# A Fault Section Location Method for Distribution Networks Based on Divide-and-Conquer
## Appendix D The Proofs of Recursive Structure
Similarly, \(f(\mathbf{X}^{\prime}\cup\mathbf{R})=-\Delta_{\mathbf{R}}^{k}-\Delta_{q}+f(\varnothing)\). Because there is no T-type section on the path between \(x_{k}\) and \(x_{q}\), according to Theorem 6, \(C_{k}=C_{q}\). \(\mathbf{R}\subseteq C_{k}\), according to the conclusion of step 1, and according to Theorem 7, \(\Delta_{\mathbf{R}}^{k}=\Delta_{\mathbf{R}}^{q}\). Combined with \(\Delta_{k}>\Delta_{q}\): \(f(\mathbf{X}_{i-t}\cup\mathbf{R})<f(\mathbf{X}^{\prime}\cup\mathbf{R})\). 2. \(\mathbf{X}^{\prime}\subseteq\mathbf{E}_{u}\). Because \(\mathbf{R}\subseteq\mathbf{E}-\mathbf{E}_{i}\subseteq\mathbf{E}-\mathbf{E}_{u}\), \(\mathbf{X}_{u}\) is the fault candidate set of \(\mathbf{G}_{u}\), according to the definition of fault candidate sets, \(f(\mathbf{X}_{u}\cup\mathbf{R})<f(\mathbf{X}^{\prime}\cup\mathbf{R})\). 3. \(\mathbf{X}^{\prime}\subseteq\mathbf{E}_{v}\). Because \(\mathbf{R}\subseteq\mathbf{E}-\mathbf{E}_{i}\subseteq\mathbf{E}-\mathbf{E}_{v}\), \(\mathbf{X}_{v}\) is the fault candidate set of \(\mathbf{G}_{v}\), according to the definition of fault candidate sets, \(f(\mathbf{X}_{v}\cup\mathbf{R})<f(\mathbf{X}^{\prime}\cup\mathbf{R})\). 4. \(\mathbf{X}^{\prime}\subseteq\mathbf{E}_{u}\cup\mathbf{E}_{v}\). Let the sets of sections \(\mathbf{X}_{a}=\mathbf{X}^{\prime}\cap\mathbf{E}_{u}\), \(\mathbf{X}_{b}=\mathbf{X}^{\prime}\cap\mathbf{E}_{v}\). Because \(\mathbf{E}_{u}\cap\mathbf{E}_{v}=\varnothing\), \(\mathbf{X}^{\prime}=\mathbf{X}_{u}\cup\mathbf{X}_{b}\). \(f(\mathbf{X}^{\prime}\cup\mathbf{R})\) can be equivalently expressed as \(f(\mathbf{X}^{\prime}\cup\mathbf{R})=f(\mathbf{X}_{a}\cup\mathbf{X}_{b}\cup\mathbf{R})\). It is known that \(\mathbf{X}_{u}\) is a fault candidate set of \(\mathbf{G}_{u}\), \(\mathbf{X}_{a}\subseteq\mathbf{E}_{u}\), \(\mathbf{X}_{b}\cup\mathbf{R}\subseteq\mathbf{E}-\mathbf{E}_{u}\), and therefore, \(f(\mathbf{X}_{u}\cup\mathbf{X}_{b}\cup\mathbf{R})<f(\mathbf{X}_{a}\cup\mathbf{X}_{b}\cup\mathbf{R})\). It is known that \(\mathbf{X}_{v}\) is a fault candidate set of \(\mathbf{G}_{v}\), \(\mathbf{X}_{b}\subseteq\mathbf{E}_{v}\), \(\mathbf{X}_{u}\cup\mathbf{R}\subseteq\mathbf{E}-\mathbf{E}_{v}\); therefore, \(f(\mathbf{X}_{u}\cup\mathbf{X}_{v}\cup\mathbf{R})<f(\mathbf{X}_{u}\cup\mathbf{X}_{b}\cup\mathbf{R})\). Combined with \(\mathbf{X}_{u\mathbf{k}\mathbf{v}}=\mathbf{X}_{u}\cup\mathbf{X}_{v}\),\(\mathbf{X}^{\prime}=\mathbf{X}_{a}\cup\mathbf{X}_{b}\): \(f(\mathbf{X}_{u\mathbf{k}\mathbf{v}}\cup\mathbf{R})<f(\mathbf{X}^{\prime}\cup\mathbf{R})\). Step 3: Prove that the set corresponding to the largest of \(\Delta_{\mathbf{X}_{i-t}}\), \(\Delta_{\mathbf{X}_{u}}\), \(\Delta_{\mathbf{X}_{v}}\) and \(\Delta_{\mathbf{X}_{u\mathbf{k}\mathbf{v}}}\) is the fault candidate set \(\mathbf{X}_{i}\) for \(\mathbf{G}_{i}\). Apply Equations (31) and (33) to rewrite \(f(\mathbf{X}_{i-t}\cup\mathbf{R})\), \(f(\mathbf{X}_{u}\cup\mathbf{R})\), \(f(\mathbf{X}_{v}\cup\mathbf{R})\), and \(f(\mathbf{X}_{u\mathbf{k}\mathbf{v}}\cup\mathbf{R})\) (the derivation process is similar to Equation (A8)). \[\left\{ \begin{array}{l}f(\mathbf{X}_{i-t}\cup\mathbf{R})=-\Delta_{\mathbf{R}}^{\mathbf{X}_{i-t }}-\Delta_{\mathbf{X}_{i-t}}+f(\varnothing)\\ f(\mathbf{X}_{u}\cup\mathbf{R})=-\Delta_{\mathbf{X}_{u}}^{\mathbf{X}_{u}}-\Delta_{\mathbf{X}_{u}}+f( \varnothing)\\ f(\mathbf{X}_{v}\cup\mathbf{R})=-\Delta_{\mathbf{R}}^{\mathbf{X}_{u}}-\Delta_{\mathbf{X}_{u\mathbf{k} \mathbf{v}}}+f(\varnothing)\\ f(\mathbf{X}_{u\mathbf{k}\mathbf{v}}\cup\mathbf{R})=-\Delta_{\mathbf{R}}^{\mathbf{X}_{u\mathbf{k}\mathbf{v}}}- \Delta_{\mathbf{X}_{u\mathbf{k}\mathbf{v}}}+f(\varnothing)\\ \end{array} \right.\] (A9)  
According to the argument process in Step 1, for \(x_{k}\in\mathbf{X}_{i-t}\), \(\forall x_{w}\in\mathbf{X}_{u}\), \(\forall x_{y}\in\mathbf{X}_{v}\), and \(\forall x_{z}\in\mathbf{X}_{u\mathbf{k}\mathbf{v}}\), there are \(C_{i}\subseteq C_{k}\), \(C_{i}\subseteq C_{w}\), \(C_{i}\subseteq C_{y}\), and \(C_{i}\subseteq C_{z}\); thus, \(\mathbf{X}_{i-t},\mathbf{X}_{u},\mathbf{X}_{v}\), \(\The proof of this conclusion follows.