# Set Difference with Union

## 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

$\blacksquare$