The statement [(p rightarrow q) wedge sim q] | vedclass

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(D) To determine when the statement $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology,we analyze the expression $[(p \rightarrow q)

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$\sim q$

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(D) To determine when the statement $[(p$ $\rightarrow q) \wedge \sim q]$ $\rightarrow r$ is a tautology,we analyze the expression $[(p \rightarrow q) \wedge \sim q]$.Note that $(p \rightarrow q) \equiv (\sim p \vee q)$.Thus,$[(p \rightarrow q) \wedge \sim q] \equiv [(\sim p \vee q) \wedge \sim q]$.Using the distributive law,this becomes $(\sim p \wedge \sim q) \vee (q \wedge \sim q)$.Since $(q \wedge \sim q) \equiv F$ (a contradiction),the expression simplifies to $(\sim p \wedge \sim q) \vee F \equiv \sim p \wedge \sim q$.Now,the original statement is $(\sim p \wedge \sim q) \rightarrow r$.For this to be a tautology,the implication must be true for all truth values of $p$ and $q$.If we set $r \equiv \sim p \wedge \sim q$,the statement becomes $(\sim p \wedge \sim q) \rightarrow (\sim p \wedge \sim q)$,which is always true.However,looking at the options,we check if $r \equiv \sim q$ works.If $r \equiv \sim q$,the statement is $(\sim p \wedge \sim q) \rightarrow \sim q$.Since $(\sim p \wedge \sim q)$ is a subset of $\sim q$ (i.e.,if $\sim p \wedge \sim q$ is true,then $\sim q$ must be true),the implication is always true.Therefore,the statement is a tautology when $r \equiv \sim q$.

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

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