The statement pattern [(p rightarrow q) wedge | vedclass

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(D) To determine when the statement pattern $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology,we first simplify the antecedent $[(p

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$\sim p$

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(D) To determine when the statement pattern $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology,we first simplify the antecedent $[(p \rightarrow q) \wedge \sim q]$.Using the implication law $p \rightarrow q \equiv \sim p \vee q$,we get:$[(\sim p \vee q) \wedge \sim q] \rightarrow r$By the distributive law,this becomes $[(\sim p \wedge \sim q) \vee (q \wedge \sim q)] \rightarrow r$.Since $q \wedge \sim q \equiv F$ (a contradiction),we have:$[(\sim p \wedge \sim q) \vee F]$ $\rightarrow r \equiv (\sim p \wedge \sim q)$ $\rightarrow r$.For this to be a tautology,the expression must be true for all truth values of $p$ and $q$.If $r = \sim p$,the expression becomes $(\sim p \wedge \sim q) \rightarrow \sim p$.Since $(\sim p \wedge \sim q)$ implies $\sim p$,the implication is always true.Thus,the statement is a tautology when $r = \sim p$.

The statement pattern $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology when $r$ is equivalent to

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