DE Notes

Boolean Laws and Theorems – Complete Reference with Proofs

Comprehensive guide to all Boolean algebra laws and theorems including identity, null, idempotent, inverse, commutative, associative, distributive, absorption, and consensus theorems with proofs.

Boolean laws and theorems are the rules used to simplify Boolean expressions and optimize digital circuits. Mastering these laws allows you to reduce complex logic expressions to their minimal forms, saving gates and reducing hardware costs.

Basic Laws (Single Variable)

Identity Law

Example

A + 0 = A A · 1 = A

OR with 0 doesn't change the value. AND with 1 doesn't change the value.

Null (Domination) Law

Example

A + 1 = 1 A · 0 = 0

OR with 1 always gives 1. AND with 0 always gives 0.

Idempotent Law

Example

A + A = A A · A = A

A variable ORed or ANDed with itself gives itself.

Complement Law

Example

A + A' = 1 A · A' = 0

A variable ORed with its complement is 1. ANDed is 0.

Involution (Double Complement) Law

Example

(A')' = A

Complementing twice returns the original value.

Absorption Theorems

First Absorption Law

Example

A + AB = A

Proof: A + AB = A(1 + B) = A·1 = A

Second Absorption Law

Example

A(A + B) = A

Proof: A(A + B) = AA + AB = A + AB = A

Third Absorption Law (Redundancy)

Example

A + A'B = A + B

Proof: A + A'B = (A + A')(A + B) = 1·(A + B) = A + B

Fourth Absorption Law

Example

A(A' + B) = AB

Proof: A(A' + B) = AA' + AB = 0 + AB = AB

Consensus Theorem

Example

AB + A'C + BC = AB + A'C

The term BC is redundant (called the "consensus term").

Proof:

Example

AB + A'C + BC = AB + A'C + BC(A + A') [since A + A' = 1] = AB + A'C + ABC + A'BC = AB(1 + C) + A'C(1 + B) = AB + A'C

Dual form:

Example

(A + B)(A' + C)(B + C) = (A + B)(A' + C)

De Morgan's Theorem (Preview)

Example

(A + B)' = A' · B' (Break the bar, change the sign) (A · B)' = A' + B'

This is covered in detail in the dedicated De Morgan's Theorem page.

Complete Laws Summary Table

Law	OR Form	AND Form
Identity	A + 0 = A	A · 1 = A
Null	A + 1 = 1	A · 0 = 0
Idempotent	A + A = A	A · A = A
Complement	A + A' = 1	A · A' = 0
Commutative	A + B = B + A	A · B = B · A
Associative	(A+B)+C = A+(B+C)	(AB)C = A(BC)
Distributive	A+BC = (A+B)(A+C)	A(B+C) = AB+AC
Absorption 1	A + AB = A	A(A+B) = A
Absorption 2	A + A'B = A+B	A(A'+B) = AB
De Morgan's	(A+B)' = A'B'	(AB)' = A'+B'
Consensus	AB+A'C+BC = AB+A'C

Simplification Examples

Example 1: Simplify F = ABC + ABC' + AB'C

Example

F = ABC + ABC' + AB'C = AB(C + C') + AB'C [Factor AB] = AB(1) + AB'C [Complement law] = AB + AB'C = A(B + B'C) [Factor A] = A(B + C) [Absorption: B + B'C = B + C]

Example 2: Simplify F = A'B'C' + A'B'C + A'BC + AB'C

Example

F = A'B'C' + A'B'C + A'BC + AB'C = A'B'(C' + C) + C(A'B + AB') = A'B' + C(A ⊕ B) [XOR pattern]

Or alternatively:

Example

F = A'B'(C' + C) + A'BC + AB'C = A'B' + C(A'B + AB') = A'B' + C(A ⊕ B)

Example 3: Simplify F = (A + B)(A + B')(A' + B)

Example

F = (A + B)(A + B')(A' + B) = [A + BB'](A' + B) [Distributive] = [A + 0](A' + B) = A(A' + B) = AA' + AB = 0 + AB = AB

Tips for Simplification

Look for complements: Terms differing by one variable can combine
Apply absorption early: Reduces terms quickly
Factor common variables: Pull out shared literals
Use consensus: Remove redundant terms
Check with truth table: Verify your simplification is correct

Interview Questions

Q1: State and prove the Consensus Theorem.

AB + A'C + BC = AB + A'C. Proof: BC = BC·1 = BC(A+A') = ABC + A'BC. Then AB + A'C + ABC + A'BC = AB(1+C) + A'C(1+B) = AB + A'C. The consensus term BC is redundant because it's covered by the other two terms.

Q2: What is the difference between the Associative and Distributive laws?

Associative law deals with grouping of the same operation: (A+B)+C = A+(B+C). Distributive law deals with two different operations: A(B+C) = AB+AC. Both are needed for expression manipulation.

Q3: Simplify X + X'Y using Boolean laws.

X + X'Y = (X + X')(X + Y) = 1·(X + Y) = X + Y. This uses the distributive law A + BC = (A+B)(A+C) with A=X, B=X', C=Y.

Q4: Why is the law A + AB = A called "absorption"?

Because the variable A "absorbs" the term AB — the larger term AB contributes nothing extra since whenever AB=1, A must already be 1. The term is absorbed into the simpler expression.

Q5: Can Boolean laws be applied to expressions with more than two variables?

Absolutely. The laws generalize to any number of variables. For example, associativity means A+B+C+D can be grouped any way. Consensus works on multi-variable terms too. The key is identifying which law applies to which part of the expression.

Exam Focus

Revise definitions, diagrams, examples, and short-answer points for Boolean Laws and Theorems – Complete Reference with Proofs.

