The expression in the image is a Boolean algebra expression, which involves logic variables and their complements. Here's the expression:

$\overline{x} \overline{y} z + \overline{x} y \overline{z} + x \overline{y} \overline{z} + z y z$

To simplify this expression, we will apply Boolean algebra rules step by step.

Step-by-step Simplification:

1. Combine similar terms: We begin by looking for opportunities to factor and apply Boolean laws.

   First, let's focus on the first and second terms:
   $\overline{x} \overline{y} z + \overline{x} y \overline{z}$ These two terms share a common factor, $\overline{x}$. We factor that out: $\overline{x} (\overline{y} z + y \overline{z})$ This term can't be simplified further using Boolean identities, so we move on.

2. Analyze remaining terms: The third term is $x \overline{y} \overline{z}$ and the fourth term is $zyz$.

   However, notice that $zyz = yz$ (because $z \cdot z = z$). So the fourth term becomes: $yz$

3. Final expression: After simplifying the terms as much as possible, the simplified expression is: $\overline{x} (\overline{y} z + y \overline{z}) + x \overline{y} \overline{z} + yz$

Final Result:

$\boxed{\overline{x} (\overline{y} z + y \overline{z}) + x \overline{y} \overline{z} + yz}$

Would you like further details on Boolean algebra simplification or any particular part of this process?

Follow-up Questions:

1. What Boolean identity can simplify $z \cdot z$ in Boolean algebra?
2. Can the distributive property be applied again in this case?
3. How can De Morgan’s laws simplify Boolean expressions?
4. Why does factoring help in Boolean simplifications?
5. Can you simplify the expression further by Karnaugh maps?

Tip: In Boolean algebra, combining terms and factoring based on common elements can significantly simplify complex expressions.