On the expressive power of temporal logic

Transcription

1 On the expressive power of temporl logic Joëlle Cohen, Dominique Perrin nd Jen-Eric Pin LITP, Pris, FRANCE Astrct We study the expressive power of liner propositionl temporl logic interpreted on finite sequences or words. We first give trnsprent proof of the fct tht forml lnguge is expressile in this logic if nd only if its syntctic semigroup is finite nd periodic. This gives n effective lgorithm to decide whether given rtionl lnguge is expressile. Our min result sttes similr condition for the restricted temporl logic (RTL), otined y discrding the until opertor. A forml lnguge is RTL-expressile if nd only if its syntctic semigroup is finite nd stisfies certin simple lgeric condition. This leds to polynomil time lgorithm to check whether the forml lnguge ccepted y n n-stte deterministic utomton is RTL-expressile. Temporl logic is prticulr cse of modl logic. It ws introduced y Pnueli [16] in connection with pplictions to the specifiction, development nd verifiction of possily prllel or non-deterministic processes. This logicl lnguge dmits severl vritions, one of them eing propositionl liner temporl logic (PTL). It uses three connectives suggestively clled next, eventully nd until. In this pper we re interested in the descriptive power of propositionl liner temporl logic nd of restriction of temporl logic (RTL) otined y considering only the opertors next nd eventully. In oth cses, we interpret temporl logic on finite words only. In this cse, temporl formul defines set of words (tht is, forml lnguge) nd our prolem is to determine precisely which forml lnguges cn e specified in this wy. In the cse of PTL, the solution hs een known for some time, s consequence of series of deep results. Indeed, Kmp [6] hs shown tht PTL is expressively equivlent to first-order logic when interpreted on words. Next, McNughton [10] proved tht forml lnguge is first-order definle if nd only if it is str-free. Finlly, str-free lnguges re chrcterized y Reserch on this pper ws prtilly supported y PRC Mthémtiques et Informtique. 1

2 deep theorem of Schützenerger [17]: rtionl (or regulr) lnguge is strfree if nd only if its syntctic semigroup is group-free. Since the syntctic semigroup of given rtionl lnguge cn e effectively computed, this provides n lgorithm to determining whether rtionl lnguge is PTLdefinle. Vrious proofs of the equivlence etween first-order, str-free nd PTL-definle hve een nnounced or given in the literture [5, 6, 11, 12] ut ll these proofs re rther involved. In this pper, we give short nd simple proof of the equivlence etween str-free nd PTL-definle, sed on wek version of the Krohn-Rhodes decomposition theorem for finite semigroups. Our proof ws inspired y the work of [11], whose proof uses n interesting connection with Petri nets. Our min result concerns the descriptive power of RTL. It ws known [5, 7] tht RTL is strictly less expressive thn PTL, ut n effective chrcteriztion of RTL-definle forml lnguges ws still to e found. We show here tht RTL-definle lnguges dmit syntctic chrcteriztion nlogous to Schützenerger s theorem: rtionl lnguge is RTL-definle if nd only if its syntctic semigroup is loclly L-trivil. This provides decision procedure to determine whether forml lnguge is RTL-definle. This lgeric chrcteriztion lso leds to polynomil time lgorithm to check whether the forml lnguge ccepted y n n-stte (complete) deterministic utomton is RTL-definle. We give nother (non-effective) description of RTL-definle forml lnguges: these forml lnguges form the smllest oolen lger of forml lnguges contining the lnguges A nd closed under the opertions L L nd L A L for every letter. 1 Semigroups nd forml lnguges. In this section, we riefly review some sic fcts out finite semigroups nd rtionl lnguges. All the definitions nd results presented in this section re stndrd, nd re reproduced for the convenience of the reder. More informtion on this suject cn e found in [3, 8, 15]. For the most prt, we follow the nottions nd terminology of Eilenerg [3]. In prticulr, if ϕ : S T is function from S into T, we denote y sϕ (insted of the usul ϕ(s)) the imge of n element s of S y ϕ. We lso use the term rtionl lnguge insted of regulr lnguge for two resons: first, the term rtionl hs much etter mthemticl foundtion (rtionl lnguges re deeply connected with rtionl series), nd second the term regulr is lso used in semigroup theory with totlly different mening, nd could e misleding in our context. 2

3 1.1 Semigroups. A semigroup is set S together with n ssocitive multipliction. A monoid M is semigroup tht hs n identity element, usully denoted y 1. The free monoid (resp. semigroup) on set A is the set, usuly denoted A (resp. A + ) of ll words (resp. non-empty words) over A, equipped with the conctention of words s multipliction. Thus A = A + {1}, where 1 is the empty word. Given two semigroups S nd T, semigroup morphism ϕ : S T is function from S into T such tht, for every s, s S, (sϕ)(s ϕ) = (ss )ϕ. All semigroups considered in this pper re finite except for free semigroups nd free monoids. Therefore, we shll use in the sequel the term semigroup insted of finite semigroup. An element e of semigroup S is idempotent if e 2 = e. The set of idempotents of semigroup S is denoted y E(S). Every non-empty semigroup contins t lest one idempotent. This is prticulr cse of the following well-known result: Proposition 1.1 For ny semigroup S, there exists n integer n Crd(S) such tht, for every s S, s n is idempotent. The smllest integer n stisfying this property is clled the exponent of S nd is usully denoted ω(s) or simply ω. Thus s ω is convenient nottion for the (unique) idempotent which is power of s. For instnce, if x, y S, (x ω y ω ) ω denotes the idempotent which is power of ef, where e (resp. f) is the idempotent which is power of x (resp. y). We shll frequently use this type of nottion in the sequel. If S is semigroup, the reverse semigroup S r is the semigroup with underlying set S together with the opertion defined y s t = ts. If S is semigroup, we denote y S 1 the monoid equl to S if S is lredy monoid, nd otherwise equl to S {1}, where 1 is new identity element. We shll consider in prticulr three semigroups, denoted respectively U 1, U 2, nd B(1, 2): U 1 is the semigroup {0, 1} with the multipliction given y 1.1 = 1 nd 0.1 = 1.0 = 0.0 = 0, B(1, 2) = {, } with the multipliction given y. =. = nd. =. =, nd U 2 = B(1, 2) 1 = {1,, }. The Green s reltions R nd L on semigroup S re the equivlence reltions defined s follows: s R t if nd only if there exist u, v S 1 such tht su = t nd tv = s, s L t if nd only if there exist u, v S 1 such tht us = t nd vt = s A semigroup S is R-trivil (respectively L-trivil) if the reltion R (respectively L) is equlity. For instnce, U 1 is oth R-trivil nd L-trivil, nd B(1, 2) nd U 2 re L ut not R-trivil, since R. 3

