in the corresponding index coding instance as in the following proposition.

For a given network coding instance, the modeling of the corresponding index coding instance in [@EQU] obtains an identical result even if the receiver $\hat{t}_{\rm all}$ is removed, that is, $\hat{t}_{\rm all}$ is redundant.

From 1) of Proposition \[prop:condition\], we can see that there is no cycle in a subgraph which consists of $\{R_e | e\in E\}$ and $\{\hat{X}_e | e\in E\}$ and thus there is no $\bm{\delta}_s$-cycle. Since this subgraph is $\bm{\delta}_s$-acyclic, the optimal codelength for the subgraph is $|E|$ by Lemma \[delta\_s-cycle\], meaning that every vector $\bold{\hat{Z}}\in\mathbb{F}_q^{|S|+|E|}$ such that $\bold{\hat{Z}}_S=\bold{0}$ and $\bold{\hat{Z}}_E\neq\bold{0}$ belongs to $\mathcal{I}(q,\mathcal{G},\bm{\delta}_s)$. In [@EQU], $\hat{t}_{\rm all}$ wants to receive $\bold{\hat{X}}_{E}$ and has $\bold{\hat{X}}_S$ as side information with $\delta_s^{(\hat{t}_{\rm all})}=0$. In fact, we do not need $\hat{t}_{\rm all}$ because there is no cycle in the subgraph mentioned above. From Theorem \[encoding\], $\mathcal{I}_{\hat{t}_{\rm all}}(q,\mathcal{G},\delta_s^{(\hat{t}_{\rm all})})$ is the set of all vectors in $\mathbb{F}_q^{|E|+|S|}$ such that $\bold{\hat{Z}}_S=\bold{0}$ and $\bold{\hat{Z}}_E\neq\bold{0}$. Since every vector in $\mathcal{I}_{\hat{t}_{\rm all}}(q,\mathcal{G},\delta_s^{(\hat{t}_{\rm all})})$ is already included in $\mathcal{I}(q,\mathcal{G},\bm{\delta}_s)$, the receiver $\hat{t}_{\rm all}$ can be removed from the corresponding index coding instance.

\[prop:tall\]

Thus, we can remove the receiver $\hat{t}_{\rm all}$ from the corresponding index 