Regular Separability and Intersection Emptiness are Independent Problems

TL;DR

This paper investigates the relationship between regular separability and intersection emptiness problems, showing through counterexamples that these problems are independent and cannot be reduced to each other under mild assumptions.

Contribution

It provides counterexamples demonstrating that regular separability and intersection emptiness are independent problems, refuting the possibility of reducing one to the other.

Findings

Intersection emptiness is decidable where regular separability is undecidable.

Regular separability is decidable where intersection emptiness is undecidable.

The two problems are fundamentally independent, with no reduction possible.

Abstract

The problem of \emph{regular separability} asks, given two languages $K$ and $L$, whether there exists a regular language $S$ with $K\subseteq S$ and $S\cap L=\emptyset$. This problem has recently been studied for various classes of languages. All the results on regular separability obtained so far exhibited a noteworthy correspondence with the intersection emptiness problem: In eachcase, regular separability is decidable if and only if intersection emptiness is decidable. This raises the question whether under mild assumptions, regular separability can be reduced to intersection emptiness and vice-versa. We present counterexamples showing that none of the two problems can be reduced to the other. More specifically, we describe language classes $\mathcal{C_1}$, $\mathcal{D_1}$, $\mathcal{C_2}$, $\mathcal{D_2}$ such that (i)~intersection emptiness is decidable for $\mathcal{C_1}$ and $\mathcal{D_1}$, but regular separability is undecidable for $\mathcal{C_1}$ and $\mathcal{D_1}$ and (ii)~regular separability is decidable for $\mathcal{C_2}$ and $\mathcal{D_2}$, but intersection emptiness is undecidable for $\mathcal{C_2}$ and $\mathcal{D_2}$.

Equations

$$\mathsf{L}(\mathcal{M})=\{w\in {\Sigma }^{\ast }\mid \text{(q0,0)w(q,m) for some (q,P) in F with m∈P}\}.$$

$$\mathsf{L}(\mathcal{V})=\{w\in {\Sigma }^{\ast }\mid \text{(q0,0,…,0)w(q,m1,…,mn) for some q∈F and m1,…,mn∈N}\}.$$

$$w_1\otimes w_2\otimes \dots \otimes w_k:=\left[\begin{array}{c}{w}_{11}^{\prime }\\ w_{21}^{\prime }\\ ⋮\\ w_{k1}^{\prime }\end{array}\right]\dots \left[\begin{array}{c}{w}_{1m}^{\prime }\\ w_{2m}^{\prime }\\ ⋮\\ w_{km}^{\prime }\end{array}\right]\in ((\Sigma \cup \{\square \}{)}^k{)}^{\ast }$$

$$\varphi (\bar{x},\bar{y},\bar{z}):=$$

$${\varphi }_1(\bar{x},\bar{y})\wedge {\varphi }_2(\bar{x},\bar{z})|{\varphi }_1(\bar{x},\bar{y})\vee {\varphi }_2(\bar{x},\bar{z})|\exists y{\varphi }^{\prime }(y,\bar{x})$$

$$\exists z_1\exists z_2\dots \exists z_k(x=n_0+\sum _{i=i}^k{z}_i{n}_i)\wedge (y=m_0+\sum _{i=i}^k{z}_i{m}_i).$$

$${\varphi }_2(y):=\exists x{\varphi }_T(x,y)\wedge S_2(x)$$

$${\mathcal{S}}_0^{\Gamma }=\Gamma ,\text{leavevmode}\text{leavevmode}\text{leavevmode}\text{leavevmode}{\mathcal{S}}_{k+1}^{\Gamma }=\{[s_1\cdots s_m{]}_{k+1}\mid m\ge 1,\text{leavevmode}{s}_1,\dots ,s_m\in {\mathcal{S}}_k^{\Gamma }\}.$$

$$[s_1\cdots s_m{]}_1\cdot {\mathsf{rew}}_{\gamma }$$

$$[s_1\cdots s_m{]}_k\cdot {\mathsf{rew}}_{\gamma }$$

$$[s_1\cdots s_m{]}_i\cdot {\mathsf{push}}_i$$

$$[s_1\cdots s_m{]}_k\cdot {\mathsf{push}}_i$$

$$[s_1\cdots s_m{]}_i\cdot {\mathsf{pop}}_i$$

$$[s_1\cdots s_m{]}_k\cdot {\mathsf{pop}}_i$$

$$\mathsf{L}(\mathcal{A})=\{w\in {\Sigma }^{\ast }\mid (q_0,[[\perp {]]}_k){\stackrel{w}{\to }}_{\mathcal{A}}(q,s)\text{leavevmode}\text{for some q∈F and s∈SkΓ}\}.$$

$$t=[[[{\perp }^{\prime }{]]}_{k+1}{t}_1\cdots t_{\ell }{]}_{k+2},$$

$$t=[[[{\perp }^{\prime }{]]}_{k+1}{t}_1\cdots t_{\ell }{]}_{k+2}=[[[{\perp }^{\prime }{]]}_{k+1}{t}_1\cdots t_{\ell -1}[[[\#{]]}_k{s}_1\cdots s_m{]}_{k+1}{]}_{k+2}$$

$$t^{\prime }=[[[{\perp }^{\prime }{]]}_{k+1}{t}_1\cdots t_{\ell -1}[[[\#{]]}_k{s}_1\cdots s_{m-1}{]}_{k+1}[[[\#{]]}_k{s}_1\cdots s_{m-1}{]}_{k+1}{]}_{k+2}.$$

$$\mu (t^{\prime })=\sum _{i=1}^{\ell -1}\mu (t_i)+2\nu (w^{\prime })=\sum _{i=1}^{\ell -1}\mu (t_i)+\nu (w)=\mu (t)$$

$$L_2=\{10^n\mid n\in \mathbb{N}\setminus 2^{\mathbb{N}}\}\cup \{10^{2^{\nu (w)}}\mid w\in K_2\}.$$

$$((p,\tilde{p}),\epsilon ,i,x,(q,\tilde{q}))$$

$$((p,\tilde{p}),a,1,0,(p,\tilde{q}))$$

$$((p,\tilde{p}),\epsilon ,i,x,(q,\tilde{p}))$$

$$((p_0,p_1),a,1,0,(q_0,q_1))$$

$$((p_0,p_1),\epsilon ,i,x,(q_0,p_1))$$

$$((p_0,p_1),\epsilon ,n+i,x,(p_0,q_1))$$

$$\bar{x}{<}_{\mathsf{lex}}\bar{y}⟺x_1<y_1\text{ or }(x_1=y_1\text{ and }(x_2,\dots ,x_n){<}_{\mathsf{lex}}(y_2,\dots ,y_k))$$

$$Y\cap R=\bigcup _{k=1}^m{Y}_k\cap \bigcup _{i=1}^n\bigcap _{j=1}^m{R}_{i,j}=\bigcup _{k=1}^m\bigcup _{i=1}^n\underset{=\varnothing }{\underbrace{Y_k\cap \bigcap _{j=1}^m{R}_{i,j}}}=\varnothing .$$

Full text

Decidability of Intersection Emptiness versus Regular

Separability

Ramanathan S. Thinniyam

  

Georg Zetzsche

Regular Separability and Intersection Emptiness are Independent

Problems

Ramanathan S. Thinniyam

  

Georg Zetzsche

We present counterexamples showing that none of the two problems can be reduced to the other. More specifically, we describe language classes $\mathcal{C}_{1}$, $\mathcal{D}_{1}$, $\mathcal{C}_{2}$, $\mathcal{D}_{2}$ such that (i) intersection emptiness is decidable for $\mathcal{C}_{1}$ and $\mathcal{D}_{1}$, but regular separability is undecidable for $\mathcal{C}_{1}$ and $\mathcal{D}_{1}$ and (ii) regular separability is decidable for $\mathcal{C}_{2}$ and $\mathcal{D}_{2}$, but intersection emptiness is undecidable for $\mathcal{C}_{2}$ and $\mathcal{D}_{2}$.

1 Introduction

The intersection emptiness problem for language classes $\mathcal{C}$ and $\mathcal{D}$ asks for two given languages $K$ from $\mathcal{C}$ and $L$ from $\mathcal{D}$, whether $K\cap L=\emptyset$. If $\mathcal{C}$ and $\mathcal{D}$ are language classes associated to classes of infinite-state systems, ,then intersection emptiness corresponds to verifying safety properties in concurrent systems where one system of $\mathcal{C}$ communicates with a system of $\mathcal{D}$ via messages or shared memory[7]. The idea of separability is to decide whether two given languages are not only disjoint, but whether there exists a finite, easily verifiable, certificate for disjointness (and thus for safety). Specifically, the $\mathcal{S}$ separability problem for a fixed class $\mathcal{S}$ of separators and language classes $\mathcal{C}$ and $\mathcal{D}$ asks, for given languages $K$ from $\mathcal{C}$ and $L$ from $\mathcal{D}$, whether there exists a language $S\in\mathcal{S}$ with $K\subseteq S$ and $S\cap L=\emptyset$.

There is extensive literature dealing with the separability problem, with a range of different separators considered. One line of work concerns separability of regular languages by separators from a variety of regular languages. Here, the investigation began with a more general problem, computing pointlikes (equivalently, the covering problem) [2, 18, 32], but later also concentrated on separability (e.g. [26, 27, 28, 29, 30, 31]). Moreover, separability has been studied for regular tree languages, where separators are either piecewise testable tree languages [19] or languages of deterministic tree-walking automata [6]. For non-regular input languages, separability has been investigated with piecewise testable languages (PTL) [13] and generalizations thereof [35] as separators. Separability of subsets of trace monoids [8] and commutative monoids [10] by recognizable subsets has been studied as well.

A natural choice for the separators is the class of regular languages. On the one hand, they have relatively high separation power and on the other hand, it is usually verifiable whether a given regular language is in fact a separator. For instance, they generalize piecewise testable languages but are less powerful than context-free languages (CFL). Since the intersection problem for CFL is undecidable, it is not easy to check if a given candidate CFL is a separator.

This has motivated a recent research effort to understand for which language classes $\mathcal{C},\mathcal{D}$ regular separability is decidable [23, 10, 9]. An early result was that regular separability is undecidable for CFL (by this we mean that both input languages are context-free) [33, 22]. More recently, it was shown that regular separability is undecidable already for one-counter languages [11], but decidable for several subclasses of vector addition systems (VASS): for one-dimensional VASS [23], for commutative VASS languages [10], and for Parikh automata (equivalently, $\mathbb{Z}$-VASS) [9]. Moreover, it is decidable for languages of well-structured transition systems [12]. Furthermore, decidability still holds in many of these cases if one of the inputs is a general VASS language [14]. However, if both inputs are VASS languages, decidability of regular separability remains a challenging open problem.

These results exhibit a striking correspondence between regular separability and the intersection problem: In all the cases where decidability of regular separability has been clarified, it is decidable if and only if intersection is decidable. In fact, in the case of well-structured transition systems, it even turned out that two languages are regularly separable if and only if they are disjoint [12]. Moreover, deciding regular separability usually involves non-trivial refinements of the methods for deciding intersection. Furthermore, so far the only method to show undecidability of regular separability is to adapt undecidability proofs for the intersection problem [22, 11, 35].

In light of these observations, there was a growing interest in whether there is a deeper connection between regular separability and intersection emptiness. In other words: Is regular separability just intersection emptiness in disguise? It is conceivable that under mild assumptions, regular separability and intersection emptiness are mutually reducible. An equivalence in this spirit already exists for separability by PTL: If $\mathcal{C}$ and $\mathcal{D}$ are closed under rational transductions, then separability by PTL for $\mathcal{C}$ and $\mathcal{D}$ is decidable if and only if the simultaneous unboundedness problem is decidable for $\mathcal{C}$ and for $\mathcal{D}$ [13].

Contribution

We show that regular separability and intersection emptiness are independent problems: Each problem can be decidable while the other is undecidable. Specifically, we present language classes $\mathcal{C}_{1}$, $\mathcal{D}_{1}$, $\mathcal{C}_{2}$, $\mathcal{D}_{2}$, so that (i) for $\mathcal{C}_{1}$ and $\mathcal{D}_{1}$, regular separability is undecidable, but intersection emptiness is decidable and (ii) for $\mathcal{C}_{2}$ and $\mathcal{D}_{2}$, regular separability is decidable, but intersection emptiness is undecidable. Some of these classes have been studied before (such as the higher-order pushdown languages), but some have not (to the best of our knowledge). However, they are all natural in the sense that they are defined in terms of machine models, are closed under rational transductions, and have decidable emptiness and membership problems. We introduce two new classes defined by counter systems that accept based on certain numerical predicates. These predicates can be specified either using reset vector addition systems or higher-order pushdown automata.

2 Preliminaries

We use $\Sigma$ (sometimes $\Gamma$) to denote a finite set of alphabets and $\Sigma^{*}$ to denote the set of finite strings (aka words) over the alphabet $\Sigma$. To distinguish between expressions over natural numbers and expressions involving words, we use typewriter font to denote letters, e.g. $\mathtt{a}$, $\mathtt{0}$, $\mathtt{1}$, etc. For example, $\mathtt{0}^{n}$ is the word consisting of an $n$-fold repetition of the letter $\mathtt{0}$, whereas $0^{n}$ is the number zero. The empty string is denoted $\varepsilon$. If $S\subseteq\mathbb{N}$ we write $\mathtt{a}^{S}$ for the set $\{\mathtt{a}^{n}\mid n\in S\}\subseteq\mathtt{a}^{*}$ and $2^{S}$ for the set $\{2^{n}\mid n\in S\}\subseteq\mathbb{N}$.

We define the map $\nu\colon\{\mathtt{0},\mathtt{1}\}^{*}\rightarrow\mathbb{N}$ which takes every word to the number which it represents in binary representation: We define $\nu(\varepsilon)=0$ and $\nu(w\mathtt{1})=2\cdot\nu(w)+1$ and $\nu(w\mathtt{0})=2\cdot\nu(w)$ for $w\in\{\mathtt{0},\mathtt{1}\}^{*}$. For example, $\nu(\mathtt{1}\mathtt{1}\mathtt{0})=6$. Often we are only concerned with words of the form $\{\mathtt{0}\}\cup\mathtt{1}\{\mathtt{0},\mathtt{1}\}^{*}$.

Languages are denoted by $L,L^{\prime},K$ etc. and the language of a machine $M$ is denoted by $\mathsf{L}(M)$. Classes of languages are denoted by $\mathcal{C}$, $\mathcal{D}$, etc.

Definition 2.1**.**

An asynchronous transducer $\mathcal{T}$ is a tuple $\mathcal{T}=(Q,\Gamma,\Sigma,E,q_{0},F)$ with a set of finite states $Q$, finite output alphabet $\Gamma$, finite input alphabet $\Sigma$, a set of edges $E\subseteq Q\times\Gamma^{*}\times\Sigma^{*}\times Q$, initial state $q_{0}\in Q$ and set of final states $F\subseteq Q$. We write $p\xrightarrow{v|u}q$ if $(p,u,v,q)\in E$ and the machine reads $v$ in state $p$, outputs $u$ and moves to state $q$. We also write $p\xrightarrow[*]{w|w^{\prime}}q$ if there are states $q_{0},q_{1}\cdots q_{n}$ and words $u_{1},u_{2},\cdots,u_{n},v_{1},v_{2},\cdots,v_{n}$ such that $p=q_{0},q=q_{n},w^{\prime}=u_{1}u_{2}\cdots u_{n},w=v_{1}v_{2}\cdots v_{n}$ and $q_{i}\xrightarrow{v_{i}|u_{i}}q_{i+1}$ for all $0\leq i\leq n$.

The transduction $T\subseteq\Gamma^{*}\times\Sigma^{*}$ generated by the transducer $\mathcal{T}$ is the set of tuples $(u,v)\in\Gamma^{*}\times\Sigma^{*}$ such that $q_{0}\xrightarrow[*]{v|u}q_{f}$ for some $q_{f}\in F$. Given a language $L\subseteq\Sigma^{*}$, we define $TL:=\{u\in\Gamma^{*}\;|\;\exists v\in L\;(u,v)\in T\}$. A transduction $T\subseteq\Gamma^{*}\times\Sigma^{*}$ is rational if it is generated by some asynchronous transducer.

A language is a subset of $\Sigma^{*}$ for some alphabet $\Sigma$. A language class is a collection of languages, together with some way to finitely represent these languages, for example using machine models or grammars. We call a language class a full trio if it is effectively closed under rational transductions. This means, given a representation of $L$ in $\mathcal{C}$ and an asynchronous transducer for $T\subseteq\Sigma^{*}\times\Gamma^{*}$, the language $TL$ belongs to $\mathcal{C}$ and one can compute a representation of $TL$ in $\mathcal{C}$.

The following equivalent definition of full trios is well known (see Berstel [3]):

Lemma 2.2**.**

A language class is closed under rational transductions if and only if it is closed under (i) homomorphic image, (ii) inverse homomorphic image, and (iii) intersection with regular languages.

We are interested in decision problems where the representation of a language $L$ (or possibly multiple languages) is the input. In particular, we study the following problems.

Problem 2.3** (Intersection Emptiness).**

Given two languages classes $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$, the intersection emptiness problem $\mathsf{IE}(\mathcal{C}_{1},\mathcal{C}_{2})$ is defined as follows:

Input:

Languages $L_{1}\in\mathcal{C}_{1}$ and $L_{2}\in\mathcal{C}_{2}$.

Question:

Is $L_{1}\cap L_{2}$ empty?

Problem 2.4** (Regular Separability).**

Given two languages classes $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$, the regular separability problem $\mathsf{RS}(\mathcal{C}_{1},\mathcal{C}_{2})$ is defined as follows:

Input:

Languages $L_{1}\in\mathcal{C}_{1}$ and $L_{2}\in\mathcal{C}_{2}$.

Question:

Is there a regular language $R$ such that $L_{1}\subseteq R$ and $L_{2}\cap R=\emptyset$?

We will write $L\mathord{\mid}K$ to denote that $L$ and $K$ are regularly separable.

Problem 2.5** (Emptiness).**

The emptiness problem for a language class $\mathcal{C}$, denoted $\mathsf{Empty}(\mathcal{C})$ is defined as:

Input:

A language $L\in\mathcal{C}$.

Question:

Is $L=\emptyset$ i.e. is $L$ empty?

Problem 2.6** (Infinity).**

The infinity problem for a language class $\mathcal{C}$, denoted $\mathsf{Inf}(\mathcal{C})$ is defined as:

Input:

A language $L\in\mathcal{C}$.

Question:

Does $L$ contain infinitely many elements?

3 Incrementing automata

The counterexamples we construct are defined using special kinds of automata which can only increment a counter, which we will define formally below. The acceptance condition requires that the counter value satisfy a specific numerical predicate, in addition to reaching a final state. By a predicate class, we mean a class $\mathcal{P}$ of predicates over natural numbers (i.e. subsets $P\subseteq\mathbb{N}$) such that there is a way to finitely describe the members of $\mathcal{P}$. As an example, if $\mathcal{C}$ is a language class, then a subset $S\subseteq\mathbb{N}$ is a pseudo-$\mathcal{C}$ predicate if $S=\nu(L)$ for some $L\in\mathcal{C}$. Now the class of all pseudo-$\mathcal{C}$ predicates constitutes a predicate class, because a pseudo-$\mathcal{C}$ predicate can be described using the finite description of a language in $\mathcal{C}$. The class of all pseudo-$\mathcal{C}$ predicates is denoted $\mathsf{pseudo}{\mathcal{C}}$.

