# A Fault Section Location Method for Distribution Networks Based on Divide-and-Conquer
## Appendix B The Proofs of the Properties of Compatible Sets  
**Proof of Theorem 4.**  
1. [label=(0)]
2. Sufficiency. \(\forall x_{i},x_{k}\in\mathbf{X}_{a}\), there are obviously \(x_{i},x_{k}\in\mathbf{X}_{a}\cup\mathbf{X}_{b}\). Because \(\mathbf{X}_{a}\cup\mathbf{X}_{b}\) satisfies the inequality constraint, \(\mathbf{P}_{i}\
eq\mathbf{P}_{i}*\mathbf{P}_{k}\), i.e., \(\mathbf{X}_{a}\) satisfies the inequality constraint. Similarly, \(\mathbf{X}_{b}\) satisfies the inequality constraint. Then, \(\forall x_{i}\  
1. Suppose for the sake of contradiction that \(x_{r}\in C_{i}\) but \(\mathbf{P}_{k}=\mathbf{P}_{k}*\mathbf{P}_{r}\). It is known that section \(x_{i}\) is on the road of section \(x_{k}\), and according to Theorem 1, \(\mathbf{P}_{i}=\mathbf{P}_{i}*\mathbf{P}_{k}\), substituting the assumption \(\mathbf{P}_{i}=\mathbf{P}_{i}*\mathbf{P}_{k}=\mathbf{P}_{i}*(\mathbf{P}_{k}*\mathbf{P}_{r})=(\mathbf{P}_{i}* \mathbf{P}_{k})*\mathbf{P}_{r}=\mathbf{P}_{i}*\mathbf{P}_{r}\) contradicts the assumption \(x_{r}\in C_{i}\).
2. Suppose for the sake of contradiction that \(x_{r}\in C_{i}\) but \(\mathbf{P}_{r}=\mathbf{P}_{k}*\mathbf{P}_{r}\). According to Theorem 2, \(\mathbf{P}_{k}*\mathbf{P}_{r}*(e-\mathbf{P}_{i})=0\) or \(\mathbf{P}_{k}*\mathbf{P}_{i}*(e-\mathbf{P}_{r})=0\). If the former equation holds, then \(\mathbf{P}_{k}*\mathbf{P}_{r}=\mathbf{P}_{k}*\mathbf{P}_{r}*\mathbf{P}_{i}\), substituting the assumptions \(\mathbf{P}_{r}=\mathbf{P}_{k}*\mathbf{P}_{r}=\mathbf{P}_{k}*\mathbf{P}_{r}*\mathbf{P}_{i}=\mathbf{P}_{r}* (\mathbf{P}_{k}*\mathbf{P}_{i})=\mathbf{P}_{r}*\mathbf{P}_{i}\), and contradicting the assumption \(x_{r}\in C_{i}\). If the latter equation holds, then \(\mathbf{P}_{k}*\mathbf{P}_{i}=\mathbf{P}_{k}*\mathbf{P}_{i}*\mathbf{P}_{r}\). Because \(x_{i}\) is on the road of \(x_{k}\), \(\mathbf{P}_{i}=\mathbf{P}_{k}*\mathbf{P}_{i}\), \(\mathbf{P}_{i}{=}\mathbf{P}_{k}*\mathbf{P}_{k}=\mathbf{P}_{k}*\mathbf{P}_{i}*\mathbf{P}_{r}=\mathbf{P}_{r}* (\mathbf{P}_{k}*\mathbf{P}_{i})=\mathbf{P}_{r}*\mathbf{P}_{i}\), contradicting the assumption \(x_{r}\in C_{i}\). \(\square\)  
Proof of Theorem 6.: According to Theorem 5, \(C_{i}\subseteq C_{k}\); we need to further prove \(C_{k}\subseteq C_{i}\). Transform the problem: \(\forall x_{r}\in C_{k}\), prove \(x_{r}\in C_{i}\), i.e., \(\mathbf{P}_{r}\
eq\mathbf{P}_{i}*\mathbf{P}_{r}\) and \(\mathbf{P}_{i}\
eq\mathbf{P}_{i}*\mathbf{P}_{r}\).  
1. Suppose for the sake of contradiction that \(x_{r}\in C_{k}\) but \(\mathbf{P}_{r}=\mathbf{P}_{i}*\mathbf{P}_{r}\). It is known that section \(x_{i}\) is on the road of section \(x_{k}\), and according to Theorem 1: \(\mathbf{P}_{i}=\mathbf{P}_{k}*\mathbf{P}_{i}\). Substitute the assumption \(\mathbf{P}_{r}=\mathbf{P}_{i}*\mathbf{P}_{r}=(\mathbf{P}_{k}*\mathbf{P}_{i})*\mathbf{P}_{r}=\mathbf{P}_{k }*(\mathbf{P}_{i}*\mathbf{P}_{r})=\mathbf{P}_{k}*\mathbf{P}_{r}\), contradicting the assumption \(x_{r}\in C_{k}\).
2. Suppose for the sake of contradiction that \(x_{r}\in C_{k}\) but \(\mathbf{P}_{i}=\mathbf{P}_{i}*\mathbf{P}_{r}\). According to Theorem 1, section \(x_{i}\) is on the road of section \(x_{r}\). We claim that there is a path between section \(x_{r}\) and section \(x_{k}\). Otherwise, under the premise that section \(x_{i}\) is simultaneously on the road of \(x_{r}\) and \(x_{k}\), the only possibility that there is no path between \(x_{r}\) and \(x_{k}\), as shown in Figure A3. As can be seen in Figure A3, section \(x_{j}\) is a T-type section, contradicting the condition that the path from \(x_{k}\) to \(x_{i}\) does not contain a T-type section; thus, there is a path between section \(x_{r}\) and section \(x_{k}\). That is, \(x_{j}\) is on the road of \(x_{k}\) or \(x_{k}\) is on the road of \(x_{r}\). According to Theorem 1, \(\mathbf{P}_{r}=\mathbf{P}_{r}*\mathbf{P}_{k}\) or \(\mathbf{P}_{k}=\mathbf{P}_{r}*\mathbf{P}_{k}\), contradicting the assumption \(x_{r}\in C_{k}\). \(\square\)