Question

The Boolean expression $(p\Rightarrow q)\wedge (q\Rightarrow \sim p)$ is equivalent to

• A. $\sim p$
• B. $\sim q$
• C. $p$
• D. $q$

Answer 1

The correct option is A $\sim p$

$\because p\to q=\sim p\vee q$
$\therefore (p\Rightarrow q)\wedge (q\Rightarrow \sim p)=(\sim p\vee q)\wedge (\sim q\vee \sim p)$
Let $a=\sim p\vee q$ and $b=\sim q\vee \sim p$

$\begin{array}{ccccccc}p& q& \sim p& \sim q& \sim p\vee q& \sim q\vee \sim p& a\wedge b\\ T& T& F& F& T& F& F\\ T& F& F& T& F& T& F\\ F& T& T& F& T& T& T\\ F& F& T& T& T& T& T\end{array}$

$\therefore (p\Rightarrow q)\wedge (q\Rightarrow \sim p)$ is equivalent to $\sim p$