Equivalence relation by the symmetric difference of sets

If $A \Delta B \subseteq S$ and $B \Delta C \subseteq S$, then

$A \Delta C= (A \Delta B) \Delta (B \Delta C) \subseteq S$ as well.

We need to show theat $A\sim B$ and $B \sim C$ give $A \sim C$. This can easily be seen by visualising $A, B, C$ in a Venn diagram (try it yourself!) To put the Venn diagram proof formally, consider any element $x$ in $A$ but not in $C$. If $x$ is in $B$, it lies in the symmetric difference of $B$ and $C$ and so is in $S$. If $x$ is not in $B$, it lies in the symmetric difference of $A$ and $B$ and so is in $S$. By symmetry any element in $C$ but not $A$ is in $S$, completing the proof.

Hint:

By definition, $A\sim B$ means $A-B$ and $B-A\subset S$. So you have to show that, if $A-B, B-A, B-C, C-B\subset S$, then both $A-C$ and $C-A$ are subsets of $S$.

Consider first an element $x\in A-C$. Either it is in $B$, or it is not in $B$. What can you deduce from the hypotheses in each case?