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computers / comp.ai.philosophy / Re:_My_honest_reviewers:_André,_Ben,_Mike,_Dennis,_Richard_[_previously_undetected_semantic_incoherence_]

SubjectAuthor
* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ keyolcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
|`* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
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| | `* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
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| `- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
|`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
|`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
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+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
|`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
+* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
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|`* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
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| |`* Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
| | `* Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| |  +* Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
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| |  ||  `- Re: _My_honest_reviewers:_André,_Ben,_Mike,olcott
| |  |`- Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| |  `- Re: _My_Dishonest_reviewers:_André,_Ben,_Mikeolcott
| `- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott
`- Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [olcott

Pages:123
Subject: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.math, sci.logic
Followup: comp.theory
Date: Sun, 3 Apr 2022 17:01 UTC
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From: NoO...@NoWhere.com (olcott)
Subject: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key
missing piece in dialogue ]
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KEY MISSING PIECE IN THE DIALOGUE (that all reviewers get incorrectly)

The key missing piece in all of these dialogues is 100% perfectly and exactly does it mean for a halt decider to compute the mapping from its input finite strings to its own final states on the basis of the actual   behavior actually specified by this finite string pair.

Halting problem undecidability and infinitely nested simulation (V4)
https://www.researchgate.net/publication/359349179_Halting_problem_undecidability_and_infinitely_nested_simulation_V4 --
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.math, sci.logic
Followup: comp.theory
Date: Sun, 3 Apr 2022 20:26 UTC
References: 1 2
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
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On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

The key missing piece in all of these dialogues is 100% perfectly and
exactly does it mean for a halt decider to compute the mapping from
its input finite strings to its own final states on the basis of the
actual behavior actually specified by this finite string pair.

You certainly have trouble understanding this so I will repeat my offer
to take you through a series of student exercises that I am confident
(provided you approach them with an open mind) will lead you to fully
understanding what this means.


OK let's give it a shot. I know that I must be correct yet when you try to explain these things using your words then this will probably result in much greater mutual understanding, thus worth the effort. After we do this I can explain my ideas using your own words.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sun, 3 Apr 2022 23:38 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
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On 4/3/2022 5:57 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/3/2022 5:07 PM, olcott wrote:
On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

The key missing piece in all of these dialogues is 100% perfectly and
exactly does it mean for a halt decider to compute the mapping from
its input finite strings to its own final states on the basis of the
actual behavior actually specified by this finite string pair.
You certainly have trouble understanding this so I will repeat my offer
to take you through a series of student exercises that I am confident
(provided you approach them with an open mind) will lead you to fully
understanding what this means.

OK let's give it a shot.

Great!  Let's start with something that will force a lot of hidden
assumptions to be made explicit.

Q: Is the set of even numbers decidable?  If so, specify the TM (call it
     E) using Linz's notation (extended if you like).  If not, why not?

(This may be slow to get started, because there is a lot low-level
things to iron out first.)

That depends on how you define your terms. Any element of the set of
integers can be determined to be divisible by 2 with or without a remainder.
Can every element of this set be enumerated in finite time?
No. Can the set of all even numbers be defined in finite space?
Yes through algorithmic compression.

I forgot decidability is merely a membership algorithm, so yes.

(Didn't see you'd moved on.  Ignore my last reply.)

I would say no.  And for the reason you keep giving: any TM that decides
membership can't have a number as input.  TMs accept or reject strings,
not numbers.

But surely the iseven(n) function is computable, so how do you think we
should resolve this?  Hint: think encoding.

Can you have a stab at specifying E now using Linz's notation?


I want to very thoroughly go through all of the points completely as efficiently as possible. Because a C function could do this as a pure function of its ASCII digit sequence inputs by merely examining the last digits for: {0,2,4,8} we know that E is decidable.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, sci.logic, sci.math, comp.ai.philosophy
Followup: comp.theory
Date: Mon, 4 Apr 2022 22:32 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
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From: NoO...@NoWhere.com (olcott)
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On 4/4/2022 5:23 PM, wij wrote:
On Tuesday, 5 April 2022 at 04:47:42 UTC+8, olcott wrote:
On 4/4/2022 3:36 PM, olcott wrote:
On 4/4/2022 2:51 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/4/2022 11:23 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/4/2022 10:32 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/4/2022 5:14 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/3/2022 8:14 PM, Ben Bacarisse wrote:

It might be time to skip ahead because the next exercise is to
do the
same for P, a TM that decides if a string encodes a prime
number.  Can
you think of how to specify that without giving an algorithm?
          P.q0 ??? ⊦* P.qy    if ???
          P.q0 ??? ⊦* P.qn    otherwise.
(The three ??? won't all be the same things.)  Any idea how to
flesh
this out?  If you can, you will be able to do it for E very
easily too.

P.q0 S ⊦* P.qy    if Is-Prime-Number(S)
P.q0 S ⊦* P.qn    otherwise.
That's getting close.  We know, from how the notation works,
that S is a
string of symbols in the TM's alphabet.  But writing
Is-Prime-Number(S)
suggests that is not natural language.  That's a very hard route
to go
down.  I'd have to ask you for the definition of Is-Prime-Number.
Defining it symbolically is messy and if the definition is /not/
formal,
dressing the definition up with a formal-looking name is just
superficial.

It goes through some tedious steps to see if it is a prime number:

A prime number is a natural number greater than 1 that is not a
product of two smaller natural numbers.
We've hit a bit of a road-block rather sooner that I had expected.
First off, there's no need to define what a prime number is.  If
at some
point it turns out that your readers do not know, go ahead and define
it, but it's too widely understood by comp.theory readers to bother
about.
But writing (as I think you are suggesting)
      P.q0 S ⊦* P.qy    it goes through some tedious steps to see
if it is a
                        prime number
is not really adequate.  There are two 'it'.  To what do they refer?

You told me to make sure that I do not provide an algorithm.
Yes, that good.  You didn't.
Now, to what do the two 'it's refer?

OK, maybe things have gone too far already, but why are you ignoring my
questions?  Your phrase used 'it' twice.  What did you intend to refer
to by these two pronouns?  It was not an idle question.  I think when
you answer it, at least one problem will become clear.

This is somewhat algorithmic:
No, no.  The non-algorithmic way is best.  You should be able to
specific what a computation does even when yo have no idea how to write
the an algorithm to do it.  Sometimes there isn't ad algorithm!

and nothing in P.q0 S ⊦* P.qy is a number so there is nothing
there to
be a prime number.
Can you see how you (a) make it shorter, (b) make it clearer?
My reply has three questions in it (depending on how you count) but you
didn't answer any of them.  This will only work if you try to answer
these questions.  Sometimes the answer will be "I don't know what you
mean", but that's a perfectly good answer.

Anything besides the bare function name is somewhat algorithmic so
what you are asking for seems impossible.

Here's a supporting exercise.  Write down at least half a dozen strings
that might be passed to P.  At least one of them should be a string that
P must accept and at least one must be a string the P should reject, but
you should say, for each one, whether P accepts of rejects it.

Be prepared for me to raise questions about what the strings represent.
It's easy to assume conventions from everyday life that should be stated
explicitly.  You must have come across this in software: "the manual
said the input should be a number but it went wrong for सहस्र."

Finally, don't fuss about the prime bit.  Just use the word prime.
Everyone one knows what it means.  The key thing here is to state /what/
must be prime for P to correctly accept a string.


I am estimating that we will finally achieve closure on this.
Creating a common language between us will achieve the basis for mutual
understanding.

The process that we are doing looks like it will be effective on the
basis of eliminating all hidden assumptions.
-- Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

No use. Even all the people you find agree with you is still useless. To claim
you solved the HP problem, you have to show ALL your H first. People cannot
judge or teach 'claims'. But your P already showed wrong, no need to publish H.


My analysis is based on this model: If "an X is a Y" and Z says that "an X is a Y" then anything in the universe that disagrees is necessarily incorrect.

"an X is a Y" =
The input to embedded_H specifies a non-halting sequence of configurations. (input is non-halting)

Z says that "an X is a Y" =
embedded_H rejects its input.

If you can think of a case where
"an X is a Y" is simultaneously true and false then you have a rebuttal, otherwise not so much.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
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From: NoO...@NoWhere.com (olcott)
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On 4/4/2022 5:41 PM, wij wrote:
On Tuesday, 5 April 2022 at 06:32:34 UTC+8, olcott wrote:
On 4/4/2022 5:23 PM, wij wrote:
On Tuesday, 5 April 2022 at 04:47:42 UTC+8, olcott wrote:
On 4/4/2022 3:36 PM, olcott wrote:
On 4/4/2022 2:51 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/4/2022 11:23 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/4/2022 10:32 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/4/2022 5:14 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/3/2022 8:14 PM, Ben Bacarisse wrote:

It might be time to skip ahead because the next exercise is to
do the
same for P, a TM that decides if a string encodes a prime
number. Can
you think of how to specify that without giving an algorithm?
P.q0 ??? ⊦* P.qy if ???
P.q0 ??? ⊦* P.qn otherwise.
(The three ??? won't all be the same things.) Any idea how to
flesh
this out? If you can, you will be able to do it for E very
easily too.

