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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Infinite set of HHH/DDD pairs --- truisms
Date: Mon, 22 Jul 2024 20:16:59 +0200
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Op 22.jul.2024 om 19:51 schreef olcott:
> 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.
> 

Indeed, but dreaming of an HHH that does not abort is irrelevant when we 
consider a HHH that is encoded to abort.
The HHH that aborts, simulates itself: the HHH that aborts. No HHH that 
does not abort is involved.
When a HHH aborts, it is not needed to abort its simulation, because it 
will halt by its own.
So for any HHH, whether it aborts or not, we can say:
HHH cannot possibly simulate itself correctly up to the end.