P-NP for decision problems

Question

The final quiz problem asked whether the statement

For decision problems L1, L2 in NP, if P is not NP, L1 is at least as hard as L2, and L2 is at least as hard as L1, then L1 and L2 are NP-complete

is true or false. In fact, it seemed to me even hard to provide a counterexample for this statement, since this is intuitively a wrong assumption, but hard to prove it. I would be welcomed to know your any ideas for this problem.

Answer 1

It could be that $L1$ and $L2$ are in $P$ (since $P\subset NP$).

Then both are equally hard but neither is $NP$-complete (since $P \ne NP$ by assumption).

Hence the statement is false.