[Solved]: Is it $\mathsf{NP}$-hard to decide whether $\mathsf{P}=\mathsf{NP}$?

Problem Detail: 

Is it $\mathsf{NP}$-hard to decide whether $\mathsf{P}=\mathsf{NP}$ ?

If so, what are the implications ? Is there result suggesting that it is the case ?

Asked By : eig

Answered By : Mike B.

A similar question is this:

Is $(a \vee \neg b) \wedge (\neg a \vee b)$ $\mathsf{NP}$-hard?

The question itself doesn't make much sense, as $\mathsf{NP}$ (as well as $\mathsf{NP}$-hard) defines a set of languages, but $(a \vee \neg b) \wedge (\neg a \vee b)$ is (just like $\mathsf{P} = \mathsf{NP}$) a word, an element of a language, not a language itself.

But all in all... one can say that it is generally hard to prove (or disprove) $\mathsf{P} = \mathsf{NP}$, it might even be impossible (as discussed in the answer by Jernej).

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Question Source : http://cs.stackexchange.com/questions/7737