Set Difference is not Associative

Theorem
For any three sets $R, S, T$:
 * $R \setminus \left({S \setminus T}\right)$ does not, in general, equal $\left({R \setminus S}\right) \setminus T$

where $R \setminus S$ etc. denotes set difference.

From Set Difference with Union, we have:
 * $\left({R \setminus S}\right) \setminus T = R \setminus \left({S \cup T}\right)$

From Set Difference with Set Difference is Union of Set Difference with Intersection, we have:
 * $R \setminus \left({S \setminus T}\right) = \left({R \setminus S}\right) \cup \left({R \cap T}\right)$

These appear to be different.

In fact:


 * $\left({R \setminus S}\right) \setminus T \subseteq R \setminus \left({S \setminus T}\right)$

Thus it is clear that set difference is not associative.

The expression:
 * $\left({R \setminus S}\right) \setminus T = R \setminus \left({S \setminus T}\right)$

holds exactly when $R \cap T = \varnothing$.

Proof
We assume a universe $\mathbb U$ such that $R, S, T \subseteq \mathbb U$.

We have the identity Set Difference as Intersection with Complement:
 * $R \setminus S = R \cap \overline S$

where $\overline S$ is the set complement of $S$: $\overline S = \mathbb U \setminus S$.

Thus we can represent the two expressions as follows.

The first one is easy enough:

The second one is more involved:

As can be seen, this full expansion of the second expression can be expressed:
 * $R \setminus \left({S \setminus T}\right) = \left({\left({R \setminus S}\right) \setminus T}\right) \cup \left({R \cap T}\right)$

Hence the first result, from Subset of Union.

It directly follows from Intersection with Null that:
 * $R \setminus \left({S \setminus T}\right) = \left({R \setminus S}\right) \setminus T \iff R \cap T = \varnothing$

Also see

 * Intersection is Associative
 * Union is Associative
 * Symmetric Difference is Associative