Size of Linearly Independent Subset is at Most Size of Finite Generator/Proof 2

Theorem
Let $R$ be a division ring.

Let $V$ be an $R$-vector space.

Let $F \subseteq V$ be a finite generator of $V$ over $R$.

Let $L \subseteq V$ be linearly independent over $R$.

Then $\left\vert{L}\right\vert \le \left\vert{F}\right\vert$.

Proof
We use a proof by contradiction.

That is, we show that if $L \subseteq V$ and $\left\vert{L}\right\vert > \left\vert{F}\right\vert$, then $L$ is linearly dependent over $R$.

Let $S \subseteq \N$ be the set of all natural numbers $n \in \N$ such that:
 * For any $L \subseteq V$ satisfying $\left\vert{L}\right\vert > \left\vert{F}\right\vert$, if $\left\vert{F \setminus L}\right\vert = n$, then there exists a $v \in L \setminus F$ which is a linear combination of $L \setminus \left\{{v}\right\}$.

We use the principle of mathematical induction to prove that $S = \N$.

If $\left\vert{F \setminus L}\right\vert = 0$, then it follows by Cardinality of Empty Set that $F \setminus L = \varnothing$.

By Set Difference with Superset is Empty Set, $F \subsetneq L$.

By Set Difference with Proper Subset, there exists a $v \in L \setminus F$.

By the definition of a generator, $v$ is a linear combination of $F$.

By Set Difference with Disjoint Set, it follows that $F = F \setminus \left\{{v}\right\} \subseteq L \setminus \left\{{v}\right\}$.

Hence $0 \in S$.

This is the basis for the induction.

Now, assume the induction hypothesis that $n \in S$.

Let $\left\vert{F \setminus L}\right\vert = n + 1 \ge 1$.

Let $w \in F \setminus L$.

Let $A = L \cup \left\{{w}\right\}$.

Then $\left\vert{A}\right\vert = \left\vert{L}\right\vert + 1 > \left\vert{F}\right\vert$ and $\left\vert{F \setminus A}\right\vert = n$.

Therefore, there exists a $u \in A \setminus F = L \setminus F$ which is a linear combination of $A \setminus \left\{{u}\right\}$.

If $u$ is a linear combination of $L \setminus \left\{{u}\right\}$, then the claim holds.

Otherwise, let $B = A \setminus \left\{{u}\right\}$.

Then $\left\vert{B}\right\vert = \left\vert{L}\right\vert > \left\vert{F}\right\vert$ and $\left\vert{F \setminus B}\right\vert = n$.

Therefore, there exists a $v \in B \setminus F$ which is a linear combination of $B \setminus \left\{{v}\right\}$.

Hence, $v$ is a linear combination of $L \setminus \left\{{v}\right\}$.

That is, $n + 1 \in S$.

This completes the induction step.

$\ldots$