Mixed Nondeterministic-Probabilistic Automata

Abstract

Graphical models in probability and statistics are a core concept in the area of probabilistic reasoning and probabilistic programming—graphical models include Bayesian networks and factor graphs. For modeling and formal verification of probabilistic systems, probabilistic automata were introduced. This paper proposes a coherent suite of models consisting of Mixed Systems, Mixed Bayesian Networks, and Mixed Automata, which extend factor graphs, Bayesian networks, and probabilistic automata with the handling of nondeterminism. Each of these models comes with a parallel composition, and we establish clear relations between these three models. Also, we provide a detailed comparison between Mixed Automata and Probabilistic Automata

Appendices

Appendix A: Addendum and Proofs Regarding Mixed Systems

In this supplementary material, we further detail the comparison between our model and imperative probabilistic programming. Then, we collect all the missing proofs.

1.1 A.1: Comparison with imperative probabilistic programming, see Discussion 1

In this appendix, we compare our model of Mixed Systems with imperative probabilistic programming following the approach promoted by Mc Iver and Morgan (2005; 2020). This line of work addresses probabilistic extensions of Hoare logic for imperative programs, focusing on evaluating the probability of the weakest preconditions of properties. In addition, we like to compare our approach with one aspect of this work, namely the modeling of the blending of probability and nondeterminism—this is only a minor aspect of the work of Mc Iver and Morgan, which focuses on decidability issues and computational cost of their proposed logic.

1.1.1 A.1.1: Demonic/angelic nondeterminism

We choose to base our comparison on a different work in the same direction: Chatterjee et al. (2018), which provides the most extensive development on demonic/angelic blending of probability and nondeterminism in the language Apps. We do not claim to cover all aspects of Apps, since this reference focuses on checking almost sure termination using supermartingale techniques. Since our scope is more modest in this appendix, we will only develop an informal comparison based on the following example corresponding to Fig. 2 of Chatterjee et al. (2018), reproduced here as Figs. 6 and 7.

Fig. 6

Example of Fig. 2 of Chatterjee et al. (2018)

Fig. 7

Semantics: SGS (Stochastic Game Structure) of $Q_3$, Fig. 6 of Chatterjee et al. (2018). The execution begins with the probabilistic choice. The left branch (corresponding to $Q_1$) is selected according to demonic nondeterminism figured by a triangle, and the right branch (corresponding to $Q_2$) is selected according to angelic nondeterminism, figured by a diamond

The program and its semantics are self-speaking. A key point here is the role of demonic and angelic nondeterminisms, and their combination in this program. Let us consider the post-condition

$$\begin{aligned} P: x \text {gets increased by one by performing}\ Q_3. \end{aligned}$$

(56)

The question is: how do we assess P? Under demonic choice, P is violated if there exists some branch in the nondeterministic choice under which P is violated. Under angelic choice, P is violated if, for all branches in the nondeterministic choice, P is violated. Inspecting Fig. 7 shows that P is violated if and only if $Q_1$ is selected. Thus the probabilistic score that P is violated is 0.6—we do not use the term “probability” since P combines both probabilistic and nondeterministic features and cannot be given a true probability.

Can we cast this example into Mixed Systems?

1.1.2 A.1.2: Casting this example to Mixed Systems?

Consider the following attempt by defining the Mixed System $S_{Q_3}=\{(\Omega ,\pi ),C,\{x,x'\}\}$, where:

• $\Omega =\{Q_1,Q_2\}$ and $\pi (\omega {=}Q_1)=0.6,\pi (\omega {=}Q_2)=0.4$;

• Variable $x,x'$ correspond to the statuses of variable x of $Q_3$ from Fig. 7, before and after executing $Q_3$; the value of x is assumed and the value of $x'$ will be established by sampling $S_{Q_3}$;

• It remains to define relation C involving $\omega ,x,x'$. To mimic Fig. 7, we would like to write something like

  $$\begin{aligned} \begin{array}{rr} x'={\textbf {if}}\; \omega =Q_1 &{} {\textbf {then ~~ angel}} \; x'\in \{x-1,x+1\}\\ &{}{\textbf {else \; demon}} \; x'\in \{x-1,x+1\} \end{array} \end{aligned}$$

Unfortunately, angelic/demonic choices are not concepts of our Mixed Systems model following Definition 1. Concerning probabilistic evaluation of state properties (item 3 of Definition 1), we could specify whether we use $\overline{\pi }$ (mirroring demonic) or $\underline{\pi }$ (mirroring angelic). Still, this does not allow to combine of both alternatives for different parts of the system.

We propose to refine Definition 1 so that both types of nondeterminism can be freely combined. Let us investigate this in the above example. Consider the Mixed System

$$\begin{aligned} S=(\Omega ,\pi ,X,C), \end{aligned}$$

(57)

where:

• $\Omega =\{Q_1,Q_2\}$ and $\pi (\omega =Q_1)=0.6,\pi (\omega =Q_2)=0.4$;

• Variable $x,x'$ correspond to the statuses of variable x of $Q_3$ from Fig. 7, before and after executing $Q_3$; the value of x is assumed and the value of $x'$ will be established by sampling $S_{Q_3}$;

• Relation C is (yet informally) defined by

  $$\begin{aligned} \omega C{x'}\ \text {iff}\ \left\{ \begin{array}{lcr} \omega =Q_1 &{} \wedge &{} {\textbf {angel}} ~x'\in \{x{-}1,x{+}1\}\\ \text {or}\\ \omega =Q_2 &{} \wedge &{} {\textbf {demon}} ~ x'\in \{x{-}1,x{+}1\} \end{array}\right. \end{aligned}$$

  (58)

This definition for C is informal, since keywords demon and angel have no mathematical meaning by themselves. We will give a semantics to (58) by assigning, to each state predicate, a probabilistic score $\pi ^*$. More precisely, we define $\pi ^*(\lnot P)$, the probabilistic score of predicate $\lnot P$, by the following formula:

$$\begin{aligned} \begin{array}{rcr} \pi ^*(\lnot P) &{}\,=_{\text {def}}\,&{} \pi ^{\textsf {c}}\left( \left\{ \omega \mid \omega {=}Q_1 \wedge \exists x'{\in }\{x{-}1,x{+}1\}: \lnot P\right\} \right) \\ &{}&{} + \pi ^{\textsf {c}}\left( \left\{ \omega \mid \omega {=}Q_2 \wedge \forall x'{\in }\{x{-}1,x{+}1\}: \lnot P\right\} \right) \end{array} \end{aligned}$$

(59)

In this formula, we give a semantics to angel in (58) by using the existential quantifier, i.e., we use the outer probability to evaluate the corresponding state predicate; we give a semantics to the demon in (58) by using the universal quantifier, i.e., we use the inner probability to evaluate the corresponding state predicate. Now, for this example, $\pi ^{\textsf {c}}=\pi$ since, with relation (58), for both choices $\omega =Q_1$ and $\omega =Q_2$, corresponding values for state $x'$ exist. Formula (59) finally yields $\pi ^*(\lnot P)=0.6$.

