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Op 13.jun.2024 om 14:44 schreef olcott:No it never meant this. If H waits for some other H to abort theirOn 6/13/2024 3:15 AM, Fred. Zwarts wrote:So, you think that if H does not reach its "ret", D can still reach its "ret"?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
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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.
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.
Try to think. D does not reach its "ret", *because* "H" does not reach its "ret".
>That is true. But it also means that H aborts one execution trace too early.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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