Question

The statement $p\Rightarrow \left(q \land p\right)$ is negation of the statement

• A. $p\Rightarrow q$
• B. $p\land q$
• C. $\sim \left(p \Rightarrow q\right)$
• D. $\sim \left(p \land q\right)$

Answer 1

Answer: (C)

$\because \sim(s \Rightarrow t) \equiv s \wedge(\sim t)$
$\because \sim(p \Rightarrow (q \wedge p)) \equiv p \wedge(\sim(q \wedge p))$
$\equiv p \wedge(\sim q \vee \sim p)$
$\equiv(p \wedge \sim q) \vee(p \wedge \sim p)$
$\equiv(p \wedge \sim p) \vee f$
$\equiv p \wedge \sim q \equiv \sim(p \Rightarrow q)$