Symmetric difference = Ø This is my first post on this forum :)
I've been trying for a while to solve this problem about symmetric difference of two sets:

Given two sets $A$ and $B$, we call symmetric difference of $A$ and $B$ the following:
$$A\mathrel{\triangle}B = (A\cup B)-(A\cap B)$$
Show that if $A\mathrel{\triangle} B = \emptyset$, then $(A\subset B) \vee (B\subset A)$.

I don’t know where to start; thank you very much for the help :)
 A: If $A \Delta B= \emptyset$ than $A \cup B=A\cap B$ so you have the thesis.
A: In fact, we can say something much stronger: If $A\mathrel{\triangle} B = \emptyset$, then $A=B$.
Suppose $x\in A$. Then if $x\notin B$, we have $x\in A\cup B$ and $x\notin A\cap B$, so $x\in A\mathrel{\triangle} B$, violating $A\mathrel{\triangle} B = \emptyset$. Thus $x\in B$.
Suppose $x\in B$. Then if $x\notin A$, we have $x\in A\cup B$ and $x\notin A\cap B$, so $x\in A\mathrel{\triangle} B$, violating $A\mathrel{\triangle} B = \emptyset$. Thus $x\in A$.
Thus $A=B$, and a fortiori $A\subseteq B\vee B\subseteq A$.

Another way to do this is to use propositional manipulation.
\begin{align}
x\notin A\mathrel{\triangle} B
&\iff\neg(x\in A\cup B\land x\notin A\cap B)\\
&\iff x\notin A\cup B\lor x\in A\cap B\\
&\iff\neg(x\in A\lor x\in B)\lor(x\in A\land x\in B)\\
&\iff(x\notin A\land x\notin B)\lor(x\in A\land x\in B)\\
&\iff(x\in A\leftrightarrow x\in B)\\
\end{align}
where in the last step we use that $\phi\leftrightarrow\psi$ is true iff $\phi,\psi$ are both true or both false. Applying $\forall x$ to each side then gives $$A\mathrel{\triangle} B = \emptyset\iff\forall x,x\notin A\mathrel{\triangle} B\iff \forall x,(x\in A\leftrightarrow x\in B)\iff A=B.$$
