Set Difference is Subset

From ProofWiki
Jump to: navigation, search


Set difference is a subset of the first set:

$S \setminus T \subseteq S$

Proof 1

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

The result follows from the definition of subset.


Proof 2

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