Independent Set can be Augmented by Larger Independent Set

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

Let $X, Y \in \mathscr I$ such that:
 * $\size X < \size Y$

Then there exists $Z \subseteq Y \setminus X$ such that:
 * $X \cup Z \in \mathscr I$
 * $\size{X \cup Z} = \size Y$

Proof
Let $\mathscr Z = \set {Z \subseteq Y \setminus X : X \cup Z \in \mathscr I}$

Note that $\O \in \mathscr Z $

So $\mathscr Z \neq \O$

Let $Z_0 \in \mathscr Z : \size {Z_0} = \max \set{ \size Z : Z \in \mathscr Z}$

Aiming for a contradiction, suppose:
 * $\size{ X \cup Z_0} < \size Y$

Then:
 * $\exists Y_0 \subseteq Y : \size{Y_0} = \size{ X \cup Z_0} + 1$

By matroid axiom $(I2)$:
 * $Y_0 \in \mathscr I$

By matroid axiom $(I3)$:
 * $\exists y \in Y_0 \setminus \paren{X \cup Z_0} : X \cup Z_0 \cup \set y \in \mathscr I$

By choice of $Z_0$ and $y$:
 * $Z_0 \cup \set y \subseteq Y \setminus X$

So:
 * $Z_0 \cup \set y \in \mathscr Z$

Then:

This contradicts the choice of $Z_0$.

It follows that:

From Finite Set Contains Subset of Smaller Cardinality:
 * $\exists Z \subseteq Z_0 : \size Z = \size Y - \size X$

Then: