Equivalence of Definitions of Matroid Circuit Axioms/Lemma 3

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 satisfying the matroid rank axioms 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 $M = \struct{S, \mathscr I}$ be the matroid corresponding to the rank function $\rho$.

Then:
 * $\mathscr C$ is the set of circuits of $M$.

Proof
Let $\mathscr C_M$ be the set of circuit of the matroid $M$.

$\mathscr C_M$ is a subset of $\mathscr C$
Let $C \in \mathscr C_M$.

From Lemma 4:
 * $\exists C_1 \in \mathscr C : C_1 \subseteq C$

From Lemma 5:
 * $\exists C' \in \mathscr C_M : C' \subseteq C_1$

From Subset Relation is Transitive:
 * $C' \subseteq C$

By the minimality of a circuit:
 * $C' = C$

By definition of set equality:
 * $C = C_1$

Hence:
 * $C \in \mathscr C$

It follows that $\mathscr C_M \subseteq \mathscr C$.

$\mathscr C$ is a subset of $\mathscr C_M$
Let $C \in \mathscr C$.

From Lemma 5:
 * $\exists C_1 \in \mathscr C_M : C_1 \subseteq C$

From Lemma 4:
 * $\exists C' \in \mathscr C : C' \subseteq C_1$

From Subset Relation is Transitive:
 * $C' \subseteq C$

By circuit axiom $(\text C 2)$:
 * $C' = C$

By definition of set equality:
 * $C = C_1$

Hence:
 * $C \in \mathscr C_M$.

It follows that $\mathscr C \subseteq \mathscr C_M$.

By definition of set equality:
 * $\mathscr C = \mathscr C_M$