# MAX 2-SAT is polynomial time reducible to 2-SAT?

I know that 2-SAT is solvable in polynomial time and 2-SAT is NP-Hard.

I have issue about this statement: MAX 2-SAT is polynomial-time reducible to 2-SAT. Can you explain to me how reduction looks like? I need the only intution about that, but not proof.

Let $$\phi$$ be a 2-SAT formula with $$m$$ clauses. If you really want to reduce an instance $$\phi$$ of 2-SAT to (the decision version of) MAX-2-SAT, then this simply amounts to checking whether at least $$m$$ (i.e., all) clauses of $$\phi$$ are satisfiable.