### Proposition Analysis**Problem:**Examine the provided "proof" of the given "proposition." Is this "proposition" true? If not, identify a counterexample.#### PropositionConsider sets $ A $, $ B $, and $ C $ such that $ A \subseteq B \cup C $. This implies $ A \subseteq B $ or $ A \subseteq C $.#### Provided ProofTo prove or disprove the proposition, follow these steps:1. **Assumption:** Let $ x $ be any arbitrary object and suppose $ x \in A $. 2. **Implication:** Since $ A \subseteq B \cup C $, it implies $ x \in B \cup C $. 3. **Definition of Union:** By the definition of the union, $ x \in B $ or $ x \in C $.4. **Conclusion:** Therefore, for any object $ x $, if $ x \in A $, then $ x \in B $ or $ x \in C $.5. **Subset Definition:** Hence, by the definition of a subset, we can conclude $ A \subseteq B $ or $ A \subseteq C $.#### Analysis and Error IdentificationThe provided "proof" contains a flaw. It incorrectly assumes that if $ x $ belongs to the union $ B \cup C $, then $ x $ necessarily belongs to either $ B $ or $ C $ independently for all elements in $ A $. However, there could be elements of $ A $ that belong to $ B \cup C $ but not entirely to one set $ B $ or $ C $.#### CounterexampleConsider the following sets:- $ A = \{1, 2\} $- $ B = \{1\} $- $ C = \{2\} $Here, $ A \subseteq B \cup C $ because every element of $ A $ is in the union of $ B $ and $ C $:\[ A = \{1, 2\} \text{ and } B \cup C = \{1, 2\} \]However, $ A $ is not a subset of $ B $ (since \(

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Description: ### Proposition Analysis**Problem:**Examine the provided "proof" of the given "proposition." Is this "proposition" true? If not, identify a counterexample.#### PropositionConsider sets $ A $, $ B $, and $ C $ such that \( A \subseteq B \cup C

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24. Find the mistake in the "proof" of the following "proposition." Is this "proposition true? If not, find a counterexample. "Proposition." Let A, B and C be sets and suppose that ACBỤC Then A s Bor ACC. "proof" Let x be any object and suppose that xEA. ThenxEBUC since ACBUC Thus, by the definition of union, xE Bor xe C. Therefore, for all objects x, ifEA, then xe Bor, for all objects x, if xEA, then xEC. Hence, by the definition of subset, AC Bor ACC.

24. Find the mistake in the "proof" of the following "proposition." Is this "proposition true? If not, find a counterexample. "Proposition." Let A, B and C be sets and suppose that ACBỤC Then A s Bor ACC. "proof" Let x be any object and suppose that xEA. ThenxEBUC since ACBUC Thus, by the definition of union, xE Bor xe C. Therefore, for all objects x, ifEA, then xe Bor, for all objects x, if xEA, then xEC. Hence, by the definition of subset, AC Bor ACC.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Concept explainers

Statements and Logic

Logic is the analysis of general patterns of thoughts without regard to any specific sense of context.

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Question

### Proposition Analysis

**Problem:**
Examine the provided "proof" of the given "proposition." Is this "proposition" true? If not, identify a counterexample.

#### Proposition
Consider sets $ A $, $ B $, and $ C $ such that $ A \subseteq B \cup C $. This implies $ A \subseteq B $ or $ A \subseteq C $.

#### Provided Proof
To prove or disprove the proposition, follow these steps:

1. **Assumption:**
Let $ x $ be any arbitrary object and suppose $ x \in A $.

2. **Implication:**
Since $ A \subseteq B \cup C $, it implies $ x \in B \cup C $.

3. **Definition of Union:**
By the definition of the union, $ x \in B $ or $ x \in C $.

4. **Conclusion:**
Therefore, for any object $ x $, if $ x \in A $, then $ x \in B $ or $ x \in C $.

5. **Subset Definition:**
Hence, by the definition of a subset, we can conclude $ A \subseteq B $ or $ A \subseteq C $.

#### Analysis and Error Identification

The provided "proof" contains a flaw. It incorrectly assumes that if $ x $ belongs to the union $ B \cup C $, then $ x $ necessarily belongs to either $ B $ or $ C $ independently for all elements in $ A $. However, there could be elements of $ A $ that belong to $ B \cup C $ but not entirely to one set $ B $ or $ C $.

#### Counterexample

Consider the following sets:
- $ A = \{1, 2\} $
- $ B = \{1\} $
- $ C = \{2\} $

Here, $ A \subseteq B \cup C $ because every element of $ A $ is in the union of $ B $ and $ C $:
\[ A = \{1, 2\} \text{ and } B \cup C = \{1, 2\} \]

However, $ A $ is not a subset of $ B $ (since \(

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