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

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 ?

Statement $-1$ : The statement $A \to (B \to A)$ is equivalent to $A \to \left( {A \vee B} \right)$.

Statement $-2$ : The statement $ \sim \left[ {\left( {A \wedge B} \right) \to \left( { \sim A \vee B} \right)} \right]$ is a Tautology

The statement $(\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q}) \wedge(\mathrm{q} \rightarrow \mathrm{r})) \rightarrow \mathrm{r}$ is :

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

If $p : 5$ is not greater than $2$ and $q$ : Jaipur is capital of Rajasthan, are two statements. Then negation of statement $p \Rightarrow  q$ is the statement