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De Morgan's laws for sets describe how the complement of a union or intersection of sets can be expressed in terms of the complements of the individual sets. There are two main laws, which can be stated as follows:

1. **First Law (Union of Complements)**:  
   \[(A \cup B)^c = A^c \cap B^c\]
   This means that the complement of the union of sets \(A\) and \(B\) is equal to the intersection of their complements.

2. **Second Law (Intersection of Complements)**:  
   \[(A \cap B)^c = A^c \cup B^c\]
   This means that the complement of the intersection of sets \(A\) and \(B\) is equal to the union of their complements.

### Proof of De Morgan's Laws

#### Proof of the First Law: \((A \cup B)^c = A^c \cap B^c\)

1. Let \(x \in (A \cup B)^c\).
   - This implies that \(x\) is not in \(A \cup B\).
   - Therefore, \(x\) is neither in \(A\) nor in \(B\).
   - Hence, \(x \in A^c\) and \(x \in B^c\).
   - This implies that \(x \in A^c \cap B^c\).
   
   Thus, \((A \cup B)^c \subseteq A^c \cap B^c\).

2. Conversely, let \(x \in A^c \cap B^c\).
   - This implies that \(x \in A^c\) and \(x \in B^c\).
   - Therefore, \(x\) is not in \(A\) and not in \(B\).
   - Hence, \(x\) is not in \(A \cup B\).
   - This implies that \(x \in (A \cup B)^c\).
   
   Thus, \(A^c \cap B^c \subseteq (A \cup B)^c\).

Since we have both \((A \cup B)^c \subseteq A^c \cap B^c\) and \(A^c \cap B^c \subseteq (A \cup B)^c\), we conclude that:
\[(A \cup B)^c = A^c \cap B^c\]

#### Proof of the Second Law: \((A \cap B)^c = A^c \cup B^c\)

1. Let \(x \in (A \cap B)^c\).
   - This implies that \(x\) is not in \(A \cap B\).
   - Therefore, \(x\) is either not in \(A\) or not in \(B\).
   - Thus, \(x \in A^c\) or \(x \in B^c\).
   - This implies that \(x \in A^c \cup B^c\).
   
   Thus, \((A \cap B)^c \subseteq A^c \cup B^c\).

2. Conversely, let \(x \in A^c \cup B^c\).
   - This implies that \(x \in A^c\) or \(x \in B^c\).
   - Therefore, \(x\) is either not in \(A\) or not in \(B\).
   - Hence, \(x\) is not in \(A \cap B\).
   - This implies that \(x \in (A \cap B)^c\).
   
   Thus, \(A^c \cup B^c \subseteq (A \cap B)^c\).

Since we have both \((A \cap B)^c \subseteq A^c \cup B^c\) and \(A^c \cup B^c \subseteq (A \cap B)^c\), we conclude that:
\[(A \cap B)^c = A^c \cup B^c\]

These are De Morgan's laws for sets.

Would you like further details on these proofs or have any questions?

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Here are five related questions to explore:

1. How do De Morgan's laws apply in Boolean algebra?
2. Can De Morgan's laws be extended to three or more sets?
3. What are the practical applications of De Morgan's laws in probability theory?
4. How do De Morgan's laws relate to logic gates in digital circuits?
5. How would you apply De Morgan's laws to solve complement problems in set theory?

**Tip**: De Morgan's laws are useful in simplifying expressions, especially in problems involving complements and unions or intersections.

State and prove De Morgan's laws for sets.

Explore De Morgan's laws for sets, covering the statement and proof of these key set theory identities. Learn how the complement of a union or intersection of sets can be expressed in terms of individual complements. Suitable for high school students in Grades 10-12.

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