Exercise 1 (De Morgan’s law)
Show De Morgan’s second law (A ∩ B)C = AC ∪ BC . Use first a Venn diagram and
provide then a formal mathematical proof.

### De Morgan's Second Law: 
The law states that the complement of the intersection of two sets \( A \) and \( B \) is equal to the union of the complements of the individual sets. Formally:

\[
(A \cap B)^C = A^C \cup B^C
\]

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### Step 1: **Venn Diagram Explanation**

Let's begin by visualizing the sets \( A \), \( B \), and their complements using Venn diagrams.

1. **Intersection of A and B (\( A \cap B \))**: This represents the region where both \( A \) and \( B \) overlap.
2. **Complement of the Intersection (\( (A \cap B)^C \))**: The complement of the intersection is everything outside of the overlap (i.e., regions that do not belong to both \( A \) and \( B \) simultaneously).
3. **Union of Complements (\( A^C \cup B^C \))**: This represents the region outside of both \( A \) and \( B \) (i.e., the part not in \( A \), the part not in \( B \), or both).

By comparing these regions visually in a Venn diagram, you will observe that both expressions cover the same area, confirming that:

\[
(A \cap B)^C = A^C \cup B^C
\]

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### Step 2: **Mathematical Proof**

To prove this formally, we will use set theory notation.

#### Proof:
We need to show that both sides of the equation represent the same set.

1. **Start with the left-hand side**: \( (A \cap B)^C \)

- By definition, the complement of \( A \cap B \) contains all elements that are **not in** both \( A \) and \( B \). In other words:
  \[
  x \in (A \cap B)^C \quad \text{if and only if} \quad x \notin A \cap B
  \]

2. **Rewrite the condition for the complement of the intersection**:
   - \( x \notin A \cap B \) means that \( x \) is **not** in both \( A \) and \( B \). Therefore, \( x \) must be either:
     - \( x \notin A \) (i.e., \( x \in A^C \)), or
     - \( x \notin B \) (i.e., \( x \in B^C \)).

   Thus, we can express this as:
   \[
   x \in (A \cap B)^C \quad \text{if and only if} \quad x \in A^C \cup B^C
   \]

3. **Conclusion**: Since the left-hand side \( (A \cap B)^C \) and the right-hand side \( A^C \cup B^C \) describe the same condition for any element \( x \), we have:
   \[
   (A \cap B)^C = A^C \cup B^C
   \]

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### Summary:

The proof shows that the complement of the intersection of two sets \( A \) and \( B \) is equal to the union of the complements of the sets. Both the Venn diagram and the formal proof support this conclusion.

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Would you like to see more details on the Venn diagram representation or the logical steps in the proof? Feel free to ask!

Here are some related questions:
1. What is De Morgan's first law and how is it proved?
2. Can we apply De Morgan's laws to more than two sets?
3. How does De Morgan's law relate to Boolean algebra?
4. Can we extend De Morgan's laws to logic statements involving negations?
5. How are set complements used in probability theory?

Tip: De Morgan's laws are crucial when working with logical negations and simplifying complex set expressions.

Exercise 1 (De Morgan’s law) Show De Morgan’s second law (A ∩ B)C = AC ∪ BC. Use first a Venn diagram and provide then a formal mathematical proof.

Explore De Morgan’s second law: (A ∩ B)^C = A^C ∪ B^C. This page provides a clear explanation with a Venn diagram and formal proof using set theory, including complements, intersections, and unions. Ideal for high school students in Grades 10-12.

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