Separation and covering for group based concatenation hierarchies

TL;DR

This paper investigates the decidability of separation problems in concatenation hierarchies of regular languages, establishing conditions under which separation is decidable for levels up to 3/2, especially for hierarchies with group language bases.

Contribution

It proves that if separation is decidable for the basis of a hierarchy with group languages, then it is decidable for levels up to 3/2, extending previous results to hierarchies with possibly infinite bases.

Findings

Decidability of separation for levels up to 3/2 when basis separation is decidable.

Applicable to hierarchies with infinite bases of group languages.

Complements existing results for finite bases and low levels.

Abstract

Concatenation hierarchies are classifications of regular languages. All such hierarchies are built through the same construction process: start from an initial class of languages and build new levels using two generic operations. Concatenation hierarchies have gathered a lot of interest since the 70s, thanks to an alternate logical definition: each concatenation hierarchy is the quantification alternation hierarchy within a variant of first-order logic over words. Our goal is to understand such hierarchies. We look at two decision problems: membership and separation. For a class of languages C, C-separation takes two regular languages as input and asks whether there exists a third one in C including the first one and disjoint from the second one. Settling whether separation is decidable for the levels within a given concatenation hierarchy is among the most fundamental and challenging questions in formal language theory. In all prominent cases, it is open, or answered positively for low levels only. Recently, a breakthrough was made using a generic approach for a specific kind of hierarchy: those with a finite basis. In this case, separation is always decidable for levels 1/2, 1 and 3/2. Our main theorem is similar but independent: we consider hierarchies with possibly infinite bases, but that may only contain group languages. An example is the quantifier alternation hierarchy of first-order logic with modular predicates: its basis consists of languages counting the length of words modulo some number. Using a generic approach, we show that for any such hierarchy, if separation is decidable for the basis, then it is decidable for levels up to 3/2. This complements the aforementioned result nicely: all bases considered in the literature are either finite or made of group languages. Thus, one may handle the lower levels of any prominent hierarchy in a generic way.

Equations

$$\begin{array}[]{lll}w^{-1}L&\stackrel{{\scriptstyle\text{def}}}{{=}}&\{u\in A^{*}\mid wu\in L\},\\ Lw^{-1}&\stackrel{{\scriptstyle\text{def}}}{{=}}&\{u\in A^{*}\mid uw\in L\}.\end{array}$$

$$\exists x_1\exists x_2\forall x_3\exists x_4\phi (x_1,x_2,x_3,x_4)\text{(with φ quantifier-free)}$$

$$\text{For all }r,s\in R,r\le s\text{ when }r+s=s.$$

$${\downarrow }_{R}S=\{r\in R\mid r\le s\text{ for some s∈S}\}.$$

$${\downarrow }_{R}S=\{(x,r)\in X\times R\mid \text{∃s∈R s.t. r≤s and (x,s)∈S}\}$$

$$\mathcal{I}[\rho ](\mathbf{K})={\downarrow }_R\{\rho (K)\mid K\in \mathbf{K}\}.$$

$$\mathcal{I}[\rho ](\mathbf{K})\subseteq \mathcal{I}[\rho ]({\mathbf{K}}^{\prime })\text{for every C-cover K′ of L}.$$

$${\mathcal{I}}_{\mathcal{C}}\left[L,\rho \right]=\mathcal{I}[\rho ](\mathbf{K})\text{for any optimal C-cover K of L for ρ}.$$

$${\mathcal{P}}_{\mathcal{C}}^{\alpha }[\rho ]=\{(s,r)\in M\times R\mid r\in {\mathcal{I}}_{\mathcal{C}}\left[{\alpha }^{-1}(s),\rho \right]\}.$$

$$\rho (L)\le \rho (L^{\prime })\text{for every L′∈C such that ε∈L′}.$$

$$\text{mathbbm}{i}_{\mathcal{C}}[\rho ]=\rho (L)\text{for any C-optimal ε-approximation L for ρ}.$$

$$\{F_w^{\prime }\mid \text{w∈H and ∣w∣≤n}\}.$$

$$\begin{array}[]{lll}u&=&u_{0}a_{1}u_{1}\cdots a_{i}xa_{j+1}u_{j+1}\cdots a_{|w|}u_{|w|}\\ v&=&a_{1}\cdots a_{i}a_{j+1}\cdots a_{|u|}\end{array}$$

$$\left(1_M\cdot \prod _{1\le i\le n}(\alpha (a_i)\cdot 1_M),\rho (L)\cdot \prod _{1\le i\le n}(\rho (a_i)\cdot \rho (L))\right)\in S.$$

$$\begin{array}[]{llll}\xi_{\mathcal{D}}[\rho]:&(2^{A^{*}},\cup)&\to&(2^{R},\cup)\\ &K&\mapsto&\mathcal{I}_{\mathcal{D}}\left[K,\rho\right]\end{array}$$

$$\begin{array}[]{llll}\zeta_{\mathcal{D}}^{\alpha}[\rho]:&(2^{A^{*}},\cup)&\to&(2^{M\times R},\cup)\\ &K&\mapsto&\{(s,r)\mid r\in\mathcal{I}_{\mathcal{D}}\left[\alpha^{-1}(s)\cap K,\rho\right]\}\end{array}$$

$$\begin{array}[]{lll}\rho(H)&=&\mu_{\rho}(\rho(H))\\ \rho(HK)&=&\mu_{\rho}\big{(}\rho(H)\cdot\rho(K)\big{)}\\ \rho(HaK)&=&\mu_{\rho}\big{(}\rho(H)\cdot\rho(a)\cdot\rho(K)\big{)}\end{array}$$

$${\mu }_{{\zeta }_{\mathcal{D}}^{\alpha }[\rho ]}(T)={\downarrow }_{R}T\text{for all T∈2M×R}.$$

$${\mu }_{{\xi }_{\mathcal{D}}[\rho ]}(T)={\downarrow }_{R}T\text{for all T∈2R}.$$

$$\text{mathbbm}{i}_{\mathcal{C}}[{\xi }_{\mathcal{D}}[\rho ]]={\xi }_{\mathcal{D}}[\rho ](L)\subseteq {\xi }_{\mathcal{D}}[\rho ](A^{\ast })={\mathcal{I}}_{\mathcal{D}}\left[\rho \right].$$

$${\xi }_{\mathcal{D}}[\rho ](A^{\ast })=\bigcup _{w\in U}{\xi }_{\mathcal{D}}[\rho ](H_w).$$

$$r\in {\downarrow }_R({\xi }_{\mathcal{D}}[\rho ](L)\cdot {\xi }_{\mathcal{D}}[\rho ](a_1)\cdot {\xi }_{\mathcal{D}}[\rho ](L)\cdots {\xi }_{\mathcal{D}}[\rho ](a_{\ell })\cdot {\xi }_{\mathcal{D}}[\rho ](L))$$

$$r\in {\downarrow }_R(\text{mathbbm}{i}_{\mathcal{C}}[{\xi }_{\mathcal{D}}[\rho ]]\cdot \{\rho (a_1)\}\cdot \text{mathbbm}{i}_{\mathcal{C}}[{\xi }_{\mathcal{D}}[\rho ]]\cdots \{\rho (a_{\ell })\}\cdot \text{mathbbm}{i}_{\mathcal{C}}[{\xi }_{\mathcal{D}}[\rho ]]).$$

$${\eta }_{\rho ,S}:(2^{A^{\ast }},\cup ,\cdot )\to (2^{R\times 2^R},\cup ,\cdot ).$$

$$\eta_{\rho,S}(a)=\big{\{}(\rho(a),\quad S\cdot\{\rho(a)\}\cdot S)\big{\}}.$$

$$\begin{array}[]{c}\text{$s\leq r_{1}+\cdots+r_{k}$ and,}\\ \text{$r_{1}+\cdots+r_{k}\in\ \mathord{\downarrow_{R}}U_{i}$ for every $i\leq k$}\end{array}$$

$$\begin{array}[]{c}\text{$s\leq r_{1}+\cdots+r_{k}$ and,}\\ \text{$r_{1}+\cdots+r_{k}\in\ \mathord{\downarrow_{R}}U_{i}$ for all $i\leq k$}\end{array}$$

$${\eta }_{\alpha ,\rho ,S}:(2^{A^{\ast }},\cup ,\cdot )\to (2^{R\times 2^{M\times R}},\cup ,\cdot ).$$

$$\eta_{\alpha,\rho,S}(a)=\big{\{}(\rho(a),\quad S\cdot\{(\alpha(a),\rho(a))\}\cdot S)\big{\}}$$

$$(e,f\cdot (1_R+r)\cdot f)\in S.$$

Full text

Our goal is to understand these hierarchies. A typical approach is to look at two decision problems: membership and separation. In the paper we are interested in the latter, which is more general. For a class of languages $\mathcal{C}$, $\mathcal{C}$-separation takes two regular languages as input and asks whether there exists a third one in $\mathcal{C}$ including the first one and disjoint from the second one. Settling whether separation is decidable for the levels within a given concatenation hierarchy is among the most fundamental and challenging questions in formal language theory. In all prominent cases, it is open, or answered positively for low levels only. Recently, a breakthrough was made using a generic approach for a specific kind of hierarchy: those with a finite basis. In this case, separation is always decidable for levels 1/2, 1 and 3/2.

Our main theorem is similar but independent: we consider hierarchies with possibly infinite bases, but that may only contain group languages. An example is the group hierarchy of Pin and Margolis: its basis consists of all group languages. Another example is the quantifier alternation hierarchy of first-order logic with modular predicates ($\textup{FO}({<},\textup{MOD})$): its basis consists of the languages that count the length of words modulo some number. Using a generic approach, we show that for any such hierarchy, if separation is decidable for the basis, then it is decidable as well for levels 1/2, 1 and 3/2 (we actually solve a more general problem called covering). This complements the aforementioned result nicely: all bases considered in the literature are either finite or made of group languages. Thus, one may handle the lower levels of any prominent hierarchy in a generic way.

Both authors acknowledge support from the DeLTA project (ANR-16-CE40-0007).

1. Introduction

Context. Concatenation hierarchies are natural classifications of regular languages. They were motivated by a celebrated theorem of Schützenberger [29], McNaughton and Papert [14], which is twofold. Its first part is simple, yet important: it states that the class of regular languages that can be defined from singletons using Boolean operations and concatenation only (but no Kleene star) coincides with the class of languages that can be defined in first-order logic. In other words, star-free languages are exactly first-order definable ones. The second and more difficult part consists of an algorithm that inputs a regular language, and outputs whether it belongs to this class.

This motivated Brzozowski and Cohen [4] for introducing the dot-depth hierarchy. It stratifies star-free languages in an infinitely increasing sequence of levels [5] spanning the whole class of star-free languages. Intuitively, a level in this hierarchy captures the number of alternations between complement and concatenation that are needed to express a star-free language. The dot-depth rose to prominence following the work of Thomas [33], who proved an exact correspondence with the quantifier alternation hierarchy of first-order logic: each level in the dot-depth hierarchy consists of all languages that can be defined with a prescribed number of quantifier blocks.

One of the most famous open problems in automata theory is to settle whether the membership problem is decidable for each individual level: is there an algorithm deciding whether an input regular language belongs to this level?

The literature about this problem is rich. We refer the reader to the surveys [24, 27, 17]. The bottom line is that after more than 45 years, little is known. Let us briefly survey the most important cornerstones in this line of research.

State of the art. The dot-depth is a particular instance of a concatenation hierarchy. All such hierarchies are built through a uniform construction scheme. Levels are numbered by integers 0,1,2… or half integers 1/2, 3/2, 5/2… Level 0 is a class of languages called the basis, which is the only parameter specific to the hierarchy. New levels are then built using two operations. The first is closure under marked concatenation $K,L\mapsto KaL$ (for some letter $a$) and union. This closure operator, when applied to level $n\in\mathbb{N}$, produces level $n+1/2$. The second operator is closure under Boolean operations, which yields level $n+1$ out of level $n+1/2$.

For instance, the basis of the dot-depth hierarchy consists of 4 languages: the empty set, the singleton set consisting of the empty word, and their complements. Between 1971 and 2015, membership was shown to be decidable up to level 5/2 and for level 7/2 [2, 11, 9, 28, 22, 23, 20, 21]. The state of the art is the same for the Straubing-Thérien hierarchy [30, 32], whose basis is the empty set and the set of all words. Actually the above results for the dot-depth were obtained by reduction to this hierarchy, via transfer theorems [31, 23].

A crucial point is that all recent results are based on more general problems than membership: separation and covering. For instance, separation for a class $\mathcal{C}$ takes as input two regular languages, and asks whether there exists a third language, which belongs to $\mathcal{C}$, contains the first language, and is disjoint from the second. Membership is the particular case of separation when the input consists of a regular language and its complement. Thus, deciding membership for a class $\mathcal{C}$ reduces to deciding separation for $\mathcal{C}$, which itself reduces to deciding covering for $\mathcal{C}$.

A generic result. Recently, we proved that all known results on both the Straubing-Thérien and the dot-depth hierarchies follow from a single one. This generic theorem [28, 21] states that in any hierarchy having a finite basis satisfying mild hypotheses, covering is decidable at levels 1/2, 1 and 3/2. All results stated above follow from this via simpler observations.

Establishing such generic results is of course desirable. This avoids developing specific arguments, pinpointing the key hypotheses needed for obtaining covering or separation algorithms, and this is amenable to generalization.

Contribution. In the literature, there is another important kind of concatenation hierarchy: those with a basis made exclusively of group languages. A group language is a language recognized by an automaton whose letters act as permutations on states. The most prominent representative is the group hierarchy, whose basis consists of all group languages. Pin and Margolis [13] proved that levels 1/2 and 1 have decidable membership. An algebraic attempt to go higher was proposed in [15]. The idea was to reduce membership at a given level to covering at the same level in the Straubing-Thérien hierarchy. Unfortunately, while the connection between both hierarchies was established, it did not provide the desired algorithm. In this paper, we develop orthogonal, complementary ideas. Note that our aforementioned generic result does not apply here, since the class of group languages is infinite.

Our contribution is another generic result which nicely complements the one applying to finite bases. While both statements are similar, they are independent. Our main theorem states that covering is decidable at levels 1/2, 1 and 3/2 in any hierarchy whose basis consists of group languages, provided that this basis has decidable separation.

Applications. Let us state three applications of this result. First, the class of all group languages is known to have decidable separation (this follows from a result of Ash [3], which was connected to separation by Almeida [1]). Therefore, our generic statement applies to the group hierarchy of Pin and Margolis.

The other examples have a nice logical interpretation. Indeed, the result of Thomas [33] was generalized in [27]: every concatenation hierarchy corresponds, level by level, to the quantification alternation hierarchy within a variant of first-order logic over words. While the hierarchies differ by their bases, the variants differ by their set of available predicates.

This correspondence makes it possible to present our second example as the quantifier alternation hierarchy of first-order logic with modular predicates ($\textup{FO}({<},\textup{MOD})$). It corresponds to the concatenation hierarchy whose basis consists of the languages that count the length of words modulo some number. It is simple to show that this basis consists of group languages and has decidable separation. Hence, our generic theorem applies: it strengthens and unifies results of [6, 12], which dealt with membership at levels up to 3/2, and [34], which proved separation at level 1 with a combinatorial proof leading to a brute force algorithm, orthogonal to our techniques. Finally, when the basis consists of languages recognized by commutative groups, we obtain by [27, 8] the quantifier alternation hierarchy of first-order logic endowed with predicates counting the number of occurrences of a letter before a position, modulo some integer. This basis has decidable covering [7, 1], so our result applies to this hierarchy, which was not yet investigated.

Organization. In Section 2, we recall the basic notions that we need. The main theorem is stated in Section 3. The framework that we use is recalled in Sections 4 and 6. Finally, Sections 5, 7 and 8 are devoted to presenting the algorithms of the three parts of the main theorem: the decidability of covering at levels 1/2, 1 and 3/2, respectively.

Due to space limitations, some proofs are given in appendix.

2. Preliminaries

2.1. Classes of languages

We fix an arbitrary finite alphabet $A$ for the whole paper. As usual, $A^{*}$ denotes the set of all words over $A$, including the empty word $\varepsilon$. We let $A^{+}=A^{*}\setminus\{\varepsilon\}$. For $u,v\in A^{*}$, we write $u\cdot v$ or $uv$ the word obtained by concatenating $u$ and $v$. Finally, for $w\in A^{*}$, we write $|w|\in\mathbb{N}$ for its length.

A language is a subset of $A^{*}$. We denote the singleton language $\{u\}$ by $u$. One may lift the concatenation operation to languages: for $K,L\subseteq A^{*}$, we let $KL=\{uv\mid u\in K\text{\;and\;}v\in L\}$. Additionally, we consider marked concatenation. Given $K,L\subseteq A^{*}$, a marked concatenation of $K$ with $L$ is a language of the form $KaL$ for some letter $a\in A$.

A class of languages $\mathcal{C}$ is a set of languages. We shall work with robust classes satisfying standard closure properties:

• •

$\mathcal{C}$ is a lattice when it is closed under union and intersection, $\emptyset\in\mathcal{C}$ and $A^{*}\in\mathcal{C}$.

• •

A Boolean algebra is a lattice closed under complement.

• •

$\mathcal{C}$ is quotient-closed when for every $L\in\mathcal{C}$ and $w\in A^{*}$, the following two languages belong to $\mathcal{C}$:

[TABLE]

All classes considered in the paper are (at least) quotient-closed lattices. Furthermore, they are included in the class of regular languages. These are the languages that can be equivalently defined by monadic second-order logic, finite automata or finite monoids. We use the definition based on monoids.

Regular languages. A semigroup is a pair $(S,\cdot)$ where $S$ is a set and “$\cdot$” is an associative binary operation on $S$ (often called multiplication). It is standard to abuse terminology and make the binary operation implicit: one simply says that “$S$ is a semigroup”. A monoid $M$ is a semigroup whose multiplication has a neutral element denoted by “$1_{M}$”. Recall that an idempotent of a semigroup $S$ is an element $e\in S$ such that $ee=e$. A standard result in semigroup theory states that when $S$ is finite, there exists $\omega(S)\in\mathbb{N}$ (written $\omega$ when $S$ is understood) such that $s^{\omega}$ is idempotent for every $s\in S$.

Clearly, $A^{*}$ is a monoid whose multiplication is concatenation ($\varepsilon$ is the neutral element). Hence, given a monoid $M$, we may consider morphisms $\alpha:A^{*}\rightarrow M$. We say that a language $L\subseteq A^{*}$ is recognized by such a morphism $\alpha$ when there exists $F\subseteq M$ such that $L=\alpha^{-1}(F)$. It is well-known that the regular languages are exactly those which can be recognized by a morphism $\alpha:A^{*}\rightarrow M$ where $M$ is a finite monoid.

Group languages. A group is a monoid $G$ such that every element $g\in G$ has an inverse $g^{-1}\in G$, i.e., $gg^{-1}=g^{-1}g=1_{G}$. We call “group language” a language $L$ which is recognized by a morphism into a finite group. In the paper, we consider classes that are quotient-closed Boolean algebras of group languages (i.e., containing group languages only).

Example 1**.**

The most immediate example of quotient-closed Boolean algebra of group languages (which is also the largest one) is the class of all group languages. We write it GR.

Quotient-closed Boolean algebras of group languages are more general than the classes of group languages that are usually considered. Typically, publications on the topic consider varieties of group languages, which involve an additional closure property called “inverse morphic image” (see [15] for details). Let us present a class which is a quotient-closed Boolean algebra of group languages, but not a variety.

Example 2**.**

The class MOD containing the finite Boolean combinations of languages $\{w\in A^{*}\mid|w|=k\mod m\}$ with $k,m\in\mathbb{N}$ such that $k<m$, is a quotient-closed Boolean algebra of group languages. We use it as a running example.

2.2. Separation and covering

In the paper, we use two decision problems to investigate specific classes of languages (all built from quotient-closed Boolean algebras of group languages): separation and covering. We define them here. The former is standard while the latter was introduced in [26]. Both of them are parametrized by an arbitrary class of languages $\mathcal{C}$. We start with separation.

Separation. Given three languages $K,L_{1},L_{2}$, we say that $K$ separates $L_{1}$ from $L_{2}$ if $L_{1}\subseteq K\text{ and }L_{2}\cap K=\emptyset$. Given a class of languages $\mathcal{C}$, we say that $L_{1}$ is $\mathcal{C}$-separable from $L_{2}$ if some language in $\mathcal{C}$ separates $L_{1}$ from $L_{2}$. Observe that when $\mathcal{C}$ is not closed under complement, the definition is not symmetrical: $L_{1}$ could be $\mathcal{C}$-separable from $L_{2}$ while $L_{2}$ is not $\mathcal{C}$-separable from $L_{1}$. The separation problem associated to a given class $\mathcal{C}$ is as follows:

[TABLE]

Remark 3**.**

Separation generalizes the classical membership problem which asks whether a single regular language belongs to $\mathcal{C}$. Indeed, $L\in\mathcal{C}$ if and only if $L$ is $\mathcal{C}$-separable from $A^{*}\setminus L$.

Covering. This more general problem was introduced in [26]. Given a language $L$, a cover of $L$ is a finite set of languages $\mathbf{K}$ such that $L\subseteq\bigcup_{K\in\mathbf{K}}K$. Moreover, if $\mathcal{C}$ is a class, a $\mathcal{C}$-cover of $L$ is a cover $\mathbf{K}$ of $L$ such that all $K\in\mathbf{K}$ belong to $\mathcal{C}$.

Covering takes as input a language $L_{1}$ and a finite set of languages $\mathbf{L}_{2}$. A separating cover for the pair $(L_{1},\mathbf{L}_{2})$ is a cover $\mathbf{K}$ of $L_{1}$ such that for every $K\in\mathbf{K}$, there exists $L\in\mathbf{L}_{2}$ which satisfies $K\cap L=\emptyset$. Finally, given a class $\mathcal{C}$, we say that the pair $(L_{1},\mathbf{L}_{2})$ is $\mathcal{C}$-coverable when there exists a separating $\mathcal{C}$-cover. The $\mathcal{C}$-covering problem is now defined as follows:

[TABLE]

It is straightforward to prove that covering generalizes separation (provided that the class $\mathcal{C}$ is a lattice) as stated in the following lemma (see Theorem 3.5 in [26] for the proof).

