Match the LIST-I with LIST-II: Match the logical equivalence propositions Choose the correct answer from the options given below: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-II, B-III, C-I, D-IV
Two propositions are logically equivalent when they have the same truth value under every assignment of truth values to their variables. Each LIST-I formula is rewritten with standard laws (implication, De Morgan, absorption, biconditional) until it matches a LIST-II formula.
Step 1 — A. The implication law gives p → q ≡ ¬p ∨ q, which is II.
Step 2 — B. Distribute inside: p ∨ (¬p ∧ q) ≡ (p ∨ ¬p) ∧ (p ∨ q) ≡ T ∧ (p ∨ q) ≡ p ∨ q. So ¬(p ∨ (¬p ∧ q)) ≡ ¬(p ∨ q), which is III.
Step 3 — C. The biconditional is true exactly when p and q are both true or both false: p ↔ q ≡ (p ∧ q) ∨ (¬p ∧ ¬q), which is I.
Step 4 — D. ¬(p ↔ q) is true exactly when p and q differ (exclusive OR). ¬p ↔ q is true exactly when ¬p and q agree, i.e. when p and q differ. Hence ¬(p ↔ q) ≡ ¬p ↔ q, which is IV.
| p | q | ¬(p ↔ q) | ¬p ↔ q |
|---|---|---|---|
| T | T | F | F |
| T | F | T | T |
| F | T | T | T |
| F | F | F | F |
Any code that pairs A with I fails because p → q is false only for p = T, q = F, whereas (p ∧ q) ∨ (¬p ∧ ¬q) is also false for p = F, q = T. Pairing B with II fails because ¬(p ∨ q) is false when p = F, q = T while ¬p ∨ q is true there. Pairing C with IV and D with I swaps the biconditional with its negation.
A-II, B-III, C-I, D-IV option is correct.
Similar Questions
Consider the compound propositions given below as :
(a) \(p \vee \sim (p \wedge q)\) (b) \((p \wedge \sim q) \vee \sim (p \wedge q)\)
(c) \(p \wedge (q \vee r)\)
Which of the above propositions are tautologies ?
In propositional logic \(P \leftrightarrow Q\) is equivalent to (Where \(\sim\) denotes NOT) :
Match the following :
List - I List - II (a) \((p \to q) \Leftrightarrow (\neg q \to \neg p)\) (i) Contrapositive (b) \([(p \wedge q) \to r] \Leftrightarrow [p \to (q \to r)]\) (ii) Exportation law (c) \((p \to q) \Leftrightarrow [(p \wedge \neg q) \to o]\) (iii) Reductio ad absurdum (d) \((p \leftrightarrow q) \Leftrightarrow [(p \to q) \wedge (q \to p)]\) (iv) Equivalence Codes :
Which of the following is NOT logically equivalent to P→Q?
Consider the following if p and q are two statements.
(A) ~(p∧q) ≡ ~p∨~q
(B) ~(p∨q) ≡ ~p∧~q
(C) p∧~p ≡ T
(D) ~(p⇒q) ≡ p∧~q
(E) p∨q ≡ ~p∨~q
Choose the correct answer from the options given below :
Important Questions from Propositional Equivalence
Consider the compound propositions given below as :
(a) \(p \vee \sim (p \wedge q)\) (b) \((p \wedge \sim q) \vee \sim (p \wedge q)\)
(c) \(p \wedge (q \vee r)\)
Which of the above propositions are tautologies ?
In propositional logic \(P \leftrightarrow Q\) is equivalent to (Where \(\sim\) denotes NOT) :
Match the following :
List - I List - II (a) \((p \to q) \Leftrightarrow (\neg q \to \neg p)\) (i) Contrapositive (b) \([(p \wedge q) \to r] \Leftrightarrow [p \to (q \to r)]\) (ii) Exportation law (c) \((p \to q) \Leftrightarrow [(p \wedge \neg q) \to o]\) (iii) Reductio ad absurdum (d) \((p \leftrightarrow q) \Leftrightarrow [(p \to q) \wedge (q \to p)]\) (iv) Equivalence Codes :
Which of the following is NOT logically equivalent to P→Q?
Consider the following if p and q are two statements.
(A) ~(p∧q) ≡ ~p∨~q
(B) ~(p∨q) ≡ ~p∧~q
(C) p∧~p ≡ T
(D) ~(p⇒q) ≡ p∧~q
(E) p∨q ≡ ~p∨~q
Choose the correct answer from the options given below :
