Equivalence of Definitions of Matroid Circuit Axioms/Lemma 1

Theorem
Let $S$ be a finite set.

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

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

Define the ordered tuple $\map \theta {x_1, \ldots, x_q}$ 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}$

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

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

Proof
It is sufficient to show that:
 * $\forall 1 \le i \le q-1 : \map t {x_1, \ldots, x_i, x_{i + 1}, \ldots, x_q} = \map t {x_1, \ldots, x_{i + 1}, x_i, \ldots, x_q}$

By definition of $t$, we have:
 * $\map t {x_1, \ldots, x_i, x_{i + 1}, \ldots, x_q} = \sum_{j = 1}^{i - 1} \map \theta {x_1, \ldots, x_q}_j + \map \theta {x_1, \ldots, x_q}_i + \map \theta {x_1, \ldots, x_q}_{i + 1} + \sum_{j = {i + 2} }^q \map \theta {x_i + 2, \ldots, x_q}_j$

Similarly:
 * $\map t {x_1, \ldots, x_{i + 1}, x_i, \ldots, x_q} = \sum_{j = 1}^{i - 1} \map \theta {x_1, \ldots, x_q}_j + \map \theta {x_1, \ldots, x_q}_{i + 1} + \map \theta {x_1, \ldots, x_q}_i + \sum_{j = {i + 2} }^q \map \theta {x_i + 2, \ldots, x_q}_j$

By definition of $\theta$:
 * $\sum_{j = {i + 2} }^q \map \theta {x_i + 2, \ldots, x_q}_j$ is independent of the order of the tuple

So it is sufficient to show that:
 * $\forall 1 \le i \le q-1 : \map t {x_1, \ldots, x_i, x_{i + 1}} = \map t {x_1, \ldots, x_{i + 1}, x_i}$