# A Complete Axiom System for 1-Free Kleene Star Expressions under Bisimilarity: An Elementary Proof

Grabmayer and Fokkink recently presented a finite and complete axiomatization for 1-free process terms over the binary Kleene star under bismilarity equivalence (proceedings of LICS 2020, preprint available). A different and considerably simpler proof is detailed in this paper. This result, albeit still somewhat technical, only relies on induction and normal forms and is therefore also much closer to a potential rewriting algorithm. In addition, a complete verification in the Coq proof assistant of all results in this work is provided, but correctness does not depend upon any computer-assisted methodology.

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04/27/2020

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## 1 Introduction

A completeness proof for a 1-free process theory modulo bisimilarity was recently presented in [6]111Preprint available at: https://www.cs.vu.nl/~wanf/publications.html222Extended version available at: https://arxiv.org/abs/2004.12740. Being somewhat daunted by the complexity of this proof provided a powerful incentive to search for a simpler solution, which is presented in this work. This different and considerably simpler proof only uses induction and normal forms, and has been additionally verified by means of the Coq proof assistant.

This paper is an intermediate step in a research line towards the resolution of a question originally posed in [7]: does there exist a finite and complete axiomatization for the unary Kleene star under bisimilarity equivalence? This problem is usually considered in the context of process theories including the constants 0 (deadlock), 1 (empty process) and the operators + (non-deterministic choice) and (sequential composition). Milner himself suggested that solving this problem may involve a considerable effort [7]. It is clear that the question remains unanswered to this day.

Earlier attempts in this direction include a completeness proof in absence of both the constants 0 and 1 in [3], and a variant where every Kleene star appears as in [4]. Completeness proofs for simpler theories (e.g. without the Kleene star) can be found in any process algebra handbook (cf. [1]).

The new approach in [6] deviates from the rather syntactic treatment in earlier works and instead takes the more semantic avenue of using process charts. This toolset is applied to prove completeness for a 1-free process theory over the binary Kleene star modulo bisimilarity. While novel and innovative, this results in a quite complex proof.

This paper presents a two-fold approach. First, it is shown that every process term has a bisimulant in a normal form and of non-increased star nesting depth. In essence, our normalization requirement conforms to the following congruence property: if term reduces to both and in one or more steps then bisimilarity of and is a consequence of bisimilarity of and . The second part of the proof applies induction towards the star nesting depth and proves equality under the condition that one operand is normalized, which is sufficient due to symmetry and transitivity.

This paper is set up in self-contained form. No claim is made that the supporting Coq code is the most neat or elegant reflection of such a proof in formalized mathematics, it just serves as an additional layer of verification.

The remainder of this paper is organized as follows. Section 2 contains a number of basic definitions and section 3 concerns the soundness of the axiom system. These basic preliminaries are then followed by three sections in which the the completeness proof is built from the ground up. Section 4 defines a summation operator and proves some basic results. Normal forms of terms are defined and shown to be derivable under bisimilarity in section 5. These results are then integrated into a completeness proof in section 6. The formalization of the theory in the Coq proof assistant is detailed in section 7, which mainly serves to make this material more accessible to readers who are less familiar with such techniques. A short concluding section 8 refines Milner’s completeness problem to a more detailed conjecture.

## 2 Definitions

We will mostly follow standard nomenclature and definitions in process algebra and propose the books [1] and [5] as general reference texts. Throughout this work, we will assume to be a set of actions. There is no requirement that is finite, as this proof concerns the completeness of closed terms only. Actions form the elementary operations in the set of process terms defined inductively as

In the process algebra the constant expresses deadlock (i.e. a process exhibiting no behavior). Every action induces a step to the special termination symbol as its sole behavior (note that is not part of the algebra ). The operation models a (possibly non-deterministic) choice between and , while the sequential composition denotes the concatenation of the behaviors of and . The binary Kleene star models zero or more iterations of , possibly followed by . Sequential composition is assumed to be right-associative, while the other operators associate to the left. The Kleene star binds stronger than sequential composition, which in turn binds stronger than plus.

