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

(N/A) To show: $A = (A \cap B) \cup (A - B)$
Let $x \in A.$
We know that $x$ must either be in $B$ or not in $B$.
Case $I$: If $x \in B$,then $x \in A \cap B$,so $x \in (A \cap B) \cup (A - B).$
Case $II$: If $x \notin B$,then $x \in A - B$,so $x \in (A \cap B) \cup (A - B).$
Thus,$A \subseteq (A \cap B) \cup (A - B).$ ..........$(1)$
Since $(A \cap B) \subseteq A$ and $(A - B) \subseteq A$,their union must also be a subset of $A$.
Thus,$(A \cap B) \cup (A - B) \subseteq A.$ ..........$(2)$
From $(1)$ and $(2)$,$A = (A \cap B) \cup (A - B).$
To show: $A \cup (B - A) = A \cup B$
Let $x \in A \cup (B - A).$
This implies $x \in A$ or $(x \in B \text{ and } x \notin A).$
By distributive law,this is $(x \in A \text{ or } x \in B) \text{ and } (x \in A \text{ or } x \notin A).$
Since $(x \in A \text{ or } x \notin A)$ is always true,we get $x \in A \cup B.$
Thus,$A \cup (B - A) \subseteq A \cup B.$ ..........$(3)$
Conversely,let $y \in A \cup B.$
This implies $y \in A$ or $y \in B.$
If $y \in A$,then $y \in A \cup (B - A).$
If $y \notin A$ and $y \in B$,then $y \in B - A$,so $y \in A \cup (B - A).$
Thus,$A \cup B \subseteq A \cup (B - A).$ ..........$(4)$
From $(3)$ and $(4)$,$A \cup (B - A) = A \cup B.$