The conditional $(p \wedge q) \Rightarrow p$ is :-

Given below are two pairs of statements. Combine these two statements using "if and only if".
$p:$ If the sum of digits of a number is divisible by $3,$ then the number is divisible by $3.$
$q:$ If a number is divisible by $3,$ then the sum of its digits is divisible by $3.$

The contrapositive of $\sim q \to p$ is equivalent to

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

By giving a counterexample,show that the following statement is not true.
$q:$ The equation $x^{2}-1=0$ does not have a root lying between $0$ and $2$.

The proposition $(p$ $\Rightarrow \sim p) \wedge (\sim p$ $\Rightarrow p)$ is a