For any sets A, B and C prove that:

Solution:

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

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

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

$\Rightarrow x \in_{A}$ and $y \in(B-C)$

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

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

$\Rightarrow(x, y)^{\in}(A \times B)$ and $(x, y) \notin(A \times C)$

$\Rightarrow(\mathrm{x}, \mathrm{y})^{\in}(\mathrm{A} \times \mathrm{B})-(\mathrm{A} \times \mathrm{C})$

From this we can conclude that

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

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

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

$\Rightarrow(\mathrm{a} \in \mathrm{A}$ and $\mathrm{b} \in \mathrm{B})$ and $(\mathrm{a} \in \mathrm{A}$ and $\mathrm{b} \notin \mathrm{C})$

$\Rightarrow \mathrm{a}^{\in} \mathrm{A}$ and $\left(\mathrm{b} \in_{\mathrm{B}}\right.$ and $\left.\mathrm{b} \notin \mathrm{C}\right)$

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

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

From this, we can conclude that,

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

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

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