Revisiting bisimilarity and its modal logic for nondeterministic and probabilistic processes

Abstract

The logic PML is a probabilistic version of Hennessy–Milner logic introduced by Larsen and Skou to characterize bisimilarity over probabilistic processes without internal nondeterminism. In this paper, two alternative interpretations of PML over nondeterministic and probabilistic processes as models are considered, and two new bisimulation-based equivalences that are in full agreement with those interpretations are provided. The new equivalences include as coarsest congruences the two bisimilarities for nondeterministic and probabilistic processes proposed by Segala and Lynch. The latter equivalences are instead known to agree with two versions of Hennessy–Milner logic extended with an additional probabilistic operator interpreted over state distributions in place of individual states. The new interpretations of PML and the corresponding new bisimilarities are thus the first ones to offer a uniform framework for reasoning on processes that are purely nondeterministic or reactive probabilistic or that mix nondeterminism and probability in an alternating/nonalternating way.

Appendices

Appendix 1: Multistep variants inspired by $\sim _\mathrm{B}$

We start by formalizing the notion of computation in the NPLTS setting as a sequence of state-to-state steps each derived from a state-to-distribution transition.

Definition 11

Let $\fancyscript{L}= (S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS. A sequence $c \equiv s_{0} \, {\mathop {-\mapsto }\limits ^{a_{1}}}_{} \, s_{1} \, {\mathop {-\mapsto }\limits ^{a_{2}}}_{} \, s_{2} \dots s_{n - 1} \, {\mathop {-\mapsto }\limits ^{a_{n}}}_{} \, s_{n}$ is a computation of $\fancyscript{L}$ of length $n$ going from $s_{0}$ to $s_{n}$ iff for all $i = 1, \dots , n$ there exists a transition $s_{i - 1} {\mathop {\longrightarrow }\limits ^{a_{i}}}_{} \fancyscript{D}_{i}$ such that $s_{i} \in { supp}(\fancyscript{D}_{i})$, with $\fancyscript{D}_{i}(s_{i})$ being the execution probability of step $s_{i - 1} \, {\mathop {-\mapsto }\limits ^{a_{i}}}_{} \, s_{i}$ of $c$ conditioned on the selection of transition $s_{i - 1} {\mathop {\longrightarrow }\limits ^{a_{i}}}_{} \fancyscript{D}_{i}$ of $\fancyscript{L}$ at state $s_{i - 1}$; in this case, we write $s_{0} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{n}$ where $\alpha = a_{1} \, a_{2} \dots a_{n}$, with $s_{0} {\mathop {\Longrightarrow }\limits ^{\varepsilon }}_{} \delta _{s_{0}}$ when $\alpha = \varepsilon$. We call combined computation a computation in which every step arises from a combined transition, denoted by ${\mathop {\Longrightarrow }\limits ^{}}_\mathrm{c}$.

We now introduce the multistep variant of $\sim _\mathrm{B}$ and prove that it coincides with $\sim _\mathrm{B}$ itself.

Definition 12

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS in which the target of each transition is a Dirac distribution. A relation $\fancyscript{B}$ over $S$ is a multistep bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$:

• For each $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{1}}$ there exists $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{2}}$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$.

• For each $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{2}}$ there exists $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{1}}$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$.

We denote by $\sim _\mathrm{B,m}$ the largest multistep bisimulation.

Theorem 8

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS in which the target of each transition is a Dirac distribution.

Let $s_{1}, s_{2} \in S$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{B,m} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{B} s_{2} \end{aligned}$$

Proof

Suppose that $s_{1} \sim _\mathrm{B,m} s_{2}$. This means that there exists a multistep bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, it holds in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$:

• For each $s'_{1} {\mathop {\Longrightarrow }\limits ^{a}}_{} \delta _{s''_{1}}$ there exists $s'_{2} {\mathop {\Longrightarrow }\limits ^{a}}_{} \delta _{s''_{2}}$ such that $(s''_{1}, s''_{2}) \in \fancyscript{B}$.

• For each $s'_{2} {\mathop {\Longrightarrow }\limits ^{a}}_{} \delta _{s''_{2}}$ there exists $s'_{1} {\mathop {\Longrightarrow }\limits ^{a}}_{} \delta _{s''_{1}}$ such that $(s''_{1}, s''_{2}) \in \fancyscript{B}$.

Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$, we have that $\fancyscript{B}$ is also a bisimulation and hence $s_{1} \sim _\mathrm{B} s_{2}$.

Suppose now that $s_{1} \sim _\mathrm{B} s_{2}$. This means that there exists a bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. We prove that $\fancyscript{B}$ is also a multistep bisimulation, so that $s_{1} \sim _\mathrm{B,m} s_{2}$ will follow. Given $s'_{1}, s'_{2} \in S$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\alpha \in A^{*}$, we proceed by induction on $|\alpha |$:

• If $|\alpha | = 0$, then $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{1}}$ and $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{2}}$ are the only possible computations from $s'_{1}$ and $s'_{2}$ labeled with $\alpha$, hence the result trivially holds.

• Let $|\alpha | = n \in \mathbb {N}_{> 0}$ and suppose that the result holds for all traces of length $n - 1$. Assume $\alpha = a \, \alpha '$. Since $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{B}$ is a bisimulation, it holds that for each $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \delta _{s'''_{1}}$ there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \delta _{s'''_{2}}$ (and vice versa) such that $(s'''_{1}, s'''_{2}) \in \fancyscript{B}$. Suppose that $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s''_{1}}$ with $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \delta _{s'''_{1}}$ and $s'''_{1} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{} \delta _{s''_{1}}$. Then $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \delta _{s'''_{2}}$ with $(s'''_{1}, s'''_{2}) \in \fancyscript{B}$ and by the induction hypothesis we have that $s'''_{2} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{} \delta _{s''_{2}}$ with $(s''_{1}, s''_{2}) \in \fancyscript{B}$. As a consequence, $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s''_{2}}$ with $(s''_{1}, s''_{2}) \in \fancyscript{B}$. With a similar argument, we derive that $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s''_{2}}$ implies $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s''_{1}}$ with $(s''_{1}, s''_{2}) \in \fancyscript{B}$.

$\square$

We now provide the $\sim _\mathrm{B,m}$-inspired definition of each of the probabilistic bisimilarities considered in this paper and prove that it coincides with the original one-step equivalence. The ct-variants of the $\sim _\mathrm{B,m}$-inspired probabilistic bisimilarities can be defined similarly and satisfy an analogous coincidence property with respect to the original one-step ct-equivalences.

Definition 13

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS. An equivalence relation $\fancyscript{B}$ over $S$ is a multistep class-distribution probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ it holds that for each $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{1}$ there exists $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{2}$ such that, for all equivalence classes $C \in S / \fancyscript{B}$, $\fancyscript{D}_{1}(C) = \fancyscript{D}_{2}(C)$. We denote by $\sim _\mathrm{PB,dis,m}$ the largest multistep class-distribution probabilistic bisimulation.

Theorem 9

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and $s_{1}, s_{2} \in S$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,dis,m} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,dis} s_{2} \end{aligned}$$

Proof

Suppose that $s_{1} \sim _\mathrm{PB,dis,m} s_{2}$. This means that there exists a multistep class-distribution probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, we have in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$ it holds that for each $s'_{1} {\mathop {\Longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ there exists $s'_{2} {\mathop {\Longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ such that, for all $C \in S / \fancyscript{B}$, $\fancyscript{D}_{1}(C) = \fancyscript{D}_{2}(C)$. Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$, we have that $\fancyscript{B}$ is also a class-distribution probabilistic bisimulation and hence $s_{1} \sim _\mathrm{PB,dis} s_{2}$.

Suppose now that $s_{1} \sim _\mathrm{PB,dis} s_{2}$. This means that there exists a class-distribution probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. We prove that $\fancyscript{B}$ is also a multistep class-distribution probabilistic bisimulation, so that $s_{1} \sim _\mathrm{PB,dis,m} s_{2}$ will follow. Given $s'_{1}, s'_{2} \in S$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\alpha \in A^{*}$, we proceed by induction on $|\alpha |$:

• If $|\alpha | = 0$, then $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{1}}$ and $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{2}}$ are the only possible computations from $s'_{1}$ and $s'_{2}$ labeled with $\alpha$ and for all $C \in S / \fancyscript{B}$ it holds that:

  $$\begin{aligned} \delta _{s'_{1}}(C) \, = \, \delta _{s'_{2}}(C) \, = \, \left\{ \begin{array}{l@{\quad }l} 1 &{} \mathrm if \,\{ s'_{1}, s'_{2} \} \subseteq C \\ 0 &{} \mathrm if \,\{ s'_{1}, s'_{2} \} \cap C = \emptyset \\ \end{array} \right. \end{aligned}$$

  because $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $C$ is an equivalence class with respect to $\fancyscript{B}$.

• Let $|\alpha | = n \in \mathbb {N}_{> 0}$ and suppose that the result holds for all traces of length $n - 1$. Assume $\alpha = a \, \alpha '$. Since $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{B}$ is a class-distribution probabilistic bisimulation, it holds that for each $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{1}$ there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{2}$ such that, for all $C \in S / \fancyscript{B}$, $\fancyscript{D}'_{1}(C) = \fancyscript{D}'_{2}(C)$. Suppose that $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{1}$ with $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{1}$, $s''_{1} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{} \fancyscript{D}_{1}$, and $\fancyscript{D}'_{1}(s''_{1}) > 0$. Then there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{2}$ such that, for all $C \in S / \fancyscript{B}$, $\fancyscript{D}'_{1}(C) = \fancyscript{D}'_{2}(C)$. If we take $s''_{2}$ such that $(s''_{1}, s''_{2}) \in \fancyscript{B}$ and $\fancyscript{D}'_{2}(s''_{2}) > 0$, by the induction hypothesis there exists $s''_{2} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{} \fancyscript{D}_{2}$ such that, for all $C \in S / \fancyscript{B}$, $\fancyscript{D}_{1}(C) = \fancyscript{D}_{2}(C)$. As a consequence, there exists $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{2}$ such that, for all $C \in S / \fancyscript{B}$, $\fancyscript{D}_{1}(C) = \fancyscript{D}_{2}(C)$. $\square$

Definition 14

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and $\bowtie \, \in \{ =, \le , \ge \}$. An equivalence relation $\fancyscript{B}$ over $S$ is a multistep $\bowtie$ -group-by-group probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ and for all groups of equivalence classes $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that for each $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{1}$ there exists $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{2}$ such that $\fancyscript{D}_{1}(\bigcup \fancyscript{G}) \bowtie \fancyscript{D}_{2}(\bigcup \fancyscript{G})$. We denote by $\sim _\mathrm{PB,gbg,\bowtie ,m}$ the largest multistep $\bowtie$-group-by-group probabilistic bisimulation.

Theorem 10

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS, $s_{1}, s_{2} \in S$, and $\bowtie \, \in \{ =, \le , \ge \}$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,gbg,\bowtie ,m} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,gbg,\bowtie } s_{2} \end{aligned}$$

Proof

Suppose that $s_{1} \sim _\mathrm{PB,gbg,\bowtie ,m} s_{2}$. This means that there exists a multistep $\bowtie$-group-by-group probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, we have in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$ and $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that for each $s'_{1} {\mathop {\Longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ there exists $s'_{2} {\mathop {\Longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ such that $\fancyscript{D}_{1}(\bigcup \fancyscript{G}) \bowtie \fancyscript{D}_{2}(\bigcup \fancyscript{G})$. Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$, we have that $\fancyscript{B}$ is also a $\bowtie$-group-by-group probabilistic bisimulation and hence $s_{1} \sim _\mathrm{PB,gbg,\bowtie } s_{2}$.

Suppose now that $s_{1} \sim _\mathrm{PB,gbg,\bowtie } s_{2}$. This means that there exists a $\bowtie$-group-by-group probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. We prove that $\fancyscript{B}$ is also a multistep $\bowtie$-group-by-group probabilistic bisimulation, so that $s_{1} \sim _\mathrm{PB,gbg,\bowtie ,m} s_{2}$ will follow. Given $s'_{1}, s'_{2} \in S$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$, $\alpha \in A^{*}$, and $\fancyscript{G}\in 2^{S / \fancyscript{B}}$, we proceed by induction on $|\alpha |$:

• If $|\alpha | = 0$, then $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{1}}$ and $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}\delta _{s'_{2}}$ are the only possible computations from $s'_{1}$ and $s'_{2}$ labeled with $\alpha$ and it holds that:

  $$\begin{aligned} \delta _{s'_{1}}\left( \bigcup \fancyscript{G}\right) \, = \, \delta _{s'_{2}}\left( \bigcup \fancyscript{G}\right) \, = \, \left\{ \begin{array}{ll} 1 &{}\quad \mathrm if \,\{s'_{1},s'_{2}\}\subseteq C\,\hbox {for some}\,C \in \fancyscript{G}\\ 0 &{}\quad \mathrm if \,\{s'_{1},s'_{2}\}\cap C = \emptyset \,\hbox {for all}\,C \in \fancyscript{G}\\ \end{array} \right. \end{aligned}$$

  because $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{G}$ is a group of equivalence classes with respect to $\fancyscript{B}$.

• Let $|\alpha | = n \in \mathbb {N}_{> 0}$ and suppose that the result holds for all traces of length $n - 1$. Assume $\alpha = a \, \alpha '$. Since $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{B}$ is a $\bowtie$-group-by-group probabilistic bisimulation, for all $\fancyscript{G}' \in 2^{S / \fancyscript{B}}$ it holds that for each $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{1}$ there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{2}$ such that $\fancyscript{D}'_{1}(\bigcup \fancyscript{G}') \bowtie \fancyscript{D}'_{2}(\bigcup \fancyscript{G}')$. Suppose that $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{1}$ with $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{1}$, $s''_{1} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{} \fancyscript{D}_{1}$, and $\fancyscript{D}'_{1}(s''_{1}) > 0$. Let $\fancyscript{G}' = \{ C' \}$ with $C'$ being the equivalence class containing $s''_{1}$. Then there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{2}$ such that $\fancyscript{D}'_{1}(\bigcup \fancyscript{G}') \bowtie \fancyscript{D}'_{2}(\bigcup \fancyscript{G}')$. If we take $s''_{2}$ such that $(s''_{1}, s''_{2}) \in \fancyscript{B}$ and $\fancyscript{D}'_{2}(s''_{2}) > 0$ – it obviously exists in the case that $\bowtie \, \in \{ =, \le \}$ because $\fancyscript{D}'_{1}(s''_{1}) > 0$, and it also exists in the case that $\bowtie$ is $\ge$ because, if $s'_{2}$ had no $a$-transition reaching $\fancyscript{G}'$ with probability greater than $0$, then all $a$-transitions of $s'_{2}$ would reach $\fancyscript{G}'' = 2^{S / \fancyscript{B}} {\setminus } \fancyscript{G}'$ with probability $1$ and hence for the transition $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{1}$ we would have $\fancyscript{D}'_{1}(\bigcup \fancyscript{G}'') = 1 - \fancyscript{D}'_{1}(\bigcup \fancyscript{G}') < 1 = \fancyscript{D}'_{2}(\bigcup \fancyscript{G}'')$ for all transitions $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{2}$, i.e., $\fancyscript{B}$ would not be a $\ge$-group-by-group probabilistic bisimulation – by the induction hypothesis there exists $s''_{2} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{} \fancyscript{D}_{2}$ such that $\fancyscript{D}_{1}(\bigcup \fancyscript{G}) \bowtie \fancyscript{D}_{2}(\bigcup \fancyscript{G})$. As a consequence, there exists $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \fancyscript{D}_{2}$ such that $\fancyscript{D}_{1}(\bigcup \fancyscript{G}) \bowtie \fancyscript{D}_{2}(\bigcup \fancyscript{G})$. $\square$

Definition 15

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS. An equivalence relation $\fancyscript{B}$ over $S$ is a multistep $\sqcup \sqcap$ -group-by-group probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ and for all groups of equivalence classes $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ iff $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ and:

We denote by $\sim _\mathrm{PB,gbg,\sqcup \sqcap ,m}$ the largest multistep $\sqcup \sqcap$-group-by-group probabilistic bisimulation.

Theorem 11

Let $(S, A,{\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and $s_{1}, s_{2} \in S$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap ,m} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap } s_{2} \end{aligned}$$

Proof

Suppose that $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap ,m} s_{2}$. This means that there exists a multistep $\sqcup \sqcap$-group-by-group probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, we have in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$ and $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that $s'_{1} {\mathop {\Longrightarrow }\limits ^{a}}_{}$ iff $s'_{2} {\mathop {\Longrightarrow }\limits ^{a}}_{}$ and:

Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$, we have that $\fancyscript{B}$ is also a $\sqcup \sqcap$-group-by-group probabilistic bisimulation and hence $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap } s_{2}$.

Suppose now that $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap } s_{2}$. This means that there exists a $\sqcup \sqcap$-group-by-group probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. We prove that $\fancyscript{B}$ is also a multistep $\sqcup \sqcap$-group-by-group probabilistic bisimulation, so that $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap ,m} s_{2}$ will follow. Given $s'_{1}, s'_{2} \in S$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$, $\alpha \in A^{*}$, and $\fancyscript{G}\in 2^{S / \fancyscript{B}}$, we proceed by induction on $|\alpha |$:

• If $|\alpha | = 0$, then $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{1}}$ and $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{2}}$ are the only possible computations from $s'_{1}$ and $s'_{2}$ labeled with $\alpha$ and it holds that:

  $$\begin{aligned} \delta _{s'_{1}}\left( \bigcup \fancyscript{G}\right) \, = \, \delta _{s'_{2}}\left( \bigcup \fancyscript{G}\right) \, = \, \left\{ \begin{array}{ll} 1 &{}\quad \mathrm if \,\{ s'_{1}, s'_{2} \} \subseteq C\,\hbox {for some}\,C \in \fancyscript{G}\\ 0 &{}\quad \mathrm if \,\{ s'_{1}, s'_{2} \} \cap C = \emptyset \,\hbox {for all}\,C \in \fancyscript{G}\\ \end{array} \right. \end{aligned}$$

  because $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{G}$ is a group of equivalence classes with respect to $\fancyscript{B}$. Therefore:

  

• Let $|\alpha | = n \in \mathbb {N}_{> 0}$ and suppose that the result holds for all traces of length $n - 1$. Assume $\alpha = a \, \alpha '$. Since $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{B}$ is a $\sqcup \sqcap$-group-by-group probabilistic bisimulation, for all $\fancyscript{G}' \in 2^{S / \fancyscript{B}}$ it holds that $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{}$ iff $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{}$ and:

  

