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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: People are still trying to get away with disagreeing with the
 semantics of the x86 language
Date: Sun, 30 Jun 2024 16:41:09 -0500
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On 6/30/2024 2:31 PM, Richard Damon wrote:
> On 6/30/24 1:18 PM, olcott wrote:
>> On 6/30/2024 3:42 AM, Mikko wrote:
>>> On 2024-06-29 16:09:19 +0000, olcott said:
>>>
>>>> People are still trying to get away with disagreeing with
>>>> the semantics of the x86 language. That is isomorphic to
>>>> trying to get away with disagreeing with arithmetic.
>>>>
>>>> typedef void (*ptr)();
>>>> int H0(ptr P);
>>>>
>>>> void Infinite_Loop()
>>>> {
>>>>    HERE: goto HERE;
>>>> }
>>>>
>>>> void Infinite_Recursion()
>>>> {
>>>>    Infinite_Recursion();
>>>> }
>>>>
>>>> void DDD()
>>>> {
>>>>    H0(DDD);
>>>> }
>>>>
>>>> int main()
>>>> {
>>>>    H0(Infinite_Loop);
>>>>    H0(Infinite_Recursion);
>>>>    H0(DDD);
>>>> }
>>>>
>>>> Every C programmer that knows what an x86 emulator is knows
>>>> that when H0 emulates the machine language of Infinite_Loop,
>>>> Infinite_Recursion, and DDD that it must abort these emulations
>>>> so that itself can terminate normally.
>>>>
>>>> When this is construed as non-halting criteria then simulating
>>>> termination analyzer H0 is correct to reject these inputs as
>>>> non-halting by returning 0 to its caller.
>>>>
>>>> Simulating termination analyzers must report on the behavior
>>>> that their finite string input specifies thus H0 must report
>>>> that DDD correctly emulated by H0 remains stuck in recursive
>>>> simulation.
>>>>
>>>> <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>>>      If simulating halt decider H correctly simulates its input D
>>>>      until H correctly determines that its simulated D would never
>>>>      stop running unless aborted then
>>>>
>>>>      H can abort its simulation of D and correctly report that D
>>>>      specifies a non-halting sequence of configurations.
>>>> </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>>>>
>>>> People are trying to get away with disagreeing with the semantics
>>>> of the x86 language by disagreeing that
>>>>
>>>> The call from DDD to HHH(DDD) when N steps of DDD are correctly
>>>> emulated by any pure function x86 emulator HHH cannot possibly
>>>> return.
>>>>
>>>> _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]
>>>>
>>>>
>>>> *A 100% complete and total rewrite of the prior paper*
>>>> https://www.researchgate.net/publication/381636432_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_P
>>>
>>> Nothing above is or points to any evdence about the alleged 
>>> disagreement.
>>>
>>
>> Of course not. I only said the actual truth.
>>
>> Richard just said that he affirms that when DDD correctly
>> simulated by HHH calls HHH(DDD) that this call returns even
>> though the semantics of the x86 language disagrees.
> 
> What in the sematics of the x86 language, which INCLUDES that ever 
> instruction WILL be followed by the next instruction, says that the HHH 
> that is calld by DDD won't eventually return.
> 

So finally you quit lying.

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