If $(p \wedge \sim r) \Rightarrow (q \vee r)$ is false and $q$ and $r$ are both false,then $p$ is

Check the validity of the statement given below by the method specified against it.
$q:$ If $n$ is a real number with $n > 3$,then $n^{2} > 9$ (by contradiction method).

The simplified switching circuit for the following circuit is

The statement $(p$ $\rightarrow q)$ $\rightarrow ((\sim p$ $\rightarrow q)$ $\rightarrow q)$ is

In a switching circuit, if the combination $(S_1 \land S_2)$ is connected in parallel to the combination $(S'_1 \land S'_2)$, then the room is lit only when ...

If the statement $(p$ $\rightarrow q)$ $\rightarrow (q$ $\rightarrow r)$ is false,then the truth values of statements $p, q, r$ respectively,can be-