## 4 Identifying Subproblem Solutions and Defining the Recursive Formulas
Given \(C_{T}(G_{\alpha})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha})\
eq\emptyset\), we consider four cases:  
Case 1: \(C_{T}(G_{\alpha_{1}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{1}})\
eq\emptyset\); \(C_{T}(G_{\alpha_{2}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{2}})\
eq\emptyset\);  
Case 2: \(C_{T}(G_{\alpha_{i}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{i}})\
eq\emptyset\);  
Case 3: \(C_{T}(G_{\alpha_{i}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{i}})\
eq\emptyset\); \(C_{T}(G_{\alpha_{3-i}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{3-i}})=\emptyset\);  
Case 4: \(C_{T}(G_{\alpha_{i}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{i}})=\emptyset\); \(C_{T}(G_{\alpha_{3-i}})=\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{3-i}})\
eq\emptyset\). In the following, we prove the statement for Case 1. The statements for the other cases can be proven similarly. Let \(f_{1}\) and \(f_{2}\) be \(\Gamma_{p,2}^{d,\ell}(x,y,z)\)-clique transversal functions of \(G_{\alpha_{1}}\) and \(G_{\alpha_{2}}\), respectively. Let \(f\) be a function of \(G_{\alpha}\) such that \(f(v)=f_{1}(v)\) for every \(v\in V(G_{\alpha_{1}})\), and \(f(v)=f_{2}(v)\) for every \(v\in V(G_{\alpha_{2}})\). For every maximal clique \(C\in C_{\overline{T}}(G_{\alpha_{1}})\), either \(C\in C_{\overline{T}}(G_{\alpha_{2}})\). Therefore, either \(f(C)=f_{1}(C)\geq x\) or \(f(C)=f_{2}(C)\geq x\). Similarly, either \(f(C)=f_{1}(C)\geq y\) or \(f(C)=f_{2}(C)\geq y\) for every maximal clique \(C\in C_{T}(G_{\alpha})\). For each clique \(C\in C(G_{\alpha}[TS(G_{\alpha})])\), either \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\) or \(C\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\), and, thus, either \(f(C)=f_{1}(C)\geq z\) or \(f(C)=f_{1}(C)\geq z\). The function \(f\) is a \(\Gamma_{p,2}^{d,\ell}(x,y,z)\)-clique transversal function of \(G_{\alpha}\). Hence,  
\[\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha})\leq\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha_{ 1}})+\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha_{2}}).\]  
Conversely, let \(f\) be a \(\Gamma_{p,2}^{d,\ell}(x,y,z)\)-clique transversal function of \(G_{\alpha}\). Let \(f_{1}\) be a function of \(G_{\alpha_{1}}\) such that \(f_{1}(v)=f(v)\) for every \(v\in V(G_{\alpha_{1}})\), and let \(f_{2}\) be a function of \(G_{\alpha_{2}}\) such that \(f_{2}(v)=f(v)\) for every \(v\in V(G_{\alpha_{2}})\). Consider the function \(f_{1}\). It is straightforward that \(C\in C_{\overline{T}}(G_{\alpha})\) (respectively, \(C\in C_{T}(G_{\alpha})\) and \(C\in C(G_{\alpha}[TS(G_{\alpha})])\)) for every \(C\in C_{\overline{T}}(G_{\alpha_{1}})\) (respectively, \(C\in C_{T}(G_{\alpha_{1}})\) and \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\). Therefore, \(f_{1}(C)=f(C)\geq x\) (respectively, \(f_{1}(C)=f(C)\geq y\) and \(f_{1}(C)=f(C)\geq z\)) for every \(C\in C_{T}(G_{\alpha_{1}})\) (respectively, \(C\in C_{T}(G_{\alpha_{1}})\) and \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\)). Similarly, \(f_{2}(C)=f(C)\geq x\) (respectively, \(f_{2}(C)=f(C)\geq y\) and \(f_{2}(C)=f(C)\geq z\)) for every \(C\in C_{\overline{T}}(G_{\alpha_{2}})\) (respectively, \(C\in C_{T}(G_{\alpha_{2}})\) and \(C\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\)). Thus, the functions \(f_{1}\) and \(f_{2}\) are \(\Gamma_{p,2}^{d,\ell}(x,y,z)\)-clique transversal functions of \(G_{\alpha_{1}}\) and \(G_{\alpha_{2}}\), respectively. We have  
\[\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha_{1}})+\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha_{ 2}})\leq\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha}).\]  
The above discussion shows that \(\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha})=\tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha_{1}})+ \tau_{p,2}^{d,\ell}(x,y,z)(G_{\alpha_{2}})\). Statement (3.3.1) therefore holds. **Lemma 8**. : _Suppose that \(G\) is a distance-hereditary graph and \(G_{\alpha}=G_{\alpha_{1}}\bigotimes G_{\alpha_{2}}\) is an induced subgraph of \(G\) formed from two disjoint distance-hereditary graphs \(G_{\alpha_{1}}\) and \(G_{\alpha_{2}}\).