4 Given semigroup S, nd n idempotent e of S, the three susets es = {es s S}, ese = {ese s S}, Se = {se s S} re susemigroups of S. The susemigroup ese is clled the locl semigroup ssocited with e. It is in fct monoid, since e is clerly n identity of ese. A semigroup S is sid to hve property loclly if for every idempotent e of S, the susemigroup ese hs the property. In prticulr, semigroup S is loclly R-trivil (respectively loclly L-trivil) if, for every idempotent e of S, ese is R-trivil (respectively L-trivil). For instnce, B(1, 2) is loclly R-trivil, ut U 2 is not, since 1.U 2.1 = U 2 is not R-trivil. Proposition 1.2 Let S e semigroup. Then (1) S is loclly R-trivil if nd only if, for every e E(S), Se is R-trivil. (2) S is loclly L-trivil if nd only if, for every e E(S), es is L-trivil. Proof. Clerly, (2) is dul version of (1). Let S e loclly R-trivil semigroup. Let e E(S), nd suppose tht se R te for some s, t S. Then there exist ue, ve (Se) 1 such tht seue = te nd teve = se. Thus s(eue)(eve) = se. Furthermore [(eue)(eve)] ω R [(eue)(eve)] ω (eue) holds in ese, nd since ese is R-trivil, it follows Therefore [(eue)(eve)] ω = [(eue)(eve)] ω (eue). se = s[(eue)(eve)] ω = s[(eue)(eve)] ω (eue) = s(eue) = te. Conversely, ssume tht Se is R-trivil. Then ese, which is susemigroup of Se, is lso R-trivil. A semigroup S is periodic if for every s S, there exists n n > 0 such tht s n = s n+1. For instnce the three semigroups U 1, U 2, nd B(1, 2) re periodic, ut non-trivil group is not periodic. 1.2 Trnsformtion semigroups. Let Q e set, nd let S e semigroup. An ction of S on Q is function 1 from Q S into Q, denoted (q, s) q s, such tht, for every q Q nd every s 1, s 2 S, (q s 1 ) s 2 = q (s 1 s 2 ). Let T (Q) e the semigroup of ll functions from Q into itself, with left-toright composition of functions s the multipliction. Any ction of S on Q defines semigroup morphism ρ : S T (Q), given, for every s S, y q (sρ) = q s for every q Q 1 The definition of Eilenerg [3] llows prtil functions, ut we don t need this more generl definition. 4

5 The ction of S on Q is fithful if ρ is injective, tht is, if two elements of S hving the sme ction on Q re equl. A trnsformtion semigroup (ts for short) is pir (Q, S), where Q is set (the set of sttes) nd S is semigroup cting fithfully on Q. Two nturl exmples of trnsformtion semigroups re frequently used: first, every semigroup S defines trnsformtion semigroup (S 1, S), the ction eing simply the product in S. This trnsformtion semigroup is usully denoted simply S, nd the context suffices to decide whether one considers semigroup or trnsformtion semigroup. The second exmple is the notion of trnsformtion semigroup of n utomton. Let A = (Q, A, ) e (complete) deterministic utomton. By definition, every word w of A + defines function wρ from Q into Q, given, for every q Q, y q(wρ) = q w This defines semigroup morphism ρ : A T (Q). The rnge of ρ is susemigroup of T (Q) denoted S(A) nd clled the semigroup of A, nd the trnsformtion semigroup T S(A) = ( Q, S(A) ) is clled the trnsformtion semigroup of A. In prctice, the nottion wρ is lmost lwys simplified to w, nd the context mkes cler whether one is considering w s word or s n element of T (Q). For instnce, if A is the utomton represented elow 1 2 0, Figure 1: The utomton A. then T S(A) = ({0, 1, 2}, {,,,, }) where the ction of ech element is represented in the following tle: We shll lso use the trnsformtion semigroup 2 = ({1, 2}, B(1, 2)) where the ction is given y the formuls 1 = 2 = 1 nd 1 = 2 = 2. 5

6 1 2 Figure 2: The trnsformtion semigroup 2. A trnsformtion semigroup (P, S) divides trnsformtion semigroup (Q, T ) if there exists surjective prtil function ϕ : Q P, nd, for every s S, there exists n element ŝ T such tht, for every q Q, (qϕ) s = (q ŝ)ϕ. For instnce oth B(1, 2) nd U 1 divide U Forml lnguges. Let A + e free semigroup. The set A is clled the lphet nd the elements of A re letters. The length of word w A + is denoted y w. A suset of A + is clled (forml) lnguge. Rtionl lnguges form the smllest clss of lnguges contining letters nd closed under union, conctention nd the plus opertion (L + = n>0 Ln ). Str-free lnguges form the smllest clss of lnguges contining letters nd closed under oolen opertions (union, intersection nd complementtion) nd conctention product. The notion of the lnguge recognized y n utomton cn e esily dpted to trnsformtion semigroups s follows: trnsformtion semigroup (Q, S) recognizes lnguge L A + if there is semigroup morphism η : A + S, stte q 0 Q (the initil stte), set of sttes F (the finl sttes) such tht L = {u A + q 0 (uη) F }. When the trnsformtion semigroup is of the form S = (S 1, S), there is more convenient equivlent definition, tht does not refer to trnsformtion semigroups: semigroup S recognizes lnguge L A + if there is morphism η : A + S, nd suset P of S, such tht L = P η 1. It is esy to see tht if lnguge L is recognized y trnsformtion semigroup X, nd if X divides trnsformtion semigroup Y, then Y lso recognizes L. For instnce, if A nd B A, the lnguges A A, A B nd A re recognized y U 1, U 2 nd B(1, 2), respectively. Conversely, we hve the following lemm (see [15], chpter 2). Lemm 1.3 (1) If lnguge of A + is recognized y U 1, then it is oolen comintion of lnguges of the form A A where A. (2) If lnguge of A + is recognized y U 2, then it is oolen comintion of lnguges of the form A B where A nd B A. (3) If lnguge of A + is recognized y B(1, 2), then it is oolen comintion of lnguges of the form A where A. 6

7 The syntctic semigroup of lnguge L A +, denoted S(L), is the quotient of A + y the congruence L defined y u L v if nd only if, for every x, y A, xuy L xvy L. The syntctic semigroup of lnguge L is the smllest semigroup tht recognizes L. It is lso the semigroup of the miniml utomton of L. As is well-known, lnguge is rtionl if nd only if it cn e recognized y finite utomton. Since there re stndrd lgorithms to compute the miniml utomton of given rtionl lnguge, this provides n lgorithm to compute the syntctic semigroup of rtionl lnguge. For str-free lnguges, we hve the following importnt result, due to Schützenerger [17]. A proof cn e found in [3, 8, 15, 14]. Theorem 1.4 Let L e lnguge. The following conditions re equivlent (1) L is str-free, (2) L is recognized y n periodic semigroup, (3) the syntctic semigroup of L is periodic. 1.4 Wreth product. The wreth product of two trnsformtion semigroups X = (P, S) nd Y = (Q, T ) is the trnsformtion semigroup X Y = (P Q, S Q T ), with multipliction given y 2 (f 1, t 1 )(f 2, t 2 ) = (f, t 1 t 2 ), where, for everyq Q, qf = (qf 1 )(qt 1 )f 2 nd where the ction of n element (f, t) of S Q T on stte (p, q) of P Q is given y (p, q) (f, t) = (p (qf), q t). The wreth product is n ssocitive opertion on trnsformtion semigroups. Aperiodic, R-trivil nd loclly R-trivil semigroups dmit simple wreth-product decompositions using the three trnsformtion semigroups U 1, U 2, nd 2 defined in section 1.1 nd 1.2. For proof, see [3, Vol. B] or [20]. Theorem 1.5 (1) A semigroup is R-trivil if nd only if it divides wreth product of the form U 1 U 1. (2) A semigroup is loclly R-trivil if nd only if it divides wreth product of the form U 1 U S Q denotes the set of ll functions from Q to S. Thus if f S Q nd q Q, qf is n element of S. 7

8 (3) A semigroup is periodic if nd only if it divides wreth product of the form U 2 U 2. Wreth products re deeply relted to sequentil functions. Recll tht trnsducer T = (Q, A, B, q 0,., ) is given y finite set of sttes Q, n input lphet A, n output lphet B, n initil stte q 0, next-stte function Q A Q, denoted (q, ) q, nd n output function Q A B +, denoted (q, ) q. The next-stte function is extended to function Q A + Q y setting q (u) = (q u) for ech u A nd A. Similrly, the output function is extended to function Q A + B + y setting q u = (q u)((q u) ). The function σ : A + B + defined y uσ = q 0 u is clled the sequentil function defined y T. Then we cn stte Proposition 1.6 [3] Let σ : A + B + e sequentil function relized y trnsducer T = (Q, A, B, q 0,, ) nd let S(σ) e the trnsformtion semigroup of the utomton (Q, A, ). If lnguge L B + is recognized y semigroup S, then Lσ 1 is recognized y S S(σ). The following result is first ppliction of Proposition 1.6 to syntctic property of the opertors L LA nd L L on lnguges. Proposition 1.7 [3, 22] Let L A + e recognizle lnguge. Then (1) S(LA ) divides U 1 S(L), (2) S(L) divides B(1, 2) S(L). Proof. Let ϕ : A + S = S(L) e the syntctic morphism of L. This morphism cn e extended to monoid morphism ϕ : A S 1. Put P = Lϕ, B = S 1 A, nd let σ : A + B + e the sequentil function defined y ( 1 n )σ = (1ϕ, 1 ) (( 1 n 1 )ϕ, n ). Note tht σ is relized y trnsducer (tht is, deterministic utomton with output) with S 1 s the set of sttes nd next-stte nd output functions defined y the following digrm. s (s, ) s(ϕ) Figure 3: A trnsducer relizing σ. 8

9 In prticulr, the semigroup S(σ) is equl to S. Put C = P {}. Then C is suset of B nd we hve (B CB )σ 1 = {u A + uσ B CB } = { 1 n A + i {1,, n 1} (( 1 i )ϕ i+1 ) C} = { 1 n A + i {1,, n 1} 1 i P ϕ 1 nd i+1 = } = (P ϕ 1 )A = LA. Therefore, y Proposition 1.6, LA is recognized y S(B CB ) S(σ). Sttement (1) follows, since S(B CB ) = U 1. Similrly, we hve (B C)σ 1 = {u A + uσ B C} = { 1 n A + (( 1 n 1 )ϕ, n ) C} = { 1 n A + 1 n 1 P ϕ 1 nd n = } = (P ϕ 1 ) = L. Therefore, y Proposition 1.6, L is recognized y S(B C) S(σ). Sttement (2) follows, since S(B C) = B(1, 2). Struing s wreth product principle reclled elow gives description of the lnguges recognized y the wreth product of two trnsformtion semigroups. Let X = (P, S) nd Y = (Q, T ) e two trnsformtion semigroups, nd let Z = X Y = (P Q, R), where R = S Q T. Let L e lnguge of A + recognized y Z: then there exist n initil stte (p 0, q 0 ) P Q, set of finl sttes F in P Q nd morphism η : A + R such tht L = {u A + (p 0, q 0 ) (uη) F }. The morphism η defines n ction of A + on P Q y setting (p, q) = (p, q)(η). Let π e the nturl projection π : R = S Q T T. Define sequentil function σ : A + (Q A) + y ( 1 n )σ = (q 0, 1 )(q 0 ( 1 ηπ), 2 ) (q 0 ( 1 n 1 )ηπ, n ). We cn now stte Proposition 1.8 (Wreth product principle [21]) The lnguge L is finite union of lnguges of the form U V σ 1, where U is lnguge of A + recognized y Y nd V is lnguge of (Q A) + recognized y X. Proposition 1.8, or some similr sttement, together with Theorem 1.5, hs een used to prove Theorem 1.4 [2, 3, 9]. 9

10 1.5 Vrieties of semigroups. A vriety of semigroups is clss of semigroups closed under tking susemigroups, quotients nd finite direct products 3. The following vrieties will e used in this rticle: A, the vriety of periodic semigroups, R, the vriety of R-trivil semigroups, L, the vriety of L-trivil semigroups, LR, the vriety of loclly R-trivil semigroups, LL, the vriety of loclly L-trivil semigroups. It is often convenient to define vrities y identities. Let u, v A +. Formlly, semigroup S stisfies the identity u = v if nd only if, for every semigroup morphism ϕ : A + S, uϕ = vϕ. For instnce, semigroup is commuttive if nd only if it stisfies the identity xy = yx. The next proposition gives identities defining the vrieties A, R, L, LR nd LL. In fct, there re not identities in the strict sense 4, since they involve the exponent ω, which depends on the semigroup S. Proposition 1.9 (1) A semigroup is periodic if nd only if it stisfies the identity x ω = x ω+1, (2) A semigroup is R-trivil if nd only if it stisfies the identity (xy) ω x = (xy) ω, (3) A semigroup is L-trivil if nd only if it stisfies the identity y(xy) ω = (xy) ω, (4) A semigroup is loclly R-trivil if nd only if it stisfies the identity (ux ω vx ω ) ω ux ω = (ux ω vx ω ) ω, or, equivlently, the identity (x ω ux ω vx ω ) ω x ω ux ω = (x ω ux ω vx ω ) ω, (5) A semigroup is loclly L-trivil if nd only if it stisfies the identity x ω v(x ω ux ω v) ω = (x ω ux ω v) ω, or, equivlently, the identity x ω v(x ω ux ω vx ω ) ω = (x ω ux ω vx ω ) ω. A vriety of semigroups V is closed under wreth product if, given two trnsformtion semigroups X = (P, S) nd Y = (Q, T ) nd their wreth product (P Q, R), the conditions S, T V imply R V. The next proposition is the vriety version of Theorem 1.5. Proposition 1.10 [3, 20] 3 The correct terminology should e pseudovriety to void possile confusion with Birkhoff s vrieties. However, we hve preferred to void this rther wkwrd terminology. 4 Agin, the correct terminology should e pseudoidentity. 10

11 (1) R is the smllest vriety of semigroups closed under wreth product contining U 1. (2) LR is the smllest vriety of semigroups closed under wreth product contining U 1 nd B(1, 2). (3) A is the smllest vriety of semigroups closed under wreth product contining U 2. 2 Propositionl temporl logic. Propositionl temporl logic (PTL for short) on n lphet A is defined s follows. The voculry consists of (1) An tomic proposition p for ech letter A (2) Connectives, nd. (3) Temporl opertors ( next ), ( eventully ) nd U ( until ). nd the formuls re constructed ccording to the rules (1) For every A, p is formul, (2) If ϕ nd ψ re formuls, so re ϕ ψ, ϕ ψ, ϕ, ϕ, ϕ, ϕ U ψ. Semntics re defined y induction on the formtion rules. Given word w A +, nd n {1, 2,..., w }, we define the expression w stisfies ϕ t the instnt n (denoted (w, n) = ϕ) s follows (1) (w, n) = p if the n-th letter of w is n. (2) (w, n) = ϕ ψ (resp. ϕ ψ, ϕ) if (w, n) = ϕ or (w, n) = ψ (resp. if (w, n) = ϕ nd (w, n) = ψ, if (w, n) does not stisfy ϕ). (3) (w, n) = ϕ if (w, n + 1) stisfies ϕ. (4) (w, n) = ϕ if there exists m such tht n m w nd (w, m) = ϕ. (5) (w, n) = ϕ U ψ if there exists m such tht n m w, (w, m) = ψ nd, for every k such tht n k < m, (w, k) = ϕ. Note tht, if w = w 1 w 2 w w, (w, n) = ϕ only depends on the word w = w n w n+1 w w. Exmple 2.1 Let w = c. Then (w, 4) = p since the fourth letter of w is n, (w, 4) = p since the fifth letter of w is nd (w, 4) = (p c p ) since c is fctor of c. If ϕ is temporl formul, we sy tht w stisfies ϕ if (w, 1) = ϕ. We just hve defined future temporl formuls ut one cn define in the sme wy pst temporl formuls y reversing time: it suffices to replce next y previous (symol ), eventully y sometimes (symol ) nd until y since (symol S). The corresponding semntics re modified s follows. (3 ) (w, n) = ϕ if n > 1 nd (w, n 1) stisfies ϕ. 11

