Set Difference over Subset

Theorem

Let $A$, $B$, and $S$ be sets.

Let $A \subseteq B$.

Then:

$A \setminus S \subseteq B \setminus S$

Proof

 $\ds A \setminus S$ $=$ $\ds A \cap \map \complement S$ Set Difference as Intersection with Complement $\ds$ $\subseteq$ $\ds B \cap \map \complement S$ Corollary to Set Intersection Preserves Subsets $\ds$ $=$ $\ds B \setminus S$ Set Difference as Intersection with Complement

$\blacksquare$