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.