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# Boolean Algebra Practice Problems

Introduction to Boolean Algebra:
The theory of Boolean algebra  was given by English Mathematician George Boole in the year 1847. Boolean algebra is apart of mathematics, which is also called as abstract algebra. Boolean algebra are used to analyze the concept of logic circuits or combined circuits. Boolean algebra has  found various form of application are

Rules of Boolean Algebra:
• A + 0 = A
• A + 1 = 1
• A . 0 = 0
• A . 1  = A
• A + A = A
• A + $\overline{A}$ = 1
• A . A = A
• A . $\overline{A}$ = 0
• $\overline{A}$ = A
• A + AB = A
• A + $\overline{A}$ B = A + B
• (A + B) (A + C) = A + BC

Postulates:

1. Existence of an identity element 0 or 1 with respect to the operator '+'  or '.'
a) A + 0 =  A
b) A . 1  = A

2. Commutative property:
a) X + Y = Y + X
b) X . Y =  Y . X

3. Associative property:
a) X +(Y + Z) = (X + Y) + Z
b) X. (Y . Z) = (X . Y) . Z

4. Distributive property
a) X +(Y . Z) = (X + Y).(X + Z)
b) X. (Y + Z) = (X . Y) + (X . Z)

5.  Existence of complement
a) A + A' = 1
b) A . A'  = 0

Demorgan's Theorems:

a)  The complement of a product of variables is equal to the sum of the complements of the variables.

$\overline{AB}=\overline{A}+\overline{B}$

b)  The complement of a sum of variables is equal to the product of the complements of the variables.

$\overline{A+B}=\overline{A} \overline{B}$

Two Valued Boolean Algebra:
The set of two elements of the variable X(0.1), along with the operators AND  and OR forms a two – valued Boolean algebra.

AND Operation:

 X Y X.Y 0 0 0 0 1 0 1 0 0 1 1 1

OR Operation:

 X Y X+Y 0 0 0 0 1 1 1 0 1 1 1 1

NOT Operation:

 X X' 0 1 1 0

Boolean Algebra Practice Problems:

1.  A + $\overline{AB}$
we simplify the expression, take the common term
= A + ($\overline{A}+\overline{B}$)
= ( A + $\overline{A}$) + $\overline{B}$ commutative and Associative laws
= 1 + $\overline{B}$            Complement rule
= 1 Identity rule

2.  A + AB
= A ( 1 + B)
= A (1)
= A

3. XY + X$\overline{Y}$
= X( Y + $\overline{Y}$)
=  X (1)
= X

4.  XY + X'Z + YZ
= XY + X'Z + YZ(X + X')
= XY + X'Z + XYZ + X'YZ
= XY(1 + Z)+ X'Z(1 + Y)
= XY + X'Z

5. $\overline{A}$B + AB + A + A$\overline{B}$
= $\overline{A}$B + A(B + 1 + $\overline{B}$)
= $\overline{A}$B + A
= A + $\overline{A}$B
= (A+$\overline{A}$)(A+B)
= A+B