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

Proof
Let $S \subseteq \N$ be the set of all natural numbers $n \in \N$ such that:
 * For any finite generator $F$ of $V$ over $R$, if $\left\vert{F \cap L}\right\vert \ge n$, then $\left\vert{L}\right\vert \le \left\vert{F}\right\vert$.

By Intersection is Subset and Cardinality of Subset of Finite Set, we have $\left\vert{F \cap L}\right\vert \le \left\vert{F}\right\vert$.

Hence, it is vacuously true that $\left\vert{F}\right\vert + 1 \in S$.

Therefore, $S$ is non-empty.

From the well-ordering principle, $S$ has a smallest element $N$.

If $N = 0$, the theorem immediately follows.

Otherwise, $N \ge 1$.

Suppose that $\left\vert{F \cap L}\right\vert \ge N - 1$.

If $L \subseteq F$, the theorem follows from Cardinality of Subset of Finite Set.

Otherwise, there exists a $v \in L$ such that $v \notin F$.

Let $F' = F \cup \left\{{v}\right\}$.

By Intersection is Subset, we have $F' \cap L \subseteq L$, so it follows by Subset of Linearly Independent Set that $F' \cap L$ is linearly independent over $R$.

Also, by Intersection is Subset:
 * $F' \cap L \subseteq F'$

We have that $F'$ is a generator of $V$ over $R$.

By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, there exists a basis $B$ for $V$ such that:
 * $F' \cap L \subseteq B \subseteq F'$

Since $v \notin F$ is a linear combination of $F$, it follows that $F'$ is linearly dependent over $R$.

By the definition of a basis:
 * $B \subsetneq F'$

By Cardinality of Subset of Finite Set and Cardinality is Additive Function:
 * $\left\vert{B}\right\vert < \left\vert{F'}\right\vert = \left\vert{F}\right\vert + 1$

Hence:
 * $\left\vert{B}\right\vert \le \left\vert{F}\right\vert$

We have that:

Since $N \in S$, it follows by the definition of a basis that:
 * $\left\vert{L}\right\vert \le \left\vert{B}\right\vert \le \left\vert{F}\right\vert$

Hence:
 * $N - 1 \in S$

But this contradicts the assumption that $N$ is the smallest element of $S$.

Therefore:
 * $N = 0$

Hence the result.