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.

For any ordered tuple $\tuple{x_1, \ldots, x_q}$ of elements of $S$, let $\map \theta {x_1, \ldots, x_q}$ be the ordered tuple defined by:
 * $\forall i \in \set{1, \ldots, q} : \map \theta {x_1, \ldots, x_q}_i = \begin{cases}

0 & : \exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_i}\\ 1 & : \text {otherwise} \end{cases}$

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

Lemma 1
Let $\rho : \powerset S \to \Z$ be the mapping defined by:
 * $\forall A \subseteq S$:
 * $\map \rho A = \begin{cases}

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

From Lemma 1:
 * $\rho$ is well-defined

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 $y \in S$

Let $X = \set{x_1, \ldots, x_q}$.

We have:

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

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

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

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

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

From Lemma 2:
 * $\exists C_z \in \mathscr C : z \in C_z \subseteq X \cup \set{z} \subseteq X \cup \set{y,z}$

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

From Equivalence of Definitions of Matroid Rank Axioms:
 * $\rho$ is the rank function of a matroid $M = \struct{S, \mathscr I}$ on $S$