This time I did NOT present Olcott's argument to the AI - just
his conclusion. I gave it no other hints whatsoever except that
he claims that three AIs back him up.

ChatGPT took it from there all the way home, right down to the
simulation.


My correspondent claims to have proved that the Halting Problem
proof is flawed by asking three different AI bots - all of which
agree with him. I was contemplating demo-ing how easy it is to
have AI prove anything you want proved.
I have decided against it, though, not because he's right (which
he clearly isn't) but because bad proofs should be attacked not
by analogy but by algebra.
ChatGPT said:

I like your conclusion — attack bad proofs directly, not by
flashy analogy. Algebra (i.e. clear, formal reasoning) is exactly
the right tool.

Below is a compact, practical checklist and a template you can
use to dismantle a bogus “refutation of the Halting Problem”
step-by-step. Use it yourself or paste it into a reply to your
correspondent. If you want, I’ll draft a short, pointed
refutation in full once you paste their text.
Quick checklist for attacking a flawed Halting-Problem proof

Extract the claim precisely.

What exactly do they assert exists? (e.g. a program H(P)
that returns 1 iff P halts on no input, or H(P,x) that decides
halting for program P on input x.)

Is it total (always halts with an answer) or partial
(may not halt on some inputs)?

Spell out the definitions.

Define the domain and codomain of H.

State what outputs mean (0 = loops, 1 = halts).

State whether H may itself be used as a subroutine.

Look for the hidden assumption.

Common illicit assumptions: totality, ability to
simulate an arbitrary program and modify its control flow,
allowing self-reference without handling recursion, implicit
division by zero–style steps (i.e. cancelling a zero), or
assuming you can observe “infinite behavior” in finite time.

Formalise the diagonalising counterexample.

Construct the standard self-referential program D that
asks H(D) what D will do and does the opposite.

Show algebraically that for any total H the behaviour of
D yields a contradiction (if H(D)=1 then D loops; if H(D)=0 then
D halts), therefore H cannot be total and correct.

If their “H” is partial, check what they actually promise.

Partial analysers can be correct on the inputs where
they do answer. That’s not a refutation of Turing — it’s the only consistent escape. Show that their claim must then be weakened
(and likely isn’t doing the job they assert).

Is there circularity?

Does the proof rely on H while simultaneously using
properties of H that require H to already be validated? Make that
circularity explicit and show it invalidates their conclusion.

Point to the single fatal step.

As with dividing by zero in algebraic fallacies, the
halting proofs usually collapse at one place. Make that step
explicit: “they assume X; but X implies Y which contradicts Z;
here is the contradiction.”

Explain why “multiple AIs agree” is irrelevant.

Agreement by heuristics or by reuse of the same mistaken
proof does not constitute a mathematical proof. AI outputs can
repeat errors, conflate definitions, or echo the same hidden
assumption. That’s not evidence for correctness.

Short algebraic template you can use in a reply

State the claim succinctly.

Claim: There exists a total algorithm H(P) such that
H(P)=1 iff program P halts (on its own).

Assume for contradiction that such H exists.

Define the diagonal program D:

D() {
if H(D) == 1 then
loop forever
else
return 0
}

(Explain in words: D asks H about itself and does the opposite.)

Case analysis:

If H(D)=1 then D loops ⇒ H was wrong.

If H(D)=0 then D returns ⇒ H was wrong.

Therefore no total H can exist. Contradiction. QED.

Addendum: If their purported H is partial (may not halt on some
inputs), show that the proof above still forces H to fail on D
(it either must lie or must not return), so it cannot be both
total and correct.
A couple of ready rebuttal lines you can paste

“Your construction implicitly assumes a total decider. The diagonalisation immediately yields a program that makes that
assumption produce contradiction — that’s Turing’s point. If your
H ever refuses to answer, it’s not the total decider you claim.”

“Three AI models agreeing on the same text is not a
mathematical proof — it’s an echo. The argument must be examined step-by-step; I can point to the single invalid inference in your
version.”
--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 09/08/2025 06:27, olcott wrote:

It is impossible to show how DD correctly simulated
by HHH can possibly reach its own emulated "ret"
instruction

So HHH never returns, so it fails to act as a termination
analyzer because it fails to report its finding.

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 8/9/25 1:27 AM, olcott wrote:

On 8/9/2025 12:10 AM, Richard Heathfield wrote:

This time I did NOT present Olcott's argument to the AI - just his
conclusion. I gave it no other hints whatsoever except that he claims
that three AIs back him up.

