# Set Difference is Right Distributive over Union

## Theorem

Let $R, S, T$ be sets.

Then:

$\paren {R \cup S} \setminus T = \paren {R \setminus T} \cup \paren {S \setminus T}$

where:

$R \setminus S$ denotes set difference
$R \cup T$ denotes set union.

## Proof

 $\displaystyle \paren {R \cup S} \setminus T$ $=$ $\displaystyle \paren {R \cup S} \cap \overline T$ Set Difference as Intersection with Complement $\displaystyle$ $=$ $\displaystyle \paren {R \cap \overline T} \cup \paren {S \cap \overline T}$ Intersection Distributes over Union $\displaystyle$ $=$ $\displaystyle \paren {R \setminus T} \cup \paren {S \setminus T}$ Set Difference as Intersection with Complement

$\blacksquare$