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The expression you have provided is:

\[
\overline{A \cdot B \cdot C + x \cdot Y} \cdot \overline{\overline{A} \cdot \overline{B} \cdot \overline{C} + x \cdot Y}
\]

To simplify this expression, let's break it down using Boolean algebra rules:

### Step 1: Apply De Morgan's Theorems
First, use De Morgan's theorem to break down the complements:

\[
\overline{A \cdot B \cdot C + x \cdot Y} = \overline{A \cdot B \cdot C} \cdot \overline{x \cdot Y}
\]

\[
\overline{\overline{A} \cdot \overline{B} \cdot \overline{C} + x \cdot Y} = \overline{\overline{A} \cdot \overline{B} \cdot \overline{C}} \cdot \overline{x \cdot Y}
\]

Using De Morgan's theorem again, we simplify further:

\[
\overline{A \cdot B \cdot C} = \overline{A} + \overline{B} + \overline{C}
\]
\[
\overline{x \cdot Y} = \overline{x} + \overline{Y}
\]
\[
\overline{\overline{A} \cdot \overline{B} \cdot \overline{C}} = A + B + C
\]

### Step 2: Substitute back into the expression

Now substitute these into the original expression:

\[
(\overline{A} + \overline{B} + \overline{C}) \cdot (\overline{x} + \overline{Y}) \cdot (A + B + C) \cdot (\overline{x} + \overline{Y})
\]

### Step 3: Simplify further

Notice that \(\overline{x} + \overline{Y}\) appears twice. So the expression simplifies to:

\[
(\overline{x} + \overline{Y}) \cdot (\overline{A} + \overline{B} + \overline{C}) \cdot (A + B + C)
\]

This is the simplified form of the given Boolean expression.

Would you like more details on this solution or do you have any questions?

### Related Questions:
1. What is De Morgan's theorem and how does it apply to Boolean algebra?
2. How do you apply De Morgan's theorem in Boolean expression simplification?
3. What are the properties of the AND, OR, and NOT operations in Boolean algebra?
4. How can Boolean algebra simplify complex logical expressions?
5. What is the significance of the complement in Boolean algebra?
6. How do you recognize opportunities to use the Distributive Law in Boolean algebra?
7. Can you apply these principles to circuit design?
8. How does Boolean algebra relate to truth tables?

**Tip:** Always look for common factors in Boolean expressions to simplify them more easily.

Learn how to simplify a complex Boolean expression using De Morgan's theorem with detailed step-by-step instructions. This problem involves applying De Morgan's theorem to reduce the complexity of Boolean algebra expressions. Suitable for undergraduate students and those interested in Boolean algebra simplification.

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