Prove that if P=NP then every two non trivial languages $A,B\in coNP$ holds $A \equiv_{P} B$

Question

I need to prove that if $P=NP$ then every two non trivial languages (meaning not $\Sigma^*$ or $\varnothing$) $A,B\in coNP$ holds $A \equiv_{P} B$

(this means that $A \leq_{P} B$ and also $B \leq_{P} A$)

Heres what I've done so far:

assume $P=NP$ and let $A,B\in coNP$, so $\overline{A}, \overline{B} \in NP$,

since $P=NP$ we get that $\overline{A}, \overline{B} \in P$, and since P is closed under complement we get that $A,B\in P$

since $P \subseteq NP$ we get that $A,B\in NP$

Now I don't know how to polynomialy reduce one to the other. I thought about using a NP-complete language like $3-SAT$ but the question is too general

Answer 1

We know that if $P=NP$, then we essentially have $P = NP = coNP$ (as you have already demonstrated). Now the problem boils down to showing that any two problems in $P$ are poly-time reducible to each other.

To show that $A \le_p B$ for any $A, B \in P$, we have to essentially exhibit the existence of a mapping function $f$ that converts any input $I_A$ of $A$ into an input $I_B$ of $B$ in poly-time. Here's how we can design such a fuction $f$:

Pick two instances $I_B^{(true)}$ and $I_B^{(false)}$ of $B$ such that one of them is a true instance and the other one is a false instance. This can be determined in polytime since $B \in P$. Now take the input $I_A$ and actually solve it in poly-time (recall $A \in P$). If the result is true, return $I_B = I_B^{(true)}$; otherwise, return $I_B = I_B^{(false)}$.

Also, you may see these related discussions: Problem about $NP \neq coNP$, Can any problem in P be converted to any other problem in P in polynomial time?.