Basic Process Algebra with Iteration: Completeness of its Equational Axioms

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1 Bsic Process Algebr with Itertion: Completeness of its Equtionl Axioms Wn Fokkink CWI Kruisln 413, 1098 SJ Amsterdm, The Netherlnds Hns Zntem Utrecht University Pduln 14, 3508 TB Utrecht, The Netherlnds Abstrct Bergstr, Bethke nd Ponse proposed n xiomtiztion for Bsic Process Algebr extended with (binry) itertion. In this pper, we prove tht this xiomtiztion is complete with respect to strong bisimultion equivlence. To obtin this result, we will set up term rewriting system, bsed on the xioms, nd prove tht this term rewriting system is terminting, nd tht bisimilr norml forms re syntcticlly equl modulo AC. 1 Introduction Kleene [10] defined binry opertor x y in the context of finite utomt, which denotes the iterte of x nd y. Intuitively, the expression x y cn choose to execute either x, fter which it evolves into x y gin, or y, fter which it termintes. Kleene formulted some lgebric lws for this opertor, notbly (in our nottion) x y = x x y + y. Copi, Elgot nd Wright [6] proposed simplifiction of Kleene s setting, e.g. they defined unry version of the Kleene str in the presence of n empty word. The unry Kleene str hs been studied extensively ever since. Redko [15] (see lso [5]) proved for the unry Kleene str tht complete finite xiomtiztion for lnguge equlity does not exist. Slom [16] presented complete finite xiomtiztion which incorportes one conditionl xiom, nmely (in our nottion) x = y x + z implies x = y z if y does not incorporte the empty word. According to Kozen [11] this lst property 1

2 is not lgebric, in the sense tht it is not preserved under substitution of terms for ctions. He proposed two lterntive conditionl xioms which do not hve this drwbck. These xioms however re not sound in the setting of (strong) bisimultion equivlence. 1 Milner [12] studied the Kleene str in the setting of bisimultion equivlence, nd rised the question whether there exists complete xiomtiztion for it. Bergstr, Bethke nd Ponse [3] incorported the binry Kleene str in Bsic Process Algebr (BPA). They suggested three xioms BKS1-3 for BPA, where xiom BKS1 is the defining xiom from Kleene, while their most dvnced xiom BKS3 origintes from Troeger [18]: x (y (x + y) z + z) = (x + y) z In this pper we prove tht BKS1-3, together with the five stndrd xioms for BPA, form complete xiomtiztion for BPA with respect to bisimultion equivlence. For this purpose, we will replce itertion by proper itertion x y. This construct executes x t lest one time, or in other words, x y is equivlent to x x y. The xioms BKS1-3 re dpted to this new setting, nd we will define term rewriting system bsed on the xioms of BPA. Deducing termintion of this TRS is key step in the completeness proof; we will pply the strtegy of semntic lbelling from one of the uthors [20]. Finlly, we will show tht bisimilr norml forms re syntcticlly equl modulo AC. These results together imply tht the xiomtiztion for BPA is complete with respect to bisimultion equivlence. Moreover, the pplied method yields n efficient lgorithm to decide whether or not two terms re bisimilr. Sewell [17] proved tht if the dedlock δ is dded to BPA, then complete finite equtionl xiomtiztion does not exist. In [7] it hs been shown tht if sequentil composition nd itertion re replced by their prefix counterprts, then six simple equtionl xioms re complete for this lgebr. Acknowledgements. Jn Bergstr is thnked for his enthusistic support, nd Jos vn Wmel for mny stimulting discussions. 2 BPA with Binry Kleene Str This section introduces the bsic notions. We ssume n lphbet A of tomic ctions, together with three binry opertors: lterntive composi- 1 For exmple, one of Kozen s xioms is x + y x + z = x = x + y z = x, which induces ( + b) c + c = ( + b) c. 2

3 tion +, sequentil composition, nd binry Kleene str. Tble 1 presents n opertionl semntics for BPA in Plotkin style [14]. The specil symbol represents (successful) termintion. x + y x x x y y y + x x x y x y y x y x + y x x x y + x x x x y x y x x x y x x y y y x y y Tble 1: Action rules for BPA Our model for BPA consists of ll the closed terms tht cn be constructed from the tomic ctions nd the three binry opertors. Tht is, the BNF grmmr for the collection of process terms is s follows, where A: p ::= p + p p p p p. In the sequel the opertor will often be omitted, so pq denotes p q. As binding convention, nd bind stronger thn +. Often, p q will be bbrevited to pq. Process terms re considered modulo (strong) bisimultion equivlence from Prk [13]. Intuitively, two process terms re bisimilr if they hve the sme brnching structure. Definition 1 Two processes p nd q re bisimilr, denoted by p q, if there exists symmetric binry reltion B on processe which reltes p nd q, such tht: - if rbs nd r - if rbs nd r r, then there is trnsition s s such tht r Bs,, then s. 3

