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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: Fri, 22 Nov 2024 10:30:45 +0200
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On 2024-11-21 15:32:38 +0000, olcott said:

> On 11/21/2024 3:12 AM, Mikko wrote:
>> On 2024-11-20 22:03:43 +0000, olcott said:
>> 
>>> On 11/20/2024 3:53 AM, Mikko wrote:
>>>> On 2024-11-20 03:23:12 +0000, olcott said:
>>>> 
>>>>> On 11/19/2024 4:12 AM, Mikko wrote:
>>>>>> 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.
>>>>>> 
>>>>> 
>>>>> I specify the infinite sets with each element numbered
>>>>> on the top of page 2 of my paper. Back in April of 2023
>>>>> 
>>>>> https://www.researchgate.net/ 
>>>>> publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D 
>>>>> 
>>>> 
>>>> You have also specifed that HHH is the program in your GitHub repository.
>>>> 
>>> 
>>> Should I assume that you must be lying about
>>> this because you did not quote where I did this?
>> 
>> No, you may assume that I was confused by your lack of clarity and
>> in particular by your bad choice of names.
>> 
>> If you clearly state that HHH is not the function HHH that you have
>> in your GitHub repository then I needn't to consider the possiblity
>> that you just triying to deceive by equivcation.
>> 
> 
> HHH is one concrete example of an infinite set of instances
> such that DDD is emulated by HHH N times.

That sentence says that there is only one HHH, contradicting your
earlier statement that HHH is a generic term for every member of some
set.

-- 
Mikko