Let B = {0, 1 }. Define the operations +, and ' B follows: " 3
0 + 0 = 0 0 + 1 = 1 1 + 0 =1 1 + 1 = 1
0 . 0 = 0 0 . 1 = 0 1 . 0 = 0 1 . 1 = 1
0' = 1 1' = 0
[Think of the operators + and . as the or (v) and the and (A) operators,
respectively...
Clearly, x + 0 = x and x . 1 = x, for every x e B; therefore, 0 is the zero
element and 1 is the unit element. Besides,
0+0'=0+1=1 and 1+1'=1+0=1
0.0'=0.1=0 and 1.1 '=1.0=0
Consequently, the complement laws are also satisfied. Thus (B, +,., ', 0, 1)
is a boolean algebra.
Some fundamental facts about boolean algebras come from the above
axioms. We will find them useful in later discussions.
THEOREM : (Un i q u e I d e n t i t i e s ) The zero element and the uni t element of a boolean
algebra B are unique.
PROOF :
The zero element in B will be proved unique; the other half will be left as a routine exercise.
Suppose there are two zero elements, 01 and 02, in B. Since 02 is a zero
element, 01 + 02 : 01. Likewise, since 01 is a zero element, 02 § 01 ---- 02.
But 02 + 01 = 01 + 02, by the commutative law. Therefore, 01 = 01 + 02 -02 § 01 ---- 02,
SO the zero element is unique.
Theorem
![Picture](/uploads/9/3/7/5/9375932/7841293.jpg?488)
( Unique Complemen t )
The complement of every
element in a boolean algebra is unique.
PROOF:
Let x be an arbitrary element in a boolean algebra. Then by the complement laws, x + x' = 1 and xx' = 0. If x has a second complement y, x + y = 1 and
xy = O.
To show that y = x' (the reason for each step is given on the RHS):
y = y1 identity law
= y(x + x') complement law
= yx + yx' distributive law
= xy + x'y commutative law
= 0 + x'y complement law
= xx' + x'y complement law
= x'x + x'y commutative law
= x' (x + y) distributive law
= x'l1 complement law
= x' identity law
Thus the complement of every element in a boolean algebra is unique.