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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD simulated by HHH cannot possibly halt (Halting Problem)
Date: Thu, 10 Apr 2025 19:27:23 -0400
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On 4/10/2025 7:26 PM, olcott wrote:
> On 4/10/2025 3:55 AM, Fred. Zwarts wrote:
>> Op 10.apr.2025 om 03:47 schreef olcott:
>>> 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
>>>
>> So the proof by induction shows that for any n HHH fails to complete 
>> the simulation. So, it has been proven that no HHH exists that is able 
>> to simulate correctly. It always aborts before it sees that the 
>> simulated HHH aborts as well.
> 
> Yet again over-your-head.
> Unless the first HHH aborts

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