4 The ction rules in Tble 1 re in the pth formt of Beten nd Verhoef [2]. Hence, bisimultion equivlence is congruence with respect to ll the opertors, which mens tht if p p nd q q, then p + q p + q nd pq p q nd p q p q. See [2] for the definition of the pth formt, nd for proof of this congruence result. (This proof uses the extr ssumption tht the rules re well-founded. Fokkink nd vn Glbbeek [8] showed tht this requirement cn be dropped.) Furthermore, the ction rules for BPA re pure, which is syntctic criterion from Groote nd Vndrger (1992), nd the two rules for itertion incorporte the Kleene str in the left-hnd side of their conclusions. Hence, BPA is n opertionlly conservtive extension of BPA, i.e. the ction rules for itertion do not influence the trnsition systems of BPA terms. See Verhoef [19] for the definitions, nd for proof of this conservtivity result. Tble 2 contins n xiom system for BPA. It consists of the stndrd xioms A1-5 for BPA, together with three xioms BKS1-3 for itertion. In the sequel, p = q will men tht the equlity cn be derived from these xioms. The xiomtiztion for BPA is sound with respect to bisimultion equivlence, i.e. if p = q then p q. Since bisimultion equivlence is congruence, this cn be verified by checking soundness for ech xiom seprtely, which is left to the reder. The purpose of this pper is to prove tht the xiomtiztion is complete with respect to bisimultion, i.e. if p q then p = q. A1 x + y = y + x A2 (x + y) + z = x + (y + z) A3 x + x = x A4 (x + y)z = xz + yz A5 (xy)z = x(yz) BKS1 x(x y) + y = x y BKS2 (x y)z = x (yz) BKS3 x (y((x + y) z) + z) = (x + y) z Tble 2: Axioms for BPA 4

5 3 A Conditionl Term Rewriting System Our im is to define Term Rewriting System (TRS) for process terms in BPA tht reduces ech term to unique norml form, such tht if two terms re bisimilr, then they hve the sme norml form. However, we shll see tht one cnnot hope to find such TRS for itertion. Therefore, we will replce it by new, equivlent opertor p q, representing the behviour of p(p q), nd we will develop TRS for the lgebr BPA. We wnt our TRS to be terminting, so we cnnot dd the xioms A1,2 s rewrite rules. Therefore, process terms re considered modulo AC, tht is, modulo ssocitivity nd commuttivity of the Turning round two rules for BPA The xiom A3 yields the expected rewrite rule x + x x. Usully, in BPA, the xiom A4 s rewrite rule ims from left to right. However, in BPA we need this rewrite rule in the opposite direction. For exmple, in order to reduce the term (( + b) c) + b(( + b) c) + c to the term ( + b) c, we need the reduction ( + b) c) + b(( + b) c) ( + b)(( + b) c). Hence, we define the rewrite rule for A4 the other wy round. xz + yz (x + y)z. In BPA the xiom A5 ims from left to right too, but since we hve reversed A4, we must do the sme for A5, otherwise the TRS would not be confluent. For exmple, the term (b)d + (c)d would hve two different norml forms: (bd) + (cd) nd (b + c)d. So we opt for the rule 3.2 Proper itertion x(yz) (xy)z. Although we hve lredy defined prt of TRS tht should reduce terms tht re bisimilr to the sme norml form, we shll see now tht such TRS does not exist t ll. 5

6 Since x y + z x y if y + z y, such terms should hve the sme norml form. Therefore, one would expect rule x y + z x y if y + z y. However, this rule does not yield unique norml forms, becuse we hve reversed the rule for A4. For exmple, the term (b + ce) + ce + de would hve two different norml forms: (b + ce) + de nd (b + ce) + (c + d)e. To void this compliction, we replce itertion by n opertor x y, clled proper itertion, which displys the behviour of x(x y). 2 The opertionl semntics nd the xiomtiztion for proper itertion re given in Tbles 3 nd 4. They re obtined from the ction rules nd xioms for itertion, using the equivlences x y x y + y nd x y x(x y). Note tht BPA + (x y = x(x y)) PI1-3, BPA + (x y = x y + y) BKS1-3. So we find tht the xiomtiztion in Tble 4 is complete for BPA if nd only if the xiomtiztion in Tble 2 is complete for BPA. x x x y x (x y + y) x x y x y + y Tble 3: Action rules for proper itertion PI1 x(x y + y) = x y PI2 (x y)z = x (yz) PI3 x (y((x + y) z + z) + z) = x((x + y) z + z) Tble 4: Axioms for proper itertion 2 The stndrd nottion for this construct would be x + y, but we wnt to void mbiguous use of the +. 6

