Separation for dot-depth two

The dot-depth hierarchy of Brzozowski and Cohen is a classification of all first-order definable languages. It rose to prominence following the work of Thomas, who established an exact correspondence with the quantifier alternation hierarchy of first-order logic: each level contains languages that can be defined with a prescribed number of quantifier blocks. One of the most famous open problems in automata theory is to obtain membership algorithms for all levels in this hierarchy.


Introduction
Concatenation hierarchies. Many fundamental problems about regular languages raised in the 70s [Pin17b] led to considerable advances, not only in automata theory but also in logic and algebra, thanks to the discovery of deep connections between these areas. Even if some of these questions are now well understood, a few others remain wide open, despite a wealth 24:6

Thomas Place and Marc Zeitoun
Vol. 17:3 Covering. Our second problem is more general and was introduced in [PZ18a]. Given a language L, a cover of L is a finite set of languages K such that: We speak of universal cover to mean a cover of A * . Moreover, if C is a class, a C-cover of L is a cover K of L such that all K ∈ K belong to C.
Covering takes as input a language L and a finite set of languages L. A cover K of L is separating for L if for every K ∈ K, there exists L ∈ L that satisfies K ∩ L = ∅. In other words, no language of K may intersect all languages of L. Figure 1 shows two covers of a language L: {K 1 , K 2 } (on the left of the picture) and {K 1 , K 2 } (on the right). Let L = {L 1 , L 2 }. The cover {K 1 , K 2 } of L is separating for L, since K 1 ∩ L 2 = ∅ and K 2 ∩ L 1 = ∅. On the other hand, the cover {K 1 , K 2 } is not separating for L, as K 1 intersects both L 1 and L 2 . Figure 1: Two covers of L. The left one is separating for {L 1 , L 2 } and the right one is not.
Finally, given a class C, we say that the pair (L , L) is C-coverable when there exists a C-cover of L which is separating for L.
The C-covering problem is now defined as follows: INPUT: A regular language L and a finite set of regular languages L. OUTPUT: Is (L, L) C-coverable?
Note that in the covering problem, we are interested in the existence of covers which are separating (but notice that we do not keep this precision in the name of the problem itself, to lighten the terminology). It is straightforward to prove that covering generalizes separation provided that the class is a lattice, as stated in the following lemma (see Theorem 3.5 in [PZ18a] for the proof).
Lemma 2.3. Let D be a lattice and let L 1 , L 2 be two languages. Then L 1 is D-separable from L 2 if and only if (L 1 , {L 2 }) is D-coverable.
In the paper, we shall not work with covering directly. Instead, we use a framework introduced in [PZ18a], which is designed to formulate and handle these problems in a more convenient manner. We recall this framework in Section 4. 2.3. Finite lattices. In the paper, we work with classes built from an arbitrary finite lattice (i.e., one that contains finitely many languages) using generic two operations: polynomial closure and Boolean closure (see Section 3). Let us present standard results about such classes.
Canonical preorder relations. Consider a finite lattice D. It is classical to associate a canonical preorder relation over A * to D. Given w, w ∈ A * , we write w D w if and only if the following holds: For all L ∈ D, w ∈ L ⇒ w ∈ L.
It is immediate from the definition that D is transitive and reflexive, making it a preorder.
Example 2.4. Consider the class D consisting of all unions of intersections of languages of the form A * aA * , for a ∈ A. By definition, this class is a finite lattice. The associated preorder D is defined by w D w if and only if every letter occurring in w also occurs in w .
We shall use several results about the relation D . We omit the proofs, which are simple and available in [Pla18].
The first lemma is where we use the hypothesis that D is finite. We say that a language L ⊆ A * is an upper set (for D ) when for any two words u, v ∈ A * , if u ∈ L and u D v, then v ∈ L. Furthermore, given u ∈ A * , we let ↑ D u ⊆ A * be the least upper set containing u: Lemma 2.5. Let D be a finite lattice. Then, for any L ⊆ A * , we have L ∈ D if and only if L is an upper set for D . In particular, D has finitely many upper sets.
Let us complete these definitions with a few additional useful results. First, as we observed for AT in Example 2.7, when the finite lattice is actually a Boolean algebra, it turns out that its canonical preorder is an equivalence relation. If C is such a Boolean algebra, we shall denote this equivalence relation by ∼ C (instead of C ).
Lemma 2.6. Let C be a finite Boolean algebra. Then, for any alphabet A, the canonical preorder C is an equivalence relation ∼ C , which admits the following direct definition: w ∼ C w if and only if for all L ∈ C, w ∈ L ⇔ w ∈ L.
Thus, for any L ⊆ A * , we have L ∈ C if and only if L is a union of ∼ C -classes. In particular, ∼ C has finite index.
Example 2.7. Consider the class AT of all Boolean combinations of languages A * aA * , for a ∈ A ("AT" stands for "alphabet testable": L ∈ AT if and only if membership of a word w in L depends only on the letters occurring in w). Clearly, AT is a finite Boolean algebra-it is the Boolean closure of the class from Example 2.4 (see Section 3.1 for the definition of Boolean closure). In that case, AT is an equivalence relation ∼ AT : w ∼ AT w if and only if w and w have the same alphabet (i.e., contain the same set of letters).
Another important and useful property is that when D is quotient-closed, the canonical preorder D is compatible with word concatenation.
Lemma 2.8. A finite lattice D is quotient-closed if and only if its associated canonical preorder D is compatible with word concatenation. That is, for any words u, v, u , v , assuming additionally that C is closed under complement. The result was then extended to quotient-closed lattices by Pin [Pin13] (see also [PZ18b]).
In the paper, we consider classes of the form Bool(P ol(C)) built by applying polynomial closure and Boolean closure successively to some arbitrary quotient-closed Boolean algebra C. For the sake of avoiding clutter, we shall write BP ol(C) for Bool(P ol(C)). Note that by Lemma 3.1 and Theorem 3.2, we have the following corollary.
Corollary 3.3. Let D be a quotient-closed lattice. Then, BP ol(D) is a quotient-closed Boolean algebra.
A key remark is that in general, classes built with Boolean closure (such as BP ol(C)) are not closed under concatenation. This contrasts with polynomial closure in which closure under (marked) concatenation holds by definition. This is an issue, as most of our techniques designed for handling separation and covering rely heavily on concatenation. We cope with this problem by using the following weak concatenation principle, which holds for any class that is the Boolean closure of another class which is itself closed under concatenation.
Lemma 3.4. Let D be a lattice closed under concatenation. Consider L, L ∈ D and let K, K be Bool(D)-covers of L and L respectively. Then, there exists a Bool(D)-cover H of LL such that for every H ∈ H, we have H ⊆ KK for some K ∈ K and K ∈ K .
Proof. Every language in K ∪ K is a Boolean combination of languages in D and L, L ∈ D. Therefore, there exists a finite lattice G ⊆ D satisfying the two following properties: (1) L, L ∈ G and, (2) every language K ∈ K ∪ K belongs to Bool(G). We define F as the least lattice such that HH ∈ F for every H, H ∈ G. Since G is finite, so is F. Moreover, we know that G ⊆ D and D is a lattice closed under concatenation. Therefore, F ⊆ D. It follows that Bool(F) is a finite Boolean algebra such that Bool(F) ⊆ Bool(D).
Consider the canonical equivalence ∼ Bool(F ) associated to Bool(F) (it relates words that belong to the same languages of Bool(F), see Section 2). Since L, L ∈ G, it is immediate by definition of F that we have LL ∈ F ⊆ Bool(F). Therefore, Lemma 2.6 implies that LL is a union of ∼ Bool(F ) -classes. We let H be the set consisting of all these ∼ Bool(F ) -classes. By definition, H is a Bool(F)-cover of LL and therefore a Bool(D)-cover as well since Bool(F) ⊆ Bool(D). It remains to prove that for every H ∈ H, we have H ⊆ KK for some K ∈ K and K ∈ K . We fix H ∈ H for the proof. We use the following fact.
Fact 3.5. Consider a finite language G ⊆ H. Then, there exists K ∈ K and K ∈ K such that G ⊆ KK .
Let us first admit Fact 3.5 and apply it to finish the main proof. For every n ∈ N, we let G n ⊆ H be the finite language containing all words of length at most n in H. Clearly, we have, H = n∈N G n and G n ⊆ G n+1 for all n ∈ N.
For every n ∈ N, Fact 3.5 yields K n ∈ K and K n ∈ K such that G n ⊆ K n K n . Since K and K are finite sets, there exist K ∈ K and K ∈ K such that K n = K and K n = K for 24:10

Thomas Place and Marc Zeitoun
Vol. 17:3 infinitely many n. Since G n ⊆ G n+1 for all n, it follows that G n ⊆ KK for every n ∈ N. Finally, since H = n∈N G n , this implies H ⊆ KK , finishing the proof.
It remains to prove Fact 3.5. Consider a finite language G ⊆ H and let G = {w 1 , . . . , w n }. We exhibit K ∈ K and K ∈ K such that G ⊆ KK .
By definition, H is a ∼ Bool(F ) -class included in LL . This implies that w 1 , . . . , w n ∈ LL and w 1 ∼ Bool(F ) · · · ∼ Bool(F ) w n . Using these equivalences, we first prove the following claim which involves the canonical preorder G associated to the finite lattice G.
Claim. For every u, v ∈ A * such that w n = uv, there exist u 1 , . . . , u n , v 1 , . . . , v n ∈ A * such that Proof. We prove the existence of u 1 , v 1 ∈ A * such that w 1 = u 1 v 1 , u G u 1 and v G v 1 using the hypothesis that w n = uv and w n ∼ Bool(F ) w 1 . One may then iterate the argument to build u 2 , . . . , u n ∈ A * and v 2 , . . . , v n ∈ A * , using the fact that w 1 ∼ Bool(F ) · · · ∼ Bool(F ) w n .
Consider the languages U = ↑ G u and V = ↑ G v (the upper sets of u and v for G ). By Lemma 2.5, we have U, V ∈ G. Therefore, we have U V ∈ F by definition of F. Clearly, w n = uv ∈ U V . Therefore, w n ∼ Bool(F ) w 1 implies that w 1 ∈ U V . This yields a decomposition w 1 = u 1 v 1 with u 1 ∈ U and v 1 ∈ V . By definition of U, V , this implies u G u 1 and v G v 1 , finishing the proof.
Since w n ∈ LL , it admits at least one decomposition w n = uv with u ∈ L and v ∈ L . Therefore, we may apply the claim: there exist u 1,1 , . . . , u n,1 , v 1,1 , . . . , v n,1 ∈ A * such that w i = u i,1 v i,1 for every i ≤ n, u G u 1,1 G · · · G u n,1 and v G v 1,1 G · · · G v n,1 . Since w n = u n,1 v n,1 , one can repeat the argument any number of times, say m, to obtain words u i,j , v i,j for 1 i n and 1 j m, such that for all such i, j, we have w i = u i,j v i,j , and: u G u 1,1 G · · · G u n,1 and v G v 1,1 G · · · G v n,1 , u n,1 G u 1,2 G · · · G u n,2 and v n,1 G v 1,2 G · · · G v n,2 , · · · u n,m−1 G u 1,m G · · · G u n,m and v n,m−1 G v 1,m G · · · G v n,m .
Moreover, since w 1 is a finite word, it admits finitely many decompositions w 1 = u 1 v 1 with u 1 , v 1 ∈ A * . Therefore, the pigeonhole principle yields two integers j < k such that u 1,j = u 1,k (and therefore v 1,j = v 1,k ), which gives words u 1 , . . . , u n , v 1 , . . . , v n ∈ A * such that w i = u i v i for every i ≤ n and, One may verify from the definitions that for every x, y ∈ A * , x G y and y G x imply that x ∼ Bool(G) y. Therefore, the above implies that, Since u ∈ L, v ∈ L and L, L ∈ G by definition of G, we get that u 1 ∈ L and v 1 ∈ L by definition of G . Therefore, since K and K are covers of L and L respectively, there exist K ∈ K and K ∈ K such that u 1 ∈ K and v 1 ∈ K . By definition of G, both K and K belong to Bool(G). It follows from (3.1) that u 1 , . . . , u n ∈ K and v 1 , . . . , v n ∈ K . Altogether, we obtain that G = {u 1 v 1 , . . . , u n v n } ⊆ KK , finishing the proof. Finally, we shall need the following variant of Lemma 3.4, which considers marked concatenation instead of standard concatenation. The proof is identical to the one of Lemma 3.4 and left to the reader.
Lemma 3.6. Let D be a lattice closed under marked concatenation. Consider L, L ∈ D and a ∈ A, and let K, K be Bool(D)-covers of L and L respectively. There exists a Bool(D)cover H of LaL such that for every H ∈ H, we have H ⊆ KaK for some K ∈ K and K ∈ K .
3.2. Main theorem. We may now state the main theorem of the paper: whenever C is a finite quotient-closed Boolean algebra, BP ol(C)-separation and BP ol(C)-covering are both decidable.
Theorem 3.7. Let C be a finite quotient-closed Boolean algebra. Then, separation and covering are decidable for BP ol(C).
Before we detail the applications of Theorem 3.7, let us make an important observation. This result completes an earlier one presented in [Pla18] which applies to classes of the form P ol(C) and P ol(BP ol(C)) (when C is a finite quotient-closed Boolean algebra). Let us recall this result.
Theorem 3.8 [Pla18]. Let C be a finite quotient-closed Boolean algebra. Then, separation and covering are decidable for P ol(C) and P ol(BP ol(C)).
An important point is that while BP ol(C) is an intermediary class between P ol(C) and P ol(BP ol(C)), the proof of Theorem 3.7 involves ideas which are very different from those used in [Pla18] to prove Theorem 3.8. This is not surprising and we already mentioned the reason above: unlike P ol(C) and P ol(BP ol(C)), the class BP ol(C) is not closed under concatenation in general. This difference is significant, since most of the techniques we have for handling covering rely heavily on concatenation. In practice, this means that Boolean closure is harder to handle than polynomial closure, at least with such techniques.
However, we do reuse a result of [Pla18] to prove Theorem 3.7. More precisely, it turns out that the arguments for handling BP ol(C) (in this paper) and P ol(BP ol(C)) (in [Pla18]) both exploit the same subresult for the simpler class P ol(C) (albeit in very different ways). This subresult is stronger than the decidability of P ol(C)-covering and is proved in [Pla18]. We recall it in Section 7. However, it is important to keep in mind that from this preliminary result for P ol(C), the arguments for BP ol(C) and P ol(BP ol(C)) build in orthogonal directions.
Remark 3.9. This discussion might seem surprising. Indeed, by definition, P ol(BP ol(C)) is built from BP ol(C) using polynomial closure. Hence, intuition suggests that one needs some knowledge about the latter to handle the former. However, this is not the case as Boolean closure can be bypassed in the definition of P ol(BP ol(C)). Specifically, one may prove that P ol(BP ol(C)) = P ol(co-Pol (C)) where co-Pol (C) is the class containing all complements of languages in P ol(C). This is exactly how the class P ol(BP ol(C)) is handled in [Pla18].
The remaining sections of the paper are devoted to proving Theorem 3.7. We rely on a framework which was designed in [PZ18a] for the specific purpose of handling the covering problem. We recall it in Section 4. The algorithm for BP ol(C)-covering is presented in Section 5. The remaining sections are then devoted to the correction proof of this algorithm.

Thomas Place and Marc Zeitoun
Vol. 17:3 However, let us first conclude the current section by detailing the important applications of Theorem 3.7.
3.3. Applications of the main theorem. Classes of the form BP ol(C) are important: they are involved in natural hierarchies of classes of languages, called concatenation hierarchies. Let us briefly recall what they are (we refer the reader to [PZ18b] for a detailed presentation).
A particular concatenation hierarchy depends on a single parameter: an arbitrary quotientclosed Boolean algebra of regular languages C, called its basis. Once the basis is chosen, the construction is uniform. Languages are classified into levels of two kinds: full levels (denoted by 0, 1, 2,. . . ) and half levels (denoted by 1/2, 3/2, 5/2,. . . ): • Level 0 is the basis (i.e., our parameter class C).
• Each half level n + 1 2 , for n ∈ N, is the polynomial closure of the previous full level, i.e., of level n.
• Each full level n + 1, for n ∈ N, is the Boolean closure of the previous half level, i.e., of level n + 1 2 . The generic process is depicted in the following figure. Hence, a reformulation of Theorem 3.7 is that for any concatenation hierarchy whose basis is finite, separation is decidable for level one. There are two prominent examples of finitely based hierarchies: • The Straubing-Thérien hierarchy [Str81,Thé81], whose basis is the class {∅, A * }.
• The dot-depth hierarchy of Brzozowski and Cohen [BC71], whose basis is the class {∅, {ε}, A + , A * }. Consequently, Theorem 3.7 implies that separation and covering are decidable for level one in these two hierarchies. These are not new results. By definition, level one in the Straubing-Thérien hierarchy is exactly the class of piecewise testable languages. For this class, separation has been solved in [CMM13,PvRZ13] and covering has been solved in [PZ18a]. For dot-depth one, the decidability of covering and separation was originally obtained indirectly. Indeed, it is known [PZ20] that separation and covering for any level in the dot-depth hierarchy reduce to the same problem for the corresponding level in the Straubing-Thérien hierarchy. Therefore, while the decidability of separation and covering for dot-depth one is not a new result, an advantage of Theorem 3.7 is that we obtain a new direct proof of this result.
However, these are not the main applications of Theorem 3.7. It turns out that the theorem also applies to level two in the Straubing-Thérien hierarchy. Indeed, it is known that this level is also level one in another finitely based concatenation hierarchy. More precisely, recall the class AT presented in Example 2.7: it consists of all Boolean combinations of languages A * aA * , for some a ∈ A. It is straightforward to verify that AT is a finite quotientclosed Boolean algebra. The following theorem was shown in [PS85] (see also [PZ18a] for a recent proof). In view of Theorem 3.10, we obtain the following immediate corollary of Theorem 3.7. Corollary 3.11. Separation and covering are decidable for level two in the Straubing-Thérien hierarchy.
Finally, this result may be lifted to dot-depth two using again the generic transfer theorem proved in [PZ20]. Hence, we obtain the following additional corollary.
Corollary 3.12. Separation and covering are decidable for dot-depth two.
Remark 3.13. Logical characterizations of these two hierarchies are known (see [Tho82] for the dot-depth hierarchy and [PP86] for the Straubing-Thérien hierarchy). Each of them corresponds to a quantifier alternation hierarchy within a particular variant of first-order logic over words. The two variants differ by the set of predicates which are allowed in sentences (they have the same overall expressive power, but this changes the levels in their respective quantifier alternation hierarchies). We refer the reader to [PZ18b] for details and a recent proof of these results.
In particular, level two in the Straubing-Thérien hierarchy corresponds to a logic denoted by BΣ 2 (<) and dot-depth two corresponds to another logic denoted by BΣ 2 (<, +1). Hence, our results also imply that covering and separation are decidable for these two logics.

