Boolean Simplification Examples

1

Example 1

Simplify the following Boolean Algebra:

(A.\overline{A} )+B

hint
  • Complement Rule
  • Identity Rule
Solution

(A.\overline{A} )+B

(0)+B#Complement

B #Identity

2

Example 2

(A.B)+(\overline{A}.B)

hint

Absorption

Solution

(A.B)+(\overline{A}.B) #Absorption

B

3

Example 3

(A+B).(A+C)

hint
Solution

4

Example 4

(\overline{A} + B).(A+B)

hint
Solution

5

Example 5

A\overline{C} + ABC

hint
Solution

6

Example 6

A.\overline{B}.D + A.\overline{B}.\overline{D}

hint
Solution

7

Example 7

B + B.C + A.C + A.B + A.C.B

hint
Solution

8

Example 8

\overline{A}.\overline{B}.\overline{C}.\overline{D} + \overline{A}.\overline{B}.\overline{C}.D + \overline{A}.\overline{B}.C.D+\overline{A}.\overline{B}.C.\overline{D} + \overline{A}.B.\overline{C}.\overline{D}

hint
Solution

9

Example 9

\overline{(\overline{A}.(B+C))}

hint
  • Use De Morgan’s Law
  • Use Involution Law(Double Inversion)
Solution

\overline{(\overline{A}.(B+C))} #Original Expression

(\overline{\overline{A}} + \overline{(B + C))} #Use De Morgan’s law on the ANDed components

(A + \overline{(B + C))}  Involution Law (Double negatives cancel)

(A + (\overline{B}.\overline{C})) De Morgan’s Law again

\bf{A + \overline{B}.\overline{C}}  # BODMAS rules allow removal of brackets

10

Example 10

hint
Solution

11

Example 11

hint
Solution