Element of Matroid Base and Circuit has Substitute/Lemma 3

Lemma for Element of Matroid Base and Circuit has Substitute
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
From Set Difference is Subset:
 * $B \setminus \set x \subseteq B$

From matroid axiom $(\text I 2)$:
 * $B \setminus \set x \in \mathscr I$

We have

From matroid axiom $(\text I 3')$:
 * $\exists y \in \paren {\paren {C \setminus \set x} \cup X} \setminus \paren {B \setminus \set x} : \paren {B \setminus \set x} \cup \set y \in \mathscr I : \card {\paren {B \setminus \set x} \cup \set y} = \card {\paren {C \setminus \set x} \cup X}$