Lemma 4**.**

Let $\mathcal{C}$ be a lattice and $L_{1},L_{2}\subseteq A^{*}$. Then $L_{1}$ is $\mathcal{C}$-separable from $L_{2}$, if and only if $(L_{1},\{L_{2}\})$ is $\mathcal{C}$-coverable.

3. Concatenation hierarchies and main theorem

In this section, we present the particular classes that we investigate in the paper and outline our results.

3.1. Closure operations and concatenation hierarchies

We are interested in concatenation hierarchies. We briefly recall this notion (see [17, 27] for details). A concatenation hierarchy is an increasing sequence of classes of languages, which depends on a single parameter: an arbitrary quotient-closed Boolean algebra, called its basis. Once the basis is fixed, the construction is uniform. Languages are classified into levels: each new level is built by applying one of two generic operations to the previous one. Let us define these operations.

Boolean closure. Given a class $\mathcal{C}$, its Boolean closure, denoted by $\mathit{Bool}(\mathcal{C})$ is the least Boolean algebra containing $\mathcal{C}$. The following lemma is immediate from the definitions (it holds simply because quotients commute with Boolean operations).

Lemma 5**.**

Let $\mathcal{C}$ be a quotient-closed lattice. Then $\mathit{Bool}(\mathcal{C})$ is a quotient-closed Boolean algebra.

Polynomial closure. Given a class $\mathcal{C}$, the polynomial closure of $\mathcal{C}$, denoted by $\mathit{Pol}(\mathcal{C})$, is the least class containing $\mathcal{C}$ which is closed under both union and marked concatenation: for every $K,L\in\mathit{Pol}(\mathcal{C})$ and $a\in A$, we have $K\cup L\in\mathit{Pol}(\mathcal{C})$ and $KaL\in\mathit{Pol}(\mathcal{C})$. While this is not obvious from the definition, when the input class $\mathcal{C}$ is a quotient-closed lattice, its polynomial closure $\mathit{Pol}(\mathcal{C})$ remains a quotient-closed lattice (the difficulty is to prove closure under intersection). This was originally proved by Arfi [2] (see [16, 27] for recent proofs).

Theorem 6** (Arfi [2]).**

Let $\mathcal{C}$ be a quotient-closed lattice. Then, $\mathit{Pol}(\mathcal{C})$ is a quotient-closed lattice closed under concatenation and marked concatenation.

Concatenation hierarchies. We let $\mathcal{C}$ as a quotient-closed Boolean algebra. The concatenation hierarchy of basis $\mathcal{C}$ classifies languages into levels of two kinds: full levels (denoted by 0, 1, 2,…) and half levels (denoted by 1/2, 3/2, 5/2,…):

• •

Level [math] is the basis $\mathcal{C}$.

• •

Each half level $n+\frac{1}{2}$, for $n\in\mathbb{N}$, is the polynomial closure of the previous full level, i.e., of level $n$.

• •

Each full level $n+1$, for $n\in\mathbb{N}$, is the Boolean closure of the previous half level, i.e., of level $n+\frac{1}{2}$.

In view of Lemma 5 and Theorem 6, it is immediate that every full level is a quotient-closed Boolean algebra while every half level is a quotient-closed lattice. Moreover, we have the following useful fact (see [21, Lemma 3.4] for a proof).

Fact 7**.**

Let $\mathcal{D}$ be a level greater or equal to $1$ in some concatenation hierarchy. Then $\{w\}\in\mathcal{D}$ for every $w\in A^{*}$.

In this paper, we are mainly interested in levels 1/2, 1 and 3/2. By definition, they correspond to the classes $\mathit{Pol}(\mathcal{C})$, $\mathit{Bool}(\mathit{Pol}(\mathcal{C}))$ and $\mathit{Pol}(\mathit{Bool}(\mathit{Pol}(\mathcal{C})))$. For the sake of avoiding clutter, we shall write $\mathit{BPol}(\mathcal{C})$ for $\mathit{Bool}(\mathit{Pol}(\mathcal{C}))$ and $\mathit{PBPol}(\mathcal{C})$ for $\mathit{Pol}(\mathit{Bool}(\mathit{Pol}(\mathcal{C})))$. Our main theorem applies to those levels for every basis $\mathcal{C}$ which is a quotient-closed Boolean algebra of group languages. It is as follows.

Theorem 8** (Main result).**

Consider a concatenation hierarchy whose basis $\mathcal{C}$ contains only group languages and such that $\mathcal{C}$-separation is decidable. Then, covering and separation are decidable for levels 1/2, 1 and 3/2 of this hierarchy.

Theorem 8 is generic and applies to all “group based” hierarchies. It complements similar results which were recently proved in [20, 21] (for levels 1/2 and 3/2) and [25, 28] (for level 1) for finitely based hierarchies.

Theorem 9** ([21, 28]).**

Consider a concatenation hierarchy whose basis is finite. Then, covering and separation are decidable for levels 1/2, 1 and 3/2.

All prominent concatenation hierarchies investigated in the literature have a basis which is either finite or made of group languages. Hence, when put together, the two above theorems can be used to handle the lower levels of all prominent concatenation hierarchies in a generic way.

Additionally, one may combine Theorem 8 with a result of [27] to get information on level 5/2 of group based hierarchies. It is shown in [27] that given an arbitrary hierarchy, if separation is decidable for some half level, then so is the membership problem for the next half level. Hence, we get the following corollary of Theorem 8.

Corollary 10**.**

Consider a concatenation hierarchy whose basis $\mathcal{C}$ contains only group languages and such that $\mathcal{C}$-separation is decidable. Then, given as input a regular language $L$, one may decide whether $L$ belongs to level 5/2 of this hierarchy.

The proof of Theorem 8 spans the remaining sections of the paper. Each level mentioned in the statement is handled independently. In Section 4, we recall a framework designed to handle the covering problem. It was originally introduced in [26] and we use it to obtain all results announced in Theorem 8. We apply it for level 1/2 in Section 5. Handling levels 1 and 3/2 is much more involved and requires extending the framework. We do so in Section 6 and handle these levels in Sections 7 and 8 respectively.

However, let us first illustrate Theorem 8 by presenting two prominent concatenation hierarchies to which it applies.

3.2. Important applications of Theorem 8

We present two prominent examples of concatenation hierarchies whose bases are quotient-closed Boolean algebra of group languages (we already presented these bases in Section 2).

Group hierarchy. This is the hierarchy of basis GR (the class of all group languages). It was originally introduced by Margolis and Pin [13]. To our knowledge, the following results were known for this hierarchy:

• •

Separation is decidable for the basis GR. This is a corollary of a theorem by Ash [3] which solves a longstanding conjecture in algebra (Rhodes’ type II conjecture). We refer the reader to [10] for background on this question.

• •

Membership is decidable for levels 1/2 and 1. The former follows from the decidability of separation for GR and a transfer result of [27]. The latter is proved in [13].

Since separation is decidable for the basis GR, we obtain the following result from Theorem 8 and Corollary 10.

Corollary 11**.**

Separation and covering are decidable for levels 1/2, 1 and 3/2 in the group hierarchy. Moreover, membership is decidable for level 5/2.

The hierarchy of basis MOD. This example is quite important because it has a natural logical equivalent. It is known that every concatenation hierarchy corresponds to the quantifier alternation hierarchy within a particular variant of first-order logic over words (the variants differ by the set of predicates that are allowed in sentences). This was first observed by Thomas [33] for a particular example: the dot-depth hierarchy of Brzozowski and Cohen [4] (whose basis is the finite class $\{\emptyset,\{\varepsilon\},A^{+},A^{*}\}$). However, the ideas of Thomas can be generalized to all concatenation hierarchies (see [27]).

Of course, depending on the concatenation hierarchy, the variant of first-order logic one ends up with may or may not be natural. It turns out that for the hierarchy of basis MOD, we get a standard one: first-order logic with modular predicates ($\textup{FO}({<},\textup{MOD})$). One may view a finite word $w$ as a logical structure made of a sequence of positions numbered from [math] to $|w|-1$. Each position can be quantified and carries a label in $A$. We denote by $\textup{FO}({<},\textup{MOD})$ the variant of first-order logic equipped with the following predicates:

• •

For each $a\in A$, a unary predicate $P_{a}$ selecting positions labeled with letter “$a$”.

• •

A binary predicate “$<$” interpreted as the linear order.

• •

For each integers $0\leq i<d$, a unary predicate $\mathsf{MOD}^{d}_{i}$ selecting positions that are congruent to $i$ modulo $d$.

• •

A constant $\mathsf{D}^{d}_{i}$, which holds for words whose length is congruent to $i$ modulo $d$.

A sentence of $\textup{FO}({<},\textup{MOD})$ defines a language (it consists of all words satisfying the sentence). Thus, $\textup{FO}({<},\textup{MOD})$ defines a class of languages which we also denote by $\textup{FO}({<},\textup{MOD})$.

We are not interested in $\textup{FO}({<},\textup{MOD})$ itself: we consider its quantifier alternation hierarchy. One may classify sentences of $\textup{FO}({<},\textup{MOD})$ by counting their number of quantifier alternations. Let $n\in\mathbb{N}$. A sentence is $\Sigma_{n}({<},\textup{MOD})$, if it can be rewritten into a sentence in prenex normal form which has either, exactly $n$ blocks of quantifiers starting with an $\exists$ or strictly less than $n$ blocks of quantifiers. For example, consider the following sentence (already in prenex normal form)

[TABLE]

This sentence is $\Sigma_{3}({<},\textup{MOD})$. In general, the negation of a $\Sigma_{n}({<},\textup{MOD})$ sentence is not a $\Sigma_{n}({<},\textup{MOD})$ sentence. Hence it is relevant to define $\mathcal{B}\Sigma_{n}({<},\textup{MOD})$ sentences as the Boolean combinations of $\Sigma_{n}({<},\textup{MOD})$ sentences. This gives a hierarchy of classes of languages: for every $n\in\mathbb{N}$, we have $\Sigma_{n}({<},\textup{MOD})\subseteq\mathcal{B}\Sigma_{n}({<},\textup{MOD})\subseteq\Sigma_{n+1}({<},\textup{MOD})$.

This hierarchy has been widely investigated in the literature. For membership, it was known that the problem is decidable for $\mathcal{B}\Sigma_{1}({<},\textup{MOD})$ (see [6]) as well as $\Sigma_{1}({<},\textup{MOD})$ and $\Sigma_{2}({<},\textup{MOD})$ (see [12]). Furthermore, it was recently shown that separation is decidable for $\mathcal{B}\Sigma_{1}({<},\textup{MOD})$ [34]. We are able to reprove these results, generalize them to the covering problem and push them to $\Sigma_{3}({<},\textup{MOD})$. Indeed, the generic correspondence of [27] implies that for every $n\in\mathbb{N}$, level $n$ in the hierarchy of basis MOD corresponds to $\mathcal{B}\Sigma_{n}({<},\textup{MOD})$ while level $n+\frac{1}{2}$ corresponds to $\Sigma_{n}({<},\textup{MOD})$. Moreover, proving the decidability of MOD-separation is a simple exercise (we present a proof in appendix). Consequently, we obtain the following statement as a corollary of Theorem 8.

Corollary 12**.**

Separation and covering are decidable for the logics $\Sigma_{1}({<},\textup{MOD})$,$\mathcal{B}\Sigma_{1}({<},\textup{MOD})$ and $\Sigma_{2}({<},\textup{MOD})$. Moreover, membership is decidable for $\Sigma_{3}({<},\textup{MOD})$.

4. Framework

In this section, we present the framework which we shall use to obtain the three decidability results announced in Theorem 8. It was originally introduced in [26]. We refer the reader to this paper for details and the proofs of the statements.

4.1. Rating maps

The framework is based on an algebraic object called “rating map”. They are morphisms of commutative and idempotent monoids. We write such monoids $(R,+)$: we call the binary operation “$+$” addition and denote the neutral element by $0_{R}$. Being idempotent means that for all $r\in R$, we have $r+r=r$. For every commutative and idempotent monoid $(R,+)$, one may define a canonical ordering $\leq$ over $R$:

[TABLE]

One may verify that $\leq$ is a partial order which is compatible with addition. Moreover, every morphism between commutative and idempotent monoids is increasing for this ordering.

Example 13**.**

For every set $E$, $(2^{E},\cup)$ is an idempotent and commutative monoid. The neutral element is $\emptyset$ and the canonical ordering is inclusion.

When dealing with subsets of a commutative and idempotent monoid $(R,+)$, we shall often apply a downset operation. Given $S\subseteq R$, we write:

[TABLE]

We extend this notation to Cartesian products of arbitrary sets with $R$. Given some set $X$ and $S\subseteq X\times R$, we write,

[TABLE]

We may now define rating maps. A rating map is a morphism $\rho:(2^{A^{*}},\cup)\to(R,+)$ where $(R,+)$ is a finite idempotent and commutative monoid, called the rating set of $\rho$. That is, $\rho$ is a map from $2^{A^{*}}$ to $R$ satisfying the following properties:

1. (1)

$\rho(\emptyset)=0_{R}$. 2. (2)

For all $K_{1},K_{2}\subseteq A^{*}$, $\rho(K_{1}\cup K_{2})=\rho(K_{1})+\rho(K_{2})$.

For the sake of improved readability, when applying a rating map $\rho$ to a singleton set $\{w\}$, we shall write $\rho(w)$ for $\rho(\{w\})$. Additionally, we write $\rho_{*}:A^{*}\to R$ for the restriction of $\rho$ to $A^{*}$: for every $w\in A^{*}$, we have $\rho_{*}(w)=\rho(w)$ (this notation is useful when referring to the language $\rho_{*}^{-1}(r)\subseteq A^{*}$, which consists of all words $w\in A^{*}$ such that $\rho(w)=r$).

Most of the theory makes sense for arbitrary rating maps. However, we shall often have to work with special rating maps satisfying additional properties. We define two kinds.

Nice rating maps. We say a rating map $\rho:2^{A^{*}}\to R$ is nice when, for every language $K\subseteq A^{*}$, there exists finitely many words $w_{1},\dots,w_{n}\in K$ such that $\rho(K)=\rho(w_{1})+\cdots+\rho(w_{k})$.

When a rating map $\rho:2^{A^{*}}\to R$ is nice, it is characterized by the canonical map $\rho_{*}:A^{*}\to R$. Indeed, for $K\subseteq A^{*}$, we may consider the sum of all elements $\rho(w)$ for $w\in K$: while it may be infinite, this sum boils down to a finite one since $R$ is commutative and idempotent. The hypothesis that $\rho$ is nice implies that $\rho(K)$ is equal to this sum.

Multiplicative rating maps. A rating map $\rho:2^{A^{*}}\to R$ is multiplicative when its rating set $R$ has more structure: it needs to be an idempotent semiring. Moreover, $\rho$ has to satisfy an additional property connecting this structure to language concatenation. Namely, it has to be a morphism of semirings.

A semiring is a tuple $(R,+,\cdot)$ where $R$ is a set and “$+$” and “$\cdot$” are two binary operations called addition and multiplication, such that the following axioms are satisfied:

• •

$(R,+)$ is a commutative monoid.

• •

$(R,\cdot)$ is a monoid (the neutral element is denoted by $1_{R}$).

• •

Multiplication distributes over addition. For $r,s,t\in R$, $r\cdot(s+t)=(r\cdot s)+(r\cdot t)$ and $(r+s)\cdot t=(r\cdot t)+(s\cdot t)$.

• •

The neutral element “$0_{R}$” of $(R,+)$ is a zero for $(R,\cdot)$: $0_{R}\cdot r=r\cdot 0_{R}=0_{R}$ for every $r\in R$.

A semiring $R$ is idempotent when $r+r=r$ for every $r\in R$, i.e., when the additive monoid $(R,+)$ is idempotent (there is no additional constraint on the multiplicative monoid $(R,\cdot)$).

Example 14**.**

A key example of infinite idempotent semiring is the set $2^{A^{*}}$. Union is the addition and language concatenation is the multiplication (with $\{\varepsilon\}$ as neutral element).

Clearly, any finite idempotent semiring $(R,+,\cdot)$ is in particular a rating set: $(R,+)$ is an idempotent and commutative monoid. In particular, one may verify that the canonical ordering “$\leq$” on $R$ is compatible with multiplication as well.

We may now define multiplicative rating maps: as expected they are semiring morphisms. Let $\rho:2^{A^{*}}\to R$ be a rating map: $(R,+)$ is an idempotent commutative monoid and $\rho$ is a morphism from $(2^{A^{*}},\cup)$ to $(R,+)$. We say that $\rho$ is multiplicative when the rating set $R$ is equipped with a multiplication “$\cdot$” such that $(R,+,\cdot)$ is an idempotent semiring and $\rho$ is also a monoid morphism from $(2^{A^{*}},\cdot)$ to $(R,\cdot)$. That is, the two following additional axioms have to be satisfied:

1. (3)

$\rho(\varepsilon)=1_{R}$. 2. (4)

For all $K_{1},K_{2}\subseteq A^{*}$, we have $\rho(K_{1}K_{2})=\rho(K_{1})\cdot\rho(K_{2})$.

Altogether, this exactly says that $\rho$ must be a semiring morphism from $(2^{A^{*}},\cup,\cdot)$ to $(R,+,\cdot)$.

Remark 15**.**

The rating maps which are both nice and multiplicative are finitely representable. Indeed, as we explained above, if a rating map $\rho:2^{A^{*}}\to R$ is nice, it is characterized by the canonical map $\rho_{*}:A^{*}\to R$. When $\rho$ is additionally multiplicative, $\rho_{*}$ is finitely representable since it is a morphism into a finite monoid. Hence, we may speak about algorithms taking nice multiplicative rating maps as input.

The rating maps which are not nice and multiplicative cannot be finitely represented in general. However, they remain crucial in the paper: while our main statements consider nice multiplicative rating maps, many proofs involve auxiliary rating maps which are neither nice nor multiplicative.

4.2. Optimal imprints

Now that we have rating maps, we turn to imprints. Consider a rating map $\rho:2^{A^{*}}\to R$. Given any finite set of languages $\mathbf{K}$, we define the $\rho$-imprint of $\mathbf{K}$. Intuitively, when $\mathbf{K}$ is a cover of some language $L$, this object measures the “quality” of $\mathbf{K}$. The $\rho$-imprint of $\mathbf{K}$ is the following subset of $R$:

[TABLE]

We may now define optimality. Consider an arbitrary rating map $\rho:2^{A^{*}}\to R$ and a lattice $\mathcal{C}$. Given a language $L$, an optimal $\mathcal{C}$-cover of $L$ for $\rho$ is a $\mathcal{C}$-cover $\mathbf{K}$ of $L$ which satisfies the following property:

[TABLE]

In general, there can be infinitely many optimal $\mathcal{C}$-covers for a given rating map $\rho$. The key point is that there always exists at least one (this requires the hypothesis that $\mathcal{C}$ is a lattice).

Lemma 16**.**

Let $\mathcal{C}$ be a lattice. For every language $L$ and every rating map $\rho$, there exists an optimal $\mathcal{C}$-cover of $L$ for $\rho$.

Clearly, for a lattice $\mathcal{C}$, a language $L$ and a rating map $\rho$, all optimal $\mathcal{C}$-covers of $L$ for $\rho$ have the same $\rho$-imprint. Hence, this unique $\rho$-imprint is a canonical object for $\mathcal{C}$, $L$ and $\rho$. We call it the $\mathcal{C}$-optimal $\rho$-imprint on $L$ and we write it $\mathcal{I}_{\mathcal{C}}\left[L,\rho\right]$:

[TABLE]

An important special case is when $L=A^{*}$. In that case, we write $\mathcal{I}_{\mathcal{C}}\left[\rho\right]$ for $\mathcal{I}_{\mathcal{C}}\left[A^{*},\rho\right]$. Finally, we have the following useful fact which is immediate from the definitions.

Fact 17**.**

Let $\rho$ be a rating map and consider two languages $H,L$ such that $H\subseteq L$. Then, $\mathcal{I}_{\mathcal{C}}\left[H,\rho\right]\subseteq\mathcal{I}_{\mathcal{C}}\left[L,\rho\right]$.

4.3. Connection with covering

We may now connect these definitions to the covering problem. The key idea is that solving $\mathcal{C}$-covering boils down to computing $\mathcal{C}$-optimal imprints from input nice multiplicative rating maps. There are actually two statements (both taken from [26]). The first one is simpler but it only applies to classes $\mathcal{C}$ which are Boolean algebras while the second (more involved) one applies to all lattices. We start with the former.

Proposition 18**.**

Let $\mathcal{C}$ be a Boolean algebra. Assume that there exists an algorithm which computes $\mathcal{I}_{\mathcal{C}}\left[\rho\right]$ from an input nice multiplicative rating map $\rho$. Then, $\mathcal{C}$-covering is decidable.

In order to handle all lattices, one needs to consider several optimal imprints simultaneously. This is formalized by the following additional object. Consider a lattice $\mathcal{C}$, a morphism $\alpha:A^{*}\to M$ into a finite monoid $M$ and a rating map $\rho:2^{A^{*}}\to R$, we define the following subset of $M\times R$:

[TABLE]

We call $\mathcal{P}_{\mathcal{C}}^{\alpha}[\rho]$ the $\alpha$-pointed $\mathcal{C}$-optimal $\rho$-imprint. Clearly, $\mathcal{P}_{\mathcal{C}}^{\alpha}[\rho]$ encodes all the sets $\mathcal{I}_{\mathcal{C}}\left[\alpha^{-1}(s),\rho\right]$ for $s\in M$.

Proposition 19**.**

Let $\mathcal{C}$ be a lattice. Assume that there exists an algorithm which computes $\mathcal{P}_{\mathcal{C}}^{\alpha}[\rho]$ from an input morphism $\alpha$ and an input nice multiplicative rating map $\rho$. Then, $\mathcal{C}$-covering is decidable.

