The negation of the compound statement $^ \sim p \vee \left( {p \vee \left( {^ \sim q} \right)} \right)$ is
$\left( {^ \sim p \wedge q} \right) \wedge p$
$\left( {^ \sim p \wedge q} \right) \vee p$
$\left( {^ \sim p \wedge q} \right){ \vee \,^ \sim }p$
$\left( {^ \sim p{ \wedge ^ \sim }q} \right){ \wedge \,^ \sim }q$
The statement $( p \wedge q ) \Rightarrow( p \wedge r )$ is equivalent to.
Let $r \in\{p, q, \sim p, \sim q\}$ be such that the logical statement $r \vee(\sim p) \Rightarrow(p \wedge q) \vee r \quad$ is a tautology. Then ' $r$ ' is equal to
Contrapositive of the statement:
'If a function $f$ is differentiable at $a$, then it is also continuous at $a$', is
Which of the following is a contradiction
If $p \to ( \sim p\,\, \vee \, \sim q)$ is false, then the truth values of $p$ and $q$ are respectively .