1. Introduction
A famous theorem of Büchi [12, Theorem 2] says that the monadic secondorder theory of is decidable. What can be added to this logic while retaining decidability? This question has seen a lot of interest, and we begin by discussing some of the existing results.
What predicates can be added?
The first natural idea is to add predicates beyond the order , e.g. a unary predicate for the primes, or a binary addition function. This idea was pursued already by Robinson in [30], in what is possibly the first published paper to mention mso on . This is before Büchi’s theorem about decidability of mso, and even before the decidability results about weak mso of Büchi [11, Corollary 1], Elgot [15, Corollary 5.8] and Trakhtenbrot [35]. After describing mso, which he credits to Tarski’s lectures, Robinson shows that adding the doubling function to mso results in an undecidable logic [30, p.242]. Other examples of unary functions that lead to undecidability were given by Elgot and Rabin [16, Section 1]. One of these examples is that mso becomes undecidable after adding any function such that is infinite for all . This result was strengthened by Siefkes [32, Theorem 5] who showed that it is enough that is infinite for all with certain periodicity properties, and then by Thomas [33, Theorem 1] who showed that it is enough for to be infinite for infinitely many . Another example of undecidability is mso extended with any unary function that is monotone and satisfies for infinitely many , see [33, Theorem 2]. This line of research is summarised in [29] as follows: “for most examples of natural functions or binary relations it turned out that the corresponding monadic theory is undecidable, usually shown via an interpretation of firstorder arithmetic”.
The undecidability issues mentioned above are avoided if one considers unary predicates. The first examples of this kind were given by Elgot and Rabin, who showed that mso remains decidable after adding unary predicates for the factorials, or squares, or cubes, etc. see [16, Theorem 4]). Following this result, a lot of attention has been devoted to identifying the unary predicates that keep mso decidable. An equivalent phrasing of this question is: which words have a decidable mso theory? An interesting example is the ThueMorse word; its mso theory is decidable, which follows from [25, Theorem 3]. A general classification of words with a decidable mso theory was given by Semenov in [31, p. 165], this line of research was continued in [13, 29]. It is worth pointing out that the classification can be hard to apply to some specific cases; an important one being the case of prime numbers. It is unknown if mso extended with a predicate for the prime numbers has a decidable theory; if this were the case then one could use the algorithm to decide if there are infinitely many twin primes^{1}^{1}1This theory is known to be decidable if one assumes Schinzel’s Hypothesis – a conjecture from number theory which implies that there are infinitely many twin primes [4, Theorem 4]..
It is worth pointing out that in all of the results discussed above, it makes no difference whether one uses mso or weak mso. The undecidability proofs for unary functions in [16, 33, 32] use only weak mso. For the results about unary predicates, it makes no difference if mso or weak mso is used, because for every word, its mso theory is decidable if and only if its weak mso theory is decidable, which follows from McNaughton’s determinisation theorem [22, p. 524]^{2}^{2}2The equivalence of mso and wmso need not hold after adding nonunary predicates. For example, wmso with addition can only define languages in the Borel hierarchy, while mso with addition can easily be shown to contain the logic mso+u that will be discussed later in the paper, and mso+u can define languages beyond the Borel hierarchy [19, Theorem 2.1].. In a sense, one could say that the results discussed above are really about extending weak mso. This will no longer be true when adding quantifiers and languages.
What quantifiers can be added?
Another line of research concerns adding new quantifiers. If the added quantifier has some implicit arithemetic, such as the Härtig quantifier [18], which expresses the existence of two sets of equal size with a given property, then mso immediately becomes undecidable. This follows directly from Robinson’s result about , and it is also discussed in more detail in [20, Theorem 14]. However, there are quantifiers which describe only topological or asymptotic behaviour, and for such quantifiers proving undecidability can be much harder. One example of an asymptotic quantifier is the bounding quantifier from [5], which expresses the property “ is true for finite sets of unbounded size” [5]. The resulting logic, called mso+u, is undecidable [9, Theorem 1.1]. However, it is close to the decidability border; in particular weak mso with the bounding quantifier is decidable [6, Theorems 3 and 5]; and the same is true for its variants and extensions [10, Theorems 11 and 13]. Other extensions of mso with asymptotic quantifiers were proposed by Michalewski, Mio and Skrzypczak in [23, 24]
, including a quantifier related to Baire category and a quantifier related to probability. The Baire quantifier does not add to the expressive power of
