Browse Questions

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

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

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

edited Jan 24, 2014