Proof that complement of HAMPATH is not in NP

The Sipser book says, Complement of HAMPATH is not in NP.

I tried to find the solution for this, actually, a rigid proof like HAMPATH, which can state Complement of HAMPATH is in/not in NP, but I could not come up with any.

This is what I think, We are given a path from s to t(which can be a certificate for the problem, or not?), and check if it is not a HAMPATH. But this is easy, we can check this in P time right--Just list down the path, see that if any vertex is repeating in it, or if it is missing any vertex of the graph.

How come we are saying this problem is not in NP

I am very new to P and NP, Please correct my mistakes, and apologies if my question sounds like a silly one.

• Yes, a single path can be checked in polynomial time but this is not sufficient to show that there is no Hamilton path in the graph. – ttnick Sep 3 '19 at 7:29
• I was looking for this line only!! Thanks a lot. My mistake was, I was taking complement of HAMPATH as: if there is a path from s to t which is not a HAMPATH. But the actual complement is : there is no HAMPATH path from s to t. – Shirley Sam Sep 3 '19 at 7:38