# Is a problem in NP if it is decided by some non-deterministic, polynomial time turing machine? [duplicate]

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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?

## marked as duplicate by Hendrik Jan, Yuval Filmus complexity-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 3 '16 at 10:09

• NP is, by definition, the things that can be decided by an NDTM in poly time. So, if you take a language such as $B$ which you've explicitly said is not in NP, then it cannot be decided by an NDTM in poly time. – David Richerby May 3 '16 at 15:13
• So there is no NDTM which can decide a CoNPC problem in poly time? What does that mean for determining if a certain instance of a NPC problem is a no instance? Can that be done in poly time by some NDTM? – Cheiron May 4 '16 at 10:25
• If NP and coNP are different, no NDTM cna decide coNP-complete problems in polynomial time. But we don't know if NP and coNP are different, which means that we don't know if there are NDTMs that decide the complement of NP-complete problems. – David Richerby May 4 '16 at 15:09