The following statement $\left( {p \to q} \right) \to $ $[(\sim p\rightarrow q) \rightarrow  q ]$ is

If $p , q$ and $r$ are three propositions, then which of the following combination of truth values of $p , q$ and $r$ makes the logical expression $\{(p \vee q) \wedge((\sim p) \vee r)\} \rightarrow((\sim q) \vee r)$ false ?

If $p \to ( \sim p\,\, \vee \, \sim q)$ is false, then the truth values of  $p$ and  $q$ are respectively .

The negation of the expression $q \vee((\sim q) \wedge p)$ is equivalent to

$\sim p \wedge q$ is logically equivalent to

$\sim ((\sim p)\; \wedge q)$ is equal to