7 3.3 One rule for xiom PI2 Now tht we hve replced itertion by proper itertion, we cn continue to define rewrite rules for this new opertor. We strt with the one for xiom PI2. The question is whether it should rewrite from left to right or vice vers. If it would rewrite from left to right, it would clsh with the rule for A4. For exmple, then the term ( b)c+dc would hve two different norml forms: (bc) + dc nd ( b + d)c. Hence, PI2 yields the rule 3.4 Four rules for xiom PI1 The next rule stems from xiom PI1: x (yz) (x y)z. x(x y + y) x y. This rewrite rule cuses serious complictions concerning confluence; it turns out tht we need three extr rules to obtin this property. 1. A term x(y z + z) + y(y z + z) hs two different reductions: x(y z + z) + y z nd (x + y)(y z + z). So for the ske of confluence, one of these two reducts should reduce to the other. If we would dd the rule (x+y)(y z+z) x(y z+z)+y z to the TRS, then the term (c+bc)((bc) d+d) would hve two different norml forms: (c)((bc) d + d) + (bc) d nd (( + b)c)((bc) d + d). Hence, we opt for the rule x(y z + z) + y z (x + y)(y z + z). 2. A term x(y(y z + z)) hs two different reductions: x(y z) nd (xy)(y z + z). A rule (xy)(y z + z) x(y z) clshes with the rule for A5, becuse then the term ((bc))((bc) d + d)) would get two different norml forms: ((bc) d) nd ((b)c)((bc) d + d)). 7

8 Therefore, we define x(y z) (xy)(y z + z). 3. Finlly, term x (y(y z + z)) hs two different reductions: x (y z) nd (x y)(y z + z). Since rule (x y)(y z + z) x (y z) would clsh with the rule for PI2, we opt for x (y z) (x y)(y z + z). 3.5 Two conditionl rules for xiom PI3 The obvious interprettion of xiom PI3 s rewrite rule, x (x ((x + x ) z + z) + z) x((x + x ) z + z), obstructs confluence. For if x nd x re norml forms, while the expression x + x is not, then fter reducing x + x we cn no longer pply this rule. Therefore, we trnslte PI3 to conditionl rule: x (x (y z + z) + z) x(y z + z) if x + x y. Agin, this rule leds to TRS tht is not confluent, becuse term x (y(y z + z) + z) with x + y y hs two reductions: x (y z + z) nd x(y z + z). So in order to obtin confluence, we dd one lst conditionl rule to the TRS: x (y z + z) x(y z + z) if x + y y. 3.6 The entire TRS The entire TRS is given once gin in Tble 5. The rules re to be interpreted modulo AC. It is esy to see tht ll rules cn be deduced from BPA. The usul strtegy for deducing tht ech term hs unique norml form, is to prove tht the TRS is both wekly confluent, (i.e. if term p hs reductions p p p, then there exists q such tht p q p ), nd terminting (i.e. there re no infinite reductions). Newmn s Lemm sys tht such TRS reduces ech term to unique norml form, which does not reduce ny further. 8

9 1. x + x x 2. xz + yz (x + y)z 3. x(yz) (xy)z 4. x (yz) (x y)z 5. x(x y + y) x y 6. x(y z + z) + y z (x + y)(y z + z) 7. x(y z) (xy)(y z + z) 8. x (y z) (x y)(y z + z) 9. x (x (y z + z) + z) x(y z + z) if x + x y 10. x (y z + z) x(y z + z) if x + y y Tble 5: Rewrite rules for BPA Although our choice of rewrite rules hs been motivted by the wish for wekly confluent TRS, it is not so esy to deduce this property yet, due to the presence of conditionl rules. The next exmple shows tht the usul method for checking wek confluence of TRS, nmely verifying this property for ll overlpping redexes, does not work in conditionl setting. Exmple 2 Consider the TRS which consists of the rules f(x) b if x, c. There re no overlpping redexes, but this TRS is not wekly confluent: f(c) f() b. However, it will turn out tht the confluence property is not needed in the proof of the min theorem, which sttes tht bisimilr norml forms re equl modulo AC. Hence, confluence will simply be consequence of this theorem. 9

