The statement pattern $\mathrm{p} \rightarrow \sim(\mathrm{p} \wedge \sim \mathrm{q})$ is equivalent to
The statement pattern $\mathrm{p} \rightarrow \sim(\mathrm{p} \wedge \sim \mathrm{q})$ is equivalent to
- $\mathrm{q}$
- $(\sim p) \vee q$
- $(\sim p) \wedge q$
- $(\sim p) \vee(\sim q)$
Solution
$\begin{aligned}
& p \rightarrow \sim(p \wedge \sim q) \\
& \equiv \sim p \vee \sim(p \wedge \sim q) \\
& \equiv \sim p \vee(\sim p \vee q) \\
& \equiv(\sim p \vee \sim p) \vee q \\
& \equiv \sim p \vee q
\end{aligned}$
$\ldots[\because \mathrm{p} \rightarrow \mathrm{q} \equiv \sim \mathrm{p} \vee \mathrm{q}]$
...[De Morgan's law]
...[Associative law]
...[Idempotent law]
Asked in: MHT CET 2023 (11 May Shift 1)
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