Axiom 1:
(a) Closure with respect to +, that is if x, y
B, x + y
B.
B
x . y
B.Axiom 2:
(a) There exists element 0
B such that: x + 0 = 0 + x = x, x
B.
(b) There exists element 1
B such that: x . 1 = 1 . x = x.
For all x, y, z
B.
Axiom 3:
(a) Commutative w.r.t + : x + y = y + x
(b) Commutative w.r.t . : x . y = y . xAxiom 4:
(a) . is distributive over + : x . (y+z) = x . y + x . z
(b) + is distributive over . : x + (y . z) = (x+y) . (x+z)Axiom 5:
" x
B, $ x'
B such that (a) x + x' = 1 and (b) x.x' = 0.
Axiom 6:
There exists at least two elements x and y such that x
y.
Note:
To prove a set to be a Boolean algebra, we have to prove all the above six properties to be true.
Whenever we say B is a Boolean algebra, it should be understood that B is accompanied with two operations satisfying all the above six properties.Two-valued Boolean Algebra
A two valued Boolean algebra is defined on a set of two elements B = {0, 1}, with rules for + and . as shown in the following tables.

These rules are exactly the same as the AND, OR and NOT operations.
Note:
- Closure is obvious from the tables, since the results of each operation is either 1 or 0.
- Also observe from the tables:
(a) 0 + 0 = 0, 0 + 1 = 1 + 0 = 1
0 is the identity w.r.t +.
1 is the identity w.r.t
.
- For 0, 1
B, 0 + 1 = 1 + 0
Commutative law is satisfied for both + and .
- (a)

(b) Similarly it can be shown that
x + (y.z) = (x + y) . (x + z)- Also from the complement table

x . x' = 0
(a) x + x' = 1. since 0 + 0' = 0 + 1 = 1, 1 + 1' = 1 + 0 = 1(b) x.x' = 0, since 0.0' = 0.1 = 0, 1.1' = 1.0 = 0
Thus, we have a two valued Boolean algebra having a set of two elements 1 and 0.
