# Intersection with Set Difference is Set Difference with Intersection

## Theorem

Let $R, S, T$ be sets.

Then:

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

where:

$R \setminus S$ denotes set difference
$R \cap T$ denotes set intersection.

## Proof 1

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

Then:

 $\displaystyle \left({R \setminus S}\right) \cap T$ $=$ $\displaystyle \left({R \cap \complement \left({S}\right)}\right) \cap T$ Set Difference as Intersection with Complement $\displaystyle$ $=$ $\displaystyle \left({R \cap T}\right) \cap \complement \left({S}\right)$ Intersection is Commutative and Intersection is Associative $\displaystyle$ $=$ $\displaystyle \left({R \cap T}\right) \setminus S$ Set Difference as Intersection with Complement

$\blacksquare$

## Proof 2

 $\displaystyle$  $\displaystyle x \in \left({R \setminus S}\right) \cap T$ $\displaystyle$ $\iff$ $\displaystyle \left({x \in R \land x \notin S}\right) \land x \in T$ Definitions of Set Intersection and Set Difference $\displaystyle$ $\iff$ $\displaystyle \left({x \in R \land x \in T}\right) \land x \notin S$ Rules of Commutation and Association $\displaystyle$ $\iff$ $\displaystyle x \in \left({R \cap T}\right) \setminus S$ Definitions of Set Intersection and Set Difference

$\blacksquare$