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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic,comp.ai.philosophy
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as
 non-halting V2
Date: Sat, 13 Jul 2024 06:39:31 -0500
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On 7/13/2024 3:15 AM, Fred. Zwarts wrote:
> Op 13.jul.2024 om 01:19 schreef olcott:
>> On 7/12/2024 5:56 PM, Richard Damon wrote:
>>> On 7/12/24 10:56 AM, olcott wrote:
>>>> We stipulate that the only measure of a correct emulation is the
>>>> semantics of the x86 programming language.
>>>
>>> Which means the only "correct emulation" that tells the behavior of 
>>> the program at the input is a non-aborted one.
>>>
>>>>
>>>> _DDD()
>>>> [00002163] 55         push ebp      ; housekeeping
>>>> [00002164] 8bec       mov ebp,esp   ; housekeeping
>>>> [00002166] 6863210000 push 00002163 ; push DDD
>>>> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>>>> [00002170] 83c404     add esp,+04
>>>> [00002173] 5d         pop ebp
>>>> [00002174] c3         ret
>>>> Size in bytes:(0018) [00002174]
>>>>
>>>> When N steps of DDD are emulated by HHH according to the
>>>> semantics of the x86 language then N steps are emulated correctly.
>>>
>>> And thus HHH that do that know only the first N steps of the behavior 
>>> of DDD, which continues per the definition of the x86 instruction set 
>>> until the COMPLETE emulation (or direct execution) reaches a terminal 
>>> instruction.
>>>
>>>>
>>>> When we examine the infinite set of every HHH/DDD pair such that:
>>>> HHH₁ one step of DDD is correctly emulated by HHH.
>>>> HHH₂ two steps of DDD are correctly emulated by HHH.
>>>> HHH₃ three steps of DDD are correctly emulated by HHH.
>>>> ...
>>>> HHH∞ The emulation of DDD by HHH never stops running.
>>>
>>> And thus, the subset that only did a finite number of steps and 
>>> aborted its emulation on a non-terminal instrucition only have 
>>> partial knowledge of the behavior of their DDD, and by returning to 
>>> their caller, they establish that behavior for ALL copies of that 
>>> HHH, even the one that DDD calls, which shows that DDD will be 
>>> halting, even though HHH stopped its observation of the input before 
>>> it gets to that point.
>>>
>>>>
>>>> The above specifies the infinite set of every HHH/DDD pair
>>>> where 1 to infinity steps of DDD are correctly emulated by HHH.
>>>>
>>>> No DDD instance of each HHH/DDD pair ever reaches past its
>>>> own machine address of 0000216b and halts.
>>>
>>> Wrong. EVERY DDD of an HHH that simulated its input for only a finite 
>>> number of steps WILL halt becuase it will reach its final return.
>>>
>>> The HHH that simulated it for only a finite number of steps, only 
>>> learned that finite number of steps of the behaivor, and in EVERY 
>>> case, when we look at the behavior past that point, which DOES occur 
>>> per the definition of the x86 instruction set, as we have not reached 
>>> a "termial" instruction that stops behavior, will see the HHH(DDD) 
>>> that DDD called continuing to simulate its input to the point that 
>>> this one was defined to stop, and then returns 0 to DDDD and then DDD 
>>> returning and ending the behavior.
>>>
>>> You continue to stupidly confuse the PARTIAL observation that HHH 
>>> does of the behavior of DDD by its PARTIAL emulation with the ACTUAL 
>>> FULL behavior of DDD as defined by the full definition of the x86 
>>> insttuction set.
>>>
>>>
>>>>
>>>> Thus each HHH element of the above infinite set of HHH/DDD
>>>> pairs is necessarily correct to reject its DDD as non-halting.
>>>>
>>>
>>> Nope.
>>>
>>> NONE Of them CORRECTLY rejected itS DDD as non-halting and you are 
>>> shown to be ignorant of what you are talking about.
>>>
>>> The HHH that did a partial emulation got the wrong answer, because 
>>> THEIR DDD will halt. and the HHH that doen't abort never get around 
>>> to rejecting its DDD as non-halting.
>>
>> *Here is the gist of my proof it is irrefutable*
>> When no DDD of every HHH/DDD that can possibly exist
>> halts then each HHH that rejects its DDD as non-halting
>> is necessarily correct.
>>
>> *No double-talk and weasel words can overcome that*
>>
> 
> This is double talk, because no HHH can possibly exist that simulates 
> itself correctly.

Your definition of correct contradicts the semantics of
the x86 language making it wrong.

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