# Proof using the symmetric difference

I just started my Analysis course, and I have little idea about the method for proving these things.

The symmetric difference of two sets $X,Y$ is defined as $$X\Delta Y=(X\setminus Y)\cup(Y\setminus X)$$ Show that $$X\Delta Z \subset (X\Delta Y) \cup (Y\Delta Z)$$ What I have tried is to somehow prove that $X\Delta Z$ can be gotten from rearranging the terms of the right hand side of the expression, but didn't achieve anything.

Hint: If $x \in X \Delta Z$, then either $x \in X$ and $x \not\in Z$ or $x \in Z$ and $x \not\in X$.
Let's assume without loss of generality that $x \in X$ and $x \not\in Z$ then consider two cases:
Case 1: $x \in Y$. If $x \in Y$, then $x \in Y$ and $x \not\in Z$, so $x \in Y \Delta Z$. Therefore in this case $x \in (X \Delta Y) \cup (Y \Delta Z)$.
Case 2: $x \not\in Y$. Then $x \in X$ and $x \not\in Y$, so...