Embrace the Logical Necessities

By contraposition on $\neg A\Rightarrow C,$ we obtain $\neg C\Rightarrow A.$

$B\wedge C$ is always false is equivalent to $\neg (B\wedge C).$

Using one of De Morgan's laws, we have $\neg (B\wedge C)\Leftrightarrow \neg B \vee \neg C$ .

We then have $\neg B \vee \neg C \Leftrightarrow B\Rightarrow \neg C.$

Thus, $(B\Rightarrow \neg C) \wedge (\neg C \Rightarrow A).$

By transitivity of implication, $B\Rightarrow A.$

Starting from $\lnot A \implies C$ .

Because $B \wedge C$ is always false, $\lnot A \implies \lnot B \wedge C$ . But then $\lnot A \implies \lnot B$ ; contraposition then implies that $B\implies A$ .

Assume $B$ .

Since $B\wedge C$ must be false, we have $\neg C$ .

Since $\neg A \Rightarrow C$ is equivalent to $A\vee C$ ,
$\neg C$ gives us $A$ .

Thus, $B\Rightarrow A$