**Problem Statement:**1. Let \(A\), \(B\) and \(C\) be sets. Prove that \(A \cup (B - C) = (A \cup B) - (C - A)\).**Explanation:**This problem requires you to prove an equality between two set expressions using set theory. - **Union (\(\cup\))**: Represents all elements that are in either of the sets.- **Difference (\(-\))**: Represents elements that are in the first set but not in the second set.To prove the given equation, you can typically follow these steps:1. **Express Elements**: Break down the elements belonging to each side of the equation.2. **Logical Arguments**: Use known properties and operations of sets (such as distributivity, associativity, and De Morgan’s laws) to manipulate and compare both expressions.### Proof:**Step 1: Analyze \(A \cup (B - C)\)**An element \(x\) is in \(A \cup (B - C)\) if and only if:- \(x \in A\) or- \(x \in (B - C)\) Which means:- \(x \in (B - C)\) implies \(x \in B\) and \(x \notin C\).Therefore:- \(x \in A\) or \((x \in B \land x \notin C)\).**Step 2: Analyze \((A \cup B) - (C - A)\)**An element \(x\) is in \((A \cup B) - (C - A)\) if and only if:- \(x \in (A \cup B)\) and- \(x \notin (C - A)\)Which means:- \(x \in (A \cup B)\) implies \(x \in A\) or \(x \in B\).- \(x \notin (C - A)\) implies \(x \notin C\) or \(x \in A\).Therefore:- \((x \in A \lor x \in B) \land (x \notin C \lor x \in A)\).**Step 3: Compare Both Sides**The expressions:1. \(A \cup (B - C)\) simplifies to \(

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Answered: 1. Let A, B abd C be sets. Prove that…

1. Let A, B abd C be sets. Prove that Au (B-C) = (AUB) - (C-A).

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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