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


Ok, so what you are saying is that P and NP become not equal when the
model of computation is changed, because of how P and NP are defined?
How is this analogy: It would be like taking the words "mouse" and
"rodent" and saying that the letter "o" now means "q" (changing the
model of computation), and thus because "mquse" is not the same as a
"rqdent," a mouse and rodent are not the same thing. (Pretend that
mouse and rodent are equivalent; I was going to use cat and feline but
they don't share any letters.)

Is that about right? Also, do you think it might make sense to
reverse the notation? E.g., say "B^P != B^NP," because what you've
really done is allowed an oracle machine to enter the model of
computation, and then examined the complexity classes under the new
model.

Here's one final question. Would it be possible to prove P != NP if
someone could find somehow prove that X^A != X^B, where X is some
complexity class, and A and B are two oracles that would be equivalent
(somehow) if P = NP? In other words, does X^A != X^B imply A != B?

Thanks for the help, I think I'm getting a better understanding of the
subject.

-Phil
.