Definition 3.1**.**

Let $\mathcal{P}$ be a predicate class. An incrementing automata over $\mathcal{P}$ is a four-tuple $\mathcal{M}=(Q,\Sigma,E,q_{0},F)$ where $Q$ is a finite set of states, $\Sigma$ is its input alphabet, $E\subseteq Q\times\Sigma^{*}\times\{0,1\}\times Q$ a finite set of edges, $q_{0}\in Q$ an initial state and $F$ is a finite set of acceptance pairs $(q,P)$ where $q\in Q$ is a state and $P$ belongs to $\mathcal{P}$ .

A configuration of $\mathcal{M}$ is a pair $(q,n)\in Q\times\mathbb{N}$. For two configurations $(q,n)$, $(q^{\prime},n^{\prime})$, we write $(q,n)\xrightarrow{w}(q^{\prime},n^{\prime})$ if there are configurations $(q_{1},n_{1}),\ldots,(q_{\ell},n_{\ell})$ with $q_{1}=q$ and $q_{\ell}=q^{\prime}$ and edges $(q_{i},w_{i},m_{i},q_{i+1})$ with $n_{i+1}=n_{i}+m_{i+1}$ for $1\leq i<\ell$ and $w=w_{1}\cdots w_{\ell}$. The language accepted by $\mathcal{M}$ is

[TABLE]

The collection of all languages accepted by incrementing automata over $\mathcal{P}$ is denoted $\mathcal{I}(\mathcal{P})$.

It turns out that even with no further assumptions on the predicate class $\mathcal{P}$, the language class $\mathcal{I}(\mathcal{P})$ has some nice closure properties.

Lemma 3.2**.**

Let $\mathcal{P}$ be a predicate class. The languages of incrementing automata over $\mathcal{P}$ are precisely the finite unions of languages of the form $T\mathtt{a}^{P}$ where $P\in\mathcal{P}$ and $T\subseteq\Sigma^{*}\times\{\mathtt{a}\}^{*}$ is a rational transduction. In particular, the class of languages accepted by incrementing automata over $\mathcal{P}$ is a full trio.

Proof.

For every accepting pair $(q,P)$ of $\cal M$, we construct a transducer $T_{q,P}$, which has the same set of states as $\cal M$, accepting state set $\{q\}$ and for each edge $(q^{\prime},w,m,q^{\prime\prime})$ of $\mathcal{M}$ the transducer reads $\mathtt{a}$ if $m=1$ or $\varepsilon$ if $m=0$ and outputs $w$. Then $\mathsf{L}(\cal M)$ is the finite union of all $T_{q,P}(\mathtt{a}^{P})$.

Conversely, since the languages accepted by incrementing automata over $\mathcal{P}$ are clearly closed under union, it suffices to show that $T\mathtt{a}^{P}$ is accepted by an incrementing automaton over $\mathcal{P}$. We may assume that $T$ is given by a transducer in which every edge is of the form $(q,w,\mathtt{a}^{m},q^{\prime})$ with $m\in\{0,1\}$. Let $\mathcal{M}$ have the same state set as $T$ and turn every edge $(q,w,\mathtt{a}^{m},q^{\prime})$ into an edge $(q,w,m,q^{\prime})$ for $\mathcal{M}$. Finally, for every final state $q$ of $T$, we give $\mathcal{M}$ an accepting pair $(q,P)$. Then clearly $\mathsf{L}(\mathcal{M})=T\mathtt{a}^{P}$.

This implies that the class of incrementing automata over $\mathcal{P}$ is a full trio: If $L\subseteq\Sigma^{*}$ is accepted by a incrementing automata over $\mathcal{P}$, then we can write $L=T_{1}\mathtt{a}^{P_{1}}\cup\cdots\cup T_{\ell}\mathtt{a}^{P_{\ell}}$ with $T_{1},\ldots,T_{\ell}\subseteq\Sigma^{*}\times\mathtt{a}^{*}$. If $T\subseteq\Gamma^{*}\times\Sigma^{*}$ is a rational transduction, then $TL=(TT_{1})\mathtt{a}^{P_{1}}\cup\cdots\cup(TT_{\ell})\mathtt{a}^{P_{\ell}}$ and since $TT_{i}$ is again a rational transduction for $1\leq i\leq\ell$, the lanuage $TL$ is accepted by some incrementing automata over $\mathcal{P}$. ∎

It is obvious that the class $\mathcal{I}(\mathcal{P})$ does not always have a decidable emptiness problem: Emptiness is decidable for $\mathcal{I}(\mathcal{P})$ if and only if it is decidable whether a given predicate from $\mathcal{P}$ intersects a given arithmetic progression, i.e. given $P$ and $m,n\in\mathbb{N}$, whether $(m+n\mathbb{N})\cap P\neq\emptyset$. For all the predicate classes $\mathcal{P}$ we consider, emptiness for $\mathcal{I}(\mathcal{P})$ will always be decidable.

4 Decidable Intersection and Undecidable Regular Separability

In this section, we present a language class $\mathcal{C}$ so that intersection emptiness problem $\mathsf{IE}(\mathcal{C},\mathcal{C})$ is decidable for $\mathcal{C}$, but the regular separability problem $\mathsf{RS}(\mathcal{C},\mathcal{C})$ is undecidable for $\mathcal{C}$. The definition of $\mathcal{C}$ is based on reset vector addition systems.

Reset Vector Addition Systems

A reset vector addition system (reset VASS) is a tuple $\mathcal{V}=(Q,\Sigma,n,E,q_{0},F)$, where $Q$ is a finite set of states, $\Sigma$ is its input alphabet, $n\in\mathbb{N}$ is its number of counters, $E\subseteq Q\times\Sigma^{*}\times\{1,\ldots,n\}\times\{0,1,-1,\mathsf{r}\}\times Q$ is a finite set of edges, $q_{0}\in Q$ is its initial state, and $F\subseteq Q$ is its set of final states. A configuration if $\mathcal{V}$ is a tuple $(q,m_{1},\ldots,m_{n})$ where $q\in Q$ and $m_{1},\ldots,m_{n}\in\mathbb{N}$. We write $(q,m_{1},\ldots,m_{n})\xrightarrow{w}(q^{\prime},m^{\prime}_{1},\ldots,m^{\prime}_{n})$ if there is an edge $(q,w,k,x,q^{\prime})$ such that for every $j\neq k$, we have $m^{\prime}_{j}=m_{j}$ and

• •

if $x\in\{-1,0,1\}$, then $m^{\prime}_{k}=m_{k}+x$,

• •

if $x=\mathsf{r}$, then $m^{\prime}_{k}=0$.

If there are configurations $c_{1},\ldots,c_{\ell}$ and words $w_{1},\ldots,w_{\ell-1}$ with $c_{i}\xrightarrow{w_{i}}c_{i+1}$ for $1\leq i<\ell$, and $w=w_{1}\cdots w_{\ell}$, then we also write $c_{1}\xrightarrow{w}c_{\ell}$. The language accepted by $\mathcal{V}$ is defined as

[TABLE]

The class of languages accepted by reset VASS is denoted $\mathcal{R}$.

Our language class will be $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$, i.e. incrementing automata with access to predicates of the form $\nu(L)$ where $L$ is the language of a reset VASS.

Theorem 4.1**.**

$\mathsf{RS}(\mathcal{I}(\mathsf{pseudo}{\mathcal{R}}),\mathcal{I}(\mathsf{pseudo}{\mathcal{R}}))$ is undecidable and $\mathsf{IE}(\mathcal{I}(\mathsf{pseudo}{\mathcal{R}}),\mathcal{I}(\mathsf{pseudo}{\mathcal{R}}))$ is decidable.

Note that $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$ is a full trio (3.2) and since intersection is decidable, in particular its emptiness problem is decidable: One has $L\cap\Sigma^{*}=\emptyset$ if and only if $L=\emptyset$. Moreover, note that we could not have chosen $\mathcal{R}$ as our example class: Since reset VASS are well-structured transition systems, regular separability is decidable for them [12].

Before we begin with the proof of 4.1, let us mention that instead of $\mathcal{R}$, we could have chosen any language class $\mathcal{D}$, for which (i) $\mathcal{D}$ is closed under rational transductions, (ii) $\mathcal{D}$ is closed under intersection, (iii) $\mathsf{Empty}(\mathcal{D})$ is decidable and (iv) $\mathsf{Inf}(\mathcal{D})$ is undecidable. For example, we could have also used lossy channel systems instead of reset VASS.

We now recall some results regarding $\mathcal{R}$ from literature.

Lemma 4.2**.**

Emptiness is decidable for $\mathcal{R}$.

The Section follows from the fact that reset VASS are well-structured transition systems [15], for which the coverability problem is decidable [1, 17] and the fact that a reset VASS has a non-empty language if and only if a particular configuration is coverable.

The following can be shown using standard product constructions, please see Appendix A.

Lemma 4.3**.**

$\mathcal{R}$ is closed under rational transductions, union, and intersection.

We now show that regular separability is undecidable for $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$. We do this using a reduction from the infinity problem for $\mathcal{R}$, whose undecidability is an easy consequence from the undecidability of boundedness of reset VASS.

