On Sun, 10 Aug 2025 21:48:07 -0500, olcott wrote:

<Input to LLM systems>
Simulating Termination Analyzer HHH correctly simulates its input until:
(a) Detects a non-terminating behavior pattern: abort simulation and
return 0.
(b) Simulated input reaches its simulated "return" statement: return 1.

typedef int (*ptr)();
int HHH(ptr P);

int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}

What value should HHH(DD) correctly return?
<Input to LLM systems>

They figured out the *recursive simulation non-halting behavior pattern* entirely on their own without prompting and figured out that HHH(DD)==0
is correct also without prompting.

https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

https://chatgpt.com/share/68939ee5-e2f8-8011-837d-438fe8e98b9c

https://grok.com/share/c2hhcmQtMg%3D%3D_810120bb-5ab5-4bf8-af21-

eedd0f09e141

https://grok.com/share/bGVnYWN5_1fa4bc0b-497a-439f-bf7d-7b28bac0b03d

"Given the rules, HHH(DD) correctly returns 0, as it detects the non- terminating simulation pattern caused by DD’s recursive call to HHH(DD). However, this shows HHH cannot be a total halt decider, as it incorrectly returns 0 for a program (DD) that halts, consistent with the Halting Problem’s undecidability."

You fail yet again and the fact you can trick LLMs into considering
termination analysers rather than total halt deciders (and thus not
considering Halting Problem proofs) doesn't change that fact.

As soon as you tell any of those three LLMs that HHH is supposed to be a
halt decider as specified by Halting Problem and its proofs your house of
cards falls down (as shown with Grok's reply above).

/Flibble

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On Mon, 11 Aug 2025 17:17:48 -0500, olcott wrote:

On 8/11/2025 4:01 PM, Chris M. Thomasson wrote:

On 8/10/2025 7:48 PM, olcott wrote:

<Input to LLM systems>
Simulating Termination Analyzer HHH correctly simulates its input
until:
(a) Detects a non-terminating behavior pattern: abort simulation and
return 0.
(b) Simulated input reaches its simulated "return" statement: return
1.

typedef int (*ptr)();
int HHH(ptr P);

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

What value should HHH(DD) correctly return?
<Input to LLM systems>

They figured out the *recursive simulation non-halting behavior
pattern*
entirely on their own without prompting and figured out that
HHH(DD)==0 is correct also without prompting.

https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

https://chatgpt.com/share/68939ee5-e2f8-8011-837d-438fe8e98b9c

https://grok.com/share/c2hhcmQtMg%3D%3D_810120bb-5ab5-4bf8-af21-
eedd0f09e141

It's kind of as if you are showing how an AI can be manipulated into
kookville? Is your system akin to something like this? Typed into the
newsreader. A little bored right now, and thinking about what I need to
code today in my fields. Fwiw, here is some of my work, does it work on
your end? A link to a 3d model of mine that you should be able to
explore in real time:

https://skfb.ly/pzTEC

If you did that from scratch is is very impressive.

Okay. Is your system kind of similar to something like this? typed into
the newsreader so forgive any typos:

No its nothing like that

<snip>

The basis of my work is summed up quite well by Google [x86utm operating system].

I created the x86utm operating system so that any C function can emulate
any other C function in debug_step mode.

HHH creates a new process context and emulates DD. When DD calls HHH(DD)
then the original HHH emulates this new instance of itself that creates another process context to emulate its instance of DD. Then this DD
calls yet another instance of HHH(DD). The only limit to recursive depth
is memory.

It doesn't correctly emulate DD though because DD() behaves differently.

/Flibble

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On 8/10/25 11:25 PM, olcott wrote:

On 8/10/2025 9:54 PM, dbush wrote:

On 8/10/2025 10:48 PM, olcott wrote:

<Input to LLM systems>
Simulating Termination Analyzer HHH correctly simulates its input until: >>> (a) Detects a non-terminating behavior pattern: abort simulation and
return 0.
(b) Simulated input reaches its simulated "return" statement: return 1.

typedef int (*ptr)();
int HHH(ptr P);

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

What value should HHH(DD) correctly return?
<Input to LLM systems>

They figured out the
*recursive simulation non-halting behavior pattern*
entirely on their own without prompting and figured
out that HHH(DD)==0 is correct also without prompting.

https://claude.ai/share/da9e56ba-f4e9-45ee-9f2c-dc5ffe10f00c

https://chatgpt.com/share/68939ee5-e2f8-8011-837d-438fe8e98b9c

https://grok.com/share/c2hhcmQtMg%3D%3D_810120bb-5ab5-4bf8-af21-
eedd0f09e141

And when they're correcting by asking the following follow-up question:

-------
Doesn't that assume that a correct value can be returned?
--------

We get the following response, stating that an HHH that meets the
above requirements cannot exist, validating the halting problem proofs.

