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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2
Date: Sun, 14 Jul 2024 11:58:25 +0300
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On 2024-07-13 13:04:01 +0000, olcott said:

> On 7/13/2024 7:20 AM, Fred. Zwarts wrote:
>> Op 13.jul.2024 om 13:39 schreef olcott:
>>> 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.
>>> 
>> 
>> You have a wrong understanding of the semantics of the x86 language. 
>> You think that the x86 language specifies that skipping instructions do 
>> not change the behaviour of a program.
> 
> You have the wrong understanding of a decider.

You seem to have a wrong understanding of a decider.
Calling a program a decider does not make it halt.

-- 
Mikko