# Directed Path 3-Arc-Connectivity of Cartesian Product Digraphs
## 3 A General Lower Bound  
Now we will provide a lower bound for \(\lambda_{3}^{p}(G\square H)\). Theorem 3.1: _Let \(G\) and \(H\) be two digraphs such that \(\delta^{0}(G)\geq 4\), \(\delta^{0}(H)\geq 4\), and \(\kappa(G)\geq 2\), \(\kappa(H)\geq 2\). We have_  
\[\lambda_{3}^{p}(G\square H)\geq\min\{2\kappa(G),2\kappa(H)\}.\]  
_Furthermore, this bound is sharp._  
Proof: It suffices to show that there are at least \(2\kappa(G)\) or \(2\kappa(H)\) arc-disjoint \((S,r)\)-paths for any \(S\subseteq V(G\square H)\) with \(|S|=3\), \(r\in S\). Let \(S=\{x,y,z\}\) and let \(r=x\). Without loss of generality, we may assume \(\kappa(G)\leq\kappa(H)\) and consider the following six cases. Case 1: Let \(x\), \(y\) and \(z\) be in the same \(H(u_{i})\) or \(G(v_{j})\) for some \(i\in[n]\), \(j\in[m]\). Without loss of generality, we may assume that \(x=u_{1,1}\), \(y=u_{2,1}\), \(z=u_{3,1}\). In this case, our overall goal is that we will use arc-disjoint paths between \(x\) and \(y\) in \(G(v_{1})\), \(y\) and \(z\) in \(G(v_{1})\), \(x\) and its out-neighbors in \(H(u_{1})\), \(y\) and its in-neighbors in \(H(u_{2})\), \(z\) and its in-neighbors in \(H(u_{3})\), and combine them together to form the required arc-disjoint paths. The general idea of the proof process is briefly described in Figure 2. The vertices and paths contained in Figure 2 are explained below. Let \(S_{1}=\{x,y\}\), \(r_{1}=x\). It is known that there are at least \(\kappa(G)\) internally disjoint \((S_{1},r_{1})\)-paths in \(G(v_{1})\), denoted as \(\bar{P}_{1i}\) (\(i\in[\kappa(G)]\)). Considering \(S_{1}^{\prime}=\{y,z\}\), \(r_{1}^{\prime}=y\), there are at least \(\kappa(G)\) internally disjoint \((S_{1}^{\prime},r_{1}^{\prime})\)-paths in \(G(v_{1})\), denoted as \(\overline{P}_{2j}\) (\(j\in[\kappa(G)]\)). For each \(j\in[\kappa(G)]\), let \(u_{s_{j},1}\) be the out-neighbor of \(y\) in \(\overline{P}_{2j}\); clearly these out-neighbors are distinct. Similarly, an in-neighbor \(u_{k,j}\) (\(j\in[\kappa(G)]\)) of \(z\) in \(\overline{P}_{2j}\) can be chosen such that these in-neighbors are distinct. In \(H(u_{1})\), if there is a vertex that is not an out-neighbor of \(x\), then choose such a vertex as \(u_{1,a}\), where \(a\
eq 1\). If there is no such vertex, that is, all vertices are out-neighbours of \(x\), then choose any vertex as \(u_{1,a}\), where \(a\
eq 1\). In \(H(u_{1})\), let \(S_{2}^{\prime}=\{x,u_{1,a}\}\), \(r_{2}^{\prime}=x\), and it is established that there exist at least \(\kappa(G)\) internally disjoint \((S_{2}^{\prime},r_{2}^{\prime})\)-paths, say \(\bar{P}_{2j}\) (\(j\in[\kappa(G)]\)). In \(G(v_{a})\), let \(S_{3}^{\prime}=\{u_{1,a},u_{2,a}\}\), \(r_{3}^{\prime}=u_{1,a}\)and it is established that there exist at least \(\kappa(G)\) internally disjoint \((S^{\prime}_{3},r^{\prime}_{3})\)-paths, say \(\widehat{P}_{2j}\)\((j\in[\kappa(G)])\). In \(H(u_{2})\), let \(S^{\prime}_{4}=\{y,u_{2,a}\}\), \(r^{\prime}_{4}=u_{2,a}\), and it is established that there exist at least \(\kappa(G)\) internally disjoint \((S^{\prime}_{4},r^{\prime}_{4})\)-paths, say \(\widetilde{P}_{2j}\)\((j\in[\kappa(G)])\). In \(H(u_{1})\), if there is a vertex that is not an out-neighbor of \(x\) in \(\widetilde{P}_{2j}\), then choose such a vertex as \(u_{1,d}\), with \(d\
otin\{1,a\}\). If there is no such vertex, then choose any vertex as \(u_{1,d}\), with \(d\
otin\{1,a\}\). In \(H(u_{2})\), with \(S_{2}=\{y,u_{2,d}\}\) and \(r_{2}=y\), it is known that there are at least \(\kappa(G)\) internally disjoint \((S_{2},r_{2})\)-paths, denoted as \(\widetilde{P}_{1i}\)\((i\in[\kappa(G)])\). For each \(i\in[\kappa(G)]\), let \(u_{2,f_{i}}\) be the out-neighbor of \(y\) in \(\widetilde{P}_{1i}\); clearly these out-neighbors are distinct. For each \(i\in[\kappa(G)]\), since \(\delta^{0}(G)\geq 4\), an out-neighbor of \(u_{2,f_{i}}\) in \(G(r_{f_{j}})\), denoted by \(u_{b,f_{i}}\)\((b\in[n])\), can be chosen, with \(b\
otin\{1,3\}\). If there exists a vertex \(u_{s_{j}1}\
otin\{u_{1,},u_{3,1}\}\), let \(b=s_{j}\).