# Set Difference with Union is Set Difference

## 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

$\blacksquare$