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From: nnymous109@gmail.com (nnymous109)
Newsgroups: comp.theory
Subject: Re: Yet another contribution to the P-NP question
Date: Sun, 29 Sep 2024 17:43:38 +0000
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> With the assertion** that any recursion that decides that a computation
> satisfies some property necessarily yields at least one element of the
> computation***, I am suggesting that it is easier to check for inclusion
> in this set that it is  to check for exclusion.

> I would call the proof strategy I have come up with an "examine all the
> cases" type proof except the underlying observation as to why we must do
> that is that if x1 and x2 are different strings, unless there is some
> extra information we have been given beforehand (about x1 and x2) that
> we can take advantage of, there is in general no correspondence between
> S.(M(x1)) and S.(M(x2)).


Yet another way of expressing the same thing is that to decide L_P
'efficiently', we cannot consider the 'x's independently. But, in
general, to know that we need not consider some x1 and x2 independently,
we'd have to consider them independently after all.