Boolean Algebra
Associated, Distributive, and Commutative Law are described below individually :
 Associated Law: There is no effect of ORing or ANDing operation on the method of variable grouping.
(i)(A+B) + C= A+(B+C)
(ii) (A.B).C = A.(B.C)
 Distributive Law:
 (B+C) = A.B+A.C
 Commutative Law: The input signals are transposed without changing the outputs.
(i) A+B=B+A
(ii) A.B=B.A
Proof:
(i) A+B=B+A
If A+B=B+A
If A=1, B=1 then LSH=1+1=1=B+A=RHS
If B=0, B=0 then LHS=0+0=B+A=RHS
If A=0, B=1 then LHS=0+1=1=1+0=B+A=RHS
(ii) A.B=B.A
If A=1, B=1 then LHS=1.1=1=B.A=RHS
If B=0, B=0 then LHS=0.0=0=B.A=RHS
If A=0, B=1 then LHS=0.1=1=1.0=B.A=RHS
De Morgan’s Theorem
De Morgan was a great logician and Mathematician, as well as a friend of Charles Boole. The theorems given by De Morgan are associated with Boolean algebra.
 First Theorem: The complement of a sum equals to the product of the complements.
(A+B)’ =A’.B’
Proof:
LHS= (A+B)’ = (0+0)’ = 0’=1
RHS=A’.B’=0’.0’=1.1=1
 Second Theorem: The complement of a product equals the sum of the complements.
Proof:
LHS = (A.B)’ = (0.0)’ = 0’ = 1
RHS = A’ + B’ = 0’ + 0’ =1’ +1’ =1
Basic Boolean Theorem
George Boole, mathematician and logician, is one of the founders of computer science which concept of Boolean algebra is the foundation stone of modern computer science. The data read write process is done in computer is based on on/ off or 1/0 is nothing but Boolean logic.
These theorems are used to solve problems given in Boolean algebra. Let a, b, c be any elements in a Boolean algebra B
Some rules with proof:
(a) The Principle of Duality: In Boolean algebra, there is a precise duality between the operators. (AND) and + (OR) and digits 0 and 1. For instance, + is interchanged with, and vice versa. In the same way 0 is interchanged with 1 and vice versa. The important property is known as principle of duality.
Column1 
Column2 
Column3 

Row 1 
1+1=1 
1+0=1 
0+0=0 
Row 2 
0.0=0 
0.1=0 
1.1=1 
(b) Law of Idempotence:
(i) A+A=A
LHS=A+A=(A+A).1
=(A+A)(A+A’)
=A+A.A’
=A+0
=A
=RHS
(ii) A.A=A
LHS=A.A+0=A.A+A.A’
=A(A+A’)
=A.A
=A
=RHS
(c) Law of Boundedless:
(i) A +1=1
LHS=A+1
=(A+1).1=(A+A’0=A+1.A’=A+A’.1=A+A’=1=RHS
(ii)A,0=1
LHS=A.0=A.0+0=A.AA’+A.A’=AA’(1+A)=AA’.1=0=RHS
(d) Absorption Law:
(i) A+AB=A
LHS=A+AB=A.1+A.B=A(1+B)+A(B+1)=A.1=A=RHS
(ii) A.(A+B)=A
LHS=A.)A+B)=A.A+A.B=A+A.B=A+A.B=A(1+B)=A.1=A=RHS
(e) Involution Law:
A” =A
When A=0, A’=1, A”=1’=0=A
When A=1, A’ =0, A”=0’=1=A
Thus A”=A
(f) Complementarity’s Law:
(i) A+A’=1
LHS=A+A’
If A=0, A’=1 then LHS=0+1=1= RHS
If A=1, A’=0 then LHS=1+0=1=RHS
(ii)A.A’=0
LHS=A.A’
If A=0, A’=1 then LHS=0.1=0=RHS
If A=1, A’=0 then LHS=1.0=0=RHS
Example:
(a) A+1=1
If A=1 then LHS=A+1=0+1=1=RHS
If A=0 then LHS=0+1=1=RHS
(b) A+A=A
If A=0 then LHS=0+0=0=RHS
If A=1 then LHS=1+1=1=RHS
(c) Prove A+(A*B)=A
Proof:
A+(A.B)=A.1 +A.B=A(1+B)=A.1=A
(d) Prove that A+ (A’ B) = A+B
Proof:
A+(A’ B) = (A +A’).(A+B)
=1.(A+B)
=1.A+1.B
=A.1+B.1
=(A+B)
(e) Prove that (A+B)’ = A’B’
Proof:
LHS=(A+B)’
A 
B 
A+B 
(A+B)’ 
0 
0 
0 
1 
0 
1 
1 
0 
1 
0 
1 
0 
1 
1 
1 
0 
RHS=A’B’
A’ 
B’ 
A’ B’ 
1 
1 
1 
1 
0 
0 
0 
1 
0 
0 
0 
0 
L.H.S=R.H.S
(f) AC+ABC =AC
LHS =AC(1+B)= AC.1 =AC=RHS (Rule:b+1=1)
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