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 ?
(a) only
A tautology is true for every assignment of truth values.
(a) \(p \vee \sim (p \wedge q) \equiv p \vee \sim p \vee \sim q \equiv T\), so (a) is a tautology.
(b) Since \(p \wedge \sim q\) already implies \(\sim (p \wedge q)\), the formula reduces to \(\sim (p \wedge q)\), which is false when p and q are both true.
(c) \(p \wedge (q \vee r)\) is false whenever p is false.
Check: only (a) stays true in all rows of the truth table, so every choice containing (b) or (c) fails.
Hence, (a) only.
Similar Questions
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 :
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
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 :
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:
