Satisfiability of ECTL^∗ with Local Tree Constraints

Abstract

Recently, we have shown that satisfiability for the temporal logic E C T L ^∗ with local constraints over (ℤ, <, =) is decidable using a new technique (Carapelle et al., 2013). This approach reduces the satisfiability problem of E C T L ^∗ with constraints over some structure $\mathcal {A}$ (or class of structures) to the problem whether $\mathcal {A}$ has a certain model theoretic property that we called EHD (for “existence of homomorphisms is definable”). Here we apply this approach to structures that are tree-like and obtain several results. We show that satisfiability of E C T L ^∗ with constraints is decidable over (i) semi-linear orders (i.e., tree-like structures where branches form arbitrary linear orders), (ii) ordinal trees (semi-linear orders where the branches form ordinals), and (iii) infinitely branching trees of height h for each fixed $h\in \mathbb {N}$. We prove that all these classes of structures have the property EHD. In contrast, we introduce Ehrenfeucht-Fraïssé-games for W M S O + B (weak M S O with the bounding quantifier) and use them to show that the infinite (order) tree does not have the EHD-property. As a consequence, our technique cannot be used to establish whether satisfiability of E C T L ^∗ with constraints over the infinite (order) tree is decidable. A preliminary version of this paper has appeared as (Carapelle et al., 2015).

Appendix: A universal semi-linear order

In the following we define a semi-linear order which is universal for the class of all countable semi-linear orders. The fact that this order is universal is known to the experts in the field of semi-linear orders. Unfortunately, to our best knowledge there is no proof of this fact in the literature. Hence we provide a proof for this result.

Definition 31

Let $\mathcal {U}=(U, <, \mathrel {\bot })$ be the countable semi-linear order with:

• $U = (\mathbb {N}\mathbb {Q})^{*}$ (the set of all finite sequences n ₁ q ₁⋯n _k q _k with k ≥ 0, $n_{1}, \ldots , n_{k} \in \mathbb {N}$ and $q_{1}, \ldots , q_{k} \in \mathbb {Q}$),

• < is the strict order induced by n ₁ p ₁ n ₂ p ₂⋯n _k p _k ≤ m ₁ q ₁ m ₂ q ₂⋯m _l q _l iff

  • k ≤ l, n _i = m _i for all 1 ≤ i ≤ k, p _i = q _i for all 1 ≤ i ≤ k − 1 and p _k ≤ q _k , and

  • ⊥ = ⊥_<.

We call $\mathcal {U}$ the universal countable semi-linear order.

Note that Droste [13] has already studied this and similar orders.

For u = n ₁ p ₁ n ₂ p ₂⋯n _k p _k ∈ U (with k ≥ 1) and $q\in \mathbb {Q}$, we define

$$ u+q = n_{1}p_{1}n_{2}p_{2}{\cdots} n_{k-1}p_{k-1}n_{k}(p_{k}+q) . $$

(8)

We say that a countable semi-linear order (A, <, ⊥ ) is closed under finite infima if for each finite set $S \subseteq A$ the linear order {a ∈ A∣a ≤ s for alls ∈ S} has a maximal element, which is denoted by $\inf (S)$. Let $E=(a_{i})_{i\in \mathbb {N}}$ be a repetition-free enumeration of A. We say E is closed under infima if for each initial subset $A_{i}=\{a_{1}, a_{2}, \dots , a_{i}\}$ and each $S\subseteq A_{i}$ we have $\inf (S) \in A_{i}$.

Lemma 32

Let $\mathcal {A} = (A, < ,\mathrel {\bot })$ be a countable semi-linear order. There is a countable semi-linear order $\mathcal {B}$ that is closed under finite infima and an injective homomorphism from $\mathcal {A}$ to $\mathcal {B}$.

Proof

For a nonempty subset $S\subseteq A$ we set ↓S = {a ∈ A∣∀s ∈ S(a ≤ s)}. Let $\bar A$ be the set of finite nonempty subsets of A, which is obviously countable. We define an equivalence on $\bar A$ by setting S ∼ T iff ↓S = ↓T. For all $S\in \bar A$, [S] denotes its equivalence class. Let B be the set of all equivalence classes. We define an order ⊏ on B by [S] ⊏ [T] if and only if ↓S ⊊ ↓T.

