Postulates and Theorems of Boolean Algebra

Theorem 5, DeMorgan

(a) (x + y)' = x'y' (b) (xy)' = x' + y'

Lemma #5

(a) A+A'B = A+B (b) A(A'+B) = AB

Lemma #3

(a) A+AB=A (b) A(A+B)=A

Theorem 4, associative

(a) x + (y + z) = (x + y) + z (b) x(yz) = (xy)z

Postulate 2

(a) x + 0 = x (b) x • 1 = x

Theorem 2

(a) x + 1 = 1 (b) x • 0 = 0

Theorem 1

(a) x + x = x (b) x • x = x

Postulate 5

(a) x + x' = 1 (b) x • x' = 0

Theorem 6, absorption

(a) x + xy = x (b) x(x + y) = x

Postulate 3, commutative

(a) x + y = y + x (b) xy =yx

Postulate 4, distributive

(a) x(y + z) = xy + xz (b) x + yz = (x + y)(x + z)

Theorem 3, involution

(x')' = x