The above coding applies only to a restricted class of relations C. In formula (59), we exploited the fact that, in relation C defined by (58), a partition of $\Omega$ is performed first (probabilistic choice). Then, each branch of this choice involves a pure state predicate independent from $\omega$.

Here are some hints to extend this link beyond the particular example. Our starting point is the semantics of Apps, which is expressed in terms of Stochastic Game Structures (SGS); see Definition 2.3 of Chatterjee et al. (2018). Since Mixed Systems do not support recursion, we consider only the subclass of SGS that are DAGs. Then, picking a probabilistic location $\ell$ of this SGS, we consider the maximal subgraph of this SGS that has $\ell$ as its only minimal location, and contains no other probabilistic location. For our example (57,58,59), this yields the whole SGS. For each such subgraph, a coding similar to (57,58,59) can be given. The partially ordered execution of the whole SGS is then mapped to a Bayesian network following Definition 9. The incremental sampling of this Bayesian Network would correspond to the execution of the SGS as a game.

We preferred not to refine our Mixed System model with this additional feature since: 1) it applies only to a restricted class of relations C, and 2) we believe it to be incompatible with having a parallel composition.

1.2 A.2: Proof of Lemma 2

Proof

It is enough to prove the result for compressed systems. For $i=1,2$, let $S_i\equiv S'_i$ and let $\varphi _i$ be the bijections defining the two equivalences. We define

$$\begin{aligned} \varphi (\omega ,q_1{\,\sqcup \,}{q_2})= & {} \left( (\omega '_1,\omega '_2),q'_1{\,\sqcup \,}{q'_2} \right) \ \text {where}\ (\omega '_i,q'_i) =\varphi _i(\omega _i,q_i), i=1,2 \end{aligned}$$

and we have to verify that $\varphi$ defines the desired equivalence between $S=_{\text {def}}S_1\Vert S_2$ and $S'=_{\text {def}}S'_1\Vert S'_2$. Using the fact that $\pi =\pi _1\times \pi _2$, we get

$$\begin{aligned} \begin{array}{rl} C_\pi =&{}\{(p_1{{\,\sqcup \,}}{p_2},\omega ,q_1{{\,\sqcup \,}}{q_2})\mid q_1{{\,{\bowtie }\,}}{q_2} \,\wedge \,\omega _1C_1{q_1} \,\wedge \, \pi _1(\omega _1){>}0 \,\wedge \,\omega _2C_2{q_2} \,\wedge \, \pi _2(\omega _2){>}0\} \\ =&{} \{(p,\omega ,q_1{\,\sqcup \,}{q_2}) \mid q_1{\,{\bowtie }\,}{q_2} \,\wedge \, \omega _1C_{1\pi }q_1 \,\wedge \, \omega _2C_{2\pi }q_2\} \end{array} \end{aligned}$$

Thus, for every $(p,\omega ,q_1{\,\sqcup \,}{q_2})\in C_\pi$, we have $q'_1=q_1{\,{\bowtie }\,}{q_2}=q'_2\ \text {and}\ \omega '_iC_{i\pi }q'_i, i=1,2$, whence $\omega 'C'_{\pi }{q'}$ and $\varphi$ is a bijection. Since $\pi '=\pi '_1\times \pi '_2$ we get $\pi '(\omega ')=\pi (\omega )$, which finishes the proof.$\square$

Appendix B: Proofs Regarding Mixed Bayesian Networks

1.1 B.1: Proof of Theorem 2

Proof

We will repeatedly use notation (34). Without loss of generality we can assume that S is compressed. We first compress ${\textbf {Margin}}_{Y}\!\left( S\right)$ by considering the following equivalence relation, where $Z=X{\setminus } Y$ and $q_Y,q_Z$ are valuations for Y and Z:

$$\begin{aligned} \omega '\sim _Y\omega&\text {iff}&\forall q_Y: \left\{ \begin{array}{c} \exists q_Z:\omega C(q_Y,q_Z) \\ \Updownarrow \\ \exists q'_Z:\omega 'C(q_Y,q'_Z) \end{array} \right. ~~;\text {let}\ \omega _Y\ \text {be the equivalence class of}\ \omega . \end{aligned}$$

Let

$$\begin{aligned} C_Y\,=_{\text {def}}\,\{(\omega _Y,q_Y)\in \Omega _Y\times {Q_Y}\mid \exists \omega \in \omega _Y:\omega \,{\textbf {Pr}}_{Y\!}(C)\,q_Y\} \end{aligned}$$

be the associated relation, and let $\pi _Y$ be the compressed probability defined by $\pi _Y(\omega _Y)=\sum _{\omega \in \omega _Y}\pi (\omega )$. Let us denote by

$$\begin{aligned} S_Y=(\Omega _Y,\pi _Y,Y,C_Y) \end{aligned}$$

the resulting compressed system, and we recall that $\Omega _Y^{\textsf {c}}=\{\omega _Y\mid \exists {q_Y}:\omega _YC_Y{q_Y}\}$. In the sequel, we feel free to identify $\omega _Y\in \Omega _Y$, an element of the set of equivalence classes, with $\omega _Y$ seen as a subset of $\Omega$ saturated for $\sim _Y$. This way, a subset of $\Omega _Y$ can also be interpreted as a subset of $\Omega$.

To prove the theorem, we compare the two probabilistic semantics, namely: which state can be output and what is the outer probability of producing it. By definition of the sequential composition of kernels, ${\textbf {Margin}}_{Y}\!\left( S\right) ;{\textbf {Cond}}_{Y}\!\left( S\right)$

• 1. samples ${\textbf {Margin}}_{Y}\!\left( S\right) \,{\leadsto }\,q_Y$; and, then

• 2. given $q_Y$, samples $(Y{=}q_Y)\Vert S$.

Regarding the relations governing the nondeterministic choice, the combination of these two steps is identical to C. Let $q_*$ be such that $S\leadsto {q_*}$, implying that ${\textbf {Margin}}_{Y}\!\left( S\right) \leadsto {q_{*Y}}$, where $q_{*Y}\,=_{\text {def}}\,{\textbf {Pr}}_{Y\!}(q_*)$. Let us evaluate the outer probabilistic score of $q_*$ for the Bayesian network ${\textbf {Margin}}_{Y}\!\left( S\right) ;{\textbf {Cond}}_{Y}\!\left( S\right)$, i.e., the probability that $q_*$ is a possible outcome of sampling ${\textbf {Margin}}_{Y}\!\left( S\right) ;{\textbf {Cond}}_{Y}\!\left( S\right)$. We need to prove that it is equal to the probability that $q_*$ is a possible outcome of S, namely $\pi ^{\textsf {c}}(C_{q_*})$—we used notation (29). To show this, we note the following:

• 1. To output $q_*$ we first must output $q_{*Y}$, which amounts to selecting $\omega _Y$ such that $\omega _YC_Y{q_{*Y}}$. Using (12), (28) and notation (29), the probabilistic score of $q_{*Y}$, i.e., the probability that $q_{*Y}$ is a possible outcome of ${\textbf {Margin}}_{Y}\!\left( S\right)$, is equal to

  $$\begin{aligned} \pi ^{\textsf {c}}_Y\bigl ((C_Y)_{q_{*Y}}\bigr )\, \end{aligned}$$

  (60)

  which is $>0$ since ${\textbf {Margin}}_{Y}\!\left( S\right) \leadsto {q_{*Y}}$.

• 2. Then, we must select $\omega$ using S, under the additional constraint that ${\textbf {Pr}}_{Y\!}(q)$ ${=}q_{*Y}$, which requires that we sample $\omega \in \Omega$ under the constraint that $\omega \in \omega _Y$ for some $\omega _Y\in (C_Y)_{q_{*Y}}$. The corresponding probabilistic score is thus equal to the conditional probability

  $$\begin{aligned} \pi ^{\textsf {c}}\bigl (C_{q_*}\mid (C_Y)_{q_{*Y}}\bigr ), \end{aligned}$$

  (61)

  which is well defined since $\pi ^{\textsf {c}}_Y((C_Y)_{q_{*Y}})>0$.

• 3. By (33), the probabilistic score of $q_*$ is equal to the product of the two scores (60) and (61):

  $$\begin{aligned} \pi ^{\textsf {c}}\bigl (C_{q_*}\mid (C_Y)_{q_{*Y}}\bigr )\;\pi ^{\textsf {c}}_Y\bigl ((C_Y)_{q_{*Y}}\bigr )~=~\pi ^{\textsf {c}}\bigl (C_{q_*}\;{\cap }\;(C_Y)_{q_{*Y}}\bigr )~=~\pi ^{\textsf {c}}\bigl (C_{q_*}\bigr )\,, \end{aligned}$$

  where the last equality follows from $C_{q_*}\subseteq (C_Y)_{q_{*Y}}.$

This shows that $q_*$ possesses identical probabilistic semantics, for the left and right hand side of Bayes formula.$\square$

1.2 B.2: Proof of Corollary 1

As a prerequisite, we need the following result:

Lemma 5

Let $S_1$ and $S_2$ be any two Mixed Systems, and let Y be a set of variables containing $X_1{\cap }X_2$. Then, we have: ${\textbf {Margin}}_{X_1{\cup }Y}\!\left( S_1{{\,\mathbin {\Vert }\,}}S_2\right) \equiv S_1{{\,\mathbin {\Vert }\,}}{} {\textbf {Margin}}_{Y}\!\left( S_2\right)$.

Proof

This is immediate by observing that, first, ${\textbf {Margin}}_{X_1{\cup }{Y}}\!\left( S_1\Vert S_2\right)$ on the one hand, and $S_1{\,\mathbin {\Vert }\,}{} {\textbf {Margin}}_{Y}\!\left( S_2\right)$ on the other hand, possess identical probability spaces, namely $(\Omega _1,\pi _1){\times }(\Omega _2,\pi _2)$, and, second, they possess identical relations ${\textbf {Pr}}_{X_1\cup {Y}\!}(C_1{\wedge }C_2)=C_1\wedge {\textbf {Pr}}_{X_1\cup {Y}\!}(C_2)=C_1\wedge {\textbf {Pr}}_{Y\!}(C_2)$.$\square$

With this lemma, we can now proceed to the proof of Corollary 1. For proving formula (35), we first apply Theorem 2 with S replaced by $S_1{{\,\mathbin {\Vert }\,}}S_2$, which yields: $S_1{{\,\mathbin {\Vert }\,}}S_2\equiv _P{\textbf {Margin}}_{X_1{\cup }Y}\!\left( S_1{{\,\mathbin {\Vert }\,}}S_2\right) ;{\textbf {Cond}}_{Y}\!\left( S_1{{\,\mathbin {\Vert }\,}}S_2\right)$. Then, by Lemma 5, ${\textbf {Margin}}_{X_1{\cup }Y}\!\left( S_1{{\,\mathbin {\Vert }\,}}S_2\right) \equiv S_1{\,\mathbin {\Vert }\,}{} {\textbf {Margin}}_{Y}\!\left( S_2\right)$ and then we conclude by observing that

$$\begin{aligned} \bigl (S_1{\,\mathbin {\Vert }\,}{} {\textbf {Margin}}_{Y}\!\left( S_2\right) \bigr );{\textbf {Cond}}_{Y}\!\left( S_1{{\,\mathbin {\Vert }\,}}S_2\right) ~\equiv _P~\bigl (S_1{\,\mathbin {\Vert }\,}{} {\textbf {Margin}}_{Y}\!\left( S_2\right) \bigr );{\textbf {Cond}}_{Y}\!\left( S_2\right) \,, \end{aligned}$$

since the outcome of $S_1$ is determined by the left hand factor of “; ”.

1.3 B.3: Proof of Theorem 3

Having proved Lemma, the proof of Theorem 3 reproduces exactly the reasoning steps establishing the message-passing algorithm mapping factor graphs to Mixed Bayesian Networks in the classical setting (Loeliger 2004); thus, we only sketch the argument of the proof here.

Proof

Since ${\mathcal G}_\mathcal {S}$ is a tree, a natural distance can be defined on the set of vertices of ${\mathcal G}_\mathcal {S}$ by taking the length of the unique path linking two vertices. Select an arbitrary system $S_o$ as an origin and partially order other systems according to their distance to the origin, let $\preceq$ be this partial order. We have thus made ${\mathcal G}_\mathcal {S}$ a rooted tree, which we can see as a DAG. Then, the following two rules, known as message passing, are considered:

• R1: Pick $S\in {\mathcal G}_\mathcal {S}$, let $S^{\uparrow }$ be its (unique) ancestor in the tree and let $X^{\uparrow }$ be the set of common variables of $S^{\uparrow }$ and S. Then, let $S\,^{\!<}$ denote the parallel composition of all strict ancestors of S in ${\mathcal G}_\mathcal {S}$ and let $X^{\!<}$ be the set of variables of $S\,^{\!<}$. Using Bayes formula, factor S as

  $$\begin{aligned} S ~~\equiv _P~~{\textbf {Margin}}_{X^{\uparrow }}\!\left( S\right) ;{\textbf {Cond}}_{X^{\uparrow }}\!\left( S\right) ~~\equiv _P~~{\textbf {Margin}}_{X^{\!<}}\!\left( S\right) ;{\textbf {Cond}}_{X^{\!<}}\!\left( S\right) , \end{aligned}$$

  where the second equivalence follows from the fact that additional variables belonging to $X^{\!<}\setminus X^{\uparrow }$ are not shared with S.

• R2: Using formula (35) of Lemma 1, reorganize $\mathcal {S}$ by rewriting

  $$\begin{aligned} S\,^{\!<}\Vert S&~\equiv _P~&\left( S\,^{\!<}{\,\mathbin {\Vert }\,}{} {\textbf {Margin}}_{X^{\!<}}\!\left( S\right) \right) ;\;{\textbf {Cond}}_{X^{\!<}}\!\left( S\right) . \end{aligned}$$

Rules R1 followed by R2 are successively applied starting from the leaves of the tree, down to its root. The result is a Mixed Bayesian Network.$\square$

1.4 B.4: Proof of Theorem 4

The proof is by induction over the cardinality n of the set $\mathbb {K}$ of Mixed Kernels involved in ${\mathcal {N}}$. Its induction step uses the formula (36) of Corollary 2.

If $n=1$, then, by Condition 1, ${\mathcal {N}}$ is a Mixed Kernel K with no input, i.e., a Mixed System S by Convention 1. By (37), we have $S\equiv S_K$, hence $S_K\equiv _PK$. Thus, ${\mathcal {N}}$ and $S_{\mathcal {N}}$ are probabilistically equivalent.

We assume the theorem is true for a cardinality of $n{-}1$ and prove it for a cardinality of n. Chose $K_*\in \mathbb {K}$ such that K is maximal for the order defined by DAG ${\mathcal {N}}$, and let ${\mathcal {N}}_1$ be the restriction of ${\mathcal {N}}$ to $\mathbb {K}-\{K_*\}$ and ${\mathcal {N}}_2\,=_{\text {def}}\,\{K_*\}$. Then, ${\mathcal {N}}={\mathcal {N}}_1\cup {\mathcal {N}}_2$ by construction. Referring to (38), set $S\,=_{\text {def}}\, S_{\mathcal {N}},S_1\,=_{\text {def}}\, S_{{\mathcal {N}}_1}$, and $S_2\,=_{\text {def}}\, S_{{\mathcal {N}}_2}$.

First, $S\equiv S_1 {\,\mathbin {\Vert }\,} S_2$ holds by construction of the embedding ${\mathcal {N}}\mapsto S_{\mathcal {N}}$ defined in (38). Next, since the inputs of ${\mathcal {N}}_2$ are outputs of ${\mathcal {N}}_1$ and by Condition 2 of Theorem 4, it follows that

$$\begin{aligned} S_1{\,\mathbin {\Vert }\,}{} {\textbf {Margin}}_{Y}\!\left( S_2\right) \equiv {S_1}\,, \end{aligned}$$

where Y is the set of shared variables between $S_1$ and $S_2$. Hence, we can apply formula (36) of Corollary 2 to the pair $(S_1,S_2)$ of Mixed Systems, which yields:

$$\begin{aligned} S \equiv S_1{\,\mathbin {\Vert }\,}{S_2}&\equiv _P&{S_1};{\textbf {Cond}}_{Y}\!\left( S_2\right) . \end{aligned}$$

(62)

By induction hypothesis, we have

$$\begin{aligned} S_1\equiv _P{{\mathcal {N}}_1}\ \text {and}\ S_2\equiv _P{{\mathcal {N}}_2}. \end{aligned}$$

(63)

By definition of the binary operator “; ”, see (32), the following identity holds:

$$\begin{aligned} {\mathcal {N}}_1;{\textbf {Cond}}_{Y}\!\left( S_2\right) ~=~ {\mathcal {N}}_1\cup {\mathcal {N}}_2\,. \end{aligned}$$

Combining this identity with (62) and (63) finishes the proof of the induction step.

Appendix C: Proofs Regarding Mixed Automata

1.1 C.1: Proof of Lemma 3

Proof

The result is immediate if both $S_1$ and $S'_1$ are compressed, see Definition 4. It is thus sufficient to prove the lemma for the following two particular cases: $S_1$ compresses to $S'_1$, and the converse.

Consider first the case: $S_1$ compresses to $S'_1$. Let $w(\omega _1,\omega _2)$ be the weighting function associated to the lifting $S_1\,\rho ^\mathbb {S}\,S_2$, and let $\pi '_1(\omega '_1)=\sum _{\omega _1\in \omega '_1}\pi _1(\omega _1)$ be the relation between $\pi '_1$ and $\pi _1$ in the compression of $S_1$ to $S'_1$. Then $w'(\omega '_1,\omega _2)=\sum _{\omega _1\in \omega '_1}w(\omega _1,\omega _2)$ defines the weighting function associated to the lifting $S'_1\,\rho ^\mathbb {S}\,S_2$. The other properties required to deduce $S'_1\,\rho ^\mathbb {S}\,S_2$ are immediate to prove.

