What is the standard approach to showing that two implication statements are equivalent? Suppose there are two statements: $P⇒Q$ and $A⇒B$
If I wanted to show that these two are equivalent, is it enough to assume $P$ and show that it implies $B$ or do I need to make use of $Q$ and $A%$ as well?
 A: First of all, observe that it's sufficient to have $P\iff A$ and $Q\iff B$, but it's not always needed. Consider $A\equiv \lnot Q$ and $B\equiv\lnot P$. So it's true $(P\rightarrow Q)\iff(\lnot Q\rightarrow\lnot P)$ (contrapositive), although we haven't $P\iff\lnot Q$ (nor $Q\iff\lnot P$) in general.
So I don't know a general approach to prove $(P\rightarrow Q)\iff(A\rightarrow B)$. But there are some theorems you can use, e.g.:


*

*$P\rightarrow Q\iff\lnot P\lor Q$. It turns implication into an or-statement.

*$P\rightarrow(Q\rightarrow R)\iff P\land Q\rightarrow R$ (import/export). It turns double implication into one implication.


Others theorems may exist that can help in accomplishing the task, but those ones already do a good job: for the first way, e.g., they give us $(P\rightarrow Q)\rightarrow(A\rightarrow B)\iff(\lnot P\lor Q)\land A\rightarrow B\iff(\lnot P\land A\rightarrow B)\land(Q\land A\rightarrow B)$. So it's necessary and sufficient to prove that $\lnot P\land A\implies B$ and $Q\land A\implies B$ in order to prove the first way ($P\rightarrow Q\implies A\rightarrow B$).
Edit: Of course, this "approach" is too difficult in general and it's most preferable the traditional way: assume $P\implies Q$ and $A$, and prove $B$. There's no general shortcut.