Framework: rating maps and optimal covers
In this section, we present the framework which we use to formulate our covering algorithm for BP ol(C) (when C is a finite quotient-closed Boolean algebra) announced in Theorem 3.7. The framework itself was designed and applied to several specific classes in [PZ18a]. Moreover, it was also used in [Pla18] to formulate algorithms for P ol(C)-and P ol(BP ol(C))-covering. Here, we recall the part of this framework that we shall actually need in the paper. We refer the reader to [PZ18a] for a complete and detailed presentation.
Let D be a lattice. In D-covering, the input is a pair (L, L) where L is a regular language and L a finite set of regular languages: we have to decide whether there exists a D-cover of L which is separating for L. We design a two-step approach for tackling this problem.
The first step is to transform this decision problem into the following computational problem. Given an input (L, L), we compute a D-cover of L, which is optimal with respect to L in the following sense: for every subset H of L, this optimal cover is separating for H if and only if (L, H) is D-coverable. Thus, this single object encapsulates all answers to the D-covering problem for inputs of the form (L, H), with H ⊆ L. Such an optimal cover always exists when D is a lattice. The set L can be viewed as a parameter constraining this optimal cover, or dually, as a measure of the quality of this cover.
The second step in the approach of [PZ18a] consists in replacing the set L by a (more general) algebraic object called a rating map. Intuitively, rating maps play the same role as the set of languages L from the above paragraph. Thus, they are also designed to measure the quality of D-covers. Given a rating map ρ and a language L, we use ρ to rank the existing D-covers of L.
We are then able to reformulate D-covering with these notions. Instead of deciding whether (L, L) is D-coverable, we compute an optimal D-cover for L for a rating map ρ that we build from L. An advantage of this approach is that it yields elegant formulations for the covering algorithms which are formulated with it. We refer the reader to [PZ18a] and [Pla18] for examples (in addition to BP ol(D), which is presented in this paper). Another important 24:14

Thomas Place and Marc Zeitoun
Vol. 17:3 motivation for using this framework is that in order to handle BP ol(D), we require a result for the simpler class P ol(D) which is stronger than the decidability of covering (this result is proved in [Pla18]). The framework of [PZ18a] is designed to formulate this result.
We start by defining rating maps. Then, we explain how they are used to measure the quality of a cover and define optimal covers. Finally, we connect these notions to the covering problem. Let us point out that several statements presented here are without proof. We refer the reader to [PZ18a] for these proofs.
4.1. Rating maps. Rating maps involve commutative and idempotent monoids. We shall 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 ∈ R, we have r + r = r. Observe that for every commutative and idempotent monoid (R, +), we may define a canonical ordering ≤ over R: For all r, s ∈ R, r ≤ s when r + s = s.
It is straightforward to verify that ≤ is a partial order which is compatible with addition. Moreover, we have the following fact which is immediate from the definitions.
Example 4.2. For any set E, it is immediate that (2 E , ∪) is an idempotent and commutative monoid. The neutral element is ∅. Moreover, the canonical ordering is set inclusion.
When manipulating the subsets of a commutative and idempotent monoid (R, +) we shall often need to apply a downset operation. Given S ⊆ R, we write ↓ R S for the set, ↓ R S = {r ∈ R | r ≤ s for some s ∈ S}.
We extend this notation to Cartesian products of arbitrary sets with R. Given some set X and a subset S ⊆ X × R, we write ↓ R S for the set, We may now define rating maps. A rating map is a morphism ρ : (2 A * , ∪) → (R, +) where (R, +) is a finite idempotent and commutative monoid called the rating set of ρ. That is, ρ is a map from 2 A * to R satisfying the following properties: For the sake of improved readability, when applying a rating map ρ to a singleton set K = {w}, we shall write ρ(w) for ρ({w}). Additionally, we write ρ * : A * → R for the restriction of ρ to A * : for every w ∈ A * , we have ρ * (w) = ρ(w) (this notation allows us to write ρ −1 * (r) ⊆ A * for the language of all words w ∈ A * such that ρ(w) = r). Most of the statements involved in our framework make sense for arbitrary rating maps. However, we shall often have to work with special rating maps which satisfy additional properties. We present them now.
Nice rating maps. A rating map ρ : 2 A * → R is nice when, for every language K ⊆ A * , there exist finitely many words w 1 , . . . , w n ∈ K such that ρ(K) = ρ(w 1 ) + · · · + ρ(w k ). Observe that in this case, ρ is characterized by the canonical map ρ * : A * → R. Indeed, for every language K, we may consider the sum of all elements ρ(w) for w ∈ K: while it may be infinite, it boils down to a finite one since R is commutative and idempotent. The hypothesis that ρ is nice implies that ρ(K) is equal to this sum.
Multiplicative rating maps. A rating map ρ : 2 A * → R is multiplicative when its rating set R has more structure: it needs to be an idempotent semiring. Moreover, ρ 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, +, ·) where R is a set and "+" and "·" are two binary operations called addition and multiplication, such that the following axioms are satisfied: is a monoid (the neutral element is denoted by 1 R ).
• Multiplication distributes over addition: r·(s+t) = (r·s)+(r·t) and (r+s)·t = (r·t)+(s·t) for all r, s, t ∈ R. • The neutral element of (R, +) is a zero for (R, ·): 0 R · r = r · 0 R = 0 R for all r ∈ R. We say that a semiring R is idempotent when r + r = r for every r ∈ R, i.e., when the additive monoid (R, +) is idempotent (on the other hand, there is no additional constraint on the multiplicative monoid (R, ·)).

Example 4.3.
A key example of an infinite idempotent semiring is the set 2 A * of all languages over A. Union is the addition (with ∅ as neutral element) and language concatenation is the multiplication (with {ε} as neutral element).
Clearly, any finite idempotent semiring (R, +, ·) is in particular a rating set: (R, +) is an idempotent and commutative monoid. In particular, one may verify that the canonical ordering "≤" on R, is compatible with multiplication as well.
We may now define multiplicative rating maps: as expected they are semiring morphisms. Let ρ : 2 A * → R be a rating map. By definition, this means that the rating set (R, +) is a finite idempotent commutative monoid and that ρ is a monoid morphism from (2 A * , ∪) to (R, +). We say that ρ is multiplicative when the rating set R is equipped with a second binary operation "·" such that (R, +, ·) is an idempotent semiring and ρ is also a monoid morphism from (2 A * , ·) to (R, ·). In other words, the two following additional axioms hold: Remark 4.4. An important point is that the rating maps which are both nice and multiplicative are finitely representable. As we explained above, a nice rating map ρ : 2 A * → R is characterized by the canonical map ρ * : A * → R. Moreover, when ρ is multiplicative as well, ρ * is finitely representable: it is a morphism from A * into the finite monoid (R, ·). Thus, we may consider algorithms taking nice multiplicative rating maps as input. Let us point out that the rating maps which are not nice and multiplicative remain important. We often deal with them in our proofs.
4.2. Imprints and optimal covers. We now explain how we use rating maps to measure the quality of covers. This involves an additional notion: "imprints". Consider a rating map 24:16

Thomas Place and Marc Zeitoun
Vol. 17:3 ρ : 2 A * → R. For any finite set of languages K, the ρ-imprint of K, denoted by I[ρ](K), is the following subset of R: When using this notion, we shall have some language L ⊆ A * in hand: our goal is to find the "best possible" cover K of L. Intuitively, ρ-imprints measure the "quality" of candidate covers K (the smaller the ρ-imprint, the better the quality).
This leads to the notion of optimality. Let D be an arbitrary lattice. Given a language L, an optimal D-cover of L for ρ is a D-cover of L which has the smallest possible ρ-imprint (with respect to inclusion). That is, K is an optimal D-cover of L for ρ if and only if, Furthermore, in the special case when L = A * , we speak of optimal universal D-cover for ρ.
If D is a lattice, one can show that there always exists at least one optimal D-cover of L for ρ (see [PZ18a,Lemma 4.15]). In general, there are actually infinitely many of them.
Lemma 4.5. Let D be a lattice. Then, for any rating map ρ : 2 A * → R and any language L ⊆ A * , there exists an optimal D-cover of L for ρ.
It is important to note that the proof of Lemma 4.5 is non-constructive: given L and ρ : 2 A * → R, computing an actual optimal D-cover of L for ρ is a difficult problem in general. As we shall see below, getting such an algorithm (in the special case when ρ is nice and multiplicative) yields a procedure for D-covering. Before we can establish this connection precisely, we require a key observation about optimal D-covers.
Optimal imprints. By definition, given a language L, all optimal D-covers of L for ρ have the same ρ-imprint. Hence, this unique ρ-imprint is a canonical object for D, L and ρ. We call it the D-optimal ρ-imprint on L and we denote it by I D [L, ρ]: Additionally, in the particular case when L = A * , we shall speak of D-optimal universal ρ-imprint and write We complete these definitions with a few properties of optimal imprints. We start with a straightforward fact which compares the optimal imprints on languages which are comparable with inclusion. The proof is available in [PZ18a, Fact 4.17].
Fact 4.6. Consider a rating map ρ : 2 A * → R and a lattice D. Let H, L be two languages such that H ⊆ L. Then, More precisely, the following fact connects optimal imprints with union of languages. We let K H and K L be optimal D-covers of H and L respectively (for ρ). Clearly, , finishing the proof.

