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The given Boolean function is:

\[
F = (A + \overline{B} + A \cdot B) \cdot (A + \overline{B}) \cdot A \cdot B
\]

Let's simplify this step by step using Boolean algebra postulates and theorems:

### Step 1: Simplify the first term
\[
A + \overline{B} + A \cdot B
\]
Using the **Absorption Law**: \( A + A \cdot B = A \), we get:
\[
A + \overline{B} + A \cdot B = A + \overline{B}
\]
So the expression becomes:
\[
F = (A + \overline{B}) \cdot (A + \overline{B}) \cdot A \cdot B
\]

### Step 2: Combine identical terms
\[
(A + \overline{B}) \cdot (A + \overline{B}) = A + \overline{B}
\]
This reduces the expression to:
\[
F = (A + \overline{B}) \cdot A \cdot B
\]

### Step 3: Distribute the remaining terms
\[
(A + \overline{B}) \cdot A \cdot B = (A \cdot A \cdot B) + (\overline{B} \cdot A \cdot B)
\]

Using **Idempotent Law** \( A \cdot A = A \) and the fact that \( \overline{B} \cdot B = 0 \), the expression simplifies to:
\[
F = A \cdot B + 0 = A \cdot B
\]

### Final Simplified Function:
\[
F = A \cdot B
\]

This is the simplest form of the given Boolean function.

Would you like further clarification or more details on any step? 

Here are 5 related questions you can explore:
1. What are the key Boolean algebra laws used in this simplification?
2. How does the Absorption Law help in Boolean algebra?
3. Can you use De Morgan's law to simplify Boolean expressions?
4. What is the role of the Idempotent Law in Boolean simplifications?
5. How do we apply the Distributive Law in Boolean algebra?

**Tip:** When simplifying Boolean expressions, always look for common patterns like \( A + A \cdot B \), as laws like absorption can drastically reduce complexity.

Simplify the Boolean expression F = (A + B̅ + A.B)(A + B̅)A.B step by step using Boolean algebra laws such as the Absorption Law, Idempotent Law, and Distributive Law. This problem is suitable for undergraduate students in Computer Science or Electrical Engineering.

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