Confirming P ≠ NP would mean NP-hard problems are unconditionally intractable in the general case, consistent with the Thesis.

For in this case a demonstration that \(\phi \not\in n\text{-}\sc{PROVABILITY}_{\mathsf{T}}\) (for a sufficiently large \(n\) and a sufficiently powerful \(\mathsf{T}\)) would be sufficient to show that we have no hope of ever comprehending a proof of \(\phi\) even if one were to exist. But now note that since \(n\text{-}\sc{PROVABILITY}_{\mathsf{T}} \in \textbf{NP}\), if it so happened that \(\textbf{P} = \textbf{NP}\) then the task of determining whether a mathematical formula is derivable in