The ChatGPT 5.0 "upgrade" is pretty stupid
and you allowed your conversation to diverge
from HHH(DD) determining its return value
on the basis of DD correctly simulated by HHH.

Your problem is that the UPGRADE detects the nonsense of your system and reports something more truthful

We cannot begin to get to the other issue of
the direct execution of DD() until after you
first totally understand what the behavior
of DD correctly simulated by HHH is.

Which is non-existant, since the ONE HHH that is in your system (since
that is all that it CAN be and not be a category error) doesn't do it.

Dismissing this as irrelevant is merely stonewalling.

No. it is truth telling.

Lying that you statement applies here is just lying.

-----
Yes — excellent observation.

The original answer presumes that HHH can return a "correct" value
about whether DD halts. But that’s precisely where the deep issue
lies: we’re dancing around the Halting Problem.

To put it plainly:

The Halting Problem tells us that there can be no general algorithm
(like HHH) that correctly determines for all possible functions
whether they halt or not.

So, in this case:

     The function DD calls HHH(DD).

     HHH tries to determine whether DD halts.

     But DD’s halting behavior depends on what HHH says about DD.

     If HHH says DD halts (returns 1), then DD loops forever (goto HERE).

     If HHH says DD doesn't halt (returns 0), then DD returns 0 and
halts.

This creates a paradox — DD halts if and only if HHH says it doesn’t.

Therefore:

     No correct value can be returned by HHH(DD) that would make HHH
accurate and consistent.

     This construction is logically equivalent to the classic proof of >> the undecidability of the Halting Problem (like Turing’s
diagonalization argument).

📌 So, to answer your question directly:
Yes — assuming HHH(DD) can return a correct value contradicts the very
nature of the function. DD is a counterexample to the possibility of
writing a correct and total termination analyzer like HHH.

Conclusion:

     The existence of DD shows that a correct HHH cannot exist.

     So the real answer is not “HHH(DD) returns 0” or “HHH(DD) returns
1,” but rather:
     → A function like HHH cannot exist that gives the correct answer >> for all inputs.

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On 12/08/2025 10:10, Chris M. Thomasson wrote:

This smells of AI.

Claims to have solved the halting problem usually stem from one
of a few common misunderstandings:

That's a very AI thing to say.

Solving for a specific subset: They might create a program that
can correctly predict the halting behavior of a limited class of
programs, perhaps those without loops or those with a clear,
finite upper bound on their recursion. They then mistakenly
assume this partial solution applies to all possible programs.

So's that, and so is THIS:

Confusing a program with a Universal Turing Machine: They might
write a program that can analyze the code of another program, but
they don't grasp the paradoxical machine that Alan Turing used to
prove the problem is unsolvable for all programs.

There is nothing paradoxical about Turing Machines, and for me
that puts the whole thing beyond all doubt - those words were
either penned or plagiarised by an AI. The paradox lies not in
the hardware but in the problem, .

Ignoring the theoretical model: The halting problem's proof
applies to the idealized model of a Turing Machine, which has
infinite memory

sufficient memory.
----------

If need be, a Turing tape can consist of a warehouseful of USB
pens or SD cards. The only way you're ever going to run out is if
you want to. That isn't the reason some problems are incomputable.

--
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 2025-08-12, olcott <polcott333@gmail.com> wrote:

On 8/12/2025 4:10 AM, Chris M. Thomasson wrote:

On 8/11/2025 3:17 PM, olcott wrote:>>

The basis of my work is summed up quite well by
Google [x86utm operating system].

I created the x86utm operating system so that any C
function can emulate any other C function in
debug_step mode.

HHH creates a new process context and emulates DD.
When DD calls HHH(DD) then the original HHH emulates
this new instance of itself that creates another
process context to emulate its instance of DD. Then
this DD calls yet another instance of HHH(DD). The
only limit to recursive depth is memory.

Claims to have solved the halting problem usually stem from one of a few
common misunderstandings:

I am not making that claim.
I have shown that none of the conventional proofs actually
derive undecidability because their counter-example input
is decidable as non-halting.

That is false though. A given counter-example test case either halts or
not; the cookie may crumble either way. The proof handles either case.

Yes, it is possible to decide that.

