Automata, Games, and Verification

Transcription

1 Automt, Gmes, nd Verifiction Prof. Bernd Finkbeiner, Ph.D. Srlnd University Summer Term 2015 Lecture Notes by Bernd Finkbeiner, Felix Klein, Tobis Slzmnn These lecture notes re working document nd my contin errors. The current version of this document is vilble t: Plese report ny bugs, comments nd ides for improvement to

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3 Contents 1 Introduction Model Checking Synthesis The Logic-Automt Connection Büchi Automt Preliminries Automt over Infinite Words The Büchi Acceptnce Condition Büchi s Chrcteriztion Theorem Kleene s Theorem ω-regulr Lnguges Closure Properties of the Büchi-Recognizble Lnguges Büchi s Chrcteriztion Theorem Deterministic Büchi Automt 15 5 Complementtion of Büchi Automt 19 6 McNughton s Theorem The Muller Acceptnce Condition From Nondeterministic to Semi-Deterministic Automt From Semi-Deterministic Büchi to Deterministic Muller Sfr s Construction Logics over Infinite Sequences Liner-time Temporl Logic (LTL) Quntified Propositionl Temporl Logic (QPTL) Mondic Second-Order Logic of One Successor (S1S) Wek Mondic Second-Order Logic of One Successor (WS1S) Alternting Büchi Automt Alternting Automt From LTL to Alternting Automt Trnslting Alternting to Nondeterministic Automt Infinite Gmes Bsic Definitions Rechbility Gmes Büchi Gmes Prity Gmes Rbin s Theorem Tree Automt Complementtion of Prity Tree Automt Mondic Second-Order Logic of Two Successors (S2S) Computtion Tree Logic CTL Alternting Tree Automt From CTL to Alternting Tree Automt

4 12 Summry Automt Chrcteriztion Theorems Trnslting Brnching Modes Trnslting Acceptnce Conditions Automt nd Gmes Determincy Logics Model Checking nd Synthesis

5 1 Introduction The theory of utomt over infinite objects provides succinct, expressive nd forml frmework for resoning bout rective systems, such s communiction protocols nd control systems. Rective systems re chrcterized by their nonterminting behviour nd persistent interction with their environment. In this course we study the min ingredients of this elegnt theory, nd its ppliction to utomtic verifiction (model checking) nd progrm synthesis. 1.1 Model Checking In model checking, we check utomticlly nd exhustively whether given system model meets its specifiction. Typicl exmples for specifictions re the ccessibility nd mutul exclusion properties of certin criticl regions. Automt over infinite objects re used both to represent the system nd to represent the specifiction. The verifiction problem cn be solved by constructing the intersection of the system utomton with n utomton for the negtion of the specifiction, nd then checking whether the lnguge of the resulting utomton is empty. Exmple 1.1. Mutul execution with progrm turn locl t: boolen where initilly t = 0 loop forever do loop forever do P 0 :: 00 : wit t = 0; 01 : criticl; P 1 :: 00 : wit t = 1; 01 : criticl; 10 : t := 1; 10 : t := 0; Progrm turn is (very) simple solution to the mutul exclusion problem: it ensures tht, t ny given point of time, t most one process is in the criticl region. The following is representtion of the behvior of turn s n utomton: s 0 s 1 s 2 s s s s 4 The lphbet of the utomton consists of bitvectors of length 5, where the first two bits represent the loction of process P 0, the next two bits the loction of process P 1, nd the finl bit the vlue of t. The utomton is sfety utomton (more bout this lter), it ccepts ll infinite repetitions of the sequence In order to verify tht turn stisfies the mutul exclusion property (P 0 nd P 1 re never in loction 01 t the sme time), we build n utomton tht represents the negtion (eventully P 0 nd P 1 re simultneously in loction 01): 01010, t 0 t 1 1