Now, consider the alternative case: $S'_1$ compresses to $S_1$, with relation

$$\begin{aligned} \begin{array}{c} \pi _1(\omega _1)=\sum _{\omega '_1\in \omega _1}\pi '_1(\omega '_1) \end{array} \end{aligned}$$

(64)

between $\pi '_1$ and $\pi _1$, where $\omega '_1\in \omega _1$ means that $\omega _1$ is the equivalence class of $\omega '_1$ with respect to relation $\sim$ defined in (19) when compressing $S'_1$. This case is slightly more involved since the weighting function $w'(\omega '_1,\omega _2)$ needs to be constructed. We need $w'(\omega '_1,\omega _2)$ to satisfy the following relations:

$$\begin{aligned} \begin{array}{rl} \forall \omega '_1:&{} \pi '_1(\omega '_1)=\sum _{\omega _2}w'(\omega '_1,\omega _2) \\ \forall \omega _2:&{} \pi _2(\omega _2)=\sum _{\omega '_1}w'(\omega '_1,\omega _2)\\ \forall (\omega '_1,\omega _2;q_1):&{} \left[ \begin{array}{c} w'(\omega '_1,\omega _2)>0 \\ \omega '_1\,C'_1\,{q_1} \end{array}\right] \Rightarrow \exists q_2: \left[ \begin{array}{c} \omega _2\,C_2\,{q_2} \\ q_1\,\rho \,{q_2}\end{array}\right] \\ \end{array} \end{aligned}$$

(65)

Focus first on the first two lines of (65). The following calculation shows that

$$\begin{aligned} w'(\omega '_1,\omega _2)&\,=_{\text {def}}\,&w(\omega _1,\omega _2) \times \frac{\pi '_1(\omega '_1)}{\pi _1(\omega _1)} \times {\textbf {1}}({\pi _1(\omega _1){>}0}), \end{aligned}$$

where $\omega _1$ is such that $\omega '_1\in \omega _1$ and ${\textbf {1}}(B)$ equals 1 if predicate B is true and 0 otherwise, yields a weighting function $w'$ satisfying the first two lines of (65):

$$\begin{aligned} \sum _{\omega _2}w'(\omega '_1,\omega _2)= & {} \sum _{\omega _2}\left( w(\omega _1,\omega _2) \times \frac{\pi '_1(\omega '_1)}{\pi _1(\omega _1)} \times {\textbf {1}}({\pi _1(\omega _1){>}0})\right) \\= & {} \frac{\pi '_1(\omega '_1)}{\pi _1(\omega _1)} \times {\textbf {1}}({\pi _1(\omega _1){>}0}) \times \sum _{\omega _2} w(\omega _1,\omega _2) ~=~ \pi '_1(\omega '_1)\\ \sum _{\omega '_1}w'(\omega '_1,\omega _2)= & {} \sum _{\omega '_1}\left( w(\omega _1,\omega _2) \times \frac{\pi '_1(\omega '_1)}{\pi _1(\omega _1)} \times {\textbf {1}}({\pi _1(\omega _1){>}0})\right) \\= & {} \sum _{\omega _1}\left( w(\omega _1,\omega _2) \times \frac{1}{\pi _1(\omega _1)} \times {\textbf {1}}({\pi _1(\omega _1){>}0})\right) \underbrace{\sum _{\omega '_1\in \omega _1}\pi '_1(\omega '_1)}_{=\pi _1(\omega _1)}\\= & {} \sum _{\omega _1}w(\omega _1,\omega _2) ~=~ \pi _2(\omega _2). \end{aligned}$$

We move to the third line of (65). The conditions $w'(\omega '_1,\omega _2)>0$ and $\omega '_1\,C'_1\,{q_1}$ together imply $w(\omega _1,\omega _2)>0$ and $\omega _1\,C_1\,{q_1}$ where $\omega _1$ is the equivalence class of $\omega '_1$, i.e., $\omega '_1\in \omega _1$. The right hand side then follows since we have $S_1\,\rho ^\mathbb {S}\,S_2$. This finishes the proof.$\square$

1.2 C.2: Proof of Lemma 4

Proof

Set $M'\,=_{\text {def}}\, M'_1{\,\mathbin {\Vert }\,}{M'_2}$ and $M\,=_{\text {def}}\, M_1{\,\mathbin {\Vert }\,}{M_2}$. Define the relation $\le$ between $Q'$ and Q by: $q'\le {q}$ iff $q'_1\le _1{q_1}$ and $q'_2\le _2{q_2}$. Let us prove that $\le$ is a simulation. Let $q'$ be such that ${q'}\xrightarrow {\alpha }_{M'}{S'}$ for some consistent $S'$. Then, $q'=q'_1{\,\sqcup \,}{q'_2}$ and $S'=S'_1\Vert S'_2$. By definition of the parallel composition, we have ${q'_i}\xrightarrow {\alpha _i}_{M'_i}{S'_i}$ for $i=1,2$, with $\alpha _1{\,\bowtie _{\varvec{\Sigma }}\;}\alpha _2$ and $\alpha =\alpha _1{\,\sqcup _{\varvec{\Sigma }}\;}\alpha _2$. Since $q'_i\le {q_i}$, we derive the existence (and uniqueness) of consistent systems $S_i,i=1,2$ such that ${q_i}\xrightarrow {\alpha _i}_{M_i}{S_i}$. Since $q=q_1{\,\sqcup \,}{q_2}$ we have $q_1{\,{\bowtie }\,}{q_2}$ and, thus, by definition of the parallel composition, we deduce ${r}\xrightarrow {\alpha }_{M}{S_1\Vert S_2}$. It remains to show that $S_1\Vert S_2$ is consistent. To prove this, remember that $S'=S'_1\Vert S'_2$ is consistent. Thus, there exist compatible $q'_1$ and $q'_2$ such that $S'_i\,{\leadsto }\,q'_i, i=1,2$. By definition of the simulations $\le _i$, we deduce that $S_i\,{\leadsto }\,q_i, i=1,2$, which shows that $S_1\Vert S_2$ is consistent.$\square$

Appendix D: Proofs Regarding the comparison with Probabilistic Automata

1.1 D.1: Proof of Theorem 5 regarding Simple Probabilistic Automata

1.1.1 D.1.1: Statement 1 of Theorem 5: from SPA to Mixed Automata

Proof

The sampling of spa P is: if P is in state $q{\in }Q$, performing $\alpha {\in }\Sigma$ leads to some target set of probability distributions over Q, of which one is selected, nondeterministically, and used to draw at random the next state $q'$.

We can reinterpret this sampling as follows: performing $\alpha {\in }\Sigma$ while being in state $q{\in }Q$ leads to the same target set of probability distributions over Q, that we use differently. We form the direct product of all distributions belonging to the target set and, we perform one trial according to this distribution, i.e., we perform independent random trials for all probabilities belonging to the target set. This yields a tuple of candidate values for the next state, of which we select one nondeterministically.

