$\sim [(p \vee \sim q) \rightarrow (p \wedge \sim q)] \equiv$

$\sim [(p \vee \sim q) \rightarrow (p \wedge \sim q)] \equiv$
  1. $(p \wedge \sim q) \wedge(\sim p \vee q)$
  2. $(p \wedge \sim q) \wedge(\sim p \wedge q)$
  3. $(p \vee \sim q) \wedge(\sim p \vee q)$
  4. $(p \vee \sim q) \vee(\sim p \vee q)$

Solution

$\begin{aligned} & \sim[(p \vee \sim q) \rightarrow(p \wedge \sim q)] \\ & \equiv \sim[\sim(p \vee \sim q) \vee(p \wedge \sim q)] \ldots[\text { Conditional law }]\end{aligned}$ $\begin{array}{ll}\equiv(p \vee \sim q) \wedge \sim(p \wedge \sim q) & \ldots[\text { De Morgan's law }] \\ \equiv(p \vee \sim q) \wedge(\sim p \vee q) & \ldots[\text { De Motgan's law }]\end{array}$

Asked in: MHT CET 2024 (03 May Shift 2)

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