Prove that if $A \mathop \triangle B \subseteq A$ then $B\subseteq A$ I'm having trouble with the logic in this proof and was wondering if anyone could point me in the right direction (if I'm wrong)?
Prove that if $A\mathop\triangle B\subseteq A$ then $B\subseteq A$. (Here $\triangle$ refers to the symmetric difference).
I started by using the definition of symmetric difference that $A\mathop\triangle B = (A\setminus B)\mathop\cup \mathop(B\setminus A)$. So $A\mathop\triangle B\subseteq A$ = $\forall\psi[(\psi\in A \wedge \psi \notin B) \vee (\psi \in B \wedge \psi \notin A) \rightarrow \psi \in A$].
Here is what I have for my proof:
Suppose $x \in B$. Suppose $x \notin A.$ Then since $x \in B$ and $A\mathop\triangle B\subseteq A$, it follows that $x \in A$. But this contradicts the fact that $x \notin A$, so we can conclude that $x \in A$. Since $x$ was an arbitrary element of $B$, it follows that $B\subseteq A$.
What I'm wondering is, is it enough to use universal instantiation on $x$ from the statement $\forall\psi[(\psi\in A \wedge \psi \notin B) \vee (\psi \in B \wedge \psi \notin A) \rightarrow \psi \in A$] given that $x \in B$ and $ x \notin A$ to get my contradiction? Also, should I be giving more information about the logic used in the proof, or is it ok to leave it to the reader? Thanks for the help!
 A: Your proof is fine: you need only show the inclusion you've shown. You chose (any) arbitrary $x$ such that $x \in B, x\notin A$, and you've reached a contradiction through your assumption that $x \in B \land x\notin A$. This implies (by definition, and perhaps you want to make this explicit) that $x \in A\triangle B$. But then since $A\triangle B \subseteq A, x\in A$. This is a contradiction which is realized whatever the $x$ satisfying the initial assumption is chosen. So the proof already shows that for any (all) $x$ such that $x \in B \land x\notin A \rightarrow x\in A \triangle B$, and since $A\triangle B \subseteq A,\;$ then $\; x\in A$. 
Universal instantiation would be redundant.
A: Without contradiction, just in case, now that your proof has been checked:
$$
B\subseteq A\cup B=A\cup (A\Delta B)\subseteq A\cup A= A
$$
A: Proof by contradiction
let $x\in B$ but $x\notin A $
hence $x\in A∆B $
hence $x\in A$
which is contradiction
hence $x\in A$
A: Throwing in my $0.02, I'd simply calculate
$$
\begin{align*}
& A \Delta B \subseteq A \\
\equiv & \;\;\;\;\;\text{"expand definition of $\;\subseteq\;$"} \\
& \langle \forall x : x \in A \Delta B : x \in A \rangle \\
\equiv & \;\;\;\;\;\text{"expand definition of $\;\Delta\;$"} \\
& \langle \forall x : x \in A \not\equiv x \in B : x \in A \rangle \\
(*) \;\; \equiv & \;\;\;\;\;\text{"logic: use negation of consequent in antecedent of (implicit) $\;\Rightarrow\;$"} \\
& \langle \forall x : \textrm{false} \not\equiv x \in B : x \in A \rangle \\
\equiv & \;\;\;\;\;\text{"logic: simplify range"} \\
& \langle \forall x : x \in B : x \in A \rangle \\
\equiv & \;\;\;\;\;\text{"definition of $\;\subseteq\;$"} \\
& B \subseteq A \\
\end{align*}
$$
and discover that also $B \subseteq A \;\Rightarrow\; A \Delta B \subseteq A$ holds.
I don't remember the name for the rule used in the key step $(*)$, but at least in proofs from Edsger W. Dijkstra c.s. it is fairly often used as a well-known law.  It might be in Gries's "A Logical Approach to Discrete Math".
