For proof by elimination, if I have $A$ implies $B$ or $C$: $A \Rightarrow (B \vee C)$, then I need to show if $A$ and $\neg B$, then C.
What happens though, when we have $A$ implies $B$, or $C$, or $D$: $A \Rightarrow (B \vee C \vee D...)$ and so on?
Basically, the proof relies on the fact that $p \to q$ is equivalent to $\lnot p \lor q$.
Thus, applying it to $A \to (B \lor C)$, we have :
$\lnot A \lor B \lor C$.
So, if $A$ and $¬B$, then $C$ follows because in order that the disjunction holds, at least one of the disjuncts must hold.
For the general case :
$A \to (B_1 \lor B_2 \lor \ldots B_n)$,
in order to conclude with $B_n$ we have to show that : $A$ and $\lnot B_1$ and ... and $\lnot B_{n-1}$.
