Set difference between three sets I am trying to prove whether the following claim is true or false.

Given sets $X, Y, Z$, is the following equality of sets true or false.
$$(X \setminus Y) \setminus Z = (X \setminus Z) \setminus Y $$

I believe the claim is true.
My thinking is that the set on the LHS is simply all those elements of $X$ that are not in $Y$ and not in $Z$. The set on the RHS is also all those elements of $X$ that are not in $Z$ and not in $Y$. For example, if $X = \{1,2,3,4,5\}$, $Y = \{1,2,3\}$ and $Z = \{3,4\}$ then $(X \setminus Y) \setminus Z = \{5\}$ and $(X \setminus Z) \setminus Y = \{5\}$ as well.
I am having trouble proving the claim formally though (perhaps I am overthinking the question or missing something obvious). If anyone could help me out, that would be great!
Thanks!
 A: Hint:
Just write the difference of two sets $A$ and $B$ as intersection:


*

*$A \setminus B = A \cap \bar B$, where $\bar B$ means the complement of $B$.


Then it will be easy for you to prove your claim.
A: Indeed, the equivalence is true.  But remember, we cannot prove it is true for all sets $A, B, C$ by citing only one example in which the equivalence is true. (On the other hand,  we can prove a theorem false if we can provide a counterexample.)
We seek to prove that: $$(X - Y) - Z = (X - Z) - Y$$
I'll proceed by "element-chasing".  We show that if $x$ is an element of the set on the left-hand side, then x is an element of the set on the right-hand side, and vice-versa.  In fact,  we can do this strictly through bidirectional implications, so in the end, we can show that if $x$ is an element of the RHS, it is also an element of the LHS, as well.
\begin{align} x \in \Big((X-Y)-Z\Big) &\iff \Big(x \in (X-Y) \land x\notin Z)\Big)\tag{Def: setminus}\\ \\ 
&\iff \Big((x \in X \land x \notin Y) \land x\notin Z\Big) \tag{Def: setminus}\\ \\
&\iff \Big(x\in X \land (x \notin Y \land x\notin Z)\Big)\tag{associativity of $\land$}\\ \\
&\iff \Big(x \in X \land (x\notin Z \land x\notin Y)\Big)\tag{commutativity of $\land$}\\ \\
&\iff \Big((x\in X \land x\notin Z) \land x\notin Y\Big)\tag{associativity of $\land$}\\ \\
&\iff \Big( x\in (X-Z) \land x\notin Y\Big)\tag{Def: setminus}\\ \\
&\iff x \in \Big((X-Z)- Y\Big)\tag{Def: setminus}
\end{align}
Hence, $(X-Y)-Z = (X-Z)-Y$
