Match the LIST-I with LIST-II. Match the equivalent propositions in List-I and List-II
| LIST-I | LIST-II |
|---|---|
| A. P→(Q→P) | I. (P∨Q)∧¬(P∧Q) |
| B. P→(Q∨R) | II. (P∨R)→Q |
| C. (P→Q)∧(R→Q) | III. (P→Q)∨(P→R) |
| D. ¬(P⇔Q) | IV. ¬P→(P→Q) |
Pick an option to check yourself, or open the answer below.
Show answer & explanation
Correct answer: B. A-IV, B-III, C-II, D-I
Explanation
A: P→(Q→P) ≡ ¬P∨¬Q∨P ≡ True, and IV: ¬P→(P→Q) ≡ P∨¬P∨Q ≡ True, so both are tautologies and A≡IV. B: P→(Q∨R) ≡ ¬P∨Q∨R, and III: (P→Q)∨(P→R) ≡ (¬P∨Q)∨(¬P∨R) ≡ ¬P∨Q∨R, so B≡III. C: (P→Q)∧(R→Q) ≡ (¬P∨Q)∧(¬R∨Q) ≡ Q∨(¬P∧¬R) ≡ Q∨¬(P∨R) ≡ (P∨R)→Q, matching II. D: ¬(P⇔Q) is exclusive-or, which equals (P∨Q)∧¬(P∧Q), matching I. So the matching is A-IV, B-III, C-II, D-I.More Discrete Structures and Optimisation questions
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