Postulates and Theorems of Boolean Algebra
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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
