<Dl> <Dd> (A ∪ B) × (C ∪ D) ≠ (A × C) ∪ (B × D) (\ displaystyle (A \ cup B) \ times (C \ cup D) \ neq (A \ times C) \ cup (B \ times D)) </Dd> </Dl> <Dd> (A ∪ B) × (C ∪ D) ≠ (A × C) ∪ (B × D) (\ displaystyle (A \ cup B) \ times (C \ cup D) \ neq (A \ times C) \ cup (B \ times D)) </Dd> <P> In fact, we have that: </P> <Dl> <Dd> (A × C) ∪ (B × D) = ((A ∖ B) × C) ∪ ((A ∩ B) × (C ∪ D)) ∪ ((B ∖ A) × D) (\ displaystyle (A \ times C) \ cup (B \ times D) = ((A \ setminus B) \ times C) \ cup ((A \ cap B) \ times (C \ cup D)) \ cup ((B \ setminus A) \ times D)) </Dd> </Dl>

Let an denote the cartesian product of a set a with itself n times that is