Interview Use

Prepare one clear explanation, one practical example, and one common mistake for this Digital Electronics topic.

Search Terms

digital-electronics, digital electronics, digital, electronics, boolean, algebra, laws, and

Continue learning this concept

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DE Notes

Boolean Laws and Theorems – Complete Reference with Proofs

Comprehensive guide to all Boolean algebra laws and theorems including identity, null, idempotent, inverse, commutative, associative, distributive, absorption, and consensus theorems with proofs.

Basic Laws (Single Variable)

Identity Law

Example

A + 0 = A A · 1 = A

OR with 0 doesn't change the value. AND with 1 doesn't change the value.

Null (Domination) Law

Example

A + 1 = 1 A · 0 = 0

OR with 1 always gives 1. AND with 0 always gives 0.

Idempotent Law

Example

A + A = A A · A = A

A variable ORed or ANDed with itself gives itself.

Complement Law

Example

A + A' = 1 A · A' = 0

A variable ORed with its complement is 1. ANDed is 0.

Involution (Double Complement) Law

Example

(A')' = A

Complementing twice returns the original value.

Absorption Theorems

First Absorption Law

Example

A + AB = A

Proof: A + AB = A(1 + B) = A·1 = A

Second Absorption Law

Example

A(A + B) = A

Proof: A(A + B) = AA + AB = A + AB = A

Third Absorption Law (Redundancy)

Example

A + A'B = A + B

Proof: A + A'B = (A + A')(A + B) = 1·(A + B) = A + B

Fourth Absorption Law

Example

A(A' + B) = AB

Proof: A(A' + B) = AA' + AB = 0 + AB = AB

Consensus Theorem

Example

AB + A'C + BC = AB + A'C

The term BC is redundant (called the "consensus term").

Proof:

Example

AB + A'C + BC = AB + A'C + BC(A + A') [since A + A' = 1] = AB + A'C + ABC + A'BC = AB(1 + C) + A'C(1 + B) = AB + A'C

Dual form:

Example

(A + B)(A' + C)(B + C) = (A + B)(A' + C)

De Morgan's Theorem (Preview)

Example

(A + B)' = A' · B' (Break the bar, change the sign) (A · B)' = A' + B'

This is covered in detail in the dedicated De Morgan's Theorem page.

Complete Laws Summary Table

Law	OR Form	AND Form
Identity	A + 0 = A	A · 1 = A
Null	A + 1 = 1	A · 0 = 0
Idempotent	A + A = A	A · A = A
Complement	A + A' = 1	A · A' = 0
Commutative	A + B = B + A	A · B = B · A
Associative	(A+B)+C = A+(B+C)	(AB)C = A(BC)
Distributive	A+BC = (A+B)(A+C)	A(B+C) = AB+AC
Absorption 1	A + AB = A	A(A+B) = A
Absorption 2	A + A'B = A+B	A(A'+B) = AB
De Morgan's	(A+B)' = A'B'	(AB)' = A'+B'
Consensus	AB+A'C+BC = AB+A'C

Simplification Examples

Example 1: Simplify F = ABC + ABC' + AB'C

Example

F = ABC + ABC' + AB'C = AB(C + C') + AB'C [Factor AB] = AB(1) + AB'C [Complement law] = AB + AB'C = A(B + B'C) [Factor A] = A(B + C) [Absorption: B + B'C = B + C]

Example 2: Simplify F = A'B'C' + A'B'C + A'BC + AB'C

Example

F = A'B'C' + A'B'C + A'BC + AB'C = A'B'(C' + C) + C(A'B + AB') = A'B' + C(A ⊕ B) [XOR pattern]

Or alternatively:

Example

F = A'B'(C' + C) + A'BC + AB'C = A'B' + C(A'B + AB') = A'B' + C(A ⊕ B)

Example 3: Simplify F = (A + B)(A + B')(A' + B)

Example

F = (A + B)(A + B')(A' + B) = [A + BB'](A' + B) [Distributive] = [A + 0](A' + B) = A(A' + B) = AA' + AB = 0 + AB = AB

Tips for Simplification

Look for complements: Terms differing by one variable can combine
Apply absorption early: Reduces terms quickly
Factor common variables: Pull out shared literals
Use consensus: Remove redundant terms
Check with truth table: Verify your simplification is correct

Interview Questions

Q1: State and prove the Consensus Theorem.

Q2: What is the difference between the Associative and Distributive laws?

Associative law deals with grouping of the same operation: (A+B)+C = A+(B+C). Distributive law deals with two different operations: A(B+C) = AB+AC. Both are needed for expression manipulation.

Q3: Simplify X + X'Y using Boolean laws.

X + X'Y = (X + X')(X + Y) = 1·(X + Y) = X + Y. This uses the distributive law A + BC = (A+B)(A+C) with A=X, B=X', C=Y.

Q4: Why is the law A + AB = A called "absorption"?

Because the variable A "absorbs" the term AB — the larger term AB contributes nothing extra since whenever AB=1, A must already be 1. The term is absorbed into the simpler expression.

Q5: Can Boolean laws be applied to expressions with more than two variables?

Exam Focus

Revise definitions, diagrams, examples, and short-answer points for Boolean Laws and Theorems – Complete Reference with Proofs.

Interview Use

Prepare one clear explanation, one practical example, and one common mistake for this Digital Electronics topic.

Search Terms

digital-electronics, digital electronics, digital, electronics, boolean, algebra, laws, and

Continue learning this concept

PreviousBoolean Expression Simplification – Methods and Solved Examples NextCanonical Forms – Minterms, Maxterms, and Standard Forms