P.q0 S ⊦* P.qy if Is-Prime-Number(S)
P.q0 S ⊦* P.qn otherwise.
That's getting close. We know, from how the notation works,
that S is a
string of symbols in the TM's alphabet. But writing
Is-Prime-Number(S)
suggests that is not natural language. That's a very hard route
to go
down. I'd have to ask you for the definition of Is-Prime-Number.
Defining it symbolically is messy and if the definition is /not/
formal,
dressing the definition up with a formal-looking name is just
superficial.

It goes through some tedious steps to see if it is a prime number:

A prime number is a natural number greater than 1 that is not a
product of two smaller natural numbers.
We've hit a bit of a road-block rather sooner that I had expected.
First off, there's no need to define what a prime number is. If
at some
point it turns out that your readers do not know, go ahead and define
it, but it's too widely understood by comp.theory readers to bother
about.
But writing (as I think you are suggesting)
P.q0 S ⊦* P.qy it goes through some tedious steps to see
if it is a
prime number
is not really adequate. There are two 'it'. To what do they refer?

You told me to make sure that I do not provide an algorithm.
Yes, that good. You didn't.
Now, to what do the two 'it's refer?

OK, maybe things have gone too far already, but why are you ignoring my
questions? Your phrase used 'it' twice. What did you intend to refer
to by these two pronouns? It was not an idle question. I think when
you answer it, at least one problem will become clear.

This is somewhat algorithmic:
No, no. The non-algorithmic way is best. You should be able to
specific what a computation does even when yo have no idea how to write
the an algorithm to do it. Sometimes there isn't ad algorithm!

and nothing in P.q0 S ⊦* P.qy is a number so there is nothing
there to
be a prime number.
Can you see how you (a) make it shorter, (b) make it clearer?
My reply has three questions in it (depending on how you count) but you
didn't answer any of them. This will only work if you try to answer
these questions. Sometimes the answer will be "I don't know what you
mean", but that's a perfectly good answer.

Anything besides the bare function name is somewhat algorithmic so
what you are asking for seems impossible.

Here's a supporting exercise. Write down at least half a dozen strings
that might be passed to P. At least one of them should be a string that
P must accept and at least one must be a string the P should reject, but
you should say, for each one, whether P accepts of rejects it.

Be prepared for me to raise questions about what the strings represent.
It's easy to assume conventions from everyday life that should be stated
explicitly. You must have come across this in software: "the manual
said the input should be a number but it went wrong for सहस्र."

Finally, don't fuss about the prime bit. Just use the word prime.
Everyone one knows what it means. The key thing here is to state /what/
must be prime for P to correctly accept a string.


I am estimating that we will finally achieve closure on this.
Creating a common language between us will achieve the basis for mutual
understanding.

The process that we are doing looks like it will be effective on the
basis of eliminating all hidden assumptions.
-- Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

No use. Even all the people you find agree with you is still useless. To claim
you solved the HP problem, you have to show ALL your H first. People cannot
judge or teach 'claims'. But your P already showed wrong, no need to publish H.
My analysis is based on this model: If "an X is a Y" and Z says that "an
X is a Y" then anything in the universe that disagrees is necessarily
incorrect.

"an X is a Y" =
The input to embedded_H specifies a non-halting sequence of
configurations. (input is non-halting)

Z says that "an X is a Y" =
embedded_H rejects its input.

If you can think of a case where
"an X is a Y" is simultaneously true and false then you have a rebuttal,
otherwise not so much.
-- Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
Genius hits a target no one else can see."
Arthur Schopenhauer

You simply do not have H. What you say? If you say yes, then show it to prove.

Whether or not I have H is moot at this point in the dialogue, the Linz proof is refuted either way.

Showing that an H is possible and answering Ben's question will be postponed until after it is fully understood that embedded_H is necessarily correct to reject its input: ⟨Ĥ⟩ ⟨Ĥ⟩.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
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On 4/5/2022 11:59 AM, Andy Walker wrote:
On 04/04/2022 14:28, Ben Bacarisse wrote:
[I wrote:]
    The reasons why it's not possible to write a halt decider show
up perfectly well if you try to do so in K&R C for a PDP 11 [64K bytes
code, same for date, 2.4M discs, ...].  They're not things that can
only be done with supercomputers with hundreds of processors, etc.
These interminable threads get hung up on the behaviour of programs of
less than a page of code.
Now here there is technical wrinkle.  One can, probably, prove that
halting of bounded-resource computations can't be solved by certain
similarly bounded-resource machines.  But who'd want to do the maths to
prove that?  And if the decider has unbounded resources, then halting of
bounded-resource computations /is/ decidable -- trivially so.

     Yes, but I was thinking of real, skilled, programmers trying to do
the manifestly impossible.  I have in mind one of our [successful!] MSc
students who visited two or three years later:  "What are you doing?"  "Oh,
my boss got fed up with [programs that something-or-other, I forget what],
so he wanted me to write a program that detects [whatever-it-was]."  "But
that's not possible, it's a simple reduction from the HP!"  "Oh, no, I've
almost done it, just a couple of cases to sort out."  A few months later,
he came by again:  "Sorted those, but there are a few bugs."  "Hm, well,
good luck, but I still think it's impossible."  A few months again, and
I got a letter [these were the days before e-mail!] -- "You were right!
It could do the toy examples I gave it, but it got stuck with any real
programs, and I can't make progress, so we've given up."

     PO's problems with his "fully coded TMs" aren't caused by his PC
being finite, nor by limitations of C or his OS, nor even [as far as we
know] by bugs in his "emulator", but by his failures to understand what
the HP is really about, and his failures to produce a proper "hatted"
version of his attempts to solve it.


I say the issue is exactly the opposite of this, I am the only one with a correct understanding of what a halt decider must do.

WE ALL AGREE ON THIS:
A halt decider must compute the mapping from its input finite strings to its own accept or reject state.

HERE IS WHERE WE DIVERGE:
A halt decider must compute the mapping from its input finite strings to its own accept or reject state:

On the basis of the actual behavior actually specified by its input.

All of my reviewers (and Linz) always measure a different sequence of configurations than the one the is actually specified by the actual input.


When trying to get PO to see how to specify a computation, I don't want
to deal with all the distractions that using some abstract C with no
pointer limitations will throw up.

     Instead, you're having to deal with his apparent [but he may just
be trolling] difficulties in understanding what you want.  I wonder how he
would get on with similar exercises in C, where he's not struggling with
the concept of unary [and efficiency] but could simply count and perhaps
analyse characters in the standard input?  [FTAOD, "wonder" is something
of an exaggeration.]



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
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On 4/5/2022 4:24 PM, André G. Isaak wrote:
On 2022-04-05 15:02, olcott wrote:
On 4/5/2022 3:57 PM, André G. Isaak wrote:
On 2022-04-05 11:07, olcott wrote:
On 4/5/2022 9:40 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/4/2022 7:57 PM, Ben Bacarisse wrote:

Aside: TMs are often specified to operate on numbers in unary notation
because it is so simple.

The notation is simple making the algorithm more complex.

No.  This exercise will help you see why that is not true:

   Write a TM that adds two numbers.  The input will be strings of the
   form {n}+{m} where {x} is the unary representation of x.  The TM
   should halt with only {n+m} on the tape.

   Now do the same where the numbers are represented in the usual decimal
   notation.


I was referring to the difference between binary and unary.

Then why not construct a TM that does this using unary notation and one that does it using binary notation? You'll find the former is much simpler.

André


Unary freaks me out with its counter-intuitive nature. Ben says that it proves to be simpler than binary and I accept his word on this.

But don't you think that it would be worth your while to try to understand *why* this is the case?

One of Ben's goals seems to be to get you to actually construct a Turing Machine, a task which would almost certainly rid you of many of your misconceptions about Turing Machines and how they work.

His example of an "Eveness Decider" is trivial to construct regardless of whether you use unary or binary representation (though it is slightly easier using the former). This example is "Hello World!" territory, not some massive undertaking. You'd really benefit from actually attempting this.

André


The only thing that is actually needed is to eliminate hidden assumptions in the meanings that are being expressed so that the last sentence of the following is understood to prove that I am correct:


I say the issue is exactly the opposite of this, I am the only one with a correct understanding of what a halt decider must do.

WE ALL AGREE ON THIS:
A halt decider must compute the mapping from its input finite strings to its own accept or reject state.

HERE IS WHERE WE DIVERGE:
A halt decider must compute the mapping from its input finite strings to its own accept or reject state:

On the basis of the actual behavior actually specified by its input.

All of my reviewers (and Linz) always measure a different sequence of configurations than the one the is actually specified by the actual input.



