and vertex set $V(\overrightarrow{D})$, which is equal to $V(D)$. Given an oriented graph $\overrightarrow{D}$, the [*underlying graph*]{} $D$ is the graph such that the directions associated to the arcs of $\overrightarrow{D}$ are removed, yielding edges. An [*oriented subgraph*]{} of an oriented graph $\overrightarrow{D}$ is an oriented graph $\overrightarrow{D'}$ such that $V(\overrightarrow{D'}) \subseteq V(\overrightarrow{D})$, $A(\overrightarrow{D'}) \subseteq A(\overrightarrow{D})$ and for an arc $a = (u,v) \in A(\overrightarrow{D'})$, it must be true that $a \in A(\overrightarrow{D})$ and $ \{ u,v \} \subseteq V(\overrightarrow{D'})$. A *proper oriented subgraph* of an oriented graph $\overrightarrow{D}$ is an oriented subgraph $\overrightarrow{D'}$ of $\overrightarrow{D}$ such that either $V(\overrightarrow{D'}) \subsetneqq V(\overrightarrow{D})$ or $A(\overrightarrow{D'}) \subsetneqq A(\overrightarrow{D})$.

Given a graph $G$, if two vertices $u$ and $v$ are adjacent in $G$, i.e., if there is an edge $e = uv \in E(G)$, we say that $u$ and $v$ are [*neighbors*]{} in $G$, and denote the neighborhood of a vertex $u$ as $N(u)$. For an oriented graph $\overrightarrow{D}$ containing an arc $(u,v)$, we say that $u$ is an [*in-neighbor*]{} of $v$ and that $v$ is an [*out-neighbor*]{} of $u$. For a vertex $u \in V(\overrightarrow{D})$, we denote the set of all in-neighbors of $u$ as $N^-(u)$, and the set of all out-neighbors 