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

Corollary to De Morgan's Laws: Difference with Intersection

Let $S, T_1, T_2$ be sets.

Suppose that $T_1 \subseteq S$.

Then:

$S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {T_1 \setminus T_2}$

Proof

\(\ds S \setminus \paren {T_1 \cap T_2}\)

\(=\)

\(\ds \paren {S \setminus T_1} \cup \paren {S \setminus T_2}\)

De Morgan's Laws: Difference with Intersection

\(\ds \)

\(=\)

\(\ds \paren {S \setminus T_1} \cup \paren {\paren {\paren {S \setminus T_1} \cup \paren {S \cap T_1} } \setminus T_2}\)

Set Difference Union Intersection

\(\ds \)

\(=\)

\(\ds \paren {S \setminus T_1} \cup \paren {\paren {S \setminus T_1} \setminus T_2} \cup \paren {\paren {S \cap T_1} \setminus T_2}\)

Set Difference is Right Distributive over Union

\(\ds \)

\(=\)

\(\ds \paren {S \setminus T_1} \cup \paren {\paren {S \setminus T_1} \setminus T_2} \cup \paren {T_1 \setminus T_2}\)

Intersection with Subset is Subset: $T_1 \subseteq S$

\(\ds \)

\(=\)

\(\ds \paren {S \setminus T_1} \cup \paren {T_1 \setminus T_2}\)

Set Difference Union First Set is First Set

$\blacksquare$