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

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


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Ans : (B)
$ ^{\sim}(p\leftrightarrow q) = ^{\sim} \{(p \rightarrow q) \wedge (q \rightarrow p)\}$
$= \{^{\sim}(p \rightarrow q)\} V \{^{\sim}(q \rightarrow p)\}$…………………[ by De Morgan’s Law]
$= (p \wedge ^{\sim}q) V (q \wedge ^{\sim}p)$
answered Jan 23, 2014 by thanvigandhi_1

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