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:

$\left({S \cup T}\right) \setminus T = S \setminus T$


Proof

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


\(\displaystyle \left({S \cup T}\right) \setminus T\) \(=\) \(\displaystyle \left({S \cup T}\right) \cap \overline T\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \left({S \cap \overline T}\right) \cup \left({T \cap \overline T}\right)\) Intersection Distributes over Union
\(\displaystyle \) \(=\) \(\displaystyle \left({S \setminus T}\right) \cup \left({T \setminus T}\right)\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \left({S \setminus T}\right) \cup \varnothing\) Set Difference with Self is Empty Set
\(\displaystyle \) \(=\) \(\displaystyle S \setminus T\) Union with Empty Set

$\blacksquare$