Closure of Subset Contains Parallel Elements

Theorem
Let $M = \struct{S, \mathscr I}$ be a matroid.

Let $A \subseteq S$.

Let $x, y \in S$.

If $x \in \map \sigma A$ and $y$ is parallel to $x$ then:
 * $y \in \map \sigma A$

Proof
Let $x \in \map \sigma A$

Let $y$ be parallel to $x$.

By the definitions of the closure operator and depends depends:
 * $\map \rho A = \map \rho {A \cup \set x}$

and:
 * $y \in \map \sigma A$ $\map \rho A = \map \rho {A \cup \set y}$

From Rank Function is Increasing:
 * $\map \rho {A \cup \set x} = \map \rho A \le \map \rho {A \cup \set y} \le \map \rho {A \cup \set {x,y}}$

To show $\map \rho A = \map \rho {A \cup \set y}$ it is sufficient to show:
 * $\map \rho {A \cup \set {x,y}} = \map \rho {A \cup \set x}$

From Rank Function is Increasing:
 * $\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x,y}}$

By the definition of parallel:
 * $\set x$ is independent

From Leigh.Samphier/Sandbox/Independent Subset is Contained in Maximal Independent Subset:
 * $\exists X \in \mathscr I : \set x \subseteq X \subseteq A \cup \set x : \size X = \map \rho {A \cup \set x}$

From Subset Relation is Transitive:
 * $X \subseteq A \cup \set {x, y}$

From Leigh.Samphier/Sandbox/Independent Subset is Contained in Maximal Independent Subset:
 * $\exists Y \in \mathscr I : X \subseteq Y \subseteq A \cup \set {x,y} : \size Y = \map \rho {A \cup \set {x,y}}$

By definition of a subset:
 * $x \in Y$

By definition of parallel:
 * $\set{x, y}$ is dependent

From Superset of Dependent Set is Dependent:
 * $\set{x, y} \not \subseteq Y$

Then:
 * $y \notin Y$

So:
 * $Y \subseteq A \cup \set x$

By definition of the rank function:
 * $\size Y \le \map \rho {A \cup \set x}$

As $\size Y = \map \rho {A \cup \set {x,y}}$ then:
 * $\map \rho {A \cup \set {x,y}} \le \map \rho {A \cup \set x}$

As $\map \rho {A \cup \set x} \le \map \rho {A \cup \set {x,y}}$ then:
 * $\map \rho {A \cup \set {x,y}} = \map \rho {A \cup \set x}$

The result follows.