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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Can D simulated by H terminate normally? --- Message_ID Provided
Date: Sat, 18 May 2024 14:57:29 -0500
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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.
> 

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