Matroid Base Substitution From Fundamental Circuit

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

Let $B$ be a base of $M$.

Let $y \in S \setminus B$.

Let $\map C {y,B}$ denote the fundamental circuit of $y$ in $B$.

Let $x \in B$.

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

That is, $y$ can be substituted for $x$ in $B$ to form another base of $M$ $x \in \map C {y,B}$.

Necessary Condition
Let $\paren{B \setminus \set x} \cup \set y$ be a base of $M$.

By definition of the fundamental circuit:
 * $\map C {y,B} \subseteq B \cup \set y$

and
 * $\map C {y,B}$ is dependent

$x \notin \map C {y,B}$.

Then:
 * $\map C {y,B} \subseteq \paren{B \setminus \set x} \cup \set y$

By matroid axiom $( \text I 2)$:
 * $\map C {y,B}$ is independent

This contradicts the fact that:
 * $\map C {y,B}$ is dependent

It follows that:
 * $x \in \map C {y,B}$

Sufficient Condition
We prove the contrapositive statement.

Let $\paren{B \setminus \set x} \cup \set y$ not be a base of $M$.

We have:

From contrapositive statement of Independent Subset is Base if Cardinality Equals Cardinality of Base:
 * $\paren{B \setminus \set x} \cup \set y$ is a dependent subset of $M$

From Dependent Subset Contains a Circuit there exists a circuit $C'$:
 * $C' \subseteq \paren{B \setminus \set x} \cup \set y \subseteq B \cup \set y$

From Matroid Base Union External Element has Fundamental Circuit:
 * $C' = \map C {y,B}$

Hence:
 * $x \notin \map C {y, B}$

From Rule of Transposition we have:
 * $x \in \map C {y, B} \implies \paren{B \setminus \set x} \cup \set y$ is a base of $M$