Element of Matroid Base and Circuit has Substitute

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

Let $B \subseteq S$ be a base of $M$.

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

Let $x \in B \cap C$.

Then:
 * $\exists y \in C \setminus B : \paren{B \setminus \set x} \cup \set y$ is a base of $M$

Proof
By definition of circuit:
 * $C$ is a minimal dependent subset

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

From matroid axiom $(\text I 2)$:
 * $\set x \in \mathscr I$

Because $C \notin \mathscr I$:
 * $\set x \ne C$

Hence:
 * $\set x \subsetneq C$

From Set Difference with Proper Subset is Proper Subset:
 * $C \setminus \set x$ is a proper subset of $C$

Hence:
 * $C \setminus \set x \in \mathscr I$

From Cardinality of Independent Set of Matroid is Smaller or Equal to Base:
 * $\card{C \setminus \set x} \le \card B$

From Independent Set can be Augmented by Larger Independent Set:
 * $\exists X \subseteq B \setminus \paren{C \setminus \set x} : \paren{C \setminus \set x} \cup X \in \mathscr I : \card {\paren{ C \setminus \set x} \cup X} = \card B$

From All Bases of Matroid have same Cardinality:
 * $\card {\paren{ C \setminus \set x} \cup X}$ is the rank of $M$

From Independent Subset is Base if Cardinality Equals Rank of Matroid:
 * $\paren{ C \setminus \set x} \cup X$ is a base of $M$