## 4 Identifying Subproblem Solutions and Defining the Recursive Formulas
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 \(v\in V(G_{\alpha_{2}})\). By Lemma 3, \(C_{T}(G_{\alpha})=C_{T}(G_{\alpha_{1}})\cup C_{\overline{T}}(G_{\alpha_{2}})\). For every maximal clique \(C\in C_{\overline{T}}(G_{\alpha})\), either \(C\in C_{\overline{T}}(G_{\alpha_{1}})\) or \(C\in C_{\overline{T}}(G_{\alpha_{2}})\). Therefore, \(f(C)=f_{1}(C)\geq x\) for every \(C\in C_{\overline{T}}(G_{\alpha_{1}})\), and \(f(C)=f_{2}(C)\geq x\) for \(C\in C_{\overline{T}}(G_{\alpha_{2}})\). For each maximal clique \(C\in C_{T}(G_{\alpha})\), \(C\) is also a maximal clique of \(G_{\alpha}[TS(G_{\alpha})]\) and it is the union of a clique \(C_{1}\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\) and a clique \(C_{2}\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\). Thus, \(f(C)=f(C_{1})+f(C_{2})=f_{1}(C_{1})+f_{2}(C_{2})\geq w_{1}+w_{2}=w\). The function \(f\) is a \(\Gamma_{p,2}^{d,\ell}(x,w,w)\)-clique transversal function of \(G_{\alpha}\). Hence,  
\[\tau_{p,2}^{d,\ell}(x,w,w)(G_{\alpha})\leq\tau_{p,2}^{d,\ell}(x,w_{1},w_{1})(G_{ \alpha_{1}})+\tau_{p,2}^{d,\ell}(x,w_{2},w_{2})(G_{\alpha_{2}}).\]  
Conversely, let \(f\) be a \(\Gamma_{p,2}^{d,\ell}(x,w,w)\)-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 \(v\in V(G_{\alpha_{2}})\). By Lemma 3, \(C_{\overline{T}}(G_{\alpha})=C_{\overline{T}}(G_{\alpha_{1}})\cup C_{\overline{T} }(G_{\alpha_{2}})\). For each maximal clique \(C\in C_{\overline{T}}(G_{\alpha})\), either \(C\in C_{\overline{T}}(G_{\alpha_{1}})\) or \(C\in C_{\overline{T}}(G_{\alpha_{2}})\  
\(\hat{C}\in C(G_{\alpha_{2}})\). Thus, \(f_{1}\) is a \(\Gamma_{p,2}^{d,\ell}(x,w_{1},w_{1})\)-clique transversal function of \(G_{\alpha_{1}}\), and \(f_{2}\) is a \(\Gamma_{p,2}^{d,\ell}(x,w_{2},w_{2})\)-clique transversal function of \(G_{\alpha_{2}}\). Hence,  
\[\tau_{p,2}^{d,\ell}(x,w_{1},w_{1})(G_{\alpha_{1}})+\tau_{p,2}^{d,\ell}(x,w_{2}, w_{2})(G_{\alpha_{2}})\leq\tau_{p,2}^{d,\ell}(x,w,w)(G_{\alpha}).\]  
We have \(\tau_{p,2}^{d,\ell}(x,w,w)(G_{\alpha})=\tau_{p,2}^{d,\ell}(x,w_{1},w_{1})(G_{ \alpha_{1}})+\tau_{p,2}^{d,\ell}(x,w_{2},w_{2})(G_{\alpha_{2}})\). Statement (2.3.1) therefore holds. **Lemma 9**. : _Suppose that \(G\) is a distance-hereditary graph and \(G_{\alpha}=G_{\alpha_{1}}\bigoplus G_{\alpha_{2}}\) is an induced subgraph of \(G\) formed from two disjoint distance-hereditary graphs \(G_{\alpha_{1}}\) and \(G_{\alpha_{2}}\). Let \(p\in\{0,-1\}\), and let \(d\), \(\ell,x,y,z\) be integers such that \(d\), \(\ell\in  
_(3.2.3): If \(C_{T}(G_{\alpha_{2}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{2}})=\emptyset\), then_  
\[\tau^{d,\ell}_{p,0}(x,y)(G_{\alpha})=\min_{ \begin{subarray}{c}x_{1},x_{2}\in \mathbb{Z}\\ x_{1}+x_{2}=x\\ y\leq x_{1}\leq u_{\max}\\ w_{\min}\leq_{2}\leq x-y\end{subarray} }\Big{\{}\tau^{d,\ell}_{p,0}(x_{1},x_{1 })(G_{\alpha_{1}})+\tau^{d,\ell}_{p,1}(x_{2},x_{2})(G_{\alpha_{2}})\Big{\}};\]  
_(3.3): Assume that \(C_{T}(G_{\alpha_{1}})=\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{1}})\
eq\emptyset\); (3.3.1): If \(C_{T}(G_{\alpha_{2}})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{2}})\
eq\emptyset\), then_  
\[\tau^{d,\ell}_{p,0}(x,y)(G_{\alpha})=\min_{ \begin{subarray}{c}x_{1},x_{2}\in \mathbb{Z}\\ y\leq x_{1}\leq u_{\max}\\ w_{\min}\leq_{2}\leq x-y\end{subarray} }\Big{\{}\tau^{d,\ell}_{p,0}(x,x_{1})(G _{\alpha_{1}})+\tau^{d,\ell}_{p,2}(x,x_{2},x_{2})(G_{\alpha_{2}})\Big{\}};\]  
_(3.3.2): If \(C_{T}(G_{\alpha_{2}})=\emptyset\) and \(C_{\overline{T}}(G_{\alpha_{2}})=\emptyset\), then_  
\[\tau^{d,\ell}_{p,0}(x,y)(G_{\alpha})=\min_{ \begin{subarray}{c}x_{1},x_{2}\in \mathbb{Z}\\ x_{1}+x_{2}=x\\ y\leq x_{1}\leq u_{\max}\\ w_{\min}\leq_{2}\leq x-y\end{subarray}