Theorem. If NP ≠ CoNP then P ≠ NP.
Proof. Suppose that NP ≠ CoNP.
The proof that P ≠ NP is by contradition. Assume that P = NP.P is a deterministic complexity class, so it is closed under complementation. That is, {S | S ∈ P} = P. So
| NP | = | P | (By supposition) |
| = | {S | S ∈ P} | (since P is closed under complementation) | |
| = | {S | S ∈ NP} | (since P = NP) | |
| = | CoNP | (by the definition of CoNP) |
So NP = CoNP. But that violates the premise that NP ≠ CoNP, and is a contradiction.