PM
```On Jul 28, 12:50=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 11:51=A0am, RussellE <reaste...@gmail.com> wrote:

> > > Sure we do; we've shown you the informal version. Then a formal
> > > version is as easy as putting the matter all in terms of exact
> > > mathematics of Turing machines.
>
> > The proof I saw uses induction.
> > Induction is nothing more than the assumption
> > Ax BNTM writes x.
>
> Certain proofs use induction. Induction is not "the assumption Ax BNTM
> writes to x".

Induction is the assumption if P is true for n implies
P is true for n+1 then P is true for ALL n.
AxEy is exactly what you are suppose to prove.
You don't get to asume it is true.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/28/2009 11:22:28 PM
```On Jul 28, 4:19=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 28, 4:05=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > Can you elaborate further
>
> > AxEy x is written at step y
>
> > and
>
> > EyAx x is written at step y
>
> > You TRULY do not understand the difference between those?

> I know the difference between them in general. =A0The first one says
> each x is written at some step, the second one says all are never
> written at any step.

No! Wrong! Come on, please, are you REALLY so severely uninformed or
are you just playing as if you are?

> I don't see the connection to a difference between each and every as
> developed in "for all x, TM writes x".

Are you just playing that your are as obtuse as a bar of lead?

MoeBlee

```
 0
jazzmobe (307)
7/28/2009 11:26:17 PM
```On Jul 28, 4:26=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 4:19=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
>
>
>
>
> > On Jul 28, 4:05=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > Can you elaborate further
>
> > > AxEy x is written at step y
>
> > > and
>
> > > EyAx x is written at step y
>
> > > You TRULY do not understand the difference between those?
> > I know the difference between them in general. =A0The first one says
> > each x is written at some step, the second one says all are never
> > written at any step.
>
> No! Wrong! Come on, please, are you REALLY so severely uninformed or
> are you just playing as if you are?

Sorry, you're right, I expressed the negation of the second
expression.

karl m
```
 0
malbrain (59)
7/28/2009 11:29:39 PM
```On Jul 28, 1:08=A0pm, "Zdislav V. Kovarik" <kova...@mcmaster.ca> wrote:

> Will somebody finally tell me which numbers are most natural?
> :-)=3D

It appears we need a consistent
set theory with a largest number.
Maybe you can whip one up.

Russell
- The universe is one dimensional
```
 0
reasterly (337)
7/28/2009 11:49:07 PM
```On Jul 28, 2:56=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 2:36=A0pm, RussellE <reaste...@gmail.com> wrote:

> > Cantor also believed his theories were given
> > to him by God. He even met with the Pope.
>
> Cantor met with the Pope? Cantor wrote at least one letter to the
> Pope. What is your source that Cantor MET with the Pope?

I thought I read it in an article in Scientific American.
It is quite possible I mis-remember the article, but
I think it said he met with the Pope to discuss his
ideas on infinity. The Catholic Church was interested
in his theories.

> Also, Cantor believed that his mathematics were in accord with certain
> theology and even that such mathematics expressed certain things
> regarding theology. But did Cantor say that his work was "given to him
> by God" in some sense of divine revelation beyond what ordinarily a
> pious Christian of his time would say about God's role? (Maybe he
> did.)

Not that I know of. But, he was very pious.
I would think he saw his theories as divine revelation.

> Anyway, that Cantor saw mathematics in certain religious ways is well
> known, and well discussed both in Dauben's and Hallett's books.
> However, none of the material that Cantor developed and that has been
> incorporated into formal set theory requires any theological
> convictions.

Says who.

> > I am sure some people are conviced my ideas
> > were given to me by Satan.
>
> Why are you sure of such a thing?

The mob with torches and pitchforks outside
my door.

> Has anyone ever even suggested such
> a thing to you?

Not yet.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/28/2009 11:59:26 PM
```On Jul 28, 4:29=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> On Jul 28, 4:26=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
>
>
> > On Jul 28, 4:19=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > On Jul 28, 4:05=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > > On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
>
> > > > > Can you elaborate further
>
> > > > AxEy x is written at step y
>
> > > > and
>
> > > > EyAx x is written at step y
>
> > > > You TRULY do not understand the difference between those?
> > > I know the difference between them in general. =A0The first one says
> > > each x is written at some step, the second one says all are never
> > > written at any step.
>
> > No! Wrong! Come on, please, are you REALLY so severely uninformed or
> > are you just playing as if you are?
>
> Sorry, you're right, I expressed the negation of the second
> expression.

Actually you expressed:

Ax~Ey x is written at step y.

Anyway, so now I hope you see that there is a difference between

AxEy x is written at step y
and
EyAx x is written at step y

and it's the first one that holds, and that is not contradictory.

MoeBlee
```
 0
jazzmobe (307)
7/29/2009 12:01:06 AM
```On Jul 28, 4:05=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:

> AxEy x is written at step y
>
> and
>
> EyAx x is written at step y
>
> You TRULY do not understand the difference between those?

I understand the difference.
I am saying both statements are implied by
Ax BNTM writes x

We both agree EyAx is false. You keep saying AxEy is true.
If one statement is true, they both must be true.

So far, you haven't given a coherent proof of AxEy.
All I can make of your argument is that there
will be a step y when AxEy is true "eventually".
There will never exist such a y.

Russell
- 2 many 2 count
```
 0
reasterly (337)
7/29/2009 12:13:01 AM
```=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=
=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=
=E0=B1=87Musatov wrote: P iff P is induced. It exists a
matter of truth. Truth exists outside recognition. No matter what is
'true' to you, it may or may not be actua"ly true. As a matter
undeniable: truth exists, whether we are wrong or right. Mmm.

RussellE wrote:
=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=
=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=
=E0=B1=87> On Jul 28, 2:56=C2=A0pm, MoeBlee
<jazzm...@hotmail.com> wrote:
> > On Jul 28, =E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=
=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=
=B1=8D=E0=B0=B2=E0=B1=872:36=C2=A0pm, RussellE <reaste...@gmail.com> wrote:
>
> > > Cantor also believed his theories were given
> > > to him by God. He even met with the Pope.
> >
> > Cantor met with the Pope? Cantor wrote at least one letter to the
> > Pope. What is your source that Cantor MET with the Pope?
=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=
=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=
=E0=B1=87=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=
=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=E0=B0=B2=E0=B1=87 =C2=A0=E0=B0=95=E0=B1=8D=
=E0=B0=B2=E0=B1=87>
> I thought I read it in an article in Scientific American.
> It is quite possible I mis-remember the article, but
> I think it said he met with the Pope to discuss his
> ideas on infinity. The Catholic Church was interested
> in his theories.
>
> > Also, Cantor believed that his mathematics were in accord with certain
> > theology and even that such mathematics expressed certain things
> > regarding theology. But did Cantor say that his work was "given to him
> > by God" in some sense of divine revelation beyond what ordinarily a
> > pious Christian of his time would say about God's role? (Maybe he
> > did.)
>
> Not that I know of. But, he was very pious.
> I would think he saw his theories as divine revelation.
>
> > Anyway, that Cantor saw mathematics in certain religious ways is well
> > known, and well discussed both in Dauben's and Hallett's books.
> > However, none of the material that Cantor developed and that has been
> > incorporated into formal set theory requires any theological
> > convictions.
>
> Says who.
>
> > > I am sure some people are conviced my ideas
> > > were given to me by Satan.
> >
> > Why are you sure of such a thing?
>
> The mob with torches and pitchforks outside
> my door.
>
> > Has anyone ever even suggested such
> > a thing to you?
>
> Not yet.
>
>
> Russell
> - 2 many 2 count
```
 0
marty.musatov (1143)
7/29/2009 12:14:05 AM
```On Jul 28, 5:13=A0pm, RussellE <reaste...@gmail.com> wrote:
> On Jul 28, 4:05=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> > AxEy x is written at step y
>
> > and
>
> > EyAx x is written at step y
>
> > You TRULY do not understand the difference between those?
>
