# Set Intersection Preserves Subsets/Corollary

## Theorem

Let $A, B, S$ be sets.

Then:

$A \subseteq B \implies A \cap S \subseteq B \cap S$

## Proof 1

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

$A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$

for arbitrary sets $S$ and $T$.

Substituting $S$ for $T$:

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

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

Hence the result:

$A \subseteq B \implies A \cap S \subseteq B \cap S$

$\blacksquare$

## Proof 2

Recall the Factor Principles, themselves a corollary of the Praeclarum Theorema:

$\paren {p \implies q} \vdash \paren {p \land r} \implies \paren {q \land r}$

This is applied as:

 $\ds$  $\ds A \subseteq B$ $\ds$ $\leadsto$ $\ds \paren {x \in A \implies x \in B}$ Definition of Subset $\ds$ $\leadsto$ $\ds \paren {x \in A \land x \in S \implies x \in B \land x \in S}$ Factor Principles $\ds$ $\leadsto$ $\ds \paren {x \in A \cap S \implies x \in B \cap S}$ Definition of Set Intersection $\ds$ $\leadsto$ $\ds A \cap S \subseteq B \cap S$ Definition of Subset

$\blacksquare$