Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Circuit Axioms/Condition 1 Implies Condition 2
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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:
| \((C1)\) | $:$ | \(\ds \O \notin \mathscr C \) | ||||||
| \((C2)\) | $:$ | \(\ds \forall C_1, C_2 \in \mathscr C:\) | \(\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2 \) | |||||
| \((C3)\) | $:$ | \(\ds \forall C_1, C_2 \in \mathscr C:\) | \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \implies \exists C_3 \in \mathscr C : C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \) |
Then:
- $\mathscr C$ satisfies the circuit axioms:
| \((C1)\) | $:$ | \(\ds \O \notin \mathscr C \) | ||||||
| \((C2)\) | $:$ | \(\ds \forall C_1, C_2 \in \mathscr C:\) | \(\ds C_1 \ne C_2 \implies C_1 \nsubseteq C_2 \) | |||||
| \((C3')\) | $:$ | \(\ds \forall C_1, C_2 \in \mathscr C:\) | \(\ds C_1 \ne C_2 \land z \in C_1 \cap C_2 \land w \in C_1 \setminus C_2 \implies \exists C_3 \in \mathscr C : y \in C_3 \subseteq \paren {C_1 \cup C_2} \setminus \set z \) |
Proof
$\blacksquare$