https://file.aiquickdraw.com/mathbot/file/17268883815681pjheleh.jpg

Let's work through the logic diagram step-by-step.

### 1. Identify the gates and inputs:
- There are three inputs: \( X \), \( Y \), and \( Z \).
- Gates used: 
  - NOT gate (inverter)
  - AND gates
  - OR gates

### 2. Step-by-step analysis:

- **First AND gate** (top left):
  - Inputs: \( X' \) (NOT of \( X \)) and \( Y \).
  - Output: \( X'Y \).

- **Second AND gate** (middle left):
  - Inputs: \( X \) and \( Z \).
  - Output: \( XZ \).

- **Bottom OR gate**:
  - Inputs: \( Y \) and \( Z \).
  - Output: \( Y + Z \).

- **NOT gates on outputs**:
  - The outputs from the two AND gates (\( X'Y \) and \( XZ \)) and the OR gate (\( Y + Z \)) are fed into separate NOT gates.
  - Outputs after NOT:
    - \( (X'Y)' \)
    - \( (XZ)' \)
    - \( (Y + Z)' \).

- **Final OR gate**:
  - Inputs: \( (X'Y)' \), \( (XZ)' \), and \( (Y + Z)' \).
  - Output: \( (X'Y)' + (XZ)' + (Y + Z)' \).

- **Final NOT gate**:
  - The output of the OR gate is passed through a final NOT gate.
  - Final output expression: \( \left[(X'Y)' + (XZ)' + (Y + Z)'\right]' \).

### 3. Simplification of the expression:
The final Boolean expression is \( \left[(X'Y)' + (XZ)' + (Y + Z)'\right]' \). This can be simplified using De Morgan's laws and other Boolean algebra rules. Here's how:

- Using De Morgan’s law on the final expression:
  \[
  \left[(X'Y)' + (XZ)' + (Y + Z)'\right]' = (X'Y) \cdot (XZ) \cdot (Y + Z)
  \]
Thus, the simplified output expression is:
\[
(X'Y) \cdot (XZ) \cdot (Y + Z)
\]

### 4. Truth Table:
Let's create the truth table for the inputs \( X \), \( Y \), and \( Z \), and calculate the outputs based on the simplified expression.

\[
\begin{array}{|c|c|c|c|}
\hline
X & Y & Z & \text{Output} \\
\hline
0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\
1 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 \\
1 & 1 & 0 & 0 \\
1 & 1 & 1 & 1 \\
\hline
\end{array}
\]

### Conclusion:
- The final simplified Boolean expression is \( (X'Y) \cdot (XZ) \cdot (Y + Z) \).
- The truth table verifies this expression with the output.

Would you like to go into more details or have further clarifications?

Here are 5 questions for practice:
1. How does De Morgan's law simplify Boolean expressions?
2. What are the basic steps to create a truth table for a given Boolean expression?
3. How does the NOT gate affect the output of an AND gate?
4. Can you reduce \( (X'Y)' + (XZ)' \) further using Boolean laws?
5. How does the order of operations affect logic circuits?

**Tip:** Always simplify logic expressions step-by-step to avoid errors in complex diagrams.

For the following logic diagrams, determine the final output (2 marks), then reduce the output expression (2 marks), then verify using the truth table (2 marks).

Explore how to simplify a Boolean expression derived from a logic diagram involving AND, OR, and NOT gates. This problem covers Boolean Algebra, logic gates, and truth tables, including the use of De Morgan's law for simplification. Suitable for students studying Logic Design and Computer Engineering.

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