# Set Difference is Subset of Union of Differences

## Theorem

Let $R, S, T$ be sets.

Then:

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

where:

$S \subseteq T$ denotes subset
$S \setminus T$ denotes set difference
$S \cup T$ denotes set union.

## Proof

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

 $\displaystyle R \setminus S$ $=$ $\displaystyle R \cap \overline S$ Set Difference as Intersection with Complement $\displaystyle$ $=$ $\displaystyle \left({R \cap \overline S}\right) \cap \mathbb U$ Intersection with Universe $\displaystyle$ $=$ $\displaystyle \left({R \cap \overline S}\right) \cap \left({\overline T \cup T}\right)$ Union with Complement $\displaystyle$ $=$ $\displaystyle \left({R \cap \overline S \cap \overline T}\right) \cup \left({T \cap R \cap \overline S}\right)$ Intersection Distributes over Union $\displaystyle$ $\subseteq$ $\displaystyle \left({R \cap \overline T}\right) \cup \left({T \cap \overline S}\right)$ Union of Intersections‎ $\displaystyle$ $=$ $\displaystyle \left({R \setminus T}\right) \cup \left({T \setminus S}\right)$ Set Difference as Intersection with Complement

$\blacksquare$