 Use Boolean algebra to reduce the equation 𝐹(𝑥,𝑦,𝑧) = (𝑥′ + 𝑦′ + 𝑥′𝑦′ + 𝑥𝑦)(𝑥′ + 𝑦𝑧)

We are tasked with simplifying the Boolean expression:

\[
F(x, y, z) = (x' + y' + x'y' + xy)(x' + yz)
\]

Let's simplify it step by step using Boolean algebra rules.

### Step 1: Simplify the first term \((x' + y' + x'y' + xy)\)

- Apply **Absorption Law**: \(x' + x'y' = x'\) (because \(x' + x' = x'\), and adding the \(y'\) doesn't affect the outcome).
  
Thus, the term simplifies as:
\[
x' + y' + x' + xy = x' + y' + xy
\]

- Apply **Idempotent Law**: \(x' + x' = x'\), so the term becomes:
\[
x' + y' + xy
\]

### Step 2: Now simplify the whole expression:

We now have:
\[
F(x, y, z) = (x' + y' + xy)(x' + yz)
\]

### Step 3: Distribute the terms

Distribute the terms:
\[
F(x, y, z) = (x' + y' + xy)(x') + (x' + y' + xy)(yz)
\]

First, distribute \( (x') \):
\[
= x'x' + y'x' + xyx'
\]

Now, simplify using Boolean algebra rules:
- \( x'x' = x' \) (Idempotent Law)
- \( xyx' = 0 \) (Complement Law: \(x \cdot x' = 0\))

Thus, this term simplifies to:
\[
x' + y'x'
\]

Now, distribute the \( (yz) \) term:
\[
+ x'yz + y'yz + xy \cdot yz
\]

Simplify the terms:
- \( x'yz = x'yz \) (no further simplification)
- \( y'yz = 0 \) (Complement Law: \(y' \cdot y = 0\))
- \( xy \cdot yz = xyz \) (apply Associative Law: \( xy \cdot yz = xyz \))

Thus, the second term simplifies to:
\[
x'yz + xyz
\]

### Step 4: Combine everything

Now, we combine both simplified parts:
\[
F(x, y, z) = x' + y'x' + x'yz + xyz
\]

### Step 5: Final simplification

Observe that:
- \(x' + x'yz = x'\) (Absorption Law)
  
Thus, the expression simplifies to:
\[
F(x, y, z) = x' + y'x' + xyz
\]

This is the simplified Boolean expression.

### Final Simplified Form:

\[
F(x, y, z) = x' + y'x' + xyz
\]

---

Would you like further clarification on any steps or more details?

Here are 5 related questions for further exploration:
1. How does the Absorption Law work in Boolean algebra?
2. Can Boolean algebra expressions be simplified differently using Karnaugh maps?
3. What is the significance of the Complement Law in simplifying Boolean expressions?
4. How does Boolean algebra apply in digital circuit design?
5. How would the truth table look for the final expression?

**Tip**: Always look for common terms and use laws like Absorption and Idempotent to simplify Boolean expressions efficiently.

Use Boolean algebra to reduce the equation F(x, y, z) = (x' + y' + x'y' + xy)(x' + yz)

This step-by-step guide simplifies the Boolean expression F(x, y, z) = (x' + y' + x'y' + xy)(x' + yz) using Boolean algebra. It applies key theorems such as Absorption Law, Complement Law, and Idempotent Law, making it suitable for undergraduate students studying Computer Science or Electrical Engineering.

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