Set Difference is Subset/Proof 1

Theorem

$S \setminus T \subseteq S$

Proof

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

The result follows from the definition of subset.

$\blacksquare$