Let A and B are two disjoint sets and N be the universal set then `A'uu((AuuB)nnB')` is equal to
(i) `phi`
(ii) `xi`
(iii) `A`
(iv) `B`

Options

A. `phi`
B. `xi`
C. `A`
D. `B`

Answer

To solve the problem, we need to analyze the expression \( A' \cup ((A \cup B) \cap B') \) step by step, given that \( A \) and \( B \) are disjoint sets and \( N \) is the universal set.

### Step 1: Understand the components of the expression
- \( A' \) is the complement of set \( A \) in the universal set \( N \). This means \( A' = N - A \).
- \( B' \) is the complement of set \( B \) in the universal set \( N \). This means \( B' = N - B \).
- Since \( A \) and \( B \) are disjoint, \( A \cap B = \phi \) (the empty set).

### Step 2: Simplify \( (A \cup B) \cap B' \)
- The expression \( A \cup B \) includes all elements in either \( A \) or \( B \).
- The intersection \( (A \cup B) \cap B' \) means we are looking for elements that are in \( A \cup B \) but not in \( B \).
- Since \( B' \) contains all elements not in \( B \), the intersection will only include elements from \( A \) because \( A \) and \( B \) are disjoint.
- Therefore, \( (A \cup B) \cap B' = A \).

### Step 3: Substitute back into the expression
Now we can substitute back into the original expression:
\[ A' \cup ((A \cup B) \cap B') = A' \cup A \]

### Step 4: Analyze \( A' \cup A \)
- The union \( A' \cup A \) includes all elements that are either in \( A \) or not in \( A \).
- This means it covers the entire universal set \( N \).

### Conclusion
Thus, we conclude that:
\[ A' \cup ((A \cup B) \cap B') = N \]
Since \( N \) represents the universal set, the answer is \( N \), which corresponds to option (ii) \( \xi \).