← Logic Lesson Two statements are logically equivalent if they have the same truth value in every row of their truth tables. Symbol: ≡ \equiv ≡ .
Famous equivalences
Commutative: p ∧ q ≡ q ∧ p p \wedge q \equiv q \wedge p p ∧ q ≡ q ∧ p , same for ∨. Double negation: ¬ ¬ p ≡ p \neg\neg p \equiv p ¬¬ p ≡ p . De Morgan: ¬ ( p ∧ q ) ≡ ¬ p ∨ ¬ q \neg(p \wedge q) \equiv \neg p \vee \neg q ¬ ( p ∧ q ) ≡ ¬ p ∨ ¬ q . Conditional as OR: p → q ≡ ¬ p ∨ q p \to q \equiv \neg p \vee q p → q ≡ ¬ p ∨ q . Contrapositive: p → q ≡ ¬ q → ¬ p p \to q \equiv \neg q \to \neg p p → q ≡ ¬ q → ¬ p . Biconditional: p ↔ q ≡ ( p → q ) ∧ ( q → p ) p \leftrightarrow q \equiv (p \to q) \wedge (q \to p) p ↔ q ≡ ( p → q ) ∧ ( q → p ) . Common NON-equivalences
p → q p \to q p → q is NOT equivalent to q → p q \to p q → p (converse).p → q p \to q p → q is NOT equivalent to ¬ p → ¬ q \neg p \to \neg q ¬ p → ¬ q (inverse).Type 1 if the two statements are equivalent, 0 if not.
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Warm-Up Quick problems to get going.
p ∧ q ≡ ? q ∧ p p \wedge q \ \stackrel{?}{\equiv}\ q \wedge p p ∧ q ≡ ? q ∧ p p ∨ q ≡ ? q ∨ p p \vee q \ \stackrel{?}{\equiv}\ q \vee p p ∨ q ≡ ? q ∨ p p → q ≡ ? q → p p \to q \ \stackrel{?}{\equiv}\ q \to p p → q ≡ ? q → p ¬ ¬ p ≡ ? p \neg\neg p \ \stackrel{?}{\equiv}\ p ¬¬ p ≡ ? p Practice Standard problems matching the lesson.
¬ ( p ∧ q ) ≡ ? ¬ p ∨ ¬ q \neg(p \wedge q) \ \stackrel{?}{\equiv}\ \neg p \vee \neg q ¬ ( p ∧ q ) ≡ ? ¬ p ∨ ¬ q ¬ ( p ∨ q ) ≡ ? ¬ p ∧ ¬ q \neg(p \vee q) \ \stackrel{?}{\equiv}\ \neg p \wedge \neg q ¬ ( p ∨ q ) ≡ ? ¬ p ∧ ¬ q p → q ≡ ? ¬ p ∨ q p \to q \ \stackrel{?}{\equiv}\ \neg p \vee q p → q ≡ ? ¬ p ∨ q p → q ≡ ? ¬ q → ¬ p p \to q \ \stackrel{?}{\equiv}\ \neg q \to \neg p p → q ≡ ? ¬ q → ¬ p p ↔ q ≡ ? ( p → q ) ∧ ( q → p ) p \leftrightarrow q \ \stackrel{?}{\equiv}\ (p \to q) \wedge (q \to p) p ↔ q ≡ ? ( p → q ) ∧ ( q → p ) p ∧ p ≡ ? p p \wedge p \ \stackrel{?}{\equiv}\ p p ∧ p ≡ ? p p ∨ p ≡ ? p p \vee p \ \stackrel{?}{\equiv}\ p p ∨ p ≡ ? p p ∧ F ≡ ? F p \wedge F \ \stackrel{?}{\equiv}\ F p ∧ F ≡ ? F p ∨ T ≡ ? T p \vee T \ \stackrel{?}{\equiv}\ T p ∨ T ≡ ? T p ∧ T ≡ ? p p \wedge T \ \stackrel{?}{\equiv}\ p p ∧ T ≡ ? p p ∨ F ≡ ? p p \vee F \ \stackrel{?}{\equiv}\ p p ∨ F ≡ ? p p → q ≡ ? p ∧ ¬ q p \to q \ \stackrel{?}{\equiv}\ p \wedge \neg q p → q ≡ ? p ∧ ¬ q ¬ ( p → q ) ≡ ? p ∧ ¬ q \neg(p \to q) \ \stackrel{?}{\equiv}\ p \wedge \neg q ¬ ( p → q ) ≡ ? p ∧ ¬ q p ∧ ( q ∨ r ) ≡ ? ( p ∧ q ) ∨ ( p ∧ r ) p \wedge (q \vee r) \ \stackrel{?}{\equiv}\ (p \wedge q) \vee (p \wedge r) p ∧ ( q ∨ r ) ≡ ? ( p ∧ q ) ∨ ( p ∧ r ) Challenge Harder problems — edge cases, trickier numbers, multiple steps.
p ∨ ( q ∧ r ) ≡ ? ( p ∨ q ) ∧ ( p ∨ r ) p \vee (q \wedge r) \ \stackrel{?}{\equiv}\ (p \vee q) \wedge (p \vee r) p ∨ ( q ∧ r ) ≡ ? ( p ∨ q ) ∧ ( p ∨ r ) p → q ≡ ? ¬ p → ¬ q p \to q \ \stackrel{?}{\equiv}\ \neg p \to \neg q p → q ≡ ? ¬ p → ¬ q ¬ ( p ∧ q ) ≡ ? ¬ p ∧ ¬ q \neg(p \wedge q) \ \stackrel{?}{\equiv}\ \neg p \wedge \neg q ¬ ( p ∧ q ) ≡ ? ¬ p ∧ ¬ q ¬ ( p ∨ q ) ≡ ? ¬ p ∨ ¬ q \neg(p \vee q) \ \stackrel{?}{\equiv}\ \neg p \vee \neg q ¬ ( p ∨ q ) ≡ ? ¬ p ∨ ¬ q p ∧ ( p ∨ q ) ≡ ? p p \wedge (p \vee q) \ \stackrel{?}{\equiv}\ p p ∧ ( p ∨ q ) ≡ ? p p ∨ ( p ∧ q ) ≡ ? p p \vee (p \wedge q) \ \stackrel{?}{\equiv}\ p p ∨ ( p ∧ q ) ≡ ? p p ↔ q ≡ ? ¬ p ↔ ¬ q p \leftrightarrow q \ \stackrel{?}{\equiv}\ \neg p \leftrightarrow \neg q p ↔ q ≡ ? ¬ p ↔ ¬ q Ask the tutor Stuck on a concept? Want another example? Ask anything about this topic.
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