Computation of supervisors for reconfigurable machine tools

Abstract

The rapid reconfiguration of manufacturing systems is an important issue in today’s manufacturing technology in order to adjust the production to varying product demands and types. In this paper, we study the control of reconfigurable machine tools (RMTs) with the aim of fast reconfiguration and an easy controller implementation. We first formulate a particular reconfiguration problem for RMTs in a discrete event system setting, and then provide a necessary and sufficient condition for its solution. Moreover, we propose a polynomial-time algorithm for the construction of a reconfiguration supervisor as the composition of one modular supervisor for each separate RMT configuration. Each modular supervisor operates in three modes. In the first mode, it tracks the plant state if its corresponding configuration is inactive. In the second mode, it performs a configuration change if its corresponding configuration becomes active and in the third mode, it follows the specified behavior of its corresponding configuration if the configuration is active. An important property of the proposed reconfiguration supervisor is that it performs reconfigurations in a bounded number of event occurrences. In addition, the modular realization of our reconfiguration supervisor enables controller modifications such as adding or removing configurations during run-time. All results presented in the paper are illustrated by an RMT example.

Appendices

Appendix A: Notation table

The notation used in this paper is summarized in Table 3.

Table 3 Notation used in the paper

Appendix B: Proof of the necessary part of Theorem 1

1.1 B.1 Proof outline and supporting lemmas

Since the proof of Proposition 1 involves several steps, we first give a brief outline of the proof. We carry out a proof by contradiction, assuming that there exists a $\rho \in \mathcal {C}$ and an x ∈ X such that x ∉ ${\Omega }_{G^{\rho }}(\{{x^\rho }\})$. Then, we consider the case that a reconfiguration to configuration ρ is requested after the RMT plant reaches state x along some string in the closed loop G ^rec||S ^rec. We formulate the closed-loop behavior after reaching x as automaton $\hat S$ and show that $\hat S$ is a supervisor for the RMT plant G _x starting from x. Here, we use the supporting Lemmas 3 and 4. It is known by 3. in Problem 1 that $\hat S$ moves the RMT plant G to the state x ^ρ in a bounded number of transitions. However, there is no assumption that the state space of $\hat S$ is isomorphic to the state space of G. Hence, $\hat S$ generally need not be a state-feedback supervisor for $G^{\rho } \sqsubseteq G$ and it cannot be directly concluded that x ∈${\Omega }_{{G}^{\rho }}$({x ^ρ}). To this end, we show in the supporting Lemma 5 that a state-feedback supervisor for G ^ρ that moves the plant to state x ^ρ can always be constructed from $\hat S$. Hence, x ∈${\Omega }_{{G}^{\rho }}$({x ^ρ}), which contradicts the assumption.

We next present the three supporting lemmas that are used in the proof of Proposition 1. The first two lemmas state basic conditions for the preservation of controllability.

Lemma 3

Let K, L ⊆ Γ ^⋆ be prefix-closed languages and let Γ _u , Σ ⊆ Γ. Assume that K is controllable for L and Γ _u . Then, K∩Σ ^⋆ is controllable for L∩Σ ^⋆ and Γ _u ∩Σ.

Proof

We know that KΓ_u∩L ⊆ K. To prove the lemma, we write

$$ (K \cap \Sigma^{\star})(\Gamma_{\mathrm{u}} \cap \Sigma) \cap (L \cap \Sigma^{\star}) = K\Gamma_{\mathrm{u}} \cap \Sigma^{\star} \cap L \cap \Sigma^{\star} = (K\Gamma_{\mathrm{u}} \cap L) \cap \Sigma^{\star} \subseteq K \cap \Sigma^{\star}. $$

This shows that K∩Σ^⋆ is controllable for L∩Σ^⋆ and Γ_u∩Σ.

Lemma 4

Let K, L⊆Σ ^⋆ be prefix-closed languages, Σ _u ⊆Σ and $\hat L \subseteq L$ . Assume that K is controllable for L and Σ _u . Then, $K \cap \hat L$ is controllable for $\hat L$ and Σ _u .

Proof

We know that KΣ_u∩L ⊆ K. To prove the lemma, we write

$$(K \cap \hat L) \Sigma_{\mathrm{u}} \cap \hat L = K\Sigma_{\mathrm{u}} \cap \hat L\Sigma_{\mathrm{u}} \cap \hat L \subseteq K\Sigma_{\mathrm{u}} \cap \hat L = (K\Sigma_{\mathrm{u}} \cap L) \cap \hat L \subseteq K \cap \hat L. $$

This shows that $K \cap \hat L$ is controllable for $\hat L$ and Σ_u.

