Browse Questions

$^{\sim}(p V q)$ is equivalent to

$\begin {array} {1 1} (A)\;(^{\sim}p)\wedge (^{\sim}q) & \quad (B)\;(^{\sim}p)V (^{\sim}q) \\ (C)\;(^{\sim}p)\wedge (^{\sim}q) & \quad (D)\;.(p) V (^{\sim}q) \end {array}$

Ans : (A)
So,$^{\sim}(p V q)$ is equivalent to $(p) \wedge (^{\sim}q)$
 p q $^{\sim}p$ $^{\sim}q$ $p V q$ $^{\sim} (p V q)$ $(^{\sim}p) \wedge (^{\sim}q)$ T T F F T F F T F F T T F F F T T F T F F F F T T F T T