For any sets A, B and C prove that:

Given: A, B and C three sets are given.

Need to prove: $A \times(B \cup C)=(A \times B) \cup(A \times C)$

Let us consider, $(x, y)^{\in} A \times(B \cup C)$

$\Rightarrow x^{\in} A$ and $y \in(B \cup C)$

$\Rightarrow x^{\in}_{A}$ and $\left(y \in_{B}\right.$ or $\left.y \in_{C}\right)$

$\Rightarrow\left(x^{\in}_{A}\right.$ and $\left.y \in_{B}\right)$ or $\left(x^{\in}_{A}\right.$ and $\left.y \in_{C}\right)$

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

$\Rightarrow(x, y) \in(A \times B) \cup(A \times C)$ From this we can conclude that,

$\Rightarrow A \times(B \cup C) \subseteq(A \times B) \cup(A \times C) \cdots(1)$

Let us consider again, $(a, b)^{\in}(A \times B) \cup(A \times C)$

$\Rightarrow(a, b) \in_{(A \times B) \text { or }(a, b)} \in_{(A \times C)}$

$\Rightarrow\left(a^{E} A\right.$ and $\left.b^{E} B\right)$ or $\left(a^{E} A\right.$ and $\left.b^{E} C\right)$

$\Rightarrow a^{\in}_{A}$ and $\left(b^{\in} B\right.$ or $\left.b^{\in} C\right)$

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

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

From this, we can conclude that,

$\Rightarrow(A \times B) \cup(A \times C) \subseteq A \times(B \cup C) \cdots(2)$

Now by the definition of the set we can say that, from (1) and (2),

$A \times(B \cup C)=(A \times B) \cup(A \times C)[$ Proved $]$