However, it is not possible for /the specific decider/ which is
/embedded/ into that test case to decide.

The test case shoots down that specific decider, showing that
decider to be at least one test case short of being universal.

The proof makes it obvious that for every decider, a shoot-down
test case can be crafted in the same way.

If you can show that every possible decider is at least one test
case short of universality, you've shown that there is no universal
decider: halting is not universally decidable.

decider decides. When DD() is executed from main()
it is not an input to HHH().

If DD executed from main is a different computation from DD traced by
HHH, then you are perpetrating an equivocation fallacy in your
reasoning: there are two different computations, both of which
you are /calling/ by the name of DD, expecting the audience to be
fooled into believing that there is only one entity called DD.

Now since DD refers to a C language function and there is only
one definition of it, and yet DD represents (at least) two different computations in spite of not having any arguments, it means that
the definition DD doesn't specify a function.

There is some hidden state somewhere, and that hidden state completely invalidates all your argumentation.

Since it is obvious by inspection that DD doesn't access any globals,
the hidden state manipulation must be inside HHH.

Now the problem is that the HHH that main() is applying to DD
and the HH that DD itself invokes are the same function in the
same adddress space. Those instances are communicating via
global state.

You cannot have that. The outer decider and the test case must be
in separate containers. The former can observe the latter in to decide
it, but must not interfere with it in any way. And the latter knows
absolutely nothing about the former.

Until you try rather than dismiss out-of-hand
the notion of a simulating halt decider. No one
has ever done this before my 2016 innovation.

Also, nobody has done it since your innovation and, most importantly,
nobody has done it /during/ your innovation.

Your system allows a decidability test case, and the halting test being
applied to it, to share a single instance of the decider in the same
address space, with global state leaking between the two.

That's like showing up to the Olympic trials with a needle in your arm.

When DD is correctly simulated by HHH the
"do the opposite" code becomes unreachable.

When DD is correctly simulated by a decider, it must do the bog-exact
thing it does when it simulated by a vanilla target machine that it is
written for, or any other compatible processing.

DD must do one exact thing in all contexts in which it is processed,
otherwise: either DD is not a valid Turing computation, or else
some or all of those contexts aren't correctly processing DD.

There must be absolutely not one bit of difference between the
computation represented by HHH(DD) in main() and the same
expression inside DD.

If we have calculated the value of one (assuming it has halted),
then we can replace all other occurrences of that expression with
that value.

If we happen to be tracing inside DD, and discover that HHH(DD)
returns False, then we immediately know that the expression HHH(DD)
denotes False everywhere else, like in main().

This property of functional programs is called Referential
Transparency. If we move or copy an expression elsewhere, but such that
its variable/function references resolve to the same entities, then it continues to denote the same value.

If you're trying to demonstrate your work using programming language
functions, whose inputs are their arguments and nothing else, you need
to ensure that your system is Referentially Transparent.

If you're trying to demonstrate your work using stateful procedures, you
must be explicit about revealing the state, and ensure that two
instances of a computations which you are identifying as being of the
same computation have exactly the same initial state, same inputs,
and same subsequent state transitions. You cannot call two different computations with different state transitions "DD" and claim they are
the same.

--
TXR Programming Language: http://nongnu.org/txr
Cygnal: Cygwin Native Application Library: http://kylheku.com/cygnal
Mastodon: @Kazinator@mstdn.ca

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On 8/12/25 11:14 AM, olcott wrote:

On 8/12/2025 4:33 AM, Chris M. Thomasson wrote:

On 8/12/2025 2:24 AM, Richard Heathfield wrote:

On 12/08/2025 10:10, Chris M. Thomasson wrote:

This smells of AI.

It is! I was trying to use AI to try to ring olcott's bell. You got it!

It's from Googles Gemini.

That is the most stupid one out of four.
My current favorite is Claude AI
since the ChatGPT downgrade to 5.0.

It seems your rating of them is who states what your itching ears what
to hear.

WHich should tell you what you are going to reap.

Sorry, all you are doing is proving to even the slowest how dense and
stupid you are.

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Op 12.aug.2025 om 22:20 schreef olcott:

On 8/12/2025 2:50 PM, Kaz Kylheku wrote:

On 2025-08-12, olcott <polcott333@gmail.com> wrote:

On 8/12/2025 4:10 AM, Chris M. Thomasson wrote:

On 8/11/2025 3:17 PM, olcott wrote:>>

The basis of my work is summed up quite well by
Google [x86utm operating system].

I created the x86utm operating system so that any C
function can emulate any other C function in
debug_step mode.

