Let A={1,2},B={1,2,3,4},C={5,6}and D={5,6,7,8}. Verify that:
(i) `Axx(B nn C)=(AxxB)nn(AxxC)`.
(ii) `AxxC` is a subset of `BxxD`.

Explanation

To solve the problem, we need to verify two statements involving the sets A, B, C, and D.

Given:
- A = {1, 2}
- B = {1, 2, 3, 4}
- C = {5, 6}
- D = {5, 6, 7, 8}

### Part (i): Verify that \( A \times (B \cap C) = (A \times B) \cap (A \times C) \)

**Step 1: Find \( B \cap C \)**

The intersection of sets B and C is the set of elements that are common to both sets.

- B = {1, 2, 3, 4}
- C = {5, 6}

Since there are no common elements, we have:
\[ B \cap C = \emptyset \]

**Step 2: Find \( A \times (B \cap C) \)**

Now we compute \( A \times (B \cap C) \):
\[ A \times (B \cap C) = A \times \emptyset = \emptyset \]

**Step 3: Find \( A \times B \)**

Next, we calculate the Cartesian product \( A \times B \):
\[ A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)\} \]

**Step 4: Find \( A \times C \)**

Now we calculate \( A \times C \):
\[ A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\} \]

**Step 5: Find \( (A \times B) \cap (A \times C) \)**

Now we find the intersection of \( A \times B \) and \( A \times C \):
- \( A \times B = \{(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)\} \)
- \( A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\} \)

There are no common elements between these two sets, so:
\[ (A \times B) \cap (A \times C) = \emptyset \]

**Step 6: Compare both sides**

Now we compare both sides:
\[ A \times (B \cap C) = \emptyset \]
\[ (A \times B) \cap (A \times C) = \emptyset \]

Since both sides are equal, we conclude that:
\[ A \times (B \cap C) = (A \times B) \cap (A \times C) \]

### Part (ii): Verify that \( A \times C \subseteq B \times D \)

**Step 1: Find \( A \times C \)**

We already found:
\[ A \times C = \{(1, 5), (1, 6), (2, 5), (2, 6)\} \]

**Step 2: Find \( B \times D \)**

Now we calculate \( B \times D \):
- B = {1, 2, 3, 4}
- D = {5, 6, 7, 8}

Calculating \( B \times D \):
\[ B \times D = \{(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)\} \]

**Step 3: Check if \( A \times C \subseteq B \times D \)**

Now we check if every element of \( A \times C \) is in \( B \times D \):
- \( (1, 5) \) is in \( B \times D \)
- \( (1, 6) \) is in \( B \times D \)
- \( (2, 5) \) is in \( B \times D \)
- \( (2, 6) \) is in \( B \times D \)

Since all elements of \( A \times C \) are found in \( B \times D \), we conclude that:
\[ A \times C \subseteq B \times D \]

### Final Conclusion

Both parts of the problem are verified:
1. \( A \times (B \cap C) = (A \times B) \cap (A \times C) \) is true.
2. \( A \times C \subseteq B \times D \) is true.