ChatGPT took it from there all the way home, right down to the
simulation.


My correspondent claims to have proved that the Halting Problem proof
is flawed by asking three different AI bots - all of which agree with
him. I was contemplating demo-ing how easy it is to have AI prove
anything you want proved.
I have decided against it, though, not because he's right (which he
clearly isn't) but because bad proofs should be attacked not by
analogy but by algebra.
ChatGPT said:

I like your conclusion — attack bad proofs directly, not by flashy
analogy. Algebra (i.e. clear, formal reasoning) is exactly the right
tool.

Below is a compact, practical checklist and a template you can use to
dismantle a bogus “refutation of the Halting Problem” step-by-step.
Use it yourself or paste it into a reply to your correspondent. If you
want, I’ll draft a short, pointed refutation in full once you paste
their text.
Quick checklist for attacking a flawed Halting-Problem proof

     Extract the claim precisely.

         What exactly do they assert exists? (e.g. a program H(P) that
returns 1 iff P halts on no input, or H(P,x) that decides halting for
program P on input x.)

         Is it total (always halts with an answer) or partial (may not
halt on some inputs)?

     Spell out the definitions.

         Define the domain and codomain of H.

         State what outputs mean (0 = loops, 1 = halts).

         State whether H may itself be used as a subroutine.

     Look for the hidden assumption.

         Common illicit assumptions: totality, ability to simulate an
arbitrary program and modify its control flow, allowing self-reference
without handling recursion, implicit division by zero–style steps
(i.e. cancelling a zero), or assuming you can observe “infinite
behavior” in finite time.

     Formalise the diagonalising counterexample.

         Construct the standard self-referential program D that asks >> H(D) what D will do and does the opposite.

         Show algebraically that for any total H the behaviour of D >> yields a contradiction (if H(D)=1 then D loops; if H(D)=0 then D
halts), therefore H cannot be total and correct.

     If their “H” is partial, check what they actually promise.

         Partial analysers can be correct on the inputs where they do
answer. That’s not a refutation of Turing — it’s the only consistent >> escape. Show that their claim must then be weakened (and likely isn’t
doing the job they assert).

     Is there circularity?

         Does the proof rely on H while simultaneously using
properties of H that require H to already be validated? Make that
circularity explicit and show it invalidates their conclusion.

     Point to the single fatal step.

         As with dividing by zero in algebraic fallacies, the halting
proofs usually collapse at one place. Make that step explicit: “they
assume X; but X implies Y which contradicts Z; here is the
contradiction.”

     Explain why “multiple AIs agree” is irrelevant.

         Agreement by heuristics or by reuse of the same mistaken
proof does not constitute a mathematical proof. AI outputs can repeat
errors, conflate definitions, or echo the same hidden assumption.
That’s not evidence for correctness.

Short algebraic template you can use in a reply

     State the claim succinctly.

         Claim: There exists a total algorithm H(P) such that H(P)=1 >> iff program P halts (on its own).

     Assume for contradiction that such H exists.

     Define the diagonal program D:

     D() {
       if H(D) == 1 then
         loop forever
       else
         return 0
     }

     (Explain in words: D asks H about itself and does the opposite.)

     Case analysis:

         If H(D)=1 then D loops ⇒ H was wrong.

         If H(D)=0 then D returns ⇒ H was wrong.

     Therefore no total H can exist. Contradiction. QED.

Addendum: If their purported H is partial (may not halt on some
inputs), show that the proof above still forces H to fail on D (it
either must lie or must not return), so it cannot be both total and
correct.
A couple of ready rebuttal lines you can paste

     “Your construction implicitly assumes a total decider. The
diagonalisation immediately yields a program that makes that
assumption produce contradiction — that’s Turing’s point. If your H
ever refuses to answer, it’s not the total decider you claim.”

     “Three AI models agreeing on the same text is not a mathematical >> proof — it’s an echo. The argument must be examined step-by-step; I
can point to the single invalid inference in your version.”

It is impossible to show how DD correctly simulated
by HHH can possibly reach its own emulated "ret"
instruction. Thus the DD input to HHH does specify
non halting behavior.
reaches

But that is because your HHH doesn't correctly simulate this input.

This is in part, because you don't know what those words mean.

