Boolean Algebra

Boolean Algebra Laws

1. Identity Laws

• A + 0 = A
• A ⋅ 1= A

2. Null Laws

• A + 1 = 1
• A ⋅ 0 = 0

3. ==Idempotent Laws==

• A + A = A
• A ⋅ A = A

4. Complement Laws

• A + A′ = 1
• A ⋅ A′ = 0

5. Domination Laws

• A+1 = 1
• A ⋅ 0 = 0

6. Double Negation

• (A′)′ = A

7. Commutative Laws

• A + B = B + A
• A ⋅ B = B ⋅ A

8. Associative Laws ⭐

• ( A + B ) + C = A + (B + C)
• ( A ⋅ B ) ⋅ C = `A . (B . C)

9. Distributive Laws ⭐

• A ( B + C ) = AB + AC
• A + BC = (A + B)(A + C)

12. Special Distributive Form ⭐

• X + X′Y = X + YX + X' Y = X + Y
• Example: B′+BA = B′+AB' + BA = B' + A

10. Absorption Laws ⭐

• A + AB = A
• A ( A + B ) = A

11. De Morgan’s Laws ⭐

• ( A + B )′ = A'B'
• (AB)′ = A' + B'

Proofs of Some Important Algebra

1. Distributive Laws

• (a) $A(B+C) = AB + AC$
• (b) $(A+B)(A+C) = A + BC$

$\begin{aligned} (A + B)(A + C) &= A(A + C) + B(A + C) \\ &= A + AC + AB + BC \\ &= A + AB + AC + BC \\ &= A (1 + B + C) + BC \\ &= A + BC \end{aligned}$

2. Special Distributive Laws

• (a) $X+X′Y= X+Y$

$\begin{aligned} X + X'Y &= (X + X')(X + Y) \\ &= (1)(X + Y) \\ &= X + Y \end{aligned}$

3. Absorption Laws

• (a) $A+AB = A(1+B) = A(1) =A$
• (b) $A(A+B)=AA+AB=A+AB=A$

4. De Morgan’s Laws

• (a) $(A+B)′=A′B′$
• (b) $(AB)′=A′+B′$

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Note 2

To solve Boolean algebra problems effectively, you need a toolkit of laws that allow you to simplify complex expressions. These laws are divided into basic rules and more advanced theorems.

1. Fundamental Laws (The Basics)

These laws are the building blocks of all Boolean logic.

Name	OR Law (+)	AND Law (⋅)
Identity Law	$A + 0 = A$	$A \cdot 1 = A$
Null (Annulment) Law	$A + 1 = 1$	$A \cdot 0 = 0$
Idempotent Law	$A + A = A$	$A \cdot A = A$
Complement Law	$A + \bar{A} = 1$	$A \cdot \bar{A} = 0$
Double Negation	$\overline{(\bar{A})} = A$	—

2. Commutative, Associative, and Distributive Laws

These function similarly to standard algebra but with one major “Boolean-only” twist.

• Commutative: $A + B = B + A$ and $A \cdot B = B \cdot A$
• Associative: $(A + B) + C = A + (B + C)$ and $(A \cdot B) \cdot C = A \cdot (B \cdot C)$
• Distributive: * Rule 1: $A(B + C) = AB + AC$ (Same as normal algebra)
  • Rule 2: $A + (BC) = (A + B)(A + C)$ (Unique to Boolean Algebra)
    • Proof of Rule 2: Right Side: $(A + B)(A + C) = AA + AC + AB + BC$ $= A + AC + AB + BC$ (since $AA = A$) $= A(1 + C + B) + BC$ (since $A + AC = A$)
      $= A(1) + BC = A + BC$.

3. Absorption Laws ⭐

These are the most powerful laws for simplifying and “shrinking” expressions.

• Law 1: $A + AB = A$

  • Derivation: $A(1 + B) = A(1) = A$.

• Law 2: $A(A + B) = A$

  • Derivation: $AA + AB = A + AB = A(1 + B) = A$.

• Redundancy Law: ⭐

  • Sum Form: $A + \bar{A}B = A + B$
    • Derivation: $(A + \bar{A})(A + B) = 1 \cdot (A + B) = A + B$.
  • Product Form: $A(\bar{A} + B) = AB$
    • **Deviation: ** $(A \cdot \bar{A}) + (A \cdot B) = 0 + (A \cdot B) = A \cdot B$

De Morgan’s Theorems ⭐

Essential for breaking long “bars” (negations) over expressions.

• Theorem 1: $\overline{A + B} = \bar{A} \cdot \bar{B}$ (The complement of a sum is the product of the complements).
• Theorem 2: $\overline{A \cdot B} = \bar{A} + \bar{B}$ (The complement of a product is the sum of the complements).

Proof (Truth Table for Theorem 2):

A	B	A⋅B	A⋅B	Aˉ	Bˉ	Aˉ+Bˉ
0	0	0	1	1	1	1
0	1	0	1	1	0	1
1	0	0	1	0	1	1
1	1	1	0	0	0	0

5. Consensus Theorem

This is an “expert level” law used to eliminate redundant terms in three-variable expressions.

• Equation: $AB + \bar{A}C + BC = AB + \bar{A}C$
• Derivation:
  1. $AB + \bar{A}C + BC(1)$
  2. $AB + \bar{A}C + BC(A + \bar{A})$ (since $A + \bar{A} = 1$)
  3. $AB + \bar{A}C + ABC + \bar{A}BC$
  4. Group terms: $(AB + ABC) + (\bar{A}C + \bar{A}BC)$
  5. $AB(1 + C) + \bar{A}C(1 + B)$
  6. $AB(1) + \bar{A}C(1) = AB + \bar{A}C$.