
(Continued )
To prove this inclusion, we need to show that if t ∈ A × (B ∪ C), then t ∈ (A × B) ∪ (A × C). Because t is the
element of the Cartesian product of two sets, namely A and (B ∪ C), t has two coordinates. So, we can write
t =(x, y)withx ∈ A and y ∈ (B ∪ C). Thus, t =(x, y)withx ∈ A and either y ∈ B or y ∈ C. Therefore, t =(x, y)
with either x ∈ A and y ∈ B or x ∈ A and y ∈ C. This implies that t =(x, y) with either (x, y) ∈ A × B or
(x, y) ∈ A × C.
So, we can conclude that (x, y) ∈ (A × B) ∪ (A × C). So t ∈ (A × B) ∪ (A × C).
Second part: A × (B ∪ C) ⊇ (A × B) ∪ (A × C)
To prove this inclusion, we need to show that if t ∈ (A × B) ∪ (A ×