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 not a correct explanation for Statement-I
• B. Statement-I is true, Statement-II is false
• C. Statement-I is false, Statement-II is true
• D. Statement-I is true, Statement-II is true , Statement-II is a correct explanation for statement-I

Answer 1

The Correct Option is A

Lets prepare the truth table for the statements.

Then Statement-I is fallacy.

Then Statement-II is tautology .
$-\sim (\sim p \vee q) \wedge \sim (\sim q \vee p)$
$ \equiv \, \sim (( \sim p \vee q) \vee (\sim q \vee p))$
$ \equiv\,\sim (( p \to q) \vee (q \to p)) \equiv \sim T $
Thus Statement-I is true because its negation is false.
$\left(\left(p\rightarrow q\right)\rightarrow\left(\sim q\rightarrow\sim p\right)\wedge\left(\left(\sim q\rightarrow\sim p\right)\rightarrow\left(p\rightarrow q\right)\right) \right)$
$= \left(\left(\sim p\vee q\right)\rightarrow\left(q \vee \sim p\right)\wedge \left(\left(q \vee \sim p\right)\rightarrow\left(\sim p \vee q\right)\right)\right)$
$\equiv T \wedge T \equiv T$ Then Statement-II is true.