10 3.7 Termintion Proving termintion of the TRS in Tble 5, modulo AC, is complicted mtter. This is minly due to the presence of Rule 7, in which the left-hnd side cn be obtined from the right-hnd side by the removl of subterms. A powerful technique for proving termintion of TRSs tht incorporte such rules is semntic lbelling [20], where opertion symbols tht occur in the rewrite rules re supplied with lbels, which depend on the semntics of the rguments. Then two TRSs re involved: the originl system nd the lbelled system. The min theorem of [20] sttes tht the lbelled system termintes if nd only if the originl system termintes. The theory of semntic lbelling hs been developed for unconditionl TRSs. Therefore, we dpt the TRS in Tble 5 to n unconditionl TRS R, simply by removing the conditions from the lst two rules. We shll prove tht R is terminting, which immeditely implies termintion of the conditionl TRS in Tble 5. Proposition 3 The TRS R is terminting. Proof. The method from [20] strts with choosing model, which consists of set M, nd for ech function symbol f in the originl signture with rity n mpping f M : M n M, such tht for every rewrite rule, nd for ll possible vlues for its vribles in the model, the left-hnd side nd the right-hnd side re equl in the model. Here we choose the model to be the positive nturl numbers. Ech process p is interpreted by its norm p, being the lest number of steps in which it cn terminte. This norm cn be defined inductively s follows: = 1 p + q = min{ p, q } pq = p + q p q = p + q. Note tht norm is ssocitive nd commuttive with respect to the choice opertor, which is essentil in order to obtin the termintion result modulo AC. Clerly norm is preserved under bisimultion equivlence. Since the Rules 1-8 of R re sound with respect to bisimultion, it follows tht norm is preserved under ppliction of these rewrite rules. And it is esy to verify tht Rules 9 nd 10 of R, which re not sound becuse they lck their originl conditions, preserve norm too. Next, we select lbels for the function symbols. As lbels for the opertors sequentil composition nd proper itertion we choose the positive nturl numbers, while the toms nd the choice opertor remin unchnged. 10

11 In ech ground term, the occurrences of sequentil composition nd proper itertion re lbelled s follows: we replce p q by p q q nd p q by p[ q ]q. Finlly, for ech rule in the TRS we construct collection of lbelled rules. This is done by replcing the vribles in the originl rule by ll possible vlues in the model, nd computing the resulting lbels for the opertors. This results in the following TRS R, where the rules re defined for positive nturl numbers i nd j. x + x x x i z + y i z (x + y) i z x i + j (y j z) (x i y) j z x[i + j](y j z) (x[i]y) j z x i (x[i]y + y) x[i]y x i (y[i]z + z) + y[i]z (x + y) i (y[i]z + z) x i + j (y[j]z) (x i y) j (y[j]z + z) x[i + j](y[j]z) (x[i]y) j (y[j]z + z) x[i](x i (y[i]z + z) + z) x i (y[i]z + z) x[i](y[i]z + z) x i (y[i]z + z) Suppose tht R dmits n infinite reduction. Replce the vribles in this reduction by constnt to obtin n infinite ground reduction in R. For ech symbol nd tht occurs in this reduction, compute its corresponding lbel. This wy the infinite ground reduction in R trnsforms into n infinite ground reduction in R. Hence, termintion of R implies termintion of R. It remins to prove termintion of R. Although R is TRS with infinitely mny rules, this is much esier thn proving termintion of R. Define weight function w: w() = 1 w(p + q) = w(p) + w(q) w(p i q) = w(p) + iw(q) w(p[i]q) = w(p) + (i + 1)w(q) It is esy to verify tht for ny choice of vlues for vribles in ny rule, the weight of the left-hnd side is strictly greter thn the weight of the right-hnd side. For exmple, in the cse of Rule 7 these weights re nd w(x) + (i + j)w(y) + (i + j)(j + 1)w(z) w(x) + (i + j)w(y) + j(j + 2)w(z) 11

12 respectively. And (i + j)(j + 1) > j(j + 2) for i, j 1. Due to the strict monotonic behviour of w (here it is essentil tht i > 0) we conclude tht ech reduction step yields strict decrese of weight. Hence the system R is terminting, nd so R is terminting. 4 Norml Forms Decide Bisimilrity In the previous section we hve developed TRS for BPA tht reduces terms to norml form. Since ll rewrite rules re sound with respect to bisimultion equivlence, it follows tht ech term is bisimilr with its norml forms. So in order to determine completeness of the xiomtiztion for BPA with respect to bisimultion equivlence, it is sufficient to prove tht if two norml forms re bisimilr, then they re equl modulo AC. 4.1 An ordering on process terms As induction bse in the proof of our min theorem, we will need wellfounded ordering on process terms tht should preferbly hve the following properties: 1. p p + q p < pq p < p q q p + q q < pq q < p q. 2. The ordering is preserved under bisimultion. However, n ordering combining these properties is never well-founded, becuse for such n ordering we hve p q p q + q < p(p q + q) Since p(p q + q) p q, it follows tht p q < p q. The norm, indicting the lest number of steps process must mke before it cn terminte, induces n ordering tht lmost stisfies ll desired properties. The only serious drwbck of this ordering is tht p p + q. Therefore we dpt it to n ordering induced by L-vlue, which is defined s follows: L(p) = mx{ p p is proper substte of p} where proper substte mens tht p cn evolve into p by one or more trnsitions. Since norm is preserved under bisimultion equivlence, the sme holds for L. 12

13 Lemm 4 If p q, then L(p) = L(q). Proof. If p is proper substte of p, then bisimilrity of p nd q implies tht there is proper substte q of q such tht p q, nd so p = q. Hence, L(p) L(q), nd by symmetry L(q) L(p). We deduce the inductive definition for L-vlue. L(p+q) is the mximum of the collection { p p proper substte of p} { q q proper substte of q}, so L(p+q) = mx{l(p), L(q)}. Next, L(pq) is the mximum of the collection { p q p proper substte of p} { q } { q q proper substte of q}, so L(pq) = mx{l(p) + q, L(q)}. Finlly, L(p q) is the mximum of the collection { p (p q + q) p proper substte of p} { p q + q } { q q proper substte of q}, so p q = mx{l(p) + q, L(q)}. Recpitulting, we hve found: L() = 0 L(p + q) = mx{l(p), L(q)} L(pq) = mx{l(p) + q, L(q)} L(p q) = mx{l(p) + q, L(q)}. Hence, L-vlue too stisfies lmost ll the requirements formulted bove; only, we hve inequlities L(q) L(pq) nd L(q) L(p q), insted of the desired strict inequlities. Therefore, we introduce second weight function g on process terms, defined by: g() = 0 g(p + q) = mx{g(p), g(q)} g(pq) = g(q) + 1 g(p q) = g(q) + 1. Note tht g-vlue is not preserved under bisimultion equivlence. However, the following lemm holds. Lemm 5 If p q, then g(p) g(q). 13

14 Proof. For ech rewrite rule it is esily checked tht the g-vlue of the left-hnd side is greter thn or equl thn the g-vlue of the right-hnd side. Since the functions tht re used in the definition of g re wekly monotonous in their coordintes, we my conclude tht g-vlue is never incresed by rewrite step. In the proof of the min theorem we will pply induction on lexicogrphicl combintion of L-vlue nd g-vlue. 4.2 Some lemms We deduce three lemms tht will be used in the proof of the min theorem. The first lemm is typicl for normed processes [1], i.e. for processes tht re ble to terminte in finitely mny trnsitions. This lemm origintes from Cucl [4]. Lemm 6 If pr qr, then p q. Proof. A trnsition p r p r in pr cnnot be mimicked by trnsition q r r in qr, becuse p r > r. Hence, ech trnsition p r p r is mimicked by trnsition q r q r, nd vice vers. This induces bisimultion reltion between p nd q; the trnsition p p in p is mimicked by the trnsition q q in q, nd vice vers. Definition 7 We sy tht two process terms p nd q hve behviour in common if there re p nd q such tht p p nd q q nd p q. Lemm 8 If two terms pq nd rs hve behviour in common, nd q s, then either q ts for some t, or q s. Proof. If pq q nd rs r s with q r s, or if pq q nd rs s with q s, then we re done. And pq p q nd rs s with p q s would contrdict q s. Thus, the only interesting cse is if pq p q nd rs r s with p q r s. The inequlity q s then yields p r. We show, with induction on p, tht p q r s together with p r indicte either q ts for some t or q s. If p = 1, then p, nd so p q q. Since p q r s, this trnsition cn be mimicked by trnsition r s r s or r s s, nd so q r s or q s respectively. Next, let p = n + 1. Clerly, there is trnsition p p with p = n. Since p q r s, nd p q p q, there must be trnsition r s r s with p q r s. Since r p = n + 1 implies r n = p, the induction hypothesis lerns tht either q ts for some t, or q s. 14

15 Lemm 9 If term rs hs norml form q, then pq or p q is not norml form. Proof. Suppose tht q is norml form of term rs. Ech rule in Tble 5 tht pplies to term of the form tu or t u, reduces it to one of either forms gin. So q must be in one of either forms. But Rules 3, 4, 7 nd 8 reduce p(tu) nd p (tu) nd p(t u) nd p (t u) respectively. Hence, pq nd p q re not in norml form. 4.3 The min theorem Process terms re considered modulo AC. From now on, this equivlence is denoted by p = AC q, nd we sy tht p nd q re of the sme form. Clerly, ech process term p is sum of terms of the form nd qr nd q r, which re clled the summnds of p. Theorem 10 If two norml forms p nd q re bisimilr, then p = AC q. Proof. In order to prove the theorem, we prove three extr sttements in prllel. A. If two norml forms p = AC rs nd q = AC tu hve common behviour, then s = AC u. B. If two norml forms p = AC rs nd q = AC t u hve common behviour, then s = AC t u + u. C. If two norml forms p = AC r s nd q = AC t u hve common behviour, then r s = AC t u. The sttement in the min theorem is lbelled D. If L(p) = L(q) = 0, then both p nd q must be sums of toms. So in this cse A nd B nd C re empty sttements. And D holds too, becuse bisimilrity of p nd q indictes tht they contin exctly the sme toms, nd Rule 1 ensures tht both terms contin ech of these toms only once. Next, fix n m > 0 nd ssume tht we hve lredy proved the four sttements if L(p) nd L(q) re smller thn m. We will prove it for the cse tht they re equl to m. Let A n nd B n nd C n nd D n denote the ssertions for pirs p, q with mx{l(p), L(q)} m nd g(p) + g(q) n. They re proved by induction on n. The cse n = 0 corresponds with the cse L(p) = L(q) = 0, becuse if g(p) + g(q) = 0, then both p nd q must be sums of toms. As induction hypothesis we now ssume A n, B n, C n nd D n, nd we shll prove A n+1, B n+1, C n+1 nd D n+1. 15

16 1. A n+1 is true. Let norml forms rs nd tu hve behviour in common, with L(rs) m nd L(tu) m nd g(rs) + g(tu) = n + 1. We wnt to prove s = AC u. By symmetry we my ssume s u, so Lemm 8 offers two possibilities. 1.1 s u. L(s) L(rs) m nd L(u) L(tu) m nd g(s) + g(u) < g(rs) + g(tu) = n + 1. Hence, D n yields s = AC u. 1.2 s vu for some v. Let w be norml form of vu. According to Lemm 5 g(w) g(vu), so g(s)+g(w) < g(rs)+g(vu) = n+1. Further, since s w, L(w) = L(s) m. Hence, D n yields s = AC w. However, Lemm 9 sys tht s cnnot be norml form of term vu. Contrdiction. 2. B n+1 is true. According to the previous point we my ssume A n+1. Let norml forms rs nd t u hve behviour in common, with L(rs) m nd L(t u) m nd g(rs) + g(t u) = n + 1. We wnt to prove s = AC t u + u. Since t u t(t u + u), Lemm 8 offers three possibilities. 2.1 s t u + u. The term t u + u is norml form, becuse we cnnot pply Rules 1,2 or 6 to it. Moreover, g(s) + g(t u + u) = g(s) + g(t u) = n, so D n yields s = AC t u + u. 2.2 vs t u + u for some v. This implies v s u for some v. As in 1.2, we cn deduce tht then u is norml form of v s, which is contrdiction ccording to Lemm s v(t u + u) for some v. Note tht g(s) + g(v(t u + u)) = n + 1, so we cnnot yet pply D n. Let v be norml form. If v = AC t then s t u, so tht D n yields s = AC t u. Then Rule 7 reduces rs, which is contrdiction. So pprently v cnnot be of the form t. Thus, Rule 5 cnnot be pplied to v(t u + u), so this term is norml form. First, consider summnd αβ of s. This term nd v(t u + u) hve behviour in common, so A n+1 yields β = AC t u + u. 16

