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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: Fri, 12 Jul 2024 18:19:26 -0500
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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*

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