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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: ChatGPT agrees that I have refuted the conventional Halting
 Problem proof technique --- Full 38 page analysis
Date: Fri, 27 Jun 2025 09:59:12 +0200
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Op 27.jun.2025 om 06:26 schreef olcott:
> On 6/26/2025 3:46 AM, Fred. Zwarts wrote:
>> Op 25.jun.2025 om 17:42 schreef olcott:
>>> On 6/25/2025 2:38 AM, Mikko wrote:
>>>> On 2025-06-24 14:39:52 +0000, olcott said:
>>>>
>>>>> *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.
>>>
>>
>> Why repeating claims that have been proven incorrect.
>> The input to HHH is a pointer to code, that includes the code of HHH, 
>> including the code to abort and halt. Therefore, it specifies a 
>> halting program.
> 
> *No, you are using an incorrect measure*
> *I have addressed this too many times*

.... with invalid measures.
The measure is not whether the simulator can do its job, the measure is 
what the input specifies.

> 
> DDD correctly simulated by HHH cannot possibly
> reach its own simulated "return" statement
> final halt state *No matter what HHH does*
> Therefore the input to HHH(DD) unequivocally
> specifies non-halting behavior.
> 

If the simulator cannot analyse this specification, that is a failure of 
the simulator, not the behaviour specified in the input.
'No matter what HHH does' is irrelevant and misleading. HHH is 
programmed to do only one thing. That defines the input.
*That* input specifies a program, including the HHH that aborts and 
returns to DDD, after which DDD halts. (Or are you still cheating with 
the Root variable, to make the simulated HHH different from the 
simulating HHH?)
That HHH is programmed to be blind for that behaviour does not change 
the specification.
A failure to process the input correctly, does not change the 
specification of the input.