User:Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 3

Theorem
Let $S$ be a finite set.

Let $B_1, B_2 \subseteq S$.

Let $z \in B_1 \setminus B_2$.

Let $y \in B_2 \setminus B_1$.

Let $B_3 = \paren{B_2 \setminus \set y} \cup \set z$

Then:
 * $\card{B_1 \cap B_3} = \card{B_1 \cap B_2} + 1$

Proof
We have: