Lower Bound is Lower Bound for Subset

Theorem
Let $\left({S, \preceq}\right)$ be a preordered set.

Let $A, B$ be subsets of $S$ such that
 * $B \subseteq A$

Let $L$ be an element of $S$.

Let $L$ be a lower bound for $A$.

Then $L$ is a lower bound for $B$.

Proof
Let $L$ be a lower bound for $A$.

By definition of lower bound:
 * $\forall x \in A: L \preceq x$

By definition of subset:
 * $\forall x \in B: x \in A$

Hence:
 * $\forall x \in B: L \preceq x$

Thus by definition:
 * $L$ is a lower bound for $B$.