Re: Yet another contribution to the P-NP question

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.