17 Next, consider summnd α β of s. This term nd v(t u + u) hve behviour in common. Since α β α(α β + β), Lemm 8 offers three possibilities. - α β + β t u + u. g(α β+β)+g(t u+u) g(s)+g(t u) = n, so D n implies α β+β = AC t u + u. Since the summnds of α β + β nd t u + u with gretest size re α β nd t u respectively, it follows tht α β = AC t u. - w(α β + β) t u + u for some w. Then w (α β + β) u for some w, nd we obtin contrdiction s in α β + β w(t u + u) for some w. Then β w (t u + u) for some w, nd we obtin contrdiction s in 1.2. So we my conclude α β = AC t u. If s contins severl summnds of the form α(t u + u) or t u, then we cn pply Rule 1,2 or 6 to s. However, s is in norml form, so pprently it consists of single term α(t u + u) or t u. Then pply Rule 3 or 7 pplies to rs, which gin is contrdiction, becuse rs is in norml form. 3. C n+1 is true. Assume norml forms r s nd t u tht hve behviour in common, with L(r s) m nd L(t u) m nd g(r s) + g(t u) = n + 1. We wnt to prove r s = AC t u. By symmetry we my ssume r s t u, so Lemm 8 offers two possibilities. 3.1 r s + s v(t u + u) for some v. Then s v (t u + u) for some v. This leds to contrdiction s in r s + s t u + u. First, suppose tht s nd u hve no behviour in common with t u nd r s respectively, so tht s u nd r s t u. Since D n pplies to the first equivlence, we get s = AC u. And the second equivlence yields r(r s + s) t(t u + u) t(r s + s), so Lemm 6 implies r t. Since L(r) = L(t) < m, sttement D then implies r = AC t, nd we re done. So we cn suppose tht either s nd t u hve behviour in common, or u nd r s hve behviour in common. We deduce contrdiction. 17

18 By symmetry it is sufficient to deduce contrdiction for the first cse only, where s nd t u hve behviour in common. If summnd αβ or α β of s hs behviour in common with t u, then B n or C n implies β = AC t u+u or α β = AC t u respectively. If s contins severl summnds of the form α(t u + u) or t u, then Rules 1,2 or 6 cn be pplied to it. However, s is norml form, so pprently it contins exctly one such summnd. If u nd r s hve behviour in common too, then similrly we cn deduce tht u hs summnd of the form β(r s + s) or r s, which indictes tht u hs size greter thn s. On the other hnd, s hs summnd α(t u + u) or t u, so s hs size greter thn u. This cnnot be, so u nd r s hve no behviour in common. And if u hs behviour in common with the summnd α(t u + u) or t u of s, then it follows from A n or B n or C n tht u hs summnd of the form β(t u + u) or t u. Agin we estblish contrdiction; u hs greter size thn itself. Hence, we hve found tht - r s + s t u + u, - s hs summnd α(t u + u) or t u, nd ll other summnds of s hve no behviour in common with t u, - u hs no behviour in common with r s, nor with the summnd α(t u + u) or t u of s. From these fcts it follows tht - s = AC α(t u + u) + s or s = AC t u + s, - s u, - r s + α(t u + u) or r s + t u is bisimilr to t u. Since s u, D n yields s = AC u. We distinguish the two possible forms of s. - s = AC α(t u + u) + u. Then r s + α(t u + u) t u. Since r s + s t u + u, this yields (r + α)(t u + u) t(t u + u). Lemm 6 implies r + α t, so since L(r + α) = L(t) < m, we obtin r + α t. Then Rule 9 cn be pplied to r s = AC r (α(t u + u) + u). Since r s is norml form, this is contrdiction. 18

19 - s = AC t u + u. Then r s + t u t u. Since r s + s t u + u, this yields (r + t)(t u + u) t(t u + u). Lemm 6 implies r + t t, so since L(r + t) = L(t) < m, we obtin r + t t. Then Rule 10 cn be pplied to r s = AC r (t u + u), nd once more we hve found contrdiction. 4. D n+1 is true. We my ssume A n+1 nd B n+1 nd C n+1. Let p nd q be bisimilr norml forms with L(p) = L(q) = m nd g(p) + g(q) = n + 1. We wnt to prove p = AC q. First, we show tht ech summnd of p is bisimilr to summnd of q, nd vice vers. Clerly, ech tomic summnd of p corresponds with summnd of q. We show tht ech non-tomic summnd of p lso corresponds to summnd of q. Suppose tht summnd rs of p hs behviour in common with two summnds of q. If these summnds re of the form tu nd t u, then A n+1 implies u = AC s = AC u, so tht Rule 2 reduces this pir. If they re of the form tu nd t u, then A n+1 nd B n+1 give u = AC s = AC t u + u, so tht Rule 6 reduces this pir. Finlly, if they re of the form t u nd t u, then B n+1 implies t u + u = AC s = AC t u + u. This mens t u = AC t u, so Rule 1 reduces this pir. Similrly, if summnd r s of p hs behviour in common with two summnds of q, we find using B n+1 nd C n+1 tht Rule 1, 2 or 6 cn be pplied to this pir. So, since q is norml form, the ssumption of non-tomic summnd of p hving behviour in common with two summnds of q leds to contrdiction. By symmetry, ech non-tomic summnd of q too cn hve behviour in common with only one summnd of p. So pprently, ech non-tomic summnd of p is bisimilr to non-tomic summnd of q nd vice vers. - Suppose tht summnds rs nd tu re bisimilr. Then A n+1 implies s = AC u, so ccording to Lemm 6 r t. Since L(r) = L(t) < m, we obtin r = AC t. - If summnds rs nd t u re bisimilr, then B n+1 implies s = AC t u+u. Then r(t u+u) = AC rs t u t(t u+u), so Lemm 6 implies r t. Since L(r) = L(t) < m, this yields r = AC t. Hence rs = AC t(t u + u), so we cn pply Rule 5 to rs. Contrdiction. 19