Clearly, these two samplings produce identical outcomes. The latter is the sampling of Mixed Automaton

$$\begin{aligned} M_P= & {} (\Sigma ,\{\xi \},q_0,\rightarrow _P), \end{aligned}$$

(66)

defined as follows:

• 1. Alphabet $\Sigma$ of $M_P$ is identical to that of P;

• 2. The unique variable $\xi$ of $M_P$ enumerates the values of Q, and initial state $q_0$ is identical to that of P; hence, P and $M_P$ possess identical sets of states, related via the identity map;

• 3. $\rightarrow _P$ maps a pair $(q,\alpha )\in {Q}{\times }\Sigma$ to the mixed system $S(q)=(\Omega ,\Pi ,\xi ,q,C)$, where:

  • (a) $\Omega$ is the product of n copies of Q, where n is the cardinality of the set $\{\pi \mid (q,\alpha ,\pi ){\in }\rightarrow \}$; thus, $\omega$ is an n-tuple of states: $\omega {=}(q_1,\dots ,q_n)$.

  • (b) $\Pi$ is the product of all probabilities belonging to $\{\pi \mid (q,\alpha ,\pi ){\in }\rightarrow \}$;

  • (c) Relation C is defined by $(\omega ,q')\in C$ if and only if $q'\in \{q_1,\dots ,q_n\}$.

So, we map spa P to Mixed Automaton ${M_P}$, defined in (66).

Mapping simulation relations: Defining simulation relations for ${\textsc {pa}}$ requires lifting relations from states to distributions over states. The formal definition for this lifting, as given in Section 4.1 of Segala (2006), corresponds to our Definition 15 when restricted to purely probabilistic mixed systems. The same holds for the strong simulation relation defined in Section 4.2 of the same reference: it is verbatim our Definition 16 when restricted to purely probabilistic mixed systems. This proves the part of Theorem 5 regarding simulation.

Mapping parallel composition: We move to parallel composition, for which the reader is referred to Lynch et al. (2003), Section 3. For $P_1=(\Sigma ,Q_1,q_{0,1},\rightarrow _1)$ and $P_2=(\Sigma ,Q_2,q_{0,2},\rightarrow _2)$ two PA, their parallel composition is $P=P_1{\,\mathbin {\Vert }\,}{P_2}=(\Sigma ,Q_1\times {Q_2},(q_{0,1},q_{0,2}),\rightarrow )$, where

$$\begin{aligned} {(q_1,q_2)}\xrightarrow {\alpha }{\pi _1 {\times } \pi _2}&\text {iff}&{q_i}\xrightarrow {\alpha }_{i}{\pi _i}\ \text {for}\ i=1,2 \end{aligned}$$

(67)

So, on one hand we consider the Mixed Automaton $M_P$. On the other hand, we consider the parallel composition of their images $M_{P_1}$ and $M_{P_2}$, namely $M=M_{P_1}{\,\mathbin {\Vert }\,} M_{P_2}=(\Sigma ,\{\xi _1,\xi _2\},(q_{0,1},q_{0,2}),\rightarrow _{12})$. In M, the state space is the domain of the pair $(\xi _1,\xi _2)$, namely $Q_1\times {Q_2}$, and, since there is no shared variable between the two Mixed Automata, the transition relation $\rightarrow _{12}$ is given by:

$$\begin{aligned} {(q_1,q_2)}\xrightarrow {\alpha }_{12}{S_1 {{\,\mathbin {\Vert }\,}} S_2}&\text {iff}&{q_i}\xrightarrow {\alpha }_{i}{S_i} \text {for}\ i=1,2 \end{aligned}$$

(68)

We thus need to show that

$$\begin{aligned} M_P\ \text {and}\ M \text {are simulation equivalent.} \end{aligned}$$

(69)

We will actually show that the identity relation between the two state spaces (both are equal to $Q_1\times {Q_2}$) is a simulation relation in both directions.

Observe first that (67) and (68) differ in that the former involves a nondeterministic transition relation, whereas the latter involves a deterministic transition function, mapping states to mixed systems. Pick $(q_1,q_2)\in {Q_1}\times {Q_2}$ and consider a transition for $M_P$:

$$\begin{aligned} {(q_1,q_2)}\xrightarrow {\alpha }_{M_P}{S}=((\Omega ,\Pi ),\xi ,(q_{1},q_{2}),C) \end{aligned}$$

where we have, for S:

• $\Omega$ is the product of $n_1$ copies of $Q_1$ and $n_2$ copies of $Q_2$, where, for $i=1,2$, $n_i$ is the cardinality of the set $\{\pi _i\mid (q_i,\alpha ,\pi _i)\in \rightarrow _i\}$, so that $\omega$ identifies $n_1\times {n_2}$-tuple of states: $\omega =(q_{11},\dots ,q_{1n_1};q_{21},\dots ,q_{2n_2})$;

• $\Pi$ is the product of all probabilities belonging to set $\{\pi _1\times \pi _2\mid (q_i,\alpha ,\pi _i)\in \rightarrow _i\}$;

• $\xi$ has domain $Q_1\times {Q_2}$;

• $(\omega ,(q''_1,q''_2))\in C$ if and only

  $$\begin{aligned} (q''_1,q''_2)\in \{(q_{1i_1},q_{2i_2})\mid i_1\in \{1,\dots ,n_1\}\ \text {and}\ i_2\in \{1,\dots ,n_2\}\}\,. \end{aligned}$$

Next, pick $(q_1,q_2)\in {Q_1}\times {Q_2}$ and consider a transition for M, see (68). We need to detail what $S_1{\,\mathbin {\Vert }\,}{S_2}=((\Omega ',\Pi ' ),\xi ',(q_1,q_2),C')$ is. We have, for $S_1{\,\mathbin {\Vert }\,}{S_2}$:

• $\Omega '$ is still the product of $n_1$ copies of $Q_1$ and $n_2$ copies of $Q_2$;

• $\Pi '$ is the product $\Pi _1\times \Pi _2$, where $\Pi _i$ is the product of all probabilities belonging to set $\{\pi _i\mid (q_i,\alpha ,\pi _i)\in \,\rightarrow _i\}$;

• $\xi '$ has domain $Q_1\times {Q_2}$;

• $(\omega ,(q'_1,q'_2))\in C'$ if and only if

  $$\begin{aligned} (q'_1,q'_2)\in \{(q_{1i_1},q_{2i_2})\mid i_1\in \{1,\dots ,n_1\}\ \text {and}\ i_2\in \{1,\dots ,n_2\}\}\,. \end{aligned}$$

