1. **Problem Statement:** Explain and prove De Morgan's Laws in set theory. 2. **De Morgan's Laws:** They describe the relationship between the complement of unions and intersections of sets. The laws are: $$\overline{A \cup B} = \overline{A} \cap \overline{B}$$ $$\overline{A \cap B} = \overline{A} \cup \overline{B}$$ 3. **Explanation:** - The complement of the union of two sets is the intersection of their complements. - The complement of the intersection of two sets is the union of their complements. 4. **Proof of the first law:** - Let $x \in \overline{A \cup B}$. - This means $x \notin A \cup B$. - So, $x \notin A$ and $x \notin B$. - Therefore, $x \in \overline{A}$ and $x \in \overline{B}$. - Hence, $x \in \overline{A} \cap \overline{B}$. - Conversely, let $x \in \overline{A} \cap \overline{B}$. - Then $x \in \overline{A}$ and $x \in \overline{B}$. - So, $x \notin A$ and $x \notin B$. - Therefore, $x \notin A \cup B$. - Hence, $x \in \overline{A \cup B}$. 5. **Proof of the second law:** - Let $x \in \overline{A \cap B}$. - Then $x \notin A \cap B$. - So, $x \notin A$ or $x \notin B$. - Therefore, $x \in \overline{A}$ or $x \in \overline{B}$. - Hence, $x \in \overline{A} \cup \overline{B}$. - Conversely, let $x \in \overline{A} \cup \overline{B}$. - Then $x \in \overline{A}$ or $x \in \overline{B}$. - So, $x \notin A$ or $x \notin B$. - Therefore, $x \notin A \cap B$. - Hence, $x \in \overline{A \cap B}$. **Final answer:** De Morgan's Laws are: $$\overline{A \cup B} = \overline{A} \cap \overline{B}$$ $$\overline{A \cap B} = \overline{A} \cup \overline{B}$$