TITLE: Counterexample for $(p\rightarrow q) \longleftrightarrow (!q \rightarrow\mathord !p) $
QUESTION [5 upvotes]: Is the statement $$(p\rightarrow q) \longleftrightarrow (!q \rightarrow \mathord!p)  $$ always true? If it is not, provide a counterexample.

Till now I cannot find a counterexample nor prove that the statement is always true.
$!x$ is the negation of $x$.

REPLY [2 votes]: A footnote: The other answers given all assume that the arrow here is the classical material conditional (i.e. has the truth-table given in @amWhy's answer, i.e. makes $p \to q$ equivalent to $\sim p \lor q$ as in @Babak's answer).
It is well worth remarking that the principle that a conditional $p \to q$ is equivalent to $\sim q \to \sim p$  holds in some non-classical logics, including some relevance logics.
If you consider a pair of natural deduction proofs from $p \to q$ to and from  $\sim q \to \sim p$ you'll see you don't need to appeal e.g. to the "irrelevant" classical rule from a contradiction infer anything (or to anything which implies it).