HHH creates a new process context and emulates DD.
When DD calls HHH(DD) then the original HHH emulates
this new instance of itself that creates another
process context to emulate its instance of DD. Then
this DD calls yet another instance of HHH(DD). The
only limit to recursive depth is memory.

Claims to have solved the halting problem usually stem from one of a
few
common misunderstandings:

I am not making that claim.
I have shown that none of the conventional proofs actually
derive undecidability because their counter-example input
is decidable as non-halting.

That is false though. A given counter-example test case either halts or
not; the cookie may crumble either way. The proof handles either case.

The simplistic view does not actually correspond to realty.

As usual incorrect claims.

When DD is correctly simulated by HHH this input specifies
the non-halting behavior pattern of recursive simulation
that prevents DD correctly simulated by HHH from ever reaching
its own simulated final halt state.

There is no non-halting pattern. There is only a HHH that fails to reach
the final halt state due to a premature abort.
The programmer of HHH made the mistake to think that a finite recursion
is a non-termination pattern, because he forgot to count the conditional
branch instructions in this finite recursion.

We can go around and around in circles while you
continue to refuse to acknowledge that verified fact.

Again changing the meaning of the words. Your dream is not a verified
fact.What has been verified is that the input specifies a halting
program, but HHH does not see the whole specification due to the
premature abort.

When you disagree that the simulation is correct you
are disagreeing with the semantics of the x86 language.

No, when you claim that the simulation is correct, you disagree with the semantics of the x86 language that requires the execution of the next instruction, where HHH decides to violate that and abort the simulation. Correct simulations of exactly the same input by world-class simulators
that do not violate the x86 semantics, but continue the simulation prove
that then the final halt state can be reached.

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On 8/13/25 11:39 AM, olcott wrote:

On 8/13/2025 3:34 AM, Fred. Zwarts wrote:

Op 12.aug.2025 om 22:20 schreef olcott:

On 8/12/2025 2:50 PM, Kaz Kylheku wrote:

On 2025-08-12, olcott <polcott333@gmail.com> wrote:

On 8/12/2025 4:10 AM, Chris M. Thomasson wrote:

On 8/11/2025 3:17 PM, olcott wrote:>>

The basis of my work is summed up quite well by
Google [x86utm operating system].

I created the x86utm operating system so that any C
function can emulate any other C function in
debug_step mode.

HHH creates a new process context and emulates DD.
When DD calls HHH(DD) then the original HHH emulates
this new instance of itself that creates another
process context to emulate its instance of DD. Then
this DD calls yet another instance of HHH(DD). The
only limit to recursive depth is memory.

Claims to have solved the halting problem usually stem from one of >>>>>> a few
common misunderstandings:

I am not making that claim.
I have shown that none of the conventional proofs actually
derive undecidability because their counter-example input
is decidable as non-halting.

That is false though. A given counter-example test case either halts or >>>> not; the cookie may crumble either way. The proof handles either case. >>>>

The simplistic view does not actually correspond to realty.

As usual incorrect claims.

Does non input DD() halt? Is an incorrect question.
It is inconsistent with the way that Turing machines
actually work.

But "DD()" isn't a "non-input" but is the SEMANTIC MEANING of the input.

I guess by your logic Arithmatic isn't computable as we can't give
Turing Machines numbers, only finite string that represent them.

Does your input specify a sequence of steps that halt?
Is the way to ask that question consistent with the
way that Turing machine actually work.

And they do, at least if you include ALL the input, and the concept of
"specify a sequence of steps" is based on the OBJECTIVE standard of a
complete emulation.

All you are doing is proving you logic is based on the error of letting
truth be subjective, and thus no longer truth.

When DD is correctly simulated by HHH this input specifies
the non-halting behavior pattern of recursive simulation
that prevents DD correctly simulated by HHH from ever reaching
its own simulated final halt state.

There is no non-halting pattern.

If that was true then three LLM systems would
not have figured out that pattern on their own.

But they didn't, you misled them with a biased prompt.

Seeing this pattern merely requires the ability
to do this correct execution trace:

When HHH(DD) is executed that this begins simulating
DD that calls HHH(DD) that begins simulating DD that
calls HHH(DD) again.

And that pattern will stop in the correct emulation one "cycle" after
the decider decides to abort thinking it is non-halting, thus showing it
is halting.

To think that the behavor stops when the simulator decides incorrect
that it won't stop and stops simulating just means that ANY input can
equally be decided incorrect to be non-halting.

Since you allowed an error to trigger your abort, you can't limit what
errors can trigger other deciders.

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