Re: D correctly simulated by H proved for THREE YEARS --- rewritten

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:

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On 5/29/2021 2:26 PM, olcott wrote:
https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
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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.

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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.

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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.
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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.

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That has always been totally irrelevant.

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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".
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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.
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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.

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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.