Set Intersection Preserves Subsets

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

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

Proof
Let $A \subseteq B$ and $S \subseteq T$.

Then:

Now we invoke the Praeclarum Theorema of propositional logic:
 * $\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$

applying it as:
 * $\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \land x \in S \implies x \in B \land x \in T}\right)$

The result follows directly from the definition of set intersection:
 * $\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \cap S \implies x \in B \cap T}\right)$

and from the definition of subset:
 * $A \subseteq B, \ S \subseteq T \implies A \cap S \subseteq B \cap T$