Equivalence of Definitions of Matroid Circuit Axioms/Formulation 2 Implies Formulation 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 (formulation 2):

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

Then:

$\mathscr C$ satisfies the circuit axioms (formulation 1):

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

Proof

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 $z \in C_1 \cap C_2$.

From circuit axiom $(\text C 2)$:

$C_2 \nsubseteq C_1$

By definition of subset and set difference:

$\exists w \in C_2 \setminus C_1$

From circuit axiom $(\text C 4)$:

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

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

$\blacksquare$