We have looked at relative computability, where problem A is computable relative to problem B if A can be solved on a machine with oracle B (a B-machine).
Relative computability is interesting partly because proofs that are done by diagonalization tend to extend to relativized settings.
For example, we showed that the halting problem is not computable. The same proof shows that the halting problem for B-machines is not solvable on a B-machine, for any B.
Definition. PB is the class of all decision problems solvable in polynomial time on a B-machine.
Definition. NPB is the class of all decision problems solvable nondeterministically in polynomial time on a B-machine.
If it is possible to prove that P ≠ NP using a proof by diagonalization, you would expect to find that you could use the same proof to show that PB ≠ NPB for every B.
But that can't work. Here are two theorems.
Theorem. PTQBF = NPTQBF.
Proof. Since TQBF is PSPACE-complete, PTQBF is the same as PSPACE.
Similarly, NPTQBF is the same as NPSPACE.
But we have seen that NPSPACE = PSPACE.
Theorem. There exists a set B such that PB ≠ NPB.
So whether PB = NPB depends on what B is.