Principle of Duality

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The dual of any statement in a Boolean algebra is the statement obtained by interchanging + and ., and simultaneously inter-changing the elements 0 and 1 in the statement.

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 is

x.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, then

x + 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 have

0 + 1 = 1 …(1)

x.0 = 0 (Theorem 2b)

In particular, for x =1

1.0 = 0

0.1 = 0 (3a) …(2)

From (1) and (2), we have

For 0 B, 0 + 1 = 1

0.1 = 0

1 is the complement of 0.

0' = 1 (5)

Theorem 5(b):

In a Boolean algebra B, we have

1' = 0

Proof:

x + 1 = 1 (Theorem 2a)

In particular, for x = 0

0 + 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 have

0 is the complement of 1. (Axiom 5)