MHT CET · Maths · Mathematical Reasoning
The expression \(((p \wedge q) \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q)\) is equivalent to
- A \(\mathrm{p} \wedge \mathrm{q}\)
- B \(\mathrm{p} \vee \sim \mathrm{q}\)
- C \(\mathrm{p} \wedge \sim \mathrm{q}\)
- D \((\sim p) \wedge(\sim q)\)
Answer & Solution
Correct Answer
(D) \((\sim p) \wedge(\sim q)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& ((p \wedge q) \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q) \\
& \equiv(p \vee(p \vee \sim q)) \wedge(q \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q) \\
& \text {...[Distributive Law] } \\
& \equiv(p \vee \sim q) \wedge(p \vee T) \wedge(\sim p \wedge \sim q) \\
& \text {...[Complement Law] } \\
& \equiv(p \vee \sim q) \wedge T \wedge(\sim p \wedge \sim q) \\
& \text {...[Identity Law] } \\
& \equiv(p \vee \sim q) \wedge(\sim p \wedge \sim q) \quad \ldots \text { [Identity Law] } \\
& \equiv(p \wedge(\sim p \wedge \sim q)) \vee(\sim q \wedge(\sim p \wedge \sim q)) \\
& \equiv F \vee(\sim p \wedge \sim q) \\
& \text {...[Distributive Law] } \\
& \equiv \sim \mathrm{p} \wedge \sim \mathrm{q} \\
& \text {...[Complement Law] }
\end{aligned}\)
& ((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}\)
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