I'm working through Smullyan's "First-Order Logic." One exercise is to prove that $$\{[(p\supset r)\wedge (q\supset r)]\wedge(p\vee q)\}\supset r$$ I'm not sure how to make a nice-looking tableau with LaTeX but here goes: $$\sim(\{[(p\supset r)\wedge (q\supset r)]\wedge(p\vee q)\}\supset r)$$ $$\sim\{[(p\supset r)\wedge (q\supset r)]\wedge(p\vee q)\}$$ $$r$$ $$\sim[(p\supset r)\wedge (q\supset r)] \;\;|\;\; \sim(p\vee q)$$ Where I'm using | to represent a branching. The first branch will lead to the negation of both of the conditionals, which implies $\sim r$, which contradicts the 3rd line. The second branch, on the other hand, has the direct consequences $\sim p$ and $\sim q$. This doesn't contradict anything and I can't see anything else to which I could apply the inference rules. How do I complete the proof?
I feel like I'm making an obvious mistake so hints would be appreciated. Also anyone is certainly welcome to edit the question if they know how to make a nicer-looking tableau.
I think there is a problem with your tree: $$\sim(\{[(p\supset r)\wedge (q\supset r)]\wedge(p\vee q)\}\supset r)$$ $$\{[(p\supset r)\wedge (q\supset r)]\wedge(p\vee q)\}$$ $$\sim r$$
Isn't it this the correct step? If you continue with the tree you will find it closed.