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Match List - I with List - II. List - IList - II (A) P ∧ (P → Q) (I) P ∨ Q (B) ¬(P ∨ Q) → (P ∧ Q) (II)…

(A) \(P \wedge (P \to Q) = P \wedge (\neg P \vee Q) = (P \wedge \neg P) \vee (P \wedge Q)\), which is II. (B) \(\neg(P \vee Q) \to (P \wedge Q)\) equals \((P \vee Q) \vee (P…

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Match List - I with List - II.

List - IList - II
(A) P ∧ (P → Q)(I) P ∨ Q
(B) ¬(P ∨ Q) → (P ∧ Q)(II) (P ∧ ¬P) ∨ (P ∧ Q)
(C) P → Q(III) (¬P) ∨ Q
(D) P ∨ (Q ∧ R)(IV) (P ∧ P) ∨ (P ∧ Q) ∨ (P ∧ R) ∨ (Q ∧ R)
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Answer and explanation

Correct answer B

(A) P∧(P→Q)=P∧(¬P∨Q)=(P∧¬P)∨(P∧Q)P \wedge (P \to Q) = P \wedge (\neg P \vee Q) = (P \wedge \neg P) \vee (P \wedge Q), which is II. (B) ¬(P∨Q)→(P∧Q)\neg(P \vee Q) \to (P \wedge Q) equals (P∨Q)∨(P∧Q)(P \vee Q) \vee (P \wedge Q), which absorbs to P∨QP \vee Q, so it matches I. (C) P→Q=¬P∨QP \to Q = \neg P \vee Q, which is III, and the expression in IV reduces to P∨(Q∧R)P \vee (Q \wedge R) because P∧P=PP \wedge P = P absorbs P∧QP \wedge Q and P∧RP \wedge R, so (D) matches IV.

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  • UGC NET Computer Science and Applications

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