De Morgan's Laws (Set Theory)/Set Difference/Difference with Intersection/Corollary

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Corollary to De Morgan's Laws: Difference with Intersection

Let $S, T_1, T_2$ be sets.

Suppose that $T_1 \subseteq S$.


Then:

$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({T_1 \setminus T_2}\right)$


Proof

\(\displaystyle S \setminus \left({T_1 \cap T_2}\right)\) \(=\) \(\displaystyle \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)\) De Morgan's Laws: Difference with Intersection
\(\displaystyle \) \(=\) \(\displaystyle \left({S \setminus T_1}\right) \cup \left({\left({\left({S \setminus T_1}\right) \cup \left({S \cap T_1}\right)}\right) \setminus T_2}\right)\) Set Difference Union Intersection
\(\displaystyle \) \(=\) \(\displaystyle \left({S \setminus T_1}\right) \cup \left({\left({S \setminus T_1}\right) \setminus T_2}\right) \cup \left({\left({S \cap T_1}\right) \setminus T_2}\right)\) Set Difference is Right Distributive over Union
\(\displaystyle \) \(=\) \(\displaystyle \left({S \setminus T_1}\right) \cup \left({\left({S \setminus T_1}\right) \setminus T_2}\right) \cup \left({T_1 \setminus T_2}\right)\) Intersection with Subset is Subset: $T_1 \subseteq S$
\(\displaystyle \) \(=\) \(\displaystyle \left({S \setminus T_1}\right) \cup \left({T_1 \setminus T_2}\right)\) Set Difference Union First Set is First Set

$\blacksquare$