Re: obvious ("dumb") question about oracles and P vs. NP

In article <c235cdee-7963-4dd2-884f-71e51ffd9535@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<cplxphil@xxxxxxxxx> wrote:
From the Baker-Gill-Solovay proof, we find that there is an oracle A
s.t. NP^A = P^A, and an oracle B s.t. NP^B != P^B.

Here is my rather dumb question: Why is it that the second inequality
doesn't prove that P != NP?

This is one of my pet peeves about the standard notation for oracles.

The notation "P^B" makes it look like you're starting with the object P and
then doing something to it. If that were the case, then your argument would
be correct. For if P = NP, then if you apply some operation to P, the
result should be the same as if you apply that same operation to NP, since
after all P and NP are (by assumption) the same thing.

However, that's not what "P^B" means. "P^B" is not obtained by starting
with the language P and then performing some operation on the language.
Instead, you're performing an operation on the *model of computation*.
Thus P^B is just something that is *analogous* to P---nothing more.

Here's an analogy that may help. Two companies A and B make the same amount
of money. Each company then hires an extra person. It doesn't follow that
A and B's profits will change by the same amount. Since A and B are
structured differently, the impact of hiring an extra person may differ.
--
Tim Chow tchow-at-alum-dot-mit-dot-edu
The range of our projectiles---even ... the artillery---however great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. ---Galileo, Dialogues Concerning Two New Sciences
.

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