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$(p \wedge q) \vee \sim p$ is equivalent to
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Verified Answer
The correct answer is:
$\sim p \vee q$
$(p \wedge q) \vee \sim p=\sim p \vee(p \wedge q)$
[By commutative law] $\begin{array}{lr}=(\sim p \vee p) \wedge(\sim p \vee q) & \text { [By distributive law] } \\ =(p \vee \sim p) \wedge(\sim p \vee q) & \text { [By commutative law] }\end{array}$
$=t \wedge(\sim p \vee q)$
[By complement law]
$=\sim p \vee q$
[By identity law]
[By commutative law] $\begin{array}{lr}=(\sim p \vee p) \wedge(\sim p \vee q) & \text { [By distributive law] } \\ =(p \vee \sim p) \wedge(\sim p \vee q) & \text { [By commutative law] }\end{array}$
$=t \wedge(\sim p \vee q)$
[By complement law]
$=\sim p \vee q$
[By identity law]
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