ConsiderStatement-1: (p wedge sim q) wedge (s | vedclass

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(D) Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$Since $(\sim q \rightarrow \sim p)$ is the contrapositive of $(p \ri

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true; Statement-$2$ is not a correct explanation for Statement-$1$

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(D) Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$Since $(\sim q \rightarrow \sim p)$ is the contrapositive of $(p \rightarrow q)$,they have the same truth values.Thus,$(p \rightarrow q) \leftrightarrow (p \rightarrow q)$ is always true,which means it is a tautology.So,Statement-$2$ is true.Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$Using the associative and commutative laws:$= (p \wedge \sim p) \wedge (\sim q \wedge q)$$= F \wedge F = F$Since the result is always false,it is a fallacy.So,Statement-$1$ is true.Both statements are true,and Statement-$2$ is not an explanation for Statement-$1$.

Consider
Statement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a fallacy.
Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$ is a tautology.

Similar Questions

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ConsiderStatement-$1$: $(p \wedge \sim q) \wedge (\sim p \wedge q)$ is a fallacy.Statement-$2$: $(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$ is a tautology.