Equivalence of Definitions of Matroid Rank Axioms/Condition 1 Implies Condition 3

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 the rank axioms:

Then $\rho$ is the rank function of a matroid on $S$.

Lemma
Let:
 * $\mathscr I = \set{X \subseteq S : \map \rho X = \card X}$

It is to be shown that:
 * $\quad \mathscr I$ satisfies the matroid axioms

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

Matroid Axiom $(\text I 1)$
From Cardinality of Empty Set:
 * $\card \O = 0$

By rank axiom $(\text R 1)$:
 * $\map \rho \O = 0$

Hence:
 * $\map \rho \O = \card \O$

So:
 * $\O \in \mathscr I$

Hence: $M$ satisfies matroid axiom $(\text I 1)$.

Matroid Axiom $(\text I 2)$
Let
 * $X \in \mathscr I$


 * $\exists Y \subseteq X : Y \notin \mathscr I$

Let:
 * $Y_0 \subseteq X : \card {Y_0} = \max \set{\card Z : Z \subseteq X \land Z \notin \mathscr I}$

From lemma:
 * $\map \rho {Y_0} < \card {Y_0}$

As $\map \rho X = \card X$:
 * $Y_0 \ne X$

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

Let $y \in X \setminus Y_0$.

We have:

This is a contradiction.

So:
 * $\forall Y \subseteq X : Y \in \mathscr I$

Hence:
 * $M$ satisfies matroid axiom $(\text I 2)$.

Matroid Axiom $(\text I 3)$
Let
 * $U \in \mathscr I$
 * $V \subseteq S$
 * $\card U < \card V$

We prove the contrapositive statement:
 * $\paren{\forall x \in V \setminus U : U \cup \set x \notin \mathscr I} \implies V \notin \mathscr I$

Let:
 * $\forall x \in V \setminus U : U \cup \set x \notin \mathscr I$