Statement-I: (p wedge sim q) wedge (sim p wed | vedclass

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(B) For Statement-$I$: $(p \wedge \sim q) \wedge (\sim p \wedge q) = p \wedge (\sim q \wedge \sim p) \wedge q = (p \wedge \sim p) \wedge (\sim q \wedg

Vedclass · Accepted Answer

Statement-$I$ is true,Statement-$II$ is true. Statement-$II$ is not a correct explanation for Statement-$I$.

Vedclass · Answer

(B) For Statement-$I$: $(p \wedge \sim q) \wedge (\sim p \wedge q) = p \wedge (\sim q \wedge \sim p) \wedge q = (p \wedge \sim p) \wedge (\sim q \wedge q) = F \wedge F = F$. Since the result is always false,it is a contradiction. Thus,Statement-$I$ is true.For Statement-$II$: The expression $(p$ $\rightarrow q) \Leftrightarrow (\sim q$ $\rightarrow \sim p)$ represents the equivalence between a conditional statement and its contrapositive. Since a conditional statement is always logically equivalent to its contrapositive,this is a tautology. Thus,Statement-$II$ is true.Since Statement-$II$ does not explain why Statement-$I$ is a contradiction,option $B$ is correct.

Statement-$I$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a contradiction.
Statement-$II$: $(p$ $\rightarrow q) \Leftrightarrow (\sim q$ $\rightarrow \sim p)$ is a tautology.

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Statement-$I$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a contradiction.Statement-$II$: $(p$ $\rightarrow q) \Leftrightarrow (\sim q$ $\rightarrow \sim p)$ is a tautology.