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

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

 

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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$

 

answered Jan 24, 2014 by thanvigandhi_1
edited Jan 24, 2014 by thanvigandhi_1
 

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