# Leigh.Samphier/Sandbox/Equivalence of Definitions of Matroid Base Axiom/Lemma

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## Theorem

Let $S$ be a finite set.

Let $B_1, B_2 \subseteq S$.

Let $x \in B_1 \setminus B_2$.

Let $y \in B_2 \setminus B_1$.

Then:

$\paren{B_1 \setminus \set x} \cup \set y = \paren{B_1 \cup \set y} \setminus \set x$

## Proof

$\set x \subseteq B_1 \setminus B_2$

and

$\set y \subseteq B_2 \setminus B_1$
$\paren{B_1 \setminus B_2} \cap \paren{B_2 \setminus B_1} = \O$
$\set x \cap \set y = \O$

We have:

 $\ds \paren{B_1 \cup \set y} \setminus \set x$ $=$ $\ds \paren {B_1 \setminus \set x} \cup \paren {\set y \setminus \set x}$ Set Difference is Right Distributive over Union $\ds$ $=$ $\ds \paren {B_1 \setminus \set x} \cup \set y$ Set Difference with Disjoint Set

$\blacksquare$