# Set Difference is Subset/Proof 1

## Theorem

$S \setminus T \subseteq S$

## Proof

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

The result follows from the definition of subset.

$\blacksquare$