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Given below is an attempted proof of a result. Proof First, we show that $A \subseteq(A \cup B)-B$. Let $x \in A$. Since $A \cap B=\emptyset$, it follows that $x \notin B$. Therefore, $x \in A \cup B$ and $x \notin B$; so $x \in(A \cup B)-B$. Thus $A \subseteq(A \cup B)-B$. Next, we show that $(A \cup B)-B \subseteq A$. Let $x \in(A \cup B)-B$. Then $x \in A \cup B$ and $x \notin B$. From this, it follows that $x \in A$. Hence $(A \cup B)-B \subseteq A$. (a) What result is being proved above? (b) What change (or changes) in this proof would make it better (from your point of view)?

   Given below is an attempted proof of a result.
Proof First, we show that $A \subseteq(A \cup B)-B$. Let $x \in A$. Since $A \cap B=\emptyset$, it follows that $x \notin B$. Therefore, $x \in A \cup B$ and $x \notin B$; so $x \in(A \cup B)-B$. Thus $A \subseteq(A \cup B)-B$.
Next, we show that $(A \cup B)-B \subseteq A$. Let $x \in(A \cup B)-B$. Then $x \in A \cup B$ and $x \notin B$. From this, it follows that $x \in A$. Hence $(A \cup B)-B \subseteq A$.
(a) What result is being proved above?
(b) What change (or changes) in this proof would make it better (from your point of view)?
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Mathematical Proofs: A Transition to Advanced Mathematics
Mathematical Proofs: A Transition to Advanced Mathematics
Gary Chartrand,… 3rd Edition
Chapter 4, Problem 85 ↓

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Step 1: The result being proved above is that $A \subseteq (A \cup B) - B$ and $(A \cup B) - B \subseteq A$.  Show more…

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Given below is an attempted proof of a result. Proof First, we show that $A \subseteq(A \cup B)-B$. Let $x \in A$. Since $A \cap B=\emptyset$, it follows that $x \notin B$. Therefore, $x \in A \cup B$ and $x \notin B$; so $x \in(A \cup B)-B$. Thus $A \subseteq(A \cup B)-B$. Next, we show that $(A \cup B)-B \subseteq A$. Let $x \in(A \cup B)-B$. Then $x \in A \cup B$ and $x \notin B$. From this, it follows that $x \in A$. Hence $(A \cup B)-B \subseteq A$. (a) What result is being proved above? (b) What change (or changes) in this proof would make it better (from your point of view)?
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Key Concepts

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Set Operations
Understanding basic set operations such as union, difference, and intersection is crucial. The union of two sets combines all elements from both sets, while the set difference removes elements belonging to a specified set. These operations are the building blocks for many arguments and proofs in set theory.
Disjoint Sets
Disjoint sets have no elements in common; their intersection is empty. This property is essential in proofs where the absence of shared elements directly influences the set relationships, as it can simplify arguments by ruling out potential overlaps between sets.
Set Equality
Set equality asserts that two sets are the same if and only if every element of one set is an element of the other, and vice versa. This concept is fundamental in set theory and is typically demonstrated by proving mutual subset relations—showing that each set is contained within the other.
Double Inclusion Method
The double inclusion method is a proof technique where one establishes the equality of two sets by proving both that the first set is a subset of the second and that the second is a subset of the first. This method is widely used because it breaks the overall proof into two manageable parts, each addressing one direction of the containment.

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The following are two proofs that for all sets A and B, A - B ⊆ A. The first is less formal, and the second is more formal. Fill in the blanks. a) Proof: Suppose A and B are any sets. To show that A - B ⊆ A, we must show that every element in A - B is in A. But any element in A - B is in A and not in B (by definition of A - B). In particular, such an element is in A. b) Proof: Suppose A and B are any sets and x ∈ A - B. We must show that x ∈ A. By definition of set difference, x ∈ A and x ∉ B. In particular, x ∈ A [which is what was to be shown]. a) Proof option choices: 1st blank choices: a) B b) A - B c) A ∪ B d) A ∩ B 2nd blank choices: a) A b) A - B c) A ∪ B d) A ∩ B 3rd blank choice: a) A b) B c) A - B d) A ∪ B 4th blank choice: a) A b) B c) A ∪ B d) A ∩ B b) Proof option choices: 1st option blank: a) x ∈ A b) x ∈ B c) x ∈ B - A d) x ∈ A ∩ B 2nd option blank: a) x ∈ A b) x ∈ B c) x ∈ A ∩ B d) x ∈ A ∪ B 3rd option blank: a) x ∈ A b) x ∈ B c) x ∈ A ∩ B d) x ∈ A ∪ B 4th option blank: a) x ∈ A ∪ B b) x ∈ A ∩ B c) x ∈ A d) x ∈ B

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