# Leigh.Samphier/Sandbox/Matroid Satisfies Base Axiom/Sufficient Condition/Lemma/Lemma 2

## Theorem

Let $S$ be a finite set.

Let $B_1, B_2, V \subseteq S$.

Let $V \subseteq B_2$.

Then:

 $\ds \card {B_1}$ $=$ $\ds \card{B_1 \cap B_2} + \card{B_1 \setminus B_2}$ $\ds \card {B_2}$ $=$ $\ds \card{B_2 \cap B_1} + \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V}$

## Proof

We have:

 $\ds \card {B_1}$ $=$ $\ds \card{ \paren{B_1 \cap B_2} \cup \paren{B_1 \setminus B_2} }$ Set Difference Union Intersection $\ds$ $=$ $\ds \card{B_1 \cap B_2} + \card{B_1 \setminus B_2}$ Set Difference and Intersection are Disjoint and Corollary to Cardinality of Set Union

and

 $\ds \card {B_2}$ $=$ $\ds \card{ \paren{B_2 \cap B_1} \cup \paren{B_2 \setminus B_1} }$ Set Difference Union Intersection $\ds$ $=$ $\ds \card{B_2 \cap B_1} + \card{B_2 \setminus B_1}$ Set Difference and Intersection are Disjoint and Corollary to Cardinality of Set Union $\ds$ $=$ $\ds \card{B_2 \cap B_1} + \card{\paren{\paren{B_2 \setminus B_1} \cap V} \cup \paren{\paren{B_2 \setminus B_1} \setminus V} }$ Set Difference Union Intersection $\ds$ $=$ $\ds \card{B_2 \cap B_1} + \card{\paren{B_2 \setminus B_1} \cap V} + \card{\paren{B_2 \setminus B_1} \setminus V}$ Set Difference and Intersection are Disjoint and Corollary to Cardinality of Set Union $\ds$ $=$ $\ds \card{B_2 \cap B_1} + \card{V \setminus B_1} + \card{\paren{B_2 \setminus B_1} \setminus V}$ Intersection with Set Difference is Set Difference with Intersection

$\blacksquare$