The third lemma concerns the existence of a state-feedback supervisor for state attraction in case a given supervisor for state attraction is not a state-feedback supervisor for the plant under consideration.

Lemma 5

Let G = (X, Σ, δ, x ₀ , X _m ) be an automaton, A⊆X be an invariant set in G and Σ _u ⊆Σ be a set of uncontrollable events. Let S = (Q, Σ, ν, q ₀ , Q _m ) be a supervisor for G with Σ _u , and define R = (Z, Σ, α, z ₀ , Z _m ) := G||S and A _R := {z = (x, q)∈Z|x∈A}. Assume that A _R is a strong attractor for Z in R. Then, there is a state-feedback supervisor $T = (W, \Sigma , \omega , -, -)\sqsubseteq G$ such that A is a strong attractor for W in T and x ₀ ∈W.

Proof

We assume that A _R is a strong attractor for Z in R. Moreover, we introduce the map $d : Z \rightarrow \mathbb {N}_{0}$ such that for each z ∈ Z, d(z) := max{|u||u ∈ Σ^⋆ and α(z, u) ∈ A _R but for all u′ < u, α(z, u′) ∉ A _R } as the maximum length path from z to A _R in R.³ For each x ∈ X∖A, we choose a state z _x ∈ Z such that z _x = (x, q) for some q ∈ Q and d(z _x ) ≤ d(z) for all other states with z = (x, q′) for some q′ ∈ Q. If there is no such z = (x, q), z _x is undefined. Then, we propose the following procedure in order to construct a state-feedback supervisor $T = (W, \Sigma , \omega , -, -) \sqsubseteq G$ such that A is a strong attractor for W in T. First, for each state x ∈ A and each event σ ∈ Σ, we define ω(x, σ) = x′ if there is z = (x, q) ∈ Z such that α(z, σ) = (x′, q′) for some x′ ∈ X and q′ ∈ Q. Next, we present an algorithm that adds additional states and transitions to T.

Algorithm 2

Initialize: W = A∪{x 0}, set of waiting states $M = \{z_{x_{0}}\}$  1. if M = ∅                           1        terminate with the result T               2     else                             3       take a state z = (x, q) from M                4  2. for all σ such that α(z, σ)!                      5       let α(z, σ) = (x′, q′)                       6       insert z x′ in M if x′ ∉ W                   7       insert x′ in W and set ω(x, σ) = x′                8        go to 1.                          9

The idea of the algorithm is to follow the transitions of R, starting from z′₀, and adding a corresponding transition in T. Most importantly, it is considered that, for each x ∈ W, there might be multiple corresponding states in R. In that case, always a state with the smallest distance to A _R is chosen, in order to make sure, that no cycles are introduced in T (line 7). An example of this procedure is shown in Fig. 14 with the plant G, the automaton R = G||S and the resulting state-feedback supervisor T. The set of uncontrollable events is Σ_u = {a, b, f} and the set of states A = {4}. Hence, A _R = {6}. Then, the distance values for each state in R are shown in Table 4.

Fig. 14

Computation of the supervisor T

Table 4 Distance values for the states in R

It is readily observed that the algorithm terminates, since each state z ∈ Z∖A _R is inserted at most once in M (line 7) and one state is removed from M in each iteration (line 4). In addition, it holds that, for each state z that is chosen in line 4 of Algorithm 2, there is at least one event σ ∈ Σ such that α(z, σ)!, since A _R is a strong attractor for Z in R. Likewise, it is clear that T is a state-feedback supervisor for G since W ⊆ X, and at each state x ∈ W, transitions from a state z = (x, q) ∈ Z that corresponds to x are inserted (line 8). That is, if T disables an event at a state x ∈ W, then also R disables this event at the same state x. Considering that R is a supervisor, no uncontrollable events are disabled by T.