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
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<t2il0q$tok$1@dont-email.me>
Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/5/2022 6:57 PM, André G. Isaak wrote:
On 2022-04-05 17:49, olcott wrote:
On 4/5/2022 6:39 PM, André G. Isaak wrote:
On 2022-04-05 17:31, olcott wrote:
On 4/5/2022 6:25 PM, André G. Isaak wrote:
On 2022-04-05 17:05, olcott wrote:
On 4/5/2022 5:32 PM, Jeff Barnett wrote:
On 4/5/2022 3:33 PM, olcott wrote:
On 4/5/2022 4:24 PM, André G. Isaak wrote:
On 2022-04-05 15:02, olcott wrote:
On 4/5/2022 3:57 PM, André G. Isaak wrote:
On 2022-04-05 11:07, olcott wrote:
On 4/5/2022 9:40 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/4/2022 7:57 PM, Ben Bacarisse wrote:

Aside: TMs are often specified to operate on numbers in unary notation
because it is so simple.

The notation is simple making the algorithm more complex.

No.  This exercise will help you see why that is not true:

   Write a TM that adds two numbers.  The input will be strings of the
   form {n}+{m} where {x} is the unary representation of x. The TM
   should halt with only {n+m} on the tape.

   Now do the same where the numbers are represented in the usual decimal
   notation.


I was referring to the difference between binary and unary.

Then why not construct a TM that does this using unary notation and one that does it using binary notation? You'll find the former is much simpler.

And while you are at it, try base three numbers (or any odd base greater than one). But no matter what, try writing a TM. Lose your fear of public failure and dive in somewhere.

It is not a fear of failure nitwit.
It is like asking a brain surgeon do you know what a bed pan is?

Except that you have consistently demonstrated that you *don't* have even the vaguest understanding of how TMs work, so the above analogy is hardly apropos.

The most direct path forward on this might be to implement a base-2 even-number decider in this system:

http://www.lns.mit.edu/~dsw/turing/turing.html
(a) Go to the end of the input (space delimited)
(b) Test last input tape cell for "0" digit
(c) Accept or Reject input.

So why not just demonstrate how this works by constructing the actual TM?


It is self-evident that I know exactly how this works by my specification.

What you give above is not a specification. A specification is what Ben has been asking you to provide but which you have been unable or unwilling to do.


If neither of you can see how the above would be translated into a TM of
this kind: http://www.lns.mit.edu/~dsw/turing/turing.html
You are not too bright.

The task I suggested for you was to construct both an evenness-decider that uses binary representations and one that uses unary representations so you could see why the latter is simpler. Giving an outline of a single algorithm without actually constructing the two Turing Machines is not going to achieve that. As I said, these TMs are of the "Hello World!" level of difficulty, so your reluctance is a bit mystifying.

André

That may or may not prove helpful to achieve mutual understanding of the last sentence shown here: (Its pointless if it doesn't help with this).

WE ALL AGREE ON THIS:
A halt decider must compute the mapping from its input finite strings to its own accept or reject state.

HERE IS WHERE WE DIVERGE:
A halt decider must compute the mapping from its input finite strings to its own accept or reject state:

On the basis of the actual behavior actually specified by its input.

THIS IS EVERYONE'S MISTAKE
All of my reviewers (and Linz) always measure a different sequence of configurations than the one that is actually specified by the actual input.



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ][ H is correct ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 6 Apr 2022 15:28 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ][ H is correct ]
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On 4/6/2022 9:19 AM, Ben Bacarisse wrote:
Andy Walker <anw@cuboid.co.uk> writes:

On 06/04/2022 02:57, Ben Bacarisse wrote:
[I wrote:]
Yes, but I was thinking of real, skilled, programmers trying to do
the manifestly impossible.  I have in mind one of our [successful!] MSc
students who visited two or three years later: [...].

What used to be called a "conversion MSc"?

Yes, and no.  We didn't think of it that way, as we treated CS
as just another branch of maths -- all of our students did some, just
as they all did some statistics, and quite a lot chose to do CS options
in the third year.  From that PoV, it was a continuation rather than a
conversion MSc, offering more advanced CS [and statistics, ...] than
had been in the BSc.  But it could have been a conversion for students
coming with [maths] BScs from other places.

Ours was more genuinely conversion as we took students with, in theory,
any background at all -- philosophy, English, history, whatever.  In
general those with maths, physics and EE backgrounds has fewer problems,
but the MSc (in contrast to the BSc) was deliberately designed to have
less theory in it.

[...]
Instead, you're having to deal with his apparent [but he may just
be trolling] difficulties in understanding what you want.  I wonder how he
would get on with similar exercises in C, [...].
Right.  I think TMs may be a step too far.  I don't think I'll get even
one TM accurately specified, let alone written.

You [and others] know this, yet persist!  I suppose there is
some interest in knowing what the next wriggle will be?

I just wonder when he'll stop the "tutorial".

As for the main mistake, I know enough about cranks to aim for only one
of two things: can they be persuaded to say enough to show others that
they are wrong (for example PO admission that H(P,P) == false is correct
despite the fact that P(P) halts),

If it is the case that the simulated input to H cannot possibly reach its own final state under any condition what-so-ever then H correctly maps this finite string input to its reject state and nothing in the universe can correctly contradict that H is correct.

If you have a white dog in your living room and everyone in the universe disagrees, you still have a white dog in your living room.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ][ H is correct ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 6 Apr 2022 16:59 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ][ H is correct ]
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On 4/6/2022 11:43 AM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 11:29:06 AM UTC-4, olcott wrote:
On 4/6/2022 9:19 AM, Ben Bacarisse wrote:
Andy Walker <a...@cuboid.co.uk> writes:

On 06/04/2022 02:57, Ben Bacarisse wrote:
[I wrote:]
Yes, but I was thinking of real, skilled, programmers trying to do
the manifestly impossible. I have in mind one of our [successful!] MSc
students who visited two or three years later: [...].

What used to be called a "conversion MSc"?

Yes, and no. We didn't think of it that way, as we treated CS
as just another branch of maths -- all of our students did some, just
as they all did some statistics, and quite a lot chose to do CS options
in the third year. From that PoV, it was a continuation rather than a
conversion MSc, offering more advanced CS [and statistics, ...] than
had been in the BSc. But it could have been a conversion for students
coming with [maths] BScs from other places.

Ours was more genuinely conversion as we took students with, in theory,
any background at all -- philosophy, English, history, whatever. In
general those with maths, physics and EE backgrounds has fewer problems,
but the MSc (in contrast to the BSc) was deliberately designed to have
less theory in it.

[...]
Instead, you're having to deal with his apparent [but he may just
be trolling] difficulties in understanding what you want. I wonder how he
would get on with similar exercises in C, [...].
Right. I think TMs may be a step too far. I don't think I'll get even
one TM accurately specified, let alone written.

You [and others] know this, yet persist! I suppose there is
some interest in knowing what the next wriggle will be?

I just wonder when he'll stop the "tutorial".

As for the main mistake, I know enough about cranks to aim for only one
of two things: can they be persuaded to say enough to show others that
they are wrong (for example PO admission that H(P,P) == false is correct
despite the fact that P(P) halts),
If it is the case that the simulated input to H cannot possibly reach
its own final state under any condition what-so-ever then H correctly
maps this finite string input to its reject state and nothing in the
universe can correctly contradict that H is correct.

If you have a white dog in your living room and everyone in the universe
disagrees, you still have a white dog in your living room.

What you've actually shown is that for any simulating halt decider H, H^ built from it, and input <H^><H^> which represents H^ applied to <H^>, no H can simulate H^ applied to <H^> to its final state.  This says nothing of whether H^ applied to <H^> halts, which is the actual question *as stated in the Linz proof*.


 >> If it is the case that the

CORRECTLY

 >> simulated input to H cannot possibly reach
 >> its own final state under any condition what-so-ever then

computation that halts … the Turing machine will halt whenever it enters a final state. (Linz:1990:234) cannot possibly be met therefore

 >> H correctly
 >> maps this finite string input to its reject state and nothing in the
 >> universe can correctly contradict that H is correct.



You may in fact have a white dog in your living room, and people do agree with that, but no one cares because they want to know if there is a black cat in your kitchen.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ][ H is correct ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 6 Apr 2022 18:05 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ][ H is correct ]
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On 4/6/2022 12:55 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 1:49:43 PM UTC-4, olcott wrote:
On 4/6/2022 12:07 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 12:59:15 PM UTC-4, olcott wrote:
On 4/6/2022 11:43 AM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 11:29:06 AM UTC-4, olcott wrote:
On 4/6/2022 9:19 AM, Ben Bacarisse wrote:
Andy Walker <a...@cuboid.co.uk> writes:

On 06/04/2022 02:57, Ben Bacarisse wrote:
[I wrote:]
Yes, but I was thinking of real, skilled, programmers trying to do
the manifestly impossible. I have in mind one of our [successful!] MSc
students who visited two or three years later: [...].

