Equivalence of Definitions of Matroid Rank Axioms/Lemma 4

Theorem
Let $S$ be a finite set.

Let $\rho : \powerset S \to \Z$ be a mapping from the power set of $S$ to the integers.

Let $\rho$ satisfy definition 1 of the rank axioms:

Let $M = \struct{S, \mathscr I}$ be the matroid where:
 * $\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$

Let $\rho_M$ be the rank function of the matroid $M = \struct{S, \mathscr I}$.

Then $\rho_M = \rho$.

Proof
Let $X \subseteq S$.

By definition of the rank function:
 * $\map {\rho_M} X = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$

Let $Y_0 \subseteq X$:
 * $\card {Y_0} = \max \set{\card Y : Y \subseteq X, Y \in \mathscr I}$

We have:

So it remains to show:
 * $\map \rho {Y_0} = \map \rho X$

Case 1 : $Y_0 = X$
Let $Y_0 = X$.

Then:
 * $\map \rho {Y_0} = \map \rho X$

Case 2 : $Y_0 \ne X$
Let $Y_0 \ne X$.

Then:
 * $Y_0 \subsetneq X$

From Set Difference with Proper Subset:
 * $X \setminus Y_0 \ne \O$

By choice of $Y_0$:
 * $\forall y \in X \setminus Y_0 : Y_0 \cup \set y \notin \mathscr I$

That is:
 * $\forall y \in X \setminus Y_0 : \map \rho {Y_0 \cup \set y} \ne \card {Y_0 \cup \set y}$

From Lemma 3:
 * $\map \rho {Y_0} = \map \rho {Y_0 \cup X} = \map \rho X$

In either case:
 * $\map \rho {Y_0} = \map \rho X$

It follows that:
 * $\forall X \subseteq S : \map {\rho_M} X = \map \rho X$

Hence $\rho$ is the rank function of the matroid $M = \struct{S, \mathscr I}$.