The boundedness problem for reset VASS is defined below and was shown to be undecidable by Dufourd, Finkel, and Schnoebelen [15] (and a simple and more general proof was given by Mayr [24]). A configuration $(q,x_{1},\ldots,x_{n})$ is reachable if there is a $w\in\Sigma^{*}$ with $(q_{0},0,\ldots,0)\xrightarrow{w}(q,x_{1},\ldots,w_{n})$. A reset VASS $\mathcal{V}$ is called bounded if there is a $B\in\mathbb{N}$ such that for every reachable $(q,x_{1},\ldots,x_{n})$, we have $x_{1}+\cdots+x_{n}\leq B$. Hence, the boundedness problem is the following.

Input:

A reset VASS $\mathcal{V}$.

Question:

Is $\mathcal{V}$ bounded?

Lemma 4.4**.**

The infinity problem for $\mathcal{R}$ is undecidable.

Proof.

From an input reset VASS $\mathcal{V}=(Q,\Sigma,n,E,q_{0},F)$ , we construct a reset VASS $\mathcal{V}^{\prime}$ over the alphabet $\Sigma^{\prime}=\{\mathtt{a}\}$ as follows. In every edge of $\mathcal{V}$, we replace the input word by the empty word $\varepsilon$. Moreover, we add a fresh state $s$, which is the only final state of $\mathcal{V}^{\prime}$. Then, we add an edge $(q,\varepsilon,1,0,s)$ for every state $q$ of $\mathcal{V}$. Finally, we add a loop $(s,\mathtt{a},i,-1,s)$ for every $i\in\{1,\ldots,n\}$. This means $\mathcal{V}^{\prime}$ simulates a computation of $\mathcal{V}$ (but disregarding the input) and can spontaneously jump into the state $s$, from where it can decrement counters. Each time it decrements a counter in $s$, it reads an $\mathtt{a}$ from the input. Thus, clearly, $\mathsf{L}(\mathcal{V}^{\prime})\subseteq\mathtt{a}^{*}$. Moreover, we have $\mathtt{a}^{m}\in\mathsf{L}(\mathcal{V}^{\prime})$ if and only if there is a reachable configuration $(q,x_{1},\ldots,x_{n})$ of $\mathcal{V}$ with $x_{1}+\cdots+x_{n}\geq m$. Thus, $\mathsf{L}(\mathcal{V}^{\prime})$ is finite if and only if $\mathcal{V}$ is bounded. ∎

Note that infinity is already undecidable for languages that are subsets of $\mathtt{1}\mathtt{0}^{*}$. This is because given $L$ from $\mathcal{R}$, a rational transduction yields $L^{\prime}=\{\mathtt{1}\mathtt{0}^{|w|}\mid w\in L\}$ and $L^{\prime}$ is infinite if and only if $L$ is.

Our reduction from the infinity problem works because the input languages have a particular shape, for which regular separability has a simple characterization.

Lemma 4.5**.**

Let $S_{0},S_{1}\subseteq\mathbb{N}$ and $\mathbb{N}\setminus 2^{\mathbb{N}}\subseteq S_{1}$. Then $\mathtt{a}^{S_{0}}$ and $\mathtt{a}^{S_{1}}$ are regularly separable if and only if $S_{0}$ is finite and disjoint from $S_{1}$.

Proof.

If $S_{0}$ is finite and disjoint from $S_{1}$, then clearly $\mathtt{a}^{S_{0}}$ is a regular separator. For the only if direction, consider any infinite regular language $R\subseteq\mathtt{a}^{*}$. It has to include an arithmetic progession, meaning that there exist $m,n\in\mathbb{N}$ with $\mathtt{a}^{m+n\mathbb{N}}\subseteq R$. Hence, for sufficiently large $\ell$, the language $\{\mathtt{a}^{x}\mid 2^{\ell}<x<2^{\ell+1}\}\subseteq S_{1}$ must intersect with $R$. In other words, no infinite $R$ can be a regular separator of $\mathtt{a}^{S_{0}}$ and $\mathtt{a}^{S_{1}}$ i.e. $S_{0}$ must be finite (and disjoint from $S_{1}$). ∎

Lemma 4.6**.**

Regular separability is undecidable for $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$.

Proof.

We reduce the infinity problem for $\mathcal{R}$ (which is undecidable by 4.4) to regular separability in $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$. Suppose we are given $L$ from $\mathcal{R}$. Since $\mathcal{R}$ is effectively closed under rational transductions, we also have $K=\{\mathtt{1}\mathtt{0}^{|w|}\mid w\in L\}$ in $\mathcal{R}$. Note that $K$ is infinite if and only if $L$ is infinite. Then $\nu(K)\subseteq 2^{\mathbb{N}}$ and $K_{1}:=\mathtt{a}^{\nu(K)}$ belongs to $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$. Let $K_{2}=\mathtt{a}^{\mathbb{N}\setminus 2^{\mathbb{N}}}=\mathtt{a}^{\nu(\mathtt{1}\{\mathtt{0},\mathtt{1}\}^{*}\mathtt{1}\{\mathtt{0},\mathtt{1}\}^{*})}$, which also belongs to $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$, because $\mathtt{1}\{\mathtt{0},\mathtt{1}\}^{*}\mathtt{1}\{\mathtt{0},\mathtt{1}\}^{*}$ is regular and thus a member of $\mathcal{R}$.

By Lemma 4.4, $K_{1}$ and $K_{2}$ are regularly separable if and only if $K_{1}$ is finite and disjoint from $K_{2}$. Since $K_{1}\cap K_{2}=\emptyset$ by construction, we have regular separability if and only if $K_{1}$ is finite, which happens if and only if $K$ is finite. ∎

For 4.1, it remains to show that intersection is decidable for $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$. We do this by expressing intersection non-emptiness in the logic $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$, which is the positive $\Sigma_{1}$ fragment of Presburger arithmetic extended with pseudo-$\mathcal{R}$ predicates. Moreover, we show that this logic has a decidable truth problem.

We begin with some notions from first-order logic (please see [16] for syntax and semantics of first-order logic). First-order formulae will be denoted by $\phi(\bar{x}),\psi(y)$ etc. where $\bar{x}$ is a tuple of (possibly superset of the) free variables and $y$ is a single free variable. For a formula $\phi(\bar{x})$, we denote by $\llbracket\phi(\bar{x})\rrbracket$ the set of its solutions (in our case, the domain is $\mathbb{N}$).

Our decision procedure for $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ is essentially the same as the procedure to decide the first-order theory of automatic structures [4], except that instead of regular languages, we use $\mathcal{R}$. For $\bar{w}=(w_{1},w_{2},\ldots,w_{k})\in(\Sigma^{*})^{k}$, the convolution $w_{1}\otimes w_{2}\otimes\ldots\otimes w_{k}$ is a word over the alphabet $(\Sigma\cup\{\Box\})^{k}$ where $\Box$ is a padding symbol not present in $\Sigma$. If $w_{i}=w_{i1}w_{i2}\ldots w_{im_{i}}$ and $m=max\{m_{1},m_{2},\ldots,m_{k}\}$ then

[TABLE]

where $w^{\prime}_{ij}=w_{ij}$ if $j\leq m_{i}$ else $w^{\prime}_{ij}=\Box$. The reversal ${w}^{\mathsf{rev}}$ of a word $w=w_{1}w_{2}w_{3}\cdots w_{n}$ (where each $w_{i}$ is a letter) is ${w}^{\mathsf{rev}}=w_{n}w_{n-1}\cdots w_{1}$. We say that a $k$-ary (arithmetic) relation $R\subseteq\mathbb{N}^{k}$ is a pseudo-$\mathcal{R}$ relation if the set of words $L_{R}=\{{w_{1}}^{\mathsf{rev}}\otimes{w_{2}}^{\mathsf{rev}}\otimes\cdots\otimes{w_{k}}^{\mathsf{rev}}\;|\;(\nu(w_{1}),\ldots,\nu(w_{k}))\in R\}$ 111This definition is stated using the reversal purely for technical purposes: the addition relation is automatic on the reverse convolution. belongs to $\mathcal{R}$. In our decision procedure for $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$, we will show inductively that every formula defines a pseudo-$\mathcal{R}$ relation.

Formally, we consider the theory $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ where $(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ is the structure with domain $\mathbb{N}$ of natural numbers, the constant symbol $1$ and the binary symbols $+$ and $\leq$ taking their canonical interpretations and $\mathsf{pseudo}{\mathcal{R}}$ is a set of predicate symbols, one for each $\mathsf{pseudo}{\mathcal{R}}$ predicate. By $\Sigma^{+}_{1}$ we mean the fragment of first order formulae obtained by using only the boolean operations $\wedge,\vee$ and existential quantification.

Definition 4.7**.**

Let $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ be the set of first order logic formulae given by the following grammar:

[TABLE]

where $S\in\mathsf{pseudo}{\mathcal{R}}$ and $t_{1},t_{2}$ are terms obtained from using variables, $1$ and $+$.

Lemma 4.8**.**

The truth problem for $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ is decidable.

Proof.

It is clear that by introducing new existentially quantified variables, one can transform each formula from $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ into an equivalent formula that is generated by the simpler grammar

[TABLE]

We want to show that given any input sentence $\psi$ from $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$, we can decide if it is true or not. If the sentence has no variables, then it is trivial to decide. Otherwise, $\psi=\exists\bar{x}\;\phi(\bar{x})$ for some formula $\phi(\bar{x})$. We claim that the solution set $R=\llbracket\phi(\bar{x})\rrbracket$ belongs to $\mathsf{pseudo}{\mathcal{R}}$ and a reset VASS for $L_{R}$ can be effectively computed. Assuming the claim, the truth of $\psi$ reduces to the emptiness of $\llbracket\phi(\bar{x})\rrbracket$ or equivalently the emptiness of $L_{R}$, which is decidable by Lemma 4.2.

We prove the claim by structural induction on the defining formula $\phi(\bar{x})$, please see Appendix B for details. ∎

Remark 4.9**.**

The truth problem for $\Pi_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ is undecidable by reduction from the infinity problem for $\mathcal{R}$. Given $L\subseteq\mathtt{1}\mathtt{0}^{*}$, let $R_{L}=\nu(L)\subseteq\mathbb{N}$ be the predicate corresponding to $L$. Now the downward closure $D:=\{x\in\mathbb{N}\mid\exists y\colon x\leq y\wedge R_{L}(y)\}$ is definable in $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ and therefore we have $K\in\mathcal{R}$ with $K=\nu(D)$ by the proof of 4.8. Then the $\Pi_{1}^{+}$-sentence $\forall x\colon R_{K}(x)$ is true if and only if $L$ is infinite.

Having established that $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ is decidable, we are ready to show that intersection emptiness is decidable for $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$.

Lemma 4.10**.**

The intersection problem is decidable for $\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$.

Proof.

Given $L_{1},L_{2}\in\mathcal{I}(\mathsf{pseudo}{\mathcal{R}})$, by Lemma 3.2, we know that both $L_{1}$ and $L_{2}$ are finite unions of languages of the form $T\mathtt{a}^{S}$, where $S$ is a pseudo-$\mathcal{R}$ predicate. Therefore, it suffices to decide the emptiness of intersections of the form $T_{1}\mathtt{a}^{S_{1}}\cap T_{2}\mathtt{a}^{S_{2}}$ where $S_{1}$ and $S_{2}$ are pseudo-$\mathcal{R}$ predicates. Note that $T_{1}\mathtt{a}^{S_{1}}\cap T_{2}\mathtt{a}^{S_{2}}=\emptyset$ iff $T_{2}^{-1}T_{1}\mathtt{a}^{S_{1}}\cap\mathtt{a}^{S_{2}}=\emptyset$. Since $T_{2}^{-1}T_{1}$ is again a rational transduction, it suffices to check emptiness of languages of the form $T\mathtt{a}^{S_{1}}\cap\mathtt{a}^{S_{2}}$ where $T\subseteq\mathtt{a}^{*}\times\mathtt{a}^{*}$ is a rational transduction. Notice that we can construct an automaton $\mathcal{A}$ over the alphabet $\Sigma^{\prime}=\{b,c\}$ with the same states as the transducer $\mathcal{M}_{T}$ for $T$ and where for any transition $p\xrightarrow{\mathtt{a}^{m}|\mathtt{a}^{n}}q$ of $\mathcal{M}_{T}$ we have a transition $p\xrightarrow{\mathtt{b}^{m}\mathtt{c}^{n}}q$ in $\mathcal{A}$. It is clear that $(\mathtt{a}^{x},\mathtt{a}^{y})\in T$ iff there exists a word $w\in\mathsf{L}(\mathcal{A})$ such that $w$ contains exactly $x$ occurrences of $\mathtt{b}$ and $y$ occurrences of $\mathtt{c}$. Now it follows from Parikh’s theorem [25] that the set $\{(x,y)\in\mathbb{N}\times\mathbb{N}\mid(\mathtt{a}^{x},\mathtt{a}^{y})\in T\}$ is semilinear, meaning that there are numbers $n_{0},\ldots,n_{k}$ and $m_{0},\ldots,m_{k}$ such that $(\mathtt{a}^{x},\mathtt{a}^{y})\in T$ if and only if

[TABLE]

In particular, there is a formula $\phi_{T}(x,y)$ in $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ such that $(\mathtt{a}^{x},\mathtt{a}^{y})\in T$ if and only if $\phi_{T}(x,y)$ is satisfied. We can now write a formula $\phi_{2}(y)$ in $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ such that $\phi_{2}(y)$ is satisfied if and only if $\mathtt{a}^{y}\in T\mathtt{a}^{S_{2}}$:

[TABLE]

In the same way, the formula $\phi_{1}(x):=S_{1}(x)$ defines $\mathtt{a}^{S_{1}}$. Now define $\phi=\exists x\;\phi_{1}(x)\;\wedge\;\phi_{2}(x)$. Then clearly $\phi\text{ is true }$ if and only if $T\mathtt{a}^{S_{2}}\cap\mathtt{a}^{S_{1}}\neq\emptyset$. Decidability of $\mathsf{IE}(\mathcal{I}(\mathsf{pseudo}{\mathcal{R}}),\mathcal{I}(\mathsf{pseudo}{\mathcal{R}}))$ follows from 4.8. ∎

5 Decidable Regular Separability and Undecidable Intersection

In this section, we present language classes $\mathcal{C}$ and $\mathcal{D}$ so that $\mathsf{IE}(\mathcal{C},\mathcal{D})$ is undecidable, but $\mathsf{RS}(\mathcal{C},\mathcal{D})$ is decidable. These classes are constructed using higher-order pushdown automata, which we define first.

We follow the definition of [20]. Higher-order pushdown automata are a generalization of pushdown automata where instead of manipulating a stack, one can manipulate a stack of stacks (order-$2$), a stack of stacks of stacks (order-$3$), etc. Therefore, we begin by defining these higher-order stacks. While for ordinary (i.e. order-$1$) pushdown automata, stacks are words over the stack alphabet $\Gamma$, order-$(k+1)$ stacks are sequences of order-$k$ stacks. Let $\Gamma$ be an alphabet and $k\in\mathbb{N}$. The set of order-$k$ stacks $\mathcal{S}^{\Gamma}_{k}$ is inductively defined as follows:

[TABLE]

For a word $v\in\Gamma^{+}$, the stack $[\cdots[[v]_{1}]_{2}\cdots]_{k}$ is also denoted $\llbracket v\rrbracket_{k}$. The function $\mathsf{top}$ yields the topmost symbol from $\Gamma$. This means, we have $\mathsf{top}([s_{1}\cdots s_{m}]_{1})=s_{m}$ and $\mathsf{top}([s_{1}\cdots s_{m}]_{k})=\mathsf{top}(s_{m})$ for $k>1$.

Higher-order pushdown automata operate on higher-order stacks by way of instructions. For the stack alphabet $\Gamma$ and for order-$k$ stacks, we have the instruction set $I^{\Gamma}_{k}=\{\mathsf{push}_{i},\mathsf{pop}_{i}\mid 1\leq i\leq k\}\cup\{\mathsf{rew}_{\gamma}\mid\gamma\in\Gamma\}$. These instructions act on $\mathcal{S}^{\Gamma}_{k}$ as follows:

[TABLE]

and in all other cases, the result is undefined. For a word $w\in(I^{\Gamma}_{k})^{*}$ and a stack $s\in\mathcal{S}^{\Gamma}_{k}$, the stack $s\cdot w$ is defined inductively by $s\cdot\varepsilon=s$ and $s\cdot(wx)=(s\cdot w)\cdot x$ for $x\in I^{\Gamma}_{k}$.

An (order-$k$) higher-order pushdown automaton (short HOPA) is a tuple $\mathcal{A}=(Q,\Sigma,\Gamma,\bot,E,q_{0},F)$, where $Q$ is a finite set of states, $\Sigma$ is its input alphabet, $\Gamma$ is its stack alphabet, $\bot\in\Gamma$ is its stack bottom symbol, $E$ is a finite subset of $Q\times\Sigma^{*}\times\Gamma\times(I^{\Gamma}_{k})^{*}\times Q$ whose elements are called edges, $q_{0}\in Q$ is its initial state, and $F\subseteq Q$ is its set of final states. A configuration is a pair $(q,s)\in Q\times\mathcal{S}^{\Gamma}_{k}$. When drawing a higher-order pushdown automaton, an edge $(q,u,\gamma,v,q^{\prime})$ is represented by an arc $q\xrightarrow{u|\gamma|v}q^{\prime}$. An arc $q\xrightarrow{u|v}q^{\prime}$ means that for each $\gamma\in\Gamma$, there is an edge $(q,u,\gamma,v,q^{\prime})$.

For configurations $(q,s),(q^{\prime},s^{\prime})$ and a word $u\in\Sigma^{*}$, we write $(q,s)\xrightarrow{u}_{\mathcal{A}}(q^{\prime},s^{\prime})$ if there are edges $(q_{1},u_{1},\gamma_{1},v_{1},q_{2}),(q_{2},u_{2},\gamma_{2},v_{2},q_{3}),\ldots,(q_{n-1},u_{n-1},\gamma_{n-1},v_{n-1},q_{n})$ in $E$ and stacks $s_{1},\ldots,s_{n}\in\mathcal{S}^{\Gamma}_{k}$ with $\mathsf{top}(s_{i})=\gamma_{i}$ and $s_{i}\cdot v_{i}=s_{i+1}$ for $1\leq i\leq n-1$ such that $(q,s)=(q_{1},s_{1})$ and $(q^{\prime},s^{\prime})=(q_{n},s_{n})$ and $u=u_{1}\cdots u_{n}$. The language accepted by $\mathcal{A}$ is defined as

[TABLE]

The languages accepted by order-$k$ pushdown automata are called order-$k$ pushdown languages. By $\mathcal{H}$, we denote the class of languages accepted by an order-$k$ pushdown automaton for some $k\in\mathbb{N}$. In our example of classes with decidable regular separability and undecidable intersection, one of the two classes is $\mathcal{H}$. The other class will again be defined using incrementing automata.

Definition 5.1**.**

Let $\mathcal{C}$ be a language class. A predicate $P\subseteq\mathbb{N}$ is a power-$\mathcal{C}$ predicate if $P=\mathbb{N}\setminus 2^{\mathbb{N}}\leavevmode\nobreak\ \cup\leavevmode\nobreak\ \{2^{\nu(w)}\mid w\in L\}$ for some language $L$ from $\mathcal{C}$. The class of power-$\mathcal{C}$ predicates is denoted $\mathsf{power}{\mathcal{C}}$.

Our example of classes with decidable regular separability but undecidable intersection is $\mathcal{H}$ on the one hand and $\mathcal{I}(\mathsf{power}{\mathcal{H}})$ on the other hand.

Theorem 5.2**.**

$\mathsf{RS}(\mathcal{H},\mathcal{I}(\mathsf{power}{\mathcal{H}}))$ is decidable, whereas $\mathsf{IE}(\mathcal{H},\mathcal{I}(\mathsf{power}{\mathcal{H}}))$ is undecidable.

Note that decidable regular separability implies that $\mathcal{I}(\mathsf{power}{\mathcal{H}})$ has a decidable emptiness problem: For $L\subseteq\Sigma^{*}$, one has $\Sigma^{*}\mathord{\mid}L$ if and only if $L=\emptyset$. Moreover, note that we could not have chosen $\mathcal{H}$ as our counterexample, because regular separability is undecidable for $\mathcal{H}$ (already for context-free languages) [33, 22].

For showing 5.2, we rely on two ingredients. The first is that infinity is decidable for higher-order pushdown languages. This is a direct consequence of the decidability of the more general simultaneous unboundedness problem [34] or diagonal problem [13], which were shown decidable for higher-order pushdown automata by Hague, Kochems and Ong [20].

Lemma 5.3** ([20]).**

$\mathsf{Inf}(\mathcal{H})$ is decidable.

The other ingredient is that turning binary representations into unary ones can be achieved in higher-order pushdown automata.

Lemma 5.4**.**

If $L\subseteq\{\mathtt{0},\mathtt{1}\}^{*}$ is an order-$k$ pushdown language, then $L^{\prime}=\{\mathtt{1}\mathtt{0}^{\nu(w)}\;|\;w\in L\}$ is an order-$(k+2)$ pushdown language.

Proof.

Let $\mathcal{A}$ be an order-$k$ HOPA accepting $L\subseteq\{\mathtt{0},\mathtt{1}\}^{*}$. We construct an order-$(k+2)$ HOPA $\mathcal{A}^{\prime}$ for $L^{\prime}$. We may clearly assume that $\mathcal{A}$ has only one final state $q_{f}$. The following diagram describes $\mathcal{A}^{\prime}$:

q^{\prime}_{0}$$q_{0}$$q_{f}$$p$$q_{f}^{\prime}$$\mathtt{1}|\mathsf{push}_{k+2}\mathsf{rew}_{\#}\mathsf{push}_{k+1}\mathsf{rew}_{\bot}$$\varepsilon|\mathsf{pop}_{k+1}

[TABLE] \varepsilon|\#|\mathsf{pop}_{k+2}$$\varepsilon|\bot^{\prime}|\varepsilon

The HOPA $\mathcal{A}^{\prime}$ starts in the configuration $(q^{\prime}_{0},\llbracket\bot^{\prime}\rrbracket_{k+2})$ and in moving to $q_{0}$, it reads $\mathtt{1}$ and goes to $(q_{0},[\llbracket\bot^{\prime}\rrbracket_{k+1}[\llbracket\#\rrbracket_{k}\llbracket\bot\rrbracket_{k}]_{k+1}]_{k+2})$. In the part in the dashed rectangle, $\mathcal{A}^{\prime}$ simulates $\mathcal{A}$. However, instead of reading an input symbol $a\in\{\mathtt{0},\mathtt{1}\}$, $\mathcal{A}$ stores that symbol on the stack. In order not to interfer with the simulation of $\mathcal{A}$, this is done by copying the order-$k$ stack used by $\mathcal{A}$ and storing $a$ in the copy below. This is achieved as follows. For every edge $p\xrightarrow{a|\gamma|v}q$ with $v\in(I^{\Gamma}_{k})^{*}$, $\mathcal{A}^{\prime}$ instead has an edge

p$$q$$\varepsilon|\gamma|\mathsf{push}_{1}\mathsf{rew}_{a}\mathsf{push}_{k+1}\mathsf{pop}_{1}v

This pushes the input symbol $a$ on the (topmost order-$k$) stack, makes a copy of the topmost order-$k$ stack, removes the $a$ from this fresh copy, and then excutes $v$. Edges $p\xrightarrow{\varepsilon|\gamma|v}$ (i.e. ones that read $\varepsilon$ from the input) are kept.

When $\mathcal{A}^{\prime}$ arrives in $q_{f}$, it has a stack $[\llbracket\bot^{\prime}\rrbracket_{k+1}[\llbracket\#\rrbracket_{k}s_{1}\cdots s_{m}s]_{k+1}]_{k+2}$, where $s$ is the order-$k$ stack reached in the computation of $\mathcal{A}$, and $s_{1},\ldots,s_{m}$ store the input word $w\in\Sigma^{*}$ read by $\mathcal{A}$, meaning $\mathsf{top}(s_{1})\cdots\mathsf{top}(s_{m})=w$. When moving to $p$, $\mathcal{A}^{\prime}$ removes $s$ so as to obtain $[\llbracket\bot^{\prime}\rrbracket_{k+1}[\llbracket\#\rrbracket_{k}s_{1}\cdots s_{m}]_{k+1}]_{k+2}$ as a stack.

In $p$, $\mathcal{A}^{\prime}$ reads the input word $\mathtt{0}^{\nu(w)}$ as follows. While in $p$, the stack always has the form

[TABLE]

where each $t_{i}$ is an order-$(k+1)$ stack of the form $[\llbracket\#\rrbracket_{k}s_{1}\cdots s_{m}]_{k+1}$ for some order-$k$ stacks $s_{1},\ldots,s_{m}\in\mathcal{S}^{\Gamma}_{k}$. To formulate an invariant that holds in state $p$, we define a function $\mu$ on the stacks as in (1). First, if $t_{i}=[\llbracket\#\rrbracket_{k}s_{1}\cdots s_{m}]_{k+1}$, then let $\mu(t_{i})=\nu(\mathsf{top}(s_{1})\cdots\mathsf{top}(s_{m}))$. Next, let $\mu(t)=\mu(t_{1})+\cdots+\mu(t_{\ell})$. It is not hard to see that the loops on $p$ preserve the following invariant: If $\mathtt{0}^{r}$ is the input word read from configuration $(p,t)$ to $(p,t^{\prime})$, then $\mu(t)=r+\mu(t^{\prime})$. To see this, consider a one step transition $(p,t)\xrightarrow{\varepsilon|\mathtt{0}|\mathsf{pop}_{k+1}\mathsf{push}_{k+2}}(p,t^{\prime})$. If

[TABLE]

then

[TABLE]

If $w=\mathsf{top}(s_{1})\ldots\mathsf{top}(s_{m})$ then $w=w^{\prime}\mathtt{0}$ where $w^{\prime}=\mathsf{top}(s_{1})\cdots\mathsf{top}(s_{m-1})$ since we popped $s_{m}$ off the stack. Moreover,

[TABLE]

Similarly we see that if the transition taken is $\mathtt{0}|\mathtt{1}|\mathsf{pop}_{k+1}\mathsf{push}_{k+2}$ then we get $\mu(t^{\prime})=\mu(t)+1$. By induction on the length of the run, we get $\mu(t)=r+\mu(t^{\prime})$ when $\mathtt{0}^{r}$ is read.

Now observe that when $\mathcal{A}^{\prime}$ first arrives in $p$ with stack $t$, then by construction we have $\ell=1$ and $\mu(t)=\mu(t_{1})=\nu(w)$. Moreover, when $\mathcal{A}^{\prime}$ moves on to $q^{\prime}_{f}$ with a stack as in (1), then $\ell=0$ and thus $\mu(t)=0$. Thus, the invariant implies that if $\mathcal{A}^{\prime}$ reads $\mathtt{0}^{r}$ while in $p$, then $r=\nu(w)$. This means, $\mathcal{A}^{\prime}$ has read $\mathtt{1}\mathtt{0}^{\nu(w)}$ in total.

Finally, from a stack $t$ as in (1), $\mathcal{A}^{\prime}$ reaches $q^{\prime}_{f}$ in finitely many steps, please see Appendix C. ∎

Lemma 5.5**.**

The problem $\mathsf{IE}(\mathcal{H},\mathcal{I}(\mathsf{power}{\mathcal{H}}))$ is undecidable.

Proof.

We reduce intersection emptiness for context-free languages, which is well-known to be undecidable [21], to $\mathsf{IE}(\mathcal{H},\mathcal{I}(\mathsf{power}{\mathcal{H}}))$. Let $K_{1},K_{2}\subseteq\{\mathtt{0},\mathtt{1}\}^{*}$ be context-free. Since $K_{1}\cap K_{2}\neq\emptyset$ if and only if $\mathtt{1}K_{1}\cap\mathtt{1}K_{2}\neq\emptyset$ and $\mathtt{1}K_{i}$ is context-free for $i=0,1$, we may assume that $K_{1},K_{2}\subseteq\mathtt{1}\{\mathtt{0},\mathtt{1}\}^{*}$. This implies $K_{1}\cap K_{2}\neq\emptyset$ if and only if $\nu(K_{1})\cap\nu(K_{2})\neq\emptyset$.

