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 $M$

That is, there exists $y \in C \setminus B$ such that substituting $y$ for $x$ in $B$ is another base.

Proof
By definition of a circuit we have:

Lemma 1
From Independent Subset of Matroid is Augmented by Base:
 * $\exists X \subseteq B \setminus \paren{C \setminus \set x} : \paren{C \setminus \set x} \cup X$ is a base of $M$

By definition of independent and dependent subsets we have:

Lemma 2
From matroid axiom $(\text I 3')$ we have:

Lemma 3
From Independent Subset is Base if Cardinality Equals Cardinality of Base:


 * $\paren {B \setminus \set x} \cup \set y$ is a base of $M$

Because $x \notin \paren {C \setminus \set x} \cup X$:
 * $y \ne x$

By definition of set difference:
 * $y \notin B \setminus \set x$:

By definition of set union:
 * $y \notin \paren {B \setminus \set x} \cup \set x = B$

By the definition of a subset:
 * $y \notin X$

By definition of set union:
 * $y \in C \setminus \set x \subseteq C$

By definition of set difference:
 * $y \in C \setminus B$