What is the contrary - not the contradictory - of $(A\rightarrow B)$? In traditional logic, there was a distinction between " contradictory" sentences and " contrary" sentences. The relation holding between contrary sentences meant maximal opposition. 
For example, the contradictory of "all $A$ is $B$" is simply  "not all $A$ is $B$", but the contrary is " all $A$ is not-$B$" ( or no $A$ is $B$). 
I'm wondering whether this distinction also applies to modern propositional logic. 
Can one say that the contrary of $(A \to B)$ is $(A \to \lnot B)$? 
Normally contrary sentences can be false at the same time ( but not true at the same time). 
 A: Two statements in propositional logic are contrary when they cannot both be true at the same time.
Thus, as pointed our in the comments, $A \to B$ and $A \to \neg B$ are certainly not contrary, since they can both be true, simply by setting $A$ to False.
OK, but you can certainly always find some contrary statement to an propositional logic statement by negating it ... this would also be its contradictory statement.
But any statement can have many non-equivalent contrary statements, e.g. both $A \land \neg B$ as well as $A \land \neg A$ are contrary to $A \to B$, but these are not equivalent to each other. Indeed, while $A \land \neg B$ and $A \to B$ are contradictory, $A \land \neg A$ and $A \to B$ are not.
Ok, but out of all contrary statements, is there some 'maximally' contrary statement? Frankly I have not heard of any such thing in the context of propositional logic ... intuitively one could maybe point to a contradiction like $A \land \neg A$ as such a statement that 'maximally goes against' any statement ... but at the same time that is not at all very interesting or useful. So, I think such a notion simply does not exist in propositional logic.
A: Note, that in classical logic, $\phi\rightarrow\psi$ fails exactly when $\phi$ is true and $\psi$ is false. Formally, it thus holds that $\phi\rightarrow\psi\equiv\neg\phi\lor\psi$. 
Thus, the formal negation of $\phi\rightarrow\psi$ is given by
$$\neg(\phi\rightarrow\psi)\equiv\neg(\neg\phi\lor\psi)\equiv\phi\land\neg\psi$$