It remains to show that A is a strong attractor for W in T according to Definition 1 in the main manuscript. We first observe that A is an invariant set in T. According to the definition of ω, for each state x ∈ A, it holds for each σ ∈ Σ that if ω(x, σ)!, then there must be a z = (x, q) ∈ Z such that α(z, σ)! (line 4, 5 and 8 of Algorithm 2). Since x ∈ A, we know that z ∈ A _R , and since A _R is an invariant set for Z in R, also z′ = (x′, q′) = α(z, σ) ∈ A _R . Hence, ω(x, σ) = x′ ∈ A. Furthermore, it holds for each x ∈ W, that there is a u ∈ Σ^⋆ such that ω(x, u) ∈ A. To prove this assertion, we first note that, for each x ∈ W, transitions to states with a smaller distance to the set A are inserted (line 8). Hence, the set A is reached in a finite number of steps for each x ∈ W. Finally, we show that the set W∖A is acyclic in T. Assume that there is a cycle in W∖A. That is, there is a sequence x ₁, x ₂, …, x _k of states and a sequence σ ₁, σ ₂, …, σ _k of events such that ω(x _i , σ _i ) = x _i+1 for i = 1, …, k−1, and ω(x _k , σ _k ) = x ₁. According to the construction in the algorithm, this cycle corresponds to the sequence $z_{x_{1}}, z_{x_{2}}, \ldots , z_{x_{k}}, z_{x_{1}}$ in R. Since the distance to the set A _R must decrease along this sequence, according to the construction, we have that $d(z_{x_{1}}) > d(z_{x_{2}}) > \cdots > d(z_{x_{k}}) > d(z_{x_{1}})$ which leads to contradiction.

Together, we have that A is an invariant set for W in T, the strict subautomaton of T with the state set W∖A is acyclic, and, for all x ∈ W∖A, there is a u ∈ Σ^⋆ such that ω(x, u) ∈ A. Moreover, T is a state-feedback supervisor for G with Σ_u. Hence, A is a strong attractor for W in T and x ₀ ∈ W according to the initialization of Algorithm 2.

1.2 B.2 Proof of proposition 1

Using Lemma 3, Lemma 4 and Lemma 5, it is now possible to prove Proposition 1.

Proof

We assume that a solution supervisor S ^rec for Problem 1 exists. It has to be shown that it holds for each $\rho \in \mathcal {C}$ that${\Omega }_{{G}^{\rho }}$({x ^ρ}) = X.

We prove the assertion by contradiction. Assume that a solution supervisor S ^rec for Problem 1 exists but for some ρ ∈ C,${\Omega }_{{G}^{\rho }}$({x ^ρ}) ≠ X. This means that there must be a $\rho \in \mathcal {C}$ and an x ∈ X such that x ∉ ${\Omega }_{G^{\rho }}(\{{x^\rho }\})$.

We first construct a string in L(G ^rec||S ^rec) that moves the RMT plant G to state x. Let u ∈ Σ^⋆ such that δ(x ₀, u) = x. Since L(G ^rec||S ^rec) ∩ Σ^⋆ = L(G) by 1. in Problem 1, it holds that u ∈ L(G ^rec||S ^rec).

Next, we consider that a reconfiguration to ρ happens and analyze the closed-loop behavior until reaching the configuration start state x ^ρ. Since $\rho_{text{st}}\in\Sigma _{\text {u}}^{\text {rec}}$, controllability of L(S ^rec) for L(G ^rec) and $\Sigma _{\text {u}}^{\text {rec}}$ implies that uρ _st ∈ L(G ^rec||S ^rec). That is, we have that s := uρ _st ∈ L(G ^rec||S ^rec) ∩ (Σ^rec)^⋆ ρ _stΣ^⋆ and it holds that δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x, z ^ρ), respecting (1).

We next introduce the subautomaton $\hat S = (\hat Q, \Sigma , \hat \nu , \hat q_{0}, -) \sqsubseteq G^{\text {rec}} || S^{\text {rec}}$ that represents the behavior of G ^rec||S ^rec after the string s = uρ _st (reconfiguration request for ρ) and until reaching the configuration start state (x ^ρ, z ^ρ) in G ^rec. To this end, we write R ^rec = (Z ^rec, Σ^rec, α ^rec, ${z}_{0}^{rec}$, ${Z}_\text{m}^{rec}$) = G ^rec||S ^rec for the closed-loop system. Then,

• $\hat q_{0} = \alpha ^{\text {rec}}(z_{0}^{\text {rec}}, s)$

• $\hat Q = \{z \in Z | z = \alpha ^{\text {rec}}(\hat q_{0}, u) \mbox { for some } u \in L(R^{\text {rec}})/s \cap \Sigma ^{\star } \mbox { such that for all } u^{\prime } < u$, δ ^rec((x, z ^ρ), u′) ≠ (x ^ρ, z ^ρ)},

• $Q_{\text {st}, \rho } = \{z \in Z^{\text {rec}} | z = \alpha ^{\text {rec}}(\hat q_{0}, u)$ for some u ∈ Σ^⋆ such that δ ^rec((x, z ^ρ), u) = (x ^ρ, z ^ρ) and for all u′ < u, δ ^rec((x, z ^ρ), u′) ≠ (x ^ρ, z ^ρ)}

• For all $q \in \hat Q \setminus Q_{\text {st}, \rho }$ and σ ∈ Σ: α ^rec(q, σ)! and $\alpha ^{\text {rec}}(q, \sigma ) \in \hat Q$ $\Rightarrow \hat \nu (q, \sigma ) = \alpha ^{\text {rec}}(q, \sigma )$.

We next consider $\hat S$ as a supervisor for the RMT plant after string s and until reaching the configuration start state x ^ρ. Consider ${G}_{x}^{\rho}$ as the automaton G ^ρ starting from state x ∈ X. Then, it holds that $L(\hat S) \subseteq L(G^{\rho }_{x})$:

$$\begin{array}{@{}rcl@{}}u \in L(\hat S) && \Rightarrow u \in \Sigma^{\star} \mbox{ and } \delta^{\text{rec}}(x, u)! \mbox{ and }\forall u^{\prime} < u, \delta^{\text{rec}}(x, u) \not= ({x^{\rho}}, z^{\rho})\\ &&\Rightarrow u \in L(G_{x}) \mbox{ and }\forall u^{\prime} < u, \delta(x, u) \not= {x^{\rho}}\\ &&\Rightarrow u \in L(G^{\rho}_{x}) \end{array} $$

We next establish that $L(\hat S)$ is controllable for L(${G}_{x}^{\rho}$) and Σ_u. To this end, we first note that $L(R^{\text {rec}}_{\hat q_{0}})$ is controllable for L(G(x, z ^ρ)rec) and $\Sigma _{\text {u}}^{\text {rec}}$ since L(R ^rec) is controllable for L(G ^rec) and $\Sigma _{\text {u}}^{\text {rec}}$. Then, Lemma 3 implies that $L(R^{\text {rec}}_{\hat q_{0}}) \cap \Sigma ^{\star }$ is controllable for L(G(x, z ^ρ)rec) ∩ Σ^⋆ and Σ_u. Next, Lemma 4 shows that $L(R^{\text {rec}}_{\hat q_{0}}) \cap \Sigma ^{\star } \cap L(G^{\rho }_{x})$ is controllable for L(G(x, z ^ρ)rec) ∩ Σ^⋆∩L(${G}_{x}^{\rho}$) and Σ_u. Noting that $L(R^{\text {rec}}_{\hat q_{0}}) \cap \Sigma ^{\star } \cap L(G^{\rho }_{x}) = L(\hat S)$ and L(G(x, z ^ρ)rec) ∩ Σ^⋆∩L(${G}_{x}^{\rho}$) = L(${G}_{x}^{\rho}$), it holds that $L(\hat S)$ is controllable for L(${G}_{x}^{\rho}$) and Σ_u.

Finally, we employ Lemma 5. We consider that $\hat S$ is a supervisor for ${G}_{x}^{\rho}$ and write $R = (Z, \Sigma , \alpha , z_{0}, Z_{\mathrm {m}}) = G^{\rho }_{x} || \hat S$. In addition, we define the set A _R = {x ^ρ}×Q _{st, ρ} and note that A _R = {(x, q) ∈ Z|x ∈ {x ^ρ}} by definition of Q _{st, ρ} . Furthermore, we know that A _R is an invariant set, since no transitions are defined at any state in Q _{st, ρ} in $\hat S$ as well as at x ^ρ in ${G}_{x}^{\rho}$. In order to show that A _R is a strong attractor for Z in R, we verify the conditions in Definition 1.

• 1. Let z ∈ Z. If z ∈ A _R , then the condition is trivially fulfilled. Otherwise, we observe that $z = (\hat x, \hat q)$ for some $\hat x\in X$ and $\hat q \in \hat Q$ and there is a u ∈ Σ^⋆ such that $\hat q = \hat \nu (\hat q_{0}, u)$. Then, 2. in Problem 1 implies that there is a u′ ∈ Σ^⋆ such that $\hat \nu (\hat q, u^{\prime }) \in Q_{\text {st}, \rho }$. Since $L(\hat S) \subseteq L(G^{\rho }_{x})$ and $\hat \nu (\hat q_{0}, uu^{\prime }) \in Q_{\text {st}, \rho } \Rightarrow \delta ^{\rho }(x, uu^{\prime }) = {x^{\rho }}$, it holds that α(z, uu′) ∈ A _R .

