I have two problems, CNF-SAT and CNF-UNSAT, that decide if a formula $\phi$ on Conjunctive Normal Form is satisfiable and unsatisfiable, respectively. I have already shown that CNF-SAT is $NP$-complete by showing that it is in $NP$ and then reduced SAT to CNF-SAT.
What I wonder is if it follows that CNF-UNSAT is $NP$-complete as well, seeing as the new algorithm only has to accept whenever the algorithm for CNF-SAT rejects and vice-versa, or is it not that simple?