Process terms exhibit behavior defined as taking actions resulting in either a new process term or the termination symbol . We set and define a relation to formally capture this behavior. Assume that , , and in the set of derivation rules listed below, using the notation for .

 aa⟶√   pa⟶vp+qa⟶v   qa⟶vp+qa⟶v   pa⟶p′p⋅qa⟶p′⋅q pa⟶√p⋅qa⟶q   pa⟶p′p\scalebox1.1$∗$qa⟶p′⋅p\scalebox1.1$∗$q   pa⟶√p\scalebox1.1$∗$qa⟶p\scalebox1.1$∗$q   qa⟶vp\scalebox1.1$∗$qa⟶v

Bisimilarity is a coinductively defined relationship which relates process terms in a lock-step fashion. The proof in [6] defines bisimilarity in terms of process charts. As these constructs are not used in this proof, we will employ a definition in terms of .

Elements are bisimilar (notation ) if there exists a relation such that and for all the following are satisfied

1. if and only if ,

2. for all there exists a such that and , and

3. for all there exists an such that and .

Examples of pairs of bisimilar terms include and , but the terms and are not bisimilar.

The following set of axioms will be shown to be sound and complete with regard to bisimilarity in this paper. In a very strict context, these should be interpreted as axiom-schemes, in the sense that for each closed instance of the variables, the corresponding axiom is defined.

 (B1) x+y = y+x (B6) x+0 = x (B2) (x+y)+z = x+(y+z) (B7) 0⋅x = 0 (B3) x+x = x (BKS1) x⋅x\scalebox1.1$∗$y+y = x\scalebox1.1$∗$y (B4) (x+y)⋅z = x⋅z+y⋅z (BKS2) (x\scalebox1.1$∗$y)⋅z = x\scalebox1.1$∗$(y⋅z) (B5) (x⋅y)⋅z = x⋅(y⋅z) (RSP) x = y⋅x+z    implies x = y\scalebox1.1$∗$z

The axiomatization is a straightforward adaptation of the set of axioms originally proposed in [7], which was in turn adapted from a language-theoretic setting based on work by Salomaa [8]). It is well-known that at least one higher-order construct is required as shown in [9]. Variants of the axiom of the recursive specification principle (RSP) have been studied extensively (cf. [1]).

For the remainder of this paper, the notation will be used to denote axiomatic equality whereas will be used to denote exact syntactic equality (e.g. but ).

The definitions listed here are sufficient to formulate the main result of this work, the proof of which is divided over several succeeding sections.

###### Theorem 1.

For all such that it holds that .

## 3 Soundness

We briefly consider soundness of the theory to fulfill the objective of being self-contained and to state a simple lemma which serves as a very useful building block in the remainder of the proof.

###### Lemma 1.

For all such that

1. for all there exists a such that and

2. for all there exists a such that ,

it holds that .

Soundness of the axiom RSP is neither deep nor straightforward.

###### Lemma 2.

If then , for all .

###### Proof.

Assume is the witnessing relation for . It is straightforward to show that the transitive closure of is again a bisimulation. If is defined as:

then can be chosen as a witnessing relation for . ∎

Soundness of the theory is required as a lemma in the completeness proof.

###### Lemma 3.

For all such that it holds that .

###### Proof.

Assume that and apply induction towards the derivation tree of . Most of the cases can be resolved easily via Lemma 1. Soundness for the axioms B5 and BKS2 is only slightly more complicated. Soundness for the axiom RSP is shown in Lemma 2. ∎

## 4 Summation

Expressing process terms as sums forms a crucial step between algebraic and semantic reasoning in the succeeding proofs. For example, the proof of Lemma 5 becomes much easier once we are able to obtain such sums under equality.

For finite sets we define the summation operator recursively by setting and

where the expression refers to the disjoint union (i.e. ).

###### Lemma 4.

For all there exists an such that and for all it holds that if and only if .

###### Proof.

Apply induction towards the structure of . In case , choose for . If for some , choose for . For the case , observe that in general always holds.

If , first use induction to obtain such that . Then, apply the following projection to each element : (1) if equals then project to , (2) if then project to and name the result . Subsequently, we have using B5.

For the case , use BKS1 to rewrite as and treat the part in the same way as for . ∎

We apply Lemma 4 to obtain a useful intermediate result stated in Lemma 5, which is very similar in form to Lemma 1. We define the predicate for all as shown below, where stands for termination-or-(axiomatically)equal.

and define as in all other cases.

