I'm trying to attempt a proof which I think is quite similar to disjunctive syllogism, if not equivalent: $a \, \lor \, b \vdash \neg a \rightarrow b$
I've started with the first case of the disjunction by assuming $a$ and $\neg a$ to get a contradiction and therefore you can get $b$, which leads to the implication $\neg a \rightarrow b$. My issue is with the second case of just assuming $b$.
I tried just assuming $\neg a$ at that point but you can't build an implication out of that. Am I on the right track?
Hint: disjunction elimination rule in action. We'll show $P\lor Q\vdash Q\lor P$. The numbers in curly braces, $\{\}$, are the set of dependencies and the numbers in brackets, $(\;)$, are showing which dependency gets discharged.
$\begin{array}{} \{1\}&1.&P\lor Q&\text{P}\\ \{2\}&2.&P&\text{A for $\lor$E}\\ \{2\}&3.&Q\lor P&\text{2 $\lor$I}\\ \{4\}&4.&Q&\text{A for $\lor$E}\\ \{4\}&5.&Q\lor P&\text{4 $\lor$I}\\ \{1\}&6.&Q\lor P&\text{1, (2), 3, (4), 5 $\lor$E}\\ \end{array}$