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
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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.

 

### 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 \( x \in B \cap C \Rightarrow x \in A \cup (B \cap C) \). \( \square \)

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**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
Transcribed Image Text:### 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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