# Leigh.Samphier/Sandbox/Rank of Matroid Circuit is One Less Than Cardinality/Lemma

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## 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:

- $X \ne C$

Aiming for a contradiction, suppose:

- $x \in X$

From Singleton of Element is Subset:

- $\set x \subseteq X$
- $\set x \subseteq C$

We have:

\(\ds X\) | \(=\) | \(\ds X \cup \set x\) | Union with Superset is Superset | |||||||||||

\(\ds \) | \(\supseteq\) | \(\ds \paren{C \setminus \set x} \cup \set x\) | Set Difference over Subset | |||||||||||

\(\ds \) | \(=\) | \(\ds C \cup \set x\) | Set Difference with Union | |||||||||||

\(\ds \) | \(=\) | \(\ds C\) | Union with Superset is Superset |

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:

\(\ds X\) | \(=\) | \(\ds X \setminus \set x\) | Set Difference with Disjoint Set | |||||||||||

\(\ds \) | \(\subseteq\) | \(\ds C \setminus \set x\) | Set Difference over Subset |

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.

$\blacksquare$