# Set Difference Intersection with First Set is Set Difference

## Theorem

The intersection of the set difference with the first set is the set difference.

Let $S, T$ be sets.

Then:

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

## Proof 1

 $\displaystyle \left({S \setminus T}\right)$ $\subseteq$ $\displaystyle S$ Set Difference is Subset $\displaystyle \implies \ \$ $\displaystyle \left({S \setminus T}\right) \cap S$ $=$ $\displaystyle S \setminus T$ Intersection with Subset is Subset‎

$\blacksquare$

## Proof 2

 $\displaystyle \left({S \setminus T}\right) \cap S$ $=$ $\displaystyle \left({S \cap S}\right) \setminus T$ Intersection with Set Difference is Set Difference with Intersection $\displaystyle$ $=$ $\displaystyle S \setminus T$ Intersection is Idempotent

$\blacksquare$