Rank of Matroid Circuit is One Less Than Cardinality/Lemma

Theorem
Let $M = \struct {S, \mathscr I}$ be a matroid.

Let $C \subseteq S$ be a circuit of $M$.

Let $x \in C$.

Then:
 * $C \setminus \set x$ is a maximal independent subset of $C$

Proof
From Set Difference is Subset:
 * $C \setminus \set x \subseteq C$

Because $x \in C$ and $x \notin C \setminus \set x$:
 * $C \setminus \set x \ne C$

From Leigh.Samphier/Sandbox/Proper Subset of Matroid Circuit is Independent and matroid axiom $(\text I 1)$:
 * $C \setminus \set x \in \mathscr I$

Let $X$ be an independent subset such that:
 * $C \setminus \set x \subseteq X \subseteq C$

As $C$ is dependent:
 * $X \ne C$


 * $x \in X$
 * $x \in X$

From Singleton of Element is Subset:
 * $\set x \subseteq X$
 * $\set x \subseteq C$

We have:

Hence $C = X$ by definition of set equality.

This contradicts the fact that:
 * $X \ne C$

Hence:
 * $x \notin X$

From Intersection With Singleton is Disjoint if Not Element:
 * $X \cap \set x = \O$

We have:

By definition of set equality:
 * $X = C \setminus \set x$

It follows that $C \setminus \set x$ is a maximal independent subset of $C$ by definition.