Union of Relative Complements of Nested Subsets

Theorem
Let $R \subseteq S \subseteq T$ be sets with the indicated inclusions.

Then:


 * $\complement_T \left({S}\right) \cup \complement_S \left({R}\right) = \complement_T \left({R}\right)$

where $\complement$ denotes relative complement.

Phrased via Set Difference as Intersection with Relative Complement:


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

where $\setminus$ denotes set difference.

Proof
From Union with Set Difference:


 * $T = T \setminus S \cup S$

and therefore by Set Difference is Right Distributive over Union:


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

Now, by Set Difference with Union and Union with Superset is Superset:


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

Combining the above yields:


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

Also see

 * Set Difference is Subset of Union of Differences is the inclusion retained when the subset conditions are dropped