# Is finding a Polytime reduction from $L_1$ to $L_2$ equivalent to proving $L_2 \in P \Rightarrow L_1 \in P$

I often hear NP-completeness as problems such that, if they were in $$P$$ all problems in $$NP$$ are in $$P$$. The true definition, though, is that NP-complete is a set of languages in NP that all languages in NP reduce to in polynomial time.

But is this idea that NP-completeness implies all problems are in P if they are in P all encompassing? Of course, it works for other problems in NP, but let's take a problem in, say, EXPTIME, which P is known to be a proper subset of.

Is it correct to say that there is a polynomial time reduction from a language $$A$$ and a language $$B$$ in EXPTIME, if and only if $$B \in P \Rightarrow A \in P$$? How about the inverse? $$B$$ (in EXPTIME) reduces to $$A$$ iff $$A \in P \Rightarrow B \in P$$? And of course, what about the general case?

• Also note: I am talking about Turing reductions in polytime. Apr 14, 2023 at 5:22

If there is a polynomial time reduction from $$A$$ to $$B$$, then $$B \in P \implies A \in P$$. This works irrespective of the what kind of languages $$A$$ and $$B$$ are.
The converse is not true. For example, take two languages $$A, B \notin P$$ such that there is no polynomial time reduction from $$A$$ to $$B$$ (these exist by the time hierarchy theorem). Then it still holds that $$B \in P \implies A \in P$$, since both sides of the implication are false.