# Element of Matroid Base and Circuit has a 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 a circuit we have:

### Lemma 1

$C \setminus \set x$ is an independent proper subset of $C$

$\Box$

$\exists X \subseteq B \setminus \paren{C \setminus \set x} : \paren{C \setminus \set x} \cup X$ is a base of $M$

By definition of independent and dependent subsets we have:

### Lemma 2

$x \notin \paren{ C \setminus \set x} \cup X$

$\Box$

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

### Lemma 3

$\exists y \in \paren{\paren{C \setminus \set x} \cup X} \setminus \paren{B \setminus \set x} : \paren{B \setminus \set x} \cup \set y \in \mathscr I : \card{\paren{B \setminus \set x} \cup \set y} = \card {\paren{ C \setminus \set x} \cup X}$

$\Box$

$\paren{B \setminus \set x} \cup \set y$ 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:

$y \in C \setminus B$

$\blacksquare$