Well, here's the proof you've all been waiting for, straight from your incipiently senile instructor.

~P->Q |- PvQ (right-left side of S22)

Behold the proof:

Actually, this is sort of a tricky proof, not an easy one to discover without just a little experience with the rules. The strategy is (as I was trying to say in class, but the old neurons went on strike): this looks like a job for the indirect approach. We don't have PvQ as a constituent of any premise, and we don't have any way to get P or Q out of the single premise ~P->Q. So, we're going to try something like this:

Here's where it gets a little harder. With any RAA proof, the trick is figuring out just which contradictory pair to deduce. A good idea (sometimes) is to try for the denial of something you already have; another good idea (sometimes) is just to try for the very thing you're trying to prove. The strategy is this: if you want to get X, see if you can assume ~X and still get X; then, you can get rid of the assumption. This is the strategy I'll try here:

So, we try to get PvQ. How could we get it? One way is to get P and use vI; another is to get Q and use vI. Which shall we use?

Since Q is the right side of a conditional in (1), we might try for that:

Here is where the proof gets clever: how shall we get ~P? We can't get it from (1), but since it's a negation we might try RAA:

All we need now is a contradictory pair: either a denial of ~(PvQ), for instance PvQ, or a denial of ~P->Q. But the first of those is easy to get, since we've assumed P: so, we have our contradictory pair, and we discharge assumption 3, and we're done!

Now, that wasn't so hard, was it?