Equivalence of Definitions of Matroid Rank Axioms/Lemma 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:
 * $\forall Y \subseteq S: \map \rho Y \le \card Y$

Proof

 * $\exists Y \subseteq S : \map \rho Y > \card Y$
 * $\exists Y \subseteq S : \map \rho Y > \card Y$

Let:
 * $Y_0 \subseteq S : \card{Y_0} = \min \set{\card Y : \map \rho Y > \card Y}$

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

From Cardinality of Empty Set:
 * $\card \O = 0$

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

Hence:
 * $Y_0 \neq \O$

Let $y \in Y_0$.

Let $Y = Y_0 \setminus \set y$.

From Cardinality of Set Difference:
 * $\card Y = \card{Y_0} - 1$

By choice of $Y_0$:
 * $\map \rho Y \le \card Y$

By rank axiom $(\text R 2)$:
 * $\map \rho {Y_0} = \map \rho {Y \cup \set y} \le \map \rho Y + 1 \le \card Y + 1 = \card{Y_0}$

This contradicts the choice of $Y_0$:
 * $\card{Y_0} = \min \set{\card Y : \map \rho Y > \card Y}$

It follows that:
 * $\forall Y \subseteq S: \map \rho Y \le \card Y$