• 2. Consider any uu′ as constructed in 1. Then, we know from 3. in Problem 1 that |uu′| < N. Hence, Z∖A _R must be acyclic in R.

That is, we conclude that all conditions in Lemma 5 are fulfilled. Hence, there is a state-feedback supervisor $S^{\prime } = (Q^{\prime }, \Sigma , \nu ^{\prime }, q_{0}^{\prime }, Q_{\mathrm {m}}^{\prime }) \sqsubseteq G^{\rho }_{x}$ such that {x ^ρ} is a strong attractor for Q′ in S′ and x ∈ Q′. But this means that $x \in \Omega _{G^{\rho }_{x}}(\{{x^{\rho }}\}) = \Omega _{G^{\rho }}(\{{x^{\rho }}\})$, which contradicts the assumption.

Together, this shows that indeed${\Omega }_{{G}^{\rho }}$({x ^ρ}) = X is necessary for the existence of S ^rec.

Appendix C: Proof of Lemma 2

In order to prove Lemma 2, we first state three properties of the modular reconfiguration supervisor $\overline S^{\rho }$ for each $\rho \in \mathcal {C}$.

Property 1 Let s ∈ L(G ^rec||S ^rec) ∩ Σ^⋆ and assume that δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x, z ⁰). Then, it holds for all $\rho \in \mathcal {C}$ that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = x$.

Proof

The proof directly follows from construction rule 3. and 5. in Algorithm 1.

Property 2 Let s ∈ L(G ^rec||S ^rec) ∩ (Σ^rec)^⋆ ρ _st and assume that δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x, z ^ρ). Then,

• 1. $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = q^{\rho }_{0}$ if x = x ^ρ and $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = \hat x$ otherwise

• 2. For all $\rho ^{\prime } \in \mathcal {C} \setminus \{\rho \}$, $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s) = x$

Proof

We prove the assertion by induction using the construction rules in Algorithm 1. Let s = u ₁ ρ ₁ u ₂ ρ ₂⋯u _n ρ _n with u _i ∈Σ^⋆ and ρ _i ∈Σ_st for i = 1, …, n and ρ _n = ρ _st. In addition, write s _i = u ₁ ρ ₁⋯u _i ρ _i and (x _i , ${z}^{\rho _{i}}$) = δ ^rec(${x}_{0}^{\mathrm{rec}}$, s _i ) for i = 1, …, n.

As the base case, we consider s ₁ = u ₁ ρ ₁ and (x ₁, ${z}^{\rho _{i}}$) = δ ^rec(${x}_{0}^{\mathrm{rec}}$, u ₁ ρ ₁). Then, G ^rec = G||F implies that δ ^rec(${x}_{0}^{\mathrm{rec}}$, u ₁) = (x ₁, z ⁰). That is, Property 1 implies that $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, u_{1}) = x_{1}$ for all $\rho ^{\prime } \in \mathcal {C}$. Next, we look at s ₁ = u ₁ ρ ₁. First consider ρ ₁ = ρ. If x ₁ = ${x}^{\rho _{i}}$, then $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s_{1}) = q^{\rho }_{0}$ with rule 8. Otherwise, $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s_{1}) = \hat x_{1}$ with rule 8. Second, consider ρ ₁ ≠ ρ′. Then, $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho \prime }_{0}, s_{1}) = x_{1}$ with rule 10. That is, Property 2 holds for s = s ₁ and δ ^rec(${x}_{0}^{\mathrm{rec}}$, s ₁) = (x ₁, ${z}^{\rho _{i}}$).

For the induction step, assume that the condition in Property 2 is fulfilled for i = k < n. We show that Property 2 also holds for i = k+1. Consider s _k+1 and δ ^rec(${x}_{0}^{\mathrm{rec}}$, s _k+1) = (x _k+1, $z^{\rho _{k+1}}$). We know because of G ^rec = G||F that ρ _k+1 ≠ ρ _k . For ρ = ρ _k+1, we know that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s_{k}) = x_{k}$ and $\overline \nu ^{\rho }(\hat x_{k}, u_{k+1}) = x_{k+1}$. If x _k+1 = $x^{\rho _{k+1}}$ rule 8. implies that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s_{k+1}) = q^{\rho _{k+1}}_{0}$. Otherwise, rule 8. implies that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s_{k+1}) = \hat x_{k+1}$. Now let ρ′ ≠ ρ _k+1. If ρ′ = ρ _k , then either $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s_{k}) = q^{\rho _{k}}_{0}$ or $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s_{k}) = \hat x_{k}$. Hence, $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s_{k}u_{k+1}) = q \in Q^{\rho ^{\prime }}$ (rule 4.) or $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s_{k}u_{k}) = \hat x_{k+1}$ (rule 6.). In the first case, the definition of m ^ρ′ implies that m ^ρ′(q) = x _k+1. That is, with rule 7., it holds that $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s_{k+1}) = x_{k+1}$. In the second case, rule 9. shows that $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s_{k+1}) = x_{k+1}$. This concludes the induction step.