What used to be called a "conversion MSc"?

Yes, and no. We didn't think of it that way, as we treated CS
as just another branch of maths -- all of our students did some, just
as they all did some statistics, and quite a lot chose to do CS options
in the third year. From that PoV, it was a continuation rather than a
conversion MSc, offering more advanced CS [and statistics, ...] than
had been in the BSc. But it could have been a conversion for students
coming with [maths] BScs from other places.

Ours was more genuinely conversion as we took students with, in theory,
any background at all -- philosophy, English, history, whatever. In
general those with maths, physics and EE backgrounds has fewer problems,
but the MSc (in contrast to the BSc) was deliberately designed to have
less theory in it.

[...]
Instead, you're having to deal with his apparent [but he may just
be trolling] difficulties in understanding what you want. I wonder how he
would get on with similar exercises in C, [...].
Right. I think TMs may be a step too far. I don't think I'll get even
one TM accurately specified, let alone written.

You [and others] know this, yet persist! I suppose there is
some interest in knowing what the next wriggle will be?

I just wonder when he'll stop the "tutorial".

As for the main mistake, I know enough about cranks to aim for only one
of two things: can they be persuaded to say enough to show others that
they are wrong (for example PO admission that H(P,P) == false is correct
despite the fact that P(P) halts),
If it is the case that the simulated input to H cannot possibly reach
its own final state under any condition what-so-ever then H correctly
maps this finite string input to its reject state and nothing in the
universe can correctly contradict that H is correct.

If you have a white dog in your living room and everyone in the universe
disagrees, you still have a white dog in your living room.

What you've actually shown is that for any simulating halt decider H, H^ built from it, and input <H^><H^> which represents H^ applied to <H^>, no H can simulate H^ applied to <H^> to its final state. This says nothing of whether H^ applied to <H^> halts, which is the actual question *as stated in the Linz proof*.


If it is the case that the
CORRECTLY
simulated input to H cannot possibly reach
its own final state under any condition what-so-ever then
computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234) cannot possibly be met therefore

The *turing machine*, not an inaccurate simulation. The measurement of correct is what H^ applied to <H^> does.

Ĥ applied to ⟨Ĥ⟩ specifies a different sequence of configurations than
the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H.

Then you're answering the wrong question.  The question being asked is "Does H^ applied to <H^> halt?",

No that is not the freaking question you freaking nitwit.

The question is: Does the input specify a sequence of configurations that would reach their own final state?


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ][ H is correct ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 6 Apr 2022 18:29 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ][ H is correct ]
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/6/2022 1:18 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 2:05:36 PM UTC-4, olcott wrote:
On 4/6/2022 12:55 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 1:49:43 PM UTC-4, olcott wrote:
On 4/6/2022 12:07 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 12:59:15 PM UTC-4, olcott wrote:
On 4/6/2022 11:43 AM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 11:29:06 AM UTC-4, olcott wrote:
On 4/6/2022 9:19 AM, Ben Bacarisse wrote:
Andy Walker <a...@cuboid.co.uk> writes:

On 06/04/2022 02:57, Ben Bacarisse wrote:
[I wrote:]
Yes, but I was thinking of real, skilled, programmers trying to do
the manifestly impossible. I have in mind one of our [successful!] MSc
students who visited two or three years later: [...].

What used to be called a "conversion MSc"?

Yes, and no. We didn't think of it that way, as we treated CS
as just another branch of maths -- all of our students did some, just
as they all did some statistics, and quite a lot chose to do CS options
in the third year. From that PoV, it was a continuation rather than a
conversion MSc, offering more advanced CS [and statistics, ...] than
had been in the BSc. But it could have been a conversion for students
coming with [maths] BScs from other places.

Ours was more genuinely conversion as we took students with, in theory,
any background at all -- philosophy, English, history, whatever. In
general those with maths, physics and EE backgrounds has fewer problems,
but the MSc (in contrast to the BSc) was deliberately designed to have
less theory in it.

[...]
Instead, you're having to deal with his apparent [but he may just
be trolling] difficulties in understanding what you want. I wonder how he
would get on with similar exercises in C, [...].
Right. I think TMs may be a step too far. I don't think I'll get even
one TM accurately specified, let alone written.

You [and others] know this, yet persist! I suppose there is
some interest in knowing what the next wriggle will be?

I just wonder when he'll stop the "tutorial".

As for the main mistake, I know enough about cranks to aim for only one
of two things: can they be persuaded to say enough to show others that
they are wrong (for example PO admission that H(P,P) == false is correct
despite the fact that P(P) halts),
If it is the case that the simulated input to H cannot possibly reach
its own final state under any condition what-so-ever then H correctly
maps this finite string input to its reject state and nothing in the
universe can correctly contradict that H is correct.

If you have a white dog in your living room and everyone in the universe
disagrees, you still have a white dog in your living room.

What you've actually shown is that for any simulating halt decider H, H^ built from it, and input <H^><H^> which represents H^ applied to <H^>, no H can simulate H^ applied to <H^> to its final state. This says nothing of whether H^ applied to <H^> halts, which is the actual question *as stated in the Linz proof*.


If it is the case that the
CORRECTLY
simulated input to H cannot possibly reach
its own final state under any condition what-so-ever then
computation that halts … the Turing machine will halt whenever it enters
a final state. (Linz:1990:234) cannot possibly be met therefore

The *turing machine*, not an inaccurate simulation. The measurement of correct is what H^ applied to <H^> does.

Ĥ applied to ⟨Ĥ⟩ specifies a different sequence of configurations than
the simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H.

Then you're answering the wrong question. The question being asked is "Does H^ applied to <H^> halt?",
No that is not the freaking question you freaking nitwit.

The question is: Does the input specify a sequence of configurations
that would reach their own final state?

FALSE.  From Linz:

Linz is wrong too.
Because we know that a halt decider must compute the mapping

FROM ITS INPUTS
FROM ITS INPUTS
FROM ITS INPUTS
FROM ITS INPUTS

Anyone that says it must compute the mapping from non-inputs such as Ĥ applied to ⟨Ĥ⟩ IS FREAKING WRONG !!!

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Wed, 6 Apr 2022 19:06 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/6/2022 1:59 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 2:56:33 PM UTC-4, olcott wrote:
On 4/6/2022 1:49 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

I thought you wanted to learn how TMs are
specified so you had the knowledge to read and understand Linz's
specifications.

Not at all. I already understand this better than you do.

Ah, let's call it a day then.
We have to get to the point where you understand that I aleady know this
better than you so I am willing to proceed with the E TM.

Bold words from someone who can't write what's effectively a "hello world" TM.

It is not the tedious details about writing a TM that are most important.

It is exactly how all of the fundamental concepts of the theory of computation fit together relative to TM's that is most crucial.

If you get the first part perfectly and in the second part you have at least one large error then you know enormously less about TM's than someone that only gets the second part correctly.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 7 Apr 2022 01:01 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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From: NoO...@NoWhere.com (olcott)
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On 4/6/2022 7:34 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/6/2022 6:35 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/6/2022 4:36 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/6/2022 9:19 AM, Ben Bacarisse wrote:

As for the main mistake, I know enough about cranks to aim for only one
of two things: can they be persuaded to say enough to show others that
they are wrong (for example PO admission that H(P,P) == false is correct
despite the fact that P(P) halts),

If it is the case that the simulated input to H cannot possibly reach
its own final state under any condition what-so-ever then H correctly
maps this finite string input to its reject state and nothing in the
universe can correctly contradict that H is correct.

If you have a white dog in your living room and everyone in the
universe disagrees, you still have a white dog in your living room.

Good to see that you are still asserting that false is the correct
result from a halt decider for at least one halting computation.

If the input to the halt decider specifies a non-halting sequence of
configurations then any damn thing anywhere else is totally
irrelevant.
If P(P) halts, H(P,P) should be true.

Like I said any damn thing else is actually 100% perfectly totally
irrelevant.

Yes!  The only thing that matters is whether the "input", (P,P),
specifies a halting computation or not.

The "input" to H is two
parameters that specify the halting computation P(P).

A halting computation that cannot possibly reach its own final state
under any condition what-so-ever?

Either P(P) halts or it does not.  Did you tell a fib when you said it
does?  Since it halts, H(P,P) == false is wrong.

The input to H(P,P) cannot possibly reach its own final state under
any condition what-so-ever, thus if God and all his angels and every
being great and small said that the input to H specifies a halting
computation they would all be liars.

You told that us P(P) halts.  Until you retract that, I will take it to
be true.  You also told us that H(P,P) == false.  Do you need to correct
one or other of these statements?


As long as the input to H(P,P) never reaches its final state under any condition what-so-ever then no matter what P(P) does H was still correct because P(P) is not an input and H is only accountable for getting its inputs correctly.

If a guard is assigned to watch the front door and no one comes in the front door and thousands of people come in the back door the guard is correct to say that no one came in the front door.

The input to H is its front door that it must guard. What P(P) does when it is not an input is all back door stuff and none of the business of any decider.



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 7 Apr 2022 03:55 UTC
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key missing piece in dialogue ]
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From: NoO...@NoWhere.com (olcott)
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On 4/6/2022 10:27 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 10:55:06 PM UTC-4, olcott wrote:
On 4/6/2022 9:45 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 10:40:15 PM UTC-4, olcott wrote:
On 4/6/2022 9:26 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 10:15:18 PM UTC-4, olcott wrote:
On 4/6/2022 9:12 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 10:10:45 PM UTC-4, olcott wrote:
On 4/6/2022 9:03 PM, André G. Isaak wrote:
On 2022-04-06 19:47, olcott wrote:
On 4/6/2022 8:41 PM, André G. Isaak wrote:
On 2022-04-06 19:25, olcott wrote:
On 4/6/2022 8:17 PM, André G. Isaak wrote:
On 2022-04-06 19:10, olcott wrote:
On 4/6/2022 7:50 PM, André G. Isaak wrote:

But, now that we've got that out of the way, here's a simple
question for you: If you want your evenness decider to decide
whether seven is even, which string would you pass to it? [yes, I
know this is trivially obvious, just humour me]

André

111[]

I'm assuming that you're using [] to indicate a blank.

Presumably your E would *reject* this string since seven is an odd
number rather than an even one.

But notice that in the above case your Turing Machine is rejecting
a finite string "111␣" based on the fact that the *integer* seven,
which is neither an input nor a finite string, is not even.

So your decider is mapping from a finite string (its input) to
reject/accept based on something which is a "non-input non-finite
string" (to use an expression you've often used).


That seems totally incoherent.

The TM maps its input finite string to its reject state based on a
property of this finite string.


But 'evenness' is not a property of strings; it is a property of
numbers, and strings are not numbers.


It can be correctly construed as the property of the string.

Not by any normal definition.

A string is an *uninterpreted* sequence of symbols.

Your decider bases its decision on a property of the string (is the
final digit 0?), but that property only corresponds to evenness by
virtue of the encoding you have chosen.

A decider which uses a unary representation (which would be slightly
simpler than your binary one) couldn't just look at a single digit. Nor

Thus not actually simpler.

If you actually attempted to write the two versions of E, you'd discover
that you are simply wrong in this assessment. The fact that you don't
realize this is merely a testament to the fact that you really don't
understand how Turing Machines work. At all.

could one that used trinary representation (which would be only
slightly more complex than your binary one).

What makes all three of these valid evenness deciders is that they
conform to the specification of an evenness decider:

E.q0 S ⊦* E.qy if the *number* represented by the string S is even
E.q0 S ⊦* E.qn otherwise.

Yes, the TM maps a finite string to an accept/reject state, but this
mapping is based on the property of the *number* which that string
encodes. That number is not an input, but because it can be encoded
as a string we can still legitimately expect a Turing Machine to
answer questions about that number.


It can be construed as a property of the string.

No. It cannot be. Properties of a string would include things like
'length', 'number of distinct symbols', 'is it a palindrome?', etc.
Evenness is not one of those properties.

Talking about a finite string as being even or odd is completely
meaningless. Only numbers can be even or odd. Yet there is no problem
in constructing such a decider.


The string has the "even binary integer" property.

No. As soon as you start assigning numerical values to a string (or even
to the individual symbols of a string) you are no longer treating it
purely as a string. You are considering the information which is encoded
in that string, which is a separate matter.

The all has to do with mathematicaly fomrmalizing semantic meanings.
Finite strings can have semantic properties.

And just like the meaning assigned to the string "111␣" is the number 7, the meaning assigned to the string <H^><H^> is the turing machine H^ applied to <H^>.
Yes.

Ah, so you're finally agreeing that the input <H^><H^> represents H^ applied to <H^>, and that therefore H applied to <H^><H^> is supposed determine if H^ applied to <H^> halts?
It represents a sequence of configurations.

And that sequence of configurations is H^ applied to <H^> as per the stipulative definition of a halt decider:
The actual sequence of configurations specified by these three is not
the same thus the behavior can vary.

H ⟨Ĥ⟩ ⟨Ĥ⟩
Ĥ ⟨Ĥ⟩
embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
You can simply assume that they must be the same, yet when you list them
you can see that they are not the same.

By definition the input to H and embedded_H is the representation of Ĥ ⟨Ĥ⟩

None-the-less the above three are not the same sequence of steps and this directly causes a significant difference in their behavior.

It is simpler to understand that this is the only thing that matters:

It is the case that the correctly simulated input to embedded_H can never possibly reach its own final state under any condition at all. Therefore embedded_H is necessarily correct to reject its input.

Even if you simply assume that it is not true:

You can verify the correctness of my reasoning by simply seeing what this would prove if it was true. If embedded_H correctly decides ⟨Ĥ⟩ ⟨Ĥ⟩ then it refutes what Linz said was impossible.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ only look at the front door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.math, sci.logic
Followup: comp.theory
Date: Thu, 7 Apr 2022 17:43 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
only look at the front door ]
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From: NoO...@NoWhere.com (olcott)
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On 4/7/2022 12:31 PM, Mike Terry wrote:
On 07/04/2022 01:24, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 8:20:16 PM UTC-4, olcott wrote:
On 4/6/2022 7:17 PM, Dennis Bush wrote:
On Wednesday, April 6, 2022 at 8:10:14 PM UTC-4, olcott wrote:
On 4/6/2022 6:44 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 6:27 PM, Ben Bacarisse wrote:

... Your hobby seems to be posting here. Are you having fun
posting here?

I am enjoying posting here because progress continues to occur.

That's fine. I'd like to think I am helping to entertain you.

I have
my whole proof boiled down to the correct understanding of a single
(very difficult to understand) sentence.

Except for the two questions you can't answer without it all
unravelling!

You merely continue to greatly disrespectfully refuse to pay enough
attention:

It is the case that the correctly simulated input to embedded_H can
never possibly reach its own final state under any condition at all.
Therefore embedded_H is necessarily correct to reject its input.

But how can we verify that the input was correctly simulated?
The exection trace that is specified by its TM description.

But another halt decider simulates the same input to completion.  So the claim that the input "can never possibly reach its own final state under any condition at all" is false:

Given an embedded_H that aborts its simulation which we'll call embedded_Ha, is embedded_Ha (and therefore Ha) correct to reject <Ha^><Ha^>?

Now we have Hb, which has the exact same halting criteria as Ha except it defers aborting for k steps. Hb simulates <Ha^><Ha^> and is able to reach the input's final state of <Ha^.qn> while remaining in UTM mode and accepts this input. This tells us that embedded_Ha is not correct to reject <Ha^><Ha^> because it aborted too soon.


PO really really really really really believes that when his simulator observes his "infinite recursive behaviour" pattern, that really really really really means that the simulation is exhibiting infinite recursive behaviour, and so would never halt however far it is continued.

It does not matter to PO that he has actually run the computation outside of the simulator, and observed himself that the computation is halting!!!!!!

If you are a guard assigned to watch the front door any nothing comes in the front door then you are correct to report that nothing has come in the front door no matter what comes in anywhere else.

The actual behavior of the actual input to embedded_H is the front door.

The actual behavior of any damn thing else anywhere else IS NOT THE FRONT DOOR.

embedded_H computes the mapping from its inputs
from its inputs
from its inputs
from its inputs
from its inputs
from its inputs  not any damn other place else.

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 7 Apr 2022 22:57 UTC
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key missing piece in dialogue ][ back door ]
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From: NoO...@NoWhere.com (olcott)
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On 4/7/2022 5:51 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 6:46:37 PM UTC-4, olcott wrote:
On 4/7/2022 5:18 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 5:51:41 PM UTC-4, olcott wrote:
On 4/7/2022 4:37 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 5:17:44 PM UTC-4, olcott wrote:
On 4/7/2022 3:21 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 4:04:48 PM UTC-4, olcott wrote:
On 4/7/2022 3:00 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 3:58:03 PM UTC-4, olcott wrote:
On 4/7/2022 2:38 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 3:19:03 PM UTC-4, olcott wrote:
On 4/7/2022 2:07 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:54:57 PM UTC-4, olcott wrote:
On 4/7/2022 1:51 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:47:53 PM UTC-4, olcott wrote:
On 4/7/2022 1:45 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:24:01 PM UTC-4, olcott wrote:
On 4/7/2022 1:08 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:04:41 PM UTC-4, olcott wrote:
On 4/7/2022 1:00 PM, olcott wrote:
On 4/7/2022 12:59 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 1:37:20 PM UTC-4, olcott wrote:
On 4/7/2022 12:09 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 1:02:27 PM UTC-4, olcott wrote:
On 4/7/2022 11:52 AM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 12:16:56 PM UTC-4, olcott wrote:
On 4/7/2022 9:45 AM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 10:35:31 AM UTC-4, olcott wrote:
On 4/7/2022 5:58 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 8:49 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 7:34 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 6:35 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 4:36 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 9:19 AM, Ben Bacarisse wrote:

As for the main mistake, I know enough about cranks
to aim for only one
of two things: can they be persuaded to say enough
to show others that
they are wrong (for example PO admission that H(P,P)
== false is correct
despite the fact that P(P) halts),

If it is the case that the simulated input to H
cannot possibly reach
its own final state under any condition what-so-ever
then H correctly
maps this finite string input to its reject state and
nothing in the
universe can correctly contradict that H is correct.

If you have a white dog in your living room and
everyone in the
universe disagrees, you still have a white dog in
your living room.

Good to see that you are still asserting that false is
the correct
result from a halt decider for at least one halting
computation.

If the input to the halt decider specifies a
non-halting sequence of
configurations then any damn thing anywhere else is
totally
irrelevant.
If P(P) halts, H(P,P) should be true.

Like I said any damn thing else is actually 100%
perfectly totally
irrelevant.
Yes! The only thing that matters is whether the "input",
(P,P),
specifies a halting computation or not.

The "input" to H is two
parameters that specify the halting computation P(P).

A halting computation that cannot possibly reach its own
final state
under any condition what-so-ever?
Either P(P) halts or it does not. Did you tell a fib when
you said it
does? Since it halts, H(P,P) == false is wrong.

The input to H(P,P) cannot possibly reach its own final
state under
any condition what-so-ever, thus if God and all his
angels and every
being great and small said that the input to H specifies
a halting
computation they would all be liars.
You told that us P(P) halts. Until you retract that, I
will take it to
be true. You also told us that H(P,P) == false. Do you
need to correct
one or other of these statements?

As long as the input to H(P,P) never reaches its final
state under any
condition what-so-ever then no matter what P(P) does H was
still
correct because P(P) is not an input and H is only
accountable for
getting its inputs correctly.

So what two arguments must be passed to H to get H to tell
us whether
P(P) halts or not? (Already asked, of course, but you a
dodging this
issue for obvious reasons.)

You won't understand what I am saying until you first
understand that
your question has nothing to do with the correctness of the
rejection
of the input.

I am referring to a point that is so subtle that no one ever
noticed
this subtle point for 90 years.

I WILL KEEP REPEATING THIS UNTIL YOU RESPOND

Of course you will. You can't answer the question without being
obviously wrong,
THIS PROVES THAT I AM CORRECT
It is the case that the correctly simulated input to embedded_H
can
never possibly reach its own final state under any condition at
all.
Therefore embedded_H is necessarily correct to reject its input.
I will not talk to you about anything besides that.

The input to UTM applied to <H^><H^>
Is not what I am talking about.


You said "under any condition at all",
Within the scope of embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
Should I just ignore your next 20 replies?

So embedded_H, and therefore H, is the sole source of truth for if
it's input reaches a final state?
The scope only includes embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ and explicitly
excludes everything else in the whole universe.

So you're saying and embedded_H and H give different output for the
same input?

I am saying that H is off topic bitch.

STFU about it.

In other words,
I absolutely positively will not tolerate the most microscopic
divergence from: embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩

Any replies with microscopic divergences will simply be ignored.

So you've implicitly agreed that embedded_H and H are the same,
I have done no such thing.

Until you provide an example of H and embedded_H giving different results from the same input, yes you have.
Liar !!!

This is when everyone watching sees that you know you don't have a case.
If I tolerate the slightest microscopic divergence from the point at
hand you will never understand what I am saying in a million years.

STFU about H !!!
It is the case that the correctly simulated input to embedded_H can
never possibly reach its own final state under any condition at all.
Therefore embedded_H is necessarily correct to reject its input.

Because embedded_H is the same as H
Because the input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H specifies a non-halting
sequence of configurations

It does not:

So the simulated input can possibly reach its own final state?

Yep.
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

Show exactly where in this execution trace that the simulated ⟨Ĥ0⟩ would
transition to ⟨Ĥ0.y⟩ or ⟨Ĥ0.n⟩.

Ĥ is applied to ⟨Ĥ0⟩
(a) Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then
(b) H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩
Then these steps would keep repeating:
(c) Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
(d) Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩

Now you're talking about Hn which never aborts.
All that I am saying is that if the simulated ⟨Ĥ0⟩ cannot possibly reach
its own final state of ⟨Ĥ0.y⟩ or ⟨Ĥ0.n⟩ then that proves that it is not
a halting computation.

You are saying know I must be wrong because that goes against your
intuition.

SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩

⟨Ĥn0⟩ never does transition to a final state. And yes Ĥn applies to ⟨Ĥn⟩ does not halt. But Hn is unable to report that fact because it can't abort its simulation and is therefore wrong by default.
The fact that the input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H cannot possibly reach its
final state under any condition what-so-ever conclusively proves that it
is not a halting computation.

Yes, we agree that ⟨Ĥn⟩ ⟨Ĥn⟩ is non-halting.  But Hn can't report that.

When embedded_H rejects its input it is necessary correct.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Thu, 7 Apr 2022 23:23 UTC
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On 4/7/2022 6:16 PM, André G. Isaak wrote:
On 2022-04-07 17:04, olcott wrote:
On 4/7/2022 6:00 PM, André G. Isaak wrote:
On 2022-04-07 16:55, olcott wrote:
On 4/7/2022 5:36 PM, André G. Isaak wrote:
On 2022-04-07 15:17, olcott wrote:

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

Show exactly where in this execution trace that the simulated ⟨Ĥ0⟩ would transition to ⟨Ĥ0.y⟩ or ⟨Ĥ0.n⟩.

That isn't an 'execution trace'. It's a bare-bones outline of what happens. Moreover, it is not an *accurate* outline of what happens.

Ĥ is applied to ⟨Ĥ0⟩
    (a) Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then
    (b) H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

Your (b) should read H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩ until either Ĥ0 ⟨Ĥ1⟩ is completed (in which case the computation goes to H.qy) or until H decides to discontinue the simulation (in which case the computation goes to H.qn and halts)

Then these steps would keep repeating:
    (c) Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩

Again, your (c) should read Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩ until either Ĥ1 ⟨Ĥ2⟩ is completed (in which case the computation goes to H0.qy) or until H0 decides to abort the simulation. (in which case the computation goes to H0.qn and halts)

    (d) Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩

And again, this is inaccurate. I'll leave fixing it as an exercise for you.


Your "corrections" are totally incorrect.

How so? They are based entirely on *your* description of how your alleged halt decider works.

André


These are the actual first four steps.

Ĥ is applied to ⟨Ĥ0⟩
    Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then
    H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩
Then these steps would keep repeating:
    Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩

 From these four steps we can see that ⟨Ĥ0⟩ never reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩ thus never halts.

If those are the actual steps, then how on earth does the topmost H manage to "correctly" decide that its input is non-halting?

According to what you write above the topmost H simply simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩ which, according to you, never ends, meaning the topmost H also never ends.

The "trace" you give above is what you would get if H were an actual UTM. But H *isn't* a UTM, it is a simulating halt decider. Ergo its behaviour won't match that of a UTM.

André


The whole point here is that the simulated ⟨Ĥ0⟩ never reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩ thus never halts.

All of the other things that you bring up are mere distractions away from this actual point.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
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On 4/7/2022 6:37 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/7/2022 10:51 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

THIS PROVES THAT I AM CORRECT
It is the case that the correctly simulated input to embedded_H can
never possibly reach its own final state under any condition at all.
Therefore embedded_H is necessarily correct to reject its input.

Yet you won't answer two simple questions!  Why?

Because I absolutely positively will not tolerate divergence from
validating my 17 years worth of work.

But you have no choice but to tolerate it.  If someone wants to talk
about why you are wrong, they will do so.

You are wrong (for the C version of H) because H(P,P) == false but P(P)
halts.  You are wrong about your TM H because H <Ĥ> <Ĥ> transitions to
qn, but Ĥ applied to <Ĥ> is a halting computation. (Feel free to deny
any of these facts if the mood takes you.)


If you believe (against the verified facts) that the simulated ⟨Ĥ0⟩ reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩ then you must show how this occurs.

Ĥ is applied to ⟨Ĥ0⟩
    Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then
    H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩
Then these steps would keep repeating:
    Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩

 From these four steps we can see that the simulated ⟨Ĥ0⟩ never reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩ thus never halts.



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
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On 4/7/2022 5:51 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 6:46:37 PM UTC-4, olcott wrote:
On 4/7/2022 5:18 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 5:51:41 PM UTC-4, olcott wrote:
On 4/7/2022 4:37 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 5:17:44 PM UTC-4, olcott wrote:
On 4/7/2022 3:21 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 4:04:48 PM UTC-4, olcott wrote:
On 4/7/2022 3:00 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 3:58:03 PM UTC-4, olcott wrote:
On 4/7/2022 2:38 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 3:19:03 PM UTC-4, olcott wrote:
On 4/7/2022 2:07 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:54:57 PM UTC-4, olcott wrote:
On 4/7/2022 1:51 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:47:53 PM UTC-4, olcott wrote:
On 4/7/2022 1:45 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:24:01 PM UTC-4, olcott wrote:
On 4/7/2022 1:08 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 2:04:41 PM UTC-4, olcott wrote:
On 4/7/2022 1:00 PM, olcott wrote:
On 4/7/2022 12:59 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 1:37:20 PM UTC-4, olcott wrote:
On 4/7/2022 12:09 PM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 1:02:27 PM UTC-4, olcott wrote:
On 4/7/2022 11:52 AM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 12:16:56 PM UTC-4, olcott wrote:
On 4/7/2022 9:45 AM, Dennis Bush wrote:
On Thursday, April 7, 2022 at 10:35:31 AM UTC-4, olcott wrote:
On 4/7/2022 5:58 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 8:49 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 7:34 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 6:35 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 4:36 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/6/2022 9:19 AM, Ben Bacarisse wrote:

As for the main mistake, I know enough about cranks
to aim for only one
of two things: can they be persuaded to say enough
to show others that
they are wrong (for example PO admission that H(P,P)
== false is correct
despite the fact that P(P) halts),

If it is the case that the simulated input to H
cannot possibly reach
its own final state under any condition what-so-ever
then H correctly
maps this finite string input to its reject state and
nothing in the
universe can correctly contradict that H is correct.

If you have a white dog in your living room and
everyone in the
universe disagrees, you still have a white dog in
your living room.

Good to see that you are still asserting that false is
the correct
result from a halt decider for at least one halting
computation.

If the input to the halt decider specifies a
non-halting sequence of
configurations then any damn thing anywhere else is
totally
irrelevant.
If P(P) halts, H(P,P) should be true.

Like I said any damn thing else is actually 100%
perfectly totally
irrelevant.
Yes! The only thing that matters is whether the "input",
(P,P),
specifies a halting computation or not.

The "input" to H is two
parameters that specify the halting computation P(P).

A halting computation that cannot possibly reach its own
final state
under any condition what-so-ever?
Either P(P) halts or it does not. Did you tell a fib when
you said it
does? Since it halts, H(P,P) == false is wrong.

The input to H(P,P) cannot possibly reach its own final
state under
any condition what-so-ever, thus if God and all his
angels and every
being great and small said that the input to H specifies
a halting
computation they would all be liars.
You told that us P(P) halts. Until you retract that, I
will take it to
be true. You also told us that H(P,P) == false. Do you
need to correct
one or other of these statements?

As long as the input to H(P,P) never reaches its final
state under any
condition what-so-ever then no matter what P(P) does H was
still
correct because P(P) is not an input and H is only
accountable for
getting its inputs correctly.

So what two arguments must be passed to H to get H to tell
us whether
P(P) halts or not? (Already asked, of course, but you a
dodging this
issue for obvious reasons.)

You won't understand what I am saying until you first
understand that
your question has nothing to do with the correctness of the
rejection
of the input.

I am referring to a point that is so subtle that no one ever
noticed
this subtle point for 90 years.

I WILL KEEP REPEATING THIS UNTIL YOU RESPOND

Of course you will. You can't answer the question without being
obviously wrong,
THIS PROVES THAT I AM CORRECT
It is the case that the correctly simulated input to embedded_H
can
never possibly reach its own final state under any condition at
all.
Therefore embedded_H is necessarily correct to reject its input.
I will not talk to you about anything besides that.

The input to UTM applied to <H^><H^>
Is not what I am talking about.


You said "under any condition at all",
Within the scope of embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩
Should I just ignore your next 20 replies?

So embedded_H, and therefore H, is the sole source of truth for if
it's input reaches a final state?
The scope only includes embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ and explicitly
excludes everything else in the whole universe.

So you're saying and embedded_H and H give different output for the
same input?

I am saying that H is off topic bitch.

STFU about it.

In other words,
I absolutely positively will not tolerate the most microscopic
divergence from: embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩

Any replies with microscopic divergences will simply be ignored.

So you've implicitly agreed that embedded_H and H are the same,
I have done no such thing.

Until you provide an example of H and embedded_H giving different results from the same input, yes you have.
Liar !!!

This is when everyone watching sees that you know you don't have a case.
If I tolerate the slightest microscopic divergence from the point at
hand you will never understand what I am saying in a million years.

STFU about H !!!
It is the case that the correctly simulated input to embedded_H can
never possibly reach its own final state under any condition at all.
Therefore embedded_H is necessarily correct to reject its input.

Because embedded_H is the same as H
Because the input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H specifies a non-halting
sequence of configurations

It does not:

So the simulated input can possibly reach its own final state?

Yep.
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

Show exactly where in this execution trace that the simulated ⟨Ĥ0⟩ would
transition to ⟨Ĥ0.y⟩ or ⟨Ĥ0.n⟩.

Ĥ is applied to ⟨Ĥ0⟩
(a) Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then
(b) H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩
Then these steps would keep repeating:
(c) Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
(d) Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩

Now you're talking about Hn which never aborts.
All that I am saying is that if the simulated ⟨Ĥ0⟩ cannot possibly reach
its own final state of ⟨Ĥ0.y⟩ or ⟨Ĥ0.n⟩ then that proves that it is not
a halting computation.

You are saying know I must be wrong because that goes against your
intuition.

SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩
SHOW ME WHERE ⟨Ĥ0⟩ TRANSITIONS TO ⟨Ĥ0.y⟩ OR ⟨Ĥ0.n⟩

⟨Ĥn0⟩ never does transition to a final state. And yes Ĥn applies to ⟨Ĥn⟩ does not halt. But Hn is unable to report that fact because it can't abort its simulation and is therefore wrong by default.
The fact that the input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H cannot possibly reach its
final state under any condition what-so-ever conclusively proves that it
is not a halting computation.

Yes, we agree that ⟨Ĥn⟩ ⟨Ĥn⟩ is non-halting. 

Conclusively proving that embedded_H would be correct when it rejects its input.

If {an X is a Y} then when {Z says} that {an X is a Y} Z is necessarily correct.

{an X is a Y} = "the input to embedded_H is non-halting."
{Z says} = "embedded_H rejects its input."

--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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On 4/8/2022 12:44 PM, André G. Isaak wrote:
On 2022-04-08 11:02, olcott wrote:

I asked about Ĥ0 and you answered with Ĥn which includes Ĥ[0...n].

You need to go back to the point where Dennis defined his Ha and Hn. They don't mean what you seem to think they mean.

André


The fact that they do not mean embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ is enough to know that they must be utterly rejected out-of-hand.

That the input: ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H is non-halting is the only thing that is relevant to the correctness of embedded_H rejecting this input.

Everything else is the God damned lie of a God damned liar.

When I say "God damned" I mean in the sense of being eternally incinerated in actual Hell.

Revelation 21:8 King James Version
....all liars, shall have their part in the lake which burneth with fire and brimstone: which is the second death.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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On 4/8/2022 1:29 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

Linz makes this difficult to understand because he simply erases key
elements of the definition of Ĥ:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

You have erased them.  Linz specifies Ĥ properly based on what H is
supposed to do:

   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞   if Ĥ applied to ⟨Ĥ⟩ halts, and
   Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn     if Ĥ applied to ⟨Ĥ⟩ does not halt.

You have spent an inordinate amount of time over the years copying out
those lines and dishonestly removing the key conditions.  We all know
why.


<Linz:1990:320>
Now Ĥ is a Turing machine, so that it will have some description in Σ*, say ŵ. This string, in addition to being the description of Ĥ can also be used as input string. We can therefore legitimately ask what would happen if Ĥ is applied to ŵ.

q0ŵ ⊢* Ĥ ∞
if Ĥ applied to ŵ halts, and

q0ŵ ⊢* Ĥy1qny2
if Ĥ applied to ŵ does not halt. This is clearly nonsense. The contradiction tells us that...
</Linz:1990:320>

In other words the copy of H embedded within Ĥ is incorrect to either reject or accept its input.

When I show that embedded_H correctly rejects its input ⟨Ĥ⟩ ⟨Ĥ⟩ I have correctly refuted Linz.

Because the simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H is non-halting when embedded_H rejects this input as non-halting it is necessarily correct.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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On 4/8/2022 2:38 PM, Dennis Bush wrote:
On Friday, April 8, 2022 at 3:20:35 PM UTC-4, olcott wrote:
On 4/8/2022 2:16 PM, Dennis Bush wrote:
On Friday, April 8, 2022 at 2:49:36 PM UTC-4, olcott wrote:
On 4/8/2022 1:29 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

Linz makes this difficult to understand because he simply erases key
elements of the definition of Ĥ:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

You have erased them. Linz specifies Ĥ properly based on what H is
supposed to do:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞ if Ĥ applied to ⟨Ĥ⟩ halts, and
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn if Ĥ applied to ⟨Ĥ⟩ does not halt.

