The expression you provided involves set operations, particularly complements and intersections. Here's how to interpret and simplify the expression step by step:

Given:

\[
(B \cup C)' \cap A
\]

Where:
- \( B \cup C \) is the union of sets \( B \) and \( C \), meaning all elements in either \( B \) or \( C \).
- \( (B \cup C)' \) is the complement of \( B \cup C \), meaning all elements **not** in \( B \cup C \).
- \( \cap A \) is the intersection with set \( A \), meaning we are looking for elements that are both in \( A \) and in \( (B \cup C)' \).

### Step-by-Step Simplification:
1. **Find the complement**: \( (B \cup C)' \) represents all elements that are **not** in the union of \( B \) and \( C \). This gives us the set of elements that are outside both \( B \) and \( C \).

2. **Intersection with \( A \)**: Now we find the intersection of \( A \) with the complement of \( B \cup C \), i.e., the elements that are in \( A \) but not in \( B \) or \( C \).

Thus, the simplified expression can be written as:

\[
(B \cup C)' \cap A = A \setminus (B \cup C)
\]

This means: all elements that are in \( A \) but **not** in \( B \) or \( C \).

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Would you like more details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How does the complement of a union differ from the union of complements?
2. Can the intersection of two sets ever be larger than either set?
3. How would the result change if we had \( (B \cap C)' \) instead of \( (B \cup C)' \)?
4. What is the result if \( A \subseteq B \cup C \)?
5. How does the distributive property of sets apply to intersections and unions?

**Tip**: The complement of a union is the intersection of the complements (De Morgan's Law). This can simplify complex expressions!

Understand how to solve the set theory expression (B ∪ C)’ ⋂ A with a step-by-step guide. Learn about complements, intersections, and how De Morgan's Law applies. This problem is ideal for high school students in Grades 9-12.

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