The expression $((p \wedge q) \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q)$ is equivalent to

The expression $((p \wedge q) \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q)$ is equivalent to
  1. $\mathrm{p} \wedge \mathrm{q}$
  2. $\mathrm{p} \vee \sim \mathrm{q}$
  3. $\mathrm{p} \wedge \sim \mathrm{q}$
  4. $(\sim p) \wedge(\sim q)$

Solution

$\begin{aligned} & ((p \wedge q) \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q) \\ & \equiv(p \vee(p \vee \sim q)) \wedge(q \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q) \\ & \text {...[Distributive Law] } \\ & \equiv(p \vee \sim q) \wedge(p \vee T) \wedge(\sim p \wedge \sim q) \\ & \text {...[Complement Law] } \\ & \equiv(p \vee \sim q) \wedge T \wedge(\sim p \wedge \sim q) \\ & \text {...[Identity Law] } \\ & \equiv(p \vee \sim q) \wedge(\sim p \wedge \sim q) \quad \ldots \text { [Identity Law] } \\ & \equiv(p \wedge(\sim p \wedge \sim q)) \vee(\sim q \wedge(\sim p \wedge \sim q)) \\ & \equiv F \vee(\sim p \wedge \sim q) \\ & \text {...[Distributive Law] } \\ & \equiv \sim \mathrm{p} \wedge \sim \mathrm{q} \\ & \text {...[Complement Law] } \end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

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