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Consider

statement I

$(p\wedge\sim q)\wedge(\sim p\wedge q)$ is a fallacy.

Statement II

$(p\rightarrow q)\leftrightarrow(\sim q \rightarrow \sim p)$ is a tautology

A) Statement I is true, statement II is true; statement II is a correct explanation for statement I

B) Statement I is true, statement II is true; statement II is not a correct explanation for statement I

C) statement I is true ; Statement II is false

D) statement I is false ; Statement II is true

Option B

statement II

$(p\rightarrow q)\leftrightarrow (\sim q\rightarrow \sim p)$

$\equiv (p\rightarrow q)\leftrightarrow (p\rightarrow q)$

which is always true, so statement II is true.

Statement I $(p\wedge \sim q)\wedge (\sim p \wedge q)$

$\equiv p \wedge \sim q \wedge \sim p \wedge q$

$\equiv p \wedge \sim p \wedge \sim q \wedge q$

$\equiv f\wedge f\equiv f$

Hence, it is a fallacy statement

So, statement I is true

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