If A, B and C are non-empty sets such that `AnnC=phi`, then what is `(AxxB)nn(CxxB)` equalt ot?

Options

A. `AxxC`
B. `AxxB`
C. `BxxC`
D. `phi`

Explanation

To solve the problem, we need to find the intersection of the Cartesian products of the sets \( A \) and \( B \) with \( C \) and \( B \). The given information is that \( A \cap C = \emptyset \).

Let's denote the Cartesian products as follows:
- \( A \times B \) is the set of all ordered pairs \( (a, b) \) where \( a \in A \) and \( b \in B \).
- \( C \times B \) is the set of all ordered pairs \( (c, b) \) where \( c \in C \) and \( b \in B \).

We want to find:
\[ (A \times B) \cap (C \times B) \]

### Step 1: Understand the intersection of the Cartesian products
The intersection \( (A \times B) \cap (C \times B) \) consists of all ordered pairs \( (x, y) \) such that:
- \( x \in A \) and \( y \in B \) (from \( A \times B \))
- \( x \in C \) and \( y \in B \) (from \( C \times B \))

### Step 2: Analyze the conditions
For an ordered pair \( (x, y) \) to be in both \( A \times B \) and \( C \times B \), the first component \( x \) must belong to both \( A \) and \( C \). However, we know from the problem statement that:
\[ A \cap C = \emptyset \]
This means there are no elements that are common to both sets \( A \) and \( C \).

### Step 3: Conclude the intersection
Since there are no elements \( x \) that can satisfy both conditions (being in \( A \) and in \( C \)), it follows that:
\[ (A \times B) \cap (C \times B) = \emptyset \]

### Final Answer
Thus, we conclude that:
\[ (A \times B) \cap (C \times B) = \emptyset \]