[alderz]

How does the paper of Hauptmann prove P != NP?

Sigma_2^p != NP as far as I know and after a brief skimming the paper does not mention P != NP.

Edit: the paper does indeed mention P != NP in the form of P != Sigma_2^p => P != Sigma_1^p = NP. Please disregard my comment.

[modalduality]

P = NP implies the polynomial hierarchy collapses, thus P = Sigma2, so by contradiction P != NP.