Decentralized diagnosis of discrete event systems subject to permanent sensor failures

Abstract

In this paper, we consider a decentralized failure diagnosis problem for discrete event systems. Each local diagnoser makes a diagnosis decision based on local event observations. A sensor that detects the occurrence of an event may possibly fail due to, for example, aging degradation. It is desirable that the occurrence of any failure string should be correctly detected in the presence of sensor failures. We introduce a new notion of codiagnosability subject to permanent sensor failures, which is defined with respect to not only the set of nondeterministic local observation masks but also the global nondeterministic observation mask. Although the global observation mask is necessary to define codiagnosability, it is not used for performing decentralized diagnosis. The introduced notion of codiagnosability guarantees that the occurrence of any failure string can be correctly detected by a decentralized diagnoser within a bounded number of steps even if permanent sensor failures occur. We develop a method for verifying the codiagnosability property subject to permanent sensor failures. In addition, we compute the delay bound within which the occurrence of any failure string can be detected.

Appendices

Appendix : A: Proof of Proposition 1

(⇐) We consider the decentralized diagnoser $\{D_{i}\}_{i \in I}: {\mathscr{M}}^{f}(L(G)) \to \{0,1\}$ that consists of the local diagnosers $D_{i}: {\mathscr{M}}_{i}^{f}(L(G)) \to \{0,1\}$ ($i=1,2,\dots ,n$) given by Eq. 5. We first show that the decentralized diagnoser {D_i}_i∈I satisfies C1). Since G is $\{{\mathscr{M}}_{i}^{f}\}_{i \in I}$-codiagnosable with respect to K, there exists $m \in \mathbb {N}$ such that

$$ \begin{array}{@{}rcl@{}} \lefteqn{\forall s \in L(G)-K, \forall t \in L(G)/s:}\\ & & [|t|\geq m \vee st \in L_{d}(G)] \\ & & \ \Rightarrow [\forall (\tau_{1},\tau_{2}, \dots, \tau_{n}) \in \mathcal{M}^{f} (st), \exists i \in I, \forall u \in L(G) : \tau_{i} \in \mathcal{M}_{i}^{f}(u) \Rightarrow u\notin K]. \end{array} $$

We consider any s ∈ L(G) − K and any t ∈ L(G)/s such that |t|≥ m or st ∈ L_d(G). For any $(\tau _{1},\tau _{2}, \dots , \tau _{n}) \in {\mathscr{M}}^{f} (st)$, there exists i ∈ I such that, for any u ∈ L(G), if $\tau _{i} \in {\mathscr{M}}_{i}^{f}(u)$, then u∉K. By Eq. 5, we have D_i(τ_i) = 1, which implies together with Eq. 3 that $\{D_{i}\}_{i \in I}(\tau _{1},\tau _{2}, \dots , \tau _{n})=1$.

We next show that {D_i}_i∈I satisfies C2). We consider any s ∈ K and any $(\tau _{1},\tau _{2}, \dots ,$ $\tau _{n}) \in {\mathscr{M}}^{f} (s)$. We have $\tau _{i} \in {\mathscr{M}}_{i}^{f} (s)$ for each i ∈ I. By Eq. 5, we have D_i(τ_i) = 0 for each i ∈ I. It follows from Eq. 3 that $\{D_{i}\}_{i \in I}(\tau _{1},\tau _{2}, \dots , \tau _{n})=0$.

(⇒) We consider any decentralized diagnoser $\{D_{i}\}_{i \in I}: {\mathscr{M}}^{f}(L(G)) \to \{0,1\}$ that satisfies C1) and C2). By C1), there exists $m \in \mathbb {N}$ such that

$$ \begin{array}{@{}rcl@{}} \lefteqn{\forall s\in L(G)-K,\forall t\in L(G)/s:}\\ & & [|t|\geq m \vee st \in L_{d}(G)]\\ & & \ \Rightarrow [\forall (\tau_{1},\tau_{2}, \dots, \tau_{n}) \in \mathcal{M}^{f} (st): \{D_{i}\}_{i \in I}(\tau_{1},\tau_{2}, \dots, \tau_{n})=1 ]. \end{array} $$

To prove $\{{\mathscr{M}}_{i}^{f}\}_{i \in I}$-codiagnosability with respect to K, we consider any s ∈ L(G) − K and any t ∈ L(G)/s such that |t|≥ m or st ∈ L_d(G). For any $(\tau _{1},\tau _{2}, \dots , \tau _{n}) \in {\mathscr{M}}^{f} (st)$, we have $\{D_{i}\}_{i \in I}(\tau _{1},\tau _{2}, \dots , \tau _{n})=1$, which implies together with Eq. 3 that there exists i ∈ I such that D_i(τ_i) = 1. For any u ∈ L(G) such that $\tau _{i} \in {\mathscr{M}}_{i}^{f}(u)$, there exists $\tau \in {\mathscr{M}}^{f}(u)$ with π_i(τ) = τ_i. Letting $\tau =(\tau _{1}^{\prime }, \dots , \tau _{i}, {\dots } \tau _{n}^{\prime })$, it follows from Eq. 3 that $\{D_{i}\}_{i \in I}(\tau _{1}^{\prime }, \dots , \tau _{i}, {\dots } \tau _{n}^{\prime })=1$. By C2), we have u∉K. Thus, G is $\{{\mathscr{M}}_{i}^{f}\}_{i \in I}$-codiagnosable with respect to K.

Appendix : B: Proof of Lemma 5

We prove that, for any s_T ∈ L(T), the three conditions of the lemma hold by the induction on the length of s_T.

As the base step, we consider the case where |s_T| = 0, that is, s_T = ε. Then, we have

• $P(s_{T})=\varepsilon \in L(\tilde {G})$ and $P_{i}(s_{T}) =\varepsilon \in \tilde {K}$ for any i ∈ I,

• $\tilde {M}_{i}(P(s_{T}))=\varepsilon =\tilde {M}_{i}(P_{i}(s_{T}))$ for any i ∈ I, and

• $\alpha (z_{0},s_{T})=((q_{0} ,q_{R0}),q_{R0},q_{R0},\dots ,q_{R0})$, $q_{0}=\tilde {\delta }(q_{0},P(s_{T}))$, $q_{R0}=\delta _{R_{d}}(q_{R0},{\varTheta }(P(s_{T})))$, and $q_{R0}=\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}))$ for any i ∈ I.

For the induction step, we suppose that, for any s_T ∈ L(T) with |s_T| = l ≥ 0, the three conditions of the lemma hold. We consider any s_T ∈ L(T) with |s_T| = l + 1. Then, s_T can be written as $s_{T}=s_{T}^{\prime }\sigma _{T}$, where $|s_{T}^{\prime }|=l$ and σ_T ∈Σ_T.

By the inductive assumption, we have

• $P(s_{T}^{\prime }) \in L(\tilde {G})$ and $P_{i}(s_{T}^{\prime }) \in \tilde {K}$ for any i ∈ I,

• $\tilde {M}_{i}(P(s_{T}^{\prime }))=\tilde {M}_{i}(P_{i}(s_{T}^{\prime }))$ for any i ∈ I, and

• $\alpha (z_{0},s_{T}^{\prime })=((\tilde {q}^{\prime },q_{R_{d}}^{\prime }),\tilde {q}_{R_{1}}^{\prime },\tilde {q}_{R_{2}}^{\prime }, \dots ,\tilde {q}_{R_{n}}^{\prime })$, where $\tilde {q}^{\prime }=\tilde {\delta }(q_{0},P(s_{T}^{\prime }))$, $q_{R_{d}}^{\prime }=\delta _{R_{d}}(q_{R0},{\varTheta }(P(s_{T}^{\prime })))$, and $\tilde {q}_{R_{i}}^{\prime }=\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}^{\prime }))$ for any i ∈ I.

Let $z^{\prime }=\alpha (z_{0},s_{T}^{\prime })$ and $\sigma _{T}=(\tilde {\sigma }, \tilde {\sigma }_{1},\tilde {\sigma }_{2}, \dots , \tilde {\sigma }_{n})$. It follows from $s_{T}^{\prime }\sigma _{T} \in L(T)$ that $\alpha (z^{\prime },\sigma _{T})!$, which implies

• $\tilde {\sigma } \neq \varepsilon \Rightarrow \tilde {\delta } (\tilde {q}^{\prime },\tilde {\sigma } )!$,

• $\forall i \in I: \tilde {\sigma }_{i} \neq \varepsilon \Rightarrow \tilde {\delta }_{R} (\tilde {q}_{R_{i}}^{\prime },\tilde {\sigma }_{i} )!$,

• $\forall i \in I: \tilde {\lambda }_{i} (\tilde {q}^{\prime },\tilde {\sigma } )= \tilde {\lambda }_{R_{i}} (\tilde {q}_{R_{i}}^{\prime },\tilde {\sigma }_{i} )$.

Since $\tilde {q}^{\prime }=\tilde {\delta }(q_{0},P(s_{T}^{\prime }))$ and $\tilde {\delta } (\tilde {q}^{\prime },\tilde {\sigma })!$ if $\tilde {\sigma } \neq \varepsilon$, we have $P(s_{T}^{\prime }\sigma _{T})=P(s_{T}^{\prime })\tilde {\sigma } \in L(\tilde {G})$. Similarly, for any i ∈ I, since $\tilde {q}_{R_{i}}^{\prime }=\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}^{\prime }))$ and $\tilde {\delta }_{R} (\tilde {q}_{R_{i}}^{\prime },\tilde {\sigma }_{i} )!$ if $\tilde {\sigma }_{i} \neq \varepsilon$, we have $P_{i}(s_{T}^{\prime }\sigma _{T})=P_{i}(s_{T}^{\prime })\tilde {\sigma }_{i} \in \tilde {K}$.

For any i ∈ I, since $\tilde {M}_{i}(P(s_{T}^{\prime }))=\tilde {M}_{i}(P_{i}(s_{T}^{\prime }))$, $\tilde {q}^{\prime }=\tilde {\delta }(q_{0},P(s_{T}^{\prime }))$, $\tilde {q}_{R_{i}}^{\prime }=\tilde {\delta }_{R}(q_{R0},P_{i}$ $(s_{T}^{\prime }))$, and $\tilde {\lambda }_{i} (\tilde {q}^{\prime },\tilde {\sigma } )= \tilde {\lambda }_{R_{i}} (\tilde {q}_{R_{i}}^{\prime },\tilde {\sigma }_{i} )$, we have $\tilde {M}_{i}(P(s_{T}^{\prime }\sigma _{T})) =\tilde {M}_{i}(P(s_{T}^{\prime })\tilde {\sigma }) =\tilde {M}_{i}(P_{i}$ $(s_{T}^{\prime })\tilde {\sigma }_{i}) =\tilde {M}_{i}(P_{i}(s_{T}^{\prime }\sigma _{T}))$.

Finally, letting $\alpha (z^{\prime },\sigma _{T}) =((\tilde {q},q_{R_{d}}),\tilde {q}_{R_{1}},\tilde {q}_{R_{2}},\dots , \tilde {q}_{R_{n}})$, it holds that $\alpha (z_{0},s_{T}^{\prime }\sigma _{T}) =((\tilde {q},q_{R_{d}}),\tilde {q}_{R_{1}},\tilde {q}_{R_{2}},\dots , \tilde {q}_{R_{n}})$. Since

$$ \tilde{q} = \left\{ \begin{array}{l l } \tilde{\delta} (\tilde{q}^{\prime}, \tilde{\sigma}), & \text{if} \ \tilde{\sigma} \neq \varepsilon \\ \tilde{q}^{\prime}, & \text{otherwise}, \end{array} \right. $$

$$ q_{R_{d}} = \left\{ \begin{array}{l l } \delta_{R_{d}} (q_{R_{d}}^{\prime}, \tilde{\sigma} ), & \text{if} \ \tilde{\sigma} \notin \{\varepsilon\} \cup F\\ q_{R_{d}}^{\prime}, & \text{otherwise}, \end{array} \right. $$

and

$$ \tilde{q}_{R_{i}} = \left\{ \begin{array}{l l } \tilde{\delta}_{R} (\tilde{q}_{R_{i}}^{\prime}, \tilde{\sigma}_{i} ), & \text{if} \ \tilde{\sigma}_{i} \neq \varepsilon \\ \tilde{q}_{R_{i}}^{\prime}, & \text{otherwise} \end{array} \right. (i \in I), $$

we have $\tilde {q}=\tilde {\delta }(q_{0},P(s_{T}^{\prime })\tilde {\sigma })=\tilde {\delta }(q_{0},P(s_{T}^{\prime }\sigma _{T}))$, $q_{R_{d}}=\delta _{R_{d}}(q_{R0},{\varTheta }(P(s_{T}^{\prime })){\varTheta }(\tilde {\sigma })) =\delta _{R_{d}}$ $(q_{R0},{\varTheta }(P(s_{T}^{\prime }\sigma _{T})))$, and $\tilde {q}_{R_{i}}=\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}^{\prime })\tilde {\sigma }_{i}) =\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}^{\prime }\sigma _{T}))$ for any i ∈ I.

Appendix : C: Proof of Lemma 6

For any $\tilde {s} \in L(\tilde {G})$ and any $\tilde {s}_{1}, \tilde {s}_{2}, \dots , \tilde {s}_{n} \in \tilde {K}$ such that $\tilde {M}_{i}(\tilde {s})=\tilde {M}_{i}(\tilde {s}_{i})$ for any i ∈ I, we show that there exists s_T ∈ L(T) such that $P(s_{T})=\tilde {s}$ and $P_{i}(s_{T})=\tilde {s}_{i}$ for any i ∈ I, by the induction on the length of $\tilde {s}$.

As the base step, we consider the case where $|\tilde {s}|=0$, that is, $\tilde {s}=\varepsilon$. Then, for any i ∈ I, we have $\tilde {M}_{i}(\tilde {s})=\varepsilon$. It follows from $\tilde {M}_{i}(\tilde {s})=\tilde {M}_{i}(\tilde {s}_{i})$ that $\tilde {M}_{i}(\tilde {s}_{i})=\varepsilon$. If $\tilde {s}_{i}=\varepsilon$ for any i ∈ I, then, for ε ∈ L(T), we have $P(\varepsilon )=\varepsilon =\tilde {s}$ and $P_{i}(\varepsilon )=\varepsilon =\tilde {s}_{i}$ for any i ∈ I. We consider the case where there exists i ∈ I such that $\tilde {s}_{i} \neq \varepsilon$. For any i ∈ I with $\tilde {s}_{i} \neq \varepsilon$, we let $\tilde {s}_{i}=\tilde {\sigma }_{i}^{(1)} \tilde {\sigma }_{i}^{(2)} \cdots \tilde {\sigma }_{i}^{(|\tilde {s}_{i}|)}$. Since $\tilde {M}_{i}(\tilde {s}_{i})=\varepsilon$, we have $\tilde {\lambda }_{R_{i}}(q_{R0},\tilde {\sigma }_{i}^{(1)})=\varepsilon$ and $\tilde {\lambda }_{R_{i}}(\tilde {\delta }_{R}(q_{R0}, \tilde {\sigma }_{i}^{(1)} \tilde {\sigma }_{i}^{(2)} {\cdots } \tilde {\sigma }_{i}^{(h)}), \tilde {\sigma }_{i}^{(h+1)})=\varepsilon$ ($h=1,2, \dots , |\tilde {s}_{i}|-1$). We let $l={\max \limits } \{|\tilde {s}_{i}| \in \mathbb {N} \mid i \in I\}$ and $s_{T}=\sigma _{T}^{(1)} \sigma _{T}^{(2)} {\cdots } \sigma _{T}^{(l)} \in {\varSigma }_{T}^{*}$, where $\sigma _{T}^{(h)}=(\varepsilon ,\bar {\sigma }_{1}^{(h)}, \bar {\sigma }_{2}^{(h)}, \dots , \bar {\sigma }_{n}^{(h)})$ ($h=1,2, \dots , l$) and

$$ \bar{\sigma}_{i}^{(h)}=\left\{ \begin{array}{ll} \tilde{\sigma}_{i}^{(h)}, & \text{if} \tilde{s}_{i} \neq \varepsilon \wedge h \leq |\tilde{s}_{i}| \\ \varepsilon, & \text{otherwise} \end{array}\right. $$

for each i ∈ I. By the definition of α, we have s_T ∈ L(T), $P(s_{T})=\varepsilon =\tilde {s}$ and $P_{i}(s_{T})=\tilde {s}_{i}$ for any i ∈ I.