Together, we confirm that the conditions in Property 2 are fulfilled for all i = 1, …, n. This establishes the proof, considering that s _n = s and (x _n , $x^{\rho_{n}}$) = (x, z ^ρ).

Property 3 Let s ∈ L(G ^rec||S ^rec) ∩ (Σ^rec)^⋆ ρ _stΣ^⋆ and assume that δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x, z ^ρ)and δ ^rec(${x}_{0}^{\mathrm{rec}}$, s′) ≠ (x ^ρ, z ^ρ) for any $s^{\prime } \in \overline {\{s\}} \cap (\Sigma ^{\text {rec}})^{\star }\rho _{\text {st}}\Sigma ^{\star } \setminus \{s\}$. Then,

• 1. $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = q^{\rho }_{0}$ if x = x ^ρ and $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = \hat x$ otherwise

• 2. For all $\rho ^{\prime } \in \mathcal {C} \setminus \{\rho \}$, $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s) = x$

Proof

We first write s = s′ρ _st u with s′ ∈ (Σ^rec)^⋆ and u ∈ Σ^⋆ and δ ^rec(${x}_{0}^{\mathrm{rec}}$, s′ρ _st) = (x′, z ^ρ).

• 1. From Property 2 1., we know that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s^{\prime }\rho _{\text {st}}) = q_{0}^{\rho }$ if x′ = x ^ρ and $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s^{\prime }\rho _{\text {st}}) = \hat x^{\prime }$ otherwise. In the first case, it already holds that s = s′, that is, $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = q^{\rho }_{0}$. In the second case, rule 6. implies that $\overline \nu ^{\rho }(\hat x^{\prime }, u) = q^{\rho }_{0}$ if x = x ^ρ and $\overline \nu ^{\rho }(\hat x^{\prime }, u) = \hat x$ otherwise.

• 2. From Property 2 2., we know that $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s^{\prime }\rho _{\text {st}}) = x^{\prime }$. Since δ ^rec((x′, z ^ρ′), u) = (x, z ^ρ′), it follows from rule 5. that $\overline \nu ^{\rho ^{\prime }}(x^{\prime }, u) = x$.

We are now able to prove Lemma 2.

Proof

We assume that the supervisor $S^{\text {rec}} = ||_{\rho \in \mathcal {C}} \overline S^{\rho }$ is constructed as described in Section 4.2. It has to be shown that 1. to 5. in Problem 1 are fulfilled.

• 1. Construction rule 1., 2., 3. and 5. directly imply that $L(\overline S^{\rho }) \cap \Sigma ^{\star } = L(G)$ and $L_{\mathrm {m}}(\overline S^{\rho }) \cap \Sigma ^{\star } = L_{\mathrm {m}}(G)$. In addition, L(G ^rec) ∩ Σ^⋆ = L(G) and L _m(G ^rec) ∩ Σ^⋆ = L _m(G) since G ^rec = G||F. Considering that $G^{\text {rec}} || S^{\text {rec}} = G^{\text {rec}} || (||_{\rho \in \mathcal {C}} \overline S^{\rho })$, the statement follows.

• 2. It has to be shown that the condition in Definition 3 is fulfilled. Let $\rho \in \mathcal {C}$, s = s′ρ _st v ∈ L(G ^rec||S ^rec) ∩ (Σ^rec)^⋆ ρ _stΣ^⋆ such that s′ ∈ (Σ^rec)^⋆, v ∈ Σ^⋆ and for all v′ < v, δ ^rec(${x}_{0}^{\mathrm{rec}}$, s′ρ _st v′) ≠ (x ^ρ, z ^ρ). We write δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x, z ^ρ). With Property 3, it holds that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = \hat x$ and $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s) = x$ for all $\rho ^{\prime } \in \mathcal {C} \setminus \{\rho \}$.