5. Polynomial closure

We prove the first part of Theorem 8: for every quotient-closed Boolean algebra of group languages $\mathcal{C}$, if $\mathcal{C}$-separation is decidable, then so is $\mathit{Pol}(\mathcal{C})$-covering.

We use rating maps: for every morphism $\alpha:A^{*}\to M$ and nice multiplicative rating map $\rho:2^{A^{*}}\to R$, we characterize $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$. When $\mathcal{C}$-separation is decidable, it is simple to deduce a least fixpoint algorithm for computing $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$. Proposition 19 then yields that $\mathit{Pol}(\mathcal{C})$-covering is decidable.

Remark 20**.**

Our characterization of $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ does not actually require $\rho$ to be nice. While useless for deciding $\mathit{Pol}(\mathcal{C})$-covering, this is significant. This will later be a requirement when handling $\mathit{BPol}(\mathcal{C})$ and $\mathit{PBPol}(\mathcal{C})$ (see remark 42).

Before we can present the characterization, we require some terminology. We define a new notion for rating maps: optimal $\varepsilon$-approximations. They are specifically designed to handle covering for concatenation hierarchies whose bases are made of group languages (it is through this notion that our characterizations are parameterized by $\mathcal{C}$-separation).

5.1. Optimal $\varepsilon$-approximations

When handling $\mathit{Pol}(\mathcal{C})$ for a quotient-closed Boolean algebra of group languages $\mathcal{C}$ (as well as $\mathit{BPol}(\mathcal{C})$ and $\mathit{PBPol}(\mathcal{C})$ later), we shall often encounter optimal $\mathcal{C}$-covers of the singleton $\{\varepsilon\}$ for various rating maps. The definitions presented here are based on a single key idea: there always exists an optimal $\mathcal{C}$-cover of $\{\varepsilon\}$ which consists of a single language.

Remark 21**.**

The definitions make sense for any lattice $\mathcal{C}$. However, they are mainly relevant when $\mathcal{C}$ is a class of group languages. In practice, the other important lattices contain the singleton $\{\varepsilon\}$ and $\{\{\varepsilon\}\}$ is always an optimal $\mathcal{C}$-cover of $\{\varepsilon\}$.

Let $\mathcal{C}$ be a lattice and $\rho:2^{A^{*}}\to R$ be a rating map. A $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$ is a language $L\in\mathcal{C}$ such that $\varepsilon\in L$ and,

[TABLE]

Remark 22**.**

Since $\{\varepsilon\}$ is a singleton, one may show that $L$ is a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$ if and only if $\{L\}$ is an optimal $\mathcal{C}$-cover of $\{\varepsilon\}$ for $\rho$.

As expected, there always exists a $\mathcal{C}$-optimal $\varepsilon$-approximation for any rating map $\rho$, provided that $\mathcal{C}$ is a lattice.

Lemma 23**.**

For any lattice $\mathcal{C}$ and any rating map $\rho:2^{A^{*}}\to R$, there exists a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$.

We complete the definition with a key remark. Clearly, all $\mathcal{C}$-optimal $\varepsilon$-approximations for $\rho$ have the same image under $\rho$. This is a canonical object for $\mathcal{C}$ and $\rho$. We write it $\mathbbm{i}_{\mathcal{C}}[\rho]\in R$:

[TABLE]

We turn to a crucial property of $\varepsilon$-approximations: when $\rho$ is a nice rating map, computing $\mathbbm{i}_{\mathcal{C}}[\rho]$ boils down to separation.

Lemma 24**.**

Let $\mathcal{C}$ be a lattice and $\rho:2^{A^{*}}\to R$ be a nice rating map. Then, $\mathbbm{i}_{\mathcal{C}}[\rho]$ is the sum of all $r\in R$ such that $\{\varepsilon\}$ is not $\mathcal{C}$-separable from $\rho_{*}^{-1}(r)$.

Lemma 24 has an important corollary. When $\rho:2^{A^{*}}\to R$ is a nice multiplicative rating map, the languages $\rho_{*}^{-1}(r)$ are regular (they are recognized by $\rho_{*}$). Hence, given a lattice $\mathcal{C}$, if $\mathcal{C}$-separation is decidable, we may compute $\mathbbm{i}_{\mathcal{C}}[\rho]$ from an input nice multiplicative rating map.

Corollary 25**.**

Consider a lattice $\mathcal{C}$ such that $\mathcal{C}$-separation is decidable. Then, given as input a nice multiplicative rating map $\rho:2^{A^{*}}\to R$, one may compute the subset $\mathbbm{i}_{\mathcal{C}}[\rho]$ of $R$.

Corollary 25 is crucial for our approach: this is exactly how the algorithms for $\mathit{Pol}(\mathcal{C})$-, $\mathit{BPol}(\mathcal{C})$- and $\mathit{PBPol}(\mathcal{C})$-covering announced in Theorem 8 are parametrized by $\mathcal{C}$-separation.

Let us point out that when considering a specific lattice $\mathcal{C}$, it is usually possible to write an algorithm for computing $\mathbbm{i}_{\mathcal{C}}[\rho]$ from an input nice multiplicative rating map which does not involve separation. We present an example for the class MOD.

Lemma 26**.**

Let $\rho:2^{A^{*}}\to R$ be a nice multiplicative rating map. Then, $\mathbbm{i}_{\textup{{MOD}}}[\rho]=(\rho(A))^{\omega}+\rho(\varepsilon)$.

Finally, we have the following lemma, which considers the special case when the lattice $\mathcal{C}$ is a quotient-closed Boolean algebra of group languages. We shall need this lemma for proving the three main results of the paper.

Lemma 27**.**

Let $\mathcal{C}$ be a quotient-closed Boolean algebra of group languages and $\rho:2^{A^{*}}\to R$ be a rating map. Then, $\mathbbm{i}_{\mathcal{C}}[\rho]\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{\varepsilon\},\rho\right]$.

Proof.

Let $\mathbf{K}$ be an optimal $\mathit{Pol}(\mathcal{C})$-cover of $\{\varepsilon\}$. By definition, there exists $K\in\mathbf{K}$ such that $\varepsilon\in K$. Since $K\in\mathit{Pol}(\mathcal{C})$, it is immediate from a simple induction on the definition of polynomial closure that we have $H\in\mathcal{C}$ such that $\varepsilon\in H$ and $H\subseteq K$ (the key idea is that the marked concatenation of two languages cannot contain $\varepsilon$). By definition of $\mathbbm{i}_{\mathcal{C}}[\rho]$, we have $\mathbbm{i}_{\mathcal{C}}[\rho]\leq\rho(H)$. Consequently, $\mathbbm{i}_{\mathcal{C}}[\rho]\leq\rho(K)$, which yields $\mathbbm{i}_{\mathcal{C}}[\rho]\in\mathcal{I}[\rho](\mathbf{K})$. Since $\mathbf{K}$ is an optimal $\mathit{Pol}(\mathcal{C})$-cover of $\{\varepsilon\}$, we obtain $\mathbbm{i}_{\mathcal{C}}[\rho]\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{\varepsilon\},\rho\right]$, as desired. ∎

5.2. Characterization

We characterize $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ when $\mathcal{C}$ is a quotient-closed Boolean algebra of group languages. Let $\mathcal{C}$ be such a class.

We first state the characterization. Consider some morphism $\alpha:A^{*}\to M$ and some multiplicative rating map $\rho:2^{A^{*}}\to R$. Consider a subset $S\subseteq M\times R$. We say that $S$ is $\mathit{Pol}(\mathcal{C})$-complete for $\alpha$ and $\rho$ when it satisfies the following properties:

• •

Trivial elements: For all $w\in A^{*}$, $(\alpha(w),\rho(w))\in S$.

• •

Downset: We have $\mathord{\downarrow_{R}}S=S$.

• •

Multiplication: For all, $(s,q),(t,r)\in S$, $(st,qr)\in S$.

• •

$\mathcal{C}$-operation: We have $(1_{M},\mathbbm{i}_{\mathcal{C}}[\rho])\in S$.

We may now state the main theorem of the section: the least $\mathit{Pol}(\mathcal{C})$-complete subset of $M\times R$ with respect to inclusion is exactly $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ (recall that $\mathcal{C}$ is required to be a quotient-closed Boolean algebra of group languages).

Theorem 28**.**

Consider a morphism $\alpha:A^{*}\to M$ and a multiplicative rating map $\rho:2^{A^{*}}\to R$. Then, $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ is the least $\mathit{Pol}(\mathcal{C})$-complete subset of $M\times R$.

By Theorem 28, if $\mathcal{C}$-separation is decidable, then one may compute $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ when given a morphism $\alpha:A^{*}\to M$ and a nice multiplicative rating map $\rho:2^{A^{*}}\to R$ as input. Indeed, computing the least $\mathit{Pol}(\mathcal{C})$-complete subset of $M\times R$ is achieved by an obvious least fixpoint procedure. We are able to implement $\mathcal{C}$-operation, since we may compute $\mathbbm{i}_{\mathcal{C}}[\rho]\in R$: by Corollary 25, this boils down to deciding $\mathcal{C}$-separation.

Together with Proposition 19, this yields that when $\mathcal{C}$-separation is decidable, so is $\mathit{Pol}(\mathcal{C})$-covering. Therefore, we get the first part of Theorem 8, regarding level 1/2. We conclude the section with the proof of Theorem 28.

5.3. Proof of Theorem 28

We fix a morphism $\alpha:A^{*}\to M$ and a multiplicative rating map $\rho:2^{A^{*}}\to R$ for the proof. We need to show that $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ is the least $\mathit{Pol}(\mathcal{C})$-complete subset of $M\times R$. We first prove that it is $\mathit{Pol}(\mathcal{C})$-complete. This means that the least fixpoint procedure computing $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ is sound.

Soundness. We have to show that $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ satisfies the four properties in the definition of $\mathit{Pol}(\mathcal{C})$-complete subsets. The first three (trivial elements, downset and multiplication) are generic: they are satisfied by $\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$ for any quotient-closed lattice $\mathcal{D}$ (we refer the reader to [26] for the proof). Hence, they are satisfied by $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ since $\mathit{Pol}(\mathcal{C})$ is a quotient-closed lattice by Theorem 6. We concentrate on $\mathit{Pol}(\mathcal{C})$-operation.

We show that $(1_{M},\mathbbm{i}_{\mathcal{C}}[\rho])\in\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$. In other words, we prove $\mathbbm{i}_{\mathcal{C}}[\rho]\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\alpha^{-1}(1_{M}),\rho\right]$. By Lemma 27, we know that $\mathbbm{i}_{\mathcal{C}}[\rho]\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{\varepsilon\},\rho\right]$. Moreover, $\{\varepsilon\}\subseteq\alpha^{-1}(1_{M})$ since $\alpha$ is a morphism. Thus, Fact 17 yields that $\mathbbm{i}_{\mathcal{C}}[\rho]\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\alpha^{-1}(1_{M}),\rho\right]$, finishing the soundness proof.

Completeness. We now prove that $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ is included in every $\mathit{Pol}(\mathcal{C})$-complete subset of $M\times R$. This direction corresponds to completeness of the least fixpoint procedure computing $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$. For the proof, we fix $S\subseteq M\times R$ which is $\mathit{Pol}(\mathcal{C})$-complete subset and show that $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]\subseteq S$. The argument is based on the following proposition

Proposition 29**.**

Let $t\in M$. There exists a $\mathit{Pol}(\mathcal{C})$-cover $\mathbf{K}_{t}$ of $\alpha^{-1}(t)$ such that for every $K\in\mathbf{K}_{t}$, we have $(t,\rho(K))\in S$.

Let us first use Proposition 29 to show that $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]\subseteq S$. Let $(t,r)\in\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$. We show that $(t,r)\in S$. By definition, we have $r\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\alpha^{-1}(t),\rho\right]$. Consider the $\mathit{Pol}(\mathcal{C})$-cover $\mathbf{K}_{t}$ of $\alpha^{-1}(t)$ given by Proposition 29. By hypothesis, we have $r\in\mathcal{I}[\rho](\mathbf{K}_{t})$. Thus, we get $K\in\mathbf{K}_{t}$ such that $r\leq\rho(K)$. Finally, since $(t,\rho(K))\in S$ and $S=\mathord{\downarrow_{R}}S$ ($S$ is $\mathit{Pol}(\mathcal{C})$-complete), we obtain that $(t,r)\in S$, finishing the proof.

It remains to prove Proposition 29. We rely on the following key technical result.

Lemma 30**.**

Let $H$ be a regular language and $L$ be a group language such that $\varepsilon\in L$. There is a cover $\mathbf{K}$ of $H$ such that for every language $K\in\mathbf{K}$, we have $n\in\mathbb{N}$ and $a_{1},\dots,a_{n}\in A$ such that $K=La_{1}L\cdots a_{n}L$ and $a_{1}\cdots a_{n}\in H$.

Proof.

Since $H$ is regular, we have a morphism $\gamma:A^{*}\to N$ into a finite monoid $N$ recognizing $H$. Moreover, since $L$ is a group language it is recognized by a morphism $\beta:A^{*}\to G$ where $G$ is a finite group. Since $\varepsilon\in L$, we have $\beta^{-1}(1_{G})\subseteq L$.

For every $w=a_{1}\cdots a_{n}\in A^{*}$, we let $F_{w}\subseteq A^{*}$ be the language $F_{w}=La_{1}L\cdots a_{n}L$ ($F_{\varepsilon}=L$). Moreover, we let $F^{\prime}_{w}\subseteq F_{w}$ be the language $F^{\prime}_{w}=\beta^{-1}(1_{G})a_{1}\beta^{-1}(1_{G})\cdots a_{n}\beta^{-1}(1_{G})$. Let $n=|N\times G|$. We show that the following is a cover of $H$:

[TABLE]

Clearly, this implies that \mathbf{K}=\{F_{w}\mid\text{w\in H$and$|w|\leq n}\} is a cover of $H$, finishing the proof. We use the following fact.

Fact 31**.**

Let $w\in H$ such that $|w|>n$. Then, there exists $v\in H$ such that $|v|<|w|$ and $F^{\prime}_{w}\subseteq F^{\prime}_{v}$.

We first use the fact to finish the main proof. Let $w\in H$. We have to find K\in\{F^{\prime}_{w}\mid\text{w\in H$and$|w|\leq n}\} such that $w\in K$. Observe that $w\in F^{\prime}_{w}$ since by definition, $\varepsilon\in\beta^{-1}(1_{G})$. We may then apply Fact 31 repeatedly to build $v\in H$ such that $|v|\leq n$ and $F^{\prime}_{w}\subseteq F^{\prime}_{v}$. Consequently, we have $w\in F^{\prime}_{v}$ and we may choose $K=F^{\prime}_{v}$. It remains to prove Fact 31.

We fix $w\in H$ such that $|w|>n$. We let $w=a_{1}\cdots a_{|w|}$, where each $a_{i}$ is a letter. Clearly, $N\times G$ is a monoid for the componentwise multiplication. Let $\eta:A^{*}\to N\times G$ be the morphism defined by $\eta(u)=(\gamma(u),\beta(u))$ for every $u\in A^{*}$. Since $|w|>n=|N\times G|$, it follows from the pigeonhole principle that there exist $i,j$ such that $1\leq i<j\leq|w|$ and $\eta(a_{1}\cdots a_{i})=\eta(a_{1}\cdots a_{j})$. Define $v=a_{1}\cdots a_{i}a_{j+1}\cdots a_{|w|}$. Clearly, we have $|v|<|w|$. Moreover, $\eta(w)=\eta(v)$, hence $\gamma(w)=\gamma(v)$. Consequently, $v\in H$, since $w\in H$ and $H$ is recognized by $\gamma$. It remains to show that $F^{\prime}_{w}\subseteq F^{\prime}_{v}$.

Let $u\in F^{\prime}_{w}$: we have $u=u_{0}a_{1}u_{1}a_{2}u_{2}\cdots a_{|w|}u_{|w|}$ with $u_{i}\in\beta^{-1}(1_{G})$ for every $i\leq|w|$. Let $x=u_{i}a_{i+1}u_{i+1}\cdots a_{j}u_{j}$. Clearly, $\beta(x)=\beta(a_{i+1}\cdots a_{j})$ and we know by definition of $i,j$ that $\beta(a_{1}\cdots a_{i})=\beta(a_{1}\cdots a_{j})$. Thus, $\beta(a_{1}\cdots a_{i})\cdot\beta(x)=\beta(a_{1}\cdots a_{i})$. Since $G$ is a group, it follows that $\beta(x)=1_{G}$. Consequently, $x\in\beta^{-1}(1_{G})$. Moreover, we have,

[TABLE]

It follows that $u\in F^{\prime}_{v}$, which concludes the proof. ∎

For $t\in M$, we are now ready to construct the $\mathit{Pol}(\mathcal{C})$-cover $\mathbf{K}_{t}$ of Proposition 29. Let $L$ be a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$: we have $L\in\mathcal{C}$, $\varepsilon\in L$ and $\rho(L)=\mathbbm{i}_{\mathcal{C}}[\rho]$. Since $\alpha^{-1}(t)$ is a regular language and $L$ is a group language (by hypothesis on $\mathcal{C}$), Lemma 30 yields a cover $\mathbf{K}_{t}$ of $\alpha^{-1}(t)$ such that every language $K\in\mathbf{K}_{t}$ is of the form $K=La_{1}L\cdots a_{n}L$ with $a_{1},\dots,a_{n}\in A$ such that $a_{1}\cdots a_{n}\in\alpha^{-1}(t)$.

Clearly, $\mathbf{K}_{t}$ is a $\mathit{Pol}(\mathcal{C})$-cover of $\alpha^{-1}(t)$ since $L\in\mathcal{C}$. It remains to prove that for every $K\in\mathbf{K}_{t}$, we have $(t,\rho(K))\in S$. By definition of $\mathbf{K}_{t}$, there exists $w=a_{1}\cdots a_{n}\in\alpha^{-1}(t)$ such that $K=La_{1}L\cdots a_{n}L$. By definition of $\mathit{Pol}(\mathcal{C})$-complete subsets, we have $(\alpha(a_{i}),\rho(a_{i}))\in S$ for every $i\leq n$ (these are trivial elements). Moreover, we know from $\mathit{Pol}(\mathcal{C})$-operation and the definition of $L$ as a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$ that $(1_{M},\rho(L))=(1_{M},\mathbbm{i}_{\mathcal{C}}[\rho])\in S$. It then follows from closure under multiplication that:

[TABLE]

Since $\alpha$ is a morphism and $\rho$ is a multiplicative rating map (and therefore a morphism for multiplication), this exactly says that $(\alpha(w),\rho(K))\in S$. Finally, $\alpha(w)=t$ by definition and we obtain as desired that $(t,\rho(K))\in S$, finishing the proof.∎

6. Extending the Framework

In this section, we introduce additional material about rating maps required to prove the remaining part of Theorem 8 (i.e., concerning levels 1 and 3/2). Let us first overview the situation.

In Section 4, we proved that given an arbitrary lattice $\mathcal{D}$, deciding $\mathcal{D}$-covering boils down to computing $\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$ from a morphism $\alpha$ and a nice multiplicative rating map $\rho$ (see Proposition 19). In fact, if $\mathcal{D}$ is a Boolean algebra, we may even work with the simpler set $\mathcal{I}_{\mathcal{D}}\left[\rho\right]$ (see Proposition 18). This is how we handled $\mathit{Pol}(\mathcal{C})$ in the previous section: we showed that given a fixed quotient-closed Boolean algebra of group languages $\mathcal{C}$, if $\mathcal{C}$-separation is decidable, then $\mathcal{P}_{\mathit{Pol}(\mathcal{C})}^{\alpha}[\rho]$ can be computed via a least fixpoint algorithm. We shall use the same approach for the classes $\mathit{BPol}(\mathcal{C})$ and $\mathit{PBPol}(\mathcal{C})$: we obtain algorithms for computing $\mathcal{I}_{\mathit{BPol}(\mathcal{C})}\left[\rho\right]$ and $\mathcal{P}_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]$. However, we do not compute these sets directly. Instead, we work with more involved sets, which carry more information. In this section, we introduce these other sets.

Remark 32**.**

This situation is not surprising. Typically, computing optimal imprints is achieved via fixpoint procedures (what we did for $\mathit{Pol}(\mathcal{C})$ in Section 7 is a typical example). Hence, replacing the object that we truly want to compute by another one which carries more information makes sense.

First, we present two constructions for building new rating maps out of already existing ones. They are taken from [28], where they are used as technical proof objects. Here, we use them in a more prominent way: they are central to the definitions of this section. We use these constructions to present refined variants of Proposition 18 and Proposition 19.

6.1. Nested rating maps

We present two constructions. The first one involves two objects: a lattice $\mathcal{D}$ and a rating map $\rho:2^{A^{*}}\to R$. We build a new map $\xi_{\mathcal{D}}[\rho]$ whose rating set is $2^{R}$. We let:

[TABLE]

The second construction takes an additional object as input: a map $\alpha:A^{*}\to M$ where $M$ is some arbitrary finite set (in practice, $\alpha$ will be a monoid morphism). We build a new map $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ whose rating set is $2^{M\times R}$:

[TABLE]

It is straightforward to verify that these two maps are in fact rating maps (a proof is available in [28], see Proposition 6.2). We state this result in the following proposition.

Proposition 33**.**

Given a lattice $\mathcal{D}$, a map $\alpha:A^{*}\to M$ and a rating map $\rho:2^{A^{*}}\to R$, $\xi_{\mathcal{D}}[\rho]$ and $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ are rating maps.

A key point is that the rating maps $\xi_{\mathcal{D}}[\rho]$ and $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ are not nice in general, even when the original rating map $\rho$ is (we refer the reader to [28] for a counterexample).

Another question is whether they are multiplicative. Clearly, when $\alpha:A^{*}\to M$ is a monoid morphism and $\rho:2^{A^{*}}\to R$ is a multiplicative rating map, we may lift the multiplications of $M$ and $R$ to the sets $2^{R}$ and $2^{M\times R}$ in the natural way. This makes $(2^{R},\cup,\cdot)$ and $(2^{M\times R},\cup,\cdot)$ semirings. Unfortunately, $\xi_{\mathcal{D}}[\rho]$ and $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ are not multiplicative, even under these hypotheses. However, they behave almost as multiplicative rating maps when the class $\mathcal{D}$ satisfies appropriate properties.

