# Directed Path 3-Arc-Connectivity of Cartesian Product Digraphs
## 3 A General Lower Bound
The general idea of the proof process is briefly described in Figure 2.7 The vertices and paths contained in Figure 7 are explained below. **Subcase 6.1. ** In the set \(\{u_{s_{j},2},u_{k_{j},2}\}\), there does not exist \(u_{3,2}\in\{u_{s_{j},2},u_{k_{j},2}\}\). Thus, \(u_{s_{j},b}\), \(u_{k_{j},b}\), \(\widetilde{P}_{2j}\), \(\widetilde{P}_{2j}^{\prime\prime}\) remain the same as in Case 5. In \(H(u_{3})\), with \(S_{1}=\{x,u_{3,c}\}\)\((c\in[m]\setminus\{1,2,b\})\) and \(r_{1}=x\), it is known that there exist at least \(\kappa(G)\) internally disjoint \((S_{1},r_{1})\)-paths in \(H(u_{3})\), denoted as \(\widetilde{P}_{1i}\). In \(G(v_{c})\), with \(S_{2}=\{u_{3,c},u_{1,c}\}\) and \(r_{2}=u_{3,c}\), it is known that there exist at least \(\kappa(G)\) internally disjoint \((S_{2},r_{2})\)-paths in \(G(v_{c})\), denoted as \(\widetilde{P}_{1i}\). In \(H(u_{1})\), with \(S_{3}=\{u_{1,c},y\}\) and \(r_{3}=u_{1,c}\), it is known that there exist at least \(\kappa(G)\) internally disjoint \((S_{3},r_{3})\)-paths in \(H(u_{1})\), denoted as \(\widetilde{P}_{1i}\). If \(u_{3,2}\in\widetilde{P}_{1r}\), then \(u_{3,2}\
otin\widetilde{P}_{1r}\). Let  
\(P_{1i}=\widetilde{P}_{1i}\cup\overline{P}_{1j}\cup\widetilde{P}_{1i}\cup \widetilde{P}_{1i}\),  
\(P_{2j}=\widetilde{P}_{2j}\cup\overline{P}_{2j}^{\prime\prime}\cup\overline{ P}_{2j}\cup\{zu_{k_{j},2}u_{k_{j},b},u_{s_{j},2}y\}\). If \(u_{s_{j},2}=u_{k_{j},2}\)\((t\in[\kappa(G)])\), then \(P_{2t}=\widetilde{P}_{2t}\cup\widetilde{P}_{2t}\cup\{zu_{k_{j},2}u_{s_{j},2}y\}\). And if \(u_{k_{j},2}=y\)\((l\in[\kappa(G)])\), then \(P_{2t}=\widetilde{P}_{2t}\cup\bar{P}_{2t}\cup\{zy\}\). Now we obtain \(2\kappa(G)\) arc-disjoint \((S,r)\)-paths. **Subcase 6.2. ** In the set \(\{u_{s_{j},2},u_{k_{j},2}\}\), only one vertex \(u_{k_{r},2}=u_{3,2}\)\((r\in[\kappa(G)])\) exists. Thus, \(u_{s_{j},b}\), \(u_{k_{j},b}\), \(\bar{P}_{2j}\), \(\widetilde{P}_{2j}^{\prime\prime}\) remain the same as in Case 5. If \(u_{k_{r},2}u_{k_{r},b}\
otin\widetilde{P}_{1i}\) in \(\widetilde{P}_{1i}\), then \(P_{1i}\), \(P_{2j}\) remain the same as in Subcase 6.1. If an arc \(u_{k_{r},2}u_{k_{r},b}\) is in path \(\widetilde{P}_{1i}\), since \(\delta(G)\geq 4\), then an out-neighbor \(u_{k_{r},a}\) and \(u_{k_{r},2}\) can be found in \(H(u_{3})\) such that \(u_{k_{r},2}u_{k_{r},a}\
otin\widetilde{P}_{1i}\) and \(a\in[m]\setminus\{c,1\}\). In \(G(v_{a})\), \(\widetilde{P}_{2r}^{\prime\prime\prime}\) is the \((S_{3}^{\prime},r_{3}^{\prime})\)-path corresponding to \(\widetilde{P}_{2r}^{\prime\prime}\), where \(S_{3}^{\prime}=\{u_{k_{r},a},u_{s_{r},a}\}\), \(r_{3}^{\prime}=u_{k_{r},a}\). In \(H(u_{s_{r}})\), with \(S_{4}^{\prime}=\{u_{s_{r},a},u_{s_{r},2}\}\) and \(r_{4}^{\prime}=u_{s_{r},a}\), it is known that there exist at least \(\kappa(G)\) internally disjoint \((S_{4}^{\prime},r_{4}^{\prime})\)-paths. Then in these paths, one of the paths \(\widetilde{P}_{2r}^{\prime}\), is chosen, with \(u_{s_{r},1}\
otin\widetilde{P}_{2r}^{\prime}\). \(P_{2j}(j\
eq r)\) and \(P_{1i}\) remain the same as in Subcase 6.1. \(P_{2r}\) is constructed as  
\(P_{2r}=\widetilde{P}_{2r}\cup\widetilde{P}_{2r}^{\prime\prime}\cup\bar{P}_{2r} \cup\{\widetilde{P}_{2r}^{\prime}\cup\{zu_{k_{r},2}u_{k_{r},a},u_{s_{r},2}y\}\). Figure 2.8 Depiction of the arc-disjoint paths found in Case 6 of the proof of Theorem 1. **Subcase 6.3**. : In the set \(\{u_{s_{j},2},u_{k_{j},2}\}\), there is only one vertex \(u_{s_{g},2}=u_{3,2}\) (\(g\in[\kappa(G)]\)). For each \(j\in[\kappa(G)]\), an in-neighbor of \(u_{s_{j},1}\) in \(H(u_{s_{j}})\) can be chosen, denoted by \(u_{s_{j},d}\) (\(d\in[m]\)), where \(d\
eq c,1\). In \(G(v_{d})\), let \(\overline{P}^{\prime}_{2j}\) be the \((S^{\prime}_{5},r^{\prime}_{5})\)-path corresponding to \(\overline{P}_{2j}\), where \(S^{\prime}_{5}=\{u_{2,d},u_{1,d}\}\), \(r^{\prime}_{5}=u_{2,d}\). The path from vertex \(u_{k_{j},d}\) to \(u_{s_{j},d}\) in path \(\overline{P}^{\prime}_{2j}\) is denoted as \(\overline{P}^{\prime\prime}_{2j}\).