The given Boolean expression is:

$\overline{A + (B \cdot C) + (A \cdot \overline{B})}$

Simplification Steps:

1. Expand terms and use Boolean rules: $A + (B \cdot C) + (A \cdot \overline{B})$

2. Combine terms and simplify: Since $A$ appears separately and in $A \cdot \overline{B}$, the final result for simplification depends on the negations inside .

But Clear needThe given Boolean expression is:

$\overline{A + (B \cdot C) + (A \cdot \overline{B})}$

Let me simplify this step-by-step.

Step 1: Analyze the inner terms

The expression inside the complement is: $A + (B \cdot C) + (A \cdot \overline{B})$

Step 2: Apply distribution or simplifications

• Combine terms logically:
  • $A + (A \cdot \overline{B}) = A$ (Absorption Law: $A + A \cdot X = A$)
  • The new expression becomes: $A + (B \cdot C)$

Step 3: Apply De Morgan's Law to the complemented term

Now we complement the entire term: $\overline{A + (B \cdot C)}$

By De Morgan's Law: $\overline{A + (B \cdot C)} = \overline{A} \cdot \overline{(B \cdot C)}$

Step 4: Simplify further

$\overline{A} \cdot \overline{(B \cdot C)}$

Using De Morgan's Law again on $\overline{(B \cdot C)}$: $\overline{(B \cdot C)} = \overline{B} + \overline{C}$

So the expression becomes: $\overline{A} \cdot (\overline{B} + \overline{C})$

Final Simplified Expression:

$\overline{A} \cdot (\overline{B} + \overline{C})$

Do you want further clarification or any related examples? Here are 5 related questions you can ask:

1. How does De Morgan's Law work in Boolean algebra?
2. Can you simplify another complex Boolean expression for me?
3. What are the basic Boolean algebra rules used in simplifications?
4. How is this expression implemented in digital circuits?
5. Can you show this as a truth table for all possible values of $A$, $B$, and $C$?

Tip: Always try to use De Morgan's Laws carefully for complement expressions—they are key to simplifications!