For any two sets A and B, the values of `[(A-B)uuB]^(C )` is equal to

Options

A. `A^(C )nnB^(C )`
B. `AuuB`
C. `A-B`
D. `B-A`

Answer

To solve the problem, we need to find the value of \((A - B) \cup B^c\) where \(A\) and \(B\) are any two sets, and \(B^c\) is the complement of set \(B\).

### Step-by-Step Solution:

1. **Understand the Expression**:
We need to evaluate \((A - B) \cup B^c\). Here, \(A - B\) represents the elements that are in set \(A\) but not in set \(B\), and \(B^c\) represents all elements that are not in set \(B\).

2. **Define the Sets**:
- Let \(U\) be the universal set containing all possible elements.
- \(A - B = \{x \in A \mid x \notin B\}\)
- \(B^c = \{x \in U \mid x \notin B\}\)

3. **Visualize with a Venn Diagram**:
- Draw a Venn diagram with two circles: one for set \(A\) and another for set \(B\).
- The area representing \(A - B\) will be the part of circle \(A\) that does not overlap with circle \(B\).
- The area representing \(B^c\) will include everything outside circle \(B\).

4. **Combine the Sets**:
- The union \((A - B) \cup B^c\) will include:
- All elements in \(A\) that are not in \(B\) (the pink region).
- All elements that are outside of \(B\) (the yellow region).

5. **Identify the Resulting Set**:
- The union of these two areas will cover all elements that are either in \(A\) but not in \(B\) or outside of \(B\).
- This means that the resulting set will include all elements that are either in \(A\) or not in \(B\).

6. **Conclusion**:
- Therefore, the expression \((A - B) \cup B^c\) can be simplified to \(A \cup B^c\).

### Final Answer:
\[ (A - B) \cup B^c = A \cup B^c \]