Re: ChatGPT agrees that I have refuted the conventional Halting Problem proof technique --- Full 38 page analysis

Op 29.jun.2025 om 16:47 schreef olcott:

On 6/29/2025 4:31 AM, Mikko wrote:

On 2025-06-28 23:19:11 +0000, olcott said:
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On 6/28/2025 6:10 PM, Richard Damon wrote:

On 6/28/25 5:52 PM, olcott wrote:

On 6/28/2025 12:41 PM, Richard Damon wrote:

On 6/28/25 9:54 AM, olcott wrote:

On 6/28/2025 7:04 AM, Mikko wrote:

On 2025-06-27 14:19:28 +0000, olcott said:
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On 6/27/2025 1:55 AM, Mikko wrote:

On 2025-06-27 02:58:47 +0000, olcott said:
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On 6/26/2025 5:16 AM, Mikko wrote:

On 2025-06-25 15:42:36 +0000, olcott said:
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On 6/25/2025 2:38 AM, Mikko wrote:

On 2025-06-24 14:39:52 +0000, olcott said:
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*ChatGPT and I agree that*
The directly executed DDD() is merely the first step of
otherwise infinitely recursive emulation that is terminated
at its second step.

>
No matter who agrees, the directly executed DDD is mote than
merely the first step of otherwise infinitely recursive
emulation that is terminated at its second step. Not much
more but anyway. After the return of HHH(DDD) there is the
return from DDD which is the last thing DDD does before its
termination.

>
*HHH(DDD) the input to HHH specifies non-terminating behavior*
The fact that DDD() itself halts does not contradict that
because the directly executing DDD() cannot possibly be an
input to HHH in the Turing machine model of computation,
thus is outside of the domain of HHH.

>
The input in HHH(DDD) is the same DDD that is executed in DDD()
so the behaviour specified by the input is the behavour of
directly executed DDD, a part of which is the behaour of the
HHH that DDD calls.
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If HHH does not report about DDD but instead reports about itself
or its own actions it is not a partial halt decideer nor a partial
termination analyzer, as those are not allowed to report on their
own behavour more than "cannot determine".

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Functions computed by Turing Machines are required to compute
the mapping from their inputs and not allowed to take other
executing Turing machines as inputs.

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There is no restriction on the functions.

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counter factual.

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That is not a magic spell to create a restriction on functions.
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A Turing machine is required
to compute the function identified in its specification and no other
function. For the halting problem the specification is that a halting
decider must compute the mapping that maps to "yes" if the computation
described by the input halts when directly executed.

>
No one ever bothered to notice that because directly
executed Turing machines cannot possibly be inputs to
other Turing machines that these directly executed
Turing machines have never been in the domain of any
Turing machine.

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Irrelevant. They are the domain of the halting problem.

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That they are in the domain of the halting problem
and not in the domain of any Turing machine proves
that the requirement of the halting problem is incorrect.

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No, it just says that you don't understand the concept of representation.
>

>
There exists no finite number of steps where N steps of
DDD are correctly simulated by HHH and this simulated DDD
reaches its simulated "return" statement final halts state.
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But there is no HHH that correctly simulates the DDD that the HHH that answers,

Proven to be counter-factual and over your head.
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void Infinite_Recursion()
{
   Infinite_Recursion();
   return;
}
>
The exact same code that correctly recognizes infinite
recursion sees this non-terminating pattern after one
single recursive emulation.

>
Recursive simulation is not the same as recorsive call. Consequently
what is correct about recursive calls may be incorrect about
recursive simulation.
>

 Actually from the POV of HHH it is exactly the same
as if DDD() called HHH(DDD) that simply calls DDD().
HHH has no idea that DDD is calling itself.
 It sees DDD call the same function twice in sequence
with no conditional branch instructions inbetween the
beginning of DDD and its called to HHH(DDD).
 There are conditional branch instructions in HHH
that HHH does ignore. These are irrelevant. They
cannot possibly cause the simulated DDD to reach
its own simulated final halt state, the correct
measure of halting.
 

Exactly these conditional branch instruction are the cause for the abort done by HHH, which then returns to DDD, which then halts.
This is shown by world class smulators and also by HHH1, which does count these conditional branch instructions and, therefore, is able to reach the end of the simulation.
That you do not understand it, does not make it incorrect.
It means that HHH is incorrect to abort the simulation before it can see that the simulated HHH would do the abort.
These conditional branch instructions are part of the specification in the input.
That HHH does not use them, does not change the specification, it only demonstrates the failure of HHH to reach the natural end of the simulation.
Your only rebuttal in all these years is repeating the same claims without any evidence. That does not count as rebuttal.