4.3.
Connection with the covering problem. Finally, we explain how rating maps are used for handling the covering problem. Given a lattice D, it turns out that the D-covering problem reduces to another problem whose input is a nice multiplicative rating map. Let us point out that two reductions of this kind are presented in [PZ18a]. The first one is simpler but restricted to Boolean algebras. On the other hand, the second one applies to any lattice, but requires working with more involved objects.
In the paper, we investigate classes of the form BP ol(D), which are Boolean algebras. Hence, we shall mostly work with the first variant whose statement is as follows.
Proposition 4.8 [PZ18a]. Let C be a Boolean algebra. Assume that there exists an algorithm for the following computational problem: Input: A nice multiplicative rating map ρ : Then, C-covering is decidable.
Proof sketch. The proof builds on several simple steps. The first one is the observation that when C is a Boolean algebra, deciding whether a pair (L, L) is C-coverable reduces to the case where L = A * . Indeed, one may check that (L, L) is C-coverable if and only if so is (A * , L ∪ {L}). We are left to show that one can decide whether a pair (A * , L) is C-coverable.
Second, observe that (2 L , ∪, ∅) is a rating set. One may verify that the function ρ L : 2 A * → 2 L defined by ρ L (K) = {L ∈ L | K ∩ L = ∅} is a nice rating map. Furthermore, for any subset H of L, one may prove that The third step is to reduce the computation of I C [ρ L ] to that of I C [ρ L ], where ρ L is a multiplicative nice rating map that we can build from ρ L . By the hypothesis of the statement, this last set is computable. We refer the reader to [PZ18a, Theorem 5.21] for more details.
Additionally, we shall need in Section 6 to apply a theorem of [Pla18] for classes of the form P ol(C) (which are lattices but not Boolean algebras) as a sub-result. Thus, we also recall the terminology associated to the generalized reduction which holds for arbitrary lattices, since we need it to state this theorem.
When working with an arbitrary lattice, one needs to consider slightly more involved objects. Given a lattice D, a map α : A * → M into a finite set M (in practice, α will be a monoid morphism but this is not required for the definition) and a rating map ρ : 2 A * → R, we write P α D [ρ] for the following set, The following statement is [Pla18, Proposition 5.18] (see also [PZ18a, Proposition 7.2]).
Proposition 4.9. Consider a lattice D and some finite quotient-closed Boolean algebra C.
Assume that there exists an algorithm for the following computational problem: Input: A C-compatible morphism α : A * → M and a nice multiplicative rating map ρ : 2 A * → R. Output: Compute the α-pointed D-optimal ρ-imprint, P α D [ρ]. Then, D-covering is decidable.

Characterization of BP ol(C)-optimal imprints
We present a generic characterization of BP ol(C)-optimal imprints which holds when C is a finite quotient-closed Boolean algebra. For the sake of avoiding clutter, we assume that the finite quotient-closed Boolean algebra C is fixed for the whole section.
Given a nice multiplicative rating map ρ : 2 A * → R, we want to characterize the set An important point is that we do not work directly with this set. Instead, we characterize the family of all sets Note that this family of sets record more information than just the set I BP ol(C) [ρ]. Indeed, by Fact 4.7, we have, For the sake of convenience, we shall encode this family as a set of pairs in (A * /∼ C ) × R. Given a multiplicative rating map ρ : 2 A * → R, we define: as the greatest subset of R satisfying specific properties. From the statement, it is straightforward to obtain a greatest fixpoint procedure for computing P C BP ol(C) [ρ] from ρ. In turns, this allows to compute I BP ol(C) [ρ] using the above equality. By Proposition 4.8, this yields an algorithm for BP ol(C)-covering, thus proving our main result: Theorem 3.7.
Remark 5.1. This characterization is rather unique among the results that have been obtained for other classes in [PZ18a] and [Pla18]. Typically, optimal imprints are characterized as least subsets, not greatest ones.
Notation. In our statements, we shall frequently manipulate subsets of (A * /∼ C ) × R. When doing so, the following notation will be convenient. Given Given a multiplicative rating map ρ : 2 A * → R, we define a notion of BP ol(C)-saturated subset of (A * /∼ C ) × R (for ρ). Our theorem then states that when ρ is nice, the greatest such subset is exactly P C BP ol(C) [ρ]. Remark 5.2. The definition of BP ol(C)-saturated sets makes sense regardless of whether ρ is nice. However, we need this hypothesis for the greatest one to be P C BP ol(C) [ρ]. The definition is based on an intermediary notion. With every set S ⊆ (A * /∼ C ) × R, we associate another set R ρ For the definition, we need to recall a few properties. Given a word w ∈ A * , we denote its ∼ C class by [w] C . Moreover, since C is closed under quotients, Lemma 2.8 yields that the equivalence ∼ C is a congruence. We denote by "•" the multiplication of ∼ C -classes in the monoid A * /∼ C . Additionally, since R is a semiring, 2 R is one as well for union as addition and the natural multiplication lifted from the one of R (for U, V ∈ 2 R , U V = {qr | q ∈ U and r ∈ V }).
We may now present our definition. Consider a set S ⊆ (A * /∼ C ) × R. We define R ρ S as the least subset of (A * /∼ C ) × R × 2 R (with respect to inclusion) which satisfies the following properties: We say that S is BP ol(C)-saturated for ρ if the following property holds: for every (D, r) ∈ S, there exist r 1 , . . . , r k ∈ R such that, r ≤ r 1 + · · · + r k and (D, We now state our characterization of BP ol(C)-optimal imprints in the following theorem.
Given as input a nice multiplicative rating map ρ : 2 A * → R, Theorem 5.3 yields an algorithm to compute I BP ol(C) [ρ]. Indeed, computing the greatest BP ol(C)-saturated subset of (A * /∼ C ) × R is achieved with a greatest fixpoint algorithm. One starts from the set S 0 = (A * /∼ C ) × R and computes a sequence S 0 ⊇ S 1 ⊇ S 2 ⊇ · · · of subsets. For every n ∈ N, S n+1 is the set of all pairs (D, r) ∈ S n satisfying (5.1) for S = S n . That is, there exist r 1 , . . . , r k ∈ R such that, r ≤ r 1 + · · · + r k and (D, r i , {r 1 + · · · + r k }) ∈ R ρ Sn for every i ≤ k Clearly, S n+1 ⊆ S n for every n ∈ N. Therefore, the computation eventually reaches a fixpoint which is the greatest BP ol(C)-saturated subset of (A * /∼ C ) × R by definition. Let us point out that the computation of S n+1 from S n involves computing R ρ Sn , which is achieved with a least fixpoint procedure by definition.
Altogether, it follows that Theorem 5.3 yields an algorithm for computing P C BP ol(C) [ρ] (and therefore, I BP ol(C) [ρ] as well by Fact 4.7) which alternates between a greatest fixpoint and a least fixpoint.
Remark 5.4. The hypothesis that ρ is nice in Theorem 5.3 is mandatory: the result fails otherwise. This is actually apparent on the definition of BP ol(C)-saturated sets. One may verify from the definition of R ρ S that for every triple (D, q, U ) ∈ R ρ S , there exists a word w ∈ A * such that D = [w] C and q = ρ(w). Therefore, it follows from (5.1) that for every (D, r) ∈ (A * /∼ C ) × R belonging to a BP ol(C)-saturated subset, its second component r must satisfy r ≤ ρ(w 1 ) + · · · + ρ(w k ) for some words w 1 , . . . , w k ∈ A * . Intuitively, this means that BP ol(C)-saturated subsets only depend on the image of singletons. Therefore, using the notion only makes sense when ρ is characterized by these images: this is exactly the definition of nice rating maps.

Thomas Place and Marc Zeitoun
Vol. 17:3 This might seem to be a minor observation. Indeed, by Proposition 4.8, being able to compute I BP ol(C) [ρ] from a nice multiplicative rating map suffices to meet our goal: getting an algorithm for BP ol(C)-covering. Actually, it does not even make sense to speak of an algorithm which takes arbitrary multiplicative rating maps as input since we are not able to finitely represent them. However, from a theoretical point of view, the fact that we only manage to get a description of I BP ol(C) [ρ] when ρ is nice is significant. In the proof of Theorem 5.3, we use a theorem of [Pla18] as a subresult. Specifically, this theorem is a characterization of P ol(C)-optimal pointed imprints: given a C-compatible morphism α : A * → M and a multiplicative rating map τ : A key point is that this characterization does not require τ to be nice. This is crucial: in the proof of Theorem 5.3, we consider auxiliary rating maps built from ρ which need not be nice.
In summary, we are able to handle P ol(C) for all multiplicative rating maps, including those that are not nice, which is crucial in order be able to handle BP ol(C). However, at this level, we are only able to deal with nice multiplicative rating maps (the situation is actually similar for P ol(BP ol(C)) as shown in [Pla18]). This explains why the results presented in this paper and in [Pla18] cannot be lifted to higher levels in concatenation hierarchies (at least not in a straightforward manner).
We turn to the proof of Theorem 5.3. It spans the remaining four sections of the paper. Given a nice multiplicative rating map ρ : 2 A * → R, we have to show that P C BP ol(C) [ρ] is the greatest BP ol(C)-saturated subset of (A * /∼ C ) × R for ρ. The main argument involves two directions which are proved independently. They correspond to soundness and completeness of the greatest fixpoint algorithm computing P C BP ol(C) [ρ]. • The soundness argument shows that P C BP ol(C) [ρ] contains every BP ol(C)-saturated subset (this implies that the greatest fixpoint procedure only computes elements of P C BP ol(C) [ρ]). We present it in Section 8.
• The completeness argument shows that P C BP ol(C) [ρ] itself is BP ol(C)-saturated (this implies that the greatest fixpoint procedure computes all elements of P C BP ol(C) [ρ]). We present it in Section 9.
When put together, these two results yield as desired that P C BP ol(C) [ρ] is the greatest BP ol(C)saturated subset of R, proving Theorem 5.3.
However, before presenting the main argument, we require some additional material about rating maps. For both directions, we shall introduce auxiliary rating maps (built from ρ) and apply a characterization of P ol(C)-optimal imprints to them (taken from [Pla18]). These auxiliary rating maps are built using generic constructions which are not specific to Theorem 5.3. We present them in Section 6. Then, we recall the theorem characterizing P ol(C)-optimal imprints from [Pla18] in Section 7 (actually, we slightly generalize it, since we shall apply it for rating maps that are more general than the ones considered in [Pla18]).

