# Equivalence of Definitions of Matroid Circuit Axioms

## Theorem

Let $S$ be a finite set.

Let $\mathscr C$ be a non-empty set of subsets of $S$.

The following definitions for the Matroid Circuit Axioms are equivalent:

### Formulation 1

$\mathscr C$ satisfies the circuit axioms:

 $(\text C 1)$ $:$ $\ds \O \notin \mathscr C$ $(\text C 2)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2$ $(\text C 3)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \implies \exists C_3 \in \mathscr C : C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z$

### Formulation 2

$\mathscr C$ satisfies the circuit axioms:

 $(\text C 1)$ $:$ $\ds \O \notin \mathscr C$ $(\text C 2)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2$ $(\text C 4)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \land w \in C_1 \setminus C_2 \implies \exists C_3 \in \mathscr C : w \in C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z$

### Formulation 3

$\mathscr C$ satisfies the circuit axioms:

 $(\text C 1)$ $:$ $\ds \O \notin \mathscr C$ $(\text C 2)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2$ $(\text C 5)$ $:$ $\ds \forall X \subseteq S \land \forall x \in S:$ $\ds \paren {\forall C \in \mathscr C : C \nsubseteq X} \implies \paren {\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x}$

## Proof

### Formulation 1 implies Formulation 2

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.

It has only to be shown that circuit axiom $(\text C 4)$ is satisfied by $\mathscr C$.

Let:

 $\ds F = \leftset{\tuple{C, D, x, y} }$ $:$ $\ds C, D \in \mathscr C \land C \neq D$ $\ds$ $\land$ $\ds x \in C \cap D \land y \in C \setminus D$ $\ds$ $\land$ $\ds \nexists C' \in \mathscr C : y \in C' \subseteq \paren{C \cup D} \setminus \set x \mathop {\rightset {} }$

To show that $\mathscr C$ satisfies circuit axiom $(\text C 4)$, it needs to be shown that $F = \O$.

Aiming for a contradiction, suppose :

$F \neq \O$

Let $\tuple{C_1, C_2, z, w} \in F$ :

$\size{C_1 \cup C_2} = \min \set{\size{C \cup D} : \tuple{C, D, x, y} \in F}$
$\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$

By assumptiom:

$w \notin C_3$

Consider $C_3 \cap \paren{C_2 \setminus C_1}$.

$C_3 \nsubseteq C_1$
$C_3 \cap \paren{C_2 \setminus C_1} \neq \O$

Let $x \in C_3 \cap \paren{C_2 \setminus C_1}$.

We have:

$x \in C_3 \cap C_2$

and

$z \in C_2 \setminus C_3$

and

$w \notin C_2 \cup C_3$
$C_2 \cup C_3 \subseteq \C_1 \cup C_2$

Since $w \notin C_2 \cup C_3$:

$C_2 \cup C_3$ is a proper subset of $C_1 \cup C_2$

By circuit axiom $(\text C 3)$ and the minimality of $C_1 \cup C_2$:

$\exists C_4 \in \mathscr C : z \in C_4 \subseteq \paren{C_2 \cup C_3} \setminus \set{x}$

Now consider $C_1$ and $C_4$, we have:

$z \in C_1 \cap C_4$

Since $w \notin C_2 \cup C_3$ then:

$w \in C_1 \setminus C_4$

We have:

$C_4 \subset C_2 \cup C_3 \subset C_1 \cup C_2$
$C_1 \cup C_4 \subseteq C_1 \cup C_2$

Recall $x \in C_3 \cap \paren{C_2 \setminus C_1}$, then:

$x \in C_2$

and

$x \notin C_1$

Since $C_4 \subseteq \paren{C_2 \cup C_3} \setminus \set{x}$, then:

$x \notin C_4$

It follows that:

$x \notin C_1 \cup C_4$

and

$x \in C_1 \cup C_2$

Hence:

$C_1 \cup C_4$ is a proper subset of $C_1 \cup C_2$

By circuit axiom $(\text C 3)$ and the minimality of $C_1 \cup C_2$:

$\exists C_5 \in \mathscr C : w \in C_5 \subseteq \paren{C_1 \cup C_4} \setminus \set{z}$

Since $C_1 \cup C_4 \subset C_1 \cup C_2$, then we have found $C_5$ such that:

$w \in C_5 \subseteq \paren{C_1 \cup C_2} \setminus \set{z}$

This contradicts the fact that $\tuple{C_1, C_2, z, w} \in F$.

It follows that circuit axiom $(\text C 4)$ is satisfied.

$\Box$

### Formulation 2 implies Formulation 1

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 4)$.

We need to show that $\mathscr C$ satisfies circuit axiom:

 $(\text C 3)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \implies \exists C_3 \in \mathscr C : C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z$

Let $C_1, C_2 \in \mathscr C : C_1 \ne C_2$.

Let $z \in C_1 \cap C_2$.

$C_2 \nsubseteq C_1$

By definition of subset and set difference:

$\exists w \in C_2 \setminus C_1$
$\exists C_3 \in \mathscr C : w \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$

It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3)$.

$\Box$

### Formulation 1 implies Formulation 3

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.

We need to show that $\mathscr C$ satisfies circuit axiom:

 $(\text C 5)$ $:$ $\ds \forall X \subseteq S \land \forall x \in S:$ $\ds \paren {\forall C \in \mathscr C : C \nsubseteq X} \implies \paren {\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x}$

Let $X \subset S : \forall C \in \mathscr C : C \nsubseteq X$.

Let $x \in S$.

$\exists C_1, C_2 \in \mathscr C : C_1 \neq C_2 : C_1, C_2 \subseteq X \cup \set x$.

Since $C_1, C_2 \nsubseteq X$ then:

$x \in C_1 \cap C_2$
$\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set x$

We have:

$\paren{C_1 \cup C_2} \setminus \set x \subseteq X$

Hence:

$C_3 \subseteq X$

$X \subset S : \forall C \in \mathscr C : C \nsubseteq X$

It follows that:

$\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x$

It follows that $\mathscr C$ satisfies circuit axiom $(\text C 5)$.

$\Box$

### Formulation 3 implies Formulation 1

Let $\mathscr C$ satisfy the circuit axioms $(\text C 1)$, $(\text C 2)$ and $(\text C 3)$.

We need to show that $\mathscr C$ satisfies circuit axiom:

 $(\text C 3)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \implies \exists C_3 \in \mathscr C : C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z$

In fact we prove the contrapositive statement:

 $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds z \in C_1 \cap C_2 \land \paren{\forall C \in \mathscr C : C \nsubseteq \paren {C_1 \cup C_2} \setminus \set z} \implies C1 = C2$

Let:

$C_1, C_2 \in \mathscr C$
$z \in C_1 \cap C_2$
$\forall C \in \mathscr C : C \nsubseteq \paren{C_1 \cup C_2} \setminus \set z$
$\exists \text{ at most one } C \in \mathscr C : C \subseteq \paren{\paren{C_1 \cup C_2} \setminus \set z} \cup \set z = C_1 \cup C_2$
$C_1, C_2 \subseteq C_1 \cup C_2$

Hence:

$C_1 = C_2$

It follows that $\mathscr C$ satisfies circuit axiom $(\text C 3)$.

$\blacksquare$