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The proposition $ p \Rightarrow ^{\sim}(p \wedge\: ^{\sim}q)$ is

$\begin {array} {1 1} (A)\;contradiction & \quad (B)\; tautology \\ (C)\;either A or B & \quad (D)\;None of these \end {array}$

 

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p p $ ^{\sim}q$ $ (p \wedge \: ^{\sim}q)$ $^{\sim} (p \wedge \: ^{\sim}q)$ $ p \Rightarrow \: ^{\sim}(p \wedge \: ^{\sim}q)$
T T F F T T
T F T T F F
F T F F T T
F F T F T T
Ans : (D)
It is clear from the table that given proposition
is neither tautology nor contradiction.

 

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

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