Nesting of rating maps
In this section, we present two generic constructions. Each of them builds a new rating map out of an already existing one and a lattice D. The constructions are new: they do not appear in [PZ18a] (however, one of them generalizes and streamlines a technical construction used in [Pla18]). Remark 6.1. As announced, we shall later rely on these constructions in the proof of Theorem 5.3 (we use them in the special cases when D is either P ol(C) or BP ol(C)). However, this section is independent from Theorem 5.3: all definitions are presented in a general context.
We first present the constructions and then investigate the properties of the output rating maps they produce.
6.1. Definition. We present two constructions. The first one involves two objects: a lattice D and a rating map ρ : 2 A * → R. We build a new rating map ξ D [ρ] whose rating set is (2 R , ∪), and which associates to a language its optimal D-optimal ρ-imprint.
The fact that ξ D [ρ] is indeed a rating map is shown below in Proposition 6.2. The second construction involves an additional object: a map α : A * → M where M is some arbitrary finite set (in practice, α will be a monoid morphism, but this is not required for the definition). We build a new rating map ζ α Let us prove that these two maps are indeed rating maps. We state this result in the following proposition. We conclude that ξ D [ρ] is indeed a rating map. We turn to ζ α By Fact 4.7, this yields, is a rating map.
A crucial observation is that the rating maps ξ D [ρ] and ζ α D [ρ] are not nice in general, even when the original rating map ρ is. Let us present a counter-example.
Example 6.3. Let D be the Boolean algebra consisting of all languages which are either finite or co-finite (i.e., their complement is finite). Moreover, let T = {0, 1} and R = 2 T . We define a nice rating map ρ : 2 A * → R as follows (actually, it is simple to verify from the definition that ρ is also multiplicative). Since we are defining a nice rating map, it suffices to 24:22

Thomas Place and Marc Zeitoun
Vol. 17:3 specify the evaluation of words: for any w ∈ A * , we let ρ(w) = {0} if w has even length and ρ(w) = {1} if w has odd length. We show that the rating map ξ D [ρ] : 2 A * → 2 R is not nice.
Recall that D contains only finite and co-finite languages. Moreover, covers may only contain finitely many languages. Hence, it is immediate that if K is an optimal D-cover of A * for ρ, then there exists K ∈ K containing a word of even length and a word of odd length. Therefore ρ ( 6.2. Multiplication. If α : A * → M is a morphism into a finite monoid and ρ : 2 A * → R is a multiplicative rating map, (M, ·) is a monoid and (R, +, ·) is an idempotent semiring. We may lift the multiplication of R to 2 R in the natural way: given U, V ∈ 2 R , we let U V = {qr | q ∈ U and r ∈ V }. One may verify that (2 R , ∪, ·) is an idempotent semiring. Similarly, we may lift the componentwise multiplication on M × R to 2 M ×R and (2 M ×R , ∪, ·) is an idempotent semiring. Whenever we consider semiring structures for 2 R and 2 M ×R , these are the additions and multiplications that we shall use.
Unfortunately, even though 2 R and 2 M ×R are semirings, neither ξ D [ρ] : 2 A * → 2 R nor ζ α D [ρ] : 2 A * → 2 M ×R are multiplicative: they are not monoid morphisms for multiplication. However, it turns out that they behave almost as multiplicative rating maps when the class D satisfies appropriate properties related to closure under concatenation. We formalize this with a new notion: quasi-multiplicative rating maps.
(2) For every K 1 , K 2 ⊆ A * , we have ρ(K 1 K 2 ) = µ ρ (ρ(K 1 ) · ρ(K 2 )). For the sake of improved readability, we often abuse terminology and simply say that "the rating map ρ is quasi-multiplicative", assuming implicitly that the endomorphism µ ρ is defined and fixed. Note however that this is slightly ambiguous as there might be several endomorphisms of (R, +) satisfying the above axioms for the same rating map ρ.
We have the following useful fact about quasi-multiplicative rating maps.
Additionally, we shall need the following lemma which is a straightforward adaptation of a result proved in [PZ18a, Lemma 5.8] to quasi-multiplicative rating maps. Proof. Let q ∈ I D [H, ρ] and r ∈ I D [L, ρ]. By definition, it suffices to prove that for every D-cover K of HL, we have µ ρ (qr) ∈ I[ρ](K). Let K be a D-cover of HL, we have to find K ∈ K such that µ ρ (qr) ≤ ρ(K). We use the following claim which is based on the Myhill-Nerode theorem.
Claim. There exists a language G ∈ D which satisfies the following two properties: (1) For all u ∈ H, there exists K ∈ K such that G ⊆ u −1 K.
(2) r ≤ ρ(G) Proof of the claim. For every u ∈ H, we let Q u = {u −1 K | K ∈ K}. Clearly, Q u is a D-cover of L since K is a cover of HL and D is closed under quotients. Moreover, we know by hypothesis on D that all languages in K are regular. Therefore, it follows from the Myhill-Nerode theorem that they have finitely many quotients. Thus, while there might be infinitely many words u ∈ H, there are only finitely many distinct sets Q u . It follows that we may use finitely many intersections to build a D-cover Q of L such that for every Q ∈ Q and every u ∈ H, there exists K ∈ K satisfying Q ⊆ u −1 K. This means that all Q ∈ Q satisfy the first item in the claim, we now pick one which satisfies the second one as well.
Since r ∈ I D [L, ρ], and Q is a D-cover of L, we have r ∈ I[ρ](Q). Thus, we get G ∈ Q such that r ≤ ρ(G) by definition. This concludes the proof of the claim.
We may now finish the proof of Lemma 6.7. Let G ∈ D be as defined in the claim and consider the following set: Observe that all languages in G belong to D. Indeed, by hypothesis on D, every K ∈ K is regular. Thus, it has finitely many right quotients by the Myhill-Nerode theorem and the language v∈H Kv −1 is the intersection of finitely many quotients of languages in D. By closure under intersection and quotients, it follows that v∈G Kv −1 ∈ D. Moreover, G is a D-cover of H. Indeed, given u ∈ H, we have K ∈ K such that G ⊆ u −1 K by the first assertion in the claim. Hence, for every v ∈ G, we have u ∈ Kv −1 and we obtain that u ∈ v∈G Kv −1 , which is an element of G.
Therefore, since q ∈ I D [H, ρ] by hypothesis, we have q ∈ I[ρ](G) and we obtain G ∈ G such that q ≤ ρ(G ). Hence, since r ≤ ρ(G) by the second item in the claim, we have 24:24

Thomas Place and Marc Zeitoun
Vol. 17:3 qr ≤ ρ(G ) · ρ(G). Since ρ is quasi-multiplicative over D and G, G ∈ D, it follows from Axiom 2 in the definition of quasi-multiplicative rating maps that, Moreover, since µ ρ is an endomorphism of (R, +) and qr ≤ ρ(G ) · ρ(G), we have by Fact 4.1: Altogether, we get µ ρ (qr) ≤ ρ(G G). Finally, observe that G G ⊆ K for some K ∈ K. Indeed, if w ∈ G G, we have w = uv with u ∈ G and v ∈ G. Moreover, G = v∈G Kv −1 for some K ∈ K by definition of G. Hence, u ∈ Kv −1 which yields w = uv ∈ K. Altogether, we get that µ ρ (qr) ≤ ρ(G G) ≤ ρ(K), which concludes the proof.
We now prove that when D is a quotient-closed lattice closed under concatenation, the rating map ζ α D [ρ] is quasi-multiplicative provided that α is a morphism and ρ is already quasi-multiplicative (actually, this is also true for ξ D [ρ] but we do not need this result). This result is tailored to the situation in which we shall later use ζ α D [ρ]: D = P ol(C). Lemma 6.8. Let D be a quotient-closed lattice closed under concatenation, α : A * → M be a morphism and ρ : 2 A * → R be a quasi-multiplicative rating map. Then, ζ α D [ρ] is quasi-multiplicative for the following associated endomorphism µ ζ α D [ρ] of (2 M ×R , ∪): We already know from Proposition 6.2 that ζ α D [ρ] is a rating map. Hence, we have to prove that the axioms of quasi-multiplicative rating maps hold for the endomorphism µ ζ α of (2 M ×R , ∪) described in the lemma (it is clear from the definition that this is indeed an endomorphism). For the sake of avoiding clutter, we write µ for µ ζ α D [ρ] . We start with the first axiom. Consider T, U, V ∈ 2 M ×R . We have to show that µ(T µ(U )V ) = µ(T U V ). Assume first that (s, r) ∈ µ(T µ(U )V ). By definition of µ, this yields (s 1 , r 1 ) ∈ T , (s 2 , r 2 ) ∈ µ(U ) and (s 3 , r 3 ) ∈ V such that s = s 1 s 2 s 3 and r ≤ µ ρ (r 1 r 2 r 3 ). Since (s 2 , r 2 ) ∈ µ(U ), we have (s 2 , r 2 ) ∈ U such that r 2 ≤ µ ρ (r 2 ). It follows that r 1 r 2 r 3 ≤ r 1 µ ρ (r 2 )r 2 and since µ ρ is an endomorphism of (R, +), we obtain, r ≤ µ ρ (r 1 r 2 r 3 ) ≤ µ ρ (r 1 µ ρ (r 2 )r 2 ).
We complete Lemma 6.9 with another statement which requires that D is closed under marked concatenation (this will be the case in practice since we apply these results for D = P ol(C)). In this case as well, we use our weak concatenation principle for classes of the form Bool(D). Specifically, we apply the variant for marked concatenation (i.e. Lemma 3.6).

