For any A sets A, B, C and D, prove that:
`(AxxB) nn (CxxD)=(A nnC)xx(BnnD)`

Explanation

To prove that \((A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)\), we will show that each side is a subset of the other.

### Step 1: Show that \((A \times B) \cap (C \times D) \subseteq (A \cap C) \times (B \cap D)\)

Let \((x, y) \in (A \times B) \cap (C \times D)\). By the definition of intersection, this means that \((x, y) \in A \times B\) and \((x, y) \in C \times D\).

From \((x, y) \in A \times B\), we have:
- \(x \in A\)
- \(y \in B\)

From \((x, y) \in C \times D\), we have:
- \(x \in C\)
- \(y \in D\)

Since \(x\) is in both \(A\) and \(C\), we conclude that:
- \(x \in A \cap C\)

Similarly, since \(y\) is in both \(B\) and \(D\), we conclude that:
- \(y \in B \cap D\)

Thus, we have:
\[ (x, y) \in (A \cap C) \times (B \cap D) \]
This shows that:
\[ (A \times B) \cap (C \times D) \subseteq (A \cap C) \times (B \cap D) \]

### Step 2: Show that \((A \cap C) \times (B \cap D) \subseteq (A \times B) \cap (C \times D)\)

Let \((\alpha, \beta) \in (A \cap C) \times (B \cap D)\). By the definition of Cartesian product, this means that:
- \(\alpha \in A \cap C\)
- \(\beta \in B \cap D\)

From \(\alpha \in A \cap C\), we have:
- \(\alpha \in A\)
- \(\alpha \in C\)

From \(\beta \in B \cap D\), we have:
- \(\beta \in B\)
- \(\beta \in D\)

Thus, we can conclude that:
\[ (\alpha, \beta) \in A \times B \quad \text{and} \quad (\alpha, \beta) \in C \times D \]
This shows that:
\[ (\alpha, \beta) \in (A \times B) \cap (C \times D) \]
Therefore, we have:
\[ (A \cap C) \times (B \cap D) \subseteq (A \times B) \cap (C \times D) \]

### Conclusion

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

We conclude that:
\[ (A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D) \]