We claim that $\mathcal {B} = (B, \sqsubset , \mathrel {\bot }_{\sqsubset })$ is a semi-linear order that is closed under finite infima and that the map ϕ given by ϕ(a) ↦ [{a}] is an injective homomorphism from $\mathcal {A}$ to $\mathcal {B}$.

• $\mathcal {B}$ is obviously a partial order. Moreover, note that ↓S is a linear and downwards closed suborder of $\mathcal {A}$ for every nonempty finite set $S \subseteq A$. In order to show that $\mathcal {B}$ is semi-linear, assume that $[S_{1}] \sqsubset [S]$ and $[S_{2}] \sqsubset [S]$, i.e., ${\downarrow } S_{1} \subsetneq {\downarrow } S$ and ${\downarrow } S_{2} \subsetneq {\downarrow } S$. Thus, all elements from ↓S ₁ and all elements from ↓S ₂ are comparable. Since both sets are downwards closed, this directly implies that either [S ₁]=[S ₂], $[S_{1}] \sqsubset [S_{2}]$ or $[S_{2}] \sqsubset [S_{1}]$.

• Let us show that $\mathcal {B}$ is closed under finite infima: Let S, S ₁, … , S _n be finite nonempty subsets of A and assume that $[S] \sqsubseteq [S_{i}]$ for all 1 ≤ i ≤ n. Thus, ${\downarrow } S \subseteq {\downarrow } S_{i}$. Hence, ${\downarrow } S \subseteq \bigcap _{i=1}^{n} {\downarrow } S_{i} = {\downarrow } \bigcup _{i=1}^{n} S_{i}$. Since ${\downarrow } \bigcup _{i=1}^{n} S_{i} \subseteq {\downarrow } S_{i}$ for all 1 ≤ i ≤ n, $[\bigcup _{i=1}^{n} S_{i}] = \inf (\{ [S_{1}], \ldots , [S_{n}] \})$.

• For a, b ∈ A with a ≠ b we have b ∉ ↓{a} or a ∉ ↓{b}. Thus, ϕ is an injective map from A to B. Moreover, a < b implies ${\downarrow }\{a\} \subsetneq {\downarrow }\{b\}$, i.e., $\phi (a) \sqsubset \phi (b)$. Similarly, a ⊥ b implies a ∉ ↓{b} and b ∉ ↓{a}, i.e., $\phi (a) \mathrel {\bot }_{\sqsubset } \phi (b)$.

□

Lemma 33

Let $\mathcal {A} =(A, < , \mathrel {\bot })$ be a countable semi-linear order that is closed under finite infima. There is a repetition-free enumeration of $\mathcal {A}$ , which is closed under infima.

Proof

Fix an arbitrary repetition-free enumeration $(a_{i})_{i\in \mathbb {N}}$ of A. Assume that we have constructed a sequence $b_{1}, b_{2}, \dots , b_{i}$ such that $B_{j} = \{b_{1}, b_{2}, \dots , b_{j}\}$ is closed under infima for every j ≤ i. Let k ∈ ℕ be minimal with a _k ∉ B _i . Let $b^{\prime }_{1} < b^{\prime }_{2} < {\dots } b^{\prime }_{m} < a_{k}$ be the list of all infima of the form $\inf (S\cup \{a_{k}\})$ for $S\subseteq B_{i}$ that are not contained in B _i . This list is indeed linearly ordered by < since all elements in the list are bounded by a _k . Now set $b_{i+l} = b^{\prime }_{l}$ for all 1 ≤ l ≤ m and set b _{i + m+1} = a _k . The resulting sequence b ₁, … , b _{i + m + 1} contains a _k and $B_{j} = \{b_{1}, b_{2}, \dots , b_{j}\}$ is closed under infima for every j ≤ i + m + 1. This can be easily shown using the fact that $\inf (X \cup \{ \inf (Y)\}) = \inf (X \cup Y)$ for all sets X and Y.

Repeating this construction leads to an enumeration $(b_{i})_{i\in \mathbb {N}}$ of A with the desired property. □

Lemma 34

Let $\mathcal {A}=(A, \sqsubset , \mathrel {\bot }_{\sqsubset })$ be a countable semi-linear order. There exists an injective homomorphism from $\mathcal {A}$ into $\mathcal {U}$.

Proof

Due to Lemma 32 and 33, we may assume that $\mathcal {A}$ is closed under finite infima and that $(a_{i})_{i\in \mathbb {N}}$ is a repetition-free enumeration of A which is closed under finite infima. Set $A_{i}=\{a_{1}, \dots , a_{i}\}$ for i ≥ 1. Inductively, we construct injective homomorphisms ϕ _i :A _i → U (i ≥ 1) such that

• 1. ϕ _i+1 extends ϕ _i , and

• 2. for all $u = n_{1}p_{1}n_{2}p_{2}{\dots } n_{k}p_{k} \in {\mathsf {im}}(\phi _{i})$ and all 1 ≤ j ≤ k we have $p_{j}\in \frac {1}{2^{i}}\mathbb {Z}$.

Define ϕ ₁:A ₁ → U by $\phi _{1}(a_{1}) = 00 \in \mathbb {N}\mathbb {Q}$. Assume that ϕ _i has already been constructed. We distinguish two cases.

• 1. If there is some a ∈ A _i with $a_{i+1} \sqsubset a$ let $u = \inf \{a \in A_{i}\mid a_{i+1} \sqsubset a\}$. Note that $a_{i+1} \sqsubseteq u$. Since the enumeration is closed under infima, we have u ∈ A _i (and thus $a_{i+1} \sqsubset u$) and we can define $\phi _{i+1}(a_{i+1}) = \phi _{i}(u)+ (\frac {-1}{2^{i+1}})$, where we add according to (8). Note that ϕ _i+1(a _i+1) < ϕ _i (u) = ϕ _i+1(u). In order to prove that this defines a homomorphism, we distinguish the following cases:

  • (a) If $a_{i+1} \sqsubset a$ for some a ∈ A _i then u ⊑ a. Hence, ϕ _i+1(a _i+1) < ϕ _i+1(u) = ϕ _i (u) ≤ ϕ _i (a) = ϕ _i+1(a) as desired.

  • (b) If $a \sqsubset a_{i+1}$ for some a ∈ A _i , then a ⊏ u. Hence, ϕ _i+1(a) = ϕ _i (a) < ϕ _i (u). Since ϕ _i uses only rationals from $\frac {1}{2^{i}}\mathbb {Z}$, we conclude that $\phi _{i+1}(a) \leq \phi _{i}(u) + \frac {-1}{2^{i}} < \phi _{i+1}(a_{i+1})$ as desired.

  • (c) If $a_{i+1} \mathrel {\bot }_{\sqsubset } a$ for some a ∈ A _i , then $a \mathrel {\bot }_{\sqsubset } u$. By induction, ϕ _i+1(a) = ϕ _i (a) ⊥_< ϕ _i (u) = ϕ _i+1(u). Thus, the assumption ϕ _i+1(a) ≤ ϕ _i+1(a _i+1) leads by transitivity of ≤ to the contradiction ϕ _i+1(a) ≤ ϕ _i+1(u). Similarly, the assumption ϕ _i+1(a) > ϕ _i+1(a _i+1) yields $\phi _{i+1}(a) \geq \phi _{i+1}(a_{i+1})+\frac {1}{2^{i+1}} = \phi _{i+1}(u)$. We can conclude that ϕ _i+1(a) ⊥_< ϕ _i+1(a _i+1) as desired.

• 2. Otherwise, for all j ≤ i we know that $\inf \{a_{j},a_{i+1}\}$ is strictly below a _i+1 and hence belongs to A _i (since the enumeration is closed under infima). In particular, the set {a ∈ A _i ∣a < a _i+1} is not empty. By semi-linearity, $u=\max \{ a\in A_{i}\mid a< a_{i+1}\}$ is well-defined. Since i m(ϕ _i ) is finite, there is some $n\in \mathbb {N}$ such that φ _i (u)n0 is incomparable to all elements from the set φ _i (A _i ∖{a ∈ A _i ∣a ≤ u}). Extending ϕ _i by setting ϕ _i+1(a _i+1) = φ _i (u)n0 is easily shown to be a homomorphism.

Finally, the limit of $(\phi _{i})_{i\in \mathbb {N}}$ clearly defines an injective homomorphism from $\mathcal {A}$ into $\mathcal {U}$. □