Why do I need this extra condition on a vector space basis theorem?

My course notes for an abstract algebra course include this theorem: Let $V$ be a vector space. If $L \subset V$ is a linearly independent subset, and $E$ is minimal amongst all generating sets of $V$ with the property that $L \subseteq E$, then $E$ is a basis.

By "minimal generator of $V$" we mean that for all $\vec{v}\in E$, the set $E\setminus \{\vec{v}\}$ does not generate $V$.

By question is simply why we need $L$ in this theorem? For example, if we assume the above theorem is true, we might as well take $L=\{\}$ to be the empty set. This 'special case' then reads "if $E$ is minimal amongst all generating sets of $V$ then $E$ is a basis" which is much simpler? and more general?

This question here is highly related: Why is this a useful way to prove the characterisation of bases? and indeed, somehow it is useful to keep the $L$ in the theorem, but it really seems pretty useless to me. For example, here is another way I could chuck $L$ into the statement "if $E$ is minimal amongst all generating sets of $V$ then $E$ is a basis" in a way that doesn't seem useful:

"if $L\subset E$ is a linearly independent subset, and $E$ is minimal amongst all generating sets of $V$ then $E \cup L$ is a basis". This is trivially true, and is not useful, I see no reason why the earlier statement doesn't fall into this category.

The way this theorem is phrased may be a bit confusing, so I'll put it a bit more explicitly:

The key of this theorem is that given a linearly independent set, we can find a basis which specifically includes it. This can be quite useful!