Characterization of P ol(C)-optimal imprints
In this section, we recall the theorem of [Pla18] for classes of the form P ol(C) (when C is a finite quotient-closed Boolean algebra). It states a characterization of P ol(C)-optimal imprints: for a C-compatible morphism α : A * → M and a multiplicative rating map ρ : 2 A * → R, P α P ol(C) [ρ] is characterized as the least subset of M × R satisfying specific properties. When used in the special case when ρ is nice, one obtains a least fixpoint procedure for computing P α P ol(C) [ρ] from α and ρ. By Proposition 4.9, this yields an algorithm for solving P ol(C)-covering. However, deciding P ol(C)-covering is not our motivation here: we need this characterization in order to use it as a subresult when proving Theorem 5.3.
Unfortunately, there are technical complications. When we apply the characterization as a subresult, we shall do so for rating maps which are not multiplicative, only quasi-multiplicative (as expected, we build them using the constructions presented in Section 6). This case is not covered by the statement of [Pla18], which only deals with (true) multiplicative rating maps. Consequently, we have to generalize this statement. We avoid redoing the whole 24:28

Thomas Place and Marc Zeitoun
Vol. 17:3 proof, by obtaining the generalized statement as a corollary of the original one from [Pla18]. Additionally, we take this opportunity to slightly tweak the original statement in order to better accommodate our use of the theorem in Sections 8 and 9. We first present the theorem and then focus on its proof. We fix an arbitrary finite quotient-closed Boolean algebra C for the presentation. 7.1. Statement. Consider a C-compatible morphism α : A * → M and a rating map ρ : 2 A * → R which is quasi-multiplicative (there is no other constraint on ρ, in particular, it need not be nice). Recall that since α is C-compatible, we know that for every s ∈ M , [s] C is well-defined as a ∼ C -class containing α −1 (s). We say that a subset S ⊆ M × R is P ol(C)-saturated (for α and ρ) when it satisfies the following properties: (1) Trivial elements: For every w ∈ A * , (α(w), ρ(w)) ∈ S.
Recall that a (true) multiplicative rating map ρ : 2 A * → R is also quasi-multiplicative for the endomorphism µ ρ defined as the identity on R (see Remark 6.5). Thus, in this case, Theorem 7.1 yields that P α P ol(C) [ρ] = ↓ R S = S where S is the least P ol(C)-saturated subset of M × R. This is the original statement of [Pla18]. When ρ is a nice multiplicative rating map, it is clear that one may compute the least P ol(C)-saturated subset of M × R with a least fixpoint algorithm. Therefore, we get an algorithm for P ol(C)-covering by Proposition 4.9.
Remark 7.2. There is a difference between Theorem 7.1 for P ol(C) and Theorem 5.3 for BP ol(C): the latter is restricted to nice rating maps while this is not the case for the former. As explained in Remark 5.4, while it is easy to miss, this difference is crucial. 7.2. Proof of Theorem 7.1. We fix a C-compatible morphism α : A * → M and a quasimultiplicative rating map ρ : 2 A * → R. Since ρ is quasi-multiplicative, we have a semiring structure (R, +, ·) on R and an endomorphism µ ρ of (R, +) satisfying the appropriate axioms. Finally, we let S as the least P ol(C)-saturated subset of M × R for α and ρ. We show that P α P ol(C) [ρ] = ↓ R (s, µ ρ (r)) | (s, r) ∈ S . We first prove that the surjective restriction of ρ is a true multiplicative rating map (for a new multiplication on the rating set which is distinct from "·"). This allows us to apply the theorem of [Pla18].
We define a new multiplication on R that we denote by " ". For every q, r ∈ R, we define q r = µ ρ (qr). It is immediate from Axiom 1 in the definition of quasi-multiplicative rating maps that " " is associative. Moreover, since µ ρ is an endomorphism of (R, +), one may verify that " " distributes over addition and that the element 0 R is a zero for " ". Yet, note that (R, +, ) need not be a semiring: " " might not have a neutral element. Axiom 2 in the definition of quasi-multiplicative rating maps solves this issue. It implies that for K 1 , K 2 ⊆ A * , we have ρ(K 1 K 2 ) = ρ(K 1 ) ρ(K 2 ). Thus, the rating map ρ is a semigroup morphism from (2 A * , ·) to (R, ·). Let Q = ρ(2 A * ) ⊆ R and let τ : 2 A * → Q be the surjective restriction of ρ. It follows that (Q, +, ) is an idempotent semiring (the neutral element is ρ(ε) = τ (ε) ∈ Q) and τ : (2 A * , ∪, ·) → (Q, +, ) is a semiring morphism, i.e. a multiplicative rating map. Additionally, since we defined τ as the surjective restriction of ρ, it is immediate that, Moreover, since τ is a true multiplicative rating map, we may apply the theorem of [Pla18]. Let T be the least P ol(C)-saturated of M × Q for α and τ . We obtain from [Pla18, Theorem 6.5] that P α P ol(C) [τ ] = T . Hence, it now suffices to prove that, Indeed, this clearly implies P α P ol(C) [ρ] = ↓ R (s, µ ρ (r)) | (s, r) ∈ S which concludes the proof of Theorem 7.1.
Remark 7.3. We are dealing with two strongly connected P ol(C)-saturated sets. Namely, S ⊆ M × R and T ⊆ M × Q ⊆ M × R. However, there is a subtle difference between the two. By definition, S is P ol(C)-saturated for α and ρ. This notion depends on the original multiplication "·" of R. On the other hand, T is P ol(C)-saturated for α and τ . By definition of τ , this notion depends on the new multiplication " " of Q.
It now remains to prove (7.1). We handle the two inclusions separately. First, we show that ↓ R T ⊆ ↓ R {(s, µ ρ (r)) | (s, r) ∈ S}. This boils down to proving that for every (s, q) ∈ T , we have r ∈ R such that (s, r) ∈ S and q ≤ µ ρ (r). By definition, T is the least P ol(C)-saturated subset of M × Q for α and τ . Therefore, every pair (s, q) ∈ T is built from trivial elements using downset, multiplication and P ol(C)-closure (here, the multiplication is " " on Q, see Remark 7.3). We proceed by induction on this construction.
If (s, q) is a trivial element, we have w ∈ A * such that s = α(w) and q = τ (w) = ρ(w). We have (s, q) ∈ S since S is P ol(C)-saturated for α and ρ. Moreover, q = µ ρ (q) by Fact 6.6 since q = ρ(w) which concludes this case. We turn to downset: we have q ∈ R such that (s, q ) ∈ T and q ≤ q . Induction yields r ∈ R such that (s, r) ∈ S and q ≤ µ ρ (r). Thus, q ≤ µ ρ (r) which concludes this case. We turn to multiplication. In that case, we have (s 1 , q 1 ), (s 2 , q 2 ) ∈ T such that s = s 1 s 2 and q = q 1 q 2 . For i = 1, 2, induction yields r i ∈ R such that (s i , r i ) ∈ S and q i ≤ µ ρ (r i ). Since S is P ol(C)-saturated for α and ρ, we obtain (s, r 1 r 2 ) ∈ S. Moreover, Axiom 1 in the definition of quasi-multiplicative rating maps yields, q = q 1 q 2 = µ ρ (q 1 q 2 ) ≤ µ ρ (µ ρ (r 1 )µ ρ (r 2 )) = µ ρ (r 1 r 2 ) This concludes the proof for this case. It remains to handle P ol(C)-closure. In that case, we have a pair of idempotents (e, f ) ∈ T such that s = e and q = f τ ([e] C ) f (here, f is an idempotent of (R, ) since T is P ol(C)-saturated for α and τ ). Induction yields r ∈ R such that (e, r) ∈ S and f ≤ µ ρ (r). Since (R, ·) is a finite monoid, there exists a number p ≥ 1 such that r p is an idempotent of (R, ·). Since S is P ol(C)-saturated for α and ρ, we obtain from closure under multiplication that (e, r p ) ∈ S. Together with P ol(C)-closure, this yields 24:30
We turn to the converse inclusion: ↓ R {(s, µ ρ (r)) | (s, r) ∈ S} ⊆ ↓ R T . It suffices to show that for every (s, r) ∈ S, we have (s, µ ρ (r)) ∈ ↓ R T . By definition, S is the least P ol(C)-saturated subset of M × R for α and ρ. Hence, every pair (s, r) ∈ S is built from trivial elements using downset, multiplication and P ol(C)-closure (here, the multiplication is "·" on R, see Remark 7.3). We proceed by induction on this construction.
If (s, r) is a trivial element, we have w ∈ A * such that s = α(w) and r = ρ(w) = τ (w). By Fact 6.6, we have µ ρ (r) = ρ(w). Thus, (s, µ ρ (r)) ∈ T ⊆ ↓ R T since T is P ol(C)-saturated for α and τ . We turn to downset: we have r ∈ R such that (s, r ) ∈ T and r ≤ r . Induction yields (s, µ ρ (r )) ∈ ↓ R T . Thus, since µ ρ (r) ≤ µ ρ (r ), we get (s, µ ρ (r)) ∈ ↓ R ↓ R T = ↓ R T as desired. We turn to multiplication. In that case, we have (s 1 , r 1 ), (s 2 , r 2 ) ∈ S such that s = s 1 s 2 and r = r 1 r 2 . By induction, we obtain that (s i , µ ρ (r i )) ∈ ↓ R T for i = 1, 2. This yields q 1 , q 2 ∈ Q such that (s i , q i ) ∈ T and µ ρ (r i ) ≤ q i for i = 1, 2. Since T is P ol(C)-saturated for α and τ , it follows that (s, q 1 q 2 ) ∈ T by closure under multiplication. Moreover, Altogether, we obtain (s, µ ρ (r)) ∈ ↓ R T as desired. It remains to handle P ol(C)-closure. We have a pair of multiplicative idempotents (e, f ) ∈ S such that s = e and r = f · ρ([e] C ) · f (here, f is an idempotent of (R, ·) since S is P ol(C)-saturated for α and ρ). By induction, we have (e, µ ρ (f )) ∈ ↓ R T . Thus, we obtain q ∈ Q such that (e, q) ∈ T and µ ρ (f ) ≤ q. Since (R, ) is a finite semigroup, there exists a number p ≥ 1 such the multiplication of p copies of q with " " is an idempotent of (R, ). We write g ∈ R for this idempotent. Since T is P ol(C)-saturated for α and τ , we obtain from closure under multiplication that (e, g) ∈ T . Together with P ol(C)-closure, this yields (e, g τ ([e] C ) g) ∈ T . Since µ ρ (f ) ≤ q and f is an idempotent of (R, ·), one may verify that, Altogether, we obtain (s, µ ρ (r)) = (e, µ ρ (f · ρ([e] C ) · f )) ∈ ↓ R T as desired. This concludes the proof.

