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
  1. $\mathrm{q}$
  2. $(\sim p) \vee q$
  3. $(\sim p) \wedge q$
  4. $(\sim p) \vee(\sim q)$

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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