>>7616321Assume that P = NP. Then there is a trivial isomorphism between P and NP. NP can be partitioned into subsets based on the verification problems in P that they require. So if NP is a set of subsets of P, and NP and P are isomorphic, by diagonalization there exists a subset of P that doesn't belong to NP. That is, there are some problems in P that are not used to verify any problem in NP. But this is a contradiction, for problems in P are verifications of themselves and therefore contain themselves as elements. So P ≠ NP.
