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
 * $X, Y \subseteq S$
 * $Y \subseteq X$

We prove the contrapositive statement:
 * $Y \notin \mathscr I \implies X \notin \mathscr I$

Let $Y \notin \mathscr I$.

From lemma:
 * $\map \rho Y < \card Y$

Let:
 * $X \setminus Y = \set{x_1, x_2, \dots, x_k}$

By rank axiom $(\text R 2)$:
 * $\map \rho {Y \cup \set{x_1}} \le \map \rho Y + 1 < \card Y + 1$

By repeated application of rank axiom $(\text R 2)$:
 * $\map \rho X = \map \rho {Y \cup \set{x_1, x_2, \dots, x_k}} < \card Y + k = \card X$

Hence:
 * $X \notin \mathscr I$

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