How to close the last branch of this analytic tableau?

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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.

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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.

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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.