mso [24, Theorem 4.1]. On the other hand, the probability quantifier leads to an undecidable logic [23, Theorem 1], because it can express the undecidable problem of checking if a probabilistic Büchi automaton accepts some word with nonzero probability [3, Theorem 7.2]. The theme for quantifiers seems to be that adding a wellbehaved quantifier to (nonweak) mso either does not change the expressive power, or leads to an undecidable logic; but in the latter case the undecidability proof can be hard.What languages can be added?
We now turn to the final kind of feature that can be added to mso, namely languages. This is the main topic of this paper. A language over a binary alphabet can be viewed as a secondorder unary predicate , which inputs a set and returns true if the language contains the word where positions from have label and the remaining positions have label . This can be generalized to alphabets with letters: Then, the predicate inputs sets and returns true if the sets form a partition of and contains the word encoding this partition. Let us write mso+ for the extension of mso which has the predicate described above.
Another equivalent way of describing the logic mso+ uses closure properties of languages. A folklore fact about mso is that existential monadic quantification is the same as taking the image of a language under a lettertoletter homomorphism, see [28, p. 2] or [34, Section 2.3]. It follows that mso+ is exactly the smallest class of languages of words which contains and all regular languages, and which is closed under (a) Boolean combinations; (b) images of lettertoletter homomorphisms; and (c) inverse images of lettertoletter homomorphisms. We will return to this language theoretic approach in Section 2.
Example 1.1.
Example 1.2.
Define to be the words where blocks of ’s have unbounded size:
In [8, Theorem 1.3] it is shown that adding the language to mso gives exactly the logic mso+u. Hence it is unambiguous to write mso+u, with both meanings (adding a quantifier or a language) being equivalent. As mentioned before, this logic is undecidable.
Example 1.3.
Our goal in this paper is to classify the languages
such that mso+ is undecidable. By the discussion in Example 1.1, this project is at least as difficult as classifying the words with a decidable mso theory. However, in the spirit of “asymptotic” conditions, we restrict attention to languages which have a neutral letter, which means that there is a letter in the alphabet, denoted by , such thatholds for every words where infinitely many are nonempty^{3}^{3}3In the definition of neutral letters, we require that the language is stable under inserting or deleting infinitely many neutral letters. However, this also implies that the language is stable under inserting or deleting finitely many neutral letters.. We now state the main theorem of this paper.
Theorem 1.4.
If has a neutral letter and is not definable in mso, then mso+ is undecidable.
In the above theorem, by undecidable we mean that there is no algorithm which decides the sentences of mso+ that are true in . Since mso can quantify over words, this is the same as saying that satisfiability is undecidable for mso+ for words. The proof of Theorem 1.4 will be given in Section 4.
Example 1.5.
Define to be the words such that eliminating all ’s gives a word in the language from Example 1.2. We claim that the logic is obtained by extending mso with (a) the bounding quantifier; or (b) the language ; or (c) the language . The equality of (a) and (b) was discussed in Example 1.2. The language can be defined using the bounding quantifier, hence the inclusion (c) (a). The language is the intersection of with language of words that do not contain the letter , hence the inclusion (b) (c). The equality of these three logics is discussed in more detail in [8].
The language will play an important role in the proof of Theorem 1.4. We will show that if is not definable in mso and contains a neutral letter, then is definable in mso+. Undecidability will then follow by Theorems [9, Theorem 1.1] and [8, Theorem 1.3]. In this sense, is the simplest undecidable extension of mso.
The paper is structured as follows. In Section 2, we discuss a version of our main theorem for finite words, which was proved by Zetzsche et al. in [36]. Like [36], we prove our main theorem using syntactic congruences, and therefore Section 3 is devoted to a discussion of syntactic congruences for languages. In Section 4, we prove our main result, and in Section 5 we show that the main theorem implies that the regular languages are the only Booleanclosed full trios that are decidable.
2. Finite words
In this section, we describe the starting point for our work, which is a theorem by Zetzsche et al., which says that the regular languages of finite words are the only decidable Booleanclosed full trio. To define full trios^{4}^{4}4Full trios are sometimes called cones in formal languages literature, they are meant to be a formalisation of robust classes of languages., recall that a homomorphism is a function
Define the arithmetic hierarchy, see [36, Section 2], to be the least class of languages of finite words that contains all recursively enumerable languages, and which is closed under complementation and homomorphic images.
Theorem 2.1.
[36, Corollary 3.2] Let be a class of languages of finite words which is a full trio, i.e. it is closed under:

