# P vs NP and can an oracle make P=EXPTIME?

As I understand, diagonalization cannot be used to prove or disprove P vs NP, because for some oracle $A$, $P^A = NP^A$, whereas under another oracle $B$, $P^B \neq NP^B$.

I don't fully understand this logic. Diagnolization (Time Hierarchy Theorem) proves that $P \neq EXPTIME$. Does it therefore mean that there is no oracle $X$ that could ever make $P^X = EXPTIME^X$ ?

Second, let's discuss your final question. Yes, it is indeed true that $P^X \ne EXPTIME^X$ for all $X$; there is no oracle $X$ that will make them equal. I'll talk through why that is. The standard way to prove $P \ne EXPTIME$ is to prove $TIME(p(n)) \ne TIME(p(n)^2)$ (the time hierarchy theorem), then notice that this implies $P \ne EXPTIME$ (since $TIME(2^n) \ne TIME(2^{2n})$, but $P \subseteq TIME(2^n)$ and $TIME(2^{2n}) \subseteq EXPTIME$). The proof of $TIME(p(n)) \ne TIME(p(n)^2)$ uses diagonalization. This proof does indeed relativize. In particular, for all oracles $X$, we do indeed have $TIME^X(p(n)) \ne TIME^X(p(n)^2)$. It follows that, for all oracles $X$, we have $P^X \ne EXPTIME^X$ (by the same argument as before).