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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Overview of proof that the input to HHH(DDD) specifies
 non-halting behavior --- Mike
Date: Thu, 15 Aug 2024 10:07:11 -0500
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On 8/15/2024 3:13 AM, Mikko wrote:
> On 2024-08-14 13:49:28 +0000, olcott said:
> 
>> On 8/14/2024 3:09 AM, Mikko wrote:
>>> On 2024-08-13 13:04:17 +0000, olcott said:
>>>
>>>> On 8/13/2024 5:57 AM, Mikko wrote:
>>>>> On 2024-08-13 01:43:49 +0000, olcott said:
>>>>>
>>>>>> We prove that the simulation is correct.
>>>>>> Then we prove that this simulation cannot possibly
>>>>>> reach its final halt state / ever stop running without being aborted.
>>>>>> The semantics of the x86 language conclusive proves this is true.
>>>>>>
>>>>>> Thus when we measure the behavior specified by this finite
>>>>>> string by DDD correctly simulated/emulated by HHH it specifies
>>>>>> non-halting behavior.
>>>>>>
>>>>>> https://www.researchgate.net/ 
>>>>>> publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>>>>>
>>>>> Input to HHH(DDD) is DDD. If there is any other input then the 
>>>>> proof is
>>>>> not interesting.
>>>>>
>>>>> The behviour specified by DDD on the first page of the linked article
>>>>> is halting if HHH(DDD) halts. Otherwise HHH is not interesting.
>>>>>
>>>>> Any proof of the false statement that "the input to HHH(DDD) specifies
>>>>> non-halting behaviour" is either uninteresting or unsound.
>>>>>
>>>>
>>>> void DDD()
>>>> {
>>>>    HHH(DDD);
>>>>    return;
>>>> }
>>>>
>>>> It is true that DDD correctly emulated by any HHH cannot
>>>> possibly reach its own "return" instruction final halt state.
>>>
>>> If DDD does not halt then HHH does not halt.
>>>
>>
>> _DDD()
>> [00002172] 55         push ebp      ; housekeeping
>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>> [00002175] 6872210000 push 00002172 ; push DDD
>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>> [0000217f] 83c404     add esp,+04
>> [00002182] 5d         pop ebp
>> [00002183] c3         ret
>> Size in bytes:(0018) [00002183]
>>
>> The impossibility of DDD emulated by HHH
>> (according to the semantics of the x86 language)
>> to reach its own machine address [00002183] is
>> complete proof that DDD never halts.
>>
>> This has nothing to do with whether or not HHH
>> halts.
> 
> Everone who understands either C or x86 machine code can see that
> the next thing DDD does after the return from HHH (if HHH ever
> returns) is that DDD returns. 

It is 100% impossible for the first emulated instance
of DDD to return because it is never called.

Mike might understand this. He does have the best
understanding of my code.

> There is no conditional code that
> could cause anything else. Therefore, if DDD does not return
> the inference that HHH does not return is correct.
> 


-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer