# Set Difference with Set Difference is Union of Set Difference with Intersection

## Theorem

Let $R, S, T$ be sets.

Then:

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

where:

$S \setminus T$ denotes set difference
$S \cup T$ denotes set union
$S \cap T$ denotes set intersection.

## Proof

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

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

$\blacksquare$