Simplify= A’B’C’ + A’C’(B) + AD + ABC’D’

To simplify the given Boolean expression:

\[
A'B'C' + A'C'(B) + AD + ABC'D'
\]

### Step-by-step simplification:

1. **Group terms**:
   \[
   A'B'C' + A'C'B + AD + ABC'D'
   \]
   
2. **Combine like terms** for the first two terms:
   - \( A'B'C' + A'C'B \)
   - Factor out \( A'C' \):
     \[
     A'C'(B' + B) + AD + ABC'D'
     \]
   - Since \( B' + B = 1 \):
     \[
     A'C' + AD + ABC'D'
     \]

3. **Final simplified expression**:
   \[
   A'C' + AD + ABC'D'
   \]

Thus, the simplified Boolean expression is:
\[
A'C' + AD + ABC'D'
\]

Would you like further clarification on any part of this process?

---

Here are 5 related questions to expand your understanding:
1. What is De Morgan's law, and how can it be applied to Boolean algebra?
2. How does factoring help in simplifying Boolean expressions?
3. What is the significance of combining like terms in Boolean algebra?
4. Can you explain the importance of the identity \( B' + B = 1 \)?
5. How does this simplification technique apply to logic circuit design?

**Tip:** Always look for opportunities to factor common terms when simplifying Boolean expressions to reduce the complexity.

Discover how to simplify the Boolean expression A’B’C’ + A’C’(B) + AD + ABC’D’ with a detailed step-by-step solution. This guide covers key Boolean algebra concepts, including simplification techniques, factoring, and the use of identity laws. Perfect for high school students studying Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation