Use an element argument to prove the following statement.. Proof: For all sets A, B, and C, An (B-C) ≤ (ANB) - (ANC). 1. Suppose A, B, and C are any sets. [To show that An (B-C) ≤ (ANB) - (An C), we must show that for every element x, if x € --Select--- 2. Suppose x € ---Select-- v [We must show that x E ---Select--- 3. By definition of intersection, x EA ---Select--xEB-C. 4. By the set difference law, xEB-Select--- ✓x € C. [By step 3, we also know that x € ---Select--- .] 5. Thus, x E A and both x E B and x ---Select-- 6. By step 5, x EA and x E B, and thus x E---Select--- 7. Also by step 5, x EA and x € C, and thus x ---Select--- 9. Hence, by definition of subset, ✓by definition of ---Select--- [Why? If x E An C, then, by definition of ---Select--- 8. Thus x EAN B and x @ An C, and so, by the set difference law, x E---Select--- [This shows that every element in ---Select--- is in ---Select--- -Select-- by definition of ---Select--- ✓, x would be in C, which it is not.] ---Select--- V. V. ✓, then x E-Select--
Use an element argument to prove the following statement.. Proof: For all sets A, B, and C, An (B-C) ≤ (ANB) - (ANC). 1. Suppose A, B, and C are any sets. [To show that An (B-C) ≤ (ANB) - (An C), we must show that for every element x, if x € --Select--- 2. Suppose x € ---Select-- v [We must show that x E ---Select--- 3. By definition of intersection, x EA ---Select--xEB-C. 4. By the set difference law, xEB-Select--- ✓x € C. [By step 3, we also know that x € ---Select--- .] 5. Thus, x E A and both x E B and x ---Select-- 6. By step 5, x EA and x E B, and thus x E---Select--- 7. Also by step 5, x EA and x € C, and thus x ---Select--- 9. Hence, by definition of subset, ✓by definition of ---Select--- [Why? If x E An C, then, by definition of ---Select--- 8. Thus x EAN B and x @ An C, and so, by the set difference law, x E---Select--- [This shows that every element in ---Select--- is in ---Select--- -Select-- by definition of ---Select--- ✓, x would be in C, which it is not.] ---Select--- V. V. ✓, then x E-Select--
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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![**Use an element argument to prove the following statement.**
For all sets \(A\), \(B\), and \(C\), \(A \cap (B - C) \subseteq (A \cap B) - (A \cap C)\).
**Proof:**
1. Suppose \(A\), \(B\), and \(C\) are any sets.
\[ \text{[To show that } A \cap (B - C) \subseteq (A \cap B) - (A \cap C), \text{ we must show that for every element } x, \text{ if } x \in \ldots, \text{ then } x \in \ldots]. \]
2. Suppose \(x \in \text{Select}\).
\[ \text{[We must show that } x \in \text{Select} \ldots]. \]
3. By definition of intersection, \(x \in A \text{ and } x \in B - C\).
4. By the set difference law, \(x \in B \text{ and } x \notin C\).
\[ \text{[By step 3, we also know that } x \in \text{Select} \ldots.] \]
5. Thus, \(x \in A \text{ and both } x \in B \text{ and } x \notin C\).
6. By step 5, \(x \in A \text{ and } x \in B\), and thus \(x \in \text{Select}\) by definition of \(\text{Select}\).
7. Also by step 5, \(x \in A \text{ and } x \notin C\), and thus \(x \in \text{Select}\) by definition of \(\text{Select}\).
\[ \text{[Why? If } x \in A \cap C, \text{ then, by definition of } \text{Select}, \text{ } x \text{ would be in } C, \text{ which it is not.]} \]
8. Thus \(x \in A \cap B \text{ and } x \notin A \cap C\), and so, by the set difference law, \(x \in \text{Select](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa0e38307-1ade-44bc-b712-aaeda4c58098%2F329bb073-d745-4167-97c1-f0ce488a6ba8%2Fw26wirh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Use an element argument to prove the following statement.**
For all sets \(A\), \(B\), and \(C\), \(A \cap (B - C) \subseteq (A \cap B) - (A \cap C)\).
**Proof:**
1. Suppose \(A\), \(B\), and \(C\) are any sets.
\[ \text{[To show that } A \cap (B - C) \subseteq (A \cap B) - (A \cap C), \text{ we must show that for every element } x, \text{ if } x \in \ldots, \text{ then } x \in \ldots]. \]
2. Suppose \(x \in \text{Select}\).
\[ \text{[We must show that } x \in \text{Select} \ldots]. \]
3. By definition of intersection, \(x \in A \text{ and } x \in B - C\).
4. By the set difference law, \(x \in B \text{ and } x \notin C\).
\[ \text{[By step 3, we also know that } x \in \text{Select} \ldots.] \]
5. Thus, \(x \in A \text{ and both } x \in B \text{ and } x \notin C\).
6. By step 5, \(x \in A \text{ and } x \in B\), and thus \(x \in \text{Select}\) by definition of \(\text{Select}\).
7. Also by step 5, \(x \in A \text{ and } x \notin C\), and thus \(x \in \text{Select}\) by definition of \(\text{Select}\).
\[ \text{[Why? If } x \in A \cap C, \text{ then, by definition of } \text{Select}, \text{ } x \text{ would be in } C, \text{ which it is not.]} \]
8. Thus \(x \in A \cap B \text{ and } x \notin A \cap C\), and so, by the set difference law, \(x \in \text{Select
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