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 $\tuple{x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}$

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

Let:
 * $a = \map t {x_1, \ldots, x_{i - 1}}$
 * $a_1 = \map t {x_1, \ldots, x_{i - 1}, x_i} = a + \map \theta {x_1, \ldots, x_{i - 1}, x_i}_i$
 * $a_2 = \map t {x_1, \ldots, x_{i - 1}, x_{i + 1}} = a + \map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i$
 * $a_{12} = \map t {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}} = a_1 + \map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1}$
 * $a_{21} = \map t {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i} = a_2 + \map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}_{i+1}$

We consider the following cases:
 * $\begin{array}{c|cccc}

\text{Case} & \map \theta {x_1, \ldots, x_{i - 1}, x_i}_i & \map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i & \map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} \\ \hline \text{Case } 1 & 1 & 1 & 1 \\ \text{Case } 2 & 1 & 1 & 0 \\ \text{Case } 3 & 1 & 0 & 1 \\ \text{Case } 4 & 1 & 0 & 0 \\ \text{Case } 5 & 0 & 1 & 1 \\ \text{Case } 6 & 0 & 1 & 0 \\ \text{Case } 7 & 0 & 0 & 1 \\ \text{Case } 8 & 0 & 0 & 0 \\ \end{array}$

Case 1
Let:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i}_i = 1$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 1$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 1$

We have: $a_1 = a + 1 = a_2$

$\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}_{i+1} = 0$

By definition of $\theta$:
 * $\exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}$

and
 * $\nexists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i}$

Hence: $x_{i + 1} \in C$

It follows that:
 * $\exists C \in \mathscr C : x_{i + 1} \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}$

This contradicts the assumption that:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 1$

Hence: $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}_{i+1} = 1$

It follows that:
 * $a_{12} = a_1 + 1 = a_2 + 1 = a_{21}$

Case 2
Let:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i}_i = 1$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 1$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 0$

We have: $a_1 = a + 1 = a_2$

By definition of $\theta$:
 * $\exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}$

and
 * $\nexists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i}$

Hence: $x_{i + 1} \in C$

It follows that:
 * $\exists C \in \mathscr C : x_{i + 1} \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}$

Hence: $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}_{i+1} = 0$

It follows that:
 * $a_{12} = a_1 = a_2 = a_{21}$

Cases 3 and 7
Common to cases 3 and 7 are the conditions $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 0$ and $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 1$.

Let:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 0$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 1$

By definition of $\theta$:
 * $\exists C \in \mathscr C : x_i \in C_1 \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i+1}}$

From Subset Relation is Transitive:
 * $\exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1}$

By definition of $\theta$:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 0$

This contradicts the assumption that:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 1$

Hence these cases can't occur.

Case 4
Let:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i}_i = 0$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 1$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 0$

We have:
 * $a_1 = a$ and $a_2 = a + 1$

By definition of $\theta$:
 * $\exists C_1 \in \mathscr C : x_{i+1} \in C_1 \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i+1}}$

and
 * $\nexists C \in \mathscr C : x_{i+1} \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i+1}}$

Hence: $x_i \in C_1$

By definition of $\theta$:
 * $\exists C_2 \in \mathscr C : x_i \in C_2 \subseteq \set{x_1, \ldots, x_{i - 1}, x_i}$

Hence:
 * $x_i \in C_1 \cap C_2$

and
 * $x_{i+1} \in C_2 \setminus C_1$

By matroid circuit axiom $(\text C 3)$:
 * $\exists C_3 \in \mathscr C : x_{i+1} \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set {x_i}$

From Union of Subsets is Subset:
 * $C_1 \cup C_2 \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i+1}}$

From Set Difference over Subset:
 * $\paren{C_1 \cup C_2} \setminus \set {x_i} \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i+1}} \setminus \set {x_i} = \set{x_1, \ldots, x_{i - 1}, x_{i+1}}$

From Subset Relation is Transitive:
 * $\exists C_3 \in \mathscr C : x_{i+1} \in C_3 \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i+1}}$

By definition of $\theta$:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 0$

This contradicts the assumption that:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 1$

Hence this case can't occur.

Case 5
Let:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i}_i = 0$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 1$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 1$

We have:
 * $a_1 = a$ and $a_2 = a + 1$

By definition of $\theta$:
 * $\exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i}$

From Subset Relation is Transitive:
 * $\exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i+1}, x_i}$

By definition of $\theta$:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}_{i+1} = 0$

It follows that:
 * $a_{12} = a_1 + 1 = a + 1 = a_2 + 1 = a_{21}$

Case 6
Let:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i}_i = 1$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 0$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 0$

We have:
 * $a_1 = a + 1$ and $a_2 = a$

$\map \theta {x_1, \ldots, x_{i - 1}, x_{i+1}, x_i}_{i+1} = 0$.

By definition of $\theta$:
 * $\exists C_1 \in \mathscr C : x_i \in C_1 \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i+1}, x_i}$

and
 * $\nexists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i}$

Hence: $x_{i+1} \in C_1$

By definition of $\theta$:
 * $\exists C_2 \in \mathscr C : x_{i+1} \in C_2 \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i+1}}$

Hence:
 * $x_{i+1} \in C_1 \cap C_2$

and
 * $x_i \in C_1 \setminus C_2$

By matroid circuit axiom $(\text C 3)$:
 * $\exists C_3 \in \mathscr C : x_i \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set {x_{i+1}}$

From Union of Subsets is Subset:
 * $C_1 \cup C_2 \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i+1}}$

From Set Difference over Subset:
 * $\paren{C_1 \cup C_2} \setminus \set {x_{i+1}} \subseteq \set{x_1, \ldots, x_{i - 1}, x_i, x_{i+1}} \setminus \set {x_{i+1}} = \set{x_1, \ldots, x_{i - 1}, x_i}$

From Subset Relation is Transitive:
 * $\exists C_3 \in \mathscr C : x_i \in C_3 \subseteq \set{x_1, \ldots, x_{i - 1}, x_i}$

By definition of $\theta$:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i}_i = 0$

This contradicts the assumption that:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 1$

Hence:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i+1}, x_i}_{i+1} = 1$

We have:
 * $a_{12} = a_1 + 0 = a + 1 = a_2 + 1 = a_{21}$

Case 8
Let:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i}_i = 0$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}}_i = 0$
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}}_{i+1} = 0$

We have:
 * $a_1 = a = a_2$

By definition of $\theta$:
 * $\exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_i}$

From Subset Relation is Transitive:
 * $\exists C \in \mathscr C : x_i \in C \subseteq \set{x_1, \ldots, x_{i - 1}, x_{i+1}, x_i}$

By definition of $\theta$:
 * $\map \theta {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}_{i+1} = 0$

We have:
 * $a_{12} = a_1 = a = a_2 = a_{21}$

The cases are exhaustive and in all possible cases:
 * $a_{12} = a_{21}$

That is:
 * $\map t {x_1, \ldots, x_{i - 1}, x_i, x_{i + 1}} = \map t {x_1, \ldots, x_{i - 1}, x_{i + 1}, x_i}$

The result follows.