Closure Ordinals of the Two-Way Modal $\mu$-Calculus

Abstract

The closure ordinal of a $\mu$-calculus formula $\varphi (x)$ is the least ordinal $\alpha$, if it exists, such that, in any model, the least fixed point of $\varphi (x)$ can be computed in at most $\alpha$ many steps, by iteration of the meaning function associated with $\varphi (x)$, starting from the empty set. In this paper we focus on closure ordinals of the two-way modal $\mu$-calculus. Our main technical contribution is the construction of a two-way formula $\varphi _n$ with closure ordinal $\omega ^n$ for an arbitrary $n\in \omega$. Building on this construction, as our main result we define a two-way formula $\varphi _\alpha$ with closure ordinal $\alpha$ for an arbitrary $\alpha <\omega ^\omega$.

A Proof of the Main Result in Section 3

The statement of Theorem 1 from Sect. 3 is a direct consequence of the following lemmas.

Lemma

 1. Let $0<n<\omega$ be a finite ordinal. Then there is a model $\mathbb {S}$ where $\gamma _x(\varphi _n, \mathbb {S}) = \omega ^n$.

Proof

For the rest of the proof we adopt the following notation: since every ordinal $\alpha <\omega ^n$ can be written as $\omega ^{n-1}\cdot k_1 + \ldots + \omega \cdot k_{n-1} + k_n$, we also denote $\alpha$ as $(k_1, \ldots , k_n)$. From now on, if we write $\alpha = (k_1, \ldots , k_n)$ we mean that $\alpha = \omega ^{n-1}\cdot k_1 + \ldots + \omega \cdot k_{n-1} + k_n$. Also, if a tuple $(k_1, \ldots , k_n)$ is of the form $(k_1, \ldots , k_i, 0, \ldots , 0)$, we mean that $k_j = 0$ for $i+1 \le j \le n$.

Fix $n>0$ and let $\varphi := \varphi _n$ as an abbreviation. We define $\mathbb {S}= (S,R,V)$ to be the model where:

• $S := \omega ^{n} = \{(k_1, \ldots , k_n)\;|\; k_j\in \omega \}$;

• $R := \bigcup \limits _{1\le i\le n}\{((k_1, \ldots , k_i + 1, 0,\ldots , 0 ), (k_1, \ldots , k_i, 0,\ldots , 0) ) \;|\; k_j\in \omega \}$;

• for $1\le i\le n$, $V(q_i) := \{(k_1, \ldots , k_{n-i+1} + 1, 0, \ldots , 0)\;|\; k_j\in \omega \}$.

Note that $R[(0, \ldots , 0)] = \emptyset$ and that $(0, \ldots , 0)$ falsifies $q_i$ for every $1\le i\le n$.

Before proving the key claim we make an observation about notation. Note that an ordinal $\beta <\omega ^n$ can both be seen as an element $\beta \in S = \omega ^n$ of the model and as a subset $\beta = \{\gamma \;|\; \gamma <\beta \}\subseteq S = \omega ^n$. To avoid confusion, until the end of the proof we write $\beta$ when we consider it as an element of the domain, and $S_\beta$ when we consider it as a subset of the domain ($S_\beta = \beta$ holds in any case).

Claim

For every $\alpha <\omega ^n$, $\varphi ^\alpha = S_\alpha$.

Proof of Claim. The proof goes by induction on $\alpha$. The case for $\alpha = 0$ is immediate. If $\alpha$ is a limit ordinal, then $\varphi ^\alpha = \bigcup _{\beta< \alpha } \varphi ^\beta =_{IH} \bigcup _{\beta <\alpha } S_\beta = S_\alpha$.

