Example:
(i) The dual of the statement (x . 1 )(0 + x') = 0 is given by
(x + 0)(1 . x') = 1(ii) The dual of the statement
(x + y)(y + z) = x.z + y isx.y + y.z = (x + z).y
Note:
The duality principle has many applications. If the dual of an algebraic expression is desired, we simply interchange + and . and replace 1 by 0 and 0 by 1.
Let B be a Boolean algebra. Let x, y, z
B.Theorem 1 (a) (Idempotent Law):

Proof:
x + x = (x + x) . 1 by Axiom 2(b)
(Refer to Axion in previous topic)
= (x + x) . (x + x') Axiom 5(a)
= x + x.x' Axiom 4(b)
= x + 0 Axiom 5(b)
= x Axiom 2(a)
Theorem 1(b) (Idempotent Law):
x.x = x
Proof:
x.x = x.x + 0 2(a) (Refer to Axion in previous topic)
= x.x + x.x' Axiom 5(b)= x. (x + x') Axiom 4(a)
= x.1 Axiom 5(a)= x Axiom 2(b)
Theorem 2(a) (Boundedness Law):
x + 1 = 1
Proof:
x + 1 = 1.(x + 1) 2(b) (Refer to Axion in previous topic)
= (x +x') . (x + 1) Axiom 5(a)= x + x'.1 Axiom 4(b)
= x + x' Axiom 2(b)= 1 Axiom 5(a)
Theorem 2(b) :
x.0 = 0
Proof :
x.0
= x.0 + 0 2(a) (Refer to Axion in previous topic)= x.0 + x.x' Axiom (5)
=x(0 + x') Axiom 4(a)= x.x' Axiom 2(a)
= 0 Axiom (5)Theorem 3 (Involution Law):
(x')' = x
Proof :
We know that x' is the complement of x.
If x + x' = 1 and x.x' = 0, thenx + x' = 1
x' + x =1 and x. x' = 0
x'.x = 0 (3a and 3b)
plement of x

x is the com'
(x')' = x
Theorem 4(a) (Absorption Laws):
x + x.y = x
Proof:
x + x.y = x.1 + x.y by 2(b) (Refer to Axion in previous topic)
= x.(1 + y) Axiom 4(a)= x.(y + 1) Axiom 3(a)
= x.1 Axiom 2(a)= x Axiom 2(b)
Theorem 4(b):
x.(x + y) = x
Proof:
x.(x+y)
= (x + 0)(x + y) 2(a) (Refer to Axion in previous topic)= x + (0.y) Axiom 4(a)
= x + (y.0) Axiom 3(b)= x + 0 (Theorem 2)
= x Axiom 2(a)Theorem 5(a):
If B is a Boolean algebra, for any x, y in B, then 0' = 1.
Proof:
x + 1 = 1 (Theorem 2a)
In particular for x = 0, we have0 + 1 = 1 …(1)
x.0 = 0 (Theorem 2b)In particular, for x =1
1.0 = 0
0.1 = 0 (3a) …(2)
For 0
B, 0 + 1 = 1
1 is the complement of 0.
0' = 1 (5)Theorem 5(b):
In a Boolean algebra B, we have
1' = 0Proof:
x + 1 = 1 (Theorem 2a)
In particular, for x = 00 + 1 =1
1 + 0 = 1 …(1) (Axiom 3a)
x . 0 = 0
In particular for 1
B, we have
1 . 0 = 0 …(2)
From (1) and (2), we have0 is the complement of 1. (Axiom 5)
