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Logic

Logical Equivalence

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: pqqpp \wedge q \equiv q \wedge p, same for ∨.
  • Double negation: ¬¬pp\neg\neg p \equiv p.
  • De Morgan: ¬(pq)¬p¬q\neg(p \wedge q) \equiv \neg p \vee \neg q.
  • Conditional as OR: pq¬pqp \to q \equiv \neg p \vee q.
  • Contrapositive: pq¬q¬pp \to q \equiv \neg q \to \neg p.
  • Biconditional: pq(pq)(qp)p \leftrightarrow q \equiv (p \to q) \wedge (q \to p).

Common NON-equivalences

  • pqp \to q is NOT equivalent to qpq \to p (converse).
  • pqp \to q is NOT equivalent to ¬p¬q\neg p \to \neg q (inverse).

Type 1 if the two statements are equivalent, 0 if not.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

pq ? qpp \wedge q \ \stackrel{?}{\equiv}\ q \wedge p

Problem 2

pq ? qpp \vee q \ \stackrel{?}{\equiv}\ q \vee p

Problem 3

pq ? qpp \to q \ \stackrel{?}{\equiv}\ q \to p

Problem 4

¬¬p ? p\neg\neg p \ \stackrel{?}{\equiv}\ p

Practice

Standard problems matching the lesson.

Problem 5

¬(pq) ? ¬p¬q\neg(p \wedge q) \ \stackrel{?}{\equiv}\ \neg p \vee \neg q

Problem 6

¬(pq) ? ¬p¬q\neg(p \vee q) \ \stackrel{?}{\equiv}\ \neg p \wedge \neg q

Problem 7

pq ? ¬pqp \to q \ \stackrel{?}{\equiv}\ \neg p \vee q

Problem 8

pq ? ¬q¬pp \to q \ \stackrel{?}{\equiv}\ \neg q \to \neg p

Problem 9

pq ? (pq)(qp)p \leftrightarrow q \ \stackrel{?}{\equiv}\ (p \to q) \wedge (q \to p)

Problem 10

pp ? pp \wedge p \ \stackrel{?}{\equiv}\ p

Problem 11

pp ? pp \vee p \ \stackrel{?}{\equiv}\ p

Problem 12

pF ? Fp \wedge F \ \stackrel{?}{\equiv}\ F

Problem 13

pT ? Tp \vee T \ \stackrel{?}{\equiv}\ T

Problem 14

pT ? pp \wedge T \ \stackrel{?}{\equiv}\ p

Problem 15

pF ? pp \vee F \ \stackrel{?}{\equiv}\ p

Problem 16

pq ? p¬qp \to q \ \stackrel{?}{\equiv}\ p \wedge \neg q

Problem 17

¬(pq) ? p¬q\neg(p \to q) \ \stackrel{?}{\equiv}\ p \wedge \neg q

Problem 18

p(qr) ? (pq)(pr)p \wedge (q \vee r) \ \stackrel{?}{\equiv}\ (p \wedge q) \vee (p \wedge r)

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 19

p(qr) ? (pq)(pr)p \vee (q \wedge r) \ \stackrel{?}{\equiv}\ (p \vee q) \wedge (p \vee r)

Problem 20

pq ? ¬p¬qp \to q \ \stackrel{?}{\equiv}\ \neg p \to \neg q

Problem 21

¬(pq) ? ¬p¬q\neg(p \wedge q) \ \stackrel{?}{\equiv}\ \neg p \wedge \neg q

Problem 22

¬(pq) ? ¬p¬q\neg(p \vee q) \ \stackrel{?}{\equiv}\ \neg p \vee \neg q

Problem 23

p(pq) ? pp \wedge (p \vee q) \ \stackrel{?}{\equiv}\ p

Problem 24

p(pq) ? pp \vee (p \wedge q) \ \stackrel{?}{\equiv}\ p

Problem 25

pq ? ¬p¬qp \leftrightarrow q \ \stackrel{?}{\equiv}\ \neg p \leftrightarrow \neg q

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