Consider an arbitrary lattice $\mathcal{G}$ and a rating map $\rho:2^{A^{*}}\to R$ whose rating set is a semiring $(R,+,\cdot)$ (but $\rho$ need not be multiplicative). We say that $\rho$ is $\mathcal{G}$-multiplicative when there exists an endomorphism $\mu_{\rho}$ of $(R,+)$ such that:

1. (1)

For every $q,r,s\in R$, $\mu_{\rho}(q\mu_{\rho}(r)s)=\mu_{\rho}(qrs)$. 2. (2)

$1_{R}\leq\rho(\varepsilon)$. 3. (3)

For every $H,K\in\mathcal{G}$ and $a\in A$, we have,

[TABLE]

When $\mathcal{G}$ is the class of all languages (i.e., $\mathcal{G}=2^{A^{*}}$), we say that $\rho$ is quasi-multiplicative. In practice, we shall always assume implicitly that the endomorphism $\mu_{\rho}$ is fixed.

Remark 34**.**

A true multiplicative rating map is always quasi-multiplicative. Indeed, in this case, it suffices to choose $\mu_{\rho}$ as the identity; $\mu_{\rho}(r)=r$ for all $r\in R$.

We have the two following lemmas, which apply to $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ and $\xi_{\mathcal{D}}[\rho]$ respectively (see [28, Lemmas 6.7 and 6.8] for proofs). The statements are rather ad hoc: they are designed to accommodate the situations that we shall encounter. We use $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ in the case when $\mathcal{D}=\mathit{Pol}(\mathcal{C})$ while we use $\xi_{\mathcal{D}}[\rho]$ in the case when $\mathcal{D}=\mathit{BPol}(\mathcal{C})$.

Lemma 35**.**

Let $\mathcal{D}$ be a quotient-closed lattice closed under concatenation, $\alpha:A^{*}\to M$ be a morphism and $\rho:2^{A^{*}}\to~{}R$ be a multiplicative rating map. Then, $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ is quasi-multiplicative, and the endomorphism $\mu_{\zeta_{\mathcal{D}}^{\alpha}[\rho]}$ of $(2^{M\times R},\cup)$ is:

[TABLE]

Lemma 36**.**

Let $\mathcal{G}$ be a quotient-closed lattice closed under concatenation and marked concatenation and let $\mathcal{D}$ = $\mathit{Bool}(\mathcal{G})$. Let $\rho:2^{A^{*}}\to R$ a multiplicative rating map. Then, $\xi_{\mathcal{D}}[\rho]$ is $\mathcal{G}$-multiplicative and the endomorphism $\mu_{\xi_{\mathcal{D}}[\rho]}$ of $(2^{R},\cup)$ is:

[TABLE]

6.2. Application to classes of group languages

We may now present the refinements of Proposition 18 and Proposition 19 announced at the beginning. They are designed to handle the classes investigated in the paper. Here, we discuss and prove the variant for Boolean algebras and simply state the one for lattices (we prove it in appendix).

By Proposition 18, given a Boolean algebra $\mathcal{D}$, deciding $\mathcal{D}$-covering boils down to computing $\mathcal{I}_{\mathcal{D}}\left[\rho\right]\subseteq R$ from a nice multiplicative rating map $\rho:2^{A^{*}}\to R$. It turns out that when $\mathcal{D}$ is a full level in some concatenation hierarchy whose basis $\mathcal{C}$ is made of group languages, one may instead compute a specific subset of $\mathcal{I}_{\mathcal{D}}\left[\rho\right]$.

By definition, $\mathcal{I}_{\mathcal{D}}\left[\rho\right]=\mathcal{I}_{\mathcal{D}}\left[A^{*},\rho\right]=\xi_{\mathcal{D}}[\rho](A^{*})$. We may consider the element $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]\in 2^{R}$ which is equal to $\xi_{\mathcal{D}}[\rho](L)$ for a $\mathcal{C}$-optimal $\varepsilon$-approximation $L$ for $\xi_{\mathcal{D}}[\rho]$. Hence, we have,

[TABLE]

It turns out that because of our hypothesis on $\mathcal{D}$, one may compute $\mathcal{I}_{\mathcal{D}}\left[\rho\right]$ from its subset $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]$ (see below for the proof). This yields the following corollary of Proposition 18.

Proposition 37**.**

Let $\mathcal{C}$ be a quotient-closed Boolean algebra of group languages and let $\mathcal{D}$ be a strictly positive full level within the concatenation hierarchy of basis $\mathcal{C}$. Assume that there exists an algorithm computing $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]$ from an input nice multiplicative rating map $\rho$. Then, $\mathcal{D}$-covering is decidable.

Before we prove Proposition 37, let us present the generalized variant of Proposition 19 which considers all lattices instead of just Boolean algebras (we detail it in appendix).

Proposition 38**.**

Let $\mathcal{C}$ be a quotient-closed Boolean algebra of group languages and let $\mathcal{D}$ be a half level within the concatenation hierarchy of basis $\mathcal{C}$. Assume that there exists an algorithm computing $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]$ from a morphism $\alpha$ and a nice multiplicative rating map $\rho$. Then, $\mathcal{D}$-covering is decidable.

6.3. Proof of Proposition 37

We fix $\mathcal{C}$ and $\mathcal{D}$ as in the statement of the proposition. We show that for an arbitrary nice multiplicative rating map $\rho$, one may compute $\mathcal{I}_{\mathcal{D}}\left[\rho\right]$ from its subset $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]$. The result is then an immediate corollary of Proposition 18.

Fix a nice multiplicative rating map $\rho:2^{A^{*}}\to R$ for the proof. We let $S\subseteq R$ be the least subset of $R$ that satisfies the following properties:

1. (1)

We have $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]\subseteq S$. 2. (2)

We have $\rho(w)\in S$ for every $w\in A^{*}$. 3. (3)

We have $\mathord{\downarrow_{R}}S=S$. 4. (4)

For every $r,r^{\prime}\in S$, we have $rr^{\prime}\in S$.

Clearly, one may compute $S$ from $\rho$ and $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]$. Hence, Proposition 37 is now immediate from the following lemma.

Lemma 39**.**

We have $S=\mathcal{I}_{\mathcal{D}}\left[\rho\right]$.

We prove Lemma 39. Let $L$ be a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\xi_{\mathcal{D}}[\rho]$: we have $L\in\mathcal{C}$, $\varepsilon\in L$ and $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]=\xi_{\mathcal{D}}[\rho](L)$.

We first show that $S\subseteq\mathcal{I}_{\mathcal{D}}\left[\rho\right]$. This amounts to proving that $\mathcal{I}_{\mathcal{D}}\left[\rho\right]$ satisfies the four properties in the definition of $S$. That $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]\subseteq\mathcal{I}_{\mathcal{D}}\left[\rho\right]$ is immediate since $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]=\xi_{\mathcal{D}}[\rho](L)$, $\mathcal{I}_{\mathcal{D}}\left[\rho\right]=\xi_{\mathcal{D}}[\rho](A^{*})$, $L\subseteq A^{*}$ and $\xi_{\mathcal{D}}[\rho]$ is a rating map. The other three properties are generic: they are satisfied whenever $\rho$ is a multiplicative rating map and $\mathcal{D}$ is a quotient-closed Boolean algebra (which is the case here as $\mathcal{D}$ is a full level within the hierarchy of basis $\mathcal{C}$). We refer the reader to [26, Lemma 6.11] for the proof.

We turn to the converse inclusion: $\mathcal{I}_{\mathcal{D}}\left[\rho\right]\subseteq S$. Consider $r\in\mathcal{I}_{\mathcal{D}}\left[\rho\right]$, we show that $r\in S$. The argument is based on the following lemma, which is where we use the hypothesis that $\mathcal{C}$ is a class of group languages (which implies that $L\in\mathcal{C}$ is recognized by a finite group).

Lemma 40**.**

There exist $\ell\in\mathbb{N}$ and $\ell$ letters $a_{1},\dots,a_{\ell}\in A$ such that $r\in\xi_{\mathcal{D}}[\rho](La_{1}L\cdots a_{\ell}L)$.

Proof.

For every word $w=a_{1}\cdots a_{\ell}\in A^{*}$, we let $H_{w}$ be the language $H_{w}=La_{1}L\cdots a_{\ell}L$ ($H_{\varepsilon}=L$). We have to find $w\in A^{*}$ such that $r\in\xi_{\mathcal{D}}[\rho](H_{w})$. Clearly $A^{*}$ is a regular language and $L$ is a group language by hypothesis. Therefore, Lemma 30 implies that there exists a finite language $U$ such that $A^{*}\subseteq\bigcup_{w\in U}H_{w}$. Since $\xi_{\mathcal{D}}[\rho]$ is a rating map, this implies:

[TABLE]

Moreover, by hypothesis, we have $r\in\mathcal{I}_{\mathcal{D}}\left[\rho\right]=\mathcal{I}_{\mathcal{D}}\left[A^{*},\rho\right]=\xi_{\mathcal{D}}[\rho](A^{*})$. Thus, $r\in\xi_{\mathcal{D}}[\rho](H_{w})$ for some $w\in A^{*}$ and the lemma is proved. ∎

By definition $\mathcal{D}$ is a strictly positive full level within the hierarchy of basis $\mathcal{C}$. Therefore, $\mathcal{D}=\mathit{Bool}(\mathcal{G})$ where $\mathcal{G}$ is the preceding half level in the hierarchy. In particular, $\mathcal{G}$ is a quotient-closed lattice closed under concatenation and marked concatenation by Lemma 5 and Theorem 6. Consequently, we obtain from Lemma 36 that $\xi_{\mathcal{D}}[\rho]$ is $\mathcal{G}$-multiplicative and the associated endomorphism of $(2^{R},\cup)$ is defined by $\mu_{\xi_{\mathcal{D}}[\rho]}(T)=\mathord{\downarrow_{R}}T$ for all $T\in 2^{R}$. Moreover, we have $L\in\mathcal{C}\subseteq\mathcal{G}$ by definition. Hence, by Axioms 1 and 3 in the definition of $\mathcal{G}$-multiplicative rating maps, the hypothesis that $r\in\xi_{\mathcal{D}}[\rho](La_{1}L\cdots a_{\ell}L)$ given by Lemma 40 implies that,

[TABLE]

By definition of $L$, we have $\xi_{\mathcal{D}}[\rho](L)=\mathbbm{i}_{\mathcal{C}}[\xi_{\mathcal{D}}[\rho]]$. Moreover, we have the following fact.

Fact 41**.**

For every $a\in A$, we have $\xi_{\mathcal{D}}[\rho](a)\subseteq\mathord{\downarrow_{R}}\{\rho(a)\}$.

Proof.

Fact 7 yields that $\{a\}\in\mathcal{D}$ ($\mathcal{D}$ is a level $n\geq 1$ in the hierarchy of basis $\mathcal{C}$). Hence, $\{\{a\}\}$ is a $\mathcal{D}$-cover of $\{a\}$. Thus, $\xi_{\mathcal{D}}[\rho](a)=\mathcal{I}_{\mathcal{D}}\left[\{a\},\rho\right]\subseteq\mathcal{I}[\rho](\{a\})=\mathord{\downarrow_{R}}\{\rho(a)\}$, which finishes the proof. ∎

Altogether, we obtain:

[TABLE]

By definition of $S$, this implies $r\in S$, concluding the proof.

7. Boolean polynomial closure

We turn to the second part of Theorem 8: for a quotient-closed Boolean algebra of group languages $\mathcal{C}$, if $\mathcal{C}$-separation is decidable, then so is $\mathit{BPol}(\mathcal{C})$-covering. We fix $\mathcal{C}$ for the section.

We rely on rating maps and use the notions introduced in Section 6. For any nice multiplicative rating map $\rho:2^{A^{*}}\to R$, we characterize $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ as the greatest subset of $R$ satisfying specific properties. When $\mathcal{C}$-separation is decidable, this yields a fixpoint algorithm for computing $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ from $\rho$. Consequently, we get the decidability of $\mathit{BPol}(\mathcal{C})$-covering from Proposition 37.

Remark 42**.**

The conditions for applying our characterizations are more restrictive than those we had for $\mathit{Pol}(\mathcal{C})$ in the previous section. We require $\rho$ to be nice: while we are able to handle $\mathit{Pol}(\mathcal{C})$ for arbitrary multiplicative rating maps, we are restricted to nice ones for $\mathit{BPol}(\mathcal{C})$. This is irrelevant for the decidability of covering: considering nice multiplicative rating maps suffices. However, this is a key point: the proof of the $\mathit{BPol}(\mathcal{C})$-characterization involves handling the simpler class $\mathit{Pol}(\mathcal{C})$ for auxiliary $\mathit{Pol}(\mathcal{C})$-multiplicative rating maps which are not nice. This is why we are not able to get results for higher levels in concatenation hierarchies: our knowledge about level 1/2 is stronger than the decidability of covering, and we are unable to replicate it for levels 1 and 3/2 (the situation is the same for $\mathit{PBPol}(\mathcal{C})$).

We first present the characterization of $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$. For every multiplicative rating map $\rho:2^{A^{*}}\to R$, we shall define the $\mathit{BPol}(\mathcal{C})$-complete subsets of $R$ for $\rho$. Our theorem states that when $\rho$ is nice, $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ is the greatest such subset.

We fix the multiplicative rating map $\rho:2^{A^{*}}\to R$ for the definition. We successively lift the multiplication “$\cdot$” of $R$ to the sets $2^{R}$, $R\times 2^{R}$ and $2^{R\times 2^{R}}$ in the natural way. This makes $(2^{R\times 2^{R}},\cup,\cdot)$ an idempotent semiring. For every $S\subseteq R$, we use it as the rating set of a nice multiplicative rating map,

[TABLE]

Since we are defining a nice multiplicative rating map, it suffices to specify the evaluation of letters. For $a\in A$, we let,

[TABLE]

By definition, we have $\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]\subseteq R\times 2^{R}$ for every $S\subseteq R$.

We now define the $\mathit{BPol}(\mathcal{C})$-complete subsets. We say that $S\subseteq R$ is $\mathit{BPol}(\mathcal{C})$-complete for $\rho$ if for every $s\in S$, there exist $(r_{1},U_{1}),\dots,(r_{k},U_{k})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ such that,

[TABLE]

We are ready to state the main theorem of this section: when $\rho$ is nice, the greatest $\mathit{BPol}(\mathcal{C})$-complete subset of $R$ (with respect to inclusion) is exactly $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$.

Theorem 43**.**

Let $\rho:2^{A^{*}}\to R$ be a nice multiplicative rating map. Then, $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ is the greatest $\mathit{BPol}(\mathcal{C})$-complete subset of $R$ for $\rho$.

The proof of Theorem 43 is presented in Appendix 12. It is rather involved and exploits two results for rating maps that were designed in [28] to handle Boolean closure in a general context. We complete them with arguments which are specific to our setting. Here, we discuss the applications of Theorem 43.

Provided that $\mathcal{C}$-separation is decidable, Theorem 43 yields a greatest fixpoint procedure for computing $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ from an input nice multiplicative rating map $\rho:2^{A^{*}}\to R$. Indeed, consider the following sequence of subsets $S_{0}\supseteq S_{1}\supseteq S_{2}\cdots$. We let $S_{0}=R$ and for $i\geq 1$, $S_{i}$ contains all $s\in S_{i-1}$ such that there exists $(r_{1},U_{1}),\dots,(r_{k},U_{k})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S_{i-1}}]$ satisfying (1):

[TABLE]

Clearly, computing $S_{i}$ from $S_{i-1}$ boils down to computing $\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S_{i-1}}]$. By Corollary 25, this is possible since we have an algorithm for $\mathcal{C}$-separation ($\eta_{\rho,S_{i-1}}$ is a nice multiplicative rating map that we may compute from $S_{i-1}$).

Finally, since $R$ is finite, the sequence $S_{0}\supseteq S_{1}\supseteq S_{2}\cdots$ stabilizes at some point: there exists some $i\in\mathbb{N}$ such that $S_{i}=S_{j}$ for all $j\geq i$. One may verify that $S_{i}$ is the greatest $\mathit{BPol}(\mathcal{C})$-complete subset of $R$ and we may compute it. Theorem 43 then states that $S_{i}=\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$.

By Proposition 37, a procedure for computing $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ yields an algorithm for $\mathit{BPol}(\mathcal{C})$-covering. Hence, we get the part of Theorem 8 regarding $\mathit{BPol}(\mathcal{C})$ (i.e., level 1) as a corollary: when $\mathcal{C}$-separation is decidable, so is $\mathit{BPol}(\mathcal{C})$-covering (as well as $\mathit{BPol}(\mathcal{C})$-separation by Lemma 4).

8. Nested polynomial closure

This section is devoted to the final part of Theorem 8: for every quotient-closed Boolean algebra of group languages $\mathcal{C}$, if $\mathcal{C}$-separation is decidable, then so is $\mathit{PBPol}(\mathcal{C})$-covering. As usual, we fix $\mathcal{C}$ for the section.

We rely on the framework outlined in Sections 4 and 6. For every morphism $\alpha:A^{*}\to M$ and every nice multiplicative rating map $\rho:2^{A^{*}}\to R$, we characterize $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ as the least subset of $M\times R$ satisfying specific properties. When $\mathcal{C}$-separation is decidable, this yields a least fixpoint algorithm for computing $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ from $\rho$. Consequently, we get the decidability of $\mathit{PBPol}(\mathcal{C})$-covering by Proposition 38.

We start by presenting the characterization. Consider a morphism $\alpha:A^{*}\to M$ and a multiplicative rating map $\rho:2^{A^{*}}\to R$. We define a notion of $\mathit{PBPol}(\mathcal{C})$-complete subset of $M\times R$. Our main theorem then states that when $\rho$ is nice, the least such subset is exactly $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$. The definition depends on auxiliary nice multiplicative rating maps. We first present them.

We successively lift the multiplications of $M$ and $R$ to the sets $2^{M\times R}$, $R\times 2^{M\times R}$ and $2^{R\times 2^{M\times R}}$ in the natural way. This makes $(2^{R\times 2^{M\times R}},\cup,\cdot)$ an idempotent semiring. For every $S\subseteq M\times R$, we define a nice multiplicative rating map:

[TABLE]

Since we are defining a nice multiplicative rating map, it suffices to specify the evaluation of letters. For $a\in A$, we let,

[TABLE]

Observe that by definition, we have $\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]\subseteq R\times 2^{M\times R}$.

We may now define $\mathit{PBPol}(\mathcal{C})$-complete subsets. Consider $S\subseteq M\times R$. We say that $S$ is $\mathit{PBPol}(\mathcal{C})$-complete for $\alpha$ and $\rho$ when the following conditions are satisfied:

• •

Downset. We have $\mathord{\downarrow_{R}}S\subseteq S$.

• •

Multiplication. We have $S\cdot S\subseteq S$.

• •

$\mathcal{C}$-operation. For all $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$, we have $T\subseteq S$.

• •

$\mathit{PBPol}(\mathcal{C})$-operation. For all $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$ and every idempotent $(e,f)\in\mathord{\downarrow_{R}}T\subseteq M\times R$, we have:

[TABLE]

We may now state the main theorem of this section. When $\rho$ is nice, $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is the least $\mathit{PBPol}(\mathcal{C})$-complete subset of $M\times R$ (with respect to inclusion).

Theorem 44**.**

Fix a morphism $\alpha:A^{*}\to M$ and a nice multiplicative rating map $\rho:2^{A^{*}}\to R$. Then, $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is the least $\mathit{PBPol}(\mathcal{C})$-complete subset of $M\times R$.

The proof of Theorem 44 is presented in Appendix 13. Similarly to most results of this kind, it involves two directions: one needs to show that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is $\mathit{PBPol}(\mathcal{C})$-complete and then that it is included in every other $\mathit{PBPol}(\mathcal{C})$-complete subset. The former direction is proved directly. The latter one requires applying the main theorem of [21] (for classes $\mathit{PBPol}(\mathcal{D})$ with $\mathcal{D}$ a finite quotient-closed Boolean algebra).

As expected, when $\mathcal{C}$-separation is decidable, Theorem 44 yields a least fixpoint procedure for computing $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ from a morphism $\alpha:A^{*}\to M$ and a nice multiplicative rating map $\rho:2^{A^{*}}\to R$. One starts from the empty set $\emptyset\subseteq R$ and saturates it with the four operations in the definition of $\mathit{PBPol}(\mathcal{C})$-complete subsets. Clearly, they may be implemented. This is immediate for downset and multiplication. Moreover, we are able to implement $\mathcal{C}$-operation and $\mathit{PBPol}(\mathcal{C})$-operation by Corollary 25 since $\mathcal{C}$-separation is decidable. Eventually, the computation reaches a fixpoint and it is straightforward to verify that this set is the least $\mathit{PBPol}(\mathcal{C})$-complete subset of $M\times R$, i.e., $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ by Theorem 44.

By Proposition 38, we know that a procedure for computing $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ yields an algorithm for $\mathit{PBPol}(\mathcal{C})$-covering. Hence, we get the $\mathit{PBPol}(\mathcal{C})$ part of Theorem 8 as a corollary: when $\mathcal{C}$-separation is decidable, so is $\mathit{PBPol}(\mathcal{C})$-covering (as well as $\mathit{PBPol}(\mathcal{C})$-separation by Lemma 4).

9. Conclusion

We proved that for any quotient-closed Boolean algebra of group languages $\mathcal{C}$, if $\mathcal{C}$-separation is decidable, then so are separation and covering for levels 1/2, 1 and 3/2 in the concatenation hierarchy of basis $\mathcal{C}$. A corollary is that these levels enjoy decidable membership, as well as level 5/2. This result nicely complements analogous statements that apply to finitely based concatenation hierarchies [28, 21].