images under homomorphisms; and

inverse images under homomorphisms; and

intersections with regular languages.
If is additionally Booleanclosed (closed under union and complementation) and it contains at least one nonregular language, then it contains the arithmetic hierarchy.
In the above theorem, there is no assumption on neutral letters. This is because condition 2 deprecates the assumption, since a neutral letter can be added to any language by taking the inverse image under the homomorphism which eliminates the neutral letter. The closure properties used in Theorem 1.4 are weaker, and hence the assumption on neutral letters is needed. As a warmup for the case of words, we give below a proof sketch for the above theorem.
Proof sketch.
The proof uses rational relations [14, p.236]. Recall that a rational relation is a binary relation on words that is recognised by a nondeterministic automaton where each transition is labelled by a pair (input word, output word), with both words being possibly empty. By Nivat’s theorem (Propositions 1 and 2 in [26]), if a language class is a full trio, then it is closed under images under rational relations, which can be visualised as the following reasoning rule:
Because is closed under complementation, we can also use a variant of the above rule where is used instead of in the conclusion of the rule (i.e. below the line).
The key idea is to use the closure properties to formalise the syntactic right congruence of the language. Let be some nonregular language in , which exists by assumption, and let be its syntactic right congruence, i.e. the equivalence relation defined by
By the MyhillNerode theorem, has infinite index, i.e. infinitely many equivalence classes. Using the reasoning rule with rational relations, one shows that contains the language
where is a fresh separator symbol. Consider now two separator symbols and . Define to be the language
Using the closure properties, one shows . A short analysis of the conditions defining reveals that every word in satisfies
Furthermore, since has infinitely many equivalence classes, it follows that can be arbitrarily large. By projecting away the words using a homomorphisms, it follows that contains the language
A string encoding of runs of twocounter machines, see [17, Theorem 2], can be used to show that contains every recursively enumerable language. The arithmetic hierarchy follows, by closure of under homomorphic images and complementation. ∎
3. Congruences for words
Like in Theorem 2.1, the proof of Theorem 1.4 also uses congruences. However, there are several issues with congruences for words, which mean that some new ideas are needed. The main problem is that there is no good notion of syntactic congruence for languages of words.
We begin by discussing several existing approaches to congruences for words, see also [21]. In all cases, we begin with a language , and use it to define an equivalence relation on finite words. The first candidate is the right congruence, which identifies two finite words if
This right congruence does not characterise the regular languages, although it does have some use, for example in automata learning [1]. There could be finitely many equivalence classes despite a language not being regular. For example, every prefix independent language will have one equivalence class of right congruence, but there are prefix independent languages which are not regular, such as
By induction on formula size one can show that if has a right congruence of finite index, then the same is true for every language definable in mso+; which shows that right congruences will not be useful for our main result. Similar problems arise for the twosided version of right congruence.
A more useful congruence for words uses twosided environments and iteration; this leads to the Arnold congruence [2, Section 2], which identifies if
The Arnold congruence still does not characterise the regular languages. For example, the language is not regular, but it has two equivalence classes under Arnold congruence: words which contain , and words which do not contain .
Fortunately, there is a successful characterisation of regular languages via congruences. This characterisation is stated below, and it corresponds to semigroups.
