Union of Relative Complements of Nested Subsets

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

Then:


 * $\relcomp T S \cup \relcomp S R = \relcomp T R$

where $\complement$ denotes relative complement.

Phrased via Set Difference as Intersection with Relative Complement:


 * $\paren {T \setminus S} \cup \paren {S \setminus R} = T \setminus R$

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:


 * $T \setminus R = \paren {\paren {T \setminus S} \setminus R} \cup \paren {S \setminus R}$

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


 * $\paren {T \setminus S} \setminus R = T \setminus \paren {S \cup R} = T \setminus S$

Combining the above yields:


 * $T \setminus R = \paren {T \setminus S} \cup \paren {S \setminus R}$

Also see

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