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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: HHH(DDD) computes the mapping from its input to HHH emulating itself emulating DDD --- anyone that says otherwise is a liar
Date: Tue, 19 Nov 2024 12:12:11 +0200
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On 2024-11-18 20:42:02 +0000, olcott said:

> On 11/18/2024 3:41 AM, Mikko wrote:
>> The "the mapping" on the subject line is not correct. The subject line
>> does not specify which mapping and there is no larger context that could
>> specify that. Therefore it should be "a mapping".
>> 
>> On 2024-11-17 18:36:17 +0000, olcott said:
>> 
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>>    return;
>>> }
>>> 
>>> _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]
>>> 
>>> DDD emulated by any encoding of HHH that emulates N
>>> to infinity number of steps of DDD cannot possibly
>>> reach its "return" instruction final halt state.
>> 
>> Because it cannot reach the instructions before tha return.
>> Because it cannot reach the instruction after the HHH call.
>> Because it cannot reach return instruction of HHH.
>> 
>>> This applies to every DDD emulated by any HHH no
>>> matter the recursive depth of emulation. Thus it is
>>> a verified fact that the input to HHH never halts.
>> 
>> That is too vague to be regareded true or false. It is perfectly possibe
>> to define two programs and call them DDD and HHH
> 
> What a jackass. DDD and HHH have been fully specified
> for many months.

They are specified in a way that makes your "every DDD" and "any DDD"
bad (perhaps even incorrect) use of Common language.

-- 
Mikko