Use element argument to prove the statement. Assume that all sets are subsets of a universal set U, Statement: For all sets A, B, and C, A ∩ (B − C) = (A ∩ B) − (A ∩ C), I have atteched the example to use to solve this problem. Please help me becuase I could not figure it out. ### Transcriptions:**10. For all sets \( A, B, \) and \( C \), \( (A \cup B) \cap C \subseteq A \cup (B \cap C) \).****Proof:** Let \( A, B, \) and \( C \) be any sets. Let \( x \in (A \cup B) \cap C \). Then \( x \in A \cup B \) and \( x \in C \).- \((x \in A \text{ or } x \in B)\) and \(x \in C\).- If \( x \in A \), then \( x \in A \cup (B \cap C) \).If \( x \in B \), then \( x \in B \cap C \Rightarrow x \in A \cup (B \cap C) \). \( \square \)---**14. For all sets \( A \) and \( B, A \cup (A \cap B) = A \).****Proof:** Let \( A \) be any set.- \( A \cup (A \cap B) \subseteq A \): Let \( x \in A \cup (A \cap B) \). By definition of union, - \( x \in A \) or \( x \in A \) and \( x \in B \). Suppose \( x \notin A \). Then \( x \in A \) and \( x \in B \), a contradiction. So \( x \in A \).- \( A \subseteq A \cup (A \cap B) \): Let \( x \in A \). Then \( x \in A \cup (A \cap B) \) by definition of union. \( \square \)---**19. For all sets \( A, B, \) and \( C \), if \( A \subseteq B \) and \( A \subseteq C \) then \( A \subseteq B \cap C \).****Proof:** Let \( A, B, \) and \( C \) be any sets.\( A \subseteq B \) & \( A \subseteq C \). Let \( x \in A \). Then \( x \in B \) (\(A \subseteq B\)) and \( x \in

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Use element argument to prove the statement. Assume that all sets are subsets of a universal set U, Statement: For all sets A, B, and C, A ∩ (B − C) = (A ∩ B) − (A ∩ C), I have atteched the example to use to solve this problem. Please help me becuase I could not figure it out.

Use element argument to prove the statement. Assume that all sets are subsets of a universal set U, Statement: For all sets A, B, and C, A ∩ (B − C) = (A ∩ B) − (A ∩ C), I have atteched the example to use to solve this problem. Please help me becuase I could not figure it out.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

### Transcriptions:

**10. For all sets \( A, B, \) and \( C \), \( (A \cup B) \cap C \subseteq A \cup (B \cap C) \).**

**Proof:** Let \( A, B, \) and \( C \) be any sets. Let \( x \in (A \cup B) \cap C \). Then \( x \in A \cup B \) and \( x \in C \).

- \((x \in A \text{ or } x \in B)\) and \(x \in C\).
- If \( x \in A \), then \( x \in A \cup (B \cap C) \).

If \( x \in B \), then \( x \in B \cap C \Rightarrow x \in A \cup (B \cap C) \). \( \square \)

---

**14. For all sets \( A \) and \( B, A \cup (A \cap B) = A \).**

**Proof:** Let \( A \) be any set.

- \( A \cup (A \cap B) \subseteq A \): Let \( x \in A \cup (A \cap B) \). By definition of union,
- \( x \in A \) or \( x \in A \) and \( x \in B \). Suppose \( x \notin A \). Then \( x \in A \) and \( x \in B \), a contradiction. So \( x \in A \).

- \( A \subseteq A \cup (A \cap B) \): Let \( x \in A \). Then \( x \in A \cup (A \cap B) \) by definition of union. \( \square \)

---

**19. For all sets \( A, B, \) and \( C \), if \( A \subseteq B \) and \( A \subseteq C \) then \( A \subseteq B \cap C \).**

**Proof:** Let \( A, B, \) and \( C \) be any sets.

\( A \subseteq B \) & \( A \subseteq C \). Let \( x \in A \). Then \( x \in B \) (\(A \subseteq B\)) and \( x \in

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