Which of the following is NOT logically equivalent to P→Q?
~Q→P
An implication P→Q is false only when P is true and Q is false. Two formulas are logically equivalent when they take the same truth value in every row of the truth table, so each option is reduced to a disjunction and compared with P→Q ≡ ~P ∨ Q.
Step 1 — ~P ∨ Q is the material-implication law itself, so it is equivalent to P→Q.
Step 2 — ~Q → ~P is the contrapositive; rewriting gives ~~Q ∨ ~P = Q ∨ ~P, the same formula.
Step 3 — By De Morgan, ~(P ∧ ~Q) = ~P ∨ Q, again the same formula.
Step 4 — ~Q→P rewrites to ~~Q ∨ P = Q ∨ P. With P false and Q false, P→Q is true while Q ∨ P is false, so this formula is not equivalent to P→Q.
~Q→P 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 :
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
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 :
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:
