Flaw in my NP = CoNP Proof? I have this very simple "proof" for NP = CoNP and I think I did something wrongly somewhere, but I cannot find what is wrong. Can someone help me out?

Let A be some problem in NP, and let M be the decider for A. Let B be the complement, i.e. B is in CoNP. Since M is a decider, you can use it to decide B as well (just flip the answer). Doesn't that mean that we solve both NP and CoNP problems with the same M?

To put it more concretely.

Let A be some NP-complete problem, and let M be decider for A. Consider any problem B in CoNP. We consider its complement not-B, which is in NP, and then get a polynomial reduction to A. Then we run our decider M and flip our answer. We thus obtain a decider for B. This implies B is in NP as well. 

May I know what is wrong with my reasoning?
There are two possible bugs in this proof:


When you say "decider" - you mean a deterministic TM. In this case, the best translation (to our knowledge) from an NP machine to a deterministic machine may yield a machine that runs in exponential time, so after complementing you will have a decider for the complement in exponential time, proving that $co-NP\subseteq EXP$ (or, after some optimization, $co-NP\subseteq PSPACE$).
When you say "decider" you mean a nondeterministic TM. In this case, flipping the answer will not necessarily complement the language. Indeed, the language of the flipped machine will be all the words for which there exists a rejecting run of $M$ on $w$