If (p wedge sim r) Rightarrow (q vee r) is fa | vedclass

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(A) The implication $(p \wedge \sim r) \Rightarrow (q \vee r)$ is false only when the antecedent $(p \wedge \sim r)$ is true and the consequent $(q \v

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(A) The implication $(p \wedge \sim r) \Rightarrow (q \vee r)$ is false only when the antecedent $(p \wedge \sim r)$ is true and the consequent $(q \vee r)$ is false.Given that $q$ and $r$ are both false,the consequent $(q \vee r)$ becomes $(F \vee F) = F$,which is consistent with the implication being false.For the antecedent $(p \wedge \sim r)$ to be true,both $p$ and $\sim r$ must be true.Since $r$ is false,$\sim r$ is true.Therefore,$p \wedge T = T$ implies that $p$ must be true.

If $(p \wedge \sim r) \Rightarrow (q \vee r)$ is false and $q$ and $r$ are both false,then $p$ is

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