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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD simulated by HHH cannot possibly halt (Halting Problem)
Date: Wed, 9 Apr 2025 20:47:32 -0500
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On 4/9/2025 3:56 PM, dbush wrote:
> On 4/9/2025 4:35 PM, olcott wrote:
>> On 4/9/2025 1:58 PM, Fred. Zwarts wrote:
>>> 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.
>>
>> Simply over your level of technical competence.
>>
>>> But HHH failed to complete the simulation of the halting program, 
>>
>> HHH is only required to report on the behavior of its
>> own correct simulation (meaning the according to the
>> semantics of the C programming language) and would be
>> incorrect to report on any other behavior.
> 
> Which means HHH has conflicting requirements,

No, it just means that the ones that you have
been saying are f-cked up and no-one noticed this before.

 > because to perform a
 > correct simulation of its input it cannot halt itself, and therefore
 > can't report that.
In other words you simply "don't believe in" the variant
form of mathematical induction that HHH uses.

A proof by induction consists of two cases. The first, the base case, 
proves the statement for 𝑛=0 without assuming any knowledge of other 
cases. The second case, the induction step, proves that if the statement 
holds for any given case 𝑛=k, then it must also hold for the next case 
𝑛=k+1. These two steps establish that the statement holds for every 
natural number 𝑛. The base case does not necessarily begin with 𝑛=0, 
but often with 𝑛=1, and possibly with any fixed natural number 𝑛=𝒩, 
establishing the truth of the statement for all natural numbers 𝑛 ≥ 𝒩. 
  https://en.wikipedia.org/wiki/Mathematical_induction



-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer