📘 Introduction
In Boolean Algebra, simplification means reducing a complex Boolean expression to its simplest form — by applying Boolean laws and theorems.
👉 Why simplify?
- To design simpler logic circuits.
- To use fewer logic gates.
- To save cost, time, and power in hardware implementation.
🛠 Basic Steps to Simplify Boolean Expressions
- Write the given expression clearly.
- Apply Boolean theorems (like Identity Law, Null Law, De Morgan’s Law, etc.).
- Group similar terms wherever possible.
- Repeat simplification until no further reduction is possible.
🔥 Most Common Theorems used in Simplification
| Law / Theorem | Example |
|---|---|
| Idempotent Law | A+A=AA + A = AA+A=A, A⋅A=AA \cdot A = AA⋅A=A |
| Identity Law | A+0=AA + 0 = AA+0=A, A⋅1=AA \cdot 1 = AA⋅1=A |
| Null Law | A+1=1A + 1 = 1A+1=1, A⋅0=0A \cdot 0 = 0A⋅0=0 |
| Complement Law | A+A‾=1A + \overline{A} = 1A+A=1, A⋅A‾=0A \cdot \overline{A} = 0A⋅A=0 |
| Absorption Law | A+(A⋅B)=AA + (A \cdot B) = AA+(A⋅B)=A, A⋅(A+B)=AA \cdot (A + B) = AA⋅(A+B)=A |
| Distributive Law | A(B+C)=AB+ACA(B + C) = AB + ACA(B+C)=AB+AC |
| De Morgan’s Theorem | A⋅B‾=A‾+B‾\overline{A \cdot B} = \overline{A} + \overline{B}A⋅B=A+B |
✏️ Examples of Simplification
Example 1
Simplify: A+A⋅BA + A \cdot BA+A⋅B
Solution:
Apply Absorption Law: A+A⋅B=AA + A \cdot B = AA+A⋅B=A
✅ Simplified Expression: AAA
Example 2
Simplify: A⋅A‾+BA \cdot \overline{A} + BA⋅A+B
Solution:
First, apply Complement Law: A⋅A‾=0A \cdot \overline{A} = 0A⋅A=0
Thus, the expression becomes: 0+B=B0 + B = B0+B=B
✅ Simplified Expression: BBB
Example 3
Simplify: (A+B)⋅(A+B‾)(A + B) \cdot (A + \overline{B})(A+B)⋅(A+B)
Solution:
Apply Distributive Law: =A⋅(A+B‾)+B⋅(A+B‾)= A \cdot (A + \overline{B}) + B \cdot (A + \overline{B})=A⋅(A+B)+B⋅(A+B)
Expand further: =(A⋅A)+(A⋅B‾)+(B⋅A)+(B⋅B‾)= (A \cdot A) + (A \cdot \overline{B}) + (B \cdot A) + (B \cdot \overline{B})=(A⋅A)+(A⋅B)+(B⋅A)+(B⋅B)
Simplify terms:
- A⋅A=AA \cdot A = AA⋅A=A (Idempotent Law)
- B⋅B‾=0B \cdot \overline{B} = 0B⋅B=0 (Complement Law)
Thus: =A+A⋅B‾+B⋅A+0= A + A \cdot \overline{B} + B \cdot A + 0=A+A⋅B+B⋅A+0
Group AAA terms: =A(1+B‾+B)= A (1 + \overline{B} + B)=A(1+B+B)
But B‾+B=1\overline{B} + B = 1B+B=1 (Complement Law), so: =A(1+1)=A×1=A= A (1 + 1) = A \times 1 = A=A(1+1)=A×1=A
✅ Simplified Expression: AAA
Example 4
Simplify: (A+B)‾\overline{(A+B)}(A+B)
Solution:
Apply De Morgan’s Theorem: A+B‾=A‾⋅B‾\overline{A+B} = \overline{A} \cdot \overline{B}A+B=A⋅B
✅ Simplified Expression: A‾⋅B‾\overline{A} \cdot \overline{B}A⋅B
🔥 Important Tips for Simplification
| Tip | Explanation |
|---|---|
| Group terms smartly | Always try to group common variables |
| Apply Complement Law quickly | Whenever you see a variable and its complement |
| Use Absorption Law early | It makes expressions smaller |
| Know De Morgan’s Theorems by heart | Very useful in dealing with NOT of AND/OR |
✅ Summary
- Simplifying Boolean expressions reduces circuit complexity.
- Use laws like Absorption, Complement, Identity, Distributive, De Morgan.
- Always expand carefully and group terms logically.
- The goal is minimal number of terms and variables.