I am working trough the book Introduction to the theory of computation, 3rd edition, by M. Sipser. On page 294, the book states:A problem is in NP iff it is decided by some non-deterministic,  polynomial time turing machine.I get that it is decidable by some NDPTTM if the problem is in NP, the other way around I do not quite get.This is because I feel like all problems in $CoNP-NP$ can also be decided by some Non Deterministic polynomial time turing machine. I have the following solution in mind: Lets assume $P \neq NP \land NP \neq CoNP$. Take a problem $A \in NP - CoNP$. Now there is problem $\overline{A}$, which is in $CoNP$. Lets say NDTM $N$ decides $A$ in polynomial time. Now we modify $N$ by replacing every reject with accept and every accept with reject. We now have a polynomial NDTM $N'$ which decides $\overline{A}$ in poly time. According to the theorem proposed by the book, this should imply that $\overline{A}$ is in $NP$, but it is not.Am I missing something here?Edit:So, I have seen the possible duplicates and I learned a lot: I see now that $N'$ does not necessarily decide $\overline{A}$. I Do still believe my question is subtly different, so here I go:Take problem $B \in CoNP-NP$. There is a NDTM which decides $B$ in poly time, right? So, according to the theorem, $B$ should be in NP.I firmly believe the book is correct, but does that mean that $B$ cannot be decided by any NDTM in poly time? 
Is a problem in NP if it is decided by some non-deterministic, polynomial time turing machine?
complexity theory;np