Let A,BA,B be sets. Then (A−B)∪(B−A)(A−B)∪(B−A) is equal to:

(A∩B)′∩(A∪B)(A∩B)′∩(A∪B)

(A∩B)∩(A∪B)′(A∩B)∩(A∪B)′

(A∪B)′∪(A∩B)(A∪B)′∪(A∩B)

(A∪B)∪(A∩B)′(A∪B)∪(A∩B)′

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We are asked to find which of the given expressions is equal to \( (A - B) \cup (B - A) \).

### Step 1: Simplifying \( (A - B) \cup (B - A) \)

- \( A - B \) means all the elements that are in \( A \) but not in \( B \).
- \( B - A \) means all the elements that are in \( B \) but not in \( A \).
  
Thus, \( (A - B) \cup (B - A) \) represents all the elements that are in \( A \) or in \( B \), but **not in both**. This is the **symmetric difference** of \( A \) and \( B \), often denoted as \( A \Delta B \).

\[
(A - B) \cup (B - A) = A \Delta B
\]

### Step 2: Analyze each option

#### Option 1: \( (A \cap B)' \cap (A \cup B) \)

- \( (A \cap B)' \) means the complement of \( A \cap B \), i.e., all elements **not** in both \( A \) and \( B \).
- \( (A \cup B) \) means all elements that are in \( A \) or in \( B \) or in both.

The intersection of \( (A \cap B)' \) with \( (A \cup B) \) represents the elements in \( A \cup B \) that are **not in both** sets \( A \) and \( B \). This is exactly the definition of \( A \Delta B \).

Thus, this option simplifies to \( A \Delta B \), which matches \( (A - B) \cup (B - A) \).

#### Option 2: \( (A \cap B) \cap (A \cup B)' \)

- \( (A \cup B)' \) is the complement of \( A \cup B \), i.e., all elements that are in neither \( A \) nor \( B \).
- \( A \cap B \) is the set of elements common to both \( A \) and \( B \).

The intersection of \( A \cap B \) with \( (A \cup B)' \) will always be the empty set because there are no elements that are both in \( A \cap B \) and outside \( A \cup B \). Thus, this expression is the **empty set**, which is not equal to \( A \Delta B \).

#### Option 3: \( (A \cup B)' \cup (A \cap B) \)

- \( (A \cup B)' \) is the complement of \( A \cup B \), i.e., the set of elements not in \( A \cup B \).
- \( A \cap B \) is the set of elements common to both \( A \) and \( B \).

The union of \( (A \cup B)' \) and \( A \cap B \) represents elements that are either outside \( A \cup B \) or inside both sets \( A \) and \( B \). This is **not** equivalent to the symmetric difference \( A \Delta B \).

#### Option 4: \( (A \cup B) \cup (A \cap B)' \)

- \( (A \cap B)' \) is the complement of \( A \cap B \), i.e., all elements **not** in both \( A \) and \( B \).
- \( (A \cup B) \) is the union of \( A \) and \( B \).

This expression includes all elements in \( A \cup B \) and all elements not in \( A \cap B \), which would result in the universal set \( U \) (everything), as it would contain all possible elements in or out of \( A \) and \( B \). Thus, this

Let A, B be sets. Then (A−B)∪(B−A) is equal to:

(A∩B)′∩(A∪B)
(A∩B)∩(A∪B)′
(A∪B)′∪(A∩B)
(A∪B)∪(A∩B)′

Explore the concept of the symmetric difference in set theory through the problem (A−B)∪(B−A). Learn how to identify equivalent expressions using set operations, complements, and De Morgan's laws. This solution is ideal for high school students studying set theory.

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