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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 14:37:10 -0500
Organization: i2pn2 (i2pn.org)
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On 2/9/25 2:33 PM, olcott wrote:
> On 2/9/2025 1:30 PM, Fred. Zwarts wrote:
>> Op 09.feb.2025 om 20:00 schreef olcott:
>>> On 2/9/2025 12:47 PM, Fred. Zwarts wrote:
>>>> Op 09.feb.2025 om 17:49 schreef olcott:
>>>>> 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.
>>>>
>>>> No sense to talk to people denying verified facts.
>>>>
>>>>>
>>>>> DD simulated by HHH cannot possibly terminate normally.
>>>>
>>>> Indeed, because HHH fails to simulate itself up to the end.
>>>> This is verified with:
>>>>
>>>>         int main() {
>>>>           return HHH(main);
>>>>         }
>>>>
>>>
>>> There is no simulating itself to the end with the above
>>> example either. 
>>
>> It seems Olcott misses trivial facts. HHH simulates main, which calls 
>> HHH, so HHH needs to simulate HHH. HHH is unable to complet the 
>> simulation up to the end, because it 
> 
> specifies a computation that cannot possibly terminate normally.
> 
> 

Right, and since an actual call to DD() will return since you say 
HHH(DD) "correctly" returns 0, we see that DD() halts.

Thus, HHH "simulation" must be incorrect.

Your problem is assume you can change the HHH that DD calls, because you 
think LYING is valid logic.

Sorry, all you are doing to proving that you are nothing but a ignorant 
FRAUD.