Consider the set of $n=4$ vectors from $\complex{5}$, \begin{equation*} R=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4} = \set{ \colvector{1\\2\\-1\\3\\2},\, \colvector{2\\1\\3\\1\\2},\, \colvector{0\\-7\\6\\-11\\-2},\, \colvector{4\\1\\2\\1\\6} }\\ \end{equation*} and define $V=\spn{R}$.
To employ Theorem LIVHS, we form a $5\times 4$ matrix, $D$, and row-reduce to understand solutions to the homogeneous system $\homosystem{D}$, \begin{equation*} D= \begin{bmatrix} 1&2&0&4\\ 2&1&-7&1\\ -1&3&6&2\\ 3&1&-11&1\\ 2&2&-2&6 \end{bmatrix} \rref \begin{bmatrix} \leading{1}&0&0&4\\ 0&\leading{1}&0&0\\ 0&0&\leading{1}&1\\ 0&0&0&0\\ 0&0&0&0 \end{bmatrix} \end{equation*}
We can find infinitely many solutions to this system, most of them nontrivial, and we choose any one we like to build a relation of linear dependence on $R$. Let us begin with $x_4=1$, to find the solution \begin{equation*} \colvector{-4\\0\\-1\\1} \end{equation*}
So we can write the relation of linear dependence, \begin{equation*} (-4)\vect{v}_1+0\vect{v}_2+(-1)\vect{v}_3+1\vect{v}_4=\zerovector \end{equation*}
Theorem DLDS guarantees that we can solve this relation of linear dependence for some vector in $R$, but the choice of which one is up to us. Notice however that $\vect{v}_2$ has a zero coefficient. In this case, we cannot choose to solve for $\vect{v}_2$. Maybe some other relation of linear dependence would produce a nonzero coefficient for $\vect{v}_2$ if we just had to solve for this vector. Unfortunately, this example has been engineered to always produce a zero coefficient here, as you can see from solving the homogeneous system. Every solution has $x_2=0$!
OK, if we are convinced that we cannot solve for $\vect{v}_2$, let us instead solve for $\vect{v}_3$, \begin{equation*} \vect{v}_3=(-4)\vect{v}_1+0\vect{v}_2+1\vect{v}_4=(-4)\vect{v}_1+1\vect{v}_4 \end{equation*}
We now claim that this particular equation will allow us to write \begin{equation*} V=\spn{R}= \spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3,\,\vect{v}_4}}= \spn{\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_4}} \end{equation*} in essence declaring $\vect{v}_3$ as surplus for the task of building $V$ as a span. This claim is an equality of two sets, so we will use Definition SE to establish it carefully. Let $R^\prime=\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_4}$ and $V^\prime=\spn{R^\prime}$. We want to show that $V=V^\prime$.
First show that $V^\prime\subseteq V$. Since every vector of $R^\prime$ is in $R$, any vector we can construct in $V^\prime$ as a linear combination of vectors from $R^\prime$ can also be constructed as a vector in $V$ by the same linear combination of the same vectors in $R$. That was easy, now turn it around.
Next show that $V\subseteq V^\prime$. Choose any $\vect{v}$ from $V$. So there are scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\alpha_4$ such that \begin{align*} \vect{v}&= \alpha_1\vect{v}_1+\alpha_2\vect{v}_2+\alpha_3\vect{v}_3+\alpha_4\vect{v}_4\\ &=\alpha_1\vect{v}_1+\alpha_2\vect{v}_2+ \alpha_3\left((-4)\vect{v}_1+1\vect{v}_4\right)+ \alpha_4\vect{v}_4\\ &=\alpha_1\vect{v}_1+\alpha_2\vect{v}_2+ \left((-4\alpha_3)\vect{v}_1+\alpha_3\vect{v}_4\right)+ \alpha_4\vect{v}_4\\ &=\left(\alpha_1-4\alpha_3\right)\vect{v}_1+ \alpha_2\vect{v}_2+ \left(\alpha_3+\alpha_4\right)\vect{v}_4. \end{align*}
This equation says that $\vect{v}$ can then be written as a linear combination of the vectors in $R^\prime$ and hence qualifies for membership in $V^\prime$. So $V\subseteq V^\prime$ and we have established that $V=V^\prime$.
If $R^\prime$ was also linearly dependent (it is not), we could reduce the set even further. Notice that we could have chosen to eliminate any one of $\vect{v}_1$, $\vect{v}_3$ or $\vect{v}_4$, but somehow $\vect{v}_2$ is essential to the creation of $V$ since it cannot be replaced by any linear combination of $\vect{v}_1$, $\vect{v}_3$ or $\vect{v}_4$.