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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Anyone with sufficient knowledge of C knows that DD specifies
 non-terminating behavior to HHH
Date: Sun, 9 Feb 2025 12:05:54 -0500
Organization: i2pn2 (i2pn.org)
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On 2/9/25 11:49 AM, olcott wrote:
> On 2/9/2025 10:43 AM, Fred. Zwarts wrote:
>> Op 09.feb.2025 om 17:37 schreef olcott:
>>> On 2/9/2025 9:53 AM, Fred. Zwarts wrote:
>>>> Op 09.feb.2025 om 16:15 schreef olcott:
>>>>> On 2/9/2025 2:09 AM, Fred. Zwarts wrote:
>>>>>> Op 09.feb.2025 om 07:04 schreef olcott:
>>>>>>> On 2/8/2025 3:49 PM, Fred. Zwarts wrote:
>>>>>>>> Op 08.feb.2025 om 15:43 schreef olcott:
>>>>>>>>> On 2/8/2025 3:54 AM, Fred. Zwarts wrote:
>>>>>>>>>> Op 08.feb.2025 om 00:13 schreef olcott:
>>>>>>>>>>> Experts in the C programming language will know that DD
>>>>>>>>>>> correctly simulated by HHH cannot possibly reach its own
>>>>>>>>>>> "if" statement.
>>>>>>>>>>
>>>>>>>>>> Yes, it demonstrates the incapability of HHH to correctly 
>>>>>>>>>> determine the halting behaviour of DD
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> The finite string DD specifies non-terminating recursive
>>>>>>>>>>> simulation to simulating termination analyzer HHH. This
>>>>>>>>>>> makes HHH necessarily correct to reject its input as
>>>>>>>>>>> non-halting.
>>>>>>>>>>
>>>>>>>>>> The finite string defines one behaviour. This finite string, 
>>>>>>>>>> when given to an X86 processor shows halting behaviour. This 
>>>>>>>>>> finite string,when given to a world class simulator, shows 
>>>>>>>>>> halting behaviour. Only HHH fails to see this proven halting 
>>>>>>>>>> behaviour. So it proves the failure of HHH.
>>>>>>>>>> HHH aborts the simulation on unsound grounds one cycle before 
>>>>>>>>>> the simulation would terminate normally.
>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> 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);
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> https://www.researchgate.net/ 
>>>>>>>>>>> publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>>>>>>>>>>>
>>>>>>>>>>> https://github.com/plolcott/x86utm/blob/master/Halt7.c
>>>>>>>>>>> has fully operational HHH and DD
>>>>>>>>>>>
>>>>>>>>>>> The halting problem has always been a mathematical mapping
>>>>>>>>>>> from finite strings to behaviors. 
>>>>>>>>>>
>>>>>>>>>> Yes. And the behaviour of this finite string has been proven 
>>>>>>>>>> to show halting behaviour. Only Olcott's HHH fails to see it.
>>>>>>>>>> His misunderstanding is that he thinks that the behaviour 
>>>>>>>>>> defined by the finite string depends on the simulator.
>>>>>>>>>
>>>>>>>>> When DD calls HHH(DD) in recursive simulation it is a
>>>>>>>>> verified fact that DD cannot possibly halt.
>>>>>>>>
>>>>>>>> Which proves the failure of HHH. It does not reach the end of a 
>>>>>>>> halting program. All other methods show that DD halts.
>>>>>>>>
>>>>>>>
>>>>>>> Your comment only proves that you lack sufficient
>>>>>>> understanding of the C programming language.
>>>>>>>
>>>>>>
>>>>>> This is a proof of lack of logical reasoning.
>>>>>>
>>>>>> Verified fact 1: DD halts 
>>>>>
>>>>> Fallacy of equivocation error.
>>>>> (a) All men are mortal
>>>>> (b) No woman is a man
>>>>> ∴ No woman is mortal
>>>>
>>>> Yes, the claim that DD does not halt is indeed such a fallacy:
>>>>
>>>> (a) Direct execution and all simulators show that DD halts.
>>>> (b) My simulator is different
>>>>  > ∴ DD does not halt.
>>>>
>>>>>
>>>>> The input to HHH(DD) cannot possibly terminate normally.
>>>>> Referring to some other DD does not change this verfied fact.
>>>>>
>>>>>
>>>>
>>>> That DD halts is a verified fact. 
>>>
>>> The input to HHH(DD) DOES NOT HALT !!!
>>
>> It is a verified fact that the finite string describes a halting 
>> program. Du to a bug, HHH does not see that, because it investigates 
>> only the first few instructions of DD. HHH is unable to process the 
>> call from DD to HHH correctly.
> 
> No sense talking to people that lack sufficient technical
> skill to verify facts. There is no bug and you know it.

There is, and your refuse to deal with it just shows that it is YOU that 
lacks the technical skill you need.

> 
> DD simulated by HHH cannot possibly terminate normally.
> DD simulated by HHH does specify the behavioral basis
> of the Boolean termination value of the DD input to HHH.
> 

Because HHH isn't a correct simulator, and thus that doesn't matter.