Theorem 3.1.
[27, Theorem 7.5] A language is regular if and only if there is an equivalence relation which has finite index and satisfies the following conditions for all sequences of finite words :
(1)  
(2) 
We use the name congruence for an equivalence relation that satisfies conditions (1) and (2) in the above theorem. We use the above theorem in our main result. Although promising, the characterisation in terms of congruences has one important drawback, namely nonuniqueness. For the right congruence, the defining property
gives a unique equivalence relation, and hence it makes sense to speak of the right congruence. A similar property holds for the Arnold congruence. The uniqueness of the definition of right congruence, and the fact that its definition can be formalised using rational relations, is what drives the proof of Theorem 2.1.
In contrast, there is no uniqueness in Theorem 3.1, and there cannot be. A language might not have a unique coarsest congruence (such an equivalence relation is called the syntactic congruence). An example is the language , see [7, Running Example 2]. For regular languages, the syntactic congruence exists and coincides with Arnold congruence, see [27, Proposition 8.8], but this is not very helpful in our setting, since we want to study congruences for languages that are not regular. These are issues that we will need to overcome in the proof of our main result.
We finish this section with a simple observation, which says that condition (1) in Theorem 3.1 is superfluous. This observation will be useful later on, since condition (2) will be easier to formalise.
Lemma 3.2.
Proof.
Induction on the number of equivalence classes in the equivalence relation, call it . In the base case, when has one equivalence class, condition (1) holds vacuously. Consider the induction step. For this proof, it is easier to work with the following equivalent form of (1):
If satisfies the above implication, then we are already done. Otherwise, choose a violation of the implication, i.e. words which do not satisfy the conclusion of the implication. By symmetry, assume . Define to be the equivalence relation obtained from by merging the equivalence classes of and . We claim that still satisfies condition (2), and therefore the induction assumption can be applied. We visualize (2) as follows:
By definition of and assumption (2) for , we can replace every in the equivalence class of by and every in the equivalence class of by , without affecting membership in . Therefore we can assume without loss of generality that every is either , or , or a word that is equivalent to neither of these. The same can be done for . By definition of , if then has to be one of or . We can now split each into two words and , likewise for , and then use again the assumption that satisfies (2) to finish the proof. ∎
4. Proof of the main theorem
In this section we prove Theorem 1.4. Fix a language that is not regular, and which contains a neutral letter. We will show that the logic mso+ is undecidable.
To prove undecidability, we will show that mso+ contains the language , and therefore undecidability follows thanks to the results about the logic mso+u. To explain how can be defined, we use a game called the congruence game. This game is played by two players called Spoiler and Duplicator, and it is parametrised by an word . The congruence game is designed so that player Duplicator wins if and only if , which means that has labelled intervals of unbounded size. This is achieved as follows. Roughly speaking, the goal of player Duplicator is to show that from the perspective of the language , each finite word is equivalent to some word which can fit infinitely often into labelled intervals in the word . Since the language is not regular, Duplicator needs intervals of unbounded size to win.
Define an interval to be a finite connected subset of , i.e. it contains all positions between its first and last position. If are intervals, then we write if the last position of is strictly before the first position of .
Definition 4.1 (Congruence Game).
The congruence game for is the following game played by two players, called Spoiler and Duplicator.

