Solution:
We need to Prove,
$A=(A \cap B) \cup(A-B)$
Proof: Let $x \in A$
We need to show
$X \in(A \cap B) \cup(A-B)$
In Case I, $X \in(A \cap B)$ $\Rightarrow X \in(A \cap B) \subset(A \cup B) \cup(A-B)$
In Case II, $X \notin A \cap B$ $\Rightarrow X \notin B$ or $X \notin A$ $\Rightarrow X \notin B(X \notin A)$
$\Rightarrow X \notin A-B \subset(A \cup B) \cup(A-B)$ $\therefore A \subset(A \cap B) \cup(A-B)(i)$
In Case II,
As a result, it can be concluded that, $A \cap B \subset A$ and $(A-B) \subset A$ Therefore, $(A \cap B) \cup(A-B) \subset A$ (ii)
Now equating (i) and (ii),
$A=(A \cap B) \cup(A-B)$
We also need to show,
Let’s suppose,
$\begin{array}{l}
X \in A \cup(B-A) \\
X \in A \text { or } X \in(B-A)
\end{array}$
$\begin{array}{l}
\Rightarrow X \in A \text { or }(X \in B \text { and } X \notin A) \\
\Rightarrow(X \in A \text { or } X \in B) \text { and }(X \in A \text { and } X \notin A) \\
\Rightarrow X \in(B \cup A) \\
\therefore A \cup(B-A) \subset(A \cup B)(i i i)
\end{array}$
As per the question,
We need to prove:
$\begin{array}{l}
(A \cup B) \subset A \cup(B-A) \\
\text { Let } y \in A \cup B \\
Y \in A \text { or } y \in B
\end{array}$
$\begin{array}{l}
(y \in A \text { or } y \in B) \text { and }(X \in A \text { and } X \notin A) \\
\Rightarrow y \in A \text { or }(y \in B \text { and } y \notin A) \\
\Rightarrow y \in A \cup(B-A)
\end{array}$
Therefore, $A \cup B \subset A \cup(B-A)$ (iv)
$\therefore$ from equations (iii) and (iv), we obtain:
$A \cup(B-A)=A \cup B$