# If a language is a NP-hard, is also its complement NP-hard?

I was asked the next question:

L is a language which |L (conjunction) {0,1}|=1. In other words, the number of words with length n is exactly one. I need to prove that if L is NP-Hard, also its complement is NP-hard.

The GENERAL question: When I solves it, I remarked that I have a verification algorithm for L and witness y. I suggest to run y on A, and answer the opposite. This will be the verification algorithm for the complement language. But is this always true, for all languages?

What did I miss? What this language is differ? Thanks a lot.

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Now for the general question. In general this is an open question; for NP-complete languages it is exactly the question of whether $NP=coNP$. If $P=NP$ then indeed $NP=coNP$, and hence, proving that $NP\ne coNP$ is "at least as hard as proving $P\ne NP$", so we don't expect an answer soon (the belief is that $NP\ne coNP$).