Soundness in Theorem 5.3
We may now start the proof of Theorem 5.3. In this section, we establish that the statement is sound. The proof is divided in two parts. First, we present a preliminary result, which applies to the Boolean closure operation in general, i.e., to classes of the form Bool(D) when D is an arbitrary lattice. Then, we apply this preliminary result in the special case when D = P ol(C) (for C a finite quotient-closed Boolean algebra) to establish the soundness direction in Theorem 5.3. 8.1. Preliminary result. We first introduce terminology that we need to state our result. We fix a lattice D. Moreover, we let ρ : 2 A * → R be a rating map. Let us recall the definition of the rating map ζ α D [ρ], defined page 21, which also depends on a map α : Using induction, we define a rating map τ n : 2 A * → Q n for every n ∈ N. When n = 0, the rating set Q 0 is (2 R , ∪) and τ 0 is defined as follows, It is immediate by definition that τ 0 is indeed a rating map (i.e., a monoid morphism). Assume now that n ≥ 1 and that τ n−1 : 2 A * → Q n−1 is defined. Recall that ρ * : A * → R denotes the canonical map associated to the rating map ρ. We define τ n as: . By Proposition 6.2, τ n is indeed a rating map. By definition, this means that for all n ≥ 1, the rating set Q n of τ n is Q n = (2 R×Q n−1 , ∪). We complete this definition with maps f n : Q n → 2 R for n ∈ N, defined by induction on n.
The following fact is immediate from the definition.
Fact 8.1. For every n ∈ N and U, U ∈ Q n such that U ⊆ U , we have f n (U ) ⊆ f n (U ).
We may now state the preliminary result that we shall use in our soundness direction of Theorem 5.3.
Proposition 8.2. Consider a language L ∈ D. Then, the following inclusion holds: Proof. The proof is based on the following more involved statement which is proved by induction on n ∈ N.
Lemma 8.3. Let n ∈ N and L, K 0 , . . . , K n , H 0 , . . . , H n ∈ D such that {K i \ H i | i ≤ n} is a cover of L. Then, for every s ∈ f 2n (τ 2n (L)), there exists j ≤ n such that s ≤ ρ(K j \ H j ).
Before proving the lemma, let us use it to prove the first property described in Proposition 8.2. Let L ∈ D be a language. We write S for the set, S = n∈N f n (τ n (L)).
We show that S ⊆ I Bool(D) [L, ρ]. First, we prove the following fact which describes a special optimal Bool(D)-cover of L for ρ.