  Suppose that $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ with $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{1}$, $s''_{1} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{}$, and $\fancyscript{D}'_{1}(s''_{1}) > 0$. Let $\fancyscript{G}' = \{ C' \}$ with $C'$ being the equivalence class containing $s''_{1}$. Then $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{}$ with:

  

  If we take $s''_{2}$ and $\fancyscript{D}'_{2}$ such that $(s''_{1}, s''_{2}) \in \fancyscript{B}$, $\fancyscript{D}'_{2}(s''_{2}) > 0$, and $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{2}$, by the induction hypothesis we have that $s''_{2} {\mathop {\Longrightarrow }\limits ^{\alpha '}}_{}$ with:

  

  As a consequence, $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ with:

  

$\square$

Definition 16

Let $(S, A,{\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and . An equivalence relation $\fancyscript{B}$ over $S$ is a multistep $\#$ -group-by-group probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ and for all groups of equivalence classes $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ iff $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ and:

$$\begin{aligned} {\mathop {\#}\limits _{s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} {\fancyscript{D}}_{1}}} \fancyscript{D}_{1}\left( \bigcup \fancyscript{G}\right) \, = \, {\mathop {\#}\limits _{s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} {\fancyscript{D}}_{2}}} \fancyscript{D}_{2}\left( \bigcup \fancyscript{G}\right) \end{aligned}$$

We denote by $\sim _\mathrm{PB,gbg,\#,m}$ the largest multistep $\#$-group-by-group probabilistic bisimulation.

Theorem 12

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS, $s_{1}, s_{2} \in S$, and . Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,gbg,\#,m} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,gbg,\#} s_{2} \end{aligned}$$

Proof

Similar to the proof of Theorem 11. With regard to the induction step of the proof that $s_{1} \sim _\mathrm{PB,gbg,\#} s_{2}$ implies $s_{1} \sim _\mathrm{PB,gbg,\#,m} s_{2}$, we observe that $s''_{2}$ and $\fancyscript{D}'_{2}$ such that $(s''_{1}, s''_{2}) \in \fancyscript{B}$, $\fancyscript{D}'_{2}(s''_{2}) > 0$, and $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'_{2}$ obviously exist in the case that $\#$ is $\sqcup$ because $\fancyscript{D}'_{1}(s''_{1}) > 0$. They also exist in the case that $\#$ is $\sqcap$ because, if $s'_{2}$ had no $a$-transition reaching $\fancyscript{G}'$ (the group composed only of the equivalence class containing $s''_{1}$) with probability greater than $0$, then all $a$-transitions of $s'_{2}$ would reach $\fancyscript{G}'' = 2^{S / \fancyscript{B}}{\setminus } \fancyscript{G}'$ with probability $1$ and hence we would have:

$$\begin{aligned} {\mathop {\sqcap }\limits _{s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} {\fancyscript{D}}_{1}^{\prime }}} \fancyscript{D}'_{1}\left( \bigcup \fancyscript{G}''\right) \, < \, 1 \, = \, {\mathop {\sqcap }\limits _{s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} {\fancyscript{D}}_{2}^{\prime }}} \fancyscript{D}'_{2}\left( \bigcup \fancyscript{G}''\right) \end{aligned}$$

i.e., the considered relation $\fancyscript{B}$ would not be a $\sqcap$-group-by-group probabilistic bisimulation $\square$

We conclude by showing that all the considered $\sim _\mathrm{B,m}$-inspired probabilistic bisimilarities collapse into $\sim _\mathrm{B,m}$ when restricting attention to fully nondeterministic processes. An analogous result holds for their ct-variants.

Theorem 13

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS in which the target of each transition is a Dirac distribution. Let $s_{1}, s_{2} \in S$ and $\circ \in \{ =, \le , \ge , \sqcup \sqcap , \sqcup , \sqcap \}$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,dis,m} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,gbg,\circ ,m} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{B,m} s_{2} \end{aligned}$$

Proof

Since every multistep transition of this specific NPLTS can reach with probability greater than $0$ a single state and hence a single class of any equivalence relation—which are thus reached with probability $1$—the reflexive, symmetric, and transitive closure of a multistep bisimulation is trivially a multistep class-distribution probabilistic bisimulation and a multistep $\circ$-group-by-group probabilistic bisimulation. $\square$

Appendix 2: Multistep variants inspired by $\sim _\mathrm{PB}$

We start by introducing the multistep variant of $\sim _\mathrm{PB}$ and proving that it coincides with $\sim _\mathrm{PB}$ itself. Given an NPLTS $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ in which the transitions of each state have different labels and given $s \in S$, $\alpha \in A^{*}$, and $S' \subseteq S$, we inductively define the multistep probability of reaching a state in $S'$ from $s$ via $\alpha$ as follows:

Definition 17

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS in which the transitions of each state have different labels. An equivalence relation $\fancyscript{B}$ over $S$ is a p-multistep probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ and for all equivalence classes $C \in S / \fancyscript{B}$ it holds that:

$$\begin{aligned} { prob}_\mathrm{m}(s_{1}, \alpha , C) \, = \, { prob}_\mathrm{m}(s_{2}, \alpha , C) \end{aligned}$$

We denote by $\sim _\mathrm{PB,pm}$ the largest p-multistep probabilistic bisimulation.

Theorem 14

Let $(S, A,{\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS in which the transitions of each state have different labels.

Let $s_{1}, s_{2} \in S$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,pm} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB} s_{2} \end{aligned}$$

Proof

Suppose that $s_{1} \sim _\mathrm{PB,pm} s_{2}$. This means that there exists a p-multistep probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, we have in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$ and $C \in S / \fancyscript{B}$:

$$\begin{aligned} { prob}_\mathrm{m}(s'_{1}, a, C) \, = \, { prob}_\mathrm{m}(s'_{2}, a, C) \end{aligned}$$

Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$ and for all $s \in S$ such that $s {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}$ it holds that:

$$\begin{aligned} { prob}_\mathrm{m}(s, a, C) \, = \, \sum \limits _{s' \in C} \fancyscript{D}(s') \, = \, \fancyscript{D}(C) \end{aligned}$$

we have that the existence of $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ implies the existence of $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ and $\fancyscript{D}_{1}(C) = \fancyscript{D}_{2}(C)$. In other words, $\fancyscript{B}$ is also a probabilistic bisimulation and hence $s_{1} \sim _\mathrm{PB} s_{2}$.