These results can be instantiated for several classical bases. First, one may consider the basis made of all group languages. This yields the so-called “group hierarchy” of Pin and Margolis [13], for which decidability of membership was known only up to level 1. Another application is the quantifier alternation hierarchy of first-order logic with modular predicates. It corresponds to the basis consisting of languages counting the length of words modulo some number (separation was shown decidable at level 1 in [34] with specific techniques). A third example is the basis consisting of all languages counting the number of occurrences of letters modulo some number, which are exactly languages recognized by finite commutative groups. In all cases, we get decidability of separation and covering for levels 1/2, 1 and 3/2 (and membership for level 5/2).

There are natural follow-up questions to this work. The most immediate one is whether our results may be pushed to higher levels. This is difficult. There are no known generic results for the levels above 2, even for finitely based hierarchies. Actually, there is one hierarchy for which covering is know to be decidable up to level 5/2: the Straubing-Thérien hierarchy (its basis is $\{\emptyset,A^{*}\}$). Unfortunately, this is based on a specific property of this hierarchy: its levels 2 and 5/2 are also levels 1 and 3/2 in another finitely based hierarchy [19]. We do not have a similar property for arbitrary group based hierarchies.

One can also investigate other closure operations, such as the star-free closure $\mathcal{C}\mapsto\textup{SF}(\mathcal{C})$, which is the union of all levels in the hierarchy of basis $\mathcal{C}$ (i.e., $\textup{SF}(\mathcal{C})$ is the least class containing $\mathcal{C}$, closed under Boolean operations and marked concatenation). If $\mathcal{C}$ is a quotient-closed Boolean algebra of group languages and $\mathcal{C}$-separation is decidable, is $\textup{SF}(\mathcal{C})$-covering decidable?

10. Omitted proofs in Section 5

In this appendix, we present the arguments for the statements of Section 6 whose proofs are missing.

10.1. Proof of Lemma 23

Let us first recall the statement of Lemma 23.

Lemma 0**.**

Let $\rho:2^{A^{*}}\to R$ be a rating map and $\mathcal{C}$ a lattice. There exists a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$.

Proof.

Let $U\subseteq R$ be the set of all elements $r\in R$ such that $r=\rho(K)$ for $K\in\mathcal{C}$ such that $\varepsilon\in K$. Clearly, $\rho(A^{*})\in U$ which means that $U$ is non-empty since $A^{*}\in\mathcal{C}$ ($\mathcal{C}$ is a lattice). For all $r\in U$, we fix an arbitrary language $K_{r}\in\mathcal{C}$ such that $\varepsilon\in K_{r}$ and $r=\rho(K_{r})$ ($K_{r}$ exists by definition of $U$). Finally, we let,

[TABLE]

Since $\mathcal{C}$ is a lattice, we have $K\in\mathcal{C}$. Moreover, $\varepsilon\in K$ by definition. Since $K\subseteq K_{r}$ for all $r\in U$, it follows that $\rho(K)\leq r$ for every $r\in U$. By definition of $U$, this implies that $\rho(K)\leq\rho(K^{\prime})$ for every $K^{\prime}\in\mathcal{C}$ such that $\varepsilon\in K^{\prime}$. Hence, $K$ is a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$. ∎

10.2. Proof of Lemma 24

Let us first recall the statement of Lemma 24.

Lemma 0**.**

Let $\rho:2^{A^{*}}\to R$ be a nice rating map and $\mathcal{C}$ a lattice. Then, $\mathbbm{i}_{\mathcal{C}}[\rho]$ is the sum of all $r\in R$ such that $\{\varepsilon\}$ is not $\mathcal{C}$-separable from $\rho_{*}^{-1}(r)$.

Proof.

We let $Q\subseteq R$ as the set of all $r\in R$ such that $\{\varepsilon\}$ is not $\mathcal{C}$-separable from $\rho_{*}^{-1}(r)$. Moreover, we let $q=\sum_{r\in Q}r$. We prove that $q=\mathbbm{i}_{\mathcal{C}}[\rho]$. First, we show that $q\leq\mathbbm{i}_{\mathcal{C}}[\rho]$. This amounts to proving that given $r\in Q$, we have $r\leq\mathbbm{i}_{\mathcal{C}}[\rho]$. By definition, $\mathbbm{i}_{\mathcal{C}}[\rho]=\rho(K)$ for some $K\in\mathcal{C}$ such that $\varepsilon\in K$. Therefore, since $\{\varepsilon\}$ is not $\mathcal{C}$-separable from $\rho_{*}^{-1}(r)$ (this is the definition of $Q$), we obtain that $K\cap\rho_{*}^{-1}(r)\neq\emptyset$. We get a word $w\in K\cap\rho_{*}^{-1}(r)$. By definition $w\in K$ and $\rho(w)=r$. This implies that $r\leq\rho(K)=\mathbbm{i}_{\mathcal{C}}[\rho]$.

Conversely, we show that $\mathbbm{i}_{\mathcal{C}}[\rho]\leq q$. By definition, for every $r\in R\setminus Q$, $\{\varepsilon\}$ is $\mathcal{C}$-separable from $\rho_{*}^{-1}(r)$. Hence, we may fix $H_{r}\in\mathcal{C}$ as a separator: $\varepsilon\in H_{r}$ and $H_{r}\cap\rho_{*}^{-1}(r)=\emptyset$. We define,

[TABLE]

Clearly, $\varepsilon\in H$ which implies that $\mathbbm{i}_{\mathcal{C}}[\rho]\leq\rho(H)$. Moreover, since $\rho$ is nice, we have $w_{1},\dots,w_{n}\in H$ such that $\rho(H)=\rho(w_{1})+\cdots+\rho(w_{n})$. Finally, it is clear that $H\cap\rho_{*}^{-1}(r)=\emptyset$ for every $r\in R\setminus Q$. Therefore, that $\rho(w_{1}),\dots,\rho(w_{n})\in Q$. This yields $\rho(H)=\rho(w_{1})+\cdots+\rho(w_{n})\leq\sum_{r\in Q}r=q$. Altogether, we get that $\mathbbm{i}_{\mathcal{C}}[\rho]\leq q$. ∎

10.3. Proof of Lemma 26

Let us first recall the statement of Lemma 26.

Lemma 0**.**

Let $\rho:2^{A^{*}}\to R$ be a nice multiplicative rating map. Then, $\mathbbm{i}_{\textup{{MOD}}}[\rho]=(\rho(A))^{\omega}+\rho(\varepsilon)$.

Proof.

We first prove that $(\rho(A))^{\omega}+\rho(\varepsilon)\leq\mathbbm{i}_{\textup{{MOD}}}[\rho]$. Let $K$ be MOD-optimal $\varepsilon$-approximation for $\rho$: we have $K\in\textup{{MOD}}$, $\varepsilon\in K$ and $\mathbbm{i}_{\textup{{MOD}}}[\rho]=\rho(K)$. Therefore, we have to show that $(\rho(A))^{\omega}+\rho(\varepsilon)\leq\rho(K)$. By definition of MOD, $K$ is a Boolean combination of languages $\{w\in A^{*}\mid|w|=k\mod m\}$ with $k,m\in\mathbb{N}$ such that $k<m$. We let $d$ as the least common multiplier of the numbers $m$ involved in the definitions of all languages in this Boolean combination. It is simple to verify that given two words $w,w^{\prime}\in A^{*}$, if the lengths of $w$ and $w^{\prime}$ are congruent modulo $d$, then $w\in K$ if and only if $w^{\prime}\in K$. Hence, since $\varepsilon\in K$. We know that every $w\in A^{*}$ whose length is congruent to [math] modulo $d$ belongs to $K$. In particular, we have,

[TABLE]

This implies as desired that $(\rho(A))^{\omega}+\rho(\varepsilon)\leq\rho(K)$ since $\rho$ is a multiplicative rating map.

We now prove that $\mathbbm{i}_{\textup{{MOD}}}[\rho]\leq(\rho(A))^{\omega}+\rho(\varepsilon)$. By definition, this amounts to finding $K\in\textup{{MOD}}$ such that $\varepsilon\in K$ and $\rho(K)\leq(\rho(A))^{\omega}+\rho(\varepsilon)$. Recall that $\omega\in\mathbb{N}$ is the idempotent power of $(R,\cdot)$. Hence, we may define $K$ as the language of all words $w\in A^{*}$ whose lengths are congruent to [math] modulo $\omega$. Clearly, $K\in\textup{{MOD}}$ and $\varepsilon\in K$. Since $\rho$ is nice, we get $w_{1},\dots,w_{n}\in K$ such that $\rho(K)=\rho(w_{1})+\cdots+\rho(w_{n})$. Let $i\leq n$, by definition of $K$, one of the two following properties holds:

• •

$w_{i}=\varepsilon$ which implies that $\rho(w_{i})=\rho(\varepsilon)$, or,

• •

$w_{i}\in A^{+}$ and $|w_{i}|=k\omega$ for some $k\geq 1$ which implies that $\rho(w_{i})\leq\rho(A^{k\omega})=(\rho(A))^{\omega}$.

Altogether, this yields that $\rho(K)\leq(\rho(A))^{\omega}+\rho(\varepsilon)$, finishing the proof of this direction. ∎

11. Omitted proofs in Section 6

In this appendix, we present an additional property of quasi-multiplicative rating maps taken from [28] that we shall need later. Moreover we present a proof for Proposition 38.

11.1. A property of quasi-multiplicative rating maps

We shall need the following property of quasi-multiplicative rating maps which is proved in [28, Lemma 6.6].

Lemma 45**.**

Let $\mathcal{G}$ be a quotient-closed lattice and $\rho:2^{A^{*}}\!\!\to R$ be a $\mathcal{G}$-multiplicative rating map. Given $H,L\subseteq A^{*}$, $q\in\mathcal{I}_{\mathcal{G}}\left[H,\rho\right]$ and $r\in\mathcal{I}_{\mathcal{G}}\left[L,\rho\right]$, we have $\mu_{\rho}(qr)\in\mathcal{I}_{\mathcal{G}}\left[HL,\rho\right]$.

11.2. Proof of Proposition 38

Let us first recall the statement of Proposition 38. It is as follows.

Proposition 0**.**

Let $\mathcal{C}$ be a quotient-closed Boolean algebra of group languages and $\mathcal{D}$ a half level within the concatenation hierarchy of basis $\mathcal{C}$. Assume that there exists an algorithm which computes $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]$ from a morphism $\alpha$ and a nice multiplicative rating map $\rho$. Then, $\mathcal{D}$-covering is decidable.

We fix $\mathcal{C}$ and $\mathcal{D}$ as in the statement of the proposition. We show that given a morphism $\alpha$ and a nice multiplicative rating map $\rho$, one may compute $\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$ from its subset $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]$. The result is then an immediate corollary of Proposition 19.

We fix a morphism $\alpha:A^{*}\to M$ and a nice multiplicative rating map $\rho:2^{A^{*}}\to R$ for the proof. We let $S\subseteq M\times R$ as the least subset of $R$ which satisfies the following properties:

1. (1)

We have $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]\subseteq S$. 2. (2)

We have $(\alpha(w),\rho(w))\in S$ for every $w\in A^{*}$. 3. (3)

We have $\mathord{\downarrow_{R}}S=S$. 4. (4)

For every $(s,r),(s^{\prime},r^{\prime})\in S$, we have $(ss^{\prime},rr^{\prime})\in S$.

Clearly, one may compute $S$ from $\alpha$, $\rho$ and $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]$. Hence, Proposition 38 is now immediate from the following lemma.

Lemma 46**.**

We have $S=\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$.

We now concentrate on proving Lemma 46. For the proof, we let $L$ as a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\zeta_{\mathcal{D}}^{\alpha}[\rho]$: $L\in\mathcal{C}$, $\varepsilon\in L$ and $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]=\zeta_{\mathcal{D}}^{\alpha}[\rho](L)$.

We first prove that $S\subseteq\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$. This amounts to proving that $\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$ satisfies the four properties in the definition of $S$. That $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]\subseteq\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$ is immediate since $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]=\zeta_{\mathcal{D}}^{\alpha}[\rho](L)$, $\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]=\zeta_{\mathcal{D}}^{\alpha}[\rho](A^{*})$, $L\subseteq A^{*}$ and $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ is a rating map. The other three properties are actually generic: they are satisfied whenever $\rho$ is a multiplicative rating map and $\mathcal{D}$ is a quotient-closed lattice (which is the case here since $\mathcal{D}$ is a half level within the hierarchy of basis $\mathcal{C}$). We refer the reader to [26, Lemma 9.11] for the proof.

We turn to the converse inclusion: $\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]\subseteq S$. Consider $(s,r)\in\mathcal{P}_{\mathcal{D}}^{\alpha}[\rho]$, we show that $(s,r)\in S$. The argument is based on the following lemma which is where we use the hypothesis that $\mathcal{C}$ is a class of group languages (which implies that $L\in\mathcal{C}$ is recognized by a finite group).

Lemma 47**.**

There exist $\ell\in\mathbb{N}$ and $\ell$ letters $a_{1},\dots,a_{\ell}\in A$ such that $(s,r)\in\zeta_{\mathcal{D}}^{\alpha}[\rho](La_{1}L\cdots a_{\ell}L)$.

Proof.

Identical to the proof of Lemma 40 using the hypothesis that $L$ is a group language and $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ is a rating map. ∎

By definition $\mathcal{D}$ is a half level within the hierarchy of basis $\mathcal{C}$. This means that $\mathcal{D}$ is the polynomial closure of a quotient-closed Boolean algebra and by Theorem 6, we obtain that $\mathcal{D}$ is a quotient-closed lattice closed under concatenation and marked concatenation. Consequently, we obtain from Lemma 35 that $\zeta_{\mathcal{D}}^{\alpha}[\rho]$ is quasi-multiplicative and the associated endomorphism of $(2^{M\times R},\cup)$ is defined by $\mu_{\zeta_{\mathcal{D}}^{\alpha}[\rho]}(T)=\mathord{\downarrow_{R}}T$ for all $T\in 2^{M\times R}$. Hence, by Axioms 1 and 3 in the definition of quasi-multiplicative rating maps, the hypothesis that $(s,r)\in\zeta_{\mathcal{D}}^{\alpha}[\rho](La_{1}L\cdots a_{\ell}L)$ given by Lemma 47 implies that,

[TABLE]

By definition of $L$, we have $\zeta_{\mathcal{D}}^{\alpha}[\rho](L)=\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]$. Moreover, we know that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]\subseteq S$ by definition of $S$. Hence, since $S$ is also closed under multiplication and downset, that that $(s,r)\in S$ is now immediate from the following lemma.

Lemma 48**.**

For every $a\in A$, we have $\zeta_{\mathcal{D}}^{\alpha}[\rho](a)\subseteq S$.

Proof.

Consider an element $(t,q)\in\zeta_{\mathcal{D}}^{\alpha}[\rho](a)$. We show that $(t,q)\in S$. By definition of $\zeta_{\mathcal{D}}^{\alpha}[\rho]$, we have $q\in\mathcal{I}_{\mathcal{D}}\left[\alpha^{-1}(t)\cap\{a\},\rho\right]$. This implies that $\alpha^{-1}(t)\cap\{a\}\neq\emptyset$ which means that $t=\alpha(a)$. Hence, we get,

[TABLE]

Recall that $\mathcal{D}$ is a half level in the hierarchy of basis $\mathcal{C}$. We now consider two cases depending on whether this level is 1/2 (i.e. $\mathcal{D}=\mathit{Pol}(\mathcal{C})$) or a strictly higher one. We start with the latter case which is simpler.

If $\mathcal{D}$ is at least the level 3/2 in the hierarchy of basis $\mathcal{C}$. It is immediate from Fact 7 that $\{a\}\in\mathcal{D}$. Hence, $\{\{a\}\}$ is a $\mathcal{D}$-cover of $\{a\}$. Therefore, $\mathcal{I}_{\mathcal{D}}\left[\{a\},\rho\right]=\mathcal{I}[\rho](\{a\})=\mathord{\downarrow_{R}}\{\rho(a)\}$. It follows that $q\leq\rho(a)$. By hypothesis, we know that $(\alpha(a),\rho(a))\in S$ and $\mathord{\downarrow_{R}}S=S$. Hence, since $t=\alpha(a)$, we conclude that $(t,q)\in S$ as desired.

We now assume that is the level 1/2: $\mathcal{D}=\mathit{Pol}(\mathcal{C})$. Let $H$ be a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\rho$: $H\in\mathcal{C}$, $\varepsilon\in H$ and $\mathbbm{i}_{\mathcal{C}}[\rho]=\rho(H)$. Since $\mathcal{D}$ is a half level within the hierarchy of basis $\mathcal{C}$, it contains $\mathcal{C}$ and is closed under marked concatenation. Thus, we know that $HaH\in\mathcal{D}$. Moreover, we have $a\in HaH$ since $\varepsilon\in H$. Hence, $\{HaH\}$ is a $\mathcal{D}$-cover of $\{a\}$ which means that $\mathcal{I}_{\mathcal{D}}\left[\{a\},\rho\right]\subseteq\mathcal{I}[\rho](\{HaH\})$. Since $q\in\mathcal{I}_{\mathcal{D}}\left[\{a\},\rho\right]$, this yields $q\in\mathcal{I}[\rho](\{HaH\})$ which means that,

[TABLE]

Since $t=\alpha(a)$ and $S$ is closed under downset by definition, it remains to show that $(\alpha(a),\mathbbm{i}_{\mathcal{C}}[\rho]\cdot\rho(a)\cdot\mathbbm{i}_{\mathcal{C}}[\rho])\in S$. This will imply as desired that $(t,q)\in S$.

Since $\mathcal{D}$ = $\mathit{Pol}(\mathcal{C})$, Lemma 27 yields that $\mathbbm{i}_{\mathcal{C}}[\rho]\in\mathcal{I}_{\mathcal{D}}\left[\{\varepsilon\},\rho\right]$. Moreover $\varepsilon\in\alpha^{-1}(1_{M})\cap L$ by definition of $L$. Hence, Fact 17 yields that,

[TABLE]

This exactly says that $(1_{M},\mathbbm{i}_{\mathcal{C}}[\rho])\in\zeta_{\mathcal{D}}^{\alpha}[\rho](L)$ which is equal to $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]$ by definition of $L$. Moreover, we know that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathcal{D}}^{\alpha}[\rho]]\subseteq S$ by definition of $S$. Hence, we obtain that, $(1_{M},\mathbbm{i}_{\mathcal{C}}[\rho])\in S$. Finally, we also know that $(\alpha(a),\rho(a))\in S$ and that $S$ is closed under multiplication. Altogether, this yields,

[TABLE]

This concludes the proof of Lemma 48. ∎

12. Omitted proofs in Section 7

This appendix is devoted to the proof of Theorem 43. Let us first recall the statement. Recall that a quotient-closed Boolean algebra of group languages $\mathcal{C}$ is fixed.

Theorem 0**.**

Let $\rho:2^{A^{*}}\to R$ be a nice multiplicative rating map. Then, $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ is the greatest $\mathit{BPol}(\mathcal{C})$-complete subset of $R$.

We fix a nice multiplicative rating map $\rho:2^{A^{*}}\to R$ for the proof. As expected, the argument involves two independent directions.

• •

First, we show that for every $\mathit{BPol}(\mathcal{C})$-complete subset $S\subseteq R$, the inclusion $S\subseteq\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ holds. This corresponds to soundness in the greatest fixpoint procedure computing $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ (it only computes elements in $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$).

• •

Then, we show that $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ itself is $\mathit{BPol}(\mathcal{C})$-complete. This corresponds to completeness in the greatest fixpoint procedure computing $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ (it computes all elements in $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$).

Both directions reuse material which was originally introduced in [28]. We start with soundness.

12.1. Soundness

We have to show that for every $\mathit{BPol}(\mathcal{C})$-complete subset $S\subseteq R$, the inclusion $S\subseteq\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ holds. We first introduce an infinite sequence of auxiliary rating maps (defined from $\rho$). We then use them to formulate two independent statements. When put together they imply the desired result.

Using induction, we define a rating map $\tau_{n}:2^{A^{*}}\to Q_{n}$ for every $n\in\mathbb{N}$. When $n=0$, the rating set $Q_{0}$ is $(2^{R},\cup)$ and $\tau_{0}$ is defined as follows,

[TABLE]

It is immediate by definition that $\tau_{0}$ is indeed a rating map (i.e. a monoid morphism).

Assume now that $n\geq 1$ and that $\tau_{n-1}:2^{A^{*}}\to Q_{n-1}$ is defined. Recall that $\rho_{*}:A^{*}\to R$ denotes the canonical monoid morphism associated to the multiplicative rating map $\rho$. We define $\tau_{n}$ as the rating map $\zeta_{\mathit{Pol}(\mathcal{C})}^{\rho_{*}}[\tau_{n-1}]$ (see Section 6 for the definition). In particular, this means that for all $n\geq 1$, the rating set $Q_{n}$ of $\tau_{n}$ is $(2^{R\times Q_{n-1}},\cup)$.

We complete this definition with maps $f_{n}:Q_{n}\to 2^{R}$ for $n\in\mathbb{N}$. In this case as well, we use induction on $n$.

• •

For $n=0$, let $T\in Q_{0}=2^{R}$. Given $s\in R$, we have $s\in f_{0}(T)$ if there exists $r_{1},\dots,r_{k}\in T$ such that $s\leq r_{1}+\cdots+r_{k}$.

• •

For $n\geq 1$, let $T\in Q_{n}=2^{R\times Q_{n-1}}$. Given $s\in R$, we have $s\in f_{n}(T)$ if there exists $(r_{1},T_{1}),\dots,(r_{k},T_{k})\in T$ such that $s\leq r_{1}+\cdots+r_{k}$ and $r_{1}+\cdots+r_{k}\in f_{n-1}(T_{i})$ for every $i\leq k$.

We have the following simple fact which is immediate from the definition.

Fact 49**.**

For every $n\in\mathbb{N}$ and $T,T^{\prime}\in Q_{n}$ such that $U\subseteq U^{\prime}$, we have $f_{n}(T)\subseteq f_{n}(T^{\prime})$.

This concludes our preliminary definitions. We now come back to the soundness proof. Using the objects that we just defined, we present two propositions. The first one is taken from [28] while the second is new and specific to our setting.

