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$ ^{\sim}[(p \wedge q) \rightarrow (^{\sim}p V q)]$ is

$\begin {array} {1 1} (A)\;tautology & \quad (B)\;contradiction \\ (C)\;neither\: A \: nor\: B & \quad (D)\;either\: A\: or\: B \end {array}$

 

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

 

answered Jan 23, 2014 by thanvigandhi_1
 

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