$\sim p \wedge q$ is logically equivalent to
$p \to q$
$q \to p$
$\sim (p \to q)$
$\sim (q \to p)$
(d) $\sim p \wedge q = \sim (q \to p)$.
The negation of the expression $q \vee((\sim q) \wedge p)$ is equivalent to
Which of the following Boolean expressions is not a tautology ?
The maximum number of compound propositions, out of $p \vee r \vee s , p \vee P \vee \sim s , p \vee \sim q \vee s$,
$\sim p \vee \sim r \vee s , \sim p \vee \sim r \vee \sim s , \sim p \vee q \vee \sim s$, $q \vee r \vee \sim s , q \vee \sim r \vee \sim s , \sim p \vee \sim q \vee \sim s$
that can be made simultaneously true by an assignment of the truth values to $p , q , r$ and $s$, is equal to
The expression $ \sim ( \sim p\, \to \,q)$ is logically equivalent to
$(p\; \wedge \sim q) \wedge (\sim p \vee q)$ is
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