Proposition 50**.**

Consider a language $L\in\mathit{Pol}(\mathcal{C})$. Then, the following inclusion holds,

[TABLE]

Proposition 51**.**

Let $S\subseteq R$ which is $\mathit{BPol}(\mathcal{C})$-complete for $\rho$. Then,

[TABLE]

When combined, the two propositions imply the desired result: for every $\mathit{BPol}(\mathcal{C})$-complete subset $S\subseteq R$, the inclusion $S\subseteq\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ holds. Indeed, consider a $\mathit{BPol}(\mathcal{C})$-complete subset $S\subseteq R$. We let $L\subseteq A^{*}$ as a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\xi_{\mathit{BPol}(\mathcal{C})}[\rho]$. That is, we have $L\in\mathcal{C}$, $\varepsilon\in L$ and $\xi_{\mathit{BPol}(\mathcal{C})}[\rho](L)=\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$. Consequently, it now suffices to show that $S\subseteq\xi_{\mathit{BPol}(\mathcal{C})}[\rho](L)$. Since $L\in\mathcal{C}$ and $\varepsilon\in L$, it is immediate that $\mathbbm{i}_{\mathcal{C}}[\tau_{n}]\subseteq\tau_{n}(L)$ for every $n\in\mathbb{N}$. Thus, by Fact 49, $f_{n}(\mathbbm{i}_{\mathcal{C}}[\tau_{n}])\subseteq f_{n}(\tau_{n}(L))$ for every $n\in\mathbb{N}$. Consequently, Proposition 50 yields that,

[TABLE]

Finally, since $S$ is $\mathit{BPol}(\mathcal{C})$-complete by hypothesis, we may combine this inclusion with the one given by Proposition 51 to obtain that $S\subseteq\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ as desired. This concludes the soundness proof. It now remains to prove the two propositions.

Proposition 50 is proved in [28, Proposition 8.2]. This is actually a generic result which holds for any class built using Boolean closure (Proposition 50 remains true when replacing $\mathit{Pol}(\mathcal{C})$ by an arbitrary lattice $\mathcal{G}$ in the definition of the rating maps $\tau_{n}$ and in the statement). Here we focus on proving Proposition 51 which is specific to our setting: we require the hypothesis that we are considering the class $\mathit{BPol}(\mathcal{C})$ for $\mathcal{C}$ a quotient-closed Boolean algebra of group languages. The remainder of this subsection is devoted to this proof.

Preliminaries. We start by presenting a few additional properties of the rating maps $\tau_{n}$ (again they are taken from [28]). We know that these rating maps are quasi-multiplicative.

First observe that since $R$ is a semiring, it is immediate that we may lift its multiplication inductively to all rating sets $Q_{n}$ in the natural way. This yields semirings $(Q_{n},\cup,\cdot)$ for all $n\in\mathbb{N}$. We shall rely heavily on these multiplications. However, it is important to keep in mind that except for $\tau_{0}$, the rating maps $\tau_{n}$ are not multiplicative. Instead, there are only quasi-multiplicative. We state this property in the following lemma. The proof is a simple induction on $n\in\mathbb{N}$ and is based on a generalization of Lemma 35 (see [28, Lemma 8.8])

Lemma 52**.**

The rating map $\tau_{0}:2^{A^{*}}\to Q_{0}$ is multiplicative. Moreover, for every $n\geq 1$, the rating map $\tau_{n}:2^{A^{*}}\to Q_{n}$ is quasi-multiplicative and the associated endomorphism $\mu_{\tau_{n}}$ of $(Q_{n},\cup)$ is as follows. For every $U\in Q_{n}=2^{T\times Q_{n-1}}$,

[TABLE]

We complete this result with two lemmas which connect the hypothesis that the rating maps $\tau_{n}$ are quasi-multiplicative with the maps $f_{n}:Q_{n}\to 2^{R}$ (the proofs are straightforward and available in [28, Lemmas 8.9 and 8.10]).

Lemma 53**.**

For every $n\in\mathbb{N}$ and $T\in Q_{n}$, we have $f_{n}(T)\subseteq f_{n}(\mu_{\tau_{n}}(T))$.

Lemma 54**.**

For every $n\in\mathbb{N}$ and $U,U^{\prime}\in Q_{n}$, we have $f_{n}(U)\cdot f_{n}(U^{\prime})\subseteq f_{n}(U\cdot U^{\prime})$.

Proof of Proposition 51. We fix a $\mathit{BPol}(\mathcal{C})$-complete subset $S\subseteq R$ for the proof. We have to show that for every $n\in\mathbb{N}$, we have $S\subseteq f_{n}(\mathbbm{i}_{\mathcal{C}}[\tau_{n}])$. We fix $n\in\mathbb{N}$ and $s\in S$: we show that $s\in f_{n}(\mathbbm{i}_{\mathcal{C}}[\tau_{n}])$. The proof is an induction on $n\in\mathbb{N}$. We start with some definitions which are common to the base case ($n=0$) and the inductive step ($n\geq 1$).

Since $S$ is $\mathit{BPol}(\mathcal{C})$-complete by hypothesis and $s\in S$, we have $(r_{1},U_{1}),\dots,(r_{k},U_{k})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ which satisfy (1): $s\leq r_{1}+\cdots+r_{k}$ and $r_{1}+\cdots+r_{k}\in\mathord{\downarrow_{R}}U_{i}$ for every $i\leq k$. We now consider two cases depending on whether $n=0$ or $n\geq 1$.

Base case: $n=0$. Let $L$ be a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\tau_{0}$: we have $L\in\mathcal{C}$, $\varepsilon\in L$ and $\mathbbm{i}_{\mathcal{C}}[\tau_{0}]=\tau_{0}(L)$. We now have to prove that $s\in f_{0}(\tau_{0}(L))$. We have $(r_{i},U_{i})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ for every $i\leq k$. Since $L\in\mathcal{C}$ satisfies $\varepsilon\in L$, we know that $\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]\subseteq\eta_{\rho,S}(L)$ by definition. Consequently, we have $(r_{i},U_{i})\in\eta_{\rho,S}(L)$ for every $i\leq k$.

By definition of $\eta_{\rho,S}$, it follows that we have $w_{1},\dots,w_{k}\in L$ such that $r_{i}=\rho(w_{i})$ for every $i\leq k$. It follows that $r_{1},\dots,r_{k}\in\tau_{0}(L)$ by definition of $\tau_{0}$. Hence, since $s\leq r_{1}+\cdots+r_{k}$, we obtain that $s\in f_{0}(\tau_{0}(L))$ by definition of $f_{0}$ which concludes the proof in this case.

Inductive step: $n\geq 1$. The argument is based on the following lemma which is proved using induction on $n$.

Lemma 55**.**

For every $(r,U)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$, there exists $T\in Q_{n-1}$ such that $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\tau_{n}]$ and $\mathord{\downarrow_{R}}U\subseteq f_{n-1}(T)$.

We first use Lemma 55 to show that $s\in f_{n}(\mathbbm{i}_{\mathcal{C}}[\tau_{n}])$ and complete the main proof. For every $i\leq k$, we know that $(r_{i},U_{i})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ and $r_{1}+\cdots+r_{k}\in\mathord{\downarrow_{R}}U_{i}$. Therefore, Lemma 55 yields that for every $i\leq k$, we have $T_{i}\in Q_{n-1}$ such that $(r_{i},T_{i})\in\mathbbm{i}_{\mathcal{C}}[\tau_{n}]$ and $r_{1}+\cdots+r_{k}\in f_{n-1}(T_{i})$. Since we have $s\leq r_{1}+\cdots+r_{k}$ this yields that $s\in f_{n}(\mathbbm{i}_{\mathcal{C}}[\tau_{n}])$ by definition of $f_{n}$. This concludes the main proof. It remains to prove Lemma 55.

Proof of Lemma 55.

Consider $(r,U)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$. We have to exhibit the appropriate $T\in Q_{n-1}$ such that,

[TABLE]

For the proof, we let $L$ as a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\tau_{n}$: we have $L\in\mathcal{C}$, $\varepsilon\in L$ and $\mathbbm{i}_{\mathcal{C}}[\tau_{n}]=\tau_{n}(L)$.

Since $L\in\mathcal{C}$ and $\varepsilon\in L$, we know that $\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]\subseteq\eta_{\rho,S}(L)$. Therefore, $(r,U)\in\eta_{\rho,S}(L)$. By definition of $\eta_{\rho,S}$, this yields $w\in L$ such that $\eta_{\rho,S}(w)=\{(r,U)\}$. We now consider two cases depending on whether $w=\varepsilon$ or $w\in A^{+}$.

Case 1: Assume first that $w=\varepsilon$. It follows that $(r,U)=(1_{R},\{1_{R}\})$ (since $\{(1_{R},\{1_{R}\})\}$ is the neutral element of $2^{R\times 2^{R}}$). We prove that (2) holds for $T=\tau_{n-1}(\varepsilon)\in Q_{n-1}$.

We first show that $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\tau_{n}]$. Since $\varepsilon\in L$ by definition, we have $\tau_{n}(\varepsilon)\subseteq\tau_{n}(L)=\mathbbm{i}_{\mathcal{C}}[\tau_{n}]$. Moreover,

[TABLE]

One may verify that this yields $(r,T)=(1_{R},\tau_{n-1}(\varepsilon))\in\tau_{n}(\varepsilon)$. Altogether, we get $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\tau_{n}]$ as desired.

It remains to show that $\mathord{\downarrow_{R}}U\subseteq f_{n-1}(T)$. One may verify that $1_{R}=\rho(\varepsilon)\in f_{n-1}(\tau_{n-1}(\varepsilon))$ (this follows from a simple induction on $n$). Thus, since $U=\{1_{R}\}$ and $T=\tau_{n-1}(\varepsilon)$, we get $U\subseteq f_{n-1}(T)$. This implies $\mathord{\downarrow_{R}}U\subseteq f_{n-1}(T)$ since $\mathord{\downarrow_{R}}f_{n-1}(T)=f_{n-1}(T)$ by definition of $f_{n-1}$.

Case 2: Assume now that $w\in A^{+}$. In that case, we have $\ell\geq 1$ and $\ell$ letters $a_{1},\dots,a_{\ell}$ such that $w=a_{1}\cdots a_{\ell}$. Let $V=\mathbbm{i}_{\mathcal{C}}[\tau_{n-1}]\in Q_{n-1}$. We show that (2) holds for,

[TABLE]

We first show that $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\tau_{n}]$. Clearly, we have $\tau_{n-1}(a)\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{a\},\tau_{n-1}\right]$ for every $a\in A$. Moreover, $V=\mathbbm{i}_{\mathcal{C}}[\tau_{n-1}]\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{\varepsilon\},\tau_{n-1}\right]$ by Lemma 27. Therefore, since $w=a_{1}\cdots a_{\ell}$, it is immediate by definition of $T$ and Lemma 45 that,

[TABLE]

Moreover, $r=\rho(w)$ and $w\in L$. Thus, $w\in\rho_{*}^{-1}(r)\cap L$. By Fact 17, this yields,

[TABLE]

By definition, $\tau_{n}$ is the rating map $\zeta_{\mathit{Pol}(\mathcal{C})}^{\rho_{*}}[\tau_{n-1}]$. Hence, the above yields $(r,T)\in\tau_{n}(L)$. By choice of $L$, this exactly says that $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\tau_{n}]$.

It remains to show that $\mathord{\downarrow_{R}}U\subseteq f_{n-1}(T)$. Since $\eta_{\rho,S}(w)=\{(r,U)\}$, it follows from the definition of $\eta_{\rho,S}$ that $r=\rho(w)$ and $U$ is equal to the following set,

[TABLE]

Since $V=\mathbbm{i}_{\mathcal{C}}[\tau_{n-1}]$, we know by induction hypothesis in Proposition 51 that $S\subseteq f_{n-1}(V)$. Moreover, it can be verified using a simple induction on $n$ that for every $a\in A$, we have $\rho(a)\in f_{n-1}(\tau_{n-1}(a))$. Hence, we may combine (3) with Lemma 54 to obtain,

[TABLE]

By definition of $T$, we then obtain from Lemma 53 that,

[TABLE]

Since $\mathord{\downarrow_{R}}f_{n-1}(T)=f_{n-1}(T)$ by definition of $f_{n-1}$, this yields that $\mathord{\downarrow_{R}}U\subseteq f_{n-1}(T)$, finishing the proof. ∎

12.2. Completeness

We turn to the completeness direction in the proof of Theorem 43. Recall that a nice multiplicative rating map $\rho:2^{A^{*}}\to R$ is fixed. We need to show that $\mathbbm{i}_{\mathcal{C}}[\xi_{\mathit{BPol}(\mathcal{C})}[\rho]]$ is $\mathit{BPol}(\mathcal{C})$-complete.

In the whole section, we shall have to deal with the rating map $\xi_{\mathit{BPol}(\mathcal{C})}[\rho]$. For the sake of avoiding clutter we denote it by $\tau$. Recall that the definition is as follows,

[TABLE]

Our objective is to show that $\mathbbm{i}_{\mathcal{C}}[\tau]\subseteq R$ is $\mathit{BPol}(\mathcal{C})$-complete. For the sake of avoiding clutter, we write $S=\mathbbm{i}_{\mathcal{C}}[\tau]$.

The completeness proof is now based on two propositions which are proved independently. The first one is taken from [28] while the second is new.

Proposition 56**.**

For every $L\in\mathit{BPol}(\mathcal{C})$ and $s\in\tau(L)$, we have $r_{1},\dots,r_{k}\in R$ such that $(r_{i},\{r_{1}+\cdots+r_{k}\})\in\zeta_{\mathit{Pol}(\mathcal{C})}^{\rho_{*}}[\tau](L)$ for every $i\leq k$ and $s\leq r_{1}+\cdots+r_{k}$.

For the second proposition, we let $L^{\prime}$ as a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\eta_{\rho,S}$: we have $L^{\prime}\in\mathcal{C}$, $\varepsilon\in L^{\prime}$ and $\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]=\eta_{\rho,S}(L^{\prime})$. Furthermore, we define $L=L^{\prime}\cap A^{+}$. Note that $L\in\mathit{Pol}(\mathcal{C})\subseteq\mathit{BPol}(\mathcal{C})$ since $L^{\prime}\in\mathcal{C}\subseteq\mathit{Pol}(\mathcal{C})$ and $A^{+}=\bigcup_{a\in A}A^{*}aA^{*}\in\mathit{Pol}(\mathcal{C})$. We shall later apply Proposition 56 for this particular language $L$.

Proposition 57**.**

For every $(r,U)\in\zeta_{\mathit{Pol}(\mathcal{C})}^{\rho_{*}}[\tau](L)$, there exists $V\in 2^{R}$ such that $(r,V)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ and $U\subseteq\mathord{\downarrow_{R}}V$.

Let us first use the two propositions to finish the completeness proof. We show that $S=\mathbbm{i}_{\mathcal{C}}[\tau]$ is $\mathit{BPol}(\mathcal{C})$-complete for $\rho$. Let $s\in S$. We have to exhibit $(r_{1},U_{1}),\dots,(r_{k},U_{k})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ such that (1) is satisfied: $s\leq r_{1}+\cdots+r_{k}$ and $r_{1}+\cdots+r_{k}\in\ \mathord{\downarrow_{R}}U_{i}$ for every $i\leq k$.

By definition, $S=\mathbbm{i}_{\mathcal{C}}[\tau]$. Thus, since $L^{\prime}\in\mathcal{C}$ and $\varepsilon\in L^{\prime}$, this implies that $S\subseteq\tau(L^{\prime})$ and we obtain that $s\in\tau(L^{\prime})$. Moreover, $L^{\prime}=L\cup\{\varepsilon\}$ by definition. Therefore, we have $s\in\tau(L)\cup\tau(\varepsilon)$. We distinguish two cases depending on which set of this union $s\in S$ belongs to.

Assume first that $s\in\tau(\varepsilon)$, i.e. $s\in\mathcal{I}_{\mathit{BPol}(\mathcal{C})}\left[\{\varepsilon\},\rho\right]$ by definition of $\tau$. Since $\{\varepsilon\}\in\mathit{BPol}(\mathcal{C})$ by Fact 7, $\{\{\varepsilon\}\}$ is a $\mathit{BPol}(\mathcal{C})$-cover of $\{\varepsilon\}$. Thus, we have $\mathcal{I}_{\mathit{BPol}(\mathcal{C})}\left[\{\varepsilon\},\rho\right]\subseteq\mathcal{I}[\rho](\{\{\varepsilon\}\})=\mathord{\downarrow_{R}}\{\rho(\varepsilon)\}$. It follows that $s\leq\rho(\varepsilon)=1_{R}$ (recall that $\rho$ is a multiplicative rating map). Clearly, $(1_{R},\{1_{R}\})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ (since $(1_{R},\{1_{R}\})$ is the neutral element of $R\times 2^{R}$, we have $\eta_{\rho,S}(\varepsilon)=(1_{R},\{1_{R}\})$). Hence it is immediate that (1) is satisfied for $k=1$ and $(r_{1},U_{1})=(1_{R},\{1_{R}\})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$: we have $s\leq 1_{R}$ and $1_{R}\in\mathord{\downarrow_{R}}U_{1}$.

Assume now that $s\in\tau(L)$. In that case, since $L\in\mathit{BPol}(\mathcal{C})$, we obtain from Proposition 56 that there exist $r_{1},\dots,r_{k}\in R$ such that $(r_{i},\{r_{1}+\cdots+r_{k}\})\in\zeta_{\mathit{Pol}(\mathcal{C})}^{\rho_{*}}[\tau](L)$ for every $i\leq k$ and $s\leq r_{1}+\cdots+r_{k}$. We may now apply Proposition 57 which yields that for every $i\leq k$, there exists $U_{i}\in 2^{R}$ such that $(r_{i},U_{i})\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ and $r_{1}+\cdots+r_{k}\in\mathord{\downarrow_{R}}U_{i}$. Altogether, we get that (1) holds, finishing the proof.

It remains to prove the propositions. Similarly to what happened for the soundness direction, Proposition 56 is actually a generic result which holds for any class built using Boolean closure (it remains true when replacing $\mathit{Pol}(\mathcal{C})$ by an arbitrary lattice $\mathcal{G}$ in the definition of the rating map $\tau$ and in the statement). It is proved in [28, Proposition 9.1]. An important point is that Proposition 56 is where we require the hypothesis that $\rho$ is nice. Here, we show Proposition 57 which is specific to our setting: we need the hypothesis that we are considering the class $\mathit{BPol}(\mathcal{C})$ for $\mathcal{C}$ a quotient-closed Boolean algebra of group languages. The remainder of this subsection is devoted to this proof.

Proof of Proposition 57. Consider $(r,U)\in\zeta_{\mathit{Pol}(\mathcal{C})}^{\rho_{*}}[\tau](L)$. Our goal is to exhibit $V\in 2^{R}$ such that,

[TABLE]

By definition of the rating map $\zeta_{\mathit{Pol}(\mathcal{C})}^{\rho_{*}}[\tau]$, our hypothesis means that we have $U\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\rho_{*}^{-1}(r)\cap L,\tau\right]$. We build a well-chosen $\mathit{Pol}(\mathcal{C})$-cover of $\rho_{*}^{-1}(r)\cap L$ in order to use this hypothesis. The construction is based on Lemma 30.

Let $H$ be a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\tau=\xi_{\mathit{BPol}(\mathcal{C})}[\rho]$: we have $H\in\mathcal{C}$, $\varepsilon\in H$ and $S=\mathbbm{i}_{\mathcal{C}}[\tau]=\tau(H)$. For every $w\in A^{*}$, we write $F_{w}$ for the language $Ha_{1}H\cdots a_{n}H$ where $w=a_{1}\cdots a_{n}$ (in particular, $F_{\varepsilon}=H$). Note that by definition, $F_{w}\in\mathit{Pol}(\mathcal{C})$ for every $w\in A^{*}$. We have the following immediate corollary of Lemma 30.

Lemma 58**.**

There exists a finite subset $G\subseteq\rho_{*}^{-1}(r)\cap L$ such that $\{F_{w}\mid w\in G\}$ is a cover of $\rho_{*}^{-1}(r)\cap L$

In view of Lemma 58, we let $\mathbf{K}=\{F_{w}\mid w\in G\}$ which is a $\mathit{Pol}(\mathcal{C})$-cover of $\rho_{*}^{-1}(r)\cap L$ by definition. Therefore, we have,

[TABLE]

Consequently, we know that $U\in\mathcal{I}[\tau](\mathbf{K})$ which yields $w\in G$ such that $U\subseteq\tau(F_{w})$.

By construction, we have $G\subseteq\rho_{*}^{-1}(r)\cap L$ and $L=L^{\prime}\cap A^{+}$. Thus, $w\in A^{+}$: we have $n\geq 1$ and $n$ letters $a_{1},\dots,a_{n}$ such that $w=a_{1}\cdots a_{n}$. We let,

[TABLE]

It remains to show that (4) holds: we have $(r,V)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ and $U\subseteq\mathord{\downarrow_{R}}V$. Let us start with the former. By definition, $\rho(w)=r$ since $w\in G\subseteq\rho_{*}^{-1}(r)\cap L$. Thus, since $w=a_{1}\cdots a_{n}$ it is immediate from the definition of $\eta_{\rho,S}$ that $\eta_{\rho,S}(w)=\{(r,V)\}$. Moreover, we have $w\in G\subseteq\rho_{*}^{-1}(r)\cap L\subseteq L^{\prime}$ and $\eta_{\rho,S}(L^{\prime})=\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$ by definition of $L^{\prime}$. It follow that $\eta_{\rho,S}(w)\subseteq\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$. Altogether, we get $(r,V)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\rho,S}]$.