You have spent an inordinate amount of time over the years copying out
those lines and dishonestly removing the key conditions. We all know
why.

<Linz:1990:320>
Now Ĥ is a Turing machine, so that it will have some description in
Σ*, say ŵ. This string, in addition to being the description of Ĥ can
also be used as input string. We can therefore legitimately ask what
would happen if Ĥ is applied to ŵ.

q0ŵ ⊢* Ĥ ∞
if Ĥ applied to ŵ halts, and

q0ŵ ⊢* Ĥy1qny2
if Ĥ applied to ŵ does not halt. This is clearly nonsense. The
contradiction tells us that...
</Linz:1990:320>

In other words the copy of H embedded within Ĥ is incorrect to either
reject or accept its input.

What you fail to notice is that there is more than one H and H^ in play. For example:

A: an H that always accepts
R: an H that always rejects

What the above is saying is that R / embedded_R rejecting <R^><R^> is incorrect and that A / embedded_A accepting <A^><A^> is incorrect. So not a *single* H but two *different* H's. This also means that A accepting <R^><R^> is correct and R rejecting <A^><A^> is correct.

In other words, no H can give a correct halting/non-halting answer for an H^ built from it (even though some other H could).
So when I show that the H embedded within Ĥ does correctly decide its
input ⟨Ĥ⟩ ⟨Ĥ⟩, Linz has been refuted.


Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

When Ĥ is applied to ⟨Ĥ0⟩
    Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then
    H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩
Then these steps would keep repeating:
    Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩

Since we can see that the simulated input: ⟨Ĥ0⟩ to embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩ we know that it is non-halting.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ key missing piece in dialogue ][ back door ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sat, 9 Apr 2022 01:01 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
key missing piece in dialogue ][ back door ]
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<GF44K.31231$r_.15461@fx41.iad>
Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
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On 4/8/2022 7:44 PM, Richard Damon wrote:
On 4/8/22 7:34 PM, olcott wrote:
On 4/8/2022 6:31 PM, Richard Damon wrote:

On 4/8/22 6:51 PM, olcott wrote:
On 4/8/2022 4:49 PM, Dennis Bush wrote:
On Friday, April 8, 2022 at 5:40:42 PM UTC-4, olcott wrote:
On 4/8/2022 4:08 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/7/2022 8:14 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/7/2022 6:37 PM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

On 4/7/2022 10:51 AM, Ben Bacarisse wrote:
olcott <No...@NoWhere.com> writes:

THIS PROVES THAT I AM CORRECT
It is the case that the correctly simulated input to embedded_H can
never possibly reach its own final state under any condition at all.
Therefore embedded_H is necessarily correct to reject its input.

Yet you won't answer two simple questions! Why?

Because I absolutely positively will not tolerate divergence from
validating my 17 years worth of work.

But you have no choice but to tolerate it. If someone wants to talk
about why you are wrong, they will do so.

You are wrong (for the C version of H) because H(P,P) == false but P(P)
halts. You are wrong about your TM H because H <Ĥ> <Ĥ> transitions to
qn, but Ĥ applied to <Ĥ> is a halting computation. (Feel free to deny
any of these facts if the mood takes you.)

If you believe (against the verified facts) that the simulated ⟨Ĥ0⟩
reaches its final state of ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩...

I believe what you've told me: that you claim that H(P,P)==false is
correct despite the fact that P(P) halts. That's wrong.

If the input to H(P,P) cannot possibly reach its final state then this
input is correctly rejected and nothing in the universe can possibly
contradict this.

Agreed facts: (1) H(P,P) == false, (2) P(P) halts. You don't dispute
either (indeed they come from you).

Your new line in waffle is just an attempt to distract attention from a
very simple claim: that the wrong answer is the right one.

Even Linz got this wrong because it is counter-intuitive.

A halt decider must compute the mapping from its inputs (not any damn
thing else in the universe) to its own final state on the basis of

the behavior specified by these inputs

Which is stipulated to be H^ applied to <H^>


When Ĥ is applied to ⟨Ĥ0⟩
    Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then
    H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩
Then these steps would keep repeating:
    Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩

Explain exactly how the actual input: ⟨Ĥ0⟩ ⟨Ĥ1⟩ to embedded_H reaches its own final state: ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩.



If that IS the proper trace, then it doesn't, but H and embedded_H have also failed to decide, because they will NEVER 'abort' their simulation and return an answer.
That is a proper trace and in the next step embedded_H aborts the simulation of its input and transitions to its reject state.



Then the words "Keep on Repeating" are a LIE.


I needed my readers to get one single point from the above sequence that the simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to embedded_H never reaches its own final state: ⟨Ĥ0.qy⟩ or ⟨Ĥ0.qn⟩.

It has taken you about three months to get this one single point.

Now that you have gotten this point we move on to the next step, embedded_H aborts its simulation and transitions to its own reject state.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [ correct halt deciding criteria ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sun, 10 Apr 2022 23:58 UTC
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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(11) [
correct halt deciding criteria ]
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From: NoO...@NoWhere.com (olcott)
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On 4/10/2022 6:26 PM, Dennis Bush wrote:
On Sunday, April 10, 2022 at 7:20:44 PM UTC-4, olcott wrote:
On 4/10/2022 6:14 PM, André G. Isaak wrote:
On 2022-04-10 17:08, olcott wrote:
On 4/10/2022 5:59 PM, André G. Isaak wrote:
On 2022-04-10 16:40, olcott wrote:
On 4/10/2022 5:35 PM, André G. Isaak wrote:
On 2022-04-10 15:56, olcott wrote:
On 4/10/2022 4:49 PM, André G. Isaak wrote:
On 2022-04-10 15:00, olcott wrote:
On 4/10/2022 3:15 PM, olcott wrote:
On 4/10/2022 3:07 PM, André G. Isaak wrote:

I'm trying to get you to write using correct and coherent
notation. That's one of the things you'll need to be able to
do if you ever hope to publish. That involves remembering to
always include conditions and using the same terms in your
'equations' as in your text.

Not sure how that makes me a 'deceitful bastard'.

André


THAT you pretended to not know what I mean by embedded_H so
that you could artificially contrive a fake basis for rebuttal
when no actual basis for rebuttal exists makes you a deceitful
bastard.

IT IS THE CASE THAT the correctly simulated input ⟨Ĥ0⟩ ⟨Ĥ1⟩ to
embedded_H never reaches its own final state of ⟨Ĥ0.qy⟩ or
⟨Ĥ0.qn⟩ under any condition what-so-ever therefore ⟨Ĥ0⟩ ⟨Ĥ1⟩ is
proved to specify a non-halting sequence of configurations.

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ2⟩ ⊢* H.qn

This is now the third reply you've made to the same post.

That post didn't make any arguments whatsoever about your claims.
It simply pointed out that you are misusing your notation and
urged you to correct it.


THE NOTATION IS A STIPULATIVE DEFINITION THUS DISAGREEMENT IS
INCORRECT.

If the notation is junk, then the definition is also junk.

That's like "stipulating" that

+×yz÷² = ±z+³

It's meaningless because the notation is meaningless, much like
your notation above.

This is meaningless:

Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qy   // what's the condition?
Ĥ.q0 ⟨Ĥ0⟩ ⊢* H ⟨Ĥ0⟩ ⟨Ĥ1⟩ ⊢* H.qn   // what's the condition?

With no conditions specified, the above is just nonsense.

André


Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would reach its
final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by embedded_H would never reach
its final state.


This is still nonsense.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would reach its
own final state.

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn
If the correctly simulated input ⟨Ĥ⟩ ⟨Ĥ⟩ to embedded_H would never
reach its own final state.

And again you're still being inconsistent. You can either use H or use
embedded_H, but you can't mix the two.

Sure I can. I just did.
This means that H pretends that it is only a UTM to see what its
simulated input would do in this case. If it would never reach its own
final state then H correctly rejects this input.

A Turing Machine cannot "pretend" to be some different Turing Machine.
It can perform a pure simulation of its input until this simulated input
matches a repeating behavior pattern that proves this input never
reaches its own final state.

If that's the case, why does an actual UTM applied to the *same* input halt?

Hint: Because the result of an actual UTM applied to the input defines the correct answer, so H answers wrong.

Intuitively that would seem to be true, this intuition is incorrect.

The ultimate definition of correct is the computation of the mapping of the inputs to an accept or reject state on the basis of the behavior that these inputs specify.

That simulated inputs to embedded_H would never reach their own final state under any condition what-so-ever
IS THE ULTIMATE MEASURE OF THEIR HALTING BEHAVIOR
and conclusively proves they specify a non-halting sequence of configurations.



--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


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