> If P = NP, then P^X = NP^X for any oracle X, right? Edit: No, I think. See child comment by JadeNB. Probably wrong { Yes, since we could just not invoke the

> If P = NP, then P^X = NP^X for any oracle X, right? Edit: No, I think. See child comment by JadeNB. Probably wrong { Yes, since we could just not invoke the oracle. But people do this sort of stuff to study it backwards e.g. what do P^X and NP^X (and any class^X) tell us about P and NP (and any other class). } I hope this is clearer, let {A, B, C, ...} be the set of all oracles. For TM + A: P^A is the class of problems that can be done in P time on this machine. NP^A is the class of problems that can be done in NP time on this machine. Similarly for TM + B TM + C ... This theorem says that if we choose one of these oracle machines uniformly at random then P =/= NP in _its_ model of computation a.s. This is an entirely probabilistic statement about machines that are more "powerful" than standard TMs.