On the Expressive Power of TeamLTL and First-Order Team Logic over Hyperproperties

Abstract

In this article we study linear temporal logics with team semantics ($\mathrm {TeamLTL}$) that are novel logics for defining hyperproperties. We define Kamp-type translations of these logics into fragments of first-order team logic and second-order logic. We also characterize the expressive power and the complexity of model-checking and satisfiability of team logic and second-order logic by relating them to second- and third-order arithmetic. Our results set in a larger context the recent results of Lück showing that the extension of TeamLTL by the Boolean negation is highly undecidable under the so-called synchronous semantics. We also study stutter-invariant fragments of extensions of TeamLTL.

Appendix

Proofs

Proof

(Sketch of the proof of Theorem 1). Let $T'\subseteq T$ and $i:T'\rightarrow \mathbb {N}$. We use $T'$ and i as a means to refer to any team that might be relevant for the evaluation of $\mathrm {TeamLTL} ^a(\sim )$ formulas when starting the evaluation with T. On the first-order side the corresponding team will be

$$ S^x_{T',i} := \{ s \mid s(x)= (t,i(t)) \text{ and } t \in T' \}. $$

We can now show using simultaneous induction on $\varphi$ that for all $T'\subseteq T$, $i:T'\rightarrow \mathbb {N}$, and $u\in \{x,y,z\}$

$$ \{t[i(t),\infty )\mid t\in T'\}\models \varphi \Leftrightarrow \mathcal {M}_T\models _{S^u_{T',i}} ST_u(\varphi ).$$

• Assume $\varphi = p_i$ and $T'\subseteq T$, $i:T'\rightarrow \mathbb {N}$ are arbitrary. Now

  $$\begin{aligned} \{t[i(t),\infty )\mid t\in T'\}\models \varphi\Leftrightarrow & {} p_i \in t(i(t)) \text{ for } \text{ all } t\in T' \\\Leftrightarrow & {} (t,i(t))\in P^{ \mathcal {M}_T}_i \text{ for } \text{ all } t\in T' \\\Leftrightarrow & {} \mathcal {M}_T\models _{S^x_{T',i}} P_i(x)\\\Leftrightarrow & {} \mathcal {M}_T\models _{S^x_{T',i}} ST_x(\varphi ) \end{aligned}$$

  Note that the second equivalence holds by the definition of the structure $\mathcal {M}_T$ and the third equivalence by first-order team semantics of atomic formulas.

• Assume $\varphi = X\psi$ and $T'\subseteq T$ and $i:T'\rightarrow \mathbb {N}$ are arbitrary. Let $i^+$ be defined by $i^+(t):=i(t)+1$ for all t. Now

  $$\begin{aligned} \{t[i(t),\infty )\mid t\in T'\}\models \varphi\Leftrightarrow & {} \{t[i^+(t),\infty )\mid t\in T'\}\models \psi \\\Leftrightarrow & {} \mathcal {M}_T\models _{S^y_{T',i^+}} ST_y(\psi )\\\Leftrightarrow & {} \mathcal {M}_T\models _{S^x_{T',i}} \exists y(S(x,y)\wedge ST_y(\psi ))\\\Leftrightarrow & {} \mathcal {M}_T\models _{S^x_{T',i}} ST_x(X\varphi ) \end{aligned}$$

  The second equivalence above holds by the induction assumption for $ST_y(\psi )$. For the the third equivalence we use the facts that the supplementation function F for y is uniquely determined by the formula and x is not free in $ST_y(\psi )$. Note that by locality it holds that

  $$ \mathcal {M}_T\models _{S^x_{T',i} [F/y]} ST_y(\psi ) \Leftrightarrow \mathcal {M}_T\models _{S^y_{T',i^+}} ST_y(\psi ), $$

  since $S^y_{T',i^+}$ is the reduct of $S^x_{T',i}[F/y]$ to the team with domain $\{y\}$.

The proof for the connectives is straightforward and for the temporal operator U it is similar to the case of X.

Proof

(Proof of Corollary 1). We show that $T\models \varphi$ if and only if $\mathcal {M}_T\models \psi$, where $\psi$ is the sentence:

$$\forall x ( \exists y (y<x) \vee (\forall y (\lnot y<x )\wedge ST_x(\varphi ))). $$

Note that

$$\begin{aligned} \mathcal {M}_T\models \psi\Leftrightarrow & {} \mathcal {M}_T\models _{\{\emptyset \}[\mathrm {dom}(\mathcal {M}_T)/x]} \exists y (y<x) \vee (\forall y \lnot (y<x) \wedge ST_x(\varphi )). \\\Leftrightarrow & {} \mathcal {M}_T\models _{S^x_{T} } ST_x(\varphi ). \end{aligned}$$

In the second line the team $\{\emptyset \}[\mathrm {dom}(\mathcal {M}_T)/]x]$ (i.e., $\mathrm {dom}(\mathcal {M}_T)$) has to be split into two disjoint parts: the subset of elements having a predecessor and to those not having a predecessor ($=S^x_{T}$). The first disjunct is then trivially satisfied (by flatness it behaves classically) hence we arrive at the case which is equivalent to $T\models \varphi$ by Theorem 1.

Proof

(Proof of Lemma 1). Suppose 2. Now $T = T \cup \emptyset$, however the empty set is only stutter equivalent to itself. Thus $T \equiv _{st}^a T'$.