Spoiler chooses an infinite family of pairwise disjoint intervals.

Duplicator chooses intervals
such that are from and contain only positions with label in the word .

Spoiler chooses words
such that for every .

Duplicator chooses words
such that for every .

Spoiler chooses a sequence of natural numbers

Duplicator wins the game if and only if:
The key result about the congruence game is the following lemma. The lemma does not use the assumption that contains a neutral letter; this assumption will be used later when formalising the congruence game in mso+.
Lemma 4.2.
Assume that is not regular. Then
Proof.
() Assume . We will show a winning strategy for player Duplicator. Suppose that player Spoiler has chosen a family in round 1. Since , intervals with only labelled positions have unbounded size, and therefore in round 2, player Duplicator can choose the intervals so that for all . For every choice of words made by player Spoiler in round 3, Duplicator’s response in round 4 is to choose the words so that for all . This guarantees victory for Duplicator, regardless of Spoiler’s move in round 5.
() Assume . We will show a winning strategy for player Spoiler. In round 1, Spoiler picks so that the lengths of the intervals tend to infinity, i.e. no size appears infinitely often. Let and be the intervals that are chosen in round 2 by player Duplicator. By choice of , the lengths of the intervals tend to infinity, while by assumption that , the lengths of the intervals are bounded. In round 3, Spoiler chooses the words so that every word from appears infinitely often. This can be done because the lengths of the intervals tend to infinity. Suppose that Duplicator chooses some words in round 4. Since the intervals have bounded size, the words chosen by Duplicator come from a finite set . This means that for every , there is some such that infinitely often and . Choose some function
which realises the dependency , i.e. for every ,
(3) 
Apply Lemma 3.2 with being the kernel of , i.e. the equivalence relation that identifies two words if they have the same image under . Since is not regular, then by Lemma 3.2 there must be a violation of (2), i.e. there must be words such that
(4) 
By (3), in Round 5, Spoiler can choose the indices so that
and therefore win thanks to (4). ∎
The following corollary, and undecidability of the logic mso+u, complete the proof of Theorem 1.4.
Corollary 4.3.
If is not regular and contains a neutral letter, then mso+u mso+.
Proof.
By [8, Theorem 1.3], it is enough to show that the language is definable in mso+. By Lemma 4.2, it is enough to show that mso+ can express that Duplicator has a winning strategy in the congruence game.
A family of disjoint intervals is represented by two sets of positions: the set of leftmost positions in the intervals, and the set of rightmost positions in the intervals. The condition that all intervals are disjoint means that
Using this representation, the choices of intervals in rounds 1 and 2 can be represented by set quantification, with choices of player Duplicator using existential quantifiers and choices of player Spoiler using universal quantifiers. The words chosen in round 3 are represented by colouring the intervals with letters from ; and using the neutral letter for positions not in the intervals . The same goes for round 4. The subsequence in round 5 is represented by a subset of leftmost positions in the intervals . The winning condition in round 6 is checked by using the predicate for . ∎
5. Boolean closed full trios
We finish the paper with a corollary of our main theorem, which is an analog of Theorem 2.1 for infinite words: if a Booleanclosed full trio of languages of infinite words contains at least one nonregular language, then it contains the entire arithmetic hierarchy, subject to a certain representation.
As mentioned in the introduction, a language of words is definable in mso+ if and only if it belongs to the smallest class of languages that contains , contains all regular languages, is closed under Boolean combinations, as well as images and inverse images under lettertoletter homomorphisms. If we lift the restriction on homomorphisms being lettertoletter, then we get a Booleanclosed full trio, as discussed below.
When applying a homomorphism that may erase some letters, an word can be mapped to a finite word. Therefore, in the presence of such homomorphisms, it makes sense to consider languages of words of length , i.e. words which are either finite or words. For such words, define a regular language to be a union of two languages: a regular language of finite words, plus a regular language of words. Define a homomorphism for words of length to be a function
which is obtained by applying to each letter a function of type .
Theorem 5.1.
Let be a class of languages of words of length at most which is closed under:

images under homomorphisms; and

inverse images under homomorphisms; and

intersections with regular languages.
If is additionally Booleanclosed (closed under union and complementation) and it contains at least one nonregular language, then for every in the arithmetic hierarchy,
Proof.
By closure under inverse images of homomorphisms and under intersection with and for any , if contains some nonregular language, then it contains some nonregular language with a neutral letter. By Corollary 4.3, contains all languages definable in mso+u. By [9, Lemma 3.2], for every recursively enumerable language , the logic mso+u defines some language , over an alphabet extended with a neutral letter, such that
where is the homomorphism that eliminates the neutral letter. Since is closed under homomorphic images, it follows that contains the loop representations of all recursive enumerable languages. For the arithmetic hierarchy, it is enough to observe that the class
is closed under Boolean combinations and homomorphic images. ∎
In contrast to the finite word setting, we cannot conclude that contains itself: Standard arguments show that if a language has a finiteindex right congruence, then every language obtained from using Boolean full trio operations also has a finiteindex right congruence. Thus, for example, if we start with , all obtainable languages over finite words are regular.
One could also consider Boolean closed full trios of languages. Then, homomorphisms would be defined by functions of type and the (inverse) image of a homomorphism on a subset of would contain only those resulting words that are infinite. For this notion of (inverse) image, Theorem 5.1 follows with essentially the same proof.
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