For the induction step, we suppose that, for any $\tilde {s} \in L(\tilde {G})$ with $|\tilde {s}|=j>0$ and any $\tilde {s}_{1}, \tilde {s}_{2}, \dots , \tilde {s}_{n} \in \tilde {K}$ such that $\tilde {M}_{i}(\tilde {s})=\tilde {M}_{i}(\tilde {s}_{i})$ for any i ∈ I, there exists s_T ∈ L(T) such that $P(s_{T})=\tilde {s}$ and $P_{i}(s_{T})=\tilde {s}_{i}$ for any i ∈ I. We consider any $\tilde {s} \in L(\tilde {G})$ with $|\tilde {s}|=j+1$ and any $\tilde {s}_{1}, \tilde {s}_{2}, \dots , \tilde {s}_{n} \in \tilde {K}$ such that $\tilde {M}_{i}(\tilde {s})=\tilde {M}_{i}(\tilde {s}_{i})$ for any i ∈ I. Then, $\tilde {s}$ can be written as $\tilde {s}=\tilde {s}^{\prime }\tilde {\sigma }$, where $|\tilde {s}^{\prime }|=j$ and $\tilde {\sigma } \in \tilde {{\varSigma }}$. We consider any i ∈ I such that $\tilde {M}_{i}(\tilde {s}^{\prime }) \neq \tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })$. Then, we have $\tilde {\lambda }_{i}(\tilde {\delta }(q_{0},\tilde {s}^{\prime }),\tilde {\sigma }) \neq \varepsilon$, which implies $\tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma }) \neq \varepsilon$. Since $\tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })=\tilde {M}_{i}(\tilde {s}_{i})$, $\tilde {s}_{i}$ can be written as $\tilde {s}_{i}=\tilde {s}_{i}^{\prime }\tilde {\sigma }_{i}\tilde {s}_{i}^{\prime \prime }$, where $\tilde {M}_{i}(\tilde {s}^{\prime }) = \tilde {M}_{i}(\tilde {s}_{i}^{\prime })$, $\tilde {\lambda }_{i}(\tilde {\delta }(q_{0},\tilde {s}^{\prime }),\tilde {\sigma }) =\tilde {\lambda }_{R_{i}}(\tilde {\delta }_{R}(q_{R0},\tilde {s}_{i}^{\prime }),\tilde {\sigma }_{i})$, and $\tilde {M}_{i}(\tilde {s}_{i}^{\prime }\tilde {\sigma }_{i}) =\tilde {M}_{i}(\tilde {s}_{i}^{\prime }\tilde {\sigma }_{i}\tilde {s}_{i}^{\prime \prime })$. To apply the induction hypothesis for $\tilde {s}^{\prime } \in L(\tilde {G})$, we let

$$ \bar{s}_{i}^{\prime}=\left\{ \begin{array}{ll} \tilde{s}_{i}^{\prime}, & \text{if} \tilde{M}_{i}(\tilde{s}^{\prime}) \neq \tilde{M}_{i}(\tilde{s}^{\prime}\tilde{\sigma}) \\ \tilde{s}_{i}, & \text{otherwise} \end{array} \right. $$

and

$$ \breve{\sigma}_{i}=\left\{ \begin{array}{ll} \tilde{\sigma}_{i}, & \text{if} \tilde{M}_{i}(\tilde{s}^{\prime}) \neq \tilde{M}_{i}(\tilde{s}^{\prime}\tilde{\sigma}) \\ \varepsilon, & \text{otherwise} \end{array}\right. $$

for each i ∈ I. Then, we have $\tilde {M}_{i}(\tilde {s}^{\prime }) = \tilde {M}_{i}(\bar {s}_{i}^{\prime })$ for each i ∈ I. For $\tilde {s}^{\prime } \in L(\tilde {G})$ and $\bar {s}_{1}^{\prime }, \bar {s}_{2}^{\prime }, \dots , \bar {s}_{n}^{\prime } \in K$, by the induction hypothesis, there exists $s_{T}^{\prime } \in L(T)$ such that $P(s_{T}^{\prime })=\tilde {s}^{\prime }$ and $P_{i}(s_{T}^{\prime })=\bar {s}_{i}^{\prime }$ for each i ∈ I. Letting $\alpha (z_{0},s_{T}^{\prime })=((\tilde {q}^{\prime } ,q_{R_{d}}^{\prime }),\tilde {q}_{R_{1}}^{\prime },\tilde {q}_{R_{2}}^{\prime }, \dots ,\tilde {q}_{R_{n}}^{\prime })$, by Lemma 5, we have $\tilde {q}^{\prime }=\tilde {\delta }(q_{0},P(s_{T}^{\prime }))$ and $\tilde {q}_{R_{i}}^{\prime }=\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}^{\prime }))$ for any i ∈ I.

We show that $\alpha (\alpha (z_{0},s_{T}^{\prime }),\sigma _{T})!$, where $\sigma _{T}=(\tilde {\sigma },\breve {\sigma }_{1},\breve {\sigma }_{2},\dots , \breve {\sigma }_{n})$. Since $\tilde {s}^{\prime }\tilde {\sigma } \in L(\tilde {G})$, $P(s_{T}^{\prime })=\tilde {s}^{\prime }$, and $\tilde {q}^{\prime }=\tilde {\delta }(q_{0},P(s_{T}^{\prime }))$, we have $\tilde {\delta }(\tilde {q}^{\prime },\tilde {\sigma })!$. For each i ∈ I with $\tilde {M}_{i}(\tilde {s}^{\prime }) \neq \tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })$, we have $\breve {\sigma }_{i}=\tilde {\sigma }_{i}$. Since $\tilde {s}_{i}^{\prime }\tilde {\sigma }_{i} \in \tilde {K}$, $P_{i}(s_{T}^{\prime })=\tilde {s}_{i}^{\prime }$, and $\tilde {q}_{R_{i}}^{\prime }=\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}^{\prime }))$, we have $\tilde {\delta }_{R}(\tilde {q}_{R_{i}}^{\prime },\breve {\sigma }_{i})!$. In addition, since $\tilde {\lambda }_{i}(\tilde {\delta }(q_{0},\tilde {s}^{\prime }),\tilde {\sigma }) =\tilde {\lambda }_{R_{i}}(\tilde {\delta }_{R}(q_{R0},\tilde {s}_{i}^{\prime }),\tilde {\sigma }_{i})$, we have $\tilde {\lambda }_{i}(\tilde {q}^{\prime },\tilde {\sigma })= \tilde {\lambda }_{R_{i}}(\tilde {q}_{R_{i}}^{\prime },\breve {\sigma }_{i})$. For each i ∈ I with $\tilde {M}_{i}(\tilde {s}^{\prime }) =\tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })$, we have $\tilde {\lambda }_{i}(\tilde {\delta }(q_{0},\tilde {s}^{\prime }),\tilde {\sigma })=\varepsilon$ and $\breve {\sigma }_{i}=\varepsilon$. It follows that $\tilde {\lambda }_{i}(\tilde {q}^{\prime },\tilde {\sigma })= \tilde {\lambda }_{R_{i}}(\tilde {q}_{R_{i}}^{\prime },\breve {\sigma }_{i})$. By the definition of α, we have $\alpha (\alpha (z_{0},s_{T}^{\prime }),\sigma _{T})!$. Then, it holds that $s_{T}^{\prime }\sigma _{T} \in L(\tilde {G})$, $P(s_{T}^{\prime }\sigma _{T})=\tilde {s}^{\prime }\tilde {\sigma }$, and

$$ P_{i}(s_{T}^{\prime}\sigma_{T}) =\left\{ \begin{array}{ll} \tilde{s}_{i}^{\prime}\tilde{\sigma}_{i}, & \text{if} \tilde{M}_{i}(\tilde{s}^{\prime}) \neq \tilde{M}_{i}(\tilde{s}^{\prime}\tilde{\sigma}) \\ \tilde{s}_{i}, & \text{otherwise} \end{array}\right. $$

for each i ∈ I. If $\tilde {s}_{i}^{\prime \prime }=\varepsilon$ for any i ∈ I with $\tilde {M}_{i}(\tilde {s}^{\prime }) \neq \tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })$, the proof is completed. We consider the case where there exists i ∈ I such that $\tilde {M}_{i}(\tilde {s}^{\prime }) \neq \tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })$ and $\tilde {s}_{i}^{\prime \prime } \neq \varepsilon$. For each i ∈ I with $\tilde {M}_{i}(\tilde {s}^{\prime }) \neq \tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })$ and $\tilde {s}_{i}^{\prime \prime } \neq \varepsilon$, we let $\tilde {s}_{i}^{\prime \prime }=\hat {\sigma }_{i}^{(1)} \hat {\sigma }_{i}^{(2)} \cdots \hat {\sigma }_{i}^{(|\tilde {s}_{i}^{\prime \prime }|)}$. In addition, we let

