https://file.aiquickdraw.com/mathbot/file/1726392760446ua4gdl6w.jpg

The expression in the image is a Boolean algebra expression:

\[
A \cdot \overline{B} + A \cdot B + \overline{A} \cdot B
\]

Let's simplify it step by step.

### Step 1: Factor common terms
We can group terms to look for simplifications:
\[
A \cdot (\overline{B} + B) + \overline{A} \cdot B
\]

### Step 2: Apply Boolean identity
Using the Boolean identity \(\overline{B} + B = 1\), the expression simplifies to:
\[
A \cdot 1 + \overline{A} \cdot B
\]

This simplifies further to:
\[
A + \overline{A} \cdot B
\]

### Step 3: Apply the Distributive Law
The next step uses the Distributive Law:
\[
A + \overline{A} \cdot B = (A + \overline{A}) \cdot (A + B)
\]

Using the identity \(A + \overline{A} = 1\):
\[
1 \cdot (A + B) = A + B
\]

### Final Simplified Expression:
\[
A + B
\]

The given Boolean expression simplifies to \(A + B\).

---

Would you like more details or explanations on any part of this solution? Here are 5 related questions you might be interested in:

1. How does the Boolean identity \(\overline{B} + B = 1\) work?
2. What are the basic rules of Boolean algebra?
3. Can you explain the distributive law in Boolean algebra?
4. Are there other methods to simplify Boolean expressions?
5. How is Boolean algebra used in digital circuit design?

**Tip:** Boolean algebra is widely used in designing and optimizing digital circuits such as logic gates in computers and electronics.

Simplify the Boolean expression A·¬B + A·B + ¬A·B.

Learn how to simplify the Boolean algebra expression A·¬B + A·B + ¬A·B using Boolean identities and theorems. This problem covers the Distributive Law and Boolean Complement Law, providing a step-by-step solution. Suitable for students in Grades 10-12.

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