Let $P_{2}=\mathbb{N}\setminus 2^{\mathbb{N}}\leavevmode\nobreak\ \cup\leavevmode\nobreak\ 2^{\nu(K_{2})}$. Then $P_{2}\subseteq\mathbb{N}$ is a power-$\mathcal{H}$ predicate, because $\mathcal{H}$ includes the context-free languages. Thus, the language $L_{2}=\{\mathtt{1}\mathtt{0}^{n}\mid n\in P_{2}\}$ belongs to $\mathcal{I}(\mathsf{power}{\mathcal{H}})$ and

[TABLE]

Moreover, let $L_{1}:=\{\mathtt{1}\mathtt{0}^{2^{\nu(w)}}\;|\;w\in K_{1}\}$. Since $L_{1}=\{\mathtt{1}\mathtt{0}^{\nu(\mathtt{1}\mathtt{0}^{\nu(w)})}\;|\;w\in K_{1}\}$ and $K_{1}$ is an order-$1$ pushdown language, applying 5.4 twice yields that $L_{1}$ is an order-$5$ pushdown language and thus belongs to $\mathcal{H}$. Now clearly $L_{1}\cap L_{2}\neq\emptyset$ if and only if $\nu(K_{1})\cap\nu(K_{2})\neq\emptyset$, which is equivalent to $K_{1}\cap K_{2}=\emptyset$. ∎

For showing decidability of regular separability, we use the following well-known fact (please see Appendix D for a proof).

Lemma 5.6**.**

Let $L=\bigcup_{i=1}^{m}L_{i}$ and $K=\bigcup_{i=1}^{n}K_{i}$. Then $K\mathord{\mid}L$ if and only if $L_{i}\mathord{\mid}K_{j}$ for all $i\in\{1,\ldots,m\}$ and $j\in\{1,\ldots,n\}$.

The last ingredient for our decision procedure is the following simple but powerful observation from [14] (for the convenience of the reader, a proof can be found in Appendix E).

Lemma 5.7**.**

Let $K\subseteq\Gamma^{*}$, $L\subseteq\Sigma^{*}$ and $T\subseteq\Sigma^{*}\times\Gamma^{*}$ be a rational transduction. Then $L\mathord{\mid}TK$ if and only if $T^{-1}L\mathord{\mid}K$.

The following now completes the proof of 5.2.

Lemma 5.8**.**

The problem $\mathsf{RS}(\mathcal{H},\mathcal{I}(\mathsf{power}{\mathcal{H}}))$ is decidable.

Proof.

Suppose we are given $L_{1}\subseteq\Sigma^{*}$ from $\mathcal{H}$ and $L_{2}\subseteq\Sigma^{*}$ from $\mathcal{I}(\mathsf{power}{\mathcal{H}})$. Then we can write $L_{2}=\bigcup_{i=1}^{n}T_{i}\mathtt{a}^{P_{i}}$, where for $1\leq i\leq n$, $T_{i}\subseteq\Sigma^{*}\times\mathtt{a}^{*}$ is a rational transduction and $P_{i}\subseteq\mathbb{N}$ is a power-$\mathcal{H}$ predicate. Since $L_{1}\mathord{\mid}L_{2}$ if and only if $L_{1}\mathord{\mid}T_{i}\mathtt{a}^{P_{i}}$ for every $i$ (5.6), we may assume $L_{2}=T\mathtt{a}^{P}$ for $T\subseteq\Sigma^{*}\times\mathtt{a}^{*}$ rational and $P\subseteq\mathbb{N}$ a power-$\mathcal{H}$ predicate. According to 5.7, $L_{1}\mathord{\mid}T\mathtt{a}^{P}$ if and only if $T^{-1}L_{1}\mathord{\mid}\mathtt{a}^{P}$. Since $T^{-1}$ is also a rational transduction and $\mathcal{H}$ is a full trio, we may assume that $L_{1}$ is in $\mathcal{H}$ with $L_{1}\subseteq\mathtt{a}^{*}$ and $L_{2}=\mathtt{a}^{P}$.

By 4.5, we know that $L_{1}\mathord{\mid}\mathtt{a}^{P}$ if and only if $L_{1}$ is finite and disjoint from $\mathtt{a}^{P}$. We can decide this as follows. First, using 5.3 we check whether $L_{1}$ is finite. If it is not, then we know that $L_{1}\mathord{\mid}L_{2}$ is not the case.

If $L_{1}$ is finite, then we can compute a list of all words in $L_{1}$: We start with $F_{0}=\emptyset$ and then successively compute finite sets $F_{i}\subseteq L_{1}$. For each $i\in\mathbb{N}$, we check whether $L_{1}\subseteq F_{i}$, which is decidable because $L_{1}\cap(\mathtt{a}^{*}\setminus F_{i})$ is in $\mathcal{H}$ and emptiness is decidable for $\mathcal{H}$. If $L_{1}\not\subseteq F_{i}$, then we enumerate words in $\mathtt{a}^{*}$ until we find $\mathtt{a}^{m}$ with $\mathtt{a}^{m}\in L_{1}$ (membership in $L_{1}$ is decidable) and $\mathtt{a}^{m}\notin F_{i}$. Then, we set $F_{i+1}=F_{i}\cup\{\mathtt{a}^{m}\}$. Since $L_{1}$ is finite, this procedure must terminate with $F_{i}=L_{1}$. Now we have $L_{1}\mathord{\mid}\mathtt{a}^{P}$ if and only if $F_{i}\cap\mathtt{a}^{P}=\emptyset$. The latter can be checked because $\mathsf{power}{\mathcal{H}}$ predicates are decidable. ∎

6 Conclusion

We have presented a language class $\mathcal{C}_{1}$ for which intersection emptiness is decidable but regular separability is undecidable in Section 4. Similarly, in Section 5 we constructed $\mathcal{C}_{2},\mathcal{D}_{2}$ for which intersection emptiness is undecidable but regular separability is decidable. All three language classes enjoy good language theoretic properties in that they are full trios and have a decidable emptiness problem.

Let us provide some intuition on why these examples work. The underlying observation is that intersection emptiness of two sets is insensitive to the shape of their members: If $f\colon X\to Y$ is any injective map and $S$ disjoint from the image of $f$, then for $A,B\subseteq X$, we have $A\cap B=\emptyset$ if and only if $(f(A)\cup S)\cap f(B)=\emptyset$. Regular separability, on the other hand, is affected by such distortions: For example, if $K,L\subseteq\mathtt{1}\{\mathtt{0},\mathtt{1}\}^{*}$ are infinite, then $\mathtt{a}^{\mathbb{N}\setminus 2^{\mathbb{N}}}\cup\mathtt{a}^{2^{\nu(K)}}$ and $\mathtt{a}^{2^{\nu(L)}}$ are never regularly separable, even if $K$ and $L$ are. Hence, roughly speaking, the examples work by distorting languages (using encodings as numbers) so that intersection emptiness is preserved, but regular separability reflects infinity of the input languages. We apply this idea to language classes where intersection is decidable, but infinity is not (4.1) or the other way around (5.2). All this suggests that regular separability and intersection emptiness are fundamentally different problems.

Moreover, our results imply any simple combinatorial decision problem that characterizes regular separability has to be incomparable with intersection emptiness. Consider for example the infinite intersection problem as a candidate. It asks whether two given languages have an infinite intersection. Note that for $L\in\mathcal{C},K\in\mathcal{D}$ we have $L\cap K\neq\emptyset$ if and only if $L\#^{*}$ and $K\#^{*}$ (where $\#$ is a symbol not present in $L$ or $K$) have infinite intersection. Moreover, as full trios, $\mathcal{C}$ and $\mathcal{D}$ effectively contain $L\#^{*}$ and $K\#^{*}$, respectively. This implies a counterexample with decidable regular separability and undecidable infinite intersection.

While the example from Section 4 is symmetric (meaning: the two language classes are the same) and natural, the example in Section 5 is admittedly somewhat contrived: While pseudo-$\mathcal{C}$ predicates rely on the common conversion of binary into unary representations, power-$\mathcal{C}$ predicates are a bit artificial. It would be interesting if there were a simpler symmetric example with decidable regular separability and undecidable intersection.

Appendix A Proof of Lemma 4.3

Closure under union is easy to see, so let us show closure under rational transductions and intersection. Suppose $\mathcal{V}=(Q,\Sigma,n,E,q_{0},F)$ is a reset VASS and a transducer $\mathcal{T}=(\tilde{Q},\Gamma,\Sigma,\tilde{E},\tilde{q}_{0},\tilde{F})$ accepting $T\subseteq\Gamma^{*}\times\Sigma^{*}$. We now think of $\mathcal{V}$ producing output instead of reading input, which will be reflected by our formulations: Here, we say that an edge $(q,w,i,x,q)$ produces the word $w$. This is because we imagine that $\mathcal{V}$ is producing a word that is read and transformed by $\mathcal{T}$, which then produces some output on its own.

We may assume that every edge in $\mathcal{V}$ is of the form $(q,a,i,x,q^{\prime})$ with $a\in\Sigma\cup\{\varepsilon\}$. Moreover, we may assume that every edge in $\mathcal{T}$ is of the form $(q,a,\varepsilon,q^{\prime})$ with $a\in\Gamma$ or of the form $(q,\varepsilon,a,q^{\prime})$ with $a\in\Sigma$.

Our reset VASS $\mathcal{V}^{\prime}$ for $T(\mathsf{L}(\mathcal{V}))$ is obtained using a simple product construction. The state set of $\mathcal{V}^{\prime}$ is $Q^{\prime}=Q\times\tilde{Q}$ and in every edge of $\mathcal{V}^{\prime}$, we either (i) advance both $\mathcal{V}$ and $\mathcal{T}$—because $\mathcal{T}$ is reading a symbol produced by $\mathcal{V}$—or (ii) only $\mathcal{T}$—because $\mathcal{T}$ is producing some output—or (iii) only $\mathcal{V}$ because it makes a move that does not produce output. Formally, we include the following edges in $\mathcal{V}^{\prime}$:

[TABLE]

The state $(q_{0},\tilde{q}_{0})$ is the initial state of $\mathcal{V}^{\prime}$ and the states in $F\times\tilde{F}$ are final. Then we clearly have $\mathsf{L}(\mathcal{V}^{\prime})=T(\mathsf{L}(\mathcal{V}))$.