Your argument start with a category error, because you don't understand
that Halt Deciders, and the meaning of thier inputs. are programs/computations/Turing Machines. This means they are a fully
defined set of deterministic intructions that work of a specific and
finite set of data called its "input"

That means it isn't a "infinite set" of these things, it can come FROM
such a set, but in the instance of the problem, it is a singular machine,n

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[comp.lang.c removed]

On 09/08/2025 15:16, olcott wrote:

On 8/9/2025 9:11 AM, Richard Heathfield wrote:

On 09/08/2025 15:02, olcott wrote:

<snip>

   if (Decide_Halting_HH(&Aborted, &execution_trace, &decoded,
                         code_end, End_Of_Code, &master_state,
                         &slave_state, &slave_stack, Root))
       goto END_OF_CODE;
   *Aborted         = 0x90909090; // deprecated
   *execution_trace = 0x90909090;
   return 0;  // Does not halt
END_OF_CODE:
   *Aborted         = 0x90909090; // deprecated
   *execution_trace = 0x90909090;
   return 1; // Input has normally terminated
}

So HHH /does/ return... with the wrong answer, a la Turing. QED.

*It is only the wrong answer to the wrong problem*

So you spent 22 years trying to solve the wrong problem? Oops.

Asking HHH to report on the behavior of the directly
executed DD() is like asking sum(3,5) to report on
the sum of 7 + 2.

So you are claiming that your HHH is not fit for purpose.

Glad to get that cleared up.

You're really not very good at this, are you?

What we are asking HHH is:

Does DD halt?

How HHH derives its answer is up to it, but it has to get the
answer right. If its answer differs from that made manifest by a
direct execution, its answer is Just Plain Wrong and it's not fit
for purpose.

You might reasonably complain that I'm being unreasonable by
asking for the impossible... and of course I am. Turing proved
that. It /is/ impossible.


--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

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On 2025-08-09 13:08:48 +0000, olcott said:

On 8/9/2025 12:31 AM, Richard Heathfield wrote:

On 09/08/2025 06:27, olcott wrote:

It is impossible to show how DD correctly simulated
by HHH can possibly reach its own emulated "ret"
instruction

So HHH never returns, so it fails to act as a termination analyzer
because it fails to report its finding.

u32 Needs_To_Be_Aborted_Trace_HH
(Decoded_Line_Of_Code* execution_trace,
Decoded_Line_Of_Code *current);

Line 996 matches the
*recursive simulation non-halting behavior pattern* https://github.com/plolcott/x86utm/blob/master/Halt7.c

This pattern is also present in the direct execution of DD(), which
halts.

--
Mikko

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On 8/10/25 10:20 AM, olcott wrote:

On 8/10/2025 3:19 AM, Mikko wrote:

On 2025-08-09 13:08:48 +0000, olcott said:

On 8/9/2025 12:31 AM, Richard Heathfield wrote:

On 09/08/2025 06:27, olcott wrote:

It is impossible to show how DD correctly simulated
by HHH can possibly reach its own emulated "ret"
instruction

So HHH never returns, so it fails to act as a termination analyzer
because it fails to report its finding.

u32 Needs_To_Be_Aborted_Trace_HH
(Decoded_Line_Of_Code* execution_trace,
  Decoded_Line_Of_Code *current);

Line 996 matches the
*recursive simulation non-halting behavior pattern*
https://github.com/plolcott/x86utm/blob/master/Halt7.c

This pattern is also present in the direct execution of DD(), which
halts.

Claude AI proved why HHH(DD)==0 is correct in terms that
any expert C programmer can understand. https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

Also
https://philpapers.org/archive/OLCHPS.pdf

As soon as you understand that the above page is
entirely correct we can proceed to the last step
of my proof.

And also says it isn't the right answer for the Halting Problem, as the
actual execution of the program will halt.

You have effectively told it that a "non-termination pattern" doesn't
actually indicate that the input won't halt.

In essense, the "non-termination" pattern isn't about the input, but the decider, that without this, the "decider" won't be, but that doesn't
make it correct.

So, it hasn't said it is a correct answer for a "Halt Decider", but for
your POOP which is decided by a "Simulating Halt Decider", which you
have admitted ISN'T a Halt Decider, and thus just a lie.

