Boolean Algebra: 20 Solved Exercises
Welcome to our new article, dear followers! In this piece, we will put into practice the concepts we explored in our previous articles on Boolean algebra. You can review those foundational lessons here: Boolean Algebra and Simplification of Logical Algebraic Expressions , Digital Electronics: Laws of Boolean Algebra Here, we will work through practical exercises focused on simplifying Boolean algebraic expressions by applying the fundamental rules and theorems of Boolean algebra.
📚 Boolean Algebra Laws Quick Reference
Identity:
A + 0 = A
A · 1 = A
A + 0 = A
A · 1 = A
Domination:
A + 1 = 1
A · 0 = 0
A + 1 = 1
A · 0 = 0
Complement:
A + Ā = 1
A · Ā = 0
A + Ā = 1
A · Ā = 0
Idempotent:
A + A = A
A · A = A
A + A = A
A · A = A
Absorption:
A + AB = A
A(A+B) = A
A + AB = A
A(A+B) = A
Distributive:
A(B+C) = AB+AC
A+BC = (A+B)(A+C)
A(B+C) = AB+AC
A+BC = (A+B)(A+C)
De Morgan:
ĀB = Ā+B̄
Ā+B = Ā·B̄
ĀB = Ā+B̄
Ā+B = Ā·B̄
Consensus:
AB+ĀC+BC = AB+ĀC
(A+B)(Ā+C)(B+C) = (A+B)(Ā+C)
AB+ĀC+BC = AB+ĀC
(A+B)(Ā+C)(B+C) = (A+B)(Ā+C)
1
Simplify: \( F = A + A \cdot B \)
Law: Absorption
Solution: \( A + AB = A \)
Result: \( F = A \)
2
Simplify: \( F = A \cdot (A + B) \)
Law: Absorption
Solution: \( A(A+B) = A \)
Result: \( F = A \)
3
Simplify: \( F = (A + B) \cdot (A + \overline{B}) \)
Law: Distributive + Complement
Solution: \( A + B\overline{B} = A + 0 = A \)
Result: \( F = A \)
4
Simplify: \( F = A \cdot B + A \cdot \overline{B} \)
Law: Factoring + Complement
Solution: \( A(B+\overline{B}) = A \cdot 1 = A \)
Result: \( F = A \)
5
Simplify: \( F = A + \overline{A} \cdot B \)
Laws: Redundancy/Absorption
Solution: \( A + \overline{A}B = A + B \)
Result: \( F = A + B \)
6
Simplify: \( S_1 = \overline{A} \cdot B + A \cdot \overline{B} + A \cdot B \)
Laws: Distributive + Complement + Absorption
Solution: \( \overline{A}B + A(\overline{B}+B) =
\overline{A}B + A = A + B \)
Result: \( S_1 = A + B \)
7
Simplify: \( S_2 = B + A \cdot B + B \cdot C + C \)
Laws: Domination (twice)
Solution: \( B + AB = B \), \( B + BC = B \), so \( B + C
\)
Result: \( S_2 = B + C \)
8
Simplify: \( ABC + \overline{A} + \overline{C} \)
Laws: Absorption (twice)
Solution: \( \overline{A} + ABC = \overline{A} + BC \),
\( BC + \overline{C} = B + \overline{C} \)
Result: \( \overline{A} + B + \overline{C} \)
9
Simplify: \( \overline{A}B + C\overline{A}D + \overline{B} + \overline{D} \)
Laws: Factoring + Generalized Absorption
Solution: \( \overline{A}(B+CD) + \overline{B} +
\overline{D} = \overline{A} + \overline{B} + \overline{D} \)
Result: \( \overline{A} + \overline{B} + \overline{D} \)
10
Simplify: \( \overline{A \cdot B} + \overline{A} \cdot \overline{B} \)
Laws: De Morgan + Absorption
Solution: \( \overline{A} + \overline{B} +
\overline{A}\overline{B} = \overline{A} + \overline{B} \)
Result: \( \overline{A} + \overline{B} \)
11
Simplify: \( \overline{\overline{A+D} \cdot \overline{\overline{C}+\overline{B}} + C} \)
Laws: De Morgan (twice) + Domination
Solution: \( \overline{\overline{A}\overline{D}CB + C} =
\overline{C} \)
Result: \( \overline{C} \)
12
Simplify: \( \overline{A + B + C} \)
Laws: De Morgan (extended)
Solution: \( \overline{A} \cdot \overline{B} \cdot
\overline{C} \)
Result: \( \overline{A} \cdot \overline{B} \cdot
\overline{C} \)
13
Simplify: \( A \cdot B + \overline{A} \cdot C + B \cdot C \)
Theorem: Consensus
Solution: \( AB + \overline{A}C + BC = AB + \overline{A}C
\)
Result: \( A \cdot B + \overline{A} \cdot C \)
14
Simplify: \( (A + B) \cdot (\overline{A} + C) \cdot (B + C) \)
Theorem: Consensus (AND form)
Solution: \( (A+B)(\overline{A}+C)(B+C) =
(A+B)(\overline{A}+C) \)
Result: \( (A + B) \cdot (\overline{A} + C) \)
15
Simplify: \( S_3 = (\overline{B} + \overline{A} \cdot B) \cdot C + A \cdot B \cdot \overline{C} \)
Laws: Distributive + Factoring + Expansion
Solution: \( \overline{B}C + \overline{A}BC +
AB\overline{C} = BC + AC + AB\overline{C} \)
Result: \( B \cdot C + A \cdot C + A \cdot B \cdot
\overline{C} \)
16
Simplify: \( S_4 = (\overline{A} + C) \cdot (\overline{A} + B) \cdot (\overline{A} + \overline{C}) \)
Laws: Distributive + Complement + Absorption
Solution: \( (\overline{A}+C)(\overline{A}+\overline{C})
= \overline{A} \), then \( \overline{A}(\overline{A}+B) = \overline{A} \)
Result: \( \overline{A} \)
17
Simplify: \( A\overline{C} + AB\overline{C} + B\overline{C} + \overline{A}B \)
Laws: Absorption + Factoring
Solution: \( A\overline{C} + B\overline{C} +
\overline{A}B = A\overline{C} + \overline{A}B \)
Result: \( A\overline{C} + \overline{A}B \)
18
XOR Definition: \( A \oplus B \)
Definition: Exclusive OR (A or B but not both)
Equivalent forms:
1. \( A \cdot \overline{B} + \overline{A} \cdot B \)
2. \( (A + B) \cdot (\overline{A} + \overline{B}) \)
3. \( (A + B) \cdot \overline{A \cdot B} \)
XOR Expressions: Multiple equivalent forms
19
Simplify: \( (A+\overline{AB}+C\overline{AB})(A+B\overline{A}+\overline{B}) \)
Laws: De Morgan + Complement + Domination
Solution: First term: \( A+\overline{A}+\overline{B} = 1
\), Second term: contains B and \(\overline{B}\) = 1
Result: \( 1 \) (always true)
20
Simplify: \( (A+B)(\overline{A}+C)(B+C) \) (alternative form)
Theorem: Consensus + Expansion
Solution: \( (A+B)(\overline{A}+C) = A\overline{A} + AC +
B\overline{A} + BC = AC + B\overline{A} + BC \)
Result: \( AC + B\overline{A} + BC \) or \(
(A+B)(\overline{A}+C) \)
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