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From: Richard Damon <richard@damon-family.org>
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 23:48:30 -0400
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On 7/12/24 11:00 PM, olcott wrote:
> On 7/12/2024 6:41 PM, Richard Damon wrote:
>> On 7/12/24 7:19 PM, olcott wrote:
>>> 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*
>>>
>>
>> Which is just your double-talk to try to redefine what halting means.
>>
> 
> You try to cut my airtight proof up in little pieces and fail.
> Every rebuttal that you make has disagreeing with the semantics
> of the x86 language as its basis.
> 

How?

What Semantic of the x86 instruction set as defined by Intel dp I violate?

Where is the statement that justifies stopping the simulation in any way 
still recreates the full behavior of the program given to the simulator?

Where does it say that calling a function make the processor do ANYTHING 
other than stepping through the function itself? (So you can try to 
argue that calling HHH, a CONDITIONAL emulator, begins AT THE x86 
INSTRUCTION LEVEL, and emulation of the program given as an input?)

What "instruction" tells the processor to "emulate" the program 
specified as a parameter to the function being called.


This is just another of you many unfounded LIES where you claim 
something that you can not prove and just deflect, deflect, deflect, as 
you KNOW you have no grounds to base it on, just you own lies.

This is just like your "non-sense" and non-existent "Diagonalization" 
proof that shows Godel wrong.

Your argument is only "Air-tight" in the sense that it is just like a 3 
year old threatening to hold his breath until he turns blue.

Go ahead, keep repeating you lie, and we can keep on pointing out how 
stupid you are to continue making the claim, it just show how absolutely 
little you understand what you are talking about.