Which of the following is false?

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(D) $1$. $p \rightarrow q \equiv \sim p \vee q$ is a standard logical equivalence.$2$. For $(p \vee q) \wedge (q \vee r)$ to be true,both $(p \vee q)$

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The truth value of $p \wedge \sim (p \vee q)$ is always $T$.

Vedclass · Answer

(D) $1$. $p \rightarrow q \equiv \sim p \vee q$ is a standard logical equivalence.$2$. For $(p \vee q) \wedge (q \vee r)$ to be true,both $(p \vee q)$ and $(q \vee r)$ must be true. If $p=T, q=F, r=T$,then $(T \vee F) \wedge (F \vee T) = T \wedge T = T$. This is correct.$3$. By De Morgan's Law,$\sim (p \wedge (q \vee r)) \equiv \sim p \vee \sim (q \vee r) \equiv \sim p \vee (\sim q \wedge \sim r) \equiv (\sim p \vee \sim q) \wedge (\sim p \vee \sim r)$. This is correct.$4$. For $p \wedge \sim (p \vee q)$,we have $p \wedge (\sim p \wedge \sim q) \equiv (p \wedge \sim p) \wedge \sim q \equiv F \wedge \sim q \equiv F$. Thus,the statement that it is always $T$ is false.

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