Now suppose that $\alpha = \beta + 1$. We want to show that $\varphi ^{\beta + 1} = S_{\beta + 1}$. We have that $\varphi ^{\beta + 1} = \varphi ^\mathbb {S}_x(\varphi ^\beta ) =_{IH} \varphi ^\mathbb {S}_x(S_\beta )$: we show

$$\begin{aligned} \varphi ^\mathbb {S}_x(S_\beta ) = S_{\beta + 1}. \end{aligned}$$

(3)

For the $\supseteq$ inclusion of (3) it suffices to show that $\mathbb {S}[x\mapsto S_\beta ], \beta \Vdash \varphi$, since $S_{\beta +1} = S_\beta \cup \{\beta \}$ and $S_\beta = \varphi ^\beta \subseteq \varphi ^{\beta + 1} = \varphi ^\mathbb {S}_x(\varphi ^\beta )$. If $\beta = 0 = (0,\ldots , 0)$ we are done. If $\beta = (k_1, \ldots , k_n + 1)$, then $\beta \in V(q_1)$ and $(k_1, \ldots , k_n)\in S_\beta \cap R[\beta ]$, so $\mathbb {S}[x\mapsto S_\beta ],\beta \Vdash c_1\wedge Fx$ and $\beta \in \varphi ^\mathbb {S}_x(S_\beta )$.

Otherwise let $\beta = (k_1, \ldots , k_{i}+1, 0, \ldots , 0)$ for some $1\le i< n$, so that $\beta \in V(q_{n-i+1})$. Note that

$$\begin{aligned} (k_1, \ldots , k_i, k, 0,\ldots , 0)&\in S_\beta \text { for all } k\in \omega , \\ (k_1,\ldots , k_i, 0,0, \ldots , 0)&\in R[\beta ] \text { and } \\ (k_1, \ldots , k_i, k, 0,\ldots , 0)&\in R[(k_1, \ldots , k_i, k+1, 0,\ldots , 0)]\cap V(q_{n-i}) \text { for all } k> 0. \end{aligned}$$

By construction of the model $\beta \Vdash \pi ^\infty _{n-i}$ also holds: then , so $\beta \in \varphi ^\mathbb {S}_x(S_\beta )$.

Now we move to the $\subseteq$ inclusion of (3). Let $\gamma \in \varphi ^\mathbb {S}_x(S_\beta )$. We want to show that $\gamma \in S_{\beta + 1}$. Since $\mathbb {S}[x\mapsto S_\beta ], \gamma \Vdash \varphi$ holds, we proceed by case distinction as to which disjunct of $\varphi$ is satisfied by $\gamma$. If $\gamma \Vdash G\bot$ then $\gamma = 0\in S_{\beta + 1}$. If $\gamma \Vdash c_1\wedge Fx$, then $\gamma \in V(q_1)$, so that $\gamma = (k_1, \ldots , k_n + 1)$ and $\gamma ' = (k_1, \ldots , k_n)\in R[\gamma ]\cap S_\beta$: as $\gamma '\in S_\beta$, then $\gamma = \gamma ' + 1 \in S_{\beta +1}$.

Now suppose for some $2\le i\le n$. Then $\gamma \in V(q_i)$, so $\gamma = (k_1, \ldots , k_{n-i+1}+1, 0, \ldots , 0)$. For $j\in \omega$ let

$$\delta _j := (k_1, \ldots , k_{n-i+1}, j,0, \ldots , 0). $$

By construction $\delta _0\in R[\gamma ]$ and $\delta _{j}\in R[\delta _{j+1}]$ for all $j\ge 0$. Since then $\delta _j\in S_\beta$ for all $j>0$. Hence

$$\beta> (k_1, \ldots , k_{n-i+1}, j, 0, \ldots , 0) \text { for all } j>0,$$

implying $\beta \ge (k_1, \ldots , k_{n-i+1}+1,0, \ldots , 0) = \gamma$, so $\gamma \in S_{\beta + 1}$.   $\vartriangleleft$

Now that we have the claim, it follows that there is a $\gamma \in \varphi ^{\omega ^n}\backslash \varphi ^\beta$ for each $\beta <\omega ^n$.

