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

Let's reduce the Boolean function \( F(x, y, z) = (x' + y' + x'y' + xy)(x' + yz) \) using Boolean algebra rules step by step.

### Step 1: Simplify the first part
The expression \( x' + y' + x'y' + xy \) can be simplified by noticing that \( x'y' \) is redundant because:

\[
x' + y' + x'y' = x' + y'
\]

This follows from the **absorption law**, which states \( a + ab = a \).

Thus, the expression becomes:
\[
x' + y' + xy
\]

### Step 2: The simplified function
Now the function becomes:

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

### Step 3: Distribute the terms
Next, we will distribute the terms of the two factors:
\[
(x' + y' + xy)(x' + yz) = x'(x' + yz) + y'(x' + yz) + xy(x' + yz)
\]

This results in:
\[
= x' \cdot x' + x' \cdot yz + y' \cdot x' + y' \cdot yz + xy \cdot x' + xy \cdot yz
\]

### Step 4: Simplify each term
- \( x' \cdot x' = x' \) (idempotent law: \( a \cdot a = a \))
- \( x' \cdot yz = x'yz \)
- \( y' \cdot x' = x'y' \)
- \( y' \cdot yz = y'yz = 0 \) (since \( y' \cdot y = 0 \))
- \( xy \cdot x' = 0 \) (since \( x \cdot x' = 0 \))
- \( xy \cdot yz = xyz \)

Thus, we now have:
\[
F(x, y, z) = x' + x'yz + x'y' + xyz
\]

### Step 5: Simplify the final expression
Group the terms involving \( x' \):
\[
F(x, y, z) = x'(1 + yz + y') + xyz
\]
Since \( 1 + yz = 1 \) (anything OR'd with 1 is 1), we simplify to:
\[
F(x, y, z) = x' + xyz
\]

Thus, the reduced Boolean expression is:
\[
\boxed{F(x, y, z) = x' + xyz}
\]

### Final Answer:
The simplified Boolean function is \( F(x, y, z) = x' + xyz \).

Would you like more details or have any questions? 

Here are 5 related questions to explore this concept further:
1. How does the absorption law work in Boolean algebra?
2. What are the idempotent and complement laws in Boolean algebra?
3. Can you simplify a different Boolean expression like \( F(x, y) = (x + y)(x' + y) \)?
4. How does distributive law apply to Boolean expressions?
5. What are other methods to simplify Boolean expressions (e.g., Karnaugh maps)?

**Tip:** Always apply basic laws like absorption and idempotent laws first when simplifying Boolean expressions.

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

Explore the step-by-step simplification of the Boolean expression F(x, y, z) = (x' + y' + x'y' + xy)(x' + yz) using Boolean algebra rules, including the absorption and idempotent laws. Suitable for advanced high school or undergraduate students.

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