The expression $(p \wedge \sim q) \vee q \vee(\sim p \wedge q)$ is equivalent to
The expression $(p \wedge \sim q) \vee q \vee(\sim p \wedge q)$ is equivalent to
- $\sim p \vee q$
- $p \wedge q$
- $\mathrm{p} \vee \mathrm{q}$
- $\mathrm{p} \vee \sim \mathrm{q}$
Solution
$\begin{aligned}
& (p \wedge \sim q) \vee q \vee(\sim p \wedge q) \\
& \equiv[(p \vee q) \wedge(\sim q \vee q)] \vee(\sim p \wedge q)
\end{aligned}$
...[Distributive law]
$\begin{aligned}
& \equiv[(p \vee q) \wedge T] \vee(\sim p \wedge q) \ldots[\text { Complement law] } \\
& \equiv(p \vee q) \vee(\sim p \wedge q) \\
& \equiv(p \vee q \vee \sim p) \wedge(p \vee q \vee q)
\end{aligned}$
...[Distributive law]
$\equiv(T \vee q) \wedge(p \vee q)$
...[Complement law and Idempotent law]
$\begin{aligned}
& \equiv T \wedge(p \vee q) \\
& \equiv p \vee q
\end{aligned}$
$\therefore$ [Identity law]
...[Identity law]
Asked in: MHT CET 2023 (13 May Shift 1)
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