Equivalence of Definitions of Matroid Circuit Axioms/Condition 1 Implies Condition 2

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:

Then:
 * $\mathscr C$ satisfies the circuit axioms:

Proof
Let:


 * $F \neq \O$

Let $\tuple{C_1, C_2, z, w} \in F$ :
 * $\size{C_1 \cup C_2} = \min \set{\size{C \cup D} : \tuple{C, D, x, y} \in F}$

By circuit axiom $(\text C 3)$:
 * $\exists C_3 \in \mathscr C : C_3 \subseteq \paren{C_1 \cup C_2} \setminus \set z$

By assumptiom:
 * $w \notin C_3$

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

By circuit axiom $(\text C 2)$:
 * $C_3 \nsubseteq C_1$

From Set Difference and Intersection form Partition:
 * $C_3 \cap \paren{C_2 \setminus C_1} \neq \O$

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
 * $w \notin C_2 \cup C_3$

From Set is Subset of Union and Union of Subsets is Subset:
 * $C_2 \cup C_3 \subseteq \C_1 \cup C_2$

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:
 * $C_4 \subset C_2 \cup C_3 \subset C_1 \cup C_2$

From Set is Subset of Union and Union of Subsets is Subset:
 * $C_1 \cup C_4 \subseteq C_1 \cup C_2$

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 $(\text C 3')$ is satisfied.