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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: How do computations actually work?
Date: Fri, 30 May 2025 10:18:07 -0500
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On 5/30/2025 2:56 AM, Mikko wrote:
> On 2025-05-28 14:33:56 +0000, olcott said:
> 
>> On 5/28/2025 2:43 AM, Mikko wrote:
>>> On 2025-05-27 15:48:14 +0000, olcott said:
>>>
>>>> On 5/27/2025 3:37 AM, Mikko wrote:
>>>>> On 2025-05-26 15:57:11 +0000, olcott said:
>>>>>
>>>>>> On 5/26/2025 3:46 AM, Mikko wrote:
>>>>>>> On 2025-05-25 14:53:06 +0000, olcott said:
>>>>>>>
>>>>>>>> On 5/25/2025 4:14 AM, Mikko wrote:
>>>>>>>>> On 2025-05-24 15:18:57 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> On 5/24/2025 2:47 AM, Mikko wrote:
>>>>>>>>>>> On 2025-05-23 02:47:40 +0000, olcott changed the subject to
>>>>>>>>>>>> How do computations actually work?
>>>>>>>>>>>
>>>>>>>>>>> Each computation works differently. It does not matter how it 
>>>>>>>>>>> works as
>>>>>>>>>>> long as there are instructions that fully specify how that 
>>>>>>>>>>> computation
>>>>>>>>>>> shall be performed.
>>>>>>>>>>
>>>>>>>>>> All termination analyzers are required to report on the
>>>>>>>>>> behavior that their input finite string specifies.
>>>>>>>>>
>>>>>>>>> To report correctly. Though the input string to a termination 
>>>>>>>>> analyzer
>>>>>>>>> usially is incomlete: the input string usually specifies different
>>>>>>>>> behavours depending on the input that is not shown to the 
>>>>>>>>> termination
>>>>>>>>> analyzer, and the analyzer's report must cover all of them.
>>>>>>>>>
>>>>>>>>> A partial termination analyzer may fail to report but is not 
>>>>>>>>> allowed
>>>>>>>>> to report incorrectly.
>>>>>>>>
>>>>>>>> void DDD()
>>>>>>>> {
>>>>>>>>    HHH(DDD);
>>>>>>>>    return;
>>>>>>>> }
>>>>>>>>
>>>>>>>> DDD simulated by HHH cannot possibly reach its
>>>>>>>> "return" statement final halt state, only liars
>>>>>>>> will disagree.
>>>>>>>
>>>>>>> Again a straw man deception. Where reasonable people disgraee is the
>>>>>>> relevance of that claim. The behavour specified by DDD does reach 
>>>>>>> the
>>>>>>> final return from DDD. Whether HHH can simulate that part of the
>>>>>>> behaviour is irrelevant. Even without the simjlation HHH decides
>>>>>>> correctly if and only if it determines and reports that DDD halts.
>>>>>>
>>>>>> _DDD()
>>>>>> [00002192] 55             push ebp
>>>>>> [00002193] 8bec           mov ebp,esp
>>>>>> [00002195] 6892210000     push 00002192
>>>>>> [0000219a] e833f4ffff     call 000015d2  // call HHH
>>>>>> [0000219f] 83c404         add esp,+04
>>>>>> [000021a2] 5d             pop ebp
>>>>>> [000021a3] c3             ret
>>>>>> Size in bytes:(0018) [000021a3]
>>>>>>
>>>>>> How many recursive emulations does HHH have to
>>>>>> wait before its emulated DDD magically halts
>>>>>> on its own without ever needing to be aborted?
>>>>>
>>>>> In order to determine that you need a program that simulates DDD 
>>>>> and all
>>>>> functios DDD calls and detects and counts and reports emulation 
>>>>> levels.
>>>>
>>>> No you don't. All that you need to do is simply imagine
>>>> that HHH is an x86 emulator capable of emulating itself
>>>> emulating DDD. // *This is fully operational code*
>>>
>>> Well, you needn't if you have a good substitution. But imagination is
>>> not a good substitution of real work with a real tool.
>>
>> https://github.com/plolcott/x86utm/blob/master/Halt7.c
>> has an HHH that does simulate itself simulating DDD.
> 
> Nothing behind that pointer refutes my claim that it is not sufficent
> to imagine.
> 

When DDD is correctly emulated by HHH the first four
instructions of DDD are emulated. When HHH(DDD) is
called from DDD then HHH emulates itself emulating DDD.

No matter how many times HHH emulates itself emulating
DDD the emulated DDD cannot possibly reach its "ret"
instruction final halt state. This proves that DDD
emulated by HHH is non-halting.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer