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Use rules of inference to show that if $\forall x (P(x) \vee Q(x))$ and $\forall x ((\sim P(x) \wedge Q(x)) \rightarrow R(x))$ are true, then $\forall x (\sim R(x) \rightarrow P(x))$ is also true, where the domains of all quantifiers are the same.

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$
\begin{flalign}
\text{Proof} \\
&& &1.\ \forall x (P(x) \lor Q(x)) \tag*{Premise} \\
&& &2.\ P(a) \lor Q(a) \tag*{Universal instantiation from (1)} \\
&& &3.\ P(a) \lor Q(a) \lor R(a) \tag*{Addition from (2)} \\
&& &4.\ \forall x ((\neg P(x) \land Q(x)) \to R(x)) \tag*{Premise} \\
&& &5.\ \forall x (P(x) \lor \neg Q(x) \lor R(x)) \tag*{From (4)} \\
&& &6.\ P(a) \lor \neg Q(a) \lor R(a) \tag*{Universal instantiation from (5)} \\
&& &7.\ P(a) \lor R(a) \tag*{Resolution from (3) & (6)} \\
&& &8.\ \forall x (\neg R(x) \to P(x)) \tag*{Universal generalization from (7)} \\
\\ &&&& \blacksquare
\end{flalign}
$
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