The proposition $\left( { \sim p} \right) \vee \left( {p\, \wedge  \sim q} \right)$

The number of choices of $\Delta \in\{\wedge, \vee, \Rightarrow, \Leftrightarrow\}$, such that $( p \Delta q ) \Rightarrow(( p \Delta \sim q ) \vee((\sim p ) \Delta q ))$ is a tautology, is

Let $p$ and $q$ be two statements.Then $\sim( p \wedge( p \Rightarrow \sim q ))$ is equivalent to

Statement $-1$ : $ \sim \left( {p \leftrightarrow \, \sim q} \right)$ is equivalent to $p \leftrightarrow q$

Statement $-2$ : $ \sim \left( {p \leftrightarrow \, \sim q} \right)$ is a tautology.

Let $\Delta, \nabla \in\{\wedge, \vee\}$ be such that $p \nabla q \Rightarrow(( p \nabla$q) $\nabla r$ ) is a tautology. Then (p $\nabla q ) \Delta r$ is logically equivalent to

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