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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1 ---
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Date: Wed, 2 Apr 2025 20:47:42 +0200
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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. 
The x86 semantics have only one interpretation: The program specified by 
this finite string halts.
But you do not understand the x86 language, as you think that it can 
have different interpretations.