  Considering that {x ^ρ} is a strong attractor for X in T ^ρ, ∃u ∈ Σ^⋆ such that ω ^ρ(x, u) = x ^ρ. By rule 6., $\overline \nu ^{\rho }(\hat x, u) = q_{0}^{\rho }$ and $\overline \nu ^{\rho ^{\prime }}(x, u) = x^{\rho }$ for $\rho ^{\prime } \in \mathcal {C} \setminus \{\rho \}$. Accordingly, δ ^rec((x, z ^ρ), u) = (x ^ρ, z ^ρ). That is, su ∈ L(G ^rec||S ^rec) and δ ^rec((x, z ^ρ), u) = (x ^ρ, z ^ρ), which proves 2.

• 3. It has to be shown that the conditions in Definition 4 are fulfilled. Let $\rho \in \mathcal {C}$, s = s′ρ _st u ∈ L(G ^rec||S ^rec) ∩ (Σ^rec)^⋆ ρ _stΣ^⋆ such that s′ ∈ (Σ^rec)^⋆, u ∈ Σ^⋆, δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x ^ρ, z ^ρ) and for all u′ < u, δ ^rec(${x}_{0}^{\mathrm{rec}}$, s′ρ _st u′) ≠ (x ^ρ, z ^ρ). Property 3 1. shows that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = q^{\rho }_{0}$ and Property 3 2. shows that $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s) = x^{\rho }$. Then, rule 4. in Algorithm 1 implies that $L(\overline S^{\rho })/s \cap \Sigma ^{\star } = L(S^{\rho })$ and $L_{\mathrm {m}}(\overline S^{\rho })/s \cap \Sigma ^{\star } = L_{\mathrm {m}}(S^{\rho })$. Furthermore, $L(\overline S^{\rho ^{\prime }})/s \cap \Sigma ^{\star } = L(G_{x^{\rho }}) \supseteq L(S^{\rho })$ and $L_{\mathrm {m}}(\overline S^{\rho ^{\prime }})/s \cap \Sigma ^{\star } = L_{\mathrm {m}}(G_{x^{\rho }}) \supseteq L_{\text {m}}(S^{\rho })$ with rule 5. Together, L(G ^rec||S ^rec)/s∩Σ^⋆ = L(S ^ρ) and L _m(G ^rec||S ^rec)/s∩Σ^⋆ = L _m(S ^ρ) which confirms 1. and 2. in Definition 4.

  Finally, we know from rule 6. in Algorithm 1 that ω ^ρ(x, u)!. Since {x ^ρ} is a strong attractor for X in T ^ρ, this means that there must be an $N^{\rho } \in \mathbb {N}$ such that |u| < N ^ρ. Taking N as the maximum of all N ^ρ for $\rho \in \mathcal {C}$ confirms that also condition 3. in Definition 4 holds.

• 4. We want to show that for all s ∈ L(G ^rec||S ^rec) and σ ∈ $\Sigma _{\text {u}}^{\text {rec}}$, sσ ∈ L(G ^rec) implies sσ ∈ L(G ^rec||S ^rec). We distinguish two cases: s ∈ Σ^⋆ and s ∈ (Σ^rec)^⋆ ρ _stΣ^⋆ for some $\rho \in \mathcal {C}$.

  In the first case, the proof of part 1. implies that L(G ^rec||S ^rec) ∩ Σ^⋆ = L(G). In addition, for all $\rho \in \mathcal {C}$ and all states $x \in \overline Q^{\rho }$ that can be reached after s, all events in Σ_st are defined in $\overline S^{\rho }$ (rule 10.). Hence, controllability is trivially fulfilled. Now consider the second case. For all $\rho ^{\prime } \in \mathcal {C} \setminus \{\rho \}$, it holds that $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s) \in X$ with Property 2 2. That is, for any σ ∈ Σ_u such that sσ ∈ L(G ^rec||S ^rec), we know that $s\sigma \in L(\overline S^{\rho ^{\prime }})$ because of rule 5. In addition, ρst′ is defined because of rule 9. For ρ, it holds that either $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = \hat x$ for some x ∈ X∖{x ^ρ} or $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = q \in Q^{\rho }$ with Property 2 1. or Property 3 1. Then, we know that sρ _st ∉ L(G ^rec) since G ^rec = G||F. Moreover, because of rule 7. and 9., ρst′ is defined at any state of $\overline S^{\rho }$. Considering σ ∈ Σ_u, in the first case, δ(x, σ)! and x ∈ X∖{x ^ρ} implies that δ ^ρ(x, σ)!. Then, also ω ^ρ(x, σ)! since T ^ρ is a state-feedback supervisor for G ^ρ and Σ_u. By rule 6., $\overline \nu ^{\rho }(\hat x, \sigma )!$. In the second case, write δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x, z ^ρ). Then, m ^ρ(q) = x. Since sσ ∈ L(G ^rec), it holds that δ ^rec((x, z ^ρ), σ)!. Hence, also δ(x, σ)!. Since S ^ρ is a supervisor for ${G}_{{x}^{\rho }}$ and Σ_u, also ν ^ρ(q, σ)!. Then, rule 4. implies that $\overline \nu ^{\rho }(q, \sigma )!$. Together, it holds that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s\sigma )!$, that is, $s\sigma \in L(\overline S^{\rho })$. Considering that sσ ∈ L(G ^rec) and $s\sigma \in L(\overline S^{\rho ^{\prime }})$ for all $\rho ^{\prime } \in \mathcal {C}$, it follows that sσ ∈ L(G ^rec||S ^rec).

