Set Difference with Union is Set Difference

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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 universe.


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

$\blacksquare$