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To solve the given expression \((A - B) \cup (A \cup B)' = B'\), let’s analyze each component step-by-step.

### Step-by-Step Solution

1. **Interpret \(A - B\):**
   \[
   A - B = A \cap B'
   \]
   This represents the elements in \(A\) but not in \(B\).

2. **Interpret \((A \cup B)'\):**
   \[
   (A \cup B)' = A' \cap B'
   \]
   This is the complement of the union of \(A\) and \(B\), meaning it contains elements that are in neither \(A\) nor \(B\).

3. **Interpret \((A - B) \cup (A \cup B)'\):**
   Substitute from steps 1 and 2:
   \[
   (A - B) \cup (A \cup B)' = (A \cap B') \cup (A' \cap B')
   \]
   Using the distributive property of union and intersection:
   \[
   = (A \cup A') \cap B' = U \cap B' = B'
   \]
   where \(U\) represents the universal set, as \(A \cup A' = U\).

### Conclusion
\[
(A - B) \cup (A \cup B)' = B'
\]
Thus, the equation holds true as both sides are equal.

Would you like further details on any step, or have any other questions?

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Here are some related questions to deepen understanding:

1. What is the difference between \(A - B\) and \(B - A\) in set operations?
2. How can we interpret \((A \cup B)'\) in a Venn diagram?
3. What properties of set operations were used in the solution above?
4. How would the solution change if we had \(A \cap B\) instead of \(A - B\)?
5. How does De Morgan's Law apply to set complements and unions?

**Tip:** When working with set expressions, always remember to apply De Morgan's Laws when dealing with complements of unions and intersections.

Explore a step-by-step solution to the set theory expression (A - B) ∪ (A ∪ B)' = B'. This problem involves key set operations like union, complement, and uses De Morgan's Laws and the distributive property. Suitable for high school students in Grades 10-12.

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