Let us now show that $\mathcal{R}$ is closed under intersection and let $\mathcal{V}_{i}=(Q_{i},\Sigma,n,E,q_{0}^{(i)},F_{i})$ for $i\in\{0,1\}$ be reset VASS (it is no loss of generality to assume that they both have $n$ counters). Again, we construct a reset VASS $\mathcal{V}^{\prime}=(Q^{\prime},\Sigma,2n,E^{\prime},q_{0}^{\prime},F^{\prime})$ with $\mathsf{L}(\mathcal{V}^{\prime})=\mathsf{L}(\mathcal{V}_{0})\cap\mathsf{L}(\mathcal{V}_{1})$ using a simple product construction with $Q^{\prime}=Q_{0}\times Q_{1}$. We may assume that every edge is $\mathcal{V}_{0}$ and $\mathcal{V}_{1}$ is of the form $(q,\varepsilon,i,x,q^{\prime})$ or $(q,a,1,0,q^{\prime})$, meaning that either an edge reads input or it operates on some counter. $\mathcal{V}^{\prime}$ has $2n$ counters and has three types of edges: (i) $\mathcal{V}_{0}$ and $\mathcal{V}_{1}$ both advance and both read some letter $a\in\Sigma$, (ii) just $\mathcal{V}_{0}$ advances and (iii) just $\mathcal{V}_{1}$ advances. Formally, this means we have

[TABLE]

The initial state of $\mathcal{V}^{\prime}$ is $q^{\prime}_{0}=(q_{0}^{(0)},q_{0}^{(1)})$ and the final states of $\mathcal{V}^{\prime}$ are $F^{\prime}=F_{0}\times F_{1}\subseteq Q^{\prime}$. Then clearly $\mathsf{L}(\mathcal{V}^{\prime})=\mathsf{L}(\mathcal{V}_{0})\cap\mathsf{L}(\mathcal{V}_{1})$.

Appendix B Details of Proof of 4.8

We show by structural induction on the formula that every formula in $\Sigma_{1}^{+}(\mathbb{N},+,\leq,1,\mathsf{pseudo}{\mathcal{R}})$ defines a relation which belongs to $\mathsf{pseudo}{\mathcal{R}}$ and a reset VASS for L can be effectively computed.

**Atomic formulae: ** $S(x)$ where $S\in\mathsf{pseudo}{\mathcal{R}}$ is by definition and one can construct automata which recognise the relations $x+y=z$ and $x=1$ (see [5]).

**Induction: ** Let $\phi(\bar{x},\bar{y},\bar{z})=\phi_{1}(\bar{x},\bar{y})\wedge\phi_{2}(\bar{x},\bar{z})$. By induction hypothesis, $\phi_{1}(\bar{x},\bar{y})$ and $\phi_{2}(\bar{x},\bar{z})$ define $\mathsf{pseudo}{\mathcal{R}}$ relations $L_{1}$ and $L_{2}$ respectively.

An $m$-ary relation $P^{\prime}\subseteq\mathbb{N}^{m}$ is said to be the cyclindrification of a $k$-ary relation $P\subseteq\mathbb{N}^{k}$ if there exist indices $i_{1},\ldots,i_{k}$ with $1\leq i_{1}<i_{2}<\ldots<i_{k}\leq m$ such that $P^{\prime}=\{\bar{w}\in\mathbb{N}^{m}\mid(w_{i_{1}},w_{i_{2}},\ldots,w_{i_{k}})\in P\}$. If $P$ is a pseudo-$\mathcal{R}$ relation then any cylindrification $P^{\prime}$ of $P$ is also a pseudo-$\mathcal{R}$ relation: It is easy to construct a rational transduction $T\subseteq(\{\mathtt{0},\mathtt{1},\Box\}^{k})^{*}\times(\{\mathtt{0},\mathtt{1},\Box\}^{m})^{*}$ with $L_{P^{\prime}}=TL_{P}$ and $\mathcal{R}$ is closed under rational transductions by 4.3.

Let $L_{1}^{\prime},L_{2}^{\prime}$ respectively be the cylindrification of $L_{1},L_{2}$ w.r.t. $\bar{x},\bar{y},\bar{z}$. Then the language defined by $\phi$ is $L_{1}^{\prime}\cap L_{2}^{\prime}$, which belongs to $\mathcal{R}$ according to 4.3.

Similarly $\phi(\bar{x},\bar{y},\bar{z})=\phi_{1}(\bar{x},\bar{y})\wedge\phi_{2}(\bar{x},\bar{z})$ is the union of the appropriate cylindrifications of the languages corresponding to $\phi_{1}$ and $\phi_{2}$ and again 4.3 applies.

Let $\phi(\bar{x})=\exists y\;\phi^{\prime}(y,\bar{x})$. Let $R$ (resp. $R^{\prime}$) be the relation defined by $\phi$ (resp. $\phi^{\prime}$). By induction hypothesis $L_{R}\in\mathcal{R}$ and since $L_{R^{\prime}}$ is a homomorphic image of $L_{R}$, 4.3 it tells us that $L_{R^{\prime}}$ belongs effectively to $\mathcal{R}$.

Appendix C Details of Proof of 5.4

We want to show that the machine $\mathcal{A}^{\prime}$ reaches $q^{\prime}_{f}$ in finitely many steps. To accomplish this, we define a new parameter on the stack which is shown to strictly decrease on every transition. We start by defining the lexicographic order on finite sequence of numbers. For $\bar{x}=(x_{1},\ldots,x_{n})\in\mathbb{N}^{n}$ and $\bar{y}=(y_{1},\ldots,y_{k})\in\mathbb{N}^{k}$, the lex order $<_{\mathsf{lex}}$ is defined inductively by

[TABLE]

Moreover, the empty sequence is smaller than any non-empty sequence.

Using $<_{\mathsf{lex}}$, we can now define an order on stacks. To this end, for an order-$(k+1)$ stack $s=[\llbracket\#\rrbracket_{k},s_{1}\cdots s_{m}]_{k+1}$, we define its length as $|s|=m+1$. Then, to an order-$(k+2)$ stack $t=[\llbracket\bot^{\prime}\rrbracket t_{1}\cdots t_{\ell}]$, we associate the sequence $\sigma(t)=(|t_{1}|,\ldots,|t_{\ell}|)$. Finally, we set $t<_{\mathsf{lex}}t^{\prime}$ if and only if $\sigma(t)<_{\mathsf{lex}}\sigma(t^{\prime})$.

We can see that for any transition $(p,t)\xrightarrow{a|a^{\prime}|v}(p,t^{\prime})$ with $t=[\llbracket\bot^{\prime}\rrbracket_{k+1}t_{1}\cdots t_{\ell-1}t_{\ell}]_{k+2}$ and $t_{\ell}=[\llbracket\#\rrbracket_{k}s_{1}\cdots s_{m}]_{k+1}$ and $\sigma(t)=(x_{1},x_{2},\ldots,x_{n})$, we have $t^{\prime}<_{\mathsf{lex}}t$:

• •

If $(a|a^{\prime}|v)=(\varepsilon|\mathtt{0}|\mathsf{pop}_{k+1}\mathsf{push}_{k+2})$ then $\sigma(t^{\prime})=(x_{1},x_{2},\ldots,x_{n-1},x_{n}-1,x_{n}-1)$.

• •

If $(a|a^{\prime}|v)=(\mathtt{0}|\mathtt{1}|\mathsf{pop}_{k+1}\mathsf{push}_{k+2})$ then $\sigma(t^{\prime})=(x_{1},x_{2},\ldots,x_{n-1},x_{n}-1,x_{n}-1)$.

• •

If $(a|a^{\prime}|v)=(\varepsilon|\#|\mathsf{pop}_{k+2})$ then $\sigma(t^{\prime})=(x_{1},x_{2},\ldots,x_{n-1})$.

Hence $\mathcal{A}^{\prime}$ must reach $q^{\prime}_{f}$ in finitely many steps.

Appendix D Proof of 5.6

A separator witnessing $X\mathord{\mid}Y$ also witnesses $X_{i}\mathord{\mid}Y_{j}$ for every $i\in[1,n]$ and $j\in[1,m]$. This shows the “only if” direction.

For the “if” direction, suppose $R_{i,j}\subseteq M$ satisfies $X_{i}\subseteq R_{i,j}$ and $R_{i,j}\cap Y_{j}=\emptyset$. We claim that $R=\bigcup_{i=1}^{n}\bigcap_{j=1}^{m}R_{i,j}$ witnesses $X\mathord{\mid}Y$.

Since $X_{i}\subseteq R_{i,j}$ for every $i\in[1,n]$, we have $X_{i}\subseteq\bigcap_{j=1}^{m}R_{i,j}$ and hence $X=\bigcup_{i=1}^{n}X_{i}\subseteq R$. On the other hand, for every $i\in[1,n]$ and $k\in[1,m]$, we have $Y_{k}\cap R_{i,k}=\emptyset$ and thus $Y_{k}\cap\bigcap_{j=1}^{m}R_{i,j}=\emptyset$. This implies

[TABLE]

Appendix E Proof of 5.7

The following proof is from [14], but we include it for the convenience of the reader.

Proof.

Suppose $L\subseteq R$ and $R\cap TK=\emptyset$ for some regular $R$. Then clearly $T^{-1}L\subseteq T^{-1}R$ and $T^{-1}R\cap K=\emptyset$. Therefore, the regular set $T^{-1}R$ witnesses $T^{-1}L\mathord{\mid}K$. Conversely, if $T^{-1}L\mathord{\mid}K$, then $K\mathord{\mid}T^{-1}L$ and hence, by the first direction, $(T^{-1})^{-1}K\mathord{\mid}L$. Since $(T^{-1})^{-1}=T$, this reads $TK\mathord{\mid}L$ and thus $L\mathord{\mid}TK$. ∎