How to close the last branch of this analytic tableau? 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.
 A: 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.
A: So you wish to prove $\{[(p\supset r)\wedge (q\supset r)]\wedge(p\vee q)\}\supset r$ by showing that its negation is a contradiction.
Firstly, recall that the Tableau tree works by chaining conjunctions, branching disjuntions, and closing off contradictions (of statements earlier in the chain) .   The root statement is proven to be contradictory when all possible branches are closed.   Conditionals (Implications) are handled through equivalences.    $\def\to{\supset} A\to B\equiv \neg A\vee B$ and $\neg(A\to B)\equiv A\wedge\neg B$.   If needed, deMorgan's laws likewise handle negations of connections.   So do the thing.
$$\begin{array}{c}
\neg\Big(\big((p\to r)\wedge(q\to r)\wedge(p\vee q)\big)\to r\Big)\\
(p\to r)\wedge(q\to r)\wedge(p\vee q)\wedge\neg r\\
(\neg p\vee r)\wedge(\neg q\vee r)\wedge(p\vee q)\wedge\neg r\\\hline
\begin{array}{ccccccc}&&&&&\neg r\\
&&&&&p\vee q\\
&&&&&\neg q\vee r\\
&&&&&\neg p\vee r\\\hline
&&& \Box &&\vert & \require{cancel}\cancel \Box\\
& \Box &&\vert&\cancel \Box&\\
\cancel \Box &\vert & \cancel \Box
\end{array}\end{array}$$ Replace the $\Box$ with the appropriate statements.
