Boolean Algebra

The two-valued logic in algebra is called the Boolean Algebra. It was first given by an English mathematician by name George Boole. The arithmetic operations which are performed on the Boolean quantities have only two outcomes either true or false, 0 or 1. And hence the Boolean logic forms the basis for the computation in binary computer systems.

Any algorithm or computer circuit can be represented using a system of Boolean equations. In the Boolean system the two possible values are zero and one. The Boolean algebra symbols used in Boolean operations are,  “.” which represents the logical operation “AND”, it is also denoted by ‘^’.  The symbol ‘+’ is used to represent the logical operation ‘OR’ which is also denoted by ‘v’. Logical negation or complement or not is denoted by  (~)or  (‘), for instance NOT A is denoted as A’ or  ~A (read as ‘negation A’)
The Boolean algebra Properties are:
Commutative law: A+B=B+A; A.B=B.A
Associative Law: (A+B)+C=A+(B+C); A.(B.C)= (A.B).C
Distributive Law: A(B+C)=(A.B)+(A.C); A+(B.C) = (A+B)(A+C)
Idempotence property: A+A=A; A.A=A
Involution Law: (A’)’= A
Law of Complements (negation): A+A’=1; A.A’=0
Boolean algebra theorems are as given below:
Simplification Theorems:
A.B+A.B’=0
(A+B)(A+B’)=A
A+AB=A
A(A+B)=A
(A+B’)B=AB
AB’+B=A+B
De-Morgan’s Theorems:
(A+B+C…..)’= A’B’C’…
(ABC…..)’=A’+B’+C’….
[f(A1,A2, A3…An,0,1,+,.)]’=f(A1’,A2’,A3’….An’,0,1,+,.)
Duality Property:
(A+B+C+….)D= ABC….
(ABC…)D= A+B+C…..
[f(A1,A2, A3…An,0,1,+,.)]D = f(A1’,A2’,A3’….An’,0,1,+,.)
Multiplying out and factoring theorem:
(A+B)(A’+C)=AC+A’B
AB+A’C=(A+C)(A’+B)
Consensus Theorem:
AB+BC+A’C=AB+A’C
(A+B)(B+C)(A’+C)=(A+B)(A’+C)
Absorption: A(A+B)=A; A+AB=A
Prove that A+BC=(A+B)(A+C)
Proof: Here to prove the given Boolean equation we expand the left hand side expression, apply distributive law, absorption law and idempotence (AA=A=A+A)
(A+B)(A+C)=AA+BA+AC+BC (using distributive law)
=A+BA+AC+BC   (using idempotence property)
=A+BC           (using absorption)


A Boolean function can be represented using a Truth table. The truth table for the logical operation AND would be,
AND    0      1
0       0      0
1       0       1
The truth table for the logical operation OR would be,
OR   0      1
0     0      1
1    1       1

Is this topic formula for quadratic equation hard for you? Watch out for my coming posts.

Boolean algebra Problems
Simplify: [[A(A+B)]’+B’A’]’
Solution: [[A(A+B)]’+B’A’]’
=[A.(A+B)]’’. [B.A’]’
=[A.(A+B)].(B’+A)
=[(A+AB)(B’+A)]
=A(1+B)(B’+A)
=A(1)(B’+A)
=A(B’+A)
=AB’+ AA
=AB’+A
=A[B’+1]
=A(1)
=A

Simplify (A+C)[AD+AD’]+AC+C
Solution: Given,
(A+C)[AD+AD’]+AC+C
= (A+C)A(D+D’) + AC + C       [using distributive law]
=(A+C)A+AC+C            [using complement identity]
=A[(A+C)+C] + C            [using associative law]
=AA+AC+C            [using distributive law]
=A+(A+1)C            [using idempotent, identity and distributive laws]
= A +C                [using identity twice]

Simplify (AB)’(A’+B)(B’+B)
Solution: Given,
(AB)’(A’+B)(B’+B)
=(AB)’(A’+B)            [using complement law and identity law]
=(A’+B’)(A’+B)            [using De-Morgan’s law]
=A’A’+A’B+B’A’+B’B        [using distributive law]
=A’+B’B                [on simplification, OR distributes over AND]
=A’                [Complement and identity laws]