Although it is relatively simple to see that integer linear programming is NP-hard, whether it lies in NP is a bit harder. Therefore, I'm wondering whether the following reasoning shows that $ILP\in NP$ implies $NP=CoNP$.To be precise, I am considering the decision variant of ILP over a non-unary input encoding, so the problem is: given an integer matrix $A$; an integer vector $b$,  does there exist an integer vector $y \in Z_n$ with $Ay \leq b$?The main issue here is that if we take such an $y$ as NP-certificate, it might be of exponential size in terms of the input, but this of course does not mean a certificate cannot exist. My idea was that if we assume ILP in NP, we can use a discrete version of the Farkas lemma (see here) to transform an ILP instance with the answer NO to an ILP instance with the answer YES, which then would imply that ILP is in coNP. So we get an NP-complete problem in coNP, which implies that $NP=coNP$. Since most people think $NP\neq coNP$, this is 'evidence' for ILP not being in NP.However, I'm not sure if my reasoning is correct. In particular, does this 'discrete Farkas lemma' work in polynomial time and create an ILP of polynomial size of the original one?
Does integer programming $\in$ NP imply $NP=CoNP$?
complexity theory;np complete;np;integer programming;co np
The notion of duality suggested by the discrete Farkas lemma found in Lasserre's paper does not correspond to an integer program. There is a notion of duality for integer programs (see for example here), but strong duality does not hold, so I doubt it can be used along the lines you suggest.