# $(p \wedge\: ^{\sim}q) \wedge (^{\sim}p V q)$ is

$\begin {array} {1 1} (A)\;tautology & \quad (B)\;contradiction \\ (C)\;fallacy & \quad (D)\;none\: of \: these \end {array}$

 p q $^{\sim}p$ $^{\sim}q$ $(p \wedge \: ^{\sim}q)$ $(^{\sim}p V q)$ $(p \wedge\: ^{\sim}q)\wedge (^{\sim}p V q)$ T T F F F T F T F F T T F F F T T F F T F F F T T F T F
Ans : (B)

edited Jan 23, 2014