# Proof of Co-Problem being in NP if Problem is in NP using negated output

Given any problem $$P$$ that we know of being in $$NP-\text{complete}$$, where is the flaw in the following proof?

Given a problem $$Co-P$$ which is the co-problem of $$P \in NP-\text{complete}$$, $$Co-P$$ is at least in $$NP$$ because the following algorithm can always be given:

Co-P(J):     bool res = P(J)     return !res 

Where $$Co-P(J)$$ is the algorithm solving $$Co-P$$ and $$P(J)$$ is the nondeterministic polynomial algorithm solving P.

Why is this not correct?