User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Circuit Axioms/Formulation 3 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 3):

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

Proof
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:

In fact we prove the contrapositive statement:

Let:
 * $C_1, C_2 \in \mathscr C$
 * $z \in C_1 \cap C_2$
 * $\forall C \in \mathscr C : C \nsubseteq \paren{C_1 \cup C_2} \setminus \set z$

From circuit axiom $(\text C 3'')$:
 * $\exists \text{ at most one } C \in \mathscr C : C \subseteq \paren{\paren{C_1 \cup C_2} \setminus \set z} \cup \set z = C_1 \cup C_2$

From Set is Subset of Union:
 * $C_1, C_2 \subseteq C_1 \cup C_2$

Hence:
 * $C_1 = C_2$

It follows that $\mathscr C$ satisfies circuit axiom ${\text C 3}$.