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

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

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1 Answer

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)
So, it is a contradiction.

 

answered Jan 23, 2014 by thanvigandhi_1
edited Jan 23, 2014 by thanvigandhi_1
 

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