Proving if statement is true by proving the converse is false Can we prove an if-statement by proving the converse is false. I.e.
$\lnot (B \implies A) \implies (A \implies B)$
I've spoken to several colleagues about this and I'm still not convinced this proof is both valid and sound. The
The logical statement
$\lnot (B \rightarrow A) \implies (A \rightarrow B)$
is valid. This can be checked by means of a truth table. However,the antecedent implies $\lnot A$ which feels like it should violate soundness.

It depends on if there is an implicit quantifier or not.  For example:
$x \text { is even } \to x \text{ is odd}$
Both that and the converse is false.  So using the falseness of one to prove the truth of the other would be a fallacy. However, because there is an implicit $\forall x$ in front of it, it is language sensitive whether you want to say it even has a converse.
On the other hand, if you look at the truth table of implication:
$$\begin{array} {cc|c}
A & B & A \to B \\
\text{T} & \text{T} & \text{T} \\
\text{T} & \text{F} & \text{F} \\
\text{F} & \text{T} & \text{T} \\
\text{F} & \text{F} & \text{T} \\
\end{array}$$
You can see that the only way for a statement $A\to B$ to be false is for $A=\text{T}$ and $B=\text{F}$, which determines that $B\to A$ is in fact true.  But this only works if there are no free variables, which is to say: no implicit quantifiers.
So you could correctly use the falseness of
$$1=1 \to 2=3$$
to prove
$$2=3 \to 1=1$$
But you'll find there are very few statements like that in mathematics without free variables, which is implicit quantification.  And that type of statement is vacuous anyway which also makes it less likely to come up.
