Match the LIST-I with LIST-II: Match the logical equivalence propositions…
2025
Match the LIST-I with LIST-II: Match the logical equivalence propositions
LIST-I | LIST-II |
A. p→q | I. (p∧q)∨(¬p∧¬q) |
B. ¬(p∨(¬p∧q)) | II. ¬p∨q |
C. p↔q | III. ¬(p∨q) |
D. ¬(p↔q) | IV. ¬p↔q |
- A.
A-I, B-III, C-II, D-IV
- B.
A-II, B-II, C-III, D-IV
- C.
A-II, B-III, C-I, D-IV
- D.
A-II, B-III, C-IV, D-I
Attempted by 142 students.
Show answer & explanation
Correct answer: C
Final matching: A → ¬p∨q, B → ¬(p∨q), C → (p∧q)∨(¬p∧¬q), D → ¬p↔q
A: p→q is equivalent to ¬p∨q. This is the standard implication equivalence.
B: ¬(p∨(¬p∧q)). Simplify inside: p∨(¬p∧q) = (p∨¬p)∧(p∨q) = True ∧ (p∨q) = p∨q. Therefore B = ¬(p∨q).
C: p↔q is equivalent to (p∧q)∨(¬p∧¬q), the biconditional form expressing both true or both false.
D: ¬(p↔q) is the negation of the biconditional, which simplifies to (p∧¬q)∨(¬p∧q). This is equivalent to ¬p↔q (the exclusive-or form).
Hence the correct option is the one that matches: A → ¬p∨q; B → ¬(p∨q); C → (p∧q)∨(¬p∧¬q); D → ¬p↔q.
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