• 5. It has to be shown that S ^rec is a nonblocking supervisor. Let s ∈ L(G ^rec||S ^rec). It has to be shown that there is a u ∈ (Σ^rec)^⋆ such that su ∈ L _m(G ^rec||S ^rec). We consider the two possible cases s ∈ Σ^⋆ and s ∈ (Σ^rec)^⋆ ρ _stΣ^⋆ for some $\rho \in \mathcal {C}$. In the first case, s ∈ L(G ^rec||S ^rec) ∩ Σ^⋆. We know from 1. in Problem 1 that L(G ^rec||S ^rec) ∩ Σ^⋆ = L(G) and L _m(G ^rec||S ^rec) ∩ Σ^⋆ = L _m(G). Since G is nonblocking, there is a u ∈ Σ^⋆ such that su ∈ L _m(G) = L _m(G ^rec||S ^rec) ∩ Σ^⋆.

  In the second case, we write δ ^rec(${x}_{0}^{\mathrm{rec}}$, s) = (x, z ^ρ), s = s′ρ _st v with s′ ∈ (Σ^rec)^⋆ and v ∈ Σ^⋆. Then either, for all v′ ≤ v, it holds that δ ^rec(${x}_{0}^{\mathrm{rec}}$, s′ρ _st v′) ≠ (x ^ρ, z ^ρ) or there exists v′ ≤ v such that δ ^rec(${x}_{0}^{\mathrm{rec}}$, s′ρ _st v′) = (x ^ρ, z ^ρ). In the first case, Property 3 1. shows that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, s) = \hat x$. Then, there is a u′ ∈ Σ^⋆ such that $\nu ^{\rho }(\hat x, u^{\prime }) = q^{\rho }_{0}$ according to the proof of condition 2. in Problem 1. Since S ^ρ is a nonblocking supervisor, it holds that there is a u″ ∈ Σ^⋆ such that ν′(q0′, u″) ∈ ${Q}_{m}^{\rho }$, that is, u″ ∈ L _m(S ^ρ). By rule 2. and 4., this means for u = u′u″ that $\overline \nu ^{\rho }(\overline q^{\rho }_{0}, su) \in \overline Q^{\rho }$ and $su \in L_{\mathrm {m}}(\overline S^{\rho })$. In the second case, let v = v′v″ such that δ ^rec(${x}_{0}^{\mathrm{rec}}$, s′ρ _st v′) = (x ^ρ, z ^ρ). Then, $\overline \nu ^{\rho }(\overline q^{\rho }, s^{\prime } \rho _{\text {st}} v^{\prime }) = q^{\rho }_{0}$ with Property 3 1. and v″ ∈ L(S ^ρ) with rule 4. Since S ^ρ is a nonblocking supervisor, there is a u ∈ Σ^⋆ such that v″u ∈ L _m(S ^ρ). Again, with rule 4., $su \in L_{\mathrm {m}}(\overline S^{\rho })$. Finally, considering $\rho ^{\prime } \in \mathcal {C} \setminus \{\rho \}$, it holds that $\overline \nu ^{\rho ^{\prime }}(\overline q^{\rho ^{\prime }}_{0}, s) = x$ with Property 3 2. Considering that u ∈ L _m(S ^ρ) for both previous cases, it holds that $u \in L_{\mathrm {m}}(G_{x}) = L_{\mathrm {m}}(\overline S^{\rho ^{\prime }})/s \cap \Sigma ^{\star }$. Hence, also $su \in L_{\mathrm {m}}(\overline S^{\rho ^{\prime }})$ for all $\rho ^{\prime } \in \mathcal {C} \setminus \{\rho \}$. Together, we found u ∈ Σ^⋆ such that su ∈ L _m(G ^rec||S ^rec).