Consider the two statements :

$(\mathrm{S} 1):(\mathrm{p} \rightarrow \mathrm{q}) \vee(\sim \mathrm{q} \rightarrow \mathrm{p})$ is a tautology

$(S2): (\mathrm{p} \wedge \sim \mathrm{q}) \wedge(\sim \mathrm{p} \vee \mathrm{q})$ is a fallacy.

Then :

Statement $-1 :$ $\sim (p \leftrightarrow \sim q)$ is equivalent to $p\leftrightarrow q $

Statement $-2 :$ $\sim (p \leftrightarrow \sim q)$ s a tautology

$( S 1)( p \Rightarrow q ) \vee( p \wedge(\sim q ))$ is a tautology $( S 2)((\sim p ) \Rightarrow(\sim q )) \wedge((\sim p ) \vee q )$ is a Contradiction. Then

If the truth value of the statement $p \to \left( { \sim q \vee r} \right)$ is false $(F)$, then the truth values of the statement $p, q, r$ are respectively

The conditional $(p \wedge q)  \Rightarrow  p$ is :-