20 - Finlly, if summnds r s nd t u re bisimilr, then C n+1 sys tht they re of the sme form. Hence, p nd q contin exctly the sme summnds. Rule 1 indictes tht ech of these summnds occurs only once in both p nd q, so p = AC q. Corollry 11 The TRS in Tble 5 is confluent. Corollry 12 The xioms A1-5 + BKS1-3 for BPA re complete with respect to bisimultion equivlence. Proof. If two terms in BPA re bisimilr, then ccording to Theorem 10 their norml forms re of the sme form. Since ll the rewrite rules cn be deduced from A1-5 + PI1-3, it follows tht this is complete xiom system for BPA. Then A1-5 + BKS1-3 is complete xiomtiztion for BPA. References [1] J.C.M. Beten, J.A. Bergstr, nd J.W. Klop. Decidbility of bisimultion equivlence for processes generting context-free lnguges. Journl of the ACM, 40(3): , [2] J.C.M. Beten nd C. Verhoef. A congruence theorem for structured opertionl semntics with predictes. In E. Best, editor, Proceedings CONCUR 93, Hildesheim, LNCS 715, pges Springer-Verlg, [3] J.A. Bergstr, I. Bethke, nd A. Ponse. Process lgebr with itertion nd nesting. The Computer Journl, 37(4): , [4] D. Cucl. Grphes cnoniques de grphes lgébriques. Theoreticl Informtics nd Applictions, 24(4): , [5] J.H. Conwy. Regulr Algebr nd Finite Mchines. Chpmn nd Hll, [6] I.M. Copi, C.C. Elgot, nd J.B. Wright. Reliztion of events by logicl nets. Journl of the ACM, 5: , [7] W. J. Fokkink. A complete equtionl xiomtiztion for prefix itertion. Informtion Processing Letters, 52(6): ,

21 [8] W. J. Fokkink nd R. J. vn Glbbeek. Ntyft/tyxt rules reduce to ntree rules. Informtion nd Computtion, 126(1):1 10, [9] J. F. Groote nd F. W. Vndrger. Structured opertionl semntics nd bisimultion s congruence, Informtion nd Computtion, 100(2): , [10] S.C. Kleene. Representtion of events in nerve nets nd finite utomt. In Automt Studies, pges Princeton University Press, [11] D. Kozen. A completeness theorem for Kleene lgebrs nd the lgebr of regulr events. Informtion nd Computtion, 110(2): , [12] R. Milner. A complete inference system for clss of regulr behviours. Journl of Computer nd System Sciences, 28: , [13] D.M.R. Prk. Concurrency nd utomt on infinite sequences. In P. Deussen, editor, Proceedings 5th GI Conference, LNCS 104, pges Springer-Verlg, [14] G.D. Plotkin. A structurl pproch to opertionl semntics. Report DAIMI FN-19, Arhus University, [15] V.N. Redko. On defining reltions for the lgebr of regulr events. Ukrinskii Mtemticheskii Zhurnl, 16: , In Russin. [16] A. Slom. Two complete xiom systems for the lgebr of regulr events. Journl of the ACM, 13(1): , [17] P. Sewell. Bisimultion is not finitely (first order) equtionlly xiomtisble. In Proceedings 9th IEEE Symposium on Logic in Computer Science (LICS 94), Pris, pges IEEE Computer Society Press, [18] D.R. Troeger. Step bisimultion is pomset equivlence on prllel lnguge without explicit internl choice. Mthemticl Structures in Computer Science, 3:25 62, [19] C. Verhoef. A generl conservtive extension theorem in process lgebr. In E.-R. Olderog, editor, Proceedings IFIP Conference on Progrmming Concepts, Methods nd Clculi (PROCOMET 94), Sn Minito, IFIP Trnsctions A-56, pges Elsevier, [20] H. Zntem. Termintion of term rewriting by semntic lbelling. Fundment Informtice, 24(1,2):89 105,