Finally, we prove that $U\subseteq\mathord{\downarrow_{R}}V$. By definition, $F_{w}=Ha_{1}H\cdots a_{n}H$. Moreover, since $\varepsilon\in H$, we have $H\subseteq H^{2}$. Thus,

[TABLE]

Therefore, since we have $U\subseteq\tau(F_{w})$, we also get $U\subseteq\tau(Ha_{1}H^{2}\cdots H^{2}a_{n}H)$. Moreover, Lemma 36 yields that the rating map $\tau=\xi_{\mathit{BPol}(\mathcal{C})}[\rho]$ is $\mathit{Pol}(\mathcal{C})$-multiplicative and the associated endomorphism of $(2^{R},\cup)$ is defined by $\mu_{\tau}(W)=\mathord{\downarrow_{R}}W$. Hence, since $H\in\mathcal{C}\subseteq\mathit{Pol}(\mathcal{C})$ and $\tau(H)=S$, it follows from Axiom 3 in the definition of $\mathit{Pol}(\mathcal{C})$-multiplicative rating maps that,

[TABLE]

Additionally, observe that we have the following fact since $\tau=\xi_{\mathit{BPol}(\mathcal{C})}[\rho]$ by definition.

Fact 59**.**

For every $a\in A$, we have $\tau(a)\subseteq\mathord{\downarrow_{R}}\{\rho(a)\}$.

Proof.

By definition of $\tau$, $\tau(a)=\mathcal{I}_{\mathit{BPol}(\mathcal{C})}\left[\{a\},\rho\right]$. Moreover, we have $\{a\}\in\mathit{BPol}(\mathcal{C})$ by Fact 7. Therefore, it is immediate that $\mathbf{K}=\{\{a\}\}$ is $\mathit{BPol}(\mathcal{C})$-cover of $\{a\}$. Consequently, $\tau(a)\subseteq\mathcal{I}[\rho](\mathbf{K})=\mathord{\downarrow_{R}}\{\rho(a)\}$. ∎

In view of Fact 59, we now obtain that,

[TABLE]

This concludes the proof.

13. Omitted proofs in Section 8

This appendix is devoted to the proof of Theorem 44. The statement is as follows. Recall that a quotient-closed Boolean algebra of group languages $\mathcal{C}$ is fixed.

Theorem 0**.**

Fix a morphism $\alpha:A^{*}\to M$ and a nice multiplicative rating map $\rho:2^{A^{*}}\to R$. Then, $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is the least $\mathit{PBPol}(\mathcal{C})$-complete subset of $M\times R$.

For the proof, we fix a morphism $\alpha:A^{*}\to M$ and a nice multiplicative rating map $\rho:2^{A^{*}}\to R$. We need to show that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is the least $\mathit{PBPol}(\mathcal{C})$-complete subset of $M\times R$. The proof involves two directions which are proved independently:

• •

First, we prove that the set $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is $\mathit{PBPol}(\mathcal{C})$-complete. This corresponds to soundness of the least fixpoint procedure computing $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$. The argument is based on results of [27] about the polynomial closure operation.

• •

Then, we prove that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is included in every $\mathit{PBPol}(\mathcal{C})$-complete subset. This corresponds to completeness of the least fixpoint procedure computing $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$. The argument is based on on the main theorem of [21].

13.1. Soundness

We prove that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is $\mathit{PBPol}(\mathcal{C})$-complete. For the sake of avoiding clutter, we shall write $\tau:2^{A^{*}}\to 2^{M\times R}$ for the rating map $\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]$. Moreover, we write $S=\mathbbm{i}_{\mathcal{C}}[\tau]$: we have to show that $S$ is $\mathit{PBPol}(\mathcal{C})$-complete. For the proof we let $L$ as a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\tau$: $L\in\mathcal{C}$, $\varepsilon\in L$ and $S=\tau(L)$. Moreover, since $\mathcal{C}$ is a quotient-closed Boolean algebra, it follows from the next lemma that we may choose the language $L$ so that $LL=L$. This will be useful.

Lemma 60**.**

There exists a $\mathcal{C}$-optimal $\varepsilon$-approximation $L$ for $\tau$ such that $LL=L$.

Proof.

Using Lemma 23, we get a $\mathcal{C}$-optimal $\varepsilon$-approximation $H$ for $\tau$. Since $\mathcal{C}$ is a quotient-closed Boolean algebra, it is standard that there exists a morphism $\beta:A^{*}\to N$ recognizing $H$ and such that every language recognized by $\beta$ belongs to $\mathcal{C}$ (it suffices to choose $\beta$ as the syntactic morphism of $H$, see [18] for example). We let $L=\beta^{-1}(1_{N})$. By hypothesis, we have $\varepsilon\in L$ and $L\in\mathcal{C}$. Moreover, $L\subseteq H$ since $\varepsilon\in H$ and $H$ is recognized by $\beta$. Thus, $\tau(L)\subseteq\tau(H)$ and since $H$ was a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\tau$, $L$ is one as well. It remains to show that $LL=L$. The inclusion $L\subseteq LL$ is immediate since $\varepsilon\in L$. Additionally, if $w\in LL$, then $w=uv$ with $\beta(u)=\beta(v)=1_{N}$ which implies that $\beta(w)=1_{N}$, i.e. $w\in L$. Thus, $LL\subseteq L$ which concludes the proof. ∎

Recall that since $\mathit{PBPol}(\mathcal{C})$ is a quotient-closed lattice closed under concatenation and marked concatenation, the following fact which is immediate from Lemma 35.

Fact 61**.**

The rating map $\tau$ is quasi-multiplicative and the associated endomorphism $\mu_{\tau}$ of $(2^{M\times R},\cup)$ is defined by $\mu_{\tau}(T)=\mathord{\downarrow_{R}}T$ for every $T\in 2^{M\times R}$.

Moreover, the following property can be verified using Fact 7 (which implies that $\mathit{PBPol}(\mathcal{C})$ contains every finite language).

Fact 62**.**

For every $w\in A^{*}$, $\tau(w)=\mathord{\downarrow_{R}}\{(\alpha(w),\tau(w))\}$.

It now remains to prove that $S=\tau(L)$ is $\mathit{PBPol}(\mathcal{C})$-complete. This involves four properties. We start with downset.

Downset. We show that $\mathord{\downarrow_{R}}S=S$. The right to left inclusion is trivial. Hence, we focus on the converse one. Consider $(s,r)\in\mathord{\downarrow_{R}}S=\mathord{\downarrow_{R}}\tau(L)$. We have $r^{\prime}\in R$ such that $r\leq r^{\prime}$ and $(s,r^{\prime})\in\tau(L)$. Since $\tau$ is an alias for $\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]$, this exactly says that,

[TABLE]

Since imprints are closed under downset by definition, we get $r\in\mathcal{I}_{\mathit{PBPol}(\mathcal{C})}\left[L\cap\alpha^{-1}(s),\rho\right]$ and we conclude that $(s,r)\in\tau(L)=S$, finishing the proof.

Multiplication. Consider $(s,r),(s^{\prime},r^{\prime})\in S=\tau(L)$, we have to show that $(ss^{\prime},rr^{\prime})\in S=\tau(L)$. We have,

[TABLE]

Since $\tau$ is quasi-multiplicative and the associated endomorphism of $(2^{M\times R},\cup)$ is $T\mapsto\mathord{\downarrow_{R}}T$, the axioms of quasi-multiplicative rating maps yield that $\tau(LL)=\mathord{\downarrow_{R}}(\tau(L)\cdot\tau(L))$. We get that $(ss,rr^{\prime})\in\tau(LL)$. Since $LL=L$ by definition of $L$, this yields $(ss^{\prime},rr^{\prime})\in\tau(L)$ as desired.

It remains to handle the last two properties in the definition of $\mathit{PBPol}(\mathcal{C})$-complete subsets: $\mathcal{C}$-operation and $\mathit{PBPol}(\mathcal{C})$-operation. First, we prove the following lemma which we shall need for both properties.

Lemma 63**.**

For every $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$, we have,

[TABLE]

Proof.

Consider $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$. By definition of $\eta_{\alpha,\rho,S}$, we have $w\in L$ such that $\eta_{\alpha,\rho,S}(w)=\{(r,T)\}$. We consider two cases depending on whether $w=\varepsilon$ or $w\in A^{+}$.

Case 1: Assume first that $w=\varepsilon$. It follows that $(r,T)=(1_{R},\{(1_{M},1_{R})\})$ (since $\{(1_{R},\{(1_{M},1_{R})\})\}$ is the neutral element of $2^{R\times 2^{M\times R}}$). It follows that $L\cap\rho_{*}^{-1}(r)=\{\varepsilon\}$ (recall that $\varepsilon\in L$ by definition). Thus, we have to show that,

[TABLE]

By Fact 62, we have $(1_{M},1_{R})\in\tau(\varepsilon)$. Thus, since it is clear that $\tau(\varepsilon)\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{\varepsilon\},\tau\right]$, we get $\{(1_{M},1_{R})\}\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{\varepsilon\},\tau\right]$ which concludes this case.

Case 2: Assume now that $w\in A^{+}$. In that case, we have $\ell\geq 1$ and $\ell$ letters $a_{1},\dots,a_{\ell}$ such that $w=a_{1}\cdots a_{\ell}$.

Since $\eta_{\alpha,\rho,S}(w)=\{(r,T)\}$, it follows from the definition of $\eta_{\alpha,\rho,S}$ that $r=\rho(w)$ and $T$ is equal to the following set,

[TABLE]

By Fact 62, we have $\{(\alpha(a),\rho(a))\}\subseteq\tau(a)$ for every $a\in A$. Therefore, we obtain that,

[TABLE]

It is clear that, $\tau(a)\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{a\},\tau\right]$ for every $a\in A$. Moreover, $S=\mathbbm{i}_{\mathcal{C}}[\tau]\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\{\varepsilon\},\tau\right]$ by Lemma 27. Therefore, since $w=a_{1}\cdots a_{\ell}$, it is immediate from the above and Lemma 45 that,

[TABLE]

Moreover, since $r=\rho(w)$ and $w\in L$, we have $w\in\rho_{*}^{-1}(r)\cap L$. Thus, Fact 17 yields that $T\in\mathcal{I}_{\mathit{Pol}(\mathcal{C})}\left[\rho_{*}^{-1}(r)\cap L,\tau\right]$, concluding the proof. ∎

We may now handle the last two operations in the definition of $\mathit{PBPol}(\mathcal{C})$-complete subsets.

$\mathcal{C}$-operation. Consider $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$. we have to show that $T\subseteq S=\tau(L)$. By Lemma 63, we have,

[TABLE]

Since $L\in\mathcal{C}\subseteq\mathit{Pol}(\mathcal{C})$, $\{L\}$ is a $\mathit{Pol}(\mathcal{C})$-cover of $L\cap\rho_{*}^{-1}(r)$. Therefore, it follows that $T\subseteq\tau(L)=S$, concluding the proof.

$\mathit{PBPol}(\mathcal{C})$-operation. Consider $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$ and a pair of idempotents $(e,f)\in\mathord{\downarrow_{R}}T\subseteq M\times R$. We have to show that,

[TABLE]

Recall that $\tau=\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]$. Hence, this amounts to proving that $frf+f\in\mathcal{I}_{\mathit{PBPol}(\mathcal{C})}\left[L\cap\alpha^{-1}(e),\rho\right]$. Thus, we fix an arbitrary $\mathit{PBPol}(\mathcal{C})$-cover $\mathbf{K}$ of $L\cap\alpha^{-1}(e)$ for the proof and show that $frf+f\in\mathcal{I}[\rho](\mathbf{K})$.

First, we use $\mathbf{K}$ to construct two auxiliary classes of languages $\mathcal{D}$ and $\mathcal{F}$ that we shall need for the proof (the argument reuses some standard results which are detailed in [27]).

Fact 64**.**

There exist two finite quotient-closed lattices $\mathcal{D}$ and $\mathcal{F}$ which satisfy the following properties:

• •

$L\in\mathcal{D}$* and $\mathcal{D}\subseteq\mathit{Pol}(\mathcal{C})$, and,*

• •

$\mathcal{F}\subseteq\mathit{Pol}(\mathit{co\textup{-}}\!\mathcal{D})\subseteq\mathit{PBPol}(\mathcal{C})$* and $K\in\mathcal{F}$ for every $K\in\mathbf{K}$.*

Proof.

By definition $\mathbf{K}$ contains finitely many languages in $\mathit{PBPol}(\mathcal{C})$. Moreover, it is shown in [27, Lemma 33] that we have $\mathit{PBPol}(\mathcal{C})=\mathit{Pol}(\mathit{co\textup{-}\!Pol}(\mathcal{C}))$ where $\mathit{co\textup{-}\!Pol}(\mathcal{C})$ denotes the class containing the languages with a complement in $\mathit{Pol}(\mathcal{C})$. Consequently, there are finitely many languages in $\mathit{Pol}(\mathcal{C})$ such that every $K\in\mathbf{K}$ is built by applying unions and of marked concatenations to complements of these languages. It then follows from [27, Lemma 17] that we may build a finite quotient-closed lattice $\mathcal{D}\subseteq\mathit{Pol}(\mathcal{C})$ which contains all these languages as well as $L\in\mathcal{C}\subseteq\mathit{Pol}(\mathcal{C})$. Thus, by definition, we have $K\in\mathit{Pol}(\mathit{co\textup{-}}\!\mathcal{D})$ for every $K\in\mathbf{K}$. Finally, since $\mathbf{K}$ is finite, we may reuse [27, Lemma 17] to build another finite quotient-closed lattice $\mathcal{F}$ such that $\mathcal{F}\subseteq\mathit{Pol}(\mathit{co\textup{-}}\!\mathcal{D})$ and $K\in\mathcal{F}$ for every $K\in\mathbf{K}$. This concludes the proof. ∎

We fix $\mathcal{D}$ and $\mathcal{F}$ as the finite quotient-closed lattices described in Fact 64 for the remainder of the proof. Since $\mathcal{D}$ and $\mathcal{F}$ are finite, it is classical to associate relations $\leqslant_{\mathcal{D}}$ and $\leqslant_{\mathcal{F}}$ over $A^{*}$. Given $w,w^{\prime}\in A^{*}$, we have, write $w\leqslant_{\mathcal{D}}w^{\prime}$ if and only if the following holds:

[TABLE]

Clearly, these are preorder relations. In particular, given $u\in A^{*}$, we let ${\mathord{\uparrow}_{\mathcal{D}}\,u}\subseteq A^{*}$ as the least upper set of $\leqslant_{\mathcal{D}}$ containing $u$: ${\mathord{\uparrow}_{\mathcal{D}}\,u}=\{v\in A^{*}\mid u\leqslant_{\mathcal{D}}v\}$. Similarly, we define ${\mathord{\uparrow}_{\mathcal{F}}\,u}$ from $\leqslant_{\mathcal{D}}$. It is straightforward to verify that ${\mathord{\uparrow}_{\mathcal{D}}\,u}\in\mathcal{D}$ and ${\mathord{\uparrow}_{\mathcal{F}}\,u}\in\mathcal{F}$ for every $u\in A^{*}$ (see [27, Section 2] for the proofs). Finally, since $\mathcal{D}$ and $\mathcal{F}$ are closed under quotients, one may verify that $\leqslant_{\mathcal{D}}$ and $\leqslant_{\mathcal{F}}$ are compatible with word concatenation (again, we refer the reader to [27] for details).

Finally, we have the following property which holds because $\mathcal{F}\subseteq\mathit{Pol}(\mathit{co\textup{-}}\!\mathcal{D})$. This is a corollary of a generic property of polynomial closure presented in [27].

Lemma 65**.**

There exist natural numbers $p,k\in\mathbb{N}$ such that for every $\ell\geq k$ and $u,v\in A^{*}$ satisfying $v\leq_{\mathcal{D}}u$, we have,

[TABLE]

Proof.

Since $\mathcal{F}\subseteq\mathit{Pol}(\mathit{co\textup{-}}\!\mathcal{D})$, it follows from [27, Proposition 34] that for every $L\in\mathcal{F}$, there exists $k_{L},p_{L}\in\mathbb{N}$ such that for every $\ell\geq k_{L}$ and every $u,v\in A^{*}$ satisfying $u\leqslant_{\mathit{co\textup{-}}\!\mathcal{D}}v$, we have:

[TABLE]

We choose $k$ as the maximum of all numbers $k_{L}$ for $L\in\mathcal{F}$ and $p$ as the least common multiplier of all numbers $p_{L}$ for $L\in\mathcal{F}$ (recall that $\mathcal{F}$ is finite by definition). It then follows, that for every $\ell\geq k$ and $u,v\in A^{*}$ satisfying $u\leqslant_{\mathit{co\textup{-}}\!\mathcal{D}}v$, we have:

[TABLE]

By definition, this exactly says that $u^{p\ell+1}\leq_{\mathcal{F}}u^{p\ell}vu^{p\ell}$.

Finally, observe that for every $u,v\in A^{*}$, the hypothesis $v\leqslant_{\mathcal{D}}u$ implies that $u\leqslant_{\mathit{co\textup{-}}\!\mathcal{D}}v$. Indeed, $v\leqslant_{\mathcal{D}}u$ means that for every $L\in\mathcal{D}$, we have $v\in L\Rightarrow u\in L$. The contrapositive then states that for every $L\in\mathcal{D}$, we have $u\not\in L\Rightarrow v\not\in L$. Finally, since the languages of $\mathit{co\textup{-}}\!\mathcal{D}$ are the complements of those in $\mathcal{D}$, it follows that for every $L\in\mathit{co\textup{-}}\!\mathcal{D}$, we have $u\in L\Rightarrow v\in L$, i.e. $u\leqslant_{\mathit{co\textup{-}}\!\mathcal{D}}v$. This concludes the proof. ∎

We may now come back to the main argument. Recall that we have $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$ and an idempotent $(e,f)\in\mathord{\downarrow_{R}}T\subseteq M\times R$. Furthermore, we have a $\mathit{PBPol}(\mathcal{C})$-cover $\mathbf{K}$ of $L\cap\alpha^{-1}(e)$. Our objective is to show that $frf+f\in\mathcal{I}[\rho](\mathbf{K})$. By definition, this amounts to exhibiting $K\in\mathbf{K}$ such that $frf+f\leq\rho(K)$. We start with the following lemma which exhibits a word $v\in A^{*}$ which we shall use for the construction of $K$. This is where we use the hypothesis that $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$.

Lemma 66**.**

There exists $v\in A^{*}$ such that $r=\rho(v)$, $v\in L$ and $f\in\mathcal{I}_{\mathit{PBPol}(\mathcal{C})}\left[\left({\mathord{\uparrow}_{\mathcal{D}}\,v}\right)\cap\alpha^{-1}(e),\rho\right]$

Proof.

Since $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$, it follows from Lemma 63 that,

[TABLE]

Since $\mathcal{D}\subseteq\mathit{Pol}(\mathcal{C})$ by definition in Fact 64, it then follows from Fact 17 that,

[TABLE]

Let $\mathbf{H}=\{{\mathord{\uparrow}_{\mathcal{D}}\,v}\mid v\in L\cap\rho_{*}^{-1}(r)\}$. By definition, $\mathbf{H}$ is a finite set of languages in $\mathcal{D}$. Moreover $\mathbf{H}$ is clearly a $\mathcal{D}$-cover of $L\cap\rho_{*}^{-1}(r)$. Consequently, we have $\mathcal{I}_{\mathcal{D}}\left[L\cap\rho_{*}^{-1}(r),\tau\right]\subseteq\mathcal{I}[\tau](\mathbf{H})$. We get $H\in\mathbf{H}$ such that $T\subseteq\tau(H)$. By definition $H={\mathord{\uparrow}_{\mathcal{D}}\,v}$ for some $v\in A^{*}$ such that $\rho(v)=r$ and $v\in L$. We now unravel the definition of $\tau=\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]$ which yields,

[TABLE]

Since $(e,f)\in\mathord{\downarrow_{R}}T$ by hypothesis, we obtain as desired that $f\in\mathcal{I}_{\mathit{PBPol}(\mathcal{C})}\left[\left({\mathord{\uparrow}_{\mathcal{D}}\,v}\right)\cap\alpha^{-1}(e),\rho\right]$. This concludes the proof. ∎

We are now ready to construct the desired language $K\in\mathbf{K}$ such that $frf+f\leq\rho(K)$. We fix $v\in A^{*}$ as the word described in Lemma 66. Consider the following set $\mathbf{H}$,

[TABLE]

Since $\mathcal{F}\subseteq\mathit{PBPol}(\mathcal{C})$ it is immediate by definition that $\mathbf{H}$ is a finite set of languages in $\mathit{PBPol}(\mathcal{C})$. Therefore, $\mathbf{H}$ is clearly a $\mathit{PBPol}(\mathcal{C})$-cover of $\left({\mathord{\uparrow}_{\mathcal{D}}\,v}\right)\cap\alpha^{-1}(e)$. Consequently, since $f\in\mathcal{I}_{\mathit{PBPol}(\mathcal{C})}\left[\left({\mathord{\uparrow}_{\mathcal{D}}\,v}\right)\cap\alpha^{-1}(e),\rho\right]$ by Lemma 66, we have $f\in\mathcal{I}[\rho](\mathbf{H})$. This yields a language $H\in\mathbf{H}$ such that $f\leq\rho(H)$. Consider the natural numbers $k,p\in\mathbb{N}$ given by Lemma 65. We define,

[TABLE]

In the following lemma, we exhibit a language $K\in\mathbf{K}$ which contains $G$. This is the desired language: using the definition of $G$ we show that $frf+f\leq\rho(K)$.

Lemma 67**.**

There exists $K\in\mathbf{K}$ such that $G\subseteq K$.

Let us first use Lemma 67 to finish the main argument. Let $K\in\mathbf{K}$ be the language given by the lemma. Since $G\subseteq K$, we have $\rho(G)\subseteq\rho(K)$ by definition of rating maps. Moreover, since $f\leq\rho(H)$ and $f$ is idempotent, we get from the definition of $G$ that,

[TABLE]

By definition in Lemma 66, we have $r=\rho(v)$. Therefore, we get that $frf+f\leq\rho(G)\leq\rho(K)$ which concludes the proof: we have $frf+f\in\mathcal{I}[\rho](\mathbf{K})$. It remains to prove the lemma.

Proof of Lemma 67.

