The negation of the statement $(p \vee q)^{\wedge}(q \vee(\sim r))$ is

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

If $p \Rightarrow (q \vee r)$ is false, then the truth values of $p, q, r$ are respectively

For the statements $p$ and $q$, consider the following compound statements :

$(a)$ $(\sim q \wedge( p \rightarrow q )) \rightarrow \sim p$

$(b)$ $((p \vee q) \wedge \sim p) \rightarrow q$

Then which of the following statements is correct?

If $p, q, r$ are simple propositions with truth values $T, F, T$, then the truth value of $(\sim p \vee q)\; \wedge \sim r \Rightarrow p$ is

The conditional $(p \wedge q) ==> p$ is