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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1 ---
 STA
Date: Wed, 2 Apr 2025 22:17:28 -0400
Organization: i2pn2 (i2pn.org)
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On 4/2/25 9:19 PM, olcott wrote:
> On 4/2/2025 1:47 PM, Fred. Zwarts wrote:
>> Op 02.apr.2025 om 17:55 schreef olcott:
>>> On 4/2/2025 9:14 AM, joes wrote:
>>>> Am Mon, 31 Mar 2025 16:26:58 -0500 schrieb olcott:
>>>>> On 3/31/2025 2:10 PM, Fred. Zwarts wrote:
>>>>>> Op 31.mrt.2025 om 20:16 schreef olcott:
>>>>
>>>>>>> A simulating termination analyzer is always correct to abort the
>>>>>>> simulation and reject the input as non-halting when-so-ever this 
>>>>>>> input
>>>>>>> would otherwise prevent itself from halting.
>>>>>>>
>>>>>> But the input is halting, as proven by direct execution.
>>>>>
>>>>> Something other than the input is halting.
>>>>> HHH1(DDD) shows the same behavior as the direct execution.
>>>>>    HHH(DDD) shows the behavior of the actual input.
>>>> Why are you not passing DDD as input? Why do you not call what you're
>>>> doing HHH(HHH(DDD))? What is the difference in what is passed to HHH1?
>>>>
>>>
>>> This seems to be above your level of technical competence.
>>>
>>> _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]
>>>
>>> Anyone understanding the above code where HHH
>>> emulates DDD according to the semantics of the
>>> x86 language knows that this DDD (not some
>>> other different DDD) cannot possibly reach its
>>> own final halt state.
> 
>> Yes it fails to reach the end of the simulation of a program that 
>> according to the x86 semantics has an end as proven by direct execution. 
> 
> In other words you don't hardly know the x86
> language at all.
> 

Really? Why doesn't the direct execution of that input, when we include 
the HHH that you have defined?

SInce that *IS* the definition of the behavior of the x86 language, how 
does HHH get a different answer?

What instruction, ACTUALLY EMULATED BY THE RULE OF THE x86 differed to 
HHH then the direct execution.

This question has been asked many times before, and you refusal to give 
an answer just shows that you KNOW you are just lying here.

Sorry, but that it the truth, you are just proving that you know you are 
nothing but a pathological liar that doesn't care about what the truth 
actually is.