Independent Subset is Base if Cardinality Equals Rank of Matroid

Theorem
Let $M = \struct {S, \mathscr I}$ be a matroid.

Let $\rho: \powerset S \to \Z$ be the rank function of $M$.

Let $B \in \mathscr I$ such that:
 * $\size B = \map \rho S$

Then:
 * $B$ is a base of $M$.

Proof
Let $Z \in \mathscr I$ such that:
 * $B \subseteq Z$

From Cardinality of Subset of Finite Set:
 * $\size B \le \size Z$

By definition of the rank function:
 * $\size Z \le \map \rho S$

Then:
 * $\size Z = \size B$

From the contrapositive statement of Cardinality of Proper Subset of Finite Set:
 * $B = Z$

It follows that $B$ is a maximal independent subset by definition.

That is, $B$ is a base by definition.