$$ l^{\prime}=\max \{|\tilde{s}_{i}^{\prime\prime}| \in \mathbb{N}\mid i^{\prime} \in I \wedge \tilde{M}_{i^{\prime}}(\tilde{s}^{\prime}) \neq \tilde{M}_{i^{\prime}}(\tilde{s}^{\prime}\tilde{\sigma})\} $$

and $\breve {s}_{T}=\breve {\sigma }_{T}^{(1)} \breve {\sigma }_{T}^{(2)} {\cdots } \breve {\sigma }_{T}^{(l^{\prime })} \in {\varSigma }_{T}^{*}$, where $\breve {\sigma }_{T}^{(h)}= (\varepsilon ,\breve {\sigma }_{1}^{(h)}, \breve {\sigma }_{2}^{(h)} \dots , \breve {\sigma }_{n}^{(h)})$ ($h=1,2, \dots , l^{\prime }$) and

$$ \breve{\sigma}_{i}^{(h)} = \left\{ \begin{array}{ll} \hat{\sigma}_{i}^{(h)}, & \text{if} \tilde{M}_{i}(\tilde{s}^{\prime}) \neq \tilde{M}_{i}(\tilde{s}^{\prime}\tilde{\sigma}) \wedge s_{i}^{\prime\prime} \neq \varepsilon \wedge h \leq |\tilde{s}_{i}^{\prime\prime}| \\ \varepsilon, & \text{otherwise} \end{array}\right. $$

for each i ∈ I. By the construction of $s_{T}^{\prime }$, σ_T, and $\breve {s}_{T}$, we have $P(s_{T}^{\prime }\sigma _{T}\breve {s}_{T})=\tilde {s}^{\prime }\tilde {\sigma }$ and $P_{i}(s_{T}^{\prime }\sigma _{T}\breve {s}_{T})=\tilde {s}_{i}$ for any i ∈ I. It remains to show that $s_{T}^{\prime }\sigma _{T}\breve {s}_{T} \in L(T)$.

Letting $\alpha (z_{0},s_{T}^{\prime }\sigma _{T})=((\tilde {q} ,q_{R_{d}}),\tilde {q}_{R_{1}},\tilde {q}_{R_{2}}, \dots ,\tilde {q}_{R_{n}})$, by Lemma 5, we have $\tilde {q}=\tilde {\delta }(q_{0},P(s_{T}^{\prime }\sigma _{T}))$ and $\tilde {q}_{R_{i}}=\tilde {\delta }_{R}(q_{R0},P_{i}(s_{T}^{\prime }\sigma _{T}))$ for any i ∈ I. For each i ∈ I with $\tilde {M}_{i}(\tilde {s}^{\prime }) \neq \tilde {M}_{i}(\tilde {s}^{\prime }\tilde {\sigma })$ and $\tilde {s}_{i}^{\prime \prime } \neq \varepsilon$, since $P_{i}(s_{T}^{\prime }\sigma _{T})=\tilde {s}_{i}^{\prime }\tilde {\sigma }_{i}$ and $\tilde {M}_{i}(\tilde {s}_{i}^{\prime }\tilde {\sigma }_{i}) =\tilde {M}_{i}(\tilde {s}_{i}^{\prime }\tilde {\sigma }_{i}\tilde {s}_{i}^{\prime \prime })$, we have $\tilde {\lambda }_{R_{i}}(\tilde {q}_{R_{i}},\hat {\sigma }_{i}^{(1)})=\varepsilon$ and $\tilde {\lambda }_{R_{i}}(\tilde {\delta }_{R}(\tilde {q}_{R_{i}}, \hat {\sigma }_{i}^{(1)} \hat {\sigma }_{i}^{(2)} {\cdots } \hat {\sigma }_{i}^{(h)}), \hat {\sigma }_{i}^{(h+1)})=\varepsilon$ ($h=1,2, \dots , |\tilde {s}_{i}^{\prime \prime }|-1$). By the definition of α, we have $\alpha (\alpha (z_{0},s_{T}^{\prime }\sigma _{T}),\breve {s}_{T})!$, which implies $s_{T}^{\prime }\sigma _{T}\breve {s}_{T} \in L(T)$.