Going through Velleman's How to Prove it, I came across the problem:
Prove $(P \rightarrow R) \wedge (Q \rightarrow R) = (P \vee Q) \rightarrow R$
Solution: $$(\neg P \vee R) \wedge (\neg Q \vee R)............(1)$$ $$(\neg P \wedge \neg Q) \vee R).................(2)$$ $$\neg(P \vee Q) \vee R = (P \vee Q) \rightarrow R$$
I understand how they got implies to become a boolean formula with just $\neg$ and $\vee$, however I don't know what they did to get from (1) to (2). When I attempted it I expanede (1) to become $$[(\neg P \vee R) \wedge \neg Q] \vee [(\neg P \vee R) \wedge R]$$ and attempted to expand/ condense from there. Could someone tell me what they did to get from (1) to (2) or better yet why my attempt isn't the best way to solve the problem?
You are doing something that's very common to beginning students of logic!
Distribution says that:
$P \land (Q \lor R) \Leftrightarrow (P \land Q) \lor (P \land R)$
Now, what students often don't realize is that this is an equivalence. That is, you can go from left to right, but you can also go from right to left.
Of course, going from right to left that doesn't feel like a 'Distribution' ... more like a 'Un-Distribution', or maybe 'Reverse Distribution' ... maybe think of it as 'Factoring' or 'Collecting common terms'... but in the end, it is still called Distribution.
So, in your case, when you have $(\neg P \lor R) \land (\neg Q \lor R)$, you can see this as the result of a Distribution of $R$ over $(\neg P \land \neg Q)$, so if we do a 'Reverse' Distribution, you take out the common $R$, and end up with $(\neg P \land \neg Q) \lor R$
See how that works?