Set Difference with Union

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Theorem

Let $R, S, T$ be sets.


Then:

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

where:

$R \setminus S$ denotes set difference
$R \cup T$ denotes set union.


Proof

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


\(\displaystyle \left({R \cup T}\right) \setminus \left({S \cup T}\right)\) \(=\) \(\displaystyle \left({R \cup T}\right) \cap \overline {\left({S \cup T}\right)}\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \left({R \cup T}\right) \cap \left({\overline S \cap \overline T}\right)\) De Morgan's Laws: Complement of Union
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({R \cup T}\right) \cap \overline T}\right) \cap \overline S\) Intersection is Associative and Intersection is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left({\left({R \cup T}\right) \setminus T}\right) \setminus S\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \left({R \setminus T}\right) \setminus S\) Set Difference with Union is Set Difference

$\Box$


Then:

\(\displaystyle R \setminus \left({S \cup T}\right)\) \(=\) \(\displaystyle R \cap \overline {\left({S \cup T}\right)}\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle R \cap \left({\overline S \cap \overline T}\right)\) De Morgan's Laws: Complement of Union
\(\displaystyle \) \(=\) \(\displaystyle \left({R \cap \overline S}\right) \cap \overline T\) Intersection is Associative
\(\displaystyle \) \(=\) \(\displaystyle \left({R \setminus S}\right) \setminus T\) Set Difference as Intersection with Complement

$\Box$


Then:

\(\displaystyle R \setminus \left({S \cup T}\right)\) \(=\) \(\displaystyle R \setminus \left({T \cup S}\right)\) Union is Commutative
\(\displaystyle \) \(=\) \(\displaystyle \left({R \setminus T}\right) \setminus S\) from above

$\blacksquare$


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