> I understand the difference.
> I am saying both statements are implied by
> Ax BNTM writes x

Then you do NOT understand the difference.

> We both agree EyAx is false. You keep saying AxEy is true.
> If one statement is true, they both must be true.

As I said, you don't understand the difference.

> So far, you haven't given a coherent proof of AxEy.

Of course we have - as an informal rendering of the actual mathematics
of Turing machines.

> All I can make of your argument is that there
> will be a step y when AxEy is true "eventually".

That very construction - "a step y when AxEy is true" - AGAIN
illustrates that you don't understand quantification ITSELF.

And my argument does NOT use any such word "eventually".

> There will never exist such a y.

By saying "such a y" you AGAIN you show you don't understand the
quantifiers.

As I said, you prefer ignorance and misunderstanding to knowledge;

MoeBlee
```
 0
jazzmobe (307)
7/29/2009 12:27:58 AM
```Numbers wrote:
MoeBlee wrote:
> On Jul 28, 5:13=A0pm, RussellE <reaste...@gmail.com> wrote:
> > On Jul 28, 4:05=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> >
> > > On Jul 28, 4:01=A0pm, Karl Malbrain <malbr...@yahoo.com> wrote:
> > > AxEy x is written at step y
> >
> > > and
> >
> > > EyAx x is written at step y
> >
> > > You TRULY do not understand the difference between those?
> >
> > I understand the difference.
> > I am saying both statements are implied by
> > Ax BNTM writes x
>
> Then you do NOT understand the difference.
>
> > We both agree EyAx is false. You keep saying AxEy is true.
> > If one statement is true, they both must be true.
>
> As I said, you don't understand the difference.
>
> > So far, you haven't given a coherent proof of AxEy.
>
> Of course we have - as an informal rendering of the actual mathematics
> of Turing machines.
>
> > All I can make of your argument is that there
> > will be a step y when AxEy is true "eventually".
>
> That very construction - "a step y when AxEy is true" - AGAIN
> illustrates that you don't understand quantification ITSELF.
>
> And my argument does NOT use any such word "eventually".
>
> > There will never exist such a y.
>
> By saying "such a y" you AGAIN you show you don't understand the
> quantifiers.
>
> As I said, you prefer ignorance and misunderstanding to knowledge;
> that is a problem of yours I can't help you with.
>
> MoeBlee
V
from comprehension it follows Ex. ... Proof: axiom of membership. ...
to form sets in ... Theorem: AxEy( y=3D{x,V} ) Proof: pairing. Theorem:
Ax: ~{x,V}eV ...

Q.E.D.
Musatov
(Aka) Numbers
```
 0
marty.musatov (1143)
7/29/2009 12:37:40 AM
```On Jul 28, 5:27=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On Jul 28, 5:13=A0pm, RussellE <reaste...@gmail.com> wrote:

> > > AxEy x is written at step y
>
> > > and
>
> > > EyAx x is written at step y
>
> > > You TRULY do not understand the difference between those?
>
> > I understand the difference.
> > I am saying both statements are implied by
> > Ax BNTM writes x
>
> Then you do NOT understand the difference.
>
> > We both agree EyAx is false. You keep saying AxEy is true.
> > If one statement is true, they both must be true.
>
> As I said, you don't understand the difference.
>
> > So far, you haven't given a coherent proof of AxEy.
>
> Of course we have - as an informal rendering of the actual mathematics
> of Turing machines.

You have shown if we assume AxEy we can "prove" AxEy.

> > All I can make of your argument is that there
> > will be a step y when AxEy is true "eventually".
>
> That very construction - "a step y when AxEy is true" - AGAIN
> illustrates that you don't understand quantification ITSELF.
>
> And my argument does NOT use any such word "eventually".

I have already shown a "formal" TM where AxEy is provably false.
I add one blank cell to the end of the tape after each step,

Ax is well defined for each step.
We know the length of the tape after each step.
We also know how many steps BNTM has performed.

After the first step, x=3D101 and y=3D1. AxEy is provably false.
AxEy is provably false for every step.
Why do you keep insisting AxEy is true?

> > There will never exist such a y.
>
> By saying "such a y" you AGAIN you show you don't understand the
> quantifiers.
>
> As I said, you prefer ignorance and misunderstanding to knowledge;
> that is a problem of yours I can't help you with.

Divine revelation is a wonderful thing.

Russell
- 2 many 2 count

```
 0
reasterly (337)
7/29/2009 1:52:59 AM
```On 2009-07-29, RussellE <reasterly@gmail.com> wrote:
> After the first step, x=101 and y=1. AxEy is provably false.

This demonstrates that you don't even have the slightest idea what the
quantifier construction "AxEy" means.

After you learn that, you then need to learn how to correctly use it
in proofs.  This is just basic logic, not even starting on the
mathematics.  However, you won't even attempt to learn either of those
as you would```