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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 --- repeat until acknowledged
Date: Sun, 30 Jun 2024 17:48:27 -0500
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On 6/30/2024 5:04 PM, Richard Damon wrote:
> On 6/30/24 6:00 PM, olcott wrote:
>> 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.
>>>
>>
>> Therefore DDD correctly simulated by HHH DOES NOT HALT.
>> Thus HHH correctly reports that DDD DOES NOT HALT.
>>
>>
> 
> And HHH can not report that fact, because, to correct emulate, as 
> presuemd, it can not stop it emulation.
> 
> If it does, it changes the behavior of DDD (remember, the code of HHH is 
> PART of the code for DDD) and DDD will Halt.
> 
> You are just showing you are a stupid LIAR.

*YOU ALREADY ADMITTED THAT I AM CORRECT*

On 6/30/2024 2:31 PM, Richard Damon wrote:
 > 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.

*THIS PROVES THAT DDD CORRECTLY EMULATED BY HHH DOES NOT HALT*
The call from DDD to HHH(DDD) when N steps of DDD are
correctly emulated by any pure function x86 emulator
HHH at machine address 0000217a 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]


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