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.

*

*Logically, the non-quantified statement $$\text{Amy being a vegan implies that she eats beef}$$ and its converse $$\text{Amy eating beef implies that she is a vegan}$$ cannot both be false: if one is false, then the other must be vacuously true.


*However, the quantified statement $$\text{Every vegan eats beef}$$ and its converse $$\text{Every beef-eater is a vegan}$$ are both false.
Explanation
The open formula $$A(x)\rightarrow B(x)$$ might not have a definite truth value, but is often implicitly universally-quantified to actually mean the sentence $$\forall x\,\Big(A(x)\rightarrow B(x)\Big).$$ The crux of your dilemma, as suggested by DanielV, stems from conflating the above with the sentence $$A \rightarrow B.$$

*

*The sentence $$\lnot (B \rightarrow A) \rightarrow (A \rightarrow
   B)$$ is a tautology. Thus, $$\text{not }\Big(B{\implies}A\Big)$$
logically entails $$A{\implies}B.$$

*On the other hand, the sentence $$\lnot\forall x\,\Big(B(x)\rightarrow
   A(x)\Big)\,\rightarrow\,\forall x\,\Big(A(x)\rightarrow B(x)\Big)$$ is
(satisfiable but) invalid. Thus, $$\text{not }\forall x\,
   \Big(B(x){\implies}A(x)\Big)$$ does not logically
entail $$\forall x\,\Big(A(x){\implies}B(x)\Big).$$
