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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: DDD simulated by HHH cannot possibly halt (Halting Problem)
Date: Wed, 9 Apr 2025 20:58:54 +0200
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Op 09.apr.2025 om 19:29 schreef olcott:
> 
> On 4/8/2025 10:31 AM, Fred. Zwarts wrote:
>> Op 08.apr.2025 om 17:13 schreef olcott:
>>> On 4/8/2025 2:45 AM, Fred. Zwarts wrote:
>>>> Op 08.apr.2025 om 06:33 schreef olcott:
>>>>>
>>>>> typedef void (*ptr)();
>>>>> int HHH(ptr P);
>>>>>
>>>>> int DD()
>>>>> {
>>>>>    int Halt_Status = HHH(DD);
>>>>>    if (Halt_Status)
>>>>>      HERE: goto HERE;
>>>>>    return Halt_Status;
>>>>> }
>>>>>
>>>>> int main()
>>>>> {
>>>>>    HHH(DD);
>>>>> }
>>>>>
>>>>> *Simulating termination analyzer Principle*
>>>>> It is always correct for any simulating termination
>>>>> analyzer to stop simulating and reject any input that
>>>>> would otherwise prevent its own termination.
>>>>
>>>>
>>>> In this case there is nothing to prevent, because the finite string 
>>>> specifies a program that halts. 
>>>
>>> int DD()
>>> {
>>>    int Halt_Status = HHH(DD);
>>>    if (Halt_Status)
>>>      HERE: goto HERE;
>>>    return Halt_Status;
>>> }
>>>
>>> This stuff is simply over-your-head.
>>> HHH(DD) meets the above: *Simulating termination analyzer Principle*
>>> Anyone with sufficient competence with the C programming language
>>> will understand this.
>>>
>> Everyone with a little bit of C knowledge understands that if HHH 
>> returns with a value 0, then DDD halts. 
> 
> DDD CORRECTLY SIMULATED BY HHH
> NOT ANY OTHER DAMN DDD IN THE UNIVERSE NITWIT.
> 
If HHH would correctly simulate DD (and the functions called by DD) then 
the simulated HHH would return to DD and DD would halt.
But HHH failed to complete the simulation of the halting program, 
because the programmer was dreaming of an infinite recursion.

If I didn't have to tell you this hundreds of times and you didn't
persist in the straw-man deception I would not have called you a nitwit.