Equivalence of Definitions of Matroid Circuit Axioms/Lemma 2

Theorem
Let $S$ be a finite set.

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

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:
 * $\map t {x_1, \ldots, x_q} = \ds \sum_{i = 1}^q \map \theta {x_1, \ldots, x_q}_i$

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}$

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

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

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

We have:

Hence: