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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: ChatGPT agrees that I have refuted the conventional Halting Problem proof technique --- Full 38 page analysis
Date: Wed, 25 Jun 2025 10:32:09 +0300
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On 2025-06-24 14:09:10 +0000, olcott said:

> On 6/24/2025 4:27 AM, joes wrote:
>> Am Mon, 23 Jun 2025 16:28:23 -0500 schrieb olcott:
>>> On 6/23/2025 2:58 PM, joes wrote:
>>>> Am Mon, 23 Jun 2025 12:40:43 -0500 schrieb olcott:
>>>>> On 6/23/2025 10:34 AM, joes wrote:
>>>>>> Am Mon, 23 Jun 2025 09:30:07 -0500 schrieb olcott:
>> 
>>>>>>> If you read the 38 pages you will see how this is incorrect. ChatGPT
>>>>>>> "understands" that any program that must be aborted at some point to
>>>>>>> prevent its infinite execution is not a halting program.
>>>>>> Such as HHH, making it not a decider (when simulated).
>>>>>> 
>>>>> My claim is that
>>>> [blah blah non sequitur]
>>>> Well MY claim is that HHH simulated HHH (itself) doesn't halt.
>>>> 
>>>>> obvious
>>>> You know what, it actually IS obvious that HHH can't simulate past the
>>>> call to HHH. Thanks for coming to my Ted talk.
>>>> 
>>> Thus when HHH is simulating DDD and DDD calls HHH(DDD) the outer HHH
>>> does simulate itself simulating DDD.
>> Sure, it simulates *into* the call, but it never returns, which is
>> precisely why you abort it.
>> 
>> [more irrelevant stuff]
> 
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> *This is the question that HHH(DDD) correctly answers*
> Can DDD correctly simulated by any termination analyzer
> HHH that can possibly exist reach its own "return" statement
> final halt state?

Answering that question prevents HHH(DDD) from answering any
other question because it can only answer one question.
A termination analyzer is required to answer a different
question, which HHH(DDD) does not. Therefore HHH is not a
termination analyzer.

-- 
Mikko