RULES OF BOOLEAN
ALGEBRA
BASIC RULES OF BOOLEAN
ALGERBA
Sr. No.
Theorem
1.
0’=1  1’=0
2.
A+0=A
3.
A+1=1
4.
A+A=A
5.
A+A’=1
6.
(A’)’=A
7.
A+AB=A
8.
A+A’B=A+B
9.
(A+B)’=(A)’.(B)’
10.
(A.B)’=(A)’+(B)’
11.
(A+B).(A+C)= A+BC
12.
A.1=A  A.0=0
13.
A.A=A  A.(A)’=0
DE MORGAN’S THEOREMS
It states, that the complement of any expression
can be obtained by replacing each variable and
element with its complement and changing OR
operators (+) with AND operators(.) and AND
operator (.) with OR operators (+).
 These theorems can be expressed as follows:

(A+B)’=(A)’.(B)’ First De Morgan’s Theorem.
 (A.B)’=(A)’+(B)’ Second De Morgan’s Theorem.
De Morgan’s theorems can also applicable to
expressions in which there are more than two
variables.

THREE STEPS OF SIMPLIFICATION
(DEMORGANIZATION)
1.
Complement each of the individual variables.
2.
Change all the OR’s (+) to AND (.) and AND’s(.)
to OR’s(+).
3.
Complement the whole Boolean function.
EXAMPLES 1
Demorganize the following expression:
AB’ + A’B’C +AC
Step No. 1:
Complement each of the individual variables
Means :
AB’ + A’B’C + AC  A’B’’ + A’’B’’C’ + A’C’

Step No. 2:
Change all the OR’s (+) to AND (.) and AND’s(.) to
OR’s(+)
EXAMPLES 1
Means:
A’B’’ + A’’B’’C’ + A’C’  A’B + ABC’ + A’C’
A’B + ABC’ + A’C’  (A’+B ).(A+B+C’).(A’+C’)
Step No. 3:
Complement the whole Boolean function
Means:
(A’+B ).(A+B+C’).(A’+C’)
((A’+B ).(A+B+C’).(A’+C’))’
TRUTH TABLE FOR THIS EXAMPLE
1. Before Demorganization.
AB’ + A’B’C +AC
A
B
C
AB’
A’B’C
AC R=AB’ + A’B’C + AC
0
0
0
0
0
0
0
0
0
1
0
1
0
1
0
1
0
0
0
0
0
0
1
1
0
0
0
0
1
0
0
1
0
0
1
1
0
1
1
0
1
1
1
1
0
0
0
0
0
1
1
1
0
0
1
1
TRUTH TABLE FOR THIS EXAMPLE
2. After Demorganization.
((A’+B ).(A+B+C’).(A’+C’))’
A
B
C
A’+B A+B+C’ A’+C’
R=(A’+B).(A+B+C’).(A’+C’)
R’
0
0
0
1
1
1
1
0
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
0
0
1
1
1
1
1
1
0
1
0
0
0
1
1
0
1
1
0
1
0
1
0
0
1
1
1
0
1
1
1
1
0
1
1
1
1
1
0
0
1
TRUTH TABLE FOR THIS EXAMPLE

Now Compare the results of the expression before
and after Demorganization
R=AB’ + A’B’C + AC
R=(A’+B).(A+B+C’).(A’+C’))’
0
0
1
1
0
0
0
0
1
1
1
1
0
0
1
1
Before Demorganization
After Demorganization
The Results are identical in both the tables.
EXAMPLES 1
Step No. 1
(A+B’).(A+B+C’).(AB)
(A’+B’’).(A’+B’+C’’).(A’B’)
Step No. 2
(A’+B’’).(A’+B’+C’’).(A’B’)
A’.B+A’.B’.C+A’+B’
Step No.3
A’.B+A’.B’.C+A’+B’
(A’.B+A’.B’.C+A’+B’)’
TRUTH TABLE FOR THIS EXAMPLE
1. Before Demorganization.
(A+B’).(A+B+C’).(AB)
A
B
C
A+B’
(A+B+C’)
(AB)
R=(A+B’).(A+B+C’).(AB)
0
0
0
1
1
0
0
0
0
1
1
0
0
0
0
1
0
0
1
0
0
0
1
1
0
1
0
0
1
0
0
1
1
0
0
1
0
1
1
1
0
0
1
1
0
1
1
1
1
1
1
1
1
1
1
1
TRUTH TABLE FOR THIS EXAMPLE
2. After Demorganization.
(A’.B+A’.B’.C+A’+B’)’
A
B
C
A’.B
A’.B’.C
A’+B’
R=A’.B+A’.B’.C+A’+B’
R’
0
0
0
0
0
1
1
0
0
0
1
0
1
1
1
0
0
1
0
1
0
1
1
0
0
1
1
1
0
1
1
0
1
0
0
0
0
1
1
0
1
0
1
0
0
1
1
0
1
1
0
0
0
0
0
1
1
1
1
0
0
0
0
1
TRUTH TABLE FOR THIS EXAMPLE

Now Compare the results of the expression before and
after Demorganization
R=(A+B’).(A+B+C’).(AB)
R=(A’.B+A’.B’.C+A’+B’)’
0
0
0
0
0
0
0
0
0
0
0
0
1
1
1
1
Before Demorganization
After Demorganization
The Results are identical in both the tables.