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

$\blacksquare$