###### Lemma 5.

For all such that

1. for all there exists a such that and

2. for all there exists a such that ,

it holds that .

###### Proof.

Assume and for as derived by Lemma 4. Rewrite as and solve the two equalities and separately by induction towards the size of the set that appears once in the respective equality. ∎

## 5 Normalization

In the remainder, let denote the fact that reduces to in zero or more steps for . Similarly, we define representing a reduction in one or more steps for .

The core of this completeness proof uses the result that terms in can be normalized under bisimilarity such that for every subterm and all reductions and such that it holds that . The precise meaning of ’subterm’ will become more clear in a short while. We require a predicate to express (a premise for) this congruence property:

Informally, if there exists a such that then is false, and true if no such exists. We may now prove a crucial lemma.

###### Lemma 6.

If and and then , for all .

###### Proof.

Assume is a witnessing relation for and define as follows:

Assume there exists such that for some . If there exists a step for and a step (i.e. when ) such that then this contradicts . The argument is symmetric. ∎

The congruence property may be used to recursively define a normal form, thereby making the notion of subterm with regard to the congruence property more precise. Two remarks are important now.

First, suppose we have a term then two properties are desired:

1. for and we have: implies and

2. for and we have: implies .

Therefore, the cases for and cannot be separated into two conditions relying solely on and . This is resolved by using a binary predicate to express the fact that a term is normalized.

In general, the process algebra does not have a neutral element under sequential composition. This necessitates the definition of two slightly different predicates for normal forms. Although this makes the proof a slightly more difficult, this does not present a fundamental complication.

We define two normal form predicates and recursively as shown below. Note that there is no mutual dependence between and .

 nfmult(0,q) ⟺ true nfmult(a,q) ⟺ true nfmult(p1+p2,q) ⟺ nfmult(p1,q)andnfmult(p2,q) nfmult(p1⋅p2,q) ⟺ nfmult(p1,p2⋅q)andnfmult(p2,q) nfmult(p1\scalebox1.1$∗$p2,q) ⟺ nfmult(p1,p1\scalebox1.1$∗$p2⋅q)andnfmult(p2,q)and congr(p1,p1\scalebox1.1$∗$p2⋅q)

and

 nf(0) ⟺ true nf(a) ⟺ true nf(p1+p2) ⟺ nf(p1)andnf(p2) nf(p1⋅p2) ⟺ nfmult(p1,p2)andnf(p2) nf(p1\scalebox1.1$∗$p2) ⟺ nfmult(p1,p1\scalebox1.1$∗$p2)andnf(p2)and congr(p1,p1\scalebox1.1$∗$p2)

We consider several simple examples. Observe that holds because there does not exist a reduction sequence such that . For the term , such a reduct indeed exists. However, the bisimulant of is normalized and constructed as such in Lemma 9 and Lemma 10.

The term is an example originally proposed in [4] that is re-used in the recent result of Fokkink and Grabmayer [6]. For compactness we abbreviate these as and such that . Now observe that such that . In this case, we can obtain a normalized bisimulant . The setup of Lemma 9 and Lemma 10 is precisely tailored to construct normal forms for these types of cases.

We now prove two straightforward results concerning the normal form predicates.

###### Lemma 7.

Both and are right-compatible under bisimilarity.

###### Proof.

For the first result, use induction towards to prove that is a consequence of , given . A similar result for follows directly from the definition. ∎

###### Lemma 8.

Both and are preserved under .

###### Proof.

Use induction towards to to prove that is a consequence of and , for some . Similarly, induction towards can be applied to derive from if for some and . ∎

We are now ready to prove the two key lemmas for deriving a bisimilar term satisfying the congruence property. Note that Lemma 9 is the first point in the proof where we will use induction towards the star-depth which is defined straightforwardly as shown below.

 d(0)=d(a) = 0 d(p1+p2)=d(p1⋅p2) = max{d(p1),d(p2)} d(p1\scalebox1.1$∗$p2) = max{1+d(p1),d(p2)}
###### Lemma 9.

