User:Leigh.Samphier/Matroids/Equivalence of Definitions of Matroid Base Axioms/Lemma 6

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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

From Singleton of Element is Subset:

$\set x \subseteq B_1 \setminus B_2$

and

$\set y \subseteq B_2 \setminus B_1$

From Set Difference is Disjoint with Reverse:

$\paren{B_1 \setminus B_2} \cap \paren{B_2 \setminus B_1} = \O$

From Subsets of Disjoint Sets are Disjoint:

$\set x \cap \set y = \O$

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

$\blacksquare$