Question

Consider
Statement I: $(p\wedge \sim q)\wedge (\sim p\wedge q)$ is a fallacy.
Statement II: $(p\to q)\leftrightarrow (\sim q\to \sim p)$ 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

Answer 1

The correct option is B Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

$StatementI:(p\wedge \sim q)\wedge (\sim p\wedge q)\equiv p\wedge \sim q\wedge \sim p\wedge q\equiv p\wedge \sim p\wedge \sim q\wedge q\equiv f\wedge f\equiv f$

Hence, it is a fallacy statement.
So, Statement I is true.
Statement II
$(p\to q)\leftrightarrow (\sim q\to \sim p)\equiv (p\to q)\leftrightarrow (p\to q)$

Which is always true, so Statement II is true.


Alternate Solution
Statement I $(p\wedge \sim q)\wedge (\sim p\wedge q)$
$\begin{array}{ccccccccc}p& q& \sim p& \sim q& p\wedge \sim q& \sim p\wedge q& (p\wedge \sim q)& \wedge & (\sim p\wedge q)\\ T& T& F& F& F& F& & F& \\ T& F& F& T& T& F& & F& \\ F& T& T& F& F& T& & F& \\ F& F& T& T& F& F& & F& \end{array}$

Hence, it is a fallacy.
Statement II $(p\to q)\leftrightarrow (\sim q\to \sim p)$
$\sim q\to \sim p$ is contrapositive of $p\to q$. Hence,
$(p\to q)\leftrightarrow (p\to q)$ will be a tautology