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

Theorem
Let $S$ be a finite set.

Let $\mathscr B$ satisfy the base axiom:

Let $B_1, B_2 \in \mathscr B$.

Let $U \subseteq B_1$ and $V \subseteq B_2$ such that:
 * $\card V < \card U$

Let $B_2 \cap \paren{U \setminus V} = \O$.

Then:
 * $\exists B_3 \in \mathscr B$:
 * $V \subseteq B_3$
 * $\card{B_1 \cap B_3} > \card{B_1 \cap B_2}$

Proof
Let $U, V \in \mathscr I$ such that:
 * $\card V < \card U$

Lemma 2
From Subset of Set Difference iff Disjoint Set:
 * $U \setminus V \subseteq B_1 \setminus B_2$

We have:

From the contrapositive statement of Cardinality of Empty Set:
 * $\paren{B_2 \setminus B_1} \setminus V \ne \O$

Hence: $\exists y \in \paren{B_2 \setminus B_1} \setminus V$

From $(\text B 1)$:
 * $\exists z \in B_1 \setminus B_2 : \paren{B_2 \setminus \set y} \cup \set z \in \mathscr B$

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

We have:

Lemma 3
Hence:
 * $\card{B_1 \cap B_3} > \card{B_1 \cap B_2}$