For any three sets A, B, C, prove that: A × (B ∪ C) = (A × B) ∪ (A × C)

Let (a, b) ∈ A × (B ∪ C)

⇒ a ∈ A and b ∈ (B ∪ C)

⇒ (a ∈ A and b ∈ B) or (a ∈ A and b ∈ C)

⇒ (a, b) ∈ A × B or (a, b) ∈ A × C

⇒ (a, b) ∈ (A × B) ∪ (A × C)

⇒ A × (B ∪ C) ⊆ (A × B) ∪ (A × C)   ... (i)

Also, let (x, y) ∈ (A × B) ∪ (A × C)

⇒ (x, y) ∈ A × B or (x, y) ∈ A × C

⇒ (x ∈ A and y ∈ B) or (x ∈ A and y ∈ C)

⇒ x ∈ A and (y ∈ B or y ∈ C)

⇒ x ∈ A and y ∈ (B ∪ C)

⇒ (x, y) ∈ A × (B ∪ C)

⇒ (A × B) ∪ (A × C) ⊆ A × (B ∪ C)   ... (ii)

From (i) and (ii), we get A × (B ∪ C) = (A × B) ∪ (A × C).