Boolean Algebra


   
 
Principle of Duality
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)
 
x is the complement of x'
 
(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)
 
  
     
   
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