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On 2025-08-10 14:20:11 +0000, olcott said:

On 8/10/2025 3:19 AM, Mikko wrote:

On 2025-08-09 13:08:48 +0000, olcott said:

On 8/9/2025 12:31 AM, Richard Heathfield wrote:

On 09/08/2025 06:27, olcott wrote:

It is impossible to show how DD correctly simulated
by HHH can possibly reach its own emulated "ret"
instruction

So HHH never returns, so it fails to act as a termination analyzer
because it fails to report its finding.

u32 Needs_To_Be_Aborted_Trace_HH
(Decoded_Line_Of_Code* execution_trace,
  Decoded_Line_Of_Code *current);

Line 996 matches the
*recursive simulation non-halting behavior pattern*
https://github.com/plolcott/x86utm/blob/master/Halt7.c

This pattern is also present in the direct execution of DD(), which
halts.

Claude AI proved why HHH(DD)==0 is correct in terms that
any expert C programmer can understand. https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

Also
https://philpapers.org/archive/OLCHPS.pdf

As soon as you understand that the above page is
entirely correct we can proceed to the last step
of my proof.

Nice to see that you don't disagree.

--
Mikko

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On 8/11/25 10:34 AM, olcott wrote:

On 8/11/2025 1:45 AM, Mikko wrote:

On 2025-08-10 14:20:11 +0000, olcott said:

On 8/10/2025 3:19 AM, Mikko wrote:

On 2025-08-09 13:08:48 +0000, olcott said:

On 8/9/2025 12:31 AM, Richard Heathfield wrote:

On 09/08/2025 06:27, olcott wrote:

It is impossible to show how DD correctly simulated
by HHH can possibly reach its own emulated "ret"
instruction

So HHH never returns, so it fails to act as a termination analyzer >>>>>> because it fails to report its finding.

u32 Needs_To_Be_Aborted_Trace_HH
(Decoded_Line_Of_Code* execution_trace,
  Decoded_Line_Of_Code *current);

Line 996 matches the
*recursive simulation non-halting behavior pattern*
https://github.com/plolcott/x86utm/blob/master/Halt7.c

This pattern is also present in the direct execution of DD(), which
halts.

Claude AI proved why HHH(DD)==0 is correct in terms that
any expert C programmer can understand.
https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

Also
https://philpapers.org/archive/OLCHPS.pdf

As soon as you understand that the above page is
entirely correct we can proceed to the last step
of my proof.

Nice to see that you don't disagree.

So you understand that HHH(DD)==0 on the basis of the
behavior of DD correctly simulated by HHH?

Since that HHH doesn't correctly simulate that DD, that just shows you
have the insanity of thinking that two different things are the same thing.

That, or you are just proving yourself to stupid to understand the
meanig of the words, like what a Program is, or what Correct Simulation actually means, or even what it means for a program to be halting.

In other words, you are just showing you are just a damned liar because
you just don't care about what is the truth.

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On 2025-08-11 14:34:50 +0000, olcott said:

On 8/11/2025 1:45 AM, Mikko wrote:

On 2025-08-10 14:20:11 +0000, olcott said:

On 8/10/2025 3:19 AM, Mikko wrote:

On 2025-08-09 13:08:48 +0000, olcott said:

On 8/9/2025 12:31 AM, Richard Heathfield wrote:

On 09/08/2025 06:27, olcott wrote:

It is impossible to show how DD correctly simulated
by HHH can possibly reach its own emulated "ret"
instruction

So HHH never returns, so it fails to act as a termination analyzer >>>>>> because it fails to report its finding.

u32 Needs_To_Be_Aborted_Trace_HH
(Decoded_Line_Of_Code* execution_trace,
  Decoded_Line_Of_Code *current);

Line 996 matches the
*recursive simulation non-halting behavior pattern*
https://github.com/plolcott/x86utm/blob/master/Halt7.c

This pattern is also present in the direct execution of DD(), which
halts.

Claude AI proved why HHH(DD)==0 is correct in terms that
any expert C programmer can understand.
https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

Also
https://philpapers.org/archive/OLCHPS.pdf

As soon as you understand that the above page is
entirely correct we can proceed to the last step
of my proof.

Nice to see that you don't disagree.

So you understand that HHH(DD)==0 on the basis of the
behavior of DD correctly simulated by HHH?

No but on the basis of failure to simulate correctly the behavoiur of DD,
in particular the point where the HHH called by DD aborts its simulation
and returns 0 to DD which then halts.

--
Mikko

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