Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail
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: Tue, 1 Jul 2025 10:40:21 +0200
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Op 30.jun.2025 om 18:20 schreef olcott:
> On 6/30/2025 2:35 AM, Fred. Zwarts wrote:
>> 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:
>>>>
>>>>> 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:
>>>>>>>>>>
>>>>>>>>>>> On 6/27/2025 1:55 AM, Mikko wrote:
>>>>>>>>>>>> On 2025-06-27 02:58:47 +0000, olcott said:
>>>>>>>>>>>>
>>>>>>>>>>>>> On 6/26/2025 5:16 AM, Mikko wrote:
>>>>>>>>>>>>>> On 2025-06-25 15:42:36 +0000, olcott said:
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> 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.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> 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".
>>>>>>>>>>>>>
>>>>>>>>>>>>> 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.
>>>>>>>>>>>>
>>>>>>>>>>>> There is no restriction on the functions.
>>>>>>>>>>>
>>>>>>>>>>> counter factual.
>>>>>>>>>>
>>>>>>>>>> That is not a magic spell to create a restriction on functions.
>>>>>>>>>>
>>>>>>>>>>>> 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.
>>>>>>>>>>
>>>>>>>>>> Irrelevant. They are the domain of the halting problem.
>>>>>>>>>
>>>>>>>>> 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.
>>>>>>>>
>>>>>>>> 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.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> But there is no HHH that correctly simulates the DDD that the HHH 
>>>>>> that answers,
>>>>> Proven to be counter-factual and over your head.
>>>>>
>>>>> 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.
> 
> *Counter-factual*

Only if you claim that your dreams are facts.

> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> int main()
> {
>    HHH(DDD);
> }
> 
> *In the above nothing returns to DDD*

Only if HHH does not return. But we know HHH has code to abort and DDD 
will halt (if not prevented by a premature abort).

> *I always quit at the first counter-factual error*

Even if it is not counter-factual.

> 
> As soon as DDD aborts its outermost DDD simulation
> every recursive simulation immediately stops and
> does not return to anywhere. Then HHH returns 0 to main.

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