12 (4 ) (w, n) = ϕ if there exists m n such tht (w, m) = ϕ. (5 ) (w, n) = ϕ S ψ if there exists m n such tht (w, m) = ψ nd for every k such tht m < k n, (w, k) = ϕ. The digrm elow illustrtes the symmetry etween the opertors until nd since. ψ ψ n... m w 1 m... n ϕ ϕ ϕ ϕ ϕ ϕ Figure 4: A digrm for (w, n) = ϕ U ψ nd for (w, n) = ϕ S ψ. If ϕ is pst temporl formul, we sy tht w stisfies ϕ if (w, w ) = ϕ. The lnguge defined y formul ϕ is the set L(ϕ) of ll words of A + tht stisfy ϕ. 3 PTL-definle lnguges. In this section, we present short proof of the following result Theorem 3.1 A lnguge of A + is PTL-definle if nd only if its syntctic semigroup is periodic. Proof. Since the reverse of n periodic semigroup is lso periodic, it suffices to prove the dul version of the theorem, otined y using pst temporl logic. We first prove tht every PTL-definle lnguge is strfree (y Schützenerger s theorem, lnguge is str-free if nd only if its syntctic semigroup is periodic). This is done y induction on the formtion rules. Indeed (1) L(p ) = A (for every letter ) is str-free. (2) L( ϕ) = L(ϕ)A. Thus if L(ϕ) is str-free, so is L( ϕ). (3) L( ϕ) = L(ϕ)A. Thus if L(ϕ) is str-free, so is L( ϕ). We need similr formul for S, ut this is slightly more complicted. Assume tht L(ϕ) nd L(ψ) re str-free. In prticulr, there is semigroup morphism η : A + S, where S is n periodic semigroup, nd suset P of S such tht L(ϕ) = P η 1. Set, for every s S, s 1 P = {t S st P }. Then we hve the following lemm, in which \ denotes set difference. 12

13 Lemm 3.2 The following equlities hold L(ϕ S ψ) = {uv A + u L(ψ), v A nd for ech left fctor v 1 of v, uv L(ϕ)} = ( )( sη 1 L(ψ) A \ ( A + \ (s 1 P )η 1) A ). s S Proof. The first equlity is direct consequence of the definition. Next, if R A +, (A + \ R)A is the set of ll words v A hving left fctor v 1 in R. Therefore, tking complements, this is equivlent to sying tht A \ (A + \ R)A is the set of ll words v A such tht, for ech left fctor v 1 of v, v / R. Let w L(ϕ S ψ). Then, y the first equlity, w = uv, where u L(ψ), v A, nd for ech left fctor v 1 of v, uv L(ϕ). Putting s = uη, we otin u sη 1 L(ψ) nd (uv )η P, whence v (s 1 P )η 1. Thus, ( w sη 1 L(ψ) )(A \ ( A + \ (s 1 P )η 1) A ) y the remrk ove. Conversely, ssume tht w = uv, where, for some s S, u sη 1 L(ψ) nd v A \ ( A + \ (s 1 P )η 1) A. Then u L(ψ), nd for ech left fctor v 1 of v, v (s 1 P )η 1. Thus (uv )η = s(v η) P, whence uv L(ϕ). Therefore, y the first equlity, w L(ϕ S ψ). Now ny lnguge of the form Qη 1, where Q S, is recognized y S, nd thus is str-free y Schützenerger s theorem. Therefore, Lemm 3.2 shows tht L(ϕ S ψ) is str-free nd this concludes the first prt of the proof of Theorem 3.1. We now show tht every str-free lnguge is PTL-definle. Let C e the clss of ll trnsformtion semigroups X such tht every lnguge recognized y X is PTL-definle. By Schützenerger s theorem, it suffices to show tht ech periodic semigroup elongs to C. The clss C is certinly closed under division, ecuse if X divides Y, every lnguge recognized y X is lso recognized y Y. Next, the trivil semigroup {1} elongs to C, since the lnguges of A + recognized y {1} re A + nd the empty set. Now, y Theorem 1.5, it remins to show tht if Y = (Q, T ) C, then U 2 Y C. By the wreth-product principle, every lnguge of A + recognized y U 2 Y is finite union of lnguges of the form U V σ 1 where σ : A + B + = (Q A) + is certin sequentil function, U A + is recognized y Y nd V B + is recognized y U 2. First, the formuls L( ϕ) = A + \ L(ϕ) nd L(ϕ ψ) = L(ϕ) L(ψ) show tht PTL-definle lnguges re closed under oolen opertions. Thus it suffices to show tht every lnguge of the form U V σ 1 ove is PTL-definle. Since Y C, U is PTL-definle 13

14 y definition. Furthermore, y Lemm 1.3, V is oolen comintion of lnguges of the form B C, where B nd C B. Since σ 1 commutes with oolen opertions, it remins to show tht lnguges of the form (B C )σ 1 re PTL-definle. We clim tht (B C )σ 1 = (B C)σ 1 S (B )σ 1 (1) Indeed, let u = 1 n e word of A + nd let ( 1 n )σ = 1 n. Then uσ B C if nd only if there exists n i such tht i = nd, for every j > i, j C. This is equivlent to sying tht ( 1 i )σ B nd for every j > i, ( 1 j )σ B C, nd this proves (1). Now (B C)σ 1 = (B )σ 1 nd therefore it suffices to show tht lnguges of the form (B )σ 1 re PTL-definle. We tke gin the nottions used in the definition of σ (cf. Proposition 1.8). Set = (q, ) (recll tht B = Q A). Then we hve ( 1 n )σ = (q 0, 1 )(q 0 ( 1 ηπ), 2 ) (q 0 ( 1 n 1 )ηπ, n ). It follows tht ( 1 n )σ B if nd only if q 0 ( 1 n 1 )ηπ = q nd n =. Therefore (B )σ 1 = L, where L = {u A + q 0 (uηπ) = q}. But L is recognized y Y nd since Y C, is PTL-definle. Now, since L(ϕ) = L( ϕ p ), L = (B )σ 1 is PTL-definle nd this concludes the proof. C 4 Restricted temporl logic. If we omit the until opertor, we otin restricted temporl logic (RTL) tht ws considered in [5, 6]. Here is first description of the lnguges definle in this logic. The sutle distinction etween conditions (2) nd (3) will e used in the proof of the min theorem elow. Proposition 4.1 Let L e lnguge of A +. The following conditions re equivlent: (1) L is RTL-definle, (2) L elongs to the smllest oolen lger of lnguges contining the lnguges A nd closed under the opertions L A L nd L L for every A, (3) L elongs to the smllest oolen lger of lnguges contining the lnguges A nd closed under the opertions L A L nd L L for every A. 14

