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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Infinite set of HHH/DDD pairs --- truisms
Date: Mon, 22 Jul 2024 12:51:29 -0500
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On 7/22/2024 12:45 PM, Fred. Zwarts wrote:
> Op 22.jul.2024 om 17:08 schreef olcott:
>> On 7/22/2024 9:32 AM, joes wrote:
>>> Am Mon, 22 Jul 2024 09:13:33 -0500 schrieb olcott:
>>>> On 7/22/2024 3:01 AM, Mikko wrote:
>>>>> On 2024-07-21 13:50:17 +0000, olcott said:
>>>>>> On 7/21/2024 4:38 AM, Mikko wrote:
>>>>>>> On 2024-07-20 13:28:36 +0000, olcott said:
>>>>>>>> On 7/20/2024 3:54 AM, Mikko wrote:
>>>>>>>>> On 2024-07-19 14:39:25 +0000, olcott said:
>>>>>>>>>> On 7/19/2024 3:51 AM, Mikko wrote:
>>>
>>>>>>>>> Anyway you did not say that some HHHᵢ can simulate the
>>>>>>>>> corresponding DDDᵢ to its termination. And each DDDᵢ does
>>>>>>>>> terminate, whether simulated or not.
>>>
>>>
>>>>>> Then DDD correctly simulated by any pure function HHH cannot possibly
>>>>>> reach its own return instruction and halt, therefore every HHH is
>>>>>> correct to reject its DDD as non-halting.
>>>>> That does not follow. It is never correct to reject a halting
>>>>> comoputation as non-halting.
>>>> In each of the above instances DDD never reaches its return instruction
>>>> and halts. This proves that HHH is correct to report that its DDD never
>>>> halts.
>>> It can't return if the simulation of it is aborted.
>>>
>>>> Within the hypothetical scenario where DDD is correctly emulated by its
>>>> HHH and this HHH never aborts its simulation neither DDD nor HHH ever
>>>> stops running.
>>> In actuality HHH DOES abort simulating.
>>>
>>>> This conclusively proves that HHH is required to abort the 
>>>> simulation of
>>>> its corresponding DDD as required by the design spec that every partial
>>>> halt decider must halt and is otherwise not any kind of decider at all.
>>> Like Fred recognised a while ago, you are arguing as if HHH didn't 
>>> abort.
>>>
>>>> That HHH is required to abort its simulation of DDD conclusively proves
>>>> that this DDD never halts.
>>> You've got it the wrong way around.
>>>
>>
>> I am talking about hypothetical possible ways that HHH could be encoded.
>> (a) HHH(DDD) is encoded to abort its simulation.
>> (b) HHH(DDD) is encoded to never abort its simulation.
>>
>> Therefore (a) is correct and (b) is incorrect according to the
>> design requirements for HHH that it must halt.
> 
> Both are incorrect. An HHH, when encoded to abort does not need to be 
> aborted when simulated, because it already halts on its own.

When no HHH(DDD) ever aborts its input then HHH never halts
conclusively proving that some HHH must abort its input.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer