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!