Proposition 4

For all $m,n\in \omega$, if $m\ge n$, then $\pi ^\infty _m \models \pi ^\infty _n$. Moreover, if $\mathbb {S}$ is a model, $\mathbb {S},s\Vdash \pi ^\infty _m$ for some state s, and $t_0 t_1 \ldots$ is an $R^{-1}$-path witnessing the truth of $\pi ^\infty _m$ at s, then $t_j\Vdash \pi ^\infty _m$ for all $j\in \omega$.

Lemma

 2. Let $\mathbb {S}= (S,R,V)$ be a model and let $n\in \omega$. For $1\le i\le n$, let $t_0 t_1 t_2\ldots$ be an infinite $R^{-1}$-path such that

$$\begin{aligned} \mathbb {S},t_0\Vdash \pi ^\infty _{i-1}{} \textit{ and, for all }j>0, \mathbb {S},t_j\Vdash c_i\wedge \pi ^\infty _{i-1}. \end{aligned}$$

Then, for any ordinal $\alpha$: if $t_0\in \varphi ^\alpha _n$ then $t_j\in \varphi _n^{\alpha + \omega ^{i-1}\cdot j +1}$ for all $j\in \omega$.

Proof

We prove the statement by induction on $1\le i\le n$.

As the base case take $i = 1$, so that by assumption we have an infinite $R^{-1}$-path $t_0t_1t_2\ldots$ such that $\mathbb {S}, t_j\Vdash c_1$ for all $j>0$. Let $t_0\in \varphi ^\alpha _n$. We want to show that, for all $j\in \omega$, $t_j\in \varphi ^{\alpha + j + 1}_n$: we prove this by induction on $j\in \omega$. If $j = 0$, then $t_0\in \varphi ^\alpha _n\subseteq \varphi ^{\alpha +1}_n$. Next, inductively assume that $t_j \in \varphi _n^{\alpha + j + 1}$: then, since $t_{j}\in R[t_{j+1}]$, it follows that $\mathbb {S}[x\mapsto \varphi _n^{\alpha + j + 1}], t_{j+1}\Vdash (c_1\wedge Fx)$, so $t_{j+1}\in \varphi ^{\alpha + (j+1)+1}_n$.

For the inductive step assume that the statement holds for i. We prove it for $i+1$, where $i<n$. Suppose then that $t_0 t_1 t_2\ldots$ is an infinite $R^{-1}$-path such that $t_0\Vdash \pi ^\infty _{i}$ and for all $j>0$, $t_j\Vdash c_{i+1}\wedge \pi ^\infty _{i}$. Let $t_0\in \varphi ^\alpha _n$. We want to show that

$$\begin{aligned} \text {for every }j\in \omega , t_j\in \varphi ^{\alpha + \omega ^{i}\cdot j +1}_n. \end{aligned}$$

The proof of this last statement goes by induction on $j\in \omega$. The base case with $j = 0$ follows immediately, as by assumption $t_0\in \varphi _n^{\alpha }$.

Now suppose that $t_j\in \varphi ^{\alpha + \omega ^{i}\cdot j +1}_n$: we show that $t_{j+1}\in \varphi ^{\alpha + \omega ^{i}\cdot (j+1) +1}_n$. By assumption $t_j\in R[t_{j+1}]$ and $t_j\Vdash \pi ^\infty _i$, which in particular means that there is an infinite $R^{-1}$-path $u_0 u_1\ldots$ (with $u_0 = t_j$) such that, for all $k>0$, $u_k\Vdash c_i$. But then this path satisfies the conditions of the inductive hypothesis: by Proposition 4, since $u_0\Vdash \pi ^\infty _{i}$, then $u_0\Vdash \pi ^\infty _{i-1}$, and for every $k>0$, $u_k\Vdash c_i\wedge \pi _{i-1}^\infty$. Then, by inductive hypothesis, since $u_0=t_j\in \varphi ^{\alpha + \omega ^{i}\cdot j +1}_n$ it follows that, for every $k\in \omega$, $u_k\in \varphi _n^{\alpha + \omega ^i\cdot j + 1 + \omega ^{i-1}\cdot k + 1}$. Since for all $k\in \omega$ it holds that