For all such that and at least one of the following always holds:

1. There exists an such that and and and , or

2. There exists an such that and and .

###### Proof.

We first apply (strong) induction towards , thereby generalizing over all variables, and subsequently induction towards the structure of , thereby generalizing over and . Note that the case for can be solved directly by choosing . If for some then we distinguish between two cases: (1) if then set , (2) otherwise set . For the situations and , we first consider the cases where the -induction hypothesis for both operands corresponds with the first possibility in this lemma.

• If then by induction we can derive such that and . Choose .

• If then first use induction to derive . Then, apply induction again setting to derive .

Note that for both the cases and , the second case in the result is directly satisfied.

If then first obtain such that via induction. Now first suppose that and observe that in this case we have . Now is a witness for the second result. For the remainder of this case, suppose that does not hold.

Assume there exists a such that and observe that now holds. If there exists a such that then we have . As we may use induction to obtain an such that , which leads to a witness for the second result, otherwise, choose for the second result.

If there does not exists a such that then can be chosen as a witness for the first result of the Lemma. ∎

Lemma 10 is very similar to Lemma 9, but it is required as a separate result due to the aforementioned absence of a neutral element under sequential composition.

###### Lemma 10.

For all such that at least one of the following always holds:

1. There exists a such that and and and , or

2. There exists a such that and and .

###### Proof.

By induction towards the structure of . For the case , one simply invokes Lemma 9 directly. The cases for are almost the same. Note that induction towards is not required in this Lemma as one can use Lemma 9 to handle the -situation for the case . ∎

In order to apply the derivation of a bisimilar term satisfying the congruence property we first formulate a Lemma corresponding to the derivation of , followed by a lemma corresponding to the predicate .

###### Lemma 11.

For all , there exists a such that and and .

###### Proof.

Apply induction towards the structure of . For the case , use Lemma 10 to derive an such that , which results in , due to soundness of RSP, and if the first result of Lemma 10 holds. Otherwise, in case of the second result of Lemma 10, is satisfied directly. ∎

###### Lemma 12.

For all there exists a such that and and .

###### Proof.

Apply induction towards the structure of and handle the case as indicated in Lemma 11. ∎

## 6 Completeness

We will use the normal form obtained under bisimilarity in section 5 to finish the completeness proof. This requires some administrative work but is not very deep. Clearly the implication is our proof obligation. The completeness proof uses induction towards . Lemma 12 can be used to derive an such that and such that holds. By transitivity, these two equalities may be solved separately. This mainly comes down to two steps where deriving axiomatic equality is mostly done by invocations of Lemma 5.

1. Using Lemma 5 and induction to reduce to an equality of a form similar to for or , which in turn can be reduced to an equality of the form for and via the axiom RSP and Lemma 5.

2. Using and application of Lemma 6 to solve using the -induction hypothesis for completeness with regard to .

We require the definition of two more predicates to aid in compact formulation of the succeeding lemmas. We say that a set is a tail of (notation: ) if there exists a such that and for all it holds that . In addition, we say that a term is next-provable (notation: ) if for all and for all for it holds that implies . Using the -induction hypothesis, we can prove the following lemma.

###### Lemma 13.

For and such that , , , and it holds that

implies

###### Proof.

Apply Lemma 5 and note that the premises arising from steps from or are directly resolved. Furthermore, apply Lemma 6 and the induction hypothesis for every step arising from and . ∎

Lemma 14 is crucial in the transformation of the equality towards application of Lemma 13. We define the predicate to formalize the proof obligation as follows. Say that holds true if and only if for all such that:

1. ,

2. ,

3. ,

4. ,

5. and

6. for all such that we have implies ,

the conclusion follows.

###### Lemma 14.

For all it holds that implies .

###### Proof.

Apply induction towards the structure of , thereby generalizing over all other variables. For , use Lemma 5. For for some , use Lemma 5 and the premise . For , one must first derive sets such that: and and

The result then follows from the respective induction hypotheses for and . Note that the case is immediate due to the setup of this lemma.

The remaining case is . Apply the axiom RSP and note that the reversal of this axiom is sound, this leads to the following proof obligation

One first derives four sets such that

and

These two equalities can be resolved via induction. The premise is a result of Lemma 13. ∎

The following lemma is the analog of Lemma 14 and only required due to the aforementioned absence of a neutral element under multiplication.

###### Lemma 15.

For and such that , , , and