If $S\subseteq V$ is linearly independent and $w\mathrm{\notin }\mathrm{span}(S)$ then $S\cup \{w\}$ is linearly independent and $\mathrm{span}(S)⊊\mathrm{span}(S\cup \{w\})$; and if $w\in \mathrm{span}(S)$ then $\mathrm{span}(S\cup \{w\})=\mathrm{span}(S)$

Claim 1. From $S\subseteq S\cup \{w\}$ we get $\mathrm{span}(S)\subseteq \mathrm{span}(S\cup \{w\})$. Conversely, assume $w\in \mathrm{span}(S)$; since also $S\subseteq \mathrm{span}(S)$, the set $S\cup \{w\}$ is contained in $\mathrm{span}(S)$, which is a linear subspace of $V$, so minimality gives $\mathrm{span}(S\cup \{w\})\subseteq \mathrm{span}(S)$. The two inclusions give the claim.

The two easy parts of claim 2. Assume $w\mathrm{\notin }\mathrm{span}(S)$. Then $w\mathrm{\notin }S$, because $S\subseteq \mathrm{span}(S)$. Also $\mathrm{span}(S)\subseteq \mathrm{span}(S\cup \{w\})$ by monotonicity, and $w$ lies in the larger set and not in the smaller, so the inclusion is strict.

Now assume in addition that $S$ is independent, and let $v:n\to S\cup \{w\}$ be an injective finite list with $\lambda :n\to F$ and ${\sum }_{i<n}{\lambda }_i{v}_i=0_V$. If $w$ is not a value of $v$, then $v$ is an injective finite list into $S$, so independence of $S$ gives ${\lambda }_i=0_F$ for every $i<n$ and there is nothing more to prove.

In the remaining case $w=v_k$ for exactly one $k<n$, since $v$ is injective. Then $n\mathrm{\ne }0$, say $n=\sigma (n^{\mathrm{\prime }})$, and $y:=v\circ {\delta }_k$ is an injective finite list $n^{\mathrm{\prime }}\to S$: it is injective as a composite of injections, and its values are the $v_j$ with $j\mathrm{\ne }k$, each of which lies in $S\cup \{w\}$ and differs from $v_k=w$. Moreover, for every $\mu :n\to F$ with ${\mu }_k=0_F$ the list $i\mapsto {\mu }_i{v}_i$ has the value $0_{F}w=0_V$ at $k$, so deleting that index gives ${\sum }_{i<n}{\mu }_i{v}_i={\sum }_{i<n^{\mathrm{\prime }}}{\mu }_{{\delta }_k(i)}{y}_i$.

In that case the coefficient of $w$ vanishes. Suppose ${\lambda }_k\mathrm{\ne }{0}_F$. Applying (F3) at $k$ to the list $i\mapsto {\lambda }_i{v}_i$ gives $0_V={\lambda }_{k}w+R$, where $R={\sum }_{i<n}{\lambda }_i^{\mathrm{\prime }}{v}_i$ with ${\lambda }_k^{\mathrm{\prime }}:=0_F$ and ${\lambda }_i^{\mathrm{\prime }}:={\lambda }_i$ for $i\mathrm{\ne }k$, using $0_{F}w=0_V$ to identify the deleted entry. By step 1.4 applied to ${\lambda }^{\mathrm{\prime }}$, $R={\sum }_{i<n^{\mathrm{\prime }}}{\lambda }_{{\delta }_k(i)}^{\mathrm{\prime }}{y}_i$, which is a linear combination of elements of $S$ and therefore lies in $\mathrm{span}(S)$. Then ${\lambda }_{k}w=-R$ lies in $\mathrm{span}(S)$, that set being a linear subspace, and hence so does $w=1_{F}w=({\lambda }_k^{-1}{\lambda }_k)w={\lambda }_k^{-1}({\lambda }_{k}w)$, contradicting $w\mathrm{\notin }\mathrm{span}(S)$. So ${\lambda }_k=0_F$.

The remaining coefficients vanish too. Since ${\lambda }_k=0_F$ by step 2.1, step 1.4 applied to $\lambda$ itself gives $0_V={\sum }_{i<n}{\lambda }_i{v}_i={\sum }_{i<n^{\mathrm{\prime }}}{\lambda }_{{\delta }_k(i)}{y}_i$; the list $y$ is an injective finite list into the independent set $S$, hence independent, so ${\lambda }_{{\delta }_k(i)}=0_F$ for every $i<n^{\mathrm{\prime }}$. As ${\delta }_k$ has image $n\setminus \{k\}$, this says ${\lambda }_j=0_F$ for every $j\mathrm{\ne }k$, and with ${\lambda }_k=0_F$ every coefficient vanishes.

Steps 1.3 and 3.1 show that every injective finite list into $S\cup \{w\}$ is independent, so $S\cup \{w\}$ is linearly independent; with step 1.2 this is claim 2, and step 1.1 is claim 1.