Thomas Place and Marc Zeitoun
Vol. 17:3 Fact 8.4. There exist n ∈ N and K 0 , . . . , K n , H 0 , . . . , H n ∈ D such that {K i \ H i | i ≤ n} is an optimal Bool(D)-cover of L for ρ.
Proof. Let H be an arbitrary optimal Bool(D)-cover of L for ρ. Each V ∈ H is the Boolean combination of languages in D. We put it in disjunctive normal form. Each disjunct is an intersection languages belonging to D, or whose complement belongs to D. Since D is lattice, both D and the complement class co-D are closed under intersection. Therefore, each disjunct in the disjunctive normal form of V is actually of the form K \ H, where K, H both belong to D. We let K as the set of all languages K \ H which are a disjunct in the disjunctive normal form of some V ∈ H. Clearly, K remains a Bool(D)-cover of L since H was one. Moreover, it is immediate that I[ρ](K) ⊆ I[ρ](H) since every language in K is included in a language of H. Hence, K remains an optimal Bool(D)-cover of L for ρ since H was one.
We let n ∈ N and K 0 , . . . , K n , H 0 , . . . , H n ∈ D be as defined in Fact 8.4. We may now prove that S ⊆ I Bool(D) [L, ρ]. Let s ∈ S. By hypothesis on S, we have s ∈ f 2n (τ 2n (L)). Therefore, since {K i \H i | i ≤ n} is by definition a cover of L, it is immediate from Lemma 8.3 that there exists j ≤ n such that s ≤ ρ(K j \ H j ). Since {K i \ H i | i ≤ n} is an optimal Bool(D)-cover of L for ρ, this implies that s ∈ I Bool(D) [L, ρ] which concludes the main proof.
We turn to the proof of Lemma 8.3. The argument is an induction on n ∈ N. We start with the base case n = 0.
Base case. Consider L, K 0 , H 0 ∈ D such that {K 0 \H 0 } is a cover of L and let s ∈ f 0 (τ 0 (L)).
We have to show that s ≤ ρ(K 0 \ H 0 ). By definition of f 0 , we get r 1 , . . . , r k ∈ τ 0 (L) such that s ≤ r 1 + · · · + r k . Moreover, the definition of τ 0 yields that for every i ≤ k, r i = ρ(w i ) for some w i ∈ L. Therefore, r i ≤ ρ(L) for every i ≤ k and since R is idempotent for addition, s ≤ r 1 + · · · + r k ≤ ρ(L). Finally, since {K 0 \ H 0 } is a cover of L, we have L ⊆ K 0 \ H 0 and we get s ≤ ρ(K 0 \ H 0 ), finishing the argument for the base case.
Inductive step. We now assume that n ≥ 1. Let L, K 0 , . . . , K n , H 0 , . . . , H n ∈ D such that {K i \ H i | i ≤ n} is a cover of L and let s ∈ f 2n (τ 2n (L)). We have to exhibit j ≤ n such that s ≤ ρ(K j \ H j ). Using the hypothesis that s ∈ f 2n (τ 2n (L)), we prove the following fact.
Sub-case 1: Assume that for every m ≤ k, we have, ρ −1 * (r m ) ∩ (K \ H ) = ∅ This means that for every m ≤ k, we have w m ∈ K \ H such that ρ(w m ) = r m . In particular, r m = ρ(w m ) ≤ ρ(K \ H ) for every m ≤ k. Finally, since R is idempotent for addition, we get r 1 + · · · + r k ≤ ρ(K \ H ). Since s ≤ r 1 + · · · + r k , we get s ≤ ρ(K \ H ) and Lemma 8.3 holds for j = in this case.
Sub-case 2: Conversely, assume that there exists m ≤ k such that, Recall that (r m , U m ) ∈ U and that U ⊆ τ 2n−1 (L ∩ K ). Since τ 2n−1 is the rating map ζ ρ * D [τ 2(n−1) ] by definition, it follows that, U m ∈ I D ρ −1 * (r m ) ∩ L ∩ K , τ 2(n−1) Combined with the inclusion K ∩ ρ −1 * (r m ) ⊆ K ∩ H and Fact 4.6, this yields that, Thus, we obtain that U m ⊆ τ 2(n−1) (L ∩ K ∩ H ). Therefore, since s ∈ f 2(n−1) (U m ) by definition of U m , we obtain from Fact 8.1 that, s ∈ f 2(n−1) (τ 2(n−1) (L ∩ K ∩ H )) Finally, since {K i \ H i | i ≤ n} was a cover of L, it is clear that {K i \ H i | i ≤ n and i = } (of size n − 1) is a cover of L ∩ K ∩ H . Therefore, it follows by induction on n in Lemma 8.3 that there exists j ≤ n (with j = ) such that s ≤ ρ(K j \ H j ), finishing the proof. 8.2. Soundness proof for Theorem 5.3. We may now come back to our main objective: soundness in Theorem 5.3. We fix a finite quotient-closed Boolean algebra C and a multiplicative rating map ρ : 2 A * → R. We show that for every BP ol(C)-saturated subset . Remark 8.6. Note that we do not use the hypothesis that ρ is nice for this direction. This is only needed for completeness. By Theorem 3.2, P ol(C) is a lattice. Therefore, we may instantiate the definitions and results presented at the beginning of the section for our nice multiplicative rating map ρ : 2 A * → R in the special case when D = P ol(C). We keep using the same notation: we have the rating maps τ n : 2 A * → Q n (as we prove below, they are now quasi-multiplicative since D = P ol(C) and ρ is multiplicative) and the maps f n : Q n → 2 R .
We complete Proposition 8.2 with another statement specific to this special case. In fact, the proof of this second proposition is based on Theorem 7.1, the characterization of P ol(C)-optimal imprints (we apply it to the rating maps τ n ). Together, these two results imply soundness in Theorem 5.3. r 1 + · · · + r k ∈ f n−1 (T i ) for every i ≤ k it is immediate by induction hypothesis that r 1 + · · · + r k ∈ f n−1 (µ τ n−1 (T i )) for every i ≤ k. Altogether, we obtain that r ∈ f n (µ τn (T )), finishing the proof of the lemma.
Lemma 8.10. For every n ∈ N and T, T ∈ Q n , we have f n (T ) · f n (T ) ⊆ f n (T · T ).
Proof. We proceed by induction on n ∈ N. We first handle the case n = 0. Let r ∈ f 0 (T ) · f 0 (T ). We have s ∈ f 0 (T ) and s ∈ f 0 (T ) such that r = ss . By definition, this yields r 1 , . . . , r k ∈ T and r 1 , . . . , r k ∈ T such that s ≤ r 1 + · · · + r k and s ≤ r 1 + · · · + r k . It follows that ss ≤ i≤k j≤k r i r j . Since r i r j ∈ T · T for every i ≤ k and j ≤ k , it follows that ss ∈ f 0 (T · T ).
Clearly, we have ss ≤ i≤k j≤k r i r j . Moreover, for every i ≤ k and j ≤ k , we have, i≤k j≤k By induction hypothesis, this yields, i≤k j≤k Finally, it is immediate that for every i ≤ k and j ≤ k , we have (r i r j , T i · T j ) ∈ T · T . Altogether, this yields ss ∈ f n (T · T ) by definition.
Proof of Proposition 8.7. We now turn to the main argument. We fix S ⊆ (A * /∼ C ) × R which is BP ol(C)-saturated (for ρ). We have to show that S(D) ⊆ f n (τ n (D)) for every n ∈ N and D ∈ A * /∼ C . The argument is an induction on n ∈ N. Base case. Assume that n = 0 and let D ∈ A * /∼ C . We show that S(D) ⊆ f 0 (τ 0 (D)). Let r ∈ S(D), i.e., (D, r) ∈ S. Since S is BP ol(C)-saturated, we get from (5.1) that there exist r 1 , . . . , r k ∈ R such that r ≤ r 1 + · · · + r k and (D, r i , {r 1 + · · · + r k }) ∈ R ρ S for every i ≤ k. For i ≤ k, one may verify from the definition of R ρ S that since (D, r i , {r 1 + · · · + r k }) ∈ R ρ S , we have w i ∈ A * such that [w i ] C = D and ρ(w i ) = r i . In particular, w 1 , . . . , w k ∈ D and it is therefore immediate from the definition of τ 0 that r 1 , . . . , r k ∈ τ 0 (D). Since r ≤ r 1 + · · · + r k , we obtain r ∈ f 0 (τ 0 (D)) by definition of f 0 , which concludes the proof. Inductive step. We now assume that n ≥ 1. The argument is based on the following lemma, which is where we use the characterization of P ol(C)-optimal imprints (i.e. Theorem 7.1): we apply it to the quasi-multiplicative rating map τ n−1 . Moreover, this is also where we apply induction on n in Proposition 8.7.
Lemma 8.11. For all (D, q, U ) ∈ R ρ S , we have T ∈ Q n−1 satisfying the following properties: T ∈ I P ol(C) ρ −1 * (q) ∩ D, τ n−1 and U ⊆ f n−1 (T ). We start by explaining how Lemma 8.11 can be applied to complete the main proof. Consider D ∈ A * /∼ C . We show that S(D) ⊆ f n (τ n (D)). Let r ∈ S(D) (i.e., (D, r) ∈ S). We prove that r ∈ f n (τ n (D)).
Since we also have r ≤ r 1 + · · · + r k and r 1 + · · · + r k ∈ f n−1 (T i ) for every i ≤ k, it is immediate by definition of f n that r ∈ f n (τ n (D)), finishing the proof of Proposition 8.7. It remains to prove Lemma 8.11.
Proof of Lemma 8.11. We first apply Theorem 7.1 and then use the result to prove the lemma. Clearly, the Cartesian product (A * /∼ C ) × R is a monoid when equipped with the componentwise multiplication. Let α : A * → (A * /∼ C ) × R be the morphism defined by α(w) = ([w] C , ρ(w)). Clearly, α is a C-compatible morphism: for every pair (D, r) ∈ (A * /∼ C ) × R, it suffices to define [(D, r)] C = D. Consequently, since we also know that τ n−1 is quasi-multiplicative by Lemma 8.8, we may apply Theorem 7.1 to obtain a description of the set P α P ol(C) [τ n−1 ] ⊆ (A * /∼ C ) × R × Q n−1 . Consider the least P ol(C)-saturated subset X of (A * /∼ C ) × R × Q n−1 for α and τ n−1 . Theorem 7.1 yields that, The proof of Lemma 8.11 is now based on the following lemma which we shall prove by induction on the construction of an element of R ρ S . Lemma 8.12. For every (D, q, U ) ∈ R ρ S . There exists P ∈ Q n−1 such that (D, q, P ) ∈ X and U ⊆ f n−1 (P ).
Inductive case 1: extended downset. We have (D, r, U ) ∈ R ρ S and such that U ⊆ ↓ R U . By induction, we obtain P ∈ Q n−1 such that (D, r, P ) ∈ X and U ⊆ f n−1 (P ). Clearly, this implies U ⊆ ↓ R f n−1 (P ). Moreover, ↓ R f n−1 (P ) = f n−1 (P ) by definition. Therefore, U ⊆ f n−1 (P ) which concludes this case.
Inductive case 3: S-restricted closure. In that case, we have idempotents (E, f, F ) ∈ R ρ S such that D = E, q = f and U = F · S(E) · F . By induction, we get V ∈ Q n−1 such that (E, f, V ) ∈ X and F ⊆ f n−1 (V ). Since Q n−1 is a finite monoid, there exists a number n ≥ 1 such that V n is a multiplicative idempotent of Q n−1 . Since (E, f, V ) ∈ X, and X is closed under multiplication (as it is P ol(C)-saturated), we obtain (E, f, V n ) ∈ X. We may again use the hypothesis that X is P ol(C)-saturated to apply P ol(C)-closure and obtain, (E, f, V n · τ n−1 (E) · V n ) ∈ X.
Since D = E and q = f , it now remains to show that U = F ·S(E)·F ⊆ f n−1 (V n ·τ n−1 (E)·V n ). It will then be immediate that Lemma 8.12 holds for P = V n · τ n−1 (E) · V n . It is immediate by induction in Proposition 8.7 that, S(E) ⊆ f n−1 (τ n−1 (E)).
Moreover, since we already know that F ⊆ f n−1 (V ) is an idempotent by definition, it follows from Lemma 8.10 that, F · S(E) · F = F n · S(E) · F n ⊆ f n−1 (V n · τ n−1 (E) · V n ).
This concludes the proof of Lemma 8.12.

Thomas Place and Marc Zeitoun
Vol. 17:3 9. Completeness in Theorem 5.3 In this section, we prove the completeness direction in Theorem 5.3. As for the soundness proof, the section is divided in two parts. We start with a preliminary result which applies to Boolean closure in general, i.e., to classes of the form Bool(D) when D is an arbitrary lattice. We then use it in the special case D = P ol(C) to prove the completeness direction of Theorem 5.3.
K ∈ P ol(C), we have τ (K) = ξ BP ol(C) [ρ](K). We have the following key lemma which we need to apply Theorem 7.1.

Conclusion
We proved that separation and covering are decidable for all classes of the form BP ol(C) when C is a finite quotient-closed Boolean algebra. This yields separation and covering algorithms for a whole family of classes. Arguably, the most important one is the level two in the Straubing-Thérien hierarchy (which corresponds to the logic BΣ 2 (<)). Additionally, this result can be lifted to depth-two using an effective reduction of [PZ20] to the level two in the Straubing-Thérien hierarchy. An interesting consequence of our results is that since we proved the decidability of separation for the level two in the Straubing-Thérien hierarchy, the main theorem of [PZ19] is an immediate corollary: membership for this level is decidable. However, the algorithm of [PZ19] was actually based on a characterization theorem: languages of level two in the Straubing-Thérien hierarchy are characterized by a syntactic property of a canonical recognizer (i.e., their syntactic monoid). It turns out that one can also deduce this characterization theorem from our results (this does require some combinatorial work however). In fact, one may generalize it to all classes BP ol(C) when C is a finite quotient-closed Boolean algebra.
Finally, the main and most natural follow-up question is much harder: can our results be pushed to higher levels within concatenation hierarchies? For now, we know that given any finite quotient-closed Boolean algebra C, P ol(C), BP ol(C) and P ol(BP ol(C)) have decidable covering (the former and the latter are results of [Pla18]). Consequently, the next relevant levels are BP ol(BP ol(C)) and P ol(BP ol(BP ol(C))).