Theorem 1(i) (For any three set `A;B;C`; prove that `Axx(BuuC)=(AxxB)uu(AxxC))`

Options

A. `(AxxB)uu(AxxC)`
B. `(AuuB)xx(AuuC)`
C. both (a) and (b)
D. none of these

Answer

To prove the theorem \( A \times (B \cup C) = (A \times B) \cup (A \times C) \), we will show that both sides of the equation are equal by demonstrating that each side is a subset of the other.

### Step 1: Prove \( A \times (B \cup C) \subseteq (A \times B) \cup (A \times C) \)

1. Let \( (x, y) \) be an arbitrary element of \( A \times (B \cup C) \).
2. By the definition of Cartesian product, this means \( x \in A \) and \( y \in (B \cup C) \).
3. Since \( y \in (B \cup C) \), it follows that \( y \in B \) or \( y \in C \) (by the definition of union).
4. Therefore, we have two cases to consider:
- **Case 1**: If \( y \in B \), then \( (x, y) \in A \times B \).
- **Case 2**: If \( y \in C \), then \( (x, y) \in A \times C \).
5. In either case, \( (x, y) \) belongs to \( (A \times B) \cup (A \times C) \).
6. Hence, we conclude that \( A \times (B \cup C) \subseteq (A \times B) \cup (A \times C) \).

### Step 2: Prove \( (A \times B) \cup (A \times C) \subseteq A \times (B \cup C) \)

1. Let \( (x, y) \) be an arbitrary element of \( (A \times B) \cup (A \times C) \).
2. By the definition of union, this means \( (x, y) \in A \times B \) or \( (x, y) \in A \times C \).
3. If \( (x, y) \in A \times B \), then \( x \in A \) and \( y \in B \).
4. If \( (x, y) \in A \times C \), then \( x \in A \) and \( y \in C \).
5. In both cases, since \( y \) is either in \( B \) or in \( C \), it follows that \( y \in (B \cup C) \).
6. Therefore, we have \( x \in A \) and \( y \in (B \cup C) \), which implies \( (x, y) \in A \times (B \cup C) \).
7. Thus, we conclude that \( (A \times B) \cup (A \times C) \subseteq A \times (B \cup C) \).

### Conclusion

Since we have shown both inclusions:
1. \( A \times (B \cup C) \subseteq (A \times B) \cup (A \times C) \)
2. \( (A \times B) \cup (A \times C) \subseteq A \times (B \cup C) \)

We can conclude that:
\[ A \times (B \cup C) = (A \times B) \cup (A \times C) \]