# Equivalence of Definitions of Matroid Circuit Axioms/Condition 3 Implies Condition 1

## Theorem

Let $S$ be a finite set.

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

Let $\mathscr C$ satisfy the circuit axioms:

 $(\text C 1)$ $:$ $\ds \O \notin \mathscr C$ $(\text C 2)$ $:$ $\ds \forall C_1, C_2 \in \mathscr C:$ $\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2$ $(\text C 3'')$ $:$ $\ds \forall X \subseteq S \land \forall x \in S:$ $\ds \paren {\forall C \in \mathscr C : C \nsubseteq X} \implies \paren {\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x}$

Then:

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

## Proof

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 $(\text C 3)$.

$\blacksquare$