Question

Let $\ast ,\square \in \{\wedge ,\vee \}$ be such that Boolean expression $(p\ast \sim q)\Rightarrow \left(p\square q\right)$ is a tautology. Then:

• A. $\ast =\wedge ,\square =\wedge$
• B. $\ast =\vee ,\square =\wedge$
• C. $\ast =\vee ,\square =\vee$
• D. $\ast =\wedge ,\square =\vee$

Answer 1

The correct option is D $\ast =\wedge ,\square =\vee$

$(p\ast \sim q)\Rightarrow \left(p\square q\right)$
$\begin{array}{ccccccc}p& q& \sim q& p\wedge \sim q& p\vee \sim q& p\wedge q& p\vee q\\ T& T& F& F& T& T& T\\ T& F& T& T& T& F& T\\ F& T& F& F& F& F& T\\ F& F& T& F& T& F& F\end{array}$

$\begin{array}{cccc}(p\wedge \sim q)\to (p\wedge q)& (p\wedge \sim q)\to (p\vee q)& (p\vee \sim q)\to (p\wedge q)& (p\vee \sim q)\to (p\vee q)\\ T& T& T& T\\ F& T& F& T\\ T& T& T& T\\ T& T& F& F\end{array}$

From above Truth table, $(p\wedge \sim q)\to (p\vee q)$ is a tautology.
$\therefore \ast =\wedge ,\square =\vee$

Alternate Solution:
$(p\ast \sim q)\Rightarrow \left(p\square q\right)$
$=\sim (p\ast \sim q)\vee \left(p\square q\right)$
$=(\sim p\square q)\vee \left(p\square q\right)$
$\text{= T}$
$\Rightarrow \square =\vee$
and $\ast =\sim \square =\wedge$