# Decision variation of max sat

I am trying to prove that following decison variation of MaxSAT is both NP hard and co-NP hard. $$(\phi ,k) \in L$$ iff an assignment of $$\phi$$ satisfies k clauses and no assignment satisfies more than k clauses.

I think we can show that NP hardness by reducing SAT to L by setting k=m (number of clauses). But I can't reduce $$\overline{SAT}$$ to L. Is that reduction possible?

Given a $$\overline{SAT}$$ instance $$\psi$$ with $$m$$ clauses, let $$\phi$$ be the formula formed by adding to each clause a new variable $$x$$ (the same for all clauses), and a new clause $$\bar{x}$$. If $$\psi$$ is satisfiable then $$\phi$$ is satisfiable, whereas if $$\psi$$ is unsatisfiable then the maximum number of clauses that can be satisfied is $$m$$.