15 Proof. Let C (respectively C ) e the smllest oolen lger of lnguges closed under the opertions L A L (respectively L A L) nd L L for every letter A. In prticulr the lnguges nd A + elong to C nd C y definition. We first prove tht C = C. The inclusion C C follows directly from the formul A L = A (L). The opposite inclusion follows from the formul A L = L A L. Thus (2) nd (3) re equivlent. (1) implies (2). We show y induction on the formtion rules tht L(ϕ) C for every RTL-formul ϕ. First, if ϕ = p, then A L(p ) = A C. If ϕ nd ψ re formuls such tht L(ϕ) nd L(ψ) elong to C, then L(ϕ ψ) = L(ϕ) L(ψ) C, L(ϕ ψ) = L(ϕ) L(ψ) C, L( ϕ) = A + \ L(ϕ) C, L( ϕ) = AL(ϕ) = A L(ϕ) C, L( ϕ) = A L(ϕ) C. (2) implies (1). Let F e the set of RTL-definle lnguges. Then F contins A = L(p ), for every A. The formuls L(ϕ) L(ψ) = L(ϕ ψ) nd A + \ L(ϕ) = L( ϕ) show tht F is oolen lger nd the formul A L(ϕ) = L( ϕ) shows tht F is closed under the opertion L A L. Finlly, the formul L(ϕ) = L(p ϕ) shows tht F is closed under the opertion L L, for every letter A. Therefore F contins C. We cn now stte our min result. Theorem 4.2 Let L e lnguge of A +. The following conditions re equivlent: (1) L is RTL-definle, (2) the syntctic semigroup of L is loclly L-trivil. Proof. As for Theorem 3.1, we prove the dul version of the theorem, which sttes tht L is definle in pst restricted temporl logic if nd only if its syntctic semigroup is R-trivil. Consider the smllest oolen lger B contining the lnguges A nd closed under the opertions L LA nd L L for every A. By Proposition 4.1 nd dulity, it suffices now 15

16 to prove the following sttement: A lnguge elongs to B if nd only if its syntctic semigroup elongs to LR. First, S(A ) = B(1, 2) LR. Now, y Proposition 1.7, S(LA ) divides U 1 S(L), nd S(L) divides B(1, 2) S(L). It follows y Proposition 1.10, tht if S(L) LR, then S(LA ) LR nd S(L) LR. Therefore, if L B, then S(L) LR. In the other direction, the proof mimics the proof of Theorem 3.1. Let C e the clss of ll trnsformtion semigroups X such tht every lnguge recognized y X elongs to B. The clss C contins the trivil semigroup nd is closed under division. Therefore, to show tht C contins LR, it suffices, y Proposition 1.10, to verify tht if Y C, then U 1 Y C nd 2 Y C. By the wreth-product principle, every lnguge of A + recognized y U 1 Y (respectively 2 Y ) is finite union of lnguges of the form U V σ 1 where σ : A + B + = (Q A) + is certin sequentil function, U A + is recognized y Y nd V B + is recognized y U 1 (respectively 2). Since Y C, U elongs to B y definition. Furthermore, y Lemm 1.3, V is oolen comintion of lnguges of the form B B, (respectively B ) where B. Since σ 1 commutes with oolen opertions, it remins to show tht the lnguges of the form (B B )σ 1 (respectively (B )σ 1 ) elong to B. We tke gin the nottions used in the definition of σ (cf. Proposition 1.8). Set = (q, ) (recll tht B = Q A). Then we hve ( 1 n )σ = (q 0, 1 )(q 0 ( 1 ηπ), 2 ) (q 0 ( 1 n 1 )ηπ, n ). First ssume q q 0. Then ( 1... n )σ B B if nd only if there exists n index i such tht q 0 ( 1 i 1 )ηπ = q nd i =. Therefore (B B )σ 1 = LA, where L = {u A + q 0 (uηπ) = q}. If q = q 0, then (B B )σ 1 = LA A. But L is recognized y Y nd since Y C, L elongs to B. Furthermore, A lso elongs to B, since nd {} = A \ ( (A )A (A )A (A )A (A )A ), A = {} {}A {}A It follows tht (B B )σ 1 elongs to B. Similrly, ( 1 n )σ B if nd only if q 0 ( 1 n 1 )ηπ = q nd n =. Therefore (B )σ 1 = L or L {} (if q = q 0 ) nd (B )σ 1 lso elongs to B. Corollry 4.3 Given rtionl lnguge L, one cn effectively decide whether it is RTL-definle. 16

17 Proof. The lnguge L cn e given either y rtionl expression or y finite utomton. In oth cses, there re well-known lgorithms to compute its miniml utomton A(L), nd then its syntctic semigroup S(L), which is lso the trnsformtion semigroup of A(L). Then it suffices, y Proposition 1.9 to verify tht S(L) stisfies the identity x ω v(x ω ux ω v) ω = (x ω ux ω v) ω. Sy tht two PTL-formuls ϕ nd ψ re equivlent if L(ϕ) = L(ψ), tht is, if they gree when interpreted on finite words. Corollry 4.4 Given PTL-formul, one cn effectively decide whether it is equivlent to some RTL-formul. We conclude this section y three exmples. Exmple 4.1 Let A = {, } nd let L = () +. Then the miniml utomton of L is represented in the digrm elow. 1 2 Figure 5: The miniml utomton of () +. The syntctic semigroup of L is the semigroup S with zero presented y the reltions 2 = 2 = 0, =, =. Thus S = {,,,, 0}. There re three idempotents,, nd 0. The corresponding locl semigroups re S = {, 0}, S = {, 0} nd 0S0 = {0}, ll of which re L-trivil. Therefore L is expressile in restricted temporl logic. Indeed, we hve L = L(ϕ), where ϕ = p (p p p ) (p p ) (p p ). Exmple 4.2 Let A = {,, c} nd let L = A {, c}. Then the miniml utomton of L is represented in the digrm elow., c 1 2, c Figure 6: The miniml utomton of A {, c}. 17

18 The syntctic semigroup of L is U 2, which is loclly L-trivil. Therefore L is expressile in restricted temporl logic. Indeed, we hve L = L(ϕ), where ϕ = (p p ). Exmple 4.3 Let A = {,, c} nd let L = {,, c}. Then the miniml utomton of L is represented in the digrm elow. 1 2,, c Figure 7: The miniml utomton of {,, c} The syntctic semigroup of L is the monoid S presented y the reltions = 1, = c =, c = cc = c. This is the reverse of U 2, nd it is periodic, ut not loclly L-trivil. Therefore, ny formul ψ such tht L = L(ψ) uses the until opertor. In fct, L = L(ϕ), where ϕ = p U p. 5 Automt, vrieties nd foridden configurtions. In the two previous sections, we hve seen how to chrcterize the forml lnguges ssocited with formul of propositionl temporl logic (section 3) nd of restricted temporl logic (section 4). Both chrcteriztions re in terms of the syntctic semigroup of the forml lnguge. We shll see here how this chrcteriztion cn e expressed in terms of utomt. In the cse of restricted temporl logic, this hs the dvntge of providing polynomil lgorithm to check whether the lnguge defined y given deterministic utomton is RTL-definle. This is of interest since, on the contrry, the corresponding prolem for PTL logic is the complement of n NP-hrd prolem [19] nd is PSPACE-complete [1]. Thus, unless P = NP, checking whether the lnguge defined y given utomton is PTLdefinle cnnot e solved in polynomil time. We egin with the chrcteriztion of utomt ssocited with R-trivil semigroups. We shll then tret the cse of loclly R-trivil semigroups. This corresponds, s we hve seen, to formuls of pst temporl logic. We shll finlly come to L-trivil nd loclly L-trivil semigroups, which correspond to RTL-formuls. We shll see how these chrcteriztions led to polynomil lgorithms. Before to give the detils of our lgorithms, let us fix some convenient nottions. Given finite (complete) deterministic utomton A = (Q, A, ) 18

19 nd positive integer k, we denote y A k = (Q k, A, ) the direct product of k copies of A, where the ction of A on Q k is given y (q 1,..., q k ) = (q 1,..., q k ) We lso denote y G k (A) the trnsitive closure of the directed grph defined y A k. For instnce, if A is the utomton represented elow, 1 2 Figure 8: then A 2 is the utomton, 1, 1 2, 2 1, 2 2, 1 nd G 2 (A) is the grph Figure 9: The utomton A 2. 1, 1 2, 2 1, 2 2, 1 Figure 10: The grph G 2 (A). Given deterministic utomton A = (Q, A, ), the set of ll pths in A defines n infinite lelled grph G(A), with Q s set of vertices, nd the 19

20 triples of the form (q, w, q.w) (where w A + ) s edges. A lelled sugrph of G(A) is sid to e configurtion present in A. Two words x, y A which hve the sme ction on Q re sid to e equivlent in A (nottion x y). The following result is lredy in [15], p Theorem 5.1 The semigroup of deterministic utomton A is R-trivil if nd only if there exist no configurtions of A of the form x p q with p q. y Figure 11: Foridden configurtion for R-trivil utomt. Proof. Suppose first tht S(A) is R-trivil nd consider configurtion s ove. Let ω e the exponent of S(A). Then we hve, for every x, y A + nd therefore (xy) ω (xy) ω x p = p (xy) ω = p = p (xy) ω x = q whence p = q. Conversely, if A = (Q, A, ) contins no foridden configurtion, let us verify tht, for every u, v A +, (uv) ω (uv) ω u. Let r Q nd let p = r (uv) ω. Since (uv) ω is idempotent, we hve p (uv) ω = p. Set x = u, y = (vu) ω 1 v nd q = p x. Then q y = p xy = p (uv) ω = p. Therefore A contins the configurtion of Figure 11 nd thus p = q. Therefore p = r (uv) ω = r (uv) ω u nd thus (uv) ω (uv) ω u. The trnsposition of the previous chrcteriztion to the cse of loclly R-trivil semigroups follows generl scheme. Let V e vriety of semigroups nd ssume tht the deterministic utomt whose semigroups elong to V cn e descried y set C of foridden configurtions. Then the deterministic utomt whose semigroups elong to the vriety LV of ll semigroups which re loclly in V cn e descried y the set C of foridden configurtions otined s follows. For ech configurtion C C, we dd to ech vertex loop leled y new symol, the sme for ll vertices. Then the semigroup of deterministic utomton A elongs to LV if nd only if tht A contins no configurtion of C. In prticulr, we hve the following result. Theorem 5.2 The semigroup of deterministic utomton A is loclly R- trivil if nd only if there exist no configurtions of A of the form 20

