Element is Member of Base iff Not Loop

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

Let $\mathscr B$ denote the set of all bases of $M$.

Let $x \in S$.

Then:
 * $\exists B \in \mathscr B : x \in B$ $x$ is not a loop

Necessary Condition
Let $B \in \mathscr B$ such that $x \in B$.

From Singleton of Element is Subset:
 * $\set x \subseteq B$

By definition of a base:
 * $B \in \mathscr I$

From matroid axiom $(I2)$:
 * $\set x \in \mathscr I$

Then $\set x$ is not a dependent subset by definition.

It follows that $x$ is not a loop by definition.

Sufficient Condition
Let $x$ not be a loop.

By definition of a loop:
 * $x$ is not a dependent subset

By definition of a dependent subset:
 * $x \in \mathscr I$

From Leigh.Samphier/Sandbox/Independent Subset is Contained in Base:
 * $\exists B \in \mathscr B: \set x \subseteq B$

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