Set Intersection Preserves Subsets/Corollary/Proof 1

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

Then:
 * $A \subseteq B \implies A \cap S \subseteq B \cap S$

Proof
Let $A \subseteq B$, and let $S$ be any set.

From the Set Intersection Preserves Subsets:
 * $A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

for arbitrary sets $S$ and $T$.

Substituting $S$ for $T$:


 * $A \subseteq B, \ S \subseteq S \implies A \cap S \subseteq B \cup S$

From Subset is Subset of Itself, $S \subseteq S$ for all sets $S$.

Hence the result:
 * $A \subseteq B \implies A \cap S \subseteq B \cap S$