If the truth value of the logical statement ( | vedclass

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Question

(C) The implication $(p \leftrightarrow \sim q) \rightarrow (\sim p \wedge q)$ is false only when the antecedent is true and the consequent is false.

Vedclass · Accepted Answer

$T, F$

Vedclass · Answer

(C) The implication $(p \leftrightarrow \sim q) \rightarrow (\sim p \wedge q)$ is false only when the antecedent is true and the consequent is false. So,$(p \leftrightarrow \sim q) \equiv T$ and $(\sim p \wedge q) \equiv F$. From $(\sim p \wedge q) \equiv F$,we know that the conjunction is false. Now,consider the antecedent $(p \leftrightarrow \sim q) \equiv T$. This means $p$ and $\sim q$ must have the same truth value. If we test option $C$ $(p=T, q=F)$: $\sim q = \sim F = T$. Then $(p \leftrightarrow \sim q) = (T \leftrightarrow T) = T$. And $(\sim p \wedge q) = (\sim T \wedge F) = (F \wedge F) = F$. Since the antecedent is $T$ and the consequent is $F$,the implication is $T \rightarrow F = F$. Thus,the truth values are $p=T$ and $q=F$.

If the truth value of the logical statement $(p \leftrightarrow \sim q) \rightarrow (\sim p \wedge q)$ is false,then the truth values of $p$ and $q$ are respectively:

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