By associativity of $\times$, $\Pi '=\Pi$, whereas other items for S on the one hand and other items for $S_1{\,\mathbin {\Vert }\,}{S_2}$ on the other hand, are synctatically identical. Thus (69) follows.$\square$

1.1.2 D.1.2: Statement 2 of Theorem 5: from Mixed Automata to SPA

Proof

Consider the following reverse mapping $M{\mapsto }{P_M}$, from Mixed Automata to spa:

• 1. The alphabet $\Sigma$ of $P_M$ is identical to that of M;

• 2. The set of states Q of $P_M$ is equal to the set of states of M, namely the domain of its set X of variables;

• 3. For $S=(\Omega ,\pi ,X,p,C)$, decompose relation $\{(\omega ,q)|\omega C{q}\}$ as $\bigcup _{\psi \in \Psi _C}\text {graph}(\psi )$, where $\Psi _C$ denotes the set of all partial functions $\Omega \rightarrow {Q}$, mapping each $\omega \in \exists {q}.C$ to some q such that $\omega C{q}$. Then, we consider, for each $\psi \in \Psi _C$, the measure defined by $\psi [\pi ](q)\,=_{\text {def}}\,\pi (\psi ^{-1}(q))$, where $\psi ^{-1}(q)=\{\omega |\psi (\omega ){=}q\}$ ($\psi [\pi ]$ is the image of $\pi$ by $\psi$), and we renormalize it by considering

  $$\begin{aligned} \frac{\psi [\pi ]}{\psi [\pi ](Q)}\,, \end{aligned}$$

  thus obtaining a probability distribution over Q. This defines a subset ${\textbf {P}}_S\subseteq \mathcal {P}(Q)$ of probability distributions.

• 4. The transition relation of $P_M$ is defined as follows:

  $$\begin{aligned} \rightarrow _{P_M}= & {} \left\{ (p,\alpha ,\mu ) \mid \exists S:(p,\alpha ,S)\in \;\rightarrow _M\ \text {and}\ \mu \in {\textbf {P}}_S \right\} \end{aligned}$$

  (70)

Consider two Mixed Automata $M,M'$ and let $\le$ be a simulation relation between their state spaces Q and $Q '$: $q\le {q'}$ and ${q}\xrightarrow {\alpha }_{M}{S}$ imply the existence of $S'\in \mathbb {S}(Q')$ such that $S\,\le ^\mathbb {S}\,S'$ and ${q'}\xrightarrow {\alpha }_{M'}{S'}$. We need to show that the same relation $\le \;\subseteq {Q}{\times }{Q'}$ is also a simulation relation for spa. Let ${q}\xrightarrow {\alpha }_{P_M}{\mu }$ be a transition of spa $P_M$. By (70), there exists a Mixed System S such that ${q}\xrightarrow {\alpha }_{M}{S}$ and $\mu \in {\textbf {P}}_S$. Since $\le$ is a simulation relation for Mixed Automata, there exists $S'\in \mathbb {S}(Q')$ such that $S\,\le ^\mathbb {S}\,S'$ and ${q'}\xrightarrow {\alpha }_{M'}{S'}$. Now, $S\,\le ^\mathbb {S}\,S'$ expands as follows: There exists a weighting function $w:\Omega \times \Omega '\rightarrow [0,1]$ such that the following two conditions hold:

• 1. For every triple $(\omega ,\omega ';q)$ such that $w(\omega ,\omega ')>0$ and $\omega C{q}$, there exists $q'$ such that $\omega 'C'{q'}$ and $q\le {q'}$;

• 2. w projects to $\pi$ and $\pi '$, respectively.

Let $\psi \in \Psi _C$ be the selection function giving rise to $\mu$ following step 3, meaning that $\mu$ is obtained by renormalizing $\psi [\pi ]$. Select any $\omega \in \exists {q}.C$ and let $q=\psi (\omega )$. Select any $\omega '$ such that $w(\omega ,\omega ')>0$ and assign to it one $q'$ such that $\omega 'C'{q'}$ and $q\le {q'}$ (such an $q'$ exists by the above Condition 1). This selection procedure defines a selection function $\psi ':\exists {q'}C'\rightarrow {Q'}$, mapping the $\omega '$ of the above Condition 1 to $q'$, which in turn defines a probability distribution $\mu '$, obtained by renormalizing $\psi '[\pi ']$. Consider the following weighting function over $Q{\times }Q'$:

$$\begin{aligned} v= & {} (\psi ,\psi ').w, \text {which expands as} \\ v(q,q')= & {} w\{(\hat{\omega },\hat{\omega }')\mid \psi (\hat{\omega })=q,\psi '(\hat{\omega }')=q'\} \end{aligned}$$

In particular $v(q,q')\ge {w(\omega ,\omega ')}>0$ by construction of $\psi ,\psi '$, and v. Then, v projects to $\mu$, and to $\mu '$:

$$\begin{aligned} \forall q: \sum _{q'}v(q,q')= & {} \sum _{q'}w\{(\hat{\omega },\hat{\omega }')\mid \psi (\hat{\omega })=q,\psi '(\hat{\omega }')=q'\}\\= & {} \sum _{\omega '}w\{(\hat{\omega },\hat{\omega }')\mid \psi (\hat{\omega })=q\}=\mu (q) \end{aligned}$$

and

$$\begin{aligned} \forall q': \sum _{q}v(q,q')= & {} \sum _{q}w\{(\hat{\omega },\hat{\omega }')\mid \psi (\hat{\omega })=q,\psi '(\hat{\omega }')=q'\}\\= & {} \sum _{\omega }w\{(\hat{\omega },\hat{\omega }')\mid \psi '(\hat{\omega '})=q'\}=\mu '(q') \end{aligned}$$

To summarize, we have constructed a probability distribution $\mu '$ such that $\mu \le ^{\mathcal {P}}\mu '$ and ${q'}\xrightarrow {\alpha }_{P_{M'}}{\mu '}$, showing that $\le$ was also a simulation relation for spa.$\square$

To complete our proof, it remains to show the following lemma:

Lemma 6

There is no mapping $M\mapsto {P_M}$ that preserves the parallel composition.