Suppose now that $s_{1} \sim _\mathrm{PB} s_{2}$. This means that there exists a probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. We prove that $\fancyscript{B}$ is also a p-multistep probabilistic bisimulation, so that $s_{1} \sim _\mathrm{PB,pm} s_{2}$ will follow. Given $s'_{1}, s'_{2} \in S$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$, $\alpha \in A^{*}$, and $C \in S / \fancyscript{B}$, we proceed by induction on $|\alpha |$:

• If $|\alpha | = 0$, then $s'_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{1}}$ and $s'_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{s'_{2}}$ are the only possible computations from $s'_{1}$ and $s'_{2}$ labeled with $\alpha$ and it holds that:

  $$\begin{aligned} { prob}_\mathrm{m}(s'_{1}, \alpha , C) \, = \, { prob}_\mathrm{m}(s'_{2}, \alpha , C) \, = \, \left\{ \begin{array}{ll} 1 &{} \quad \mathrm if \,\{ s'_{1}, s'_{2} \} \subseteq C \\ 0 &{} \quad \mathrm if \,\{ s'_{1}, s'_{2} \} \cap C = \emptyset \\ \end{array} \right. \end{aligned}$$

  because $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $C$ is an equivalence class with respect to $\fancyscript{B}$.

• Let $|\alpha | = n \in \mathbb {N}_{> 0}$ and suppose that the result holds for all traces of length $n - 1$. Assume $\alpha = a \, \alpha '$. Since $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{B}$ is a probabilistic bisimulation, for all $C' \in S / \fancyscript{B}$ it holds that the existence of $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ implies the existence of $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ and $\fancyscript{D}_{1}(C') = \fancyscript{D}_{2}(C')$. Given $s \in S$ such that $s {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ with $s {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}$, it holds that:

  $$\begin{aligned} { prob}_\mathrm{m}(s, \alpha , C)&= \sum \limits _{s' \in S} \fancyscript{D}(s') \cdot { prob}_\mathrm{m}(s', \alpha ',C) \\&= \sum \limits _{C' \in S / \fancyscript{B}} \, \sum \limits _{s' \in C'} \fancyscript{D}(s') \cdot { prob}_\mathrm{m}(s', \alpha ', C) \\&= \sum \limits _{C' \in S / \fancyscript{B}} \, \sum \limits _{s' \in C'} \fancyscript{D}(s') \cdot { prob}_\mathrm{m}(s_{C'}, \alpha ', C) \\&= \sum \limits _{C' \in S / \fancyscript{B}} { prob}_\mathrm{m}(s_{C'}, \alpha ', C) \cdot \sum \limits _{s' \in C'} \fancyscript{D}(s') \\&= \sum \limits _{C' \in S / \fancyscript{B}} { prob}_\mathrm{m}(s_{C'}, \alpha ', C) \cdot \fancyscript{D}(C') \end{aligned}$$

  where $s_{C'} \in C'$ and the factorization of ${ prob}_\mathrm{m}(s_{C'}, \alpha ', C)$ stems from the application of the induction hypothesis on $\alpha '$ to all states of each equivalence class $C'$. Since the existence of $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ implies the existence of $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ and $\fancyscript{D}_{1}(C') = \fancyscript{D}_{2}(C')$ for all $C' \in S / \fancyscript{B}$ – remember that the quantification over $C'$ can be equivalently anticipated or postponed in the absence of internal nondeterminism – we derive that:

  $$\begin{aligned} { prob}_\mathrm{m}(s'_{1}, \alpha , C) \, = \, { prob}_\mathrm{m}(s'_{2}, \alpha , C) \end{aligned}$$

$\square$

When considering an arbitrary NPLTS $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$, internal nondeterminism comes into play and hence there might be several computations labeled with the same trace belonging to different resolutions of nondeterminism. In that case, their multistep probabilities have to be kept separate, otherwise they would be summed up like in the case of reactive probabilistic processes.

Since preserving the connection between each computation and the resolution of nondeterminism to which it belongs is important to define a $\sim _\mathrm{PB,m}$-inspired multistep variant of $\sim _\mathrm{PB,dis}$, we formalize below the notion of resolution. We call resolution of a state $s$ of an NPLTS $\fancyscript{L}$ the result of a possible way of resolving nondeterminism starting from $s$. A resolution is a tree-like structure, whose branching points are probabilistic choices corresponding to target distributions of transitions. This is obtained by unfolding from $s$ the graph structure underlying $\fancyscript{L}$ and by selecting at each reached state at most one transition—deterministic scheduler—or a convex combination of equally labeled transitions—randomized scheduler—among all the transitions in $\fancyscript{L}$ departing from that state. A resolution of $s$ can be formalized as an NPLTS $\fancyscript{Z}$ rooted at a state $z_{s}$ corresponding to $s$, in which every state has at most one outgoing transition, so that function ${ prob}_\mathrm{m}$ can be safely applied.

Definition 18

Let $\fancyscript{L}= (S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and $s \in S$. An NPLTS $\fancyscript{Z}= (Z, A, {\mathop {\longrightarrow }\limits ^{}}_{\fancyscript{Z}})$ is a resolution of $s$ obtained via a deterministic scheduler iff there exists a state correspondence function ${ corr} : Z \rightarrow S$ such that $s = { corr}(z_{s})$, for some $z_{s} \in Z$, and for all $z \in Z$ it holds that:

• If $z {\mathop {\longrightarrow }\limits ^{a}}_{\fancyscript{Z}} \fancyscript{D}$, then ${ corr}(z) {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}'$ with ${ corr}$ being injective over ${ supp}(\fancyscript{D})$ and $\fancyscript{D}(z') = \fancyscript{D}'({ corr}(z'))$ for all $z' \in { supp}(\fancyscript{D})$.

• If $z {\mathop {\longrightarrow }\limits ^{a_{1}}}_{\fancyscript{Z}} \fancyscript{D}_{1}$ and $z {\mathop {\longrightarrow }\limits ^{a_{2}}}_{\fancyscript{Z}} \fancyscript{D}_{2}$, then $a_{1} = a_{2}$ and $\fancyscript{D}_{1} = \fancyscript{D}_{2}$.

We denote by ${ Res}(s)$ the set of resolutions of $s$.

On the basis of the notion above, we provide a $\sim _\mathrm{PB,pm}$-inspired definition of $\sim _\mathrm{PB,dis}$ and show that it coincides with $\sim _\mathrm{PB,dis}$ itself. The ct-variant of the $\sim _\mathrm{PB,pm}$-inspired equivalence can be defined similarly by relying on resolutions obtained from randomized schedulers, and satisfies an analogous property with respect to the original one-step ct-equivalence.

