Equivalence of Definitions of Matroid Circuit Axioms/Condition 2 Implies Condition 4

Theorem
Let $S$ be a finite set.

Let $\mathscr C$ be a non-empty set of subsets of $S$ that satisfies the circuit axioms:

Then:
 * $\mathscr C$ is the set of circuits of a matroid $M = \struct{S, \mathscr I}$ on $S$

Proof
We will define a mapping $\rho$ associated with $\mathscr C$.

It will be shown that $\rho$ is the rank function of a matroid $M$ which has $\mathscr C$ as the set of circuits.

Let $\tuple{x_1, \ldots, x_q}$ be any ordered tuple of elements of $S$.

Define the ordered tuple $\map \theta {\tuple{x_1, \ldots, x_q}}$ by:
 * $\forall i \in \set{1, \ldots, q} : \map \theta {\tuple{x_1, \ldots, x_q}}_i = \begin{cases}

0 & : \exists C \in \mathscr C : x_i \in C \\ 1 & : \text {otherwise} \end{cases}$

Define a mapping $t$ from the set of ordered tuple of $S$ by:
 * $\map t {\tuple{x_1, \ldots, x_q}} = \ds \sum_{i = 1}^q \map \theta {\tuple{x_1, \ldots, x_q}}_i$

Lemma 1
Let $\tuple{x_1, \ldots, x_q}$ be any ordered tuple of elements of $S$.

Let $\pi$ be any permutation of $\tuple{x_1, \ldots, x_q}$.

Then:
 * $\map t {\tuple{x_1, \ldots, x_q}} = \map t {\tuple{x_{\map \pi 1}, \ldots, x_{\map \pi q}}}$

Proof
Define a mapping $\rho : \powerset S \to \Z$ by:
 * $\forall A \subseteq S$:
 * $\map \rho A = \begin{cases}

0 & : \text{if } A = \O \\ \map {\rho'} {\tuple{x_1, \ldots, x_q} } & : \text{if } A = \set{x_1, \ldots, x_q} \end{cases}$

From Lemma 1 then $\rho$ is well-defined.

Lemma 2
Let $X \subseteq S$ and $y \in S \setminus X$.

Then:
 * $\map \rho {X \cup \set y} = \map \rho X$ $\exists C \in \mathscr C : y \in C \subseteq X$

We now show that $\rho$ satisfies the rank function axioms:

$\rho$ satisfies $(\text R 1)$
Follows from definition of $\rho$.

$\rho$ satisfies $(\text R 2)$
Let:
 * $X \subseteq S$ and $y \in S$.

$\rho$ satisfies $(\text R 3)$
Let:
 * $X \subseteq S$ and $y, z \in S$.

Let:
 * $\map \rho {X \cup \set y} = \map \rho {X \cup \set z} = \map \rho X$.

Let:
 * $X = {x_1, \ldots, x_q}$

Case 1 : $z \in X$
Let $y \in X$.

Then:
 * $\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set z}$

Since $\map \rho {X \cup \set z} = \map \rho X$, then:
 * $\map \rho {X \cup \set {y,z}} = \map \rho X$

Case 2 : $z = y$
Let $y = z$.

Then
 * $\map \rho {X \cup \set {y,z}} = \map \rho {X \cup \set y}$

Since $\map \rho {X \cup \set y} = \map \rho X$, then:
 * $\map \rho {X \cup \set {y,z}} = \map \rho X$

Case 3 : $z \neq y$ and $z \notin X$
Let $z \neq y$ and $z \notin X$.

Then $z \notin X \cup \set y$.

By Lemma 2:
 * $\exists C_z \in \mathscr C : z \in C_z \subseteq X$

From [] and []:
 * $z \in C_z \subseteq X \subseteq X \cup \set y$

From Lemma 2:
 * $\map \rho {X \cup \set{y, z}} = \map \rho {X \cup \set y} = \map \rho X$

Proof 2
Define a mapping $\theta : \powerset S \to \powerset S$ by:
 * $\forall X \subseteq S : \map \theta X = \set{x \in X : \nexists C \in \mathscr C : x \in C \subseteq X}$

Define a mapping $\theta : \powerset S \to \powerset S$ by:
 * $\forall X \subseteq S : \map \theta X = X \setminus \bigcup \set{C \in \mathscr C : C \subseteq X}$

Define a mapping $\rho : \powerset S \to \Z$ by:
 * $\forall X \subseteq S : \map \rho X = \size{\map \theta X}$