Implications of the form $A \vee B \implies C$: what to do about the disjunction? I am aware that $A \vee B \implies C$ is logically equivalent to $(A \implies C) \wedge (B \implies C)$.  This is what is causing me an issue. If I am proving the original disjunction, is it sufficient to consider a case when $A$ alone is true, and show that the implication is true? That would mean that the disjunction is true, so hypothesis is true. Or do I need to consider the entirety of the conjunction in order to prove it, since if there are cases in which $B$ is false but $C$ is true the conjunction would also be false.
My issue is that in the case where $B$ is false but $C$ is true, the disjunction hypothesis may be true (depending on $A$), while the entire conjunction is false, so it appears as though the two are not logically equivalent.
 A: The hypothesis is true when $A\vee B$ is true. $A\vee B$ is true when $A$ is true or $B$ is true. $B$ might be true when $A$ is not true, making the hypothesis true when $A$ is false. So just considering the case $A$ true is insufficient. For example, $x=1\vee x=2$ is true when $x=1$ or $x=2$.
When $B$ is false and $C$ is true, $B\implies C$ is true (remember that the only contradiction to $B\implies C$ is when $B$ is true and $C$ is false). The disjunction hypothesis is true iff $A$ is true. Then $A\implies C$ is true so the conjunction is also true. Indeed, the two statements are equivalent.
A: 
My issue is that in the case where B is false but C is true, the disjunction hypothesis may be true (depending on A), while the entire conjunction is false, so it appears as though the two are not logically equivalent.

That is not so.
When $C$ is true, any conditional of the form $\phi\to C$ is true, whatever $\phi$ may be.   Thus too is any statement of the form $(\phi\to C)\leftrightarrow((\psi\to C)\wedge(\chi\to C))$, including:$$((A\vee B)\to C)\leftrightarrow((A\to C)\wedge (B\to C))$$

When $B$ is false, then any conditional of the form $B\to\phi$ is true. So any $\psi$ is equivalent to $\psi\wedge(B\to\phi)$. Likewise any $\chi\vee B$ is equivalent to $\chi$.
Thus $\neg B$ entails that: $(A\vee B)\to C$ is equivalent to $A\to C$ and this is equivalent to $(A\to C)\wedge (B\to C)$.
