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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1 --- STA
Date: Fri, 4 Apr 2025 10:47:13 +0300
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On 2025-04-03 22:29:00 +0000, olcott said:

> On 4/3/2025 3:53 AM, Fred. Zwarts wrote:
>> Op 03.apr.2025 om 03:19 schreef olcott:
>>> 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.
>> 
>> I see, you have your own definitions for the x86 language.
> 
> Any idiot can be a mindless naysayer.

We already know. But can you be anything else?

-- 
Mikko