By definition of $\mathbf{H}$, $H={\mathord{\uparrow}_{\mathcal{F}}\,u}$ for some $u\in\left({\mathord{\uparrow}_{\mathcal{D}}\,v}\right)\cap\alpha^{-1}(e)$. This implies that $u\in\alpha^{-1}(e)$ and since $e$ is idempotent, it follows that $u^{pk+1}\in\alpha^{-1}(e)$. Moreover, we have $v\in L$ by definition in Lemma 66 and since $L\in\mathcal{D}$, $u\in{\mathord{\uparrow}_{\mathcal{D}}\,v}$ yields $u\in L$. Thus, since $LL=L$, we get $u^{pk+1}\in L$. Altogether, this yields $u^{pk+1}\in L\cap\alpha^{-1}(e)$ and since $\mathbf{K}$ is by definition a $\mathit{PBPol}(\mathcal{C})$-cover of $L\cap\alpha^{-1}(e)$, there exists $K\in\mathbf{K}$ such that $u^{pk+1}\in K$. We prove that $G\subseteq K$ which concludes the proof.

Consider $w\in G$. We show that $u^{pk+1}\leq_{\mathcal{F}}w$. Since $K\in\mathcal{F}$ by definition of $\mathcal{F}$ in Fact 64 and $u^{pk+1}\in K$, it will follow as desired that $w\in K$. By definition, $G$ is the union of two languages. Thus, there are two cases. Assume first that $w\in H^{pk+1}$ (i.e. $w$ is the concatenation of $pk+1$ words in $H$). Since $H={\mathord{\uparrow}_{\mathcal{F}}\,u}$ and $\leq_{\mathcal{F}}$ is compatible with concatenation, it is immediate that this implies $u^{pk+1}\leq_{\mathcal{F}}w$, finishing the proof for this case.

We now assume that $w\in H^{pk}\cdot\{v\}\cdot H^{pk}$. It follows that $w=xvy$ with $x,y\in H^{pk}$. We may reuse the fact that $\leq_{\mathcal{F}}$ is compatible with concatenation to obtain $u^{pk}\leq_{\mathcal{F}}x$ and $u^{pk}\leq_{\mathcal{F}}y$. Consequently, we get $u^{pk}vu^{pk}\leq_{\mathcal{F}}xvy$ and by transitivity, it suffices to show that $u^{pk+1}\leq_{\mathcal{F}}u^{pk}vu^{pk}$. This is immediate from Lemma 65 as we have $u\in{\mathord{\uparrow}_{\mathcal{D}}\,v}$ which means that $v\leqslant_{\mathcal{D}}u$. ∎

13.2. Completeness

We have proved that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]$ is a $\mathit{PBPol}(\mathcal{C})$-complete subset of $M\times R$. It remains to show that it is the least such subset. Therefore, we fix an arbitrary $\mathit{PBPol}(\mathcal{C})$-complete set $S\subseteq M\times R$ and show that $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]\subseteq S$.

By definition, $\mathbbm{i}_{\mathcal{C}}[\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho]]\subseteq\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho](L)$ for every language $L\in\mathcal{C}$ such that $\varepsilon\in L$. Hence, it suffices to exhibit a language $L\in\mathcal{C}$ such that $\varepsilon\in L$ and $\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho](L)\subseteq S$. We first choose the appropriate language $L$.

We let $H$ as a $\mathcal{C}$-optimal $\varepsilon$-approximation for the nice multiplicative rating map $\eta_{\alpha,\rho,S}$. That is, we have $H\in\mathcal{C}$, $\varepsilon\in H$ and $\eta_{\alpha,\rho,S}(H)=\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$. Since $H\in\mathcal{C}$ is a group language and $\mathcal{C}$ is a quotient-closed Boolean algebra, it is standard that $H$ is recognized by a morphism $\psi:A^{*}\to G$ into a finite group $G$ such that every language recognized by $\psi$ belongs to $\mathcal{C}$. (it suffices to choose $\psi$ as the syntactic morphism of $H$, see [18] for example). In particular, the language $L=\psi^{-1}(1_{G})$ belongs to $\mathcal{C}$ and contains $\varepsilon$ by definition. It now remains to show the following inclusion:

[TABLE]

The argument reuses the main theorem of [21] which characterizes the set $\mathcal{P}_{\mathit{PBPol}(\mathcal{G})}^{\alpha}[\rho]$ when $\mathcal{G}$ is a finite quotient-closed Boolean algebra. Let us first define $\mathcal{G}$.

We let $\mathcal{G}$ as the class containing all languages recognized by $\psi$ (which is included in $\mathcal{C}$ by definition). It is immediate that $\mathcal{G}$ is a finite quotient-closed Boolean algebra. Moreover, the quotient set ${A^{*}}/{\sim_{\mathcal{G}}}$ is isomorphic to the group $G$ and the morphism $w\mapsto[w]_{\mathcal{G}}$ from $A^{*}$ to ${A^{*}}/{\sim_{\mathcal{G}}}$ corresponds to the morphism $\psi:A^{*}\to G$. Hence, for the sake of avoiding clutter, we identify ${A^{*}}/{\sim_{\mathcal{G}}}$ with $G$ in the proof. We may now recall the theorem of [21]. Note that we slightly adapt the formulation for the sake of convenience. In particular, the theorem of [21] requires manipulating monoid morphisms and multiplicative rating maps which are “$\mathcal{G}$-compatible” (essentially, this means that they compute $\sim_{\mathcal{G}}$-classes). Here, we avoid this requirement (which is not satisfied by $\alpha$ and $\rho$) by manipulating $\sim_{\mathcal{G}}$-classes (which we identify with elements of $G$) explicitly.

Consider two sets $S^{\prime}\subseteq G\times M\times R$ and $\mathcal{T}\subseteq G\times R\times 2^{M\times R}$. We say that the pair $(S^{\prime},\mathcal{T})$ is $\mathit{PBPol}(\mathcal{G})$-saturated if the following conditions are satisfied:

• •

Properties on $S^{\prime}\subseteq G\times M\times R$:

1. (1)

Trivial elements: For every $w\in A^{*}$, we have $(\psi(w),\alpha(w),\rho(w))\in S^{\prime}$. 2. (2)

Downset: We have $\mathord{\downarrow_{R}}S^{\prime}=S^{\prime}$. 3. (3)

Multiplication: For every $(g,s,q),(h,t,r)\in S^{\prime}$, we have $(gh,st,qr)\in S^{\prime}$. 4. (4)

$\mathit{PBPol}(\mathcal{G})$-closure. For every $(g,r,U)\in\mathcal{T}$ where $g\in G$ is idempotent (i.e, $g=1_{G}$) and every pair of idempotents $(e,f)\in\mathord{\downarrow_{R}}U\subseteq M\times R$, we have $(g,e,f\cdot(r+1_{R})\cdot f)\in S^{\prime}$.

• •

Properties on $\mathcal{T}\subseteq G\times R\times 2^{M\times R}$:

1. (5)

Trivial elements: For every $w\in A^{*}$, we have $(\psi(w),\rho(w),\{(\alpha(w),\rho(w))\})\in\mathcal{T}$. 2. (6)

Multiplication: For every $(g,q,U),(h,r,V)\in\mathcal{T}$, we have $(gh,qr,UV)\in\mathcal{T}$. 3. (7)

Nested $\mathit{Pol}(\mathcal{C})$-closure. For every triple of idempotents $(g,f,F)\in\mathcal{T}$, we have $(g,f,F\cdot S^{\prime}(g)\cdot F)\in\mathcal{T}$ (where $S^{\prime}(g)=\{(s,r)\mid M\times R\mid(g,s,r)\in S^{\prime}\}$).

The following proposition can be verified from [21, Theorem 7.11] (it is a reformulation of this result with our terminology).

Proposition 68**.**

Consider the least $\mathit{PBPol}(\mathcal{G})$-saturated pair $(S^{\prime},\mathcal{T})$ (with respect to pairwise inclusion). Then, for every $g\in G$ and every $s\in M$, the following equality holds:

[TABLE]

We now use Proposition 68 to prove that (5) holds (i.e. $\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho](L)\subseteq S$). First, we exhibit a specific $\mathit{PBPol}(\mathcal{G})$-saturated pair $(S^{\prime},\mathcal{T})$ which we build from $S$ and the nice multiplicative rating map $\eta_{\alpha,\rho,S}$. Then, we apply the proposition to this pair and use the information we get to deduce (5).

Definition of $(S^{\prime},\mathcal{T})$. We start with the definition of the set $\mathcal{T}\subseteq G\times R\times 2^{M\times R}$:

[TABLE]

We may now define $S^{\prime}\subseteq G\times M\times R$. We let,

[TABLE]

Before we prove that the pair $(S^{\prime},\mathcal{T})$ is $\mathit{PBPol}(\mathcal{G})$-saturated, let us present two simple facts about $S^{\prime}$ and $\mathcal{T}$ that will be useful.

Fact 69**.**

For every $(g,q,T)\in\mathcal{T}$, one of the two following properties hold:

• •

$(g,q,T)=(1_{G},1_{R},\{(1_{M},1_{R})\})$, or,

• •

we have $STS=ST=TS=T$.

Proof.

Let us make two preliminary observations. First, we have $(1_{M},1_{R})\in S$. Indeed, we know that $S$ is $\mathit{PBPol}(\mathcal{C})$-complete which implies that for every $(r,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]=\eta_{\alpha,\rho,S}(H)$, we have $T\subseteq S$. Since $\varepsilon\in H$ and $\eta_{\alpha,\rho,\varepsilon}=\{(1_{R},\{(1_{M},1_{R})\})\}$, it follows that $(1_{M},1_{R})\in S$. Moreover, we know that $SS=S$. Indeed, since $S$ is $\mathit{PBPol}(\mathcal{C})$-complete, we have $S\cdot SS$ and the converse inclusion is immediate since $(1_{M},1_{R})\in S$.

We may now prove the fact. Consider $(g,q,T)\in\mathcal{T}$. By definition of $\mathcal{T}$, there are two cases. First, if $(g,q,T)=(1_{G},1_{R},S)$, it is immediate that second assertion in the fact holds since $SS=S$. Otherwise, we have $(q,T)\in\eta_{\alpha,\rho,S}(\psi^{-1}(g))$. By definition of $\eta_{\alpha,\rho,S}$, it is immediate that there exists some word $w\in\psi^{-1}(g)$ such that $\eta_{\alpha,\rho,S}(w)=\{(q,T)\}$. If $w=\varepsilon$, then $g=1_{G}$, $q=1_{R}$ and $T=\{(1_{M},1_{R})\}$: the first assertion in the fact holds. Otherwise $w\in A^{+}$ and since $SS=S$ it is immediate from the definition of $\eta_{\alpha,\rho,S}$ that $STS=ST=TS=T$: the second assertion in the fact holds. ∎

Fact 70**.**

We have $S^{\prime}(1_{G})=S$.

Proof.

We first show that $S\subseteq S^{\prime}(1_{G})$. By definition, $(1_{G},1_{R},S)\in\mathcal{T}$ which implies that $S\subseteq S^{\prime}(1_{G})$ by definition of $S^{\prime}$. it remains to prove that $S^{\prime}(1_{G})\subseteq S$. let $(s,r)\in S^{\prime}(1_{G})$. By definition, we have $(q,T)\in R\times 2^{M\times R}$ such that $(1_{G},q,T)\in\mathcal{T}$ and $(s,r)\in\mathord{\downarrow_{R}}T$. By definition of $\mathcal{T}$, there are now two cases. First, if $(1_{G},q,T)=(1_{G},1_{R},S)$, it is immediate that $(s,r)\in\mathord{\downarrow_{R}}S$ which concludes the proof: since $S$ is $\mathit{PBPol}(\mathcal{C})$-saturated, we have $\mathord{\downarrow_{R}}S=S$. Otherwise, we have $(q,T)\in\eta_{\alpha,\rho,S}(\psi^{-1}(1_{G}))$. Recall that $\psi$ recognizes $H$ and $\varepsilon\in H$. Thus, $\psi^{-1}(1_{G})\subseteq H$ which yields $(q,T)\in\eta_{\alpha,\rho,S}(H)$. Since $H$ is a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\eta_{\alpha,\rho,S}$ by definition, it follows that $(q,T)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$. Thus, since $S$ is $\mathit{PBPol}(\mathcal{C})$-complete by definition, we have $\mathord{\downarrow_{R}}T\subseteq\mathord{\downarrow_{R}}S=S$ and we conclude that $(s,r)\in S$, finishing the proof. ∎

Main argument. We may now prove the inclusion described in (5). Using the hypothesis that $S\subseteq M\times R$ is $\mathit{PBPol}(\mathcal{C})$-complete, one may prove the following proposition.

Proposition 71**.**

The pair $(S^{\prime},\mathcal{T})$ is $\mathit{PBPol}(\mathcal{G})$-saturated.

Let us first use Proposition 71 to show that (5) holds and conclude the main argument. We have to show that $\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho](L)\subseteq S$. Consider $(s,r)\in\zeta_{\mathit{PBPol}(\mathcal{C})}^{\alpha}[\rho](L)$. We show that $(s,r)\in S$. By definition, we have,

[TABLE]

Since $\mathcal{G}\subseteq\mathcal{C}$, it is immediate that $\mathit{PBPol}(\mathcal{G})\subseteq\mathit{PBPol}(\mathcal{C})$. Therefore, it is clear that,

[TABLE]

Moreover, we have $L=\psi^{-1}(1_{G})$ by definition. Therefore, since $(S^{\prime},\mathcal{T})$ is $\mathit{PBPol}(\mathcal{G})$-saturated, it is immediate from Proposition 68 that,

[TABLE]

By Fact 70, this yield $(s,r)\in S$, finishing the completeness proof. It remains to present an argument for Proposition 71.

Proof of Proposition 71.

We prove that $(S^{\prime},\mathcal{T})$ is $\mathit{PBPol}(\mathcal{G})$-saturated. There are seven properties to verify: four apply to $S^{\prime}$ and three to $\mathcal{T}$. We start with the latter: we must show that $\mathcal{T}$ contains the trivial elements and is closed under multiplication and nested $\mathit{Pol}(\mathcal{G})$-closure.

Multiplication for $\mathcal{T}$. Consider $(g,q,T),(g^{\prime},q^{\prime},T^{\prime})\in\mathcal{T}$, we have to show that $(gg^{\prime},qq^{\prime},TT^{\prime})\in\mathcal{T}$. Since $\mathcal{T}$ is defined as the union of two sets, there are several cases. Assume first that either $(g,q,T)$ or $(g^{\prime},q^{\prime},T^{\prime})$ is equal to $(1_{G},1_{R},S)$. By symmetry, we consider the later case: $(g^{\prime},q^{\prime},T^{\prime})=(1_{G},1_{R},S)$. it follows that $(gg^{\prime},qq^{\prime},TT^{\prime})=(g,q,TS)$. In view of Fact 69, there are now two cases:

• •

First, if we have $(g,q,T)=(1_{G},1_{R},\{(1_{M},1_{R})\})$, then $(gg^{\prime},qq^{\prime},TT^{\prime})=(1_{G},1_{R},S)\in\mathcal{T}$.

• •

Otherwise, $TS=T$ and $(gg^{\prime},qq^{\prime},TT^{\prime})=(g,q,T)\in\mathcal{T}$.

Now assume that $(q,T)\in\eta_{\alpha,\rho,S}(\psi^{-1}(g))$ and $(q^{\prime},T^{\prime})\in\eta_{\alpha,\rho,S}(\psi^{-1}(g^{\prime}))$. Under these hypotheses, we obtain that, $(qq^{\prime},TT^{\prime})\in\eta_{\alpha,\rho,S}(\psi^{-1}(g)\psi^{-1}(g^{\prime}))$. Since $\psi$ is a morphism, we have $\psi^{-1}(g)\psi^{-1}(g^{\prime})\subseteq\psi^{-1}(gg^{\prime})$ which yields $(qq^{\prime},TT^{\prime})\in\eta_{\alpha,\rho,S}(\psi^{-1}(gg^{\prime}))$. By definition, this implies $(gg^{\prime},qq^{\prime},TT^{\prime})\in\mathcal{T}$.

Trivial elements for $\mathcal{T}$. Consider $w\in A^{*}$, we show that $(\psi(w),\rho(w),\{\alpha(w),\rho(w)\})\in\mathcal{T}$. Assume first that $w=\varepsilon$. In that case, observe that $\eta_{\alpha,\rho,S}(\varepsilon)=\{(1_{R},\{(1_{M},1_{R})\})\}$ (this is the neutral element of $2^{R\times 2^{M\times R}}$). Hence, since $\varepsilon\in\psi^{-1}(\psi(\varepsilon))$, we have $(\rho(\varepsilon),\{(\alpha(\varepsilon),\rho(\varepsilon)\})\in\eta_{\alpha,\rho,S}(\psi^{-1}(\psi(\varepsilon)))$ which yields that $(\psi(\varepsilon),\rho(\varepsilon),\{\alpha(\varepsilon),\rho(\varepsilon)\})\in\mathcal{T}$.

Assume now that $w\in A^{+}$. Since we already proved that $\mathcal{T}$ is closed under multiplication, it suffices to consider the case when $w=a\in A$. We already observed that $(1_{M},1_{R})\in S$ (see the proof of Fact 69). Therefore, it is immediate that $(\rho(a),\{(\alpha(a),\rho(a))\})\in\eta_{\alpha,\rho,S}(a)$ by definition. Hence, since $a\in\psi^{-1}(\psi(a))$, we have $(\rho(a),\{(\alpha(a),\rho(a)\})\in\eta_{\alpha,\rho,S}(\psi^{-1}(\psi(a)))$ which yields that $(\psi(a),\rho(a),\{\alpha(a),\rho(a)\})\in\mathcal{T}$.

Nested $\mathit{Pol}(\mathcal{G})$-closure. Consider a triple of idempotents $(1_{G},f,F)\in\mathcal{T}$ (note that the first element is $1_{G}$ since the only idempotent in a group is its neutral element). We have to show that,

[TABLE]

By Fact 70, we know that $S^{\prime}(1_{G})=S$. Hence, we have to show that $(1_{G},f,F\cdot S\cdot F)\in\mathcal{T}$. In view of Fact 69, there are two cases:

• •

First, if we have $(1_{G},f,F)=(1_{G},1_{R},\{(1_{M},1_{R})\})$, then $(1_{G},f,F\cdot S\cdot F)=(1_{G},1_{R},S)\in\mathcal{T}$.

• •

Otherwise, $FSF=F$ and $(1_{G},f,F\cdot S\cdot F)=(1_{G},f,F)\in\mathcal{T}$.

We turn to the properties concerning $S^{\prime}$: we have to show that $S^{\prime}$ contains the trivial elements and is closed under downset, multiplication and $\mathit{PBPol}(\mathcal{G})$-closure.

Trivial elements for $S^{\prime}$. Consider $w\in A^{*}$, we have to show that $(\psi(w),\alpha(w),\rho(w))\in S^{\prime}$. We already show that $(\psi(w),\rho(w),\{\alpha(w),\rho(w)\})\in\mathcal{T}$. Thus, it is immediate by definition of $S^{\prime}$ from $\mathcal{T}$ that $(\psi(w),\alpha(w),\rho(w))\in S^{\prime}$.

Downset for $S^{\prime}$. It is immediate from the definition of $S^{\prime}$ that $S^{\prime}=\mathord{\downarrow_{R}}S^{\prime}$ (the closure under downset is built-in).

Multiplication for $S^{\prime}$. Consider $(g_{1},s_{1},r_{1}),(g_{2},s_{2},r_{2})\in S^{\prime}$, we show that $(g_{1}g_{2},s_{1}s_{2},r_{1}r_{2})\in S^{\prime}$. By definition of $S^{\prime}$, we have $(q_{1},T_{1}),(q_{2},T_{2})\in R\times 2^{M\times R}$ such that $(g_{i},q_{i},T_{i})\in\mathcal{T}$ and $(s_{i},r_{i})\in\mathord{\downarrow_{R}}T_{i}$ for $i=1,2$. Since we already established that $\mathcal{T}$ is closed under multiplication, we have $(g_{1}q_{2},q_{1}q_{2},T_{1}T_{2})\in\mathcal{T}$. Moreover, it is clear that $(s_{1}q_{2},r_{1}r_{2})\in(\mathord{\downarrow_{R}}T_{1})(\mathord{\downarrow_{R}}T_{2})\subseteq\mathord{\downarrow_{R}}(T_{1}T_{2})$. Hence, we get as desired that $(g_{1}g_{2},s_{1}s_{2},r_{1}r_{2})\in S^{\prime}$ by definition of $S^{\prime}$.

$\mathit{PBPol}(\mathcal{G})$-closure. Consider an element $(1_{G},r,U)\in\mathcal{T}$ and a pair of idempotents $(e,f)\in\mathord{\downarrow_{R}}U$, we have to show that,

[TABLE]

By definition of $\mathcal{T}$, there are two cases. First, it $(1_{G},r,U)=(1_{G},1_{R},S)$, then $(e,f)\in\mathord{\downarrow_{R}}S$ and $(1_{G},e,f\cdot(1_{R}+r)\cdot f)=(1_{G},e,f)$. It is now immediate by definition of $S^{\prime}$ from $\mathcal{T}$ that $(1_{G},e,f)\in S^{\prime}$.

Otherwise, we have $(r,U)\in\eta_{\alpha,\rho,S}(\psi^{-1}(1_{G}))$. Recall that $\psi$ recognizes $H$ and $\varepsilon\in H$. Thus, $\psi^{-1}(1_{G})\subseteq H$ which yields $(r,U)\in\eta_{\alpha,\rho,S}(H)$. Since $H$ is a $\mathcal{C}$-optimal $\varepsilon$-approximation for $\eta_{\alpha,\rho,S}$ by definition, it follows that $(r,U)\in\mathbbm{i}_{\mathcal{C}}[\eta_{\alpha,\rho,S}]$. Thus, since $S$ is $\mathit{PBPol}(\mathcal{C})$-complete by definition, $(e,f)\in\mathord{\downarrow_{R}}U$ implies that $(e,f\cdot(1_{R}+r)\cdot f)\in S$. Since $(1_{G},1_{R},S)\in\mathcal{T}$, this implies that $(1_{G},e,f\cdot(1_{R}+r)\cdot f)\in S^{\prime}$ by definition of $S^{\prime}$ from $\mathcal{T}$, finishing the proof. ∎