For any sets A, B and C prove that:
$A \times(B \cap C)=(A \times B) \cap(A \times C)$
Given: A, B and C three sets are given.
Need to prove: $A \times(B \cap C)=(A \times B) \cap(A \times C)$
Let us consider, $(x, y)^{\in} A \times(B \cap C)$
$\Rightarrow x \in_{A}$ and $y \in_{(B \cap C)}$
$\Rightarrow x \in_{A}$ and $\left(y \in_{B}\right.$ and $\left.y \in_{C}\right)$
$\Rightarrow\left(x \in_{A}\right.$ and $\left.y \in_{B}\right)$ and $\left(x \in_{A}\right.$ and $\left.y \in C\right)$
$\Rightarrow(x, y)^{\in}(A \times B)$ and $(x, y)^{\in}(A \times C)$
$\Rightarrow(x, y)^{\in}(A \times B) \cap(A \times C)$
From this we can conclude that,
$\Rightarrow A \times(B \cap C) \subseteq(A \times B) \cap(A \times C) \cdots(1)$
Let us consider again, $(a, b)^{\in}(A \times B) \cap(A \times C)$
$\Rightarrow(a, b)^{\in}(A \times B)$ and $(a, b)^{\in}(A \times C)$
$\Rightarrow(\mathrm{a} \in \mathrm{A}$ and $\mathrm{b} \in \mathrm{B})$ and $(\mathrm{a} \in \mathrm{A}$ and $\mathrm{b} \in \mathrm{C})$
$\Rightarrow \mathrm{a}^{\in} \mathrm{A}$ and $\left(\mathrm{b}^{\in} \mathrm{B}\right.$ and $\left.\mathrm{b}^{\in} \mathrm{C}\right)$
$\Rightarrow \mathrm{a}^{\in} \mathrm{A}$ and $\mathrm{b}^{\in}(\mathrm{B} \cap \mathrm{C})$
$\Rightarrow(\mathrm{a}, \mathrm{b})^{\in} \mathrm{A} \times(\mathrm{B} \cap \mathrm{C})$
From this, we can conclude that,
$\Rightarrow(A \times B) \cap(A \times C) \subseteq A \times(B \cap C)-(2)$
Now by the definition of the set we can say that, from (1) and (2),
$A \times(B \cap C)=(A \times B) \cap(A \times C)[$ Proved $]$