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:

 $\ds \paren {R \setminus S} \cap T$ $=$ $\ds \paren {R \cap \map \complement S} \cap T$ Set Difference as Intersection with Complement $\ds$ $=$ $\ds \paren {R \cap T} \cap \map \complement S$ Intersection is Commutative and Intersection is Associative $\ds$ $=$ $\ds \paren {R \cap T} \setminus S$ Set Difference as Intersection with Complement

$\blacksquare$

Proof 2

 $\ds$  $\ds x \in \paren {R \setminus S} \cap T$ $\ds \leadstoandfrom \ \$ $\ds$  $\ds \paren {x \in R \land x \notin S} \land x \in T$ Definition of Set Intersection and Definition of Set Difference $\ds \leadstoandfrom \ \$ $\ds$  $\ds \paren {x \in R \land x \in T} \land x \notin S$ Rule of Commutation and Rule of Association $\ds \leadstoandfrom \ \$ $\ds$  $\ds x \in \paren {R \cap T} \setminus S$ Definition of Set Intersection and Definition of Set Difference

$\blacksquare$