I'm self studying from this book "How to Prove it" by Velleman (3rd edition). I don't have access to a math professor, so I need a little help from the community. Please verify my proof.
Exercise 6 pag. 150
The symmetric difference of two sets $A$ and $B$ is the set $A \Delta B$ = $(A \setminus B) \cup (B \setminus A) = (A \cup B) \setminus (B \cap A)$. Prove that if $A \Delta B \subseteq A$ then $B \subseteq A$.
Proof. Let's note $A \Delta B \subseteq A$ as (1). In order to satisfy (1), an element $x$ must be a member of both $A$ and $B$. Now (1) can be rewritten as : $\forall x(x \in B\ and\ x \in A \implies x \in A)$ or $\forall x(x \notin A \implies x \notin B\ or\ x \notin A)$ (2)
Next, (2) can be rewritten as
$$ \left\{\begin{matrix} \forall x(x \notin A \implies x \notin B) (3) \\ \forall x(x \notin A \implies x \notin A) (4) \end{matrix}\right. $$
Evidently we can drop (4), and (3) can be rewritten as $\forall x(x \in B \implies x \in A)$ $\blacksquare $