Definition 19

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS. An equivalence relation $\fancyscript{B}$ over $S$ is a p-multistep class-distribution probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ it holds that for each resolution $\fancyscript{Z}_{1} \in { Res}(s_{1})$ there exists a resolution $\fancyscript{Z}_{2} \in { Res}(s_{2})$ such that for all equivalence classes $C \in S / \fancyscript{B}$:

$$\begin{aligned} { prob}_\mathrm{m}(z_{s_{1}}, \alpha , { corr}_{\fancyscript{Z}_{1}}^{-1}(C)) \, = \, { prob}_\mathrm{m}(z_{s_{2}}, \alpha , { corr}_{\fancyscript{Z}_{2}}^{-1}(C)) \end{aligned}$$

We denote by $\sim _\mathrm{PB,dis,pm}$ the largest p-multistep class-distribution probabilistic bisimulation.

Theorem 15

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and $s_{1}, s_{2} \in S$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,dis,pm} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,dis} s_{2} \end{aligned}$$

Proof

Suppose that $s_{1} \sim _\mathrm{PB,dis,pm} s_{2}$. This means that there exists a p-multistep class-distribution probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, we have in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$ it holds that for each $\fancyscript{Z}_{1} \in { Res}(s_{1})$ there exists $\fancyscript{Z}_{2} \in { Res}(s_{2})$ such that for all $C \in S / \fancyscript{B}$:

$$\begin{aligned} { prob}_\mathrm{m}(z_{s_{1}}, a, { corr}_{\fancyscript{Z}_{1}}^{-1}(C)) \, = \, { prob}_\mathrm{m}(z_{s_{2}}, a, { corr}_{\fancyscript{Z}_{2}}^{-1}(C)) \end{aligned}$$

Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$ and for all $s \in S$ and $\fancyscript{Z}\in { Res}(s)$ it holds that:

$$\begin{aligned} { prob}_\mathrm{m}(z_{s}, a, { corr}_{\fancyscript{Z}}^{-1}(C)) \, = \, \sum \limits _{z_{s'} \in { corr}_{\fancyscript{Z}}^{-1}(C)} \fancyscript{D}(z_{s'}) \, = \, \fancyscript{D}({ corr}_{\fancyscript{Z}}^{-1}(C)) \end{aligned}$$

we have that for each $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ such that, for all $C \in S / \fancyscript{B}$, $\fancyscript{D}_{1}(C) = \fancyscript{D}_{2}(C)$. In other words, $\fancyscript{B}$ is also a class-distribution probabilistic bisimulation and hence $s_{1} \sim _\mathrm{PB,dis} s_{2}$.

Suppose now that $s_{1} \sim _\mathrm{PB,dis} s_{2}$. This means that there exists a class-distribution probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. We prove that $\fancyscript{B}$ is also a p-multistep class-distribution probabilistic bisimulation, so that $s_{1} \sim _\mathrm{PB,dis,pm} s_{2}$ will follow. Given $s'_{1}, s'_{2} \in S$ such that $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\alpha \in A^{*}$, we proceed by induction on $|\alpha |$:

• If $|\alpha | = 0$, then $z_{s'_{1}} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{z_{s'_{1}}}$ and $z_{s'_{2}} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} \delta _{z_{s'_{2}}}$ are the only possible computations labeled with $\alpha$ in any resolution $\fancyscript{Z}_{1} \in { Res}(s'_{1})$ and any resolution $\fancyscript{Z}_{2} \in { Res}(s'_{2})$, respectively, and for all $C \in S / \fancyscript{B}$ it holds that:

  $$\begin{aligned} { prob}_\mathrm{m}(z_{s'_{1}}, \alpha , { corr}_{\fancyscript{Z}_{1}}^{-1}(C)) \, = \, { prob}_\mathrm{m}(z_{s'_{2}}, \alpha , { corr}_{\fancyscript{Z}_{2}}^{-1}(C)) \, = \, \left\{ \begin{array}{ll} 1 &{}\quad \mathrm if \,\{ s'_{1}, s'_{2} \} \subseteq C \\ 0 &{}\quad \mathrm if \,\{ s'_{1}, s'_{2} \} \cap C = \emptyset \\ \end{array} \right. \end{aligned}$$

  because $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $C$ is an equivalence class with respect to $\fancyscript{B}$.

• Let $|\alpha | = n \in \mathbb {N}_{> 0}$ and suppose that the result holds for all traces of length $n - 1$. Assume $\alpha = a \, \alpha '$. Since $(s'_{1}, s'_{2}) \in \fancyscript{B}$ and $\fancyscript{B}$ is a class-distribution probabilistic bisimulation, it holds that for each $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ such that, for all $C \in S / \fancyscript{B}$, $\fancyscript{D}_{1}(C) = \fancyscript{D}_{2}(C)$. Given $s \in S$ such that $z_{s} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ with $z_{s} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}$ in a resolution $\fancyscript{Z}\in { Res}(s)$, for all $C \in S / \fancyscript{B}$ it holds that:

  $$\begin{aligned} { prob}_\mathrm{m}(z_{s}, \alpha , { corr}_{\fancyscript{Z}}^{-1}(C))&= \sum \limits _{z_{s'} \in Z} \fancyscript{D}(z_{s'}) \cdot { prob}_\mathrm{m}(z_{s'}, \alpha ', { corr}_{\fancyscript{Z}}^{-1}(C)) \\&= \sum \limits _{C' \in S / \fancyscript{B}} \, \sum \limits _{z_{s'} \in { corr}_{\fancyscript{Z}}^{-1}(C')} \fancyscript{D}(z_{s'}) \cdot { prob}_\mathrm{m}(z_{s'}, \alpha ', { corr}_{\fancyscript{Z}}^{-1}(C)) \\&= \sum \limits _{C' \in S / \fancyscript{B}} \, \sum \limits _{z_{s'} \in { corr}_{\fancyscript{Z}}^{-1}(C')} \fancyscript{D}(z_{s'}) \cdot { prob}_\mathrm{m}(z_{s_{C'}}, \alpha ', { corr}_{\fancyscript{Z}}^{-1}(C)) \\&= \sum \limits _{C' \in S / \fancyscript{B}} { prob}_\mathrm{m}(z_{s_{C'}}, \alpha ', { corr}_{\fancyscript{Z}}^{-1}(C)) \cdot \sum \limits _{z_{s'} \in { corr}_{\fancyscript{Z}}^{-1}(C')} \fancyscript{D}(z_{s'}) \\&= \sum \limits _{C' \in S / \fancyscript{B}} { prob}_\mathrm{m}(z_{s_{C'}}, \alpha ', { corr}_{\fancyscript{Z}}^{-1}(C)) \cdot \fancyscript{D}({ corr}_{\fancyscript{Z}}^{-1}(C')) \end{aligned}$$

