# Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Circuit Axioms/Condition 2 Implies Condition 3

## 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:

 $(C1)$ $:$ $\displaystyle \O \notin \mathscr C$ $(C2)$ $:$ $\displaystyle \forall C_1, C_2 \in \mathscr C:$ $\displaystyle C_1 \neq C_2 \implies C_1 \not \subseteq C_2$ $(C3')$ $:$ $\displaystyle \forall C_1, C_2 \in \mathscr C:$ $\displaystyle C_1 \neq C_2 \land z \in C_1 \cap C_2 \land w \in C_1 \setminus C_2 \implies \exists C_3 \in \mathscr C : y \in C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$

Then:

$\mathscr C$ satisfies the circuit axioms:
 $(C1)$ $:$ $\displaystyle \O \notin \mathscr C$ $(C2)$ $:$ $\displaystyle \forall C_1, C_2 \in \mathscr C:$ $\displaystyle C_1 \neq C_2 \implies C_1 \not \subseteq C_2$ $(C3'')$ $:$ $\displaystyle \forall X \subseteq S \land \forall x \in S:$ $\displaystyle \paren{\forall C \in \mathscr C : C \not \subseteq X} \implies \paren{\exists \text{ at most one } C \in \mathscr C : C \subseteq X \cup \set x}$

## Proof

$\blacksquare$