Show that for any sets $\mathrm{A}$ and $\mathrm{B}$, $A=(A \cap B) \cup(A-B)$ and $A \cup(B-A)=(A \cup B).$

To show: $A=(A \cap B) \cup(A-B)$

Let $x \in A$

We have to show that $x \in(A \cap B) \cup(A-B)$

Case $I$

$x \in A \cap B$

Then, $x \in(A \cap B) \subset(A \cup B) \cup(A-B)$

Case $II$

$x \notin A \cap B$

$\Rightarrow x \notin A$ or $x \notin B$

$\therefore x \notin B[x \notin A]$

$\therefore x \notin A-B \subset(A \cup B) \cup(A-B)$

$\therefore A \subset(A \cap B) \cup(A-B)$         ...........$(1)$

It is clear that

$A \cap B \subset A$ and $(A-B) \subset A$

$\therefore(A \cap B) \cup(A-B) \subset A$           ..........$(2)$

From $(1)$ and $(2),$ we obtain

$A=(A \cap B) \cup(A-B)$

To prove: $A \cup(B-A) \subset A \cup B$

Let $x \in A \cup(B-A)$

$\Rightarrow x \in A$ or $(x \in B$ and $x \notin A)$

$ \Rightarrow (x \in A$ or $x \in B)$ and $(x \in A$ or $x \notin A)$

$\Rightarrow x \in(A \cup B)$

$\therefore A \cup(B-A) \subset(A \cup B) $       .........$(3)$

Next, we show that $(A \cup B) \subset A \cup(B-A)$

Let $y \in A \cup B$

$\Rightarrow y \in A$ or $y \in B$

$ \Rightarrow (y \in A$ or $y \in B)$ and $(y \in A{\rm{ }}$ or $y \notin A)$

$\Rightarrow y \in A$ or $(y \in B$ and $y \notin A)$

$\Rightarrow y \in A \cup(B-A)$

$\therefore A \cup B \subset A \cup(B-A)$      ...........$(4$)

Hence, from $(3)$ and $(4)$, we obtain $A \cup(B-A)=A \cup B$.