For $L_1 \in NP, L_2 \in NPC$, does $L_1 \in P \Rightarrow L_2 \in P$ imply $L_1 \in NPC$?

Question

Let $L_1 \in NP, L_2 \in NPC$. It is known that if $L_1 \in P$ then $L_2 \in P$.

Can we formally deduce that $L_1 \in NPC$ (show that $\forall L'\in NP$ exists a polynomial reduction $L'\leq_p L_1$)?

Answer 1

The implication is true if and only if $P = NP$.

If $P = NP$, the consequent $L_1 \in NPC$ is true from the assumption $L_1 \in NP$ because $NP = NPC$.

If $P \neq NP$, there exists an NP-intermediate problem. Now, assign $L_1$ to be an NP-intermediate language, and assign $L_2$ to be an NP-complete language. That means we have $L_1 \notin P$ and $L_1 \notin NPC$. The antecedent "$L_1 \in P \implies L_2 \in P$" true, but the consequent "$L_1 \in NPC$" is false.

Answer 2

I think the answer is unknown.

In particular: suppose the way we know that "if $L_1\in P$ then $L_2 \in P$" is because we have a Cook reduction from $L_2$ to $L_1$. I believe it is not known whether this implies the existence of a Karp reduction from $L_2$ to $L_1$ (which is equivalent to $L_1$ being NP-complete). So, I think it is not known whether this implies that $L_1$ is NP-complete. See https://cstheory.stackexchange.com/a/18922/5038, Can we construct a Karp reduction from a Cook reduction between NP problems?, https://cstheory.stackexchange.com/q/138/5038.

In practice, if you ran across $L_1$ naturally (without trying to craft some special language just to mess up this heuristic) and if you can prove "if $L_1 \in P$ then $L_2 \in P$", it would be natural to strongly suspect that $L_1$ is probably NP-complete, so it is a good use of your time to try to find a concrete Karp reduction to prove it NP-complete.