## 4 Identifying Subproblem Solutions and Defining the Recursive Formulas
Note that \(C_{\overline{T}}(G_{\alpha_{1}})=C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\). Therefore, \(f_{1}(C)\geq\max\{x,y\}\) for every \(C\in C_{\overline{T}}(G_{\alpha_{1}})\) and \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\). Similarly, \(f_{2}(C)\geq\max\{x,y\}\) for every \(C\in C_{\overline{T}}(G_{\alpha_{2}})\) and \(C\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\). Let \(w=\max\{x,y\}\). Thus, \(f_{1}\) and \(f_{2}\) are \(\Gamma_{p,1}^{d,\ell}(w,w)\)-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 \(C\in C_{\overline{T}}(G_{\alpha_{1}})\) or \(C\in C_{\overline{T}}(G_{\alpha_{2}})\). Hence, either \(f(C)=f_{1}(C)\geq w\) or \(f(C)=f_{2}(C)\geq w\). Similarly, either \(f(C)=f_{1}(C)\geq w\) or \(f(C)=f_{2}(C)\geq w\) for every \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). The function \(f\) is a \(\Gamma_{p,1}^{d,\ell}(w,w)\)-clique transversal function of \(G_{\alpha}\). Consequently, \[\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})\leq\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha_{1} })+\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha_{2}}).\] Conversely, let \(f\) be a \(\Gamma_{p,1}^{d,\ell}(x,y)\)-clique transversal function of \(G_{\alpha}\), and let \(w=\max\{x,y\}\). By Definition 7, \(f(C)\geq x\) for every \(C\in C_{\overline{T}}(G_{\alpha})\), and \(f(C)\geq y\) for every \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). Note that \(C_{\overline{T}}(G_{\alpha})=C(G_{\alpha}[TS(G_{\alpha})])\). Therefore, \(f(C)\geq\max\{x,y\}\) for every \(C\in C_{\overline{T}}(G_{\alpha})\) and \(C\in C(G_{\alpha}[TS(G_{\alpha})])\). The function \(f\) is a \(\Gamma_{p,1}^{d,\ell}(w,w)\)-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 \(v\in V(G_{\alpha_{2}})\). Recall 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
* \(C_{T}(G_{\alpha})=C_{T}G_{\alpha_{1}})\cup C_{T}(G_{\alpha_{2}})\). Thus, \(f_{1}(C)=f(C)\geq w\) for every \(C\in C_{T}(G_{\alpha_{1}})\) and \(C\in C(G_{\alpha_{1}}[TS(G_{\alpha_{1}})])\). Similarly, \(f_{2}(C)=f(C)\geq w\) for every \(C\in C_{T}(G_{\alpha_{2}})\) and \(C\in C(G_{\alpha_{2}}[TS(G_{\alpha_{2}})])\). The functions \(f_{1}\) and \(f_{2}\) are \(\Gamma_{p,1}^{d,\ell}(w,w)\)-clique transversal functions of \(G_{\alpha_{1}}\) and \(G_{\alpha_{2}}\), respectively. Consequently,  
\[\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha_{1}})+\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha_{2 }})\leq\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha}).\]  
The above discussion shows that \(\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})=\tau_{p,1}^{d,\ell}(w,w)(G_{\alpha_{1}})+ \tau_{p,1}^{d,\ell}(w,w)(G_{\alpha_{2}})\). Statement (2.3) therefore holds. **(3) Assume that \(C_{T}(G_{\alpha})\
eq\emptyset\) and \(C_{\overline{T}}(G_{\alpha})\
eq\emptyset\). ** By Definition 7, \(\tau_{p,0}^{d,\ell}(x,y)(G_{\alpha})=\infty\) and \(\tau_{p,1}^{d,\ell}(x,y)(G_{\alpha})=\infty\). Statements (3.1) and (3.2) therefore hold. Next, we consider the remaining statements. Note that \(C_{T}(G_{\alpha})=C_{T}(G_{\alpha_{1}})\cup C_{T}(G_{\alpha_{2}})\) and \(C_{\overline{T}}(G_{\alpha})=C_{\overline{T}}(G_{\alpha_{1}})\cup C_{\overline {T}}(G_{\alpha_{2}})\). Let \(i=\{1,2\}\).