  where $s_{C'} \in C'$ and the factorization of ${ prob}_\mathrm{m}(z_{s_{C'}}, \alpha ', { corr}_{\fancyscript{Z}}^{-1}(C))$ stems from the application of the induction hypothesis on $\alpha '$ to all states of each equivalence class $C'$. Since for each $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ such that, for all $C' \in S / \fancyscript{B}$, $\fancyscript{D}_{1}(C') = \fancyscript{D}_{2}(C')$, we derive that for each $\fancyscript{Z}_{1} \in { Res}(s'_{1})$ there exists $\fancyscript{Z}_{2} \in { Res}(s'_{2})$ such that for all $C \in S / \fancyscript{B}$:

  $$\begin{aligned} { prob}_\mathrm{m}(z_{s'_{1}}, \alpha , { corr}_{\fancyscript{Z}}^{-1}(C)) \, = \, { prob}_\mathrm{m}(z_{s'_{2}}, \alpha , { corr}_{\fancyscript{Z}}^{-1}(C)) \end{aligned}$$

$\square$

Using the notion of resolution, we can also provide a $\sim _\mathrm{PB,pm}$-inspired definition of each of the six group-by-group probabilistic bisimilarities. The ct-variants of the six $\sim _\mathrm{PB,pm}$-inspired group-by-group probabilistic bisimilarities can be defined similarly by relying on resolutions obtained from randomized schedulers.

Definition 20

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and $\bowtie \, \in \{ =, \le , \ge \}$. An equivalence relation $\fancyscript{B}$ over $S$ is a p-multistep $\bowtie$ -group-by-group probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ and for all groups of equivalence classes $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that for each resolution $\fancyscript{Z}_{1} \in { Res}(s_{1})$ there exists a resolution $\fancyscript{Z}_{2} \in { Res}(s_{2})$ such that:

$$\begin{aligned} { prob}_\mathrm{m}\left( z_{s_{1}}, \alpha , { corr}_{\fancyscript{Z}_{1}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \, \bowtie \, { prob}_\mathrm{m}\left( z_{s_{2}}, \alpha , { corr}_{\fancyscript{Z}_{2}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \end{aligned}$$

We denote by $\sim _\mathrm{PB,gbg,\bowtie ,pm}$ the largest p-multistep $\bowtie$-group-by-group probabilistic bisimulation.

Definition 21

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS. An equivalence relation $\fancyscript{B}$ over $S$ is a p-multistep $\sqcup \sqcap$ -group-by-group probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ and for all groups of equivalence classes $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ iff $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ and:

We denote by $\sim _\mathrm{PB,gbg,\sqcup \sqcap ,pm}$ the largest p-multistep $\sqcup \sqcap$-group-by-group probabilistic bisimulation.

Definition 22

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS and . An equivalence relation $\fancyscript{B}$ over $S$ is a p-multistep $\#$ -group-by-group probabilistic bisimulation iff, whenever $(s_{1}, s_{2}) \in \fancyscript{B}$, then for all traces $\alpha \in A^{*}$ and for all groups of equivalence classes $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that $s_{1} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ iff $s_{2} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{}$ and:

$$\begin{aligned}&\mathop {\#}\limits _{\fancyscript{Z}_{1} \in { Res}(s_{1}) \, \mathrm{s.t.} \, z_{s_{1}} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} } { prob}_\mathrm{m}\left( z_{s_{1}}, \alpha , { corr}_{\fancyscript{Z}_{1}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \\&\quad = \, \mathop {\#}\limits _{\fancyscript{Z}_{2} \in { Res}(s_{2}) \, \mathrm{s.t.} \, z_{s_{2}} {\mathop {\Longrightarrow }\limits ^{\alpha }}_{} } { prob}_\mathrm{m}\left( z_{s_{2}}, \alpha , { corr}_{\fancyscript{Z}_{2}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \end{aligned}$$

We denote by $\sim _\mathrm{PB,gbg,\#,pm}$ the largest p-multistep $\#$-group-by-group probabilistic bisimulation.

The six $\sim _\mathrm{PB,pm}$-inspired group-by-group probabilistic bisimilarities can be alternatively defined without making explicit use of the notion of resolution. Given $s \in S$, $\alpha \in A^{*}$, and $S' \subseteq S$, we inductively define the set of multistep probabilities of reaching a state in $S'$ from $s$ via $\alpha$ as follows:

Since ${ probset}_\mathrm{m}(s, \alpha , S') = \{ { prob}_\mathrm{m}(z_{s}, \alpha , { corr}_{\fancyscript{Z}}^{-1}(S')) \mid \fancyscript{Z}\in { Res}(s) \}$, it is easy to see that in Definitions 20 to 22 we could have used ${ probset}_\mathrm{m}(s_{i}, \alpha , \bigcup \fancyscript{G})$ in place of ${ prob}_\mathrm{m}(z_{s_{i}}, \alpha , { corr}_{\fancyscript{Z}_{i}}^{-1}(\bigcup \fancyscript{G}))$ for $i = 1, 2$. This is not possible in Definition 19 because the use of ${ probset}_\mathrm{m}$ causes the connection between each computation and the resolution to which it belongs to be broken.

Each of the six $\sim _\mathrm{PB,pm}$-inspired group-by-group probabilistic bisimilarities is contained in the corresponding original one-step equivalence. The ct-variants of the six $\sim _\mathrm{PB,pm}$-inspired group-by-group probabilistic bisimilarities satisfy an analogous inclusion property with respect to the original one-step ct-equivalences.

