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.