21 x u q q v x Figure 12: Foridden configurtion for loclly R-trivil utomt. with q q. Proof. By Proposition 1.9, semigroup is loclly R-trivil if nd only if it stisfies the identity (ux ω vx ω ) ω ux ω = (ux ω vx ω ) ω (2) Suppose tht S(A) is loclly R-trivil nd tht A contins configurtion of the form represented in 12. Then y (2), q = q (ux ω vx ω ) ω = q (ux ω vx ω ) ω ux ω = q Conversely, suppose tht A stisfies the condition of the theorem, nd let u, v, x e ritrry words of A +. Set u = ux ω, v = vx ω nd x = x ω. Let q e stte, nd set q 1 = q (ux ω vx ω ) ω nd q 2 = q 1 ux ω. Then short computtion shows tht A contins the following configurtion: u x q 1 q 2 x v Figure 13: A configurtion contined in A. nd thus q 1 = q 2. It follows tht q (ux ω vx ω ) ω = q (ux ω vx ω ) ω ux ω for ny stte q, nd thus S stisfies the identity (2). Thus S(A) is loclly R-trivil. The previous result yields to polynomil time lgorithm to check whether the semigroup of n n-stte deterministic utomton A is loclly R-trivil or not. Indeed, one first oserve tht given two sttes q nd q, there is word w A + such tht q w = q nd q w = q if nd only if ( (q, q ), (q, q ) ) is n edge in the directed grph G 2 (A). Therefore, one cn check whether A contins configurtion of the form 12 with q q y computing G 1 nd G 2 nd y verifying there re no pirs {q, q } of sttes such tht () (q, q ) nd (q, q) re edges in G 1 (A), nd () ( (q, q ), (q, q ) ) is n edge of G 2 (A). 21

22 Since G 1 (resp. G 2 ) hs n (n 2 ) vertices, this gives polynomil lgorithm. This is in fct generl property of vrieties defined y foridden configurtions. Let indeed V e vriety of semigroups nd ssume tht the deterministic utomt whose semigroups elong to V cn e descried y finite set C of foridden configurtions. Then there is polynomil lgorithm to check whether given n-stte deterministic utomton A elongs to V. For this we hve to check whether or not some configurtion C of C is present in A. The numer of possile ssignements of sttes to the vertices of C is polynomil in n. And for ech ssignement, the existence of given set of k edges with the sme lel is solved y reduction to n ccessiility prolem in the grph G k (A). The overll lgorithm is polynomil. In prticulr, we hve the following result. Corollry 5.3 There is polynomil time lgorithm for testing whether the reverse of the lnguge ccepted y n n-stte deterministic utomton is RTL-definle. We illustrte this method on the following exmple. Exmple 5.1 Let A e the utomton given in Figure 6 nd lredy considered in Exmple 4.2. To check whether S(A) is loclly R-trivil, we construct the grph G 2 (A). It is represented in Figure 14. 1, 1 2, 2 1, 2 2, 1 Figure 14: The grph G 2 (A). Now this grph contins cycle of length 1 round (1, 2) nd 1 nd 2 re in the sme strongly connected component of G 1 (A). This indictes the presence of foridden configurtion. It is indeed otined for instnce with the lels given in Figure 15 22

23 c 1 2 c Figure 15: A foridden configurtion. It follows tht A is not R-trivil nd L(A) is not expressile in reverse restricted temporl logic. We now consider the cse of L-trivil semigroups. Proposition 5.4 The semigroup of deterministic utomton A is L-trivil if nd only if the configurtion y y y p q x r x x Figure 16: Foridden configurtion for L-trivil utomt. with p r is not present in A. Proof. Let us first suppose tht S(A) is L-trivil. We consider configurtion s ove. Since (yx) ω x(yx) ω, we hve r = q x(yx) ω = q (yx) ω = p whence p = r. Conversely, suppose tht the ove configurtion is not present in A. Let x, y A + nd let q Q e ritrry. Let r = q x(yx) ω nd p = q (yx) ω. Then p = r y the hypothesis nd therefore x(yx) ω (yx) ω. Thus S(A) is L-trivil. Note tht the chrcteriztion of Proposition 5.4, contrry to tht of Theorem 5.1 requires the hypothesis tht the utomton is complete. There is in fct no possiility of chrcteriztion y foridden configurtions of L- trivil semigroups given y deterministic utomton if it is not complete. Indeed the utomton of figure 17 (i) is sugrph of the leled grph of the utomton of figure 17 (ii). The semigroup of the first one is not L-trivil wheres the second one is. 23

24 (i) (ii) Figure 17: Two utomt. We finlly give the nnounced chrcteriztion of loclly L-trivil semigroups. It is corollry of Proposition 5.4. Proposition 5.5 The semigroup of deterministic utomton A is loclly L-trivil if nd only if the configurtion t y y p q x r x t t t Figure 18: Foridden configurtion for loclly L-trivil utomt. with p r is not present in A. Together with Theorem 4.2, we otin. Corollry 5.6 There is polynomil time lgorithm for testing whether the lnguge ccepted y n n-stte deterministic utomton is RTL-definle. This does not give, however, polynomil lgorithm to check whether given PTL-formul is equivlent with RTL formul. We presently do not know ny resonle ound on the complexity of this prolem. 6 Conclusion. We hve given n effective chrcteriztion of the lnguges definle in liner propositionl temporl logic nd in restricted temporl logic. It would e interesting to otin similr chrcteriztions when the temporl logic is interpreted on infinite words. This will e the suject of future pper. Another interesting question is to consider the temporl logic whose only opertor is eventully. Sistl nd Zuck [18] hve given description of the set of infinite words definle in this logic, ut this description doesn t seem to e effective. y x t 24

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