# Set Difference is Subset

## Theorem

Set difference is a subset of the first set:

$S \setminus T \subseteq S$

## Proof 1

 $\displaystyle x \in S \setminus T$ $\leadsto$ $\displaystyle x \in S \land x \notin T$ Definition of Set Difference $\displaystyle$ $\leadsto$ $\displaystyle x \in S$ Rule of Simplification

The result follows from the definition of subset.

$\blacksquare$

## Proof 2

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

$\blacksquare$