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From: Richard Damon <richard@damon-family.org>
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 18:04:35 -0400
Organization: i2pn2 (i2pn.org)
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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.