# On the Outer-Independent Roman Domination in Graphs
## 3 Rooted Product Graphs
Hence, \(x\in V_{0}\) and as \(V_{0}\cap V(H_{x})\) is an independent set, we deduce that \(N(x)\cap V(H_{x})\subseteq V_{1}\), which completes the proof. From Lemma 4 (i) we deduce that any \(\gamma_{oiR}(G\circ_{v}H)\)-function \(f\) induces three subsets \(\mathcal{A}_{f}\), \(\mathcal{B}_{f}\) and \(\mathcal{C}_{f}\) of \(V(G)\) as follows. \[\mathcal{A}_{f} =\{x\in V(G):\omega(f_{x})>\gamma_{oiR}(H)\},\] \[\mathcal{B}_{f} =\{x\in V(G):\omega(f_{x})=\gamma_{oiR}(H)\},\] \[\mathcal{C}_{f} =\{x\in V(G):\omega(f_{x})=\gamma_{oiR}(H)-1\}.\]  
Next, we state the four possible values of \(\gamma_{oiR}(G\circ_{v}H)\). **Theorem 8**. : _Let \(G\) and \(H\) be two graphs with no isolated vertex and \(|V(G)|=n\). If \(v\in V(H)\), then_  
\[\gamma_{oiR}(G\circ_{v}H)\in\{\alpha(G)+n\gamma_{oiR}(H),\,n\gamma_{oiR}(H), \,\gamma_{oiR}(G)+n(\gamma_{oiR}(H)-1),\,\alpha(G)+n(\gamma_{oiR}(H)-1)\}.\]  
Proof. : Let \(f(V_{0},V_{1},V_{2})\) be a \(\gamma_{oiR}(G\circ_{v}H)\)-function. By Lemma 3 (i) and (ii) we deduce the upper bound \(\gamma_{oiR}(G\circ_{v}H)\leq\alpha(G)+n\gamma_{oiR}(H)\). Now, we consider the subsets \(\mathcal{A}_{f},\mathcal{B}_{f},\mathcal{C}_{f}\subseteq V(G)\) associated to \(f\) and distinguish the following cases. Case 1. \(\mathcal{C}_{f}=\emptyset\). In this case, for any \(x\in V(G)\) we have that \(\omega(f_{x})\geq\gamma_{oiR}(H)\) and, as a consequence, \(\gamma_{oiR}(G\circ_{v}H)=\omega(f)\geq n\gamma_{oiR}(H)\). If \(\mathcal{A}_{f}=\emptyset\), then \(\gamma_{oiR}(G\circ_{v}H)=n\gamma_{oiR}(H)\). Hence, assume that \(\mathcal{A}_{f}\
eq\emptyset\). This implies that \(\omega(f)>n\gamma_{oiR}(H)\). Moreover, we note that \(\mathcal{B}_{f}\
eq\emptyset\) because \(\alpha(G)<n\) and\(\omega(f)\leq\alpha(G)+n\gamma_{oiR}(H)\). Thus, by Lemma 3 (ii) we obtain that \(\mathcal{B}_{f}\subseteq V_{0}\), and as \(V_{0}\) is an independent set, we have that \(\mathcal{A}_{f}\) is a vertex cover of \(G\). Therefore,  
\[\gamma_{oiR}(G\circ_{v}H) =\sum_{x\in\mathcal{A}_{f}}\omega(f_{x})+\sum_{x\in\mathcal{B}_{ f}}\omega(f_{x})\] \[\geq\sum_{x\in\mathcal{A}_{f}}(\gamma_{oiR}(H)+1)+\sum_{x\in \mathcal{B}_{f}}\gamma_{oiR}(H)\] \[=|\mathcal{A}_{f}|+\sum_{x\in V(G)}\gamma_{oiR}(H)\] \[\geq\alpha(G)+n\gamma_{oiR}(H).\]  
Hence, \(\gamma_{oiR}(G\circ_{v}H)=\alpha(G)+n\gamma_{oiR}(H)\). Case 2. \(\mathcal{C}_{f}\
eq\emptyset\). Let \(z\in\mathcal{C}_{f}\). By Lemma 4 (ii) we obtain that \(z\in V_{0}\) and \(N(z)\cap V(H_{z})\subseteq V_{1}\). Hence, \(f_{z}^{-}\) is an OIRDF on \(H_{z}-z\), and so \(\gamma_{oiR}(H-v)=\gamma_{oiR}(H_{z}-z)\leq\omega(f_{z}^{-})=\gamma_{oiR}(H)-1\). Thus, Lemma 2 leads to \(\gamma_{oiR}(H_{z}-z)=\gamma_{oiR}(H)-1\). This implies that \(f_{z}^{-}\) is a \(\gamma_{oiR}(H_{z}-z)\)-function which satisfies Lemma 3 (iii). Therefore, \(\gamma_{oiR}(G\circ_{v}H)\leq\gamma_{oiR}(G)+n(\gamma_{oiR}(H)-1)\). Now, observe the following inequality chain. \[\gamma_{oiR}(G\circ_{v}H)=\sum_{x\in\mathcal{A}_{f}\cup\mathcal{B}_{f}}\omega (f_{x})+\sum_{x\in\mathcal{C}_{f}}\omega(f_{x})\geq(2|\mathcal{A}_{f}|+| \mathcal{B}_{f}|)+n(\gamma_{oiR}(H)-1). \tag{1}\]  
By Lemma 4 (ii) we have that \(\mathcal{C}_{f}\subseteq V_{0}\), which implies that \(\mathcal{A}_{f}\cup\mathcal{B}_{f}\) is a vertex cover of \(G\). Thus, Inequality chain (1) leads to \(\gamma_{oiR}(G\circ_{v}H)=\omega(f)\geq\alpha(G)+n(\gamma_{oiR}(H)-1)\). Next, we consider the following two subcases. Subcase 1. There exists a \(\gamma_{oiR}(H)\)-function \(g\) such that \(g(v)=2\). Let \(D\) be an \(\alpha(G)\)-set. From \(D\), \(g\) and \(f_{z}\), we define a function \(h\) on \(G\circ_{v}H\) as follows. For every \(x\in D\), the restriction of \(h\) to \(V(H_{x})\) is induced from \(g\). Moreover, if \(x\in V(G)\setminus D\), then the restriction of \(h\) to \(V(H_{x})\) is induced from \(f_{z}\). By the construction of \(g\) and \(f_{z}\), it is straightforward to see that \(h\) is an OIRDF on \(G\circ_{v}H\).