Given A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6}.

Step 1: Find B ∩ C. B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = φ (empty set, since no common elements)

Step 2: Find A × (B ∩ C). A × (B ∩ C) = A × φ = φ

Step 3: Find A × B. A × B = {(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4)}

Step 4: Find A × C. A × C = {(1,5), (1,6), (2,5), (2,6)}

Step 5: Find (A × B) ∩ (A × C). Since A × B has second elements from {1,2,3,4} and A × C has second elements from {5,6}, there are no common ordered pairs. (A × B) ∩ (A × C) = φ

Conclusion: A × (B ∩ C) = φ = (A × B) ∩ (A × C). Hence verified.

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

Step 1: Find A × C. A × C = {(1,5), (1,6), (2,5), (2,6)}

Step 2: Find B × D. B × 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 each element of A × C belongs to B × D.

• (1,5) ∈ B × D (since 1 ∈ B and 5 ∈ D) ✓
• (1,6) ∈ B × D (since 1 ∈ B and 6 ∈ D) ✓
• (2,5) ∈ B × D (since 2 ∈ B and 5 ∈ D) ✓
• (2,6) ∈ B × D (since 2 ∈ B and 6 ∈ D) ✓

Conclusion: Every element of A × C is in B × D. Therefore, A × C ⊆ B × D. Hence verified.