Suppose 1 and suppose that $T = T_1 \cup T_2$, hence we have asynchronous stuttering functions F of T and G of $T'$, such that $T[F] = U = T'[G]$. We consider the subteams induced by the stuttering function, i.e. $T_i[F|T_i]$. Since $T[F] = T'[G]$, there exist subteams $T'_1,T'_2$ such that $T'_i[G|T'_i] = T_i[F|T_i]$. Thus $T_i \equiv _{st}^a T'_i$.

It remains to show that the subteams $T'_1$ and $T'_2$ constitute the entirety of the team $T'$. It is clear that $T'_1\cup T'_2 \subset T'$, so it remains to show the converse. Let $t' \in T'$. Now, since $T \equiv _{st}^a T'$, there exists a $t \in T$ such that $t[F|\{t\}] = t'[G|\{t'\}]$. By our assumption, the trace t belongs to either $T_1$ or $T_2$. Without loss of generality we may assume that $t \in T_1$, but then $t'[G|\{t'\}] \in T'_1[G|T'_1]$. Since the team $T'$ is a set, i.e. it does not contain duplicates, we may conclude that $t' \in T'$.

Proof

(Proof of Lemma 2). By the definition of the asynchronous stuttering function, for each coordinate of j(t) there exists a constant $a_t$ such that $f_t(a_t) \le j(t)$ and $t(f_t(a_t)) = t(j(t))$. Let i be the configuration defined by $i(t) = f_t(a_t)$. Now we can use the stuttering function F to construct stuttering functions $F'$ and $F''$ for $T[i,\infty )$ and $T[j,\infty )$ respectively. First of we define $F'$ via $f'_t(n) = f_t(n + a_t)$ for all $t \in T$, which clearly is a stuttering function of $T[i,\infty )$. Next we define

$$\begin{aligned} F''(n) = {\left\{ \begin{array}{ll} (j(t))_{t\in T} &{}\text{ if } n = 0\\ F'(n) &{}\text{ otherwise. } \end{array}\right. } \end{aligned}$$

Since $t(f_t(a_t)) = t(i(t)) = t(j(t))$ for all $t \in T$, it follows that $T[i,\infty )[F'] = T[j,\infty )[F'']$. Thus $T[i,\infty ) \equiv _{st}^a T[j,\infty )$.

For the second claim we use the assumption that $T \equiv _{st}^a T'$. We let the configuration i be as above. Now for all n, $t \in T$ and $t' \in T'$ it holds that $t(f_t(n)) = t'(g_{t'}(n))$. Thus there exists some configuration $k:T' \rightarrow \mathbb N$ such that $t'(g_{t'}(k(t'))) = t(f_t(i(t)))$, which allows us to define the stuttering function $G'$ of $T'[k,\infty )$ as $g'_{t'}(n) = g_{t'}(n + k(t'))$ for all $t' \in T'$. Clearly now $T'[k,\infty )[G'] = T[i,\infty )[F']$, and hence $T[i,\infty ) \equiv _{st}^a T'[k,\infty )$.

Proof

(Proof of Theorem 4). We define a inductive translation $\mathrm {Tr}$ from second-order logic to third order arithmetic as follows. The key ideas in the translation are:

• an element of the domain $T\times \mathbb {N}$ of the structure $\mathcal {M}_T$ can be uniquely identified by specifying a trace t and $i\in \mathbb {N}$. Hence, syntactically, a first-order variable x can be encoded by a pair of variables $(R_x, z_x)$ where $R_x$ is a binary relation and $z_x$ is a first-order variable;

• a subset of $T\times \mathbb {N}$ is a set of pairs (t, i) and hence in the translation a unary relation X is encoded by a third-order variable $\mathfrak {b}_X$ of type ((2), (1)), where the unary relation encodes i by the singleton $\{ i\}$.

Define now a formula translation $\mathrm {Tr}$ as follows. We omit below the obvious cases of the Boolean connectives and, for clarity, we consider only unary relations X on the side of SO. It is straightforward to write the corresponding translations also for relations of arbitrary arities.

$$\begin{aligned} \begin{aligned}&\mathrm {Tr}(x = y) := \forall u \forall v (R_x(u,v) \leftrightarrow \\&\quad R_y(u,v))\wedge z_x = z_y\\&\mathrm {Tr}(x \le y) := \forall u \forall v (R_x(u,v) \leftrightarrow \\&\quad R_y(u,v))\wedge z_x \le z_y\\&\mathrm {Tr}(P_i(x)) := \mathfrak {a}(R_x) \wedge R_x(z_x,i)\\ \end{aligned} \quad \begin{aligned}&\mathrm {Tr}(X(x)) := \exists Y(\mathfrak {b}_X (R_{x},Y)\wedge Y=\{z_x\}) \\&\mathrm {Tr}(\exists x\varphi ) := \exists R_x \exists z_x( \mathfrak {a}(R_x) \wedge \mathrm {Tr}(\varphi ))\\&\mathrm {Tr}(\exists X\varphi ) := \exists \mathfrak {b}_X(\forall R\forall Y( \mathfrak {b}_X(R,Y)\rightarrow \mathfrak {a}(R)\wedge \\&\quad |Y|=1)\wedge \mathrm {Tr}(\varphi )) \end{aligned} \end{aligned}$$

In the above formulas, i denotes a (definable) constant. It is now straightforward to show using induction on $\varphi$ that for all s and $s^*$:

$$\begin{aligned} \mathcal {M}_T \models _s \varphi \iff (\mathbb {N},+,\times ,\le 0,1)\models _{s^*} \mathrm {Tr}(\varphi )(\mathcal {A}_T /\mathfrak {a}), \end{aligned}$$

where the interpretations s and $s^*$ relate to each other as described above.