Prove that:

(i) (A ∪ B) × C = (A × C) ∪ (B × C)

Let (a, b) be an arbitrary element of (A ∪ B) × C.

Thus, we have:

$(a, b) \in(A \cup B) \times C$

$\Rightarrow a \in(A \cup B)$ and $b \in C$

$\Rightarrow(a \in A$ or $a \in B)$ and $b \in C$

$\Rightarrow(a \in A$ and $b \in C)$ or $(a \in B$ and $b \in C)$

$\Rightarrow(a, b) \in(A \times C)$ or $(a, b) \in(B \times C)$

$\Rightarrow(a, b) \in(A \times C) \cup(B \times C)$

$\therefore(A \cup B) \times C \subseteq(A \times C) \cup(B \times C)$          ...(i)

Again, let (x, y) be an arbitrary element of (A × C) ∪ (B × C).

Thus, we have:

$(x, y) \in(A \times C) \cup(B \times C)$

$\Rightarrow(x, y) \in(A \times C)$ or $(x, y) \in(B \times C)$

$\Rightarrow(x \in A \& y \in C)$ or $(x \in B \& y \in C)$

$\Rightarrow(x \in A$ or $x \in B)$ or $y \in C$

$\Rightarrow(x \in A \cup B) \& y \in C$

$\Rightarrow(x, y) \in(A \cup B) \times C$

$\therefore(A \times C) \cup(B \times C) \subseteq(A \cup B) \times C \quad \ldots$ (ii)

From (i) and (ii), we get:

(A ∪ B) × C = (A × C) ∪ (B × C)

(ii) (A ∩ B) × C = (A × C) ∩ (B×C)

Let (a, b) be an arbitrary element of (A ∩ B) × C.

Thus, we have:

$(a, b) \in(A \cap B) \times C$

$\Rightarrow a \in(A \cap B) \& b \in C$

$\Rightarrow(a \in A \& a \in B) \& b \in C$

$\Rightarrow(a \in A \& b \in C) \&(a \in B \& b \in C)$

$\Rightarrow(a, b) \in(A \times C) \&(a, b) \in(B \times C)$

$\Rightarrow(a, b) \in(A \times C) \cap(B \times C)$

$\therefore(A \cap B) \times C \subseteq(A \times C) \cap(B \times C) \quad \ldots$ (iii)

Again, let (x, y) be an arbitrary element of (A × C) ∩ (B × C).

Thus, we have:

$(x, y) \in(A \times C) \cap(B \times C)$

$\Rightarrow(x, y) \in(A \times C) \&(x, y) \in(B \times C)$

$\Rightarrow(x \in A \& y \in C) \&(x \in B \& y \in C)$

$\Rightarrow(x \in A \& x \in B) \& y \in C$

$\Rightarrow x \in(A \cap B) \& y \in C$

$\Rightarrow(x, y) \in(A \cap B) \times C$

$\therefore(A \times C) \cap(B \times C) \subseteq(A \cap B) \times C \quad \ldots$ (iv)

From (iii) and (iv), we get:

(A ∩ B) × C = (A × C) ∩ (B × C)