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 :
(A), (B) and (D) Only
Two compound propositions are logically equivalent (≡) when they have the same truth value for every assignment of truth values to p and q. Each statement can be checked with a known law or a truth table.
(A) is true: this is De Morgan's law, \(\neg(p \wedge q) \equiv \neg p \vee \neg q\).
(B) is true: the second De Morgan law, \(\neg(p \vee q) \equiv \neg p \wedge \neg q\).
(C) is false: \(p \wedge \neg p\) is a contradiction, always false, so it is equivalent to F, not T.
(D) is true: \(p \Rightarrow q \equiv \neg p \vee q\), so its negation is \(p \wedge \neg q\).
(E) is false: with p and q both true, p∨q is true but ~p∨~q is false.
So (A), (B) and (D) hold; the choices containing (C) or (E) are wrong.
(A), (B) and (D) Only 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?
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 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?
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 Choose the correct answer from the options given below:
