## 4 Identifying Subproblem Solutions and Defining the Recursive Formulas
Since \(G_{\alpha}=G_{\alpha_{1}}\bigodot G_{\alpha_{2}}\), by Theorem 1, we have \(G_{\alpha}=G_{\alpha_{1}}\cup G_{\alpha_{2}}\) and \(TS(G_{\alpha})=TS(G_{\alpha_{1}})\cup TS(G_{\alpha_{2}})\). Following Lemma 2, we know that\(C(G_{\alpha}[TS(G_{\alpha})])=C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\cup C(G_{ \alpha_{2}}[TS(G_{\alpha_{2}})])\); and
* \(C_{\overline{\tau}}(G_{\alpha})=C_{\overline{\tau}}(G_{\alpha_{1}})\cup C_{ \overline{\tau}}(G_{\alpha_{2}})\). Let \(f_{1}\) and \(f_{2}\) be \(\Gamma_{p,0}^{d,\ell}(x,y)\)-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 \(C\in C_{\overline{\tau}}(G_{\alpha_{1}})\), either \(C\in C_{\overline{\tau}}(G_{\alpha_{2}})\). 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 \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). The function \(f\) is a \(\Gamma_{p,0}^{d,\ell}(x,y)\)-clique transversal function of \(G_{\alpha}\). Hence, \[\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha})\leq\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha_{1} })+\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha_{2}}).\] Conversely, let \(f\) be a \(\Gamma_{p,0}^{d,\ell}(x,y)\)-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 \(v\in V(G_{\alpha_{2}})\). For every \(C\in C_{\overline{\tau}}(G_{\alpha_{1}})\) (respectively, \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\), \(C\in C_{\overline{\tau}}(G_{\alpha})\) (respectively, \(C\in C(G_{\alpha}[TS(G_{\alpha})])\)). Therefore, \(f_{1}(C)=f(C)\geq x\) (respectively, \(f_{1}(C)\geq y\)) for every \(C\in C_{\overline{\tau}}(G_{\alpha_{1}})\) (respectively, \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\)). Similarly, \(f_{2}(C)=f(C)\geq x\) (respectively, \(f_{2}(C)\geq y\)) for every \(C\in C_{\overline{\tau}}(G_{\alpha_{2}})\) (respectively, \(C\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\)). Thus, the functions \(f_{1}\) and \(f_{2}\) are \(\Gamma_{p,0}^{d,\ell}(x,y)\)-clique transversal functions of \(G_{\alpha_{1}}\) and \(G_{\alpha_{2}}\), respectively. Consequently, \[\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha_{1}})+\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha_{2 }})\leq\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha}).\] The above discussion shows that \(\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha})=\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha_{1}})+ \tau_{p,0}^{d,\ell}(x,y)(G_{\alpha_{2}})\). Statement (1.3) therefore holds. **(2) Assume that \(C_{\overline{T}}(G_{\alpha})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha})=\emptyset\). ** Following Definition 7, we can conclude that Statements (2.1) and (2.2) are true. We now consider Statement (2.3). Following Lemma 2, we know that  
* \(C(G_{\alpha}[TS(G_{\alpha})])=C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\cup C(G_{ \alpha_{2}}[TS(G_{\alpha_{2}})])\); and
* \(C_{\overline{T}}(G_{\alpha})=C_{\overline{T}}G_{\alpha_{1}})\cup C_{ \overline{T}}(G_{\alpha_{2}})\). Recall that \(G_{\alpha}=G_{\alpha_{1}}\cup G_{\alpha_{2}}\) and \(TS(G_{\alpha})=TS(G_{\alpha_{1}})\cup TS(G_{\alpha_{2}})\). Since \(C_{\overline{\tau}}(G_{\alpha})=\emptyset\), we have \(C(G_{\alpha})=C_{\overline{T}}(G_{\alpha})=C(G_{\alpha}[TS(G_{\alpha})])\). Therefore, \(C_{\overline{T}}(G_{\alpha_{1}})=C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\) and \(C_{\overline{T}}(G_{\alpha_{2}})=C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\). Let \(f_{1}\) and \(f_{2}\) be \(\Gamma_{p,1}^{d,\ell}(x,y)\)-clique transversal functions of \(G_{\alpha_{1}}\) and \(G_{\alpha_{2}}\), respectively. By Definition 7, \(f_{1}(v)\geq x\) for every \(C\in C_{\overline{T}}(G_{\alpha_{1}})\), and \(f(C_{1})\geq y\) for every \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\).