$$\begin{aligned} \omega ^i\cdot j + 1 + \omega ^{i-1}\cdot k + 1&< \omega ^i\cdot j + 1 + \omega ^i&(\text {as } \omega ^{i-1}\cdot k + 1 < \omega ^i \text { for } i>0) \\&= \omega ^i\cdot j + \omega ^i&(1 + \omega ^i = \omega ^i \text { for } i> 0) \\&= \omega ^i\cdot (j+1)&\end{aligned}$$

then also

$$\begin{aligned} \alpha + \omega ^i\cdot j + 1 + \omega ^{i-1}\cdot k + 1 <\alpha + \omega ^i\cdot (j+1). \end{aligned}$$

It follows that $u_k\in \varphi _n^{\alpha + \omega ^i\cdot (j+1)}$ for all $k\in \omega$, so that

$$\begin{aligned} \mathbb {S}[x\mapsto \varphi _n^{\alpha + \omega ^i\cdot (j+1)}],t_{j+1}\Vdash c_{i+1}\wedge \pi ^\infty _i\wedge F(\nu y. P(x\wedge y \wedge c_i)). \end{aligned}$$

We conclude that $t_{j+1}\in \varphi _n^{\alpha + \omega ^i\cdot (j+1)+1}$ as desired.

Lemma

 3. For an arbitrary model $\mathbb {S}$ and $0<n<\omega$: $\gamma _x(\varphi _n, \mathbb {S}) \le \omega ^n$.

Proof

It is sufficient to prove that $\varphi ^{\omega ^n+1}_n \subseteq \varphi ^{\omega ^n}_n$ for every model $\mathbb {S}$. Let $s\in \varphi ^{\omega ^n+1}_n$, that is, $\mathbb {S}[x\mapsto \varphi ^{\omega ^n}_n], s\Vdash \varphi _n$. We proceed by case distinction as to which disjunct of $\varphi _n$ is satisfied by s to prove that $s\in \varphi ^{\omega ^n}_n$. If $s\Vdash G\bot$ then $s\in (\varphi _n)^\mathbb {S}_x(\emptyset )\subseteq \varphi ^{\omega ^n}_n$, while if $s\Vdash c_1\wedge Fx$, then there is a $t\in R[s]$ such that $t\in \varphi ^\alpha _n$ for some $\alpha <\omega ^n$, so that $s\in \varphi ^{\alpha + 1}_n\subseteq \varphi ^{\omega ^n}_n$.

Now suppose for some $2\le i\le n$. Then in particular there is a point $t\in R[s]$ and a $R^{-1}$-path $t_0 t_1\ldots$ such that: (i) $t\in R[t_0]$, (ii) for all $j\in \omega$, $t_j\in \varphi ^{\omega ^n}_n$ and $t_j\Vdash c_{i-1}$. In particular, $t_0\in \varphi ^\alpha _n$ for some $\alpha <\omega ^n$. Observe that $\varphi _n\wedge c_{i-1}\wedge F\top \models \pi ^\infty _{i-2}$: this implies that $t_j\Vdash \pi ^\infty _{i-2}$ for all $j\in \omega$, since $t_j\in \varphi ^{\omega ^n}_n$, $t_j\Vdash c_{i-1}$ and $R[t_j]\ne \emptyset$. This means that we can apply Lemma 2 and it follows that $t_j\in \varphi _n^{\alpha + \omega ^{i-2}\cdot j +1}\subseteq \varphi _n^{\alpha + \omega ^{i-1}}$ for all $j\in \omega$. Hence $\mathbb {S}[x\mapsto \varphi ^{\alpha + \omega ^{i-1}}], s\Vdash \varphi _n$ and $s\in \varphi _n^{\alpha + \omega ^{i-1} + 1}\subseteq \varphi ^{\omega ^n}_n$ (since $i \le n$ and $\alpha <\omega ^n$ imply $\alpha + \omega ^{i-1}+1 < \omega ^n$).