# Leigh.Samphier/Sandbox/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 $M$ if and only if $x \in \map C {y,B}$

## Proof

### 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

Aiming for a contradiction, suppose $x \notin \map C {y,B}$.

Then:

$\map C {y,B} \subseteq \paren{B \setminus \set x} \cup \set y$
$\map C {y,B}$ is independent

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

It follows that:

$x \in \map C {y,B}$

$\Box$

### Sufficient Condition

We prove the contrapositive statement.

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

We have:

 $\ds \card{\paren{B \setminus \set x} \cup \set y}$ $=$ $\ds \card{B \setminus \set x} + \card{\set y}$ Corollary of Cardinality of Set Union $\ds$ $=$ $\ds \paren{\card B - \card{\set x} } + \card{\set y}$ Cardinality of Set Difference $\ds$ $=$ $\ds \paren{\card B - 1 } + 1$ Cardinality of Singleton $\ds$ $=$ $\ds \card B$
$\paren{B \setminus \set x} \cup \set y$ is a dependent subset of $M$

From Dependent Subset Contains a Circuit there exists a circuit $C'$:

$C' \subseteq \paren{B \setminus \set x} \cup \set y \subseteq B \cup \set y$
$C' = \map C {y,B}$

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 $M$

$\blacksquare$