The logic expression \((\bar{P} \wedge Q) \vee(P \wedge \bar{Q}) \vee(P \wedge…
2022
The logic expression \((\bar{P} \wedge Q) \vee(P \wedge \bar{Q}) \vee(P \wedge Q)\) is equivalent to
- A.
\(\bar{P} \vee Q\) - B.
\(P \vee \bar{Q}\) - C.
\(P \vee Q\) - D.
\(\bar{P} \vee \bar{Q}\)
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Correct answer: C
Key idea: simplify by grouping like terms and applying distributive/absorption laws.
Combine the terms that share P: (P ∧ ¬Q) ∨ (P ∧ Q) = P ∧ (¬Q ∨ Q) = P.
Substitute this back into the full expression: (¬P ∧ Q) ∨ (P ∧ ¬Q) ∨ (P ∧ Q) becomes (¬P ∧ Q) ∨ P.
Apply absorption/distributive law: P ∨ (¬P ∧ Q) = (P ∨ ¬P) ∧ (P ∨ Q) = True ∧ (P ∨ Q) = P ∨ Q.
Conclusion: the original expression is equivalent to P ∨ Q.
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