To support the above claim, we consider the following counter-example, where S(p) indicates that S has previous state p:

Example 18

Let $X=\{x_1,x,x_2\}$ be a set of three variables with finite domains $Q_{x_1},Q_{x},Q_{x_2}$. Consider the two systems $S_i(p_i)=(\Omega _i,\pi _i,X_i,p_i,C_i), i=1,2$, where: $X_1=\{x_1,x\}$, $X_2=\{x,x_2\}$; $p_1{\,{\bowtie }\,}{p_2}$; $\Omega _i=Q_i$ with $Q_1=Q_{x_1}{\times }Q_{x}$ and $Q_2=Q_{x}{\times }Q_{x_2}$; $\pi _i$ is a probability over $\Omega _i$; and $\omega _iC_i{q_i}$ iff $\omega _i={q_i}$. Define

$$\begin{aligned} {\textbf {x}}:\Omega _1\uplus \Omega _2\rightarrow {Q_x},\ \text {such that}\ \left\{ \begin{array}{l} {\textbf {x}}(\omega _1)=q \;\text {if}\ \omega _1=(q_1,q) \\ {\textbf {x}}(\omega _2)=q' \text {if}\ \omega _2=(q',q_2) \end{array}\right. \end{aligned}$$

(71)

System $S_1$ amounts to defining the pair $(x_1,x)$ as random variables with joint distribution $\pi _1$; similarly, $S_2$ amounts to defining the pair $(x,x_2)$ as random variables with joint distribution $\pi _2$. We assume that the set of all $q{\in }Q_x$ such that $\pi _1(Q_1{\times }\{q\}){>}0$ $\text {and}\ \pi _2(\{q\}{\times }Q_2){>}0$ is non empty. Forming the composition $S_1{{\,\mathbin {\Vert }\,}}S_2$ yields the system $S(p){=}(\Omega ,\pi ,X,p,C)$, where $X=X_1{\cup }X_2=\{x_1,x,x_2\}$, $Q=Q_{x_1}{\times }Q_{x}{\times }Q_{x_2}$, $p=p_1{\,\sqcup \,}{p_2}$, $\Omega =\Omega _1\times \Omega _2$, $\pi =\pi _1\times \pi _2$, and $\omega C{(q_1,q,q_2)}$ iff $\omega _1C_1(q_1,q)$ and $\omega _2C_2(q,q_2)$. According to Definition 1, the sampling of S is the following: draw $(\omega _1,\omega _2)$ at random with the conditional distribution $\pi _1\times \pi _2\bigl ((\omega _1,\omega _2)| {\textbf {x}}(\omega _1){=}{} {\textbf {x}}(\omega _2)\bigr )$, where the map ${\textbf {x}}$ was defined in (71); the resulting $(\omega _1,\omega _2)$ uniquely defines $(q_1,q,q_2)\in {Q}$ (no nondeterminism). In words, the parallel composition $S_1\Vert S_2$ amounts to making the triple of variables $(x_1,x,x_2)$ to be random with the joint distribution $\pi _1\times \pi _2\bigl ((\omega _1,\omega _2)\mid {\textbf {x}}(\omega _1)={\textbf {x}}(\omega _2)\bigr )$.

Next, consider the Mixed Automaton $M{=}(\{\alpha \},{X},q_0,\rightarrow )$, where $X{=}\{x_1,x,x_2\}$, set Q of states is defined accordingly $Q{=}Q_{x_1}{\times }Q_{x}{\times }Q_{x_2}$, and $\rightarrow$ maps, through action $\alpha$, any state $p{\in }Q$ to the above system S(p). Similarly, we consider the two Mixed Automata $M_i=(\{\alpha \},{X_i},q_{i,0},\rightarrow _i),i{=}1,2$, where $X_i$ is as above, $q_{i,0}$ is the projection of $q_0$ on $Q_i$, and $\rightarrow _i$ maps, through action $\alpha$, any state $p_i{\in }Q_i$ to the above system $S_i(p_i)$. We have $M=M_1{{\,\mathbin {\Vert }\,}}M_2$.

The only candidate way of mapping $M_i$ to a spa is by considering the two spa $P_i$ with sets of states $Q_i$ and transition relation ${p_i}\xrightarrow {\alpha }_{i}{\pi _i}$, where $\pi _i$ was defined above. Now, $P_1{{\,\mathbin {\Vert }\,}}P_2$ has transition relation ${p}\xrightarrow {\alpha }{\pi _1\times \pi _2}$, which reflects no interaction between the two spa, so it cannot represent $M_1{{\,\mathbin {\Vert }\,}}M_2$.

1.2 D.2: Proof of Theorem 6 regarding Probabilistic Automata

Proof

We consider the mapping $P\mapsto {M_P}=(\{1\},X,q_0,\rightarrow _{M_P})$, from pa to Mixed Automata, defined as follows:

• 1. Alphabet $\{1\}$ is the trivial singleton (the particular element does not matter);

• 2. $X=\{\xi _\Sigma ,\xi _Q\}$, where the variables $\xi _\Sigma$ and $\xi _Q$ enumerate $\Sigma$ and Q;

• 3. Transition $\rightarrow _{M_P}$ maps state p to system $S(p)=((\Omega ,\pi ),X,p,C)$, where

  • $\Omega =(\Sigma {\times }Q)^n$, where n is the cardinal of the image of p by transition $\rightarrow$;

  • $\pi$ is the product of all the distributions selected by transition $\rightarrow$ starting from p;

  • C is the nondeterministic selection of one component of $\omega$.

We only need to prove the positive statement related to simulation. Consider a simulation relation for pa $q\le {q'}$. We need to prove that $\le$ is also a simulation relation for Mixed Automata. Let $\mu$ be such that $(q,\mu )\in \;\rightarrow$. Since $\le$ is a simulation relation for pa, there exists $\mu '$ such that $(q',\mu ')\in \;\rightarrow '$ and $\mu \le ^\mathcal {P}\mu '$. Let S and $S'$ be the mixed systems to which q and $q'$ are mapped by step 3 of the mapping $P\mapsto {M_P}$. We have to prove that $S\,\le ^\mathbb {S}\,S'$. For each $\mu$ such that $(q,\mu )\in \;\rightarrow$, let the function $\chi :\mathcal {P}(Q)\rightarrow \mathcal {P}(Q')$ select one $\mu '$ such that $(q',\mu ')\in \;\rightarrow '$ and $\mu \le ^\mathcal {P}\mu '$ and let $v_\mu$ be a weighting function associated to relation $\mu \le ^\mathcal {P}\mu '$. The following weighting function

$$\begin{aligned} w(\omega ,\omega ')&\,=_{\text {def}}\,&\prod _{\mu :(q,\mu )\in \;\rightarrow }v_\mu (q_\mu ,q'_\mu ) \end{aligned}$$

where $(q_\mu ,q'_\mu )\in {Q{\times }Q'}$, solves the problem.$\square$