Set Difference with Union is Set Difference

Theorem

The set difference between two sets is the same as the set difference between their union and the second of the two sets:

Let $S, T$ be sets.

Then:

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

Proof

Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universal set.

\(\ds \paren {S \cup T} \setminus T\)

\(=\)

\(\ds \paren {S \cup T} \cap \overline T\)

Set Difference as Intersection with Complement

\(\ds \)

\(=\)

\(\ds \paren {S \cap \overline T} \cup \paren {T \cap \overline T}\)

Intersection Distributes over Union

\(\ds \)

\(=\)

\(\ds \paren {S \setminus T} \cup \paren {T \setminus T}\)

Set Difference as Intersection with Complement

\(\ds \)

\(=\)

\(\ds \paren {S \setminus T} \cup \O\)

Set Difference with Self is Empty Set

\(\ds \)

\(=\)

\(\ds S \setminus T\)

Union with Empty Set

$\blacksquare$