Matroid Unique Circuit Property

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

Let $X \subseteq S$ be an independent subset of $M$.

Let $x \in S$ such that:
 * $X \cup \set x$ is a dependent subset of $M$.

Then there exists a unique circuit $C$ such that:
 * $C \subseteq X \cup \set x$

Proof
From Dependent Subset Contains a Circuit:
 * there exists a circuit $C$ such that $C \subseteq X \cup \set x$

$C'$ is circuit of $M$ such that:
 * $C' \ne C$
 * $C' \subseteq X \cup \set x$

From Superset of Dependent Set is Dependent:
 * $C' \not \subseteq X$

Hence:
 * $C' \setminus X \ne \O$

From Set Difference over Subset:
 * $C' \setminus X \subseteq \paren{X \cup \set x} \setminus X = \set x$

It follows that:
 * $x \in C'$

By the definition of minimal dependent subset:
 * $C \not \subseteq C'$

By the definition of a subset:
 * $\exists y \in C \setminus C'$

By the definition of minimal dependent subset:
 * $C \setminus \set y \in \mathscr I$

From Independent Subset is Contained in Maximal Independent Subset:
 * $\exists Y \in \mathscr I : C \setminus \set y \subseteq Y \subseteq X \cup \set x : Y$ is a maximal independent subset of $X \cup \set x$

By assumption:
 * $X$ is a maximal independent subset of $X \cup \set x$

By matroid axiom $(\text I 3''')$:
 * $\card Y = \card X$

As $x \in C'$ and $y \notin C'$ then:
 * $x \ne y$

Hence:
 * $x \in C \setminus \set y$

By matroid axiom $(\text I 2)$:
 * $C \setminus \set y \cup \set y = C \not \subseteq Y$

Hence:
 * $y \notin Y$

Hence:
 * $Y \subseteq \paren{X \cup \set x} \setminus \set y$

From Cardinality of Subset of Finite Set:
 * $\card Y \le \card{\paren{X \cup \set x} \setminus \set y}$

We have:

From Cardinality of Proper Subset of Finite Set:
 * $Y = \paren{X \cup \set x} \setminus \set y$

Hence:
 * $\paren{X \cup \set x} \setminus \set y \in \mathscr I$

As $y \notin C'$ and $C' \subseteq X \cup \set x$ then:
 * $C' \subseteq \paren{X \cup \set x} \setminus \set y$

From Superset of Dependent Set is Dependent:
 * $\paren{X \cup \set x} \setminus \set y \notin \mathscr I$

This contradicts the previous statement that
 * $\paren{X \cup \set x} \setminus \set y \in \mathscr I$

It follows that $C$ is the unique circuit such that:
 * $C \subseteq X \cup \set x$