Proof through inclusion (A∆B) ∪ A = A ∪ B

We would like to prove that 
\[A\cup B= (A \Delta B)\cup A .\]
Note that $(A \Delta B)\cup A = [(A\cup B) -(A\cap B)]\cup A$. Since 
\[A\Delta B \subset A\cup B    \text{and}    A \subset A\cup B,\]
we have 
\[(A\Delta B ) \cup A\subset (A\cup B).        (1)\]
Next we prove 
\[(A\cup B) \subset  (A\Delta B ) \cup A.         \]
Let $x \in A\cup B$. We consider two cases:

(i) If $x\in A$, then $x\in (A\Delta B ) \cup A.$

(ii) If $x\not \in A$ then we must have $x\in B$. Hence $x\in (A\cup B)-(A\cap B)=A\Delta B.$ Therefore $x\in (A\Delta B)\cup A$.

Thus $x\in A\cup B$ implies $x \in (A\Delta B)\cup A$. This means 
\[(A\cup B) \subset (A\Delta B ) \cup A.    (2)\]

Finally (1) and (2) implies 
\[(A\cup B) =(A\Delta B ) \cup A. \]