# IFDA: Intermittent Fault Diagnosis Algorithm for Augmented Cubes Under the PMC Model
## Appendix A  
### Proof of the Lemma 9  
Let \(x,y,s,t,u\) be five nodes in \(AQ_{n}\), where \(n\geq 5\). Then, \(|N_{AQ_{n}}(\{x,y,s,t,u\})|\geq 10n-41\) holds.  
We use mathematical induction to conduct the proofs. The base case for \(AQ_{5}\) can be verified according to its structure diagram, i.e., let \(x,y,s,t,u\) be five nodes in \(AQ_{5}\); then, \(|N_{AQ_{5}}(\{x,y,s,t,u\})|\geq 10\times 5-41=9\) holds. Assume that the lemma holds for \(AQ_{n-1}\), where \(n\geq 6\). Now, we consider \(AQ_{n}=L\oplus R\), where \(L\cong AQ_{n-1}^{Q}\) and \(R\cong AQ_{n-1}^{1}\). For the distribution of \(x,y,s,t,u\), we consider the following three cases:  
**Case 1.** All of \(\{x,y,s,t,u\}\) are in the same subcube. We assume that \(x,y,s,t,u\in V_{L}\). According to the definition of the augmented cube, \(L\) contains at most two complement pairs among \(x,y,s,t,u\). Hence, there are the following subcases:  
**Case 1.1.**\(L\) contains two complement pairs among \(x,y,s,t,u\). We assume that \((x,y)\) and \((s,t)\) are the two complement pairs of \(L\) (see Figure A1a). By Lemma 7, we have that \(|N_{L}(\{x,y,s,t,u\})|\geq 10(n-1)-35=10n-45\). By Property 2, \(|N_{R}(x,y)|=|N_{R}(s,t)|=2\). Because \(|N_{R}(u)|=2\); Hence, \(|N_{R}(\{x,y,s,t,u\})|=|N_{R}(x,y)|+|N_{R}(s,t)|+|N_{R}(u)|=2+2=6\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u\})|=|N_{L}(\{x,y,s,t,u\})|+|N_{R}(\{x,y,s,t,u\})| \geq(10n-45)+6=10n-39\geq 10n-41\) holds.  
**Case 1.2.**\(L\) contains only one complement pair among \(x,y,s,t,u\). We assume that \((x,y)\) is the complement pair of \(L\) (see Figure A1b). By Lemma 8, we have that \(|N_{L}(\{x,y,s,t,u\})|\geq 10(n-1)-39=10n-49\). By Property 2, \(|N_{R}(x,y)|=2\). Because \(|N_{R}(s)|=|N_{R}(t)|=|N_{R}(u)|=2\); Hence, \(|N_{R}(\{x,y,s,t,u\})|=|N_{R}(x,y)|+|N_{R}(s)|+|N_{R}(t)|+|N_{R}(u)|=2+2+2+2=8\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u\})|=|N_{L}(\{x,y,s,t,u\})|+|N_{R}(\{x,y,s,t,u\})|\geq (10n-49)+8=10n-41\) holds.  
**Case 1.3.**\(L\) contains no complement pair among \(x,y,s,t,u\) (see Figure A1c). By the induction hypothesis, we have that \(|N_{L}(\{x,y,s,t,u\})|\geq 10(n-1)-41=10n-51\). Because \(|N_{R}(x)|=|N_{R}(y)|=|N_{R}(s)|=|N_{R}(t)|=|N_{R}(u)|=2\); Hence, \(|N_{R}(\{x,y,s,t,u\})|=|N_{R}(x)|+|N_{R}(y)|+|N_{R}(s)|+|N_{R}(t)|+|N_{R}(u)|=2+2+2+2 =10\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u\})|=|N_{L}(\{x,y,s,t,u\})|+|N_{R}(\{x,y,s,t,u\})|\geq (10n-51)+10=10n-41\) holds.  
**Case 2.** Four of \(\{x,y,s,t,u\}\) are in one subcube, and the remaining one is in the other subcube. We assume that \(x,y,s,t\in V_{L}\) and \(u\in V_{R}\) (see Figure A1d). By Lemma 6, we have \(|N_{L}(\{x,y,s,t\})|\geq 8(n-1)-28=8n-36\). By Property 1, we have \(|N_{R}(u)|=2(n-1)-1=2n-3\). Therefore, \(|N_{AQ_{n}}(\{x,y,s,t,u\})|\geq|N_{L}(\{x,y,s,t\})|+|N_{R}(u)|\geq(8n-36)+(2 n-3)=10n-39\geq 10n-41\) holds.  
**Case 3.** Three of \(\{x,y,s,t,u\}\) are in one subcube, and the remaining two are in the other subcube. We assume that \(x,y,s\in V_{L}\) and \(t,u\in V_{R}\) (see Figure A1e). By Lemma 5, we have \(|N_{L}(\{x,y,s\})|\geq 6(n-1)-17=6n-23\). By Lemma 4, we have \(|N_{R}(\{t,u\})|\geq 4(n-1)-8=4n-12\). Therefore, \(\left|N_{AQ_{n}}(\{x,y,s,t,u\})\right|\geq|N_{L}(\{x,y,s\})|+|N_{R}(\{t,u\})| \geq(6n-23)+(4n-12)=10n-35\geq 10n-41\) holds.  
By combining the above three cases, we complete the proof.