Re: Can D simulated by H terminate normally? --- Message_ID Provided

On 5/1/2024 7:10 PM, Richard Damon wrote:

The second method uses the fact that you have not restricted what H is allowed to do, and thus H can remember that it is simulating, and if a call to H shows that it is currently doing a simulation, just immediately return 0.

Nice try but this has no effect on any D correctly simulated by H.
When the directly executed H aborts its simulation it only returns
to whatever directly executed it.
If the directly executed outermost H does not abort then none of
the inner simulated ones abort because they are the exact same code.
When the directly executed outermost H does abort it can only return
to its own caller.

Thus, H can actually correct simulate the instruction at the call to H, as they will execute just a few instructions testing that condition and returning, and thus not run into the problem you ran into where H just couldn't simulate itself because it got bogged down.
 In this case it is actually true that the direct execution of D(D) differs from the correct simulation of the input by H, as H is no longer a "Computation" per the rules of Computation Theory, but you have admitted that you are abandoning those, so it doesn't matter (of course that make trying to get your results to apply to something similar harder, but that is why you need to try to come up with some actual definitons.)
 So, by the rules of Compuation Theory, your H is not correct, but by your lack of rules, your conclusion that H can not simulate past the call are incorrect, so you proof is also broken.
 

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