Theorem 16

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{} )$ be an NPLTS, $s_{1}, s_{2} \in S$, and $\circ \in \{ =, \le , \ge , \sqcup \sqcap , \sqcup , \sqcap \}$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,gbg,\circ ,pm} s_{2} \, \Longrightarrow \, s_{1} \sim _\mathrm{PB,gbg,\circ } s_{2} \end{aligned}$$

Proof

Let $\bowtie \, \in \{ =, \le , \ge \}$ and suppose that $s_{1} \sim _\mathrm{PB,gbg,\bowtie ,pm} s_{2}$. This means that there exists a p-multistep $\bowtie$-group-by-group probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, we have in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$ and $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that for each $\fancyscript{Z}_{1} \in { Res}(s'_{1})$ there exists $\fancyscript{Z}_{2} \in { Res}(s'_{2})$ such that:

$$\begin{aligned} { prob}_\mathrm{m}\left( z_{s'_{1}}, a, { corr}_{\fancyscript{Z}_{1}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \, \bowtie \, { prob}_\mathrm{m}\left( z_{s'_{2}}, a, { corr}_{\fancyscript{Z}_{2}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \end{aligned}$$

Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$ and for all $s \in S$ and $\fancyscript{Z}\in { Res}(s)$ it holds that:

$$\begin{aligned} { prob}_\mathrm{m}\left( z_{s}, a, { corr}_{\fancyscript{Z}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \, = \, \sum \limits _{z_{s'} \in { corr}_{\fancyscript{Z}}^{-1}\left( \bigcup \fancyscript{G}\right) } \fancyscript{D}(z_{s'}) \, = \, \fancyscript{D}\left( { corr}_{\fancyscript{Z}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \end{aligned}$$

we have that for each $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{1}$ there exists $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{} \fancyscript{D}_{2}$ such that $\fancyscript{D}_{1}(\bigcup \fancyscript{G}) \bowtie \fancyscript{D}_{2}(\bigcup \fancyscript{G})$. In other words, $\fancyscript{B}$ is also a $\bowtie$-group-by-group probabilistic bisimulation and hence $s_{1} \sim _\mathrm{PB,gbg,\bowtie } s_{2}$.

Suppose now that $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap ,pm} s_{2}$. This means that there exists a p-multistep $\sqcup \sqcap$-group-by-group probabilistic bisimulation $\fancyscript{B}$ over $S$ such that $(s_{1}, s_{2}) \in \fancyscript{B}$. As a consequence, we have in particular that, whenever $(s'_{1}, s'_{2}) \in \fancyscript{B}$, then for all $a \in A$ and $\fancyscript{G}\in 2^{S / \fancyscript{B}}$ it holds that $s'_{1} {\mathop {\Longrightarrow }\limits ^{a}}_{}$ iff $s'_{2} {\mathop {\Longrightarrow }\limits ^{a}}_{}$ and:

Since ${\mathop {\Longrightarrow }\limits ^{a}}_{}$ coincides with ${\mathop {\longrightarrow }\limits ^{a}}_{}$ and for all $s \in S$ and $\fancyscript{Z}\in { Res}(s)$ it holds that:

$$\begin{aligned} { prob}_\mathrm{m}\left( z_{s}, a, { corr}_{\fancyscript{Z}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \, = \, \sum \limits _{z_{s'} \in { corr}_{\fancyscript{Z}}^{-1}\left( \bigcup \fancyscript{G}\right) } \fancyscript{D}(z_{s'}) \, = \, \fancyscript{D}\left( { corr}_{\fancyscript{Z}}^{-1}\left( \bigcup \fancyscript{G}\right) \right) \end{aligned}$$

we have that $s'_{1} {\mathop {\longrightarrow }\limits ^{a}}_{}$ iff $s'_{2} {\mathop {\longrightarrow }\limits ^{a}}_{}$ and:

In other words, $\fancyscript{B}$ is also a $\sqcup \sqcap$-group-by-group probabilistic bisimulation and hence $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap } s_{2}$.

Finally, the proof that $s_{1} \sim _\mathrm{PB,gbg,\#,pm} s_{2}$ implies $s_{1} \sim _\mathrm{PB,gbg,\#} s_{2}$ for $\# \in \{ \sqcup , \sqcap \}$ is similar to the proof that $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap ,pm} s_{2}$ implies $s_{1} \sim _\mathrm{PB,gbg,\sqcup \sqcap } s_{2}$. $\square$

Fig. 7

Two models related by $\sim _\mathrm{PB,gbg,=}$ that are distinguished by $\sim _\mathrm{PB,gbg,=,pm}$

Unlike Theorem 15, the reverse implication of Theorem 16 does not hold in general. For example, in Fig. 7 we have that $s_{1} \sim _\mathrm{PB,gbg,=} s_{2}$ but because, for $\alpha = a \, b \, c$ and $\fancyscript{G}$ containing all the states with no outgoing transitions, it turns out that the multistep probability of reaching $\fancyscript{G}$ via $\alpha$ in the maximal resolution of $s_{1}$ starting with the rightmost $a$-transition – which is $0.1 \cdot 0.7 + 0.9 \cdot 0.6 = 0.61$ – is not matched by any of the multistep probabilities of reaching $\fancyscript{G}$ via $\alpha$ in the three maximal resolutions of $s_{2}$ starting with the three $a$-transitions – which are $0.8 \cdot 0.7 + 0.2 \cdot 0.6 = 0.68$, $0.1 \cdot 0.7 = 0.07$, and $0.9 \cdot 0.6 = 0.54$.

We conclude by showing that all the considered $\sim _\mathrm{PB,pm}$-inspired probabilistic bisimilarities collapse into $\sim _\mathrm{PB,pm}$ when restricting attention to reactive probabilistic processes. An analogous result holds for their ct-variants.

Theorem 17

Let $(S, A, {\mathop {\longrightarrow }\limits ^{}}_{})$ be an NPLTS in which the transitions of each state have different labels.

Let $s_{1}, s_{2} \in S$ and $\circ \in \{ =, \le , \ge , \sqcup \sqcap , \sqcup , \sqcap \}$. Then:

$$\begin{aligned} s_{1} \sim _\mathrm{PB,dis,pm} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,gbg,\circ ,pm} s_{2} \, \Longleftrightarrow \, s_{1} \sim _\mathrm{PB,pm} s_{2} \end{aligned}$$

Proof

Since every state of this specific NPLTS has at most one transition labeled with a certain action, a p-multistep probabilistic bisimulation is trivially a p-multistep class-distribution probabilistic bisimulation and a p-multistep $\circ$-group-by-group probabilistic bisimulation. $\square$