Statement 1: If `AnnB=phi` then `n((AxxB)nn(BxxA))=0`
Statement 2: `A-(BuuC)=AnnB^'nnC^'`

Options

A. Statement 1: is True, Statement 2 is True , Statement 2 is a correct explanation for statement 1
B. Statement 1 is True, Statement 2 is True Statement 2 is NOT a correct explanation for Statement 1
C. Statement 1 is True, Statement 2 is False
D. Statement 1 is False, Statement 2 is True

Explanation

To solve the given problem, we will analyze both statements step by step.

### Statement 1: If \( A \cap B = \phi \) then \( n((A \times B) \cap (B \times A)) = 0 \)

1. **Understanding the Intersection**:
- The statement begins with the condition \( A \cap B = \phi \), which means that sets A and B have no elements in common.

2. **Defining Cartesian Products**:
- The Cartesian product \( A \times B \) consists of all ordered pairs \( (a, b) \) where \( a \in A \) and \( b \in B \).
- Similarly, \( B \times A \) consists of all ordered pairs \( (b, a) \) where \( b \in B \) and \( a \in A \).

3. **Finding the Intersection**:
- The intersection \( (A \times B) \cap (B \times A) \) would include pairs that are in both products. For a pair \( (a, b) \) to be in both \( A \times B \) and \( B \times A \), it must satisfy:
- \( a \in A \) and \( b \in B \) (from \( A \times B \))
- \( b \in B \) and \( a \in A \) (from \( B \times A \))
- However, since \( A \cap B = \phi \), there are no elements \( a \) and \( b \) such that both conditions can be satisfied simultaneously.

4. **Conclusion**:
- Therefore, the intersection \( (A \times B) \cap (B \times A) \) contains no elements, which implies that \( n((A \times B) \cap (B \times A)) = 0 \).
- Thus, Statement 1 is **True**.

### Statement 2: \( A - (B \cup C) = A \cap (B' \cap C') \)

1. **Understanding Set Difference**:
- The left-hand side \( A - (B \cup C) \) represents the elements in A that are not in either B or C.

2. **Using Complement**:
- The expression \( B' \) represents the complement of B, which includes all elements not in B.
- Similarly, \( C' \) includes all elements not in C.

3. **Finding the Intersection**:
- The right-hand side \( A \cap (B' \cap C') \) includes elements that are in A and also not in B and not in C.

4. **Equivalence**:
- Both sides express the same condition: the elements in A that are not in B or C.
- Therefore, we can conclude that \( A - (B \cup C) = A \cap (B' \cap C') \) is indeed true.

5. **Conclusion**:
- Thus, Statement 2 is also **True**.

### Final Conclusion:
- Both Statement 1 and Statement 2 are true, but Statement 2 does not serve as a correct explanation for Statement 1.