# Element of Matroid Base and Circuit has a Substitute/Lemma 3

## 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 $x \in B \cap C$.

Let $X \subseteq S$ such that:

$\paren{ C \setminus \set x} \cup X$ is a base of $M$.

Then:

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

## Proof

$B \setminus \set x \subseteq B$
$B \setminus \set x \in \mathscr I$

We have

 $\ds \card {B \setminus \set x} + 1$ $=$ $\ds \paren{\card B - \card{\set x} } + 1$ Cardinality of Set Difference with Subset $\ds$ $=$ $\ds \paren{\card B - 1 } + 1$ Cardinality of Singleton $\ds$ $=$ $\ds \card B$ $\ds$ $=$ $\ds \card {\paren{ C \setminus \set x} \cup X}$ All Bases of Matroid have same Cardinality
$\exists y \in \paren{\paren{C \setminus \set x} \cup X} \setminus \paren{B \setminus \set x} : \paren{B \setminus \set x} \cup \set y \in \mathscr I : \card{\paren{B \setminus \set x} \cup \set y} = \card {\paren{ C \setminus \set x} \cup X}$

$\blacksquare$