P | ¬P |
---|---|
T | F |
F | T |
P | Q | P ∧ Q |
---|---|---|
T | T | T |
T | F | F |
F | T | F |
F | F | F |
P | Q | P ∨ Q |
---|---|---|
T | T | T |
T | F | T |
F | T | T |
F | F | F |
P | Q | P ⊕ Q |
---|---|---|
T | T | F |
T | F | T |
F | T | T |
F | F | F |
P | Q | P → Q |
---|---|---|
T | T | T |
T | F | F |
F | T | T |
F | F | T |
P | Q | P ↔ Q |
---|---|---|
T | T | T |
T | F | F |
F | T | F |
F | F | T |
Example: Prove that P ∨ (Q ∧ R) and (P ∨ Q) ∧ (P ∨ R) are logically equivalent.
P | Q | R | Q ∧ R | P ∨ (Q ∧ R) | P ∨ Q | P ∨ R | (P ∨ Q) ∧ (P ∨ R) |
---|---|---|---|---|---|---|---|
T | T | T | T | T | T | T | T |
T | T | F | F | T | T | T | T |
T | F | T | F | T | T | T | T |
T | F | F | F | T | T | T | T |
F | T | T | T | T | T | T | T |
F | T | F | F | F | T | F | F |
F | F | T | F | F | F | T | F |
F | F | F | F | F | F | F | F |
Equivalence | Name |
---|---|
P ↔ P | Identity laws |
P ∨ F ↔ P | Domination laws |
P ∨ T ↔ T | Domination laws |
P ∧ F ↔ F | Domination laws |
P ∧ T ↔ P | Identity laws |
P ∨ P ↔ P | Idempotent laws |
P ∧ P ↔ P | Idempotent laws |
¬(¬P) ↔ P | Double negation law |
P ∨ Q ↔ Q ∨ P | Commutative laws |
(P ∨ Q) ∨ R ↔ P ∨ (Q ∨ R) | Associative laws |
P ∧ (Q ∨ R) ↔ (P ∧ Q) ∨ (P ∧ R) | Distributive laws |
¬(P ∧ Q) ↔ ¬P ∨ ¬Q | De Morgan's laws |
P ∧ Q ↔ P ∧ (Q ∨ R) | Absorption laws |
P ∨ Q ↔ P ∨ (P ∧ Q) | Absorption laws |
Statement | When True | When False |
---|---|---|
∀x P(x) | P(x) is true for every x | There is an x for which P(x) is false. |
∃x P(x) | There is an x for which P(x) is true. | P(x) is false for every x. |
Negation | Equivalent Statement |
---|---|
¬∀x P(x) | ∃x ¬P(x) |
¬∃x P(x) | ∀x ¬P(x) |
Rule of Inference | Name |
---|---|
P, P → Q | Modus ponens |
P, ¬Q | ¬(P → Q) |
P ∧ Q, P | Simplification |
P ∨ Q, ¬P | Q |
P & Q | Conjunction |
(P ∨ Q) ∧ ¬R | Q |
Rule of Inference | Name |
---|---|
∀x P(x) | Universal Instantiation |
P(c) for arbitrary c | Universal Generalization |
∃x P(x) | Existential Instantiation |
P(c) for some element c | Existential Generalization |
If we assume √2 is rational, it can be expressed as a fraction p/q (where p and q are integers and q ≠ 0). Squaring both sides results in a contradiction…
Assume that n is even, then n can be expressed as 2k. Therefore, 3(2k) + 2 = 6k + 2, which is even. This contradicts the odd result.