To simplify the Boolean expression $x' + xy + xz' + xy'z'$, let's go step by step.

Expression:

$x' + xy + xz' + xy'z'$

Step 1: Group terms

Group similar terms for easier manipulation: $x' + (xy + xz') + xy'z'$

Step 2: Factor common terms

In the group $xy + xz'$, factor out $x$: $x' + x(y + z') + xy'z'$

Step 3: Simplify within the parentheses

Using the Distributive Law: y + z' = 1 \quad \text{(since for any Boolean value of y$or$z', this results in 1)} So the expression simplifies to: $x' + x(1) + xy'z'$

Step 4: Simplify further

Since $x(1) = x$, the expression becomes: $x' + x + xy'z'$

Step 5: Apply the Absorption Law

Using the Absorption Law, $x' + x = 1$: $1 + xy'z'$

Step 6: Apply the Identity Law

Finally, $1 + xy'z' = 1$ (since $1 + anything = 1$).

Final Simplified Expression:

$1$

The Boolean expression simplifies to 1.

Would you like further clarification on any step?

Here are some related questions to explore:

1. What is the Distributive Law in Boolean algebra, and how is it used in simplification?
2. Can you explain the Absorption Law with more examples?
3. How do you recognize when to apply the Identity Law in Boolean algebra?
4. What would happen if different Boolean operations were applied here?
5. How can Karnaugh maps help simplify Boolean expressions like this?

Tip: Always group similar terms when simplifying Boolean expressions. It can often lead to quicker simplifications!