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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: D correctly simulated by H proved for THREE YEARS --- rewritten
Date: Thu, 13 Jun 2024 14:41:10 -0500
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On 6/13/2024 2:33 PM, Fred. Zwarts wrote:
> Op 13.jun.2024 om 14:44 schreef olcott:
>> On 6/13/2024 3:15 AM, Fred. Zwarts wrote:
>>> Op 12.jun.2024 om 21:53 schreef olcott:
>>>> On 6/12/2024 2:46 PM, Fred. Zwarts wrote:
>>>>> Op 12.jun.2024 om 21:20 schreef olcott:
>>>>>>
>>>>>> On 5/29/2021 2:26 PM, olcott wrote:
>>>>>> https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
>>>>>>
>>>>>> If that was true then you could provide every step of D correctly
>>>>>> simulated by H such that D simulated by H reaches its own simulated
>>>>>> "ret" instruction.
>>>>>
>>>>> I said that each H is unable to hit its target, so how could it 
>>>>> reach the "ret" instruction of D? Please, think before you reply.
>>>>
>>>> It is a binary choice either D correctly simulated by H can
>>>> possibly terminate normally by reaching its "ret" instruction
>>>> or not. Your attempt to twist these words to make it look like
>>>> there is more than these two possibilities is either ignorant
>>>> or deceptive.
>>>>
>>>
>>> Please, take some more attention to what I said. Read, then think, 
>>> before you reply.
>>> I said that H is not able to reach its own "ret" when it is 
>>> simulating itself. 
>>
>> That has always been totally irrelevant.
> 
> So, you think that if H does not reach its "ret", D can still reach its 
> "ret"?
> Try to think. D does not reach its "ret", *because* "H" does not reach 
> its "ret".
> 
>>
>>> So, no disagreement with that. That proves that H misses its target. 
>>> The abort is too early. The target is just some steps further. It 
>>> does not mean that the target is at infinity.
>>>
>>
>> The outer H always has one more execution trace to base its halt
>> status decision on than any of the nested emulations. This means
>> that unless the outer H aborts its simulation then none of them do.
> 
> That is true. But it also means that H aborts one execution trace too 
> early. 

No it never meant this. If H waits for some other H to abort their
simulation then H waits forever. H is always at least one execution
trace ahead of every other H.

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