# Intersection with Set Difference is Set Difference with Intersection

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## 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 |

$\blacksquare$