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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory,sci.logic
Subject: Re: D correctly simulated by H proved for THREE YEARS --- rewritten
Date: Fri, 14 Jun 2024 11:59:43 +0200
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Op 13.jun.2024 om 21:41 schreef olcott:
> 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. 

Yes, it does mean this. Using another simulator shows that even the 
simulated H reaches its "ret". It is only that H simulated by itself is 
aborted too early. Is that so difficult to understand for you?

> If H waits for some other H to abort their
> simulation then H waits forever.

There is no other H. This H aborts too early. This H does not wait, so 
it does not help to dream of another H that waits. H does what it is 
programmed to do and aborts too early, because that is the fundamental 
problem of a simulator simulating itself. It will never see its final 
simulated state.

>  H is always at least one execution
> trace ahead of every other H.

Exactly! That is the reason why the abort one execution trace too early. 
It seems you start to see it. H will never see that it is only some 
steps from the final state of its simulation, because it aborts before 
it can see that.
That does not mean that there is an infinitely repeated recursion, but 
that the recursion is only repeated one time more than can be simulated 
by H. That is the fundamental problem of a simulator simulating itself.

You can try to simulate longer, but that does not help. The simulation 
invariant is that the abort is always one execution trace too early. The 
other invariant is that in an aborting simulator there is never an 
infinitely repeated recursion.