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

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

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

edited Jan 24, 2014