Circuits of Matroid iff 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:
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 $\tuple{x_1, \ldots, x_q}$ 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 $(\text C 4)$:
:$\exists C_3 \in \mathscr C : x_{i+1} \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set {x_i}$
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 $(\text C 4)$:
:$\exists C_3 \in \mathscr C : x_i \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set {x_{i+1}}$
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.
Category:Circuits of Matroid iff Matroid Circuit Axioms