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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 19:45:55 +0200
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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.
HHH, when simulated by itself, runs one cycle behind its simulator. 
Therefore, the simulating always aborts, when it is not needed, because 
the simulated HHH is encoded to abort one cycle later.
Therefore, the simulation is incomplete and, therefore, incorrect.

> 
> It is also a truism that any input that must be aborted
> is a non-halting input.

It is also a truism that a halting program does not need to be aborted.