6 This utomton is Büchi utomton with ccepting stte t 1. A word is ccepted if t 1 is visited infinitely often (more bout this lter). It is esy to see tht for every word ccepted by the system utomton, the property utomton stys in t 0 forever nd therefore does not ccept the word. This mens tht there does not exist n infinite sequence of bitvectors tht is both generted by the progrm nd ccepted by the utomton representing the negtion of the specifiction. In other words, turn is correct. 1.2 Synthesis In synthesis, we check utomticlly if there exists progrm tht stisfies given specifiction. If the nswer is yes, we lso construct such progrm. The synthesis problem cn be solved by determining the winner of two-plyer gme between system plyer nd n environment plyer. The system plyer wins ply if the specifiction is stisfied. A winning strtegy for the system plyer cn be trnslted into progrm tht is gurnteed to stisfy the specifiction. Exmple 1.2. Mutul execution by rbitrtion P 0 :: locl t 0, t 1 : boolen where initilly t 0 = t 1 = 0 loop [ forever do ] loop [ forever do 0 : wit t0 = 1 P 1 :: 0 : wit t1 = 1 1 : criticl; 1 : criticl; ] Arbiter::? In this exmple, the implementtion of the rbiter process is left open. We wish to find n implementtion tht gurntees certin desirble properties such s mutul exclusion. We build gme ren where, in ech round, first the system plyer fixes the output t 1 nd t 2 of the rbiter process nd then the environment plyer mkes move in processes P 0 or P 1, resolving ny nondeterminism, such s whether P 0 or P 1 enters the criticl section if both re in loction 0 nd both t 0 nd t 1 hve been set to 1 by the rbiter We depict sttes in which the system plyer gets to chose the next move with circles, nd sttes in which the environment plyer gets to chose the next move with rectngles. The 2

7 sttes of the system plyer consist here of ll possible combintions of loctions of processes P 0 nd P 1, i.e., bitvectors of length 2, where the first bit indictes the loction of P 0, nd the second bit indictes the loction of P 1. The sttes of the environment plyer re bitvectors of length 4, where the dditionl third nd forth bit represent the vlues of t 0 nd t 1, respectively, tht were chosen by the system plyer in the previous move. Suppose now tht our specifiction consists gin of the mutul exclusion property, i.e., we wish to ensure tht stte 11 is never visited. Our gme is now sfety gme, which consists of gme ren nd set of sfe sttes (more bout this lter). Here, the sfe sttes re ll sttes except 11. The system plyer wins ll plys in which the gme stys in the sfe sttes forever. Note tht the system plyer hs very simple strtegy to win ny gme tht strts in stte 00: simply choose the move to stte 0000, i.e., block both processes. This rbiter is indeed correct, t lest with respect to the mutul exclusion property! More comprehensive specifictions led to more interesting winning strtegies nd more useful synthesized progrms. We might, for exmple, require tht every process visits its criticl section infinitely often, i.e., the ply must visit sttes 10 or 11 infinitely often, nd lso sttes 01 or 11 infinitely often. A winning strtegy for the system plyer might then, for exmple, lternte between 0010 nd 0001 in subsequent visits to 00, nd choose 1000 from 10, nd 0100 from 01, in order to void visits to The Logic-Automt Connection In pplictions like verifiction nd synthesis, the utomt- nd gme-theoretic mchinery is usully hidden behind logicl formultion of the problem. For exmple, we might express the mutul exclusion property s formul of liner-time temporl logic (LTL) such s (t 1 (P 0 ) t 1 (P 1 )). The verifiction or synthesis lgorithm then relies on trnsltion of the formul into n equivlent utomton, nd, for synthesis, on further trnsltion into gme. The following picture shows some useful logics together with their corresponding utomt: LTL S1S word utomt CTL CTL* S2S tree utomt Liner-time temporl logic (LTL) describes sets of infinite words. Exmple: t 1 (P 0 ), mening tht P 0 is infinitely often t loction 1. Computtion-tree logic (CTL / CTL*) describes sets of infinite trees. Exmple: EF t 1 (P 0 ) EF t 1 (P 1 ), mening tht there exists computtion pth in which P 0 reches loction 1, nd there is (possibly different) computtion pth in which P 1 reches loction 1. Mondic second-order logic with one successor (S1S) is logic over infinite words. Its expressiveness exceeds tht of LTL. Exmple: x. x P S(x) P, mening tht (given) set of nturl numbers P is either empty or consists of ll positions strting from some position of the word. Mondic second-order logic with two successors (S2S) is logic over binry trees. Its expressiveness exceeds tht of CTL* (on binry trees). Exmple: x. x P S 1 (x) P S 2 (x) P, mening tht (given) set of nodes P contins from ech node n P n entire pth strting in n. 3

8 Besides verifiction nd synthesis, n immedite nd very importnt ppliction of the connection between logic nd utomt is to decide the stisfibility problem of the vrious logics. Exmple 1.3. Suppose we wish to know if there exist two nturl numbers x nd y such tht x = y + 1 nd y = x + 1. This is expressed s the S1S formul x = S(y) y = S(x). We trnslte ech conjunct into n utomton: {y} {x} s 0 s 1 s 2 {x} {y} s 0 s 1 s 2 Here the lphbet consists of subsets of {x, y}. A vrible ppers t exctly one position in word, nmely t the position tht corresponds to its vlue. It is esy to see tht there does not exist word tht is ccepted by both utomt. Hence, our formul is unstisfible. 4

9 2 Büchi Automt We now introduce our first type of utomt over infinite words. Büchi utomt re nmed fter Julius Richrd Büchi, Swiss logicin nd mthemticin ( 1924, 1984). Büchi utomt were invented during the 1960s for the purpose of developing decision procedure for S1S. Even though Büchi utomt re firly simple vrition of stndrd finite utomt, they lredy suffice to chrcterize the entire clss of ω-regulr lnguges. 2.1 Preliminries We begin with some bsic nottions. We use the symbol N to denote the set {0, 1, 2, 3,...} of nturl numbers nd N + to denote the set {1, 2, 3,...} of positive nturl numbers. For n, m N with n m we use [n, m] to denote the set {n, n + 1,..., m} nd [n] for the specil cse of [0, n 1]. An lphbet is nonempty, finite set of symbols, usully denoted by Σ. The elements of n lphbet re clled letters. Let n lphbet Σ be given, then the conctention w = w(0)w(1)...w(n 1) of finitely mny letters of Σ is clled finite word over Σ, where n defines the length of w lso denoted by w. The only word of length 0 is the empty word denoted by ɛ. The set of ll finite words over Σ is denoted by Σ. For Σ \ {ɛ} we use the shortcut Σ +. Given some word w, we hve tht for ech n [ w ] the n-th letter of w is denoted by w(n). The conctention of infinitely mny letters defines n infinite word which hs infinite length. The set of ll infinite words is denoted by Σ ω. If Σ is n lphbet, then ech subset of Σ is lnguge over finite words. Ech subset of Σ ω is lnguge over infinite words, lso referred to s n ω-lnguge. 2.2 Automt over Infinite Words The opertionl behvior of n utomton over infinite words is very similr to tht of stndrd utomton over finite words; strting with n initil stte, the utomton constructs run by reding one letter of the input lphbet t time nd trnsitioning to successor stte permitted by its trnsitions. Most components of n utomton over infinite words re thus fmilir from the stndrd utomt over finite words: n lphbet Σ, finite set Q of sttes, set of initil sttes I, set of trnsitions T. Automt over infinite words differ from utomt over finite words, nd different types of utomt over infinite words differ mong ech other in the cceptnce condition, which is responsible for determining whether n infinite run of the utomton is ccepting or not. We leve this component generic in our generl definition of utomt over infinite words, nd lter give first concrete definition for the cse of Büchi utomt. Definition 2.1. An utomton over infinite words is tuple A = (Σ, Q, I, T, Acc), where Σ is finite lphbet, Q is finite set of sttes, I Q is subset of initil sttes, T Q Σ Q is set of trnsitions, nd Acc Q ω is the ccepting condition. In the following, we refer to n utomton over infinite words (whenever cler from context) simply s n utomton. Now we define how n utomton uses n infinite word s input. 5

10 Note tht we do not refer to cceptnce in this definition. Definition 2.2. A run of n utomton A on n infinite input word α is n infinite sequence of sttes r = r(0)r(1)r(2)... such tht the following hold: r(0) I for ll i N, (r(i), α(i), r(i + 1)) T Exmple 2.1. The following is grphicl representtion of n utomton over the lphbet Σ = {, b} nd with set of sttes Q = {A, B, C, D}, initil set of sttes I = {A}, nd set of trnsitions T = {(A,, B), (B,, C), (C, b, D), (D, b, A)}: A B b D b C On the infinite input word bbbb..., the (only) run of the utomton is ABCDABCDABCD.... An utomton is deterministic if I 1 nd {(q, σ, q ) T q Q} 1 for every q Q nd σ Σ, otherwise the utomton is nondeterministic. The utomton is complete if {(q, σ, q ) T q Q} 1 for every q Q nd σ Σ. For deterministic nd complete utomt, the trnsitions re usully given s function δ : Q Σ Q. The utomton in the exmple is deterministic but not complete. Definition 2.3. An utomton A ccepts n infinite word α if: there is run r of A on α, nd r is ccepting: r Acc. The lnguge recognized by A is defined s follows: L(A) = {α Σ ω A ccepts α} We sy tht two utomt re equivlent if they hve the sme lnguge. 2.3 The Büchi Acceptnce Condition The Büchi cceptnce condition is given s set of ccepting sttes F. A run of Büchi utomton is ccepting if some stte from this set occurs infinitely often. Formlly, for n infinite word α over Σ we use Inf(α) = {σ Σ m N. n N. n m nd α(n) = σ} to denote the set of ll letters of Σ tht occur infinitely often in α, clled the infinity set of α. Definition 2.4. The Büchi condition büchi(f ) on set of sttes F Q is the set büchi(f ) = {α Q ω Inf(α) F }. An utomton A = (Σ, Q, I, T, Acc) with Acc = büchi(f ) is clled Büchi utomton. The set F is clled the set of ccepting sttes. 6

11 Exmple 2.2. Let A be the utomton from Exmple 2.1 with Büchi cceptnce condition büchi({d}). The lnguge of the utomton consists of single word: L(A) = {bbbbbb...} As first observtion bout Büchi utomt, we estblish tht for every Büchi utomton A one cn construct n equivlent complete Büchi utomton A : Construction 2.1. For Büchi utomton A = (Σ, Q, I, T, büchi(f )), we define the complete Büchi utomton A = (Σ, Q, I, T, büchi(f )) using: Q = Q {q f } where q f is fresh stte I = I T = T {(q, σ, q f ) q. (q, σ, q ) T } {(q f, σ, q f ) σ Σ} F = F We prove the correctness of the construction by the following theorem. Theorem 2.1. For every Büchi utomton A, there is complete Büchi utomton A such tht L(A) = L(A ). Proof. The runs of A re superset of those of A, becuse we hve not removed ny trnsitions during the construction of A. Furthermore, on ny infinite input word the ccepting runs of A nd A re the sme, becuse ny run tht reches the new stte q f stys in q f forever, nd, since q f F, such run is not ccepting. Exmple 2.3. Completing the utomton from the previous exmple we obtin the following Büchi utomton (the ccepting stte D is shown with double lines): A B b b b D b C f, b A complete deterministic utomton my be viewed s totl function from Σ ω to Q ω. A complete (possibly nondeterministic) utomton produces t lest one run for every input word. 7

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13 3 Büchi s Chrcteriztion Theorem Our next gol is to chrcterize the lnguges tht cn be recognized by Büchi utomt. For lnguges over finite words, Kleene s theorem sttes tht the set of lnguges over finite words tht cn be defined by regulr expressions is exctly the set of lnguges tht cn be recognized by utomt over finite words. We quickly review this result, then define ω-regulr expressions, nd finlly prove the corresponding theorem for ω-regulr lnguges: Büchi s chrcteriztion theorem. 3.1 Kleene s Theorem We give quick recp of regulr lnguges nd Kleene s theorem. This is not ment s full introduction to finite utomt theory; if you wnt to brush up some more on this, we recommend the textbook Introduction to Automt Theory, Lnguges, nd Computtion by Hopcroft, Motwni, nd Ullmn. Regulr expressions consist of constnts nd opertor symbols tht denote lnguges of finite words nd opertions over these lnguges, respectively. Definition 3.1. Regulr expressions re defined s follows: The constnts ε nd re regulr expressions. L(ε) = {ε}, L( ) =. If Σ is symbol, then is regulr expression. L() = {}. If E nd F re regulr expressions, then E + F is regulr expression: L(E + F ) = L(E) L(F ). If E nd F re regulr expressions, then E F is regulr expression: L(E F ) = {uv u L(E), v L(F )}. If E is regulr expression, then E is regulr expression. L(E ) = {u 1 u 2... u n n N, u i L(E) for ll 0 i n}. A lnguge over finite words is regulr if it is definble by regulr expression. Alterntively, regulr lnguges cn be defined s the lnguges recognized by utomt over finite words. Definition 3.2. An utomton on finite words A is tuple (Σ, Q, I, T, F ), where Σ is n input lphbet, Q is nonempty finite set of sttes, I Q is set of initil sttes, Q Σ Q is set of trnsitions, nd F Q re set of finl sttes. An utomton A ccepts finite word w Σ if there is finite sequence of sttes q(0)q(1)... q( w ) such tht q(0) I nd (q(i), w(i), q(i + 1)) for ll i < w nd with q( w ) F. The set of ll words ccepted by A is clled the lnguge of A, denoted by L(A). The equivlence of regulr expressions nd finite utomt is known s Kleene s theorem. Theorem 3.1 (Kleene s Theorem). A lnguge is regulr iff it is recognized by some finite word utomton. 9

14 3.2 ω-regulr Lnguges ω-regulr expressions denote lnguges over infinite words. In ddition to lnguge union on lnguges over infinite words, we hve two opertions tht convert lnguges over finite words into lnguges over infinite words: the infinite conctention R ω of words of regulr lnguge R nd the conctention R U of regulr lnguge R nd n ω-regulr lnguge U. Definition 3.3. The ω-regulr expressions re defined s follows. If E is regulr expression where ε L(E), then E ω is n ω-regulr expression. L(E ω ) = L(E) ω where L ω = {w 0 w 1... w i L, w i > 0 for ll i N} for L Σ. If E is regulr expression nd W is n ω-regulr expression, then E W is n ω-regulr expression. L(E W ) = L(E) L(W ) where L 1 L 2 = {wα w L 1, α L 2 } for L 1 Σ, L 2 Σ ω. If W 1 nd W 2 re ω-regulr expressions, then W 1 + W 2 is n ω-regulr expression. L(W 1 + W 2 ) = L(W 1 ) L(W 2 ). A lnguge over infinite words is ω-regulr if it is definble by n ω-regulr expression. 3.3 Closure Properties of the Büchi-Recognizble Lnguges Büchi s chrcteriztion theorem sttes tht the ω-regulr lnguges re exctly the lnguges recognized by Büchi utomt. In the following we will refer to the lnguges recognized by Büchi utomt simply s Büchi-recognizble lnguges. To prepre for the proof of Büchi s theorem, we estblish severl closure properties of the Büchi-recognizble lnguges. First off, the Büchi-recognizble lnguges re closed under lnguge union, s demonstrted by the following simple construction. Construction 3.1. Let L 1 nd L 2 be ω-lnguges recognized by the Büchi utomt A 1 = (Σ, Q 1, I 1, T 1, büchi(f 1 )) nd A 2 = (Σ, Q 2, I 2, T 2, büchi(f 2 )), respectively. We construct A = (Σ, Q, I, T, büchi(f )) with L(A ) = L 1 L 2 s follows: Q = Q 1 Q 2 (w.l.o.g. we ssume Q 1 Q 2 = ) I = I 1 I 2 T = T 1 T 2 F = F 1 F 2 The correctness of this construction is proven by the following theorem. Theorem 3.2. If L 1 nd L 2 re Büchi-recognizble, then so is L 1 L 2. Proof. We prove tht the Büchi utomton A built by Construction 3.1 from the Büchi utomt A 1 nd A 2 indeed recognizes the union of the lnguges of the two utomt. L(A ) L(A 1 ) L(A 2 ): For α L(A ), we hve n ccepting run r = r(0)r(1)r(2)... of A on α. If r(0) I 1, then r is n ccepting run of A 1, otherwise r(0) I 2 nd r is n ccepting run of A 2. L(A ) L(A 1 ) L(A 2 ): For i {1, 2} nd α L(A i ), there is n ccepting run r of A i. The run r is lso n ccepting run of A. 10

15 As we will see lter, the Büchi-recognizble lnguges re closed under complement. Together with the closure under union, the closure under complement implies the closure under intersection (becuse the intersection of two sets is the complement of the union of the complements of the two sets). Since the proof of the closure under complement is more complicted, we postpone this for now, however, nd insted give direct construction for the closure under intersection. Construction 3.2. Let L 1 nd L 2 be ω-lnguges recognized by the Büchi utomt A 1 = (Σ, Q 1, I 1, T 1, büchi(f 1 )) nd A 2 = (Σ, Q 2, I 2, T 2, büchi(f 2 )), respectively. We construct A = (Σ, Q, I, T, büchi(f )) with L(A ) = L 1 L 2 s follows: Q = Q 1 Q 2 {1, 2} I = I 1 I 2 {1} T = {((q 1, q 2, i), σ, (q 1, q 2, j)) (q 1, σ, q 1) T 1, (q 2, σ, q 2) T 2, i, j {1, 2}, q i / F i i = j q i F i i j} F = {(q 1, q 2, 2) q 1 Q 1, q 2 F 2 } The utomton for the intersection is essentilly the product of the two utomt. The sttes of the new utomton contin the sttes of the two utomt plus third component, which is used to combine the cceptnce conditions of the two utomt. The product utomton lterntes between witing for visit to the ccepting sttes of the first utomton (in which cse the third component is 1) nd witing for visit to the ccepting sttes of the second utomton (in which cse the third component is 2). The ccepting sttes of the product re those sttes where the third component is 2 nd where the second utomton visits n ccepting stte, i.e., where both utomt hve seen n ccepting stte since the lst visit to n ccepting stte of the product utomton. Theorem 3.3. If L 1 nd L 2 re Büchi-recognizble, then so is L 1 L 2. Proof. We prove tht the Büchi utomton A built by Construction 3.2 from the Büchi utomt A 1 nd A 2 indeed recognizes the intersection of the lnguges of the two utomt. r = (q 0 1, q 0 2, t 0 )(q 1 1, q 1 2, t 1 )... is run of A on n input word α iff r 1 = q 0 1q is run of A 1 on α nd r 2 = q 0 2q is run of A 2 on α. The run r is ccepting iff r 1 is ccepting nd r 2 is ccepting. Next, the conctention of regulr lnguge nd Büchi-recognizble lnguge is gin Büchi-recognizble. Construction 3.3. Let A 1 = (Σ, Q 1, I 1, T 1, F 1 ) be n utomton over finite words tht recognizes the lnguge L 1, nd let A 2 = (Σ, Q 2, I 2, T 2, büchi(f 2 )) be Büchi utomton over the sme lphbet tht recognizes L 2. We construct Büchi utomton A = (Σ, Q, I, T, büchi(f 2 )) with L(A ) = L 1 L 2 s follows: Q = Q 1 Q 2 (w.l.o.g. we ssume Q 1 Q 2 = ) { I I1 if I = 1 F 1 = I 1 I 2 otherwise T = T 1 T 2 {(q, σ, q ) (q, σ, f) T 1, f F 1, q I 2 } 11

16 The correctness of this construction is proven by the following theorem. Theorem 3.4. If L 1 is regulr lnguge nd L 2 is Büchi-recognizble, then L 1 L 2 is Büchi-recognizble. Proof. We prove tht the Büchi utomton A built in Construction 3.3 from the utomton on finite words A 1 nd the Büchi utomton A 2 indeed recognizes the conctention of the lnguges of the two utomt. L(A ) L(A 1 ) L(A 2 ): For α L(A ), let r = r(0)r(1)r(2)... be n ccepting run of A on α. If r(0) I 1, then there is n n N such tht (r(n), α(n), r(n + 1)) Q 1 Σ I 2 nd therefore, there is finl stte f F 1 such tht r(0)r(1)... r(n)f is n ccepting run of A 1 on α(0)α(1) α(n) nd r(n + 1)r(n + 2)... is n ccepting run of A 2 on α(n + 1)α(n + 2).... If r(0) I 2 then I 1 F 1 nd therefore, ɛ L(A 1 ), α L(A 2 ). L(A ) L(A 1 ) L(A 2 ): For w L(A 1 ), let r = r(0)r(1) r(n) be n ccepting run of A 1 on w. For α L(A 2 ), let s = s(0)s(1)... be n ccepting run of A 2 on α. Then, r(n) F 1 nd, by construction, r(0)r(1)... r(n 1)s(0)s(1)... is n ccepting run of A on wα. Finlly, we show tht the infinite conctention of words of regulr lnguge forms Büchi-recognizble lnguge. We construct Büchi utomton for this lnguge from n utomton over finite words in two steps. The first construction modifies given utomton over finite words into n equivlent utomton with single initil stte tht hs no incoming trnsitions. Construction 3.4. Let A = (Σ, Q, I, T, F ) be n utomton over finite words. We ssume tht ε / L(A). We construct n utomton A = (Σ, Q, I, T, F ) over finite words such tht L(A) = L(A ) nd A hs single initil stte tht hs no incoming trnsitions. Q = Q {q f } where q f is fresh stte I = {q f } T = T {(q f, σ, q ) (q, σ, q ) T for some q I} The second construction builds the Büchi utomton tht recognizes the infinite conctentions of words from the regulr lnguge by dding loop to the modified utomton. Construction 3.5. Let A be n utomton over finite words, for which we ssume tht ε / L(A). We construct Büchi utomton A = (Σ, Q, I, T, büchi(f )) such tht L(A ) = L(A) ω. Let A = (Σ, Q, I, T, F ) be the utomton of Construction 3.4. Then A is defined s follows: Q = Q I = I T = T {(q, σ, q f ) (q, σ, q ) T nd q F } F = I The correctness of this construction is proven by the following theorem. 12

17 Theorem 3.5. If L is regulr lnguge such tht ɛ L, then L ω is Büchi-recognizble. Proof. Construction 3.4 does not ffect the lnguge of A. We show tht the Büchi utomton A built in Construction 3.5 from the resulting utomton A on finite words indeed recognizes L(A ) ω. L(A ) L(A ) ω : Assume tht α L(A ) nd tht r(0)r(1)r(2)... is n ccepting run of A on α. Hence, we hve tht r(i) = q f F = I for infinitely mny indices i: i 0, i 1, i 2,.... This provides sequence of runs of A : run r(0)r(1)... r(i 0 1)q on w 0 = α(0)α(1)... α(i 0 1) for some q F run r(i 0 )r(i 0 + 1)... r(i 1 1)q on w 1 = α(i 0 )α(i 0 + 1)... α(i 1 1) for some q F nd so forth. We thus hve tht w k L(A ) for every k 0. Hence, α L(A ) ω. L(A ) L(A ) ω : Assume tht α = w 0 w 1 w 2... Σ ω such tht w k L(A ) for ll k 0. For ech k, we choose n ccepting run r k (0)r k (1)r k (2)... r k (n k ) of A on w k. Hence, r k (0) I nd r k (n k ) F for ll k > 1. Thus, r 0 (0)... r 0 (n 0 1)r 1 (0)... r 1 (n 1 1)r 2 (0)... r 2 (n 2 1)... is n ccepting run of A on α. Hence, α L(A ). 3.4 Büchi s Chrcteriztion Theorem We re now redy to prove Büchi s Chrcteriztion Theorem, result from Theorem 3.6 (Büchi s Chrcteriztion Theorem). An ω-lnguge is Büchi-recognizble iff it is ω-regulr. Proof. follows from the closure properties of the Büchi-recognizble lnguges estblished by Theorems 3.2, 3.4, nd 3.5. : Given Büchi utomton A, we consider for ech pir q, q Q the regulr lnguge W q,q = {w Σ finite-word utomton (Σ, Q, {q}, T, {q }) ccepts w }. We clim tht L(A) = q I,q F W q,q (W q,q \ {ɛ})ω. The clim is proven s follows. L(A) q I,q F W q,q (W q,q \ {ɛ})ω : Let α L(A). Then there is n ccepting run r of A on α, which begins t some q = r(0) I nd visits some q F infinitely often: r : q w0 q w 1 q w 2 q w 3..., where w i Σ for ll i 0, w i > 0 for ll i > 0 nd α = w 0 w 1 w The nottion q 0 qk+1 for some finite word w = w(0)w(1)... w(k) mens tht there exist sttes w q 1,... q k Q s.t. (q i, w(i), q i+1 ) T for ll 0 i k. Since w 0 W q,q nd w k W q,q for k > 0, we hve tht α W q,q W q,q ω for some q I, q F. L(A) q I,q F W q,q (W q,q \ {ɛ})ω : Let α = w 0 w 1 w 2... with w 0 W q,q nd w k W q,q for some q I, q F nd for ll k > 0. Then the run r : q w0 q w 1 q w 2 q w 3... exists nd is ccepting becuse q F. It follows tht α L(A). 13

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19 4 Deterministic Büchi Automt One of the most importnt constructions in the theory of utomt over finite words is the Rbin-Scott powerset construction, which converts nondeterministic utomton over finite words into deterministic utomton tht recognizes the sme lnguge. The powerset construction estblishes tht while nondeterminism mkes utomt over finite words more concise (the powerset construction produces n exponentil number of sttes), nondeterminism does not mke utomt over finite words more expressive. The sitution is different for Büchi utomt: even though the lnguge L = ( + b) b ω is clerly Büchi-recognizble, there is, s the following theorem shows, no deterministic Büchi utomton tht recognizes L. Theorem 4.1. The lnguge L = ( + b) b ω Büchi utomton. is not recognizble by deterministic Proof. Assume, by wy of contrdiction, tht L is recognizble by the deterministic Büchi utomton A = (Σ, Q, I, T, büchi(f )). Since α 0 = b ω is in L, there is unique run r 0 = r 0 (0)r 0 (1)r 0 (2)... of A on α 0 with r 0 (n 0 ) F for some n 0 N. Similrly, α 1 = b n0 b ω in L nd there is unique run r 1 = r 0 (0)r 0 (1)r 0 (2)... r 0 (n 0 )r 1 (n 0 + 1)r 1 (n 0 + 2)... of A on α 1 with r 1 (n 1 ) F for some n 1 > n 0. Becuse A is deterministic, r 0 nd r 1 re identicl up to position n 0. By repeting this rgument infinitely often, we obtin word α = b n0 b n1 b n2... nd run r with infinitely mny visits to F. Hence, α is ccepted by A. However, α is not n element of L. This contrdicts L = L(A). Wht lnguges re recognizble by deterministic Büchi utomt? A strightforwrd chrcteriztion cn be given in terms of the limit opertor. Definition 4.1 (Limit). The limit W of lnguge W Σ over finite words is the following lnguge over infinite words: W = {α Σ ω there exist infinitely mny n N s.t. α[0, n] W }. Theorem 4.2. An ω-lnguge L Σ ω is recognizble by deterministic Büchi utomton iff there is regulr lnguge W Σ s.t. L = W. Proof. We clim tht the lnguges of deterministic Büchi utomton A B nd of deterministic utomton over finite words A F, where the utomt A B = (Σ, Q, I, T, büchi(f )) nd A F = (Σ, Q, I, T, F ), constructed from the sme components, re relted s follows: L(A B ) = L(A F ). Since every regulr lnguge is recognized by deterministic utomton over finite words, the theorem follows. To prove the clim, we consider n infinite word α Σ ω. 15

20 iff iff iff α L(A B ) for the unique run r of A B on α, Inf(r) F α[0, n] L(A F ) for infinitely mny n N α L(A F ) Next on our gend is the complementtion of lnguges. For regulr lnguges, this is very simple: one trnsltes given complete deterministic utomton A tht recognizes some lnguge L Σ into n utomton A tht recognizes the complement Σ \ L by complementing the set of finl sttes, i.e., F = F \ F. For deterministic Büchi utomt, the construction is slightly more complicted, becuse it introduces nondeterminism. Construction 4.1. Let A = (Σ, Q, I, T, büchi(f )) be complete deterministic Büchi utomton. We construct Büchi utomton A = (Σ, Q, I, T, büchi(f )) with L(A ) = Σ ω \ L(A) s follows: Q = (Q {0}) ((Q \ F ) {1}) I = I {0} T = {((q, 0), σ, (q, i)) (q, σ, q ) T, i {0, 1}, (q, i) Q } {((q, 1), σ, (q, 1)) (q, σ, q ) T, q Q \ F } F = (Q \ F ) {1} The construction uses two copies of the given utomton A, where ll ccepting sttes re eliminted from the second copy. The switch from the first copy (identified by the flg 0) to the second copy (identified by the flg 1) hppens nondeterministiclly. The utomton ccepts if the run ends up in the second copy, which mens tht, on the unique run of A on the input word, the ccepting sttes of A re only visited finitely often. Note tht the resulting utomton is nondeterministic. This is, in generl, unvoidble, becuse there re lnguges, such s L = (b ) ω where the lnguge itself is recognizble by deterministic Buchi utomton, while its complement Σ ω \L cn only be recognized by nondeterministic Büchi utomton. Theorem 4.3. For every deterministic Büchi utomton A, there exists Büchi utomton A such tht L(A ) = Σ ω \ L(A). Proof. Let A be constructed from the given deterministic Büchi utomton A (which we ssume, w.l.o.g., to be complete) by Construction 4.1. We prove tht L(A ) = Σ ω \ L(A). L(A ) Σ ω \ L(A): For every word α L(A ) we hve n ccepting run r : (q 0, 0)(q 1, 0)... (q j, 0)(q 0, 1)(q 1, 1)... on A. Hence, r : q 0 q 1 q 2... q j q 0q 1... is the unique run of A on α. Since q 0, q 1,... Q \ F, we hve tht Inf(r) Q \ F. Hence, r is not ccepting nd α Σ ω \ L(A). L(A ) Σ ω \ L(A): Let α / L(A) be some word tht is not in the lnguge of A. Since A is complete nd deterministic, there exists unique run r : q 0 q 1 q 2... of A on α nd 16

21 Inf(r) F =. Thus, there exists k N such tht q j / F for ll j > k. This gives us the run r : (q 0, 0)(q 1, 0)... (q k, 0)(q k+1, 1)(q k+2, 1)... of A on α with Inf(r ) ((Q \ F ) {1}) = F. α L(A ). Hence, r is ccepting nd therefore Exmple 4.1. We use Construction 4.1 to complement the deterministic Büchi utomton A with lnguge (b ) ω. The result is the nondeterministic Büchi utomton A with lnguge ( + b) b ω. b b p b b q p, 0 q, 0 b b q, 1 b Note tht, since not ll ω-regulr lnguges cn be recognized by deterministic Büchi utomt, Construction 4.1 does not provide us with complementtion construction for ll ω-regulr lnguges. Such generl construction will be the subject of the following section. 17

22 18

23 5 Complementtion of Büchi Automt Complementtion is n importnt opertion in verifiction. In prticulr we cn, s discussed in the introduction, reduce the question whether the lnguge of n utomton A 1, representing the system, is contined in the lnguge of n utomton A 2, representing the specifiction, to the lnguge emptiness problem of n utomton tht recognizes the intersection of the lnguge of A 1 with the complement of the lnguge of A 2. While the closure of the Büchi-recognizble lnguges under complementtion hs been known since the 1960s, improvements to the complementtion construction were mde until round The construction discussed in the following is due to Kupfermn nd Vrdi. In order to determine whether given word is in the complement of the lnguge of nondeterministic utomton, we must check whether ll runs of the utomton re rejecting. We begin by representing the set of ll runs s n infinite directed cyclic grph (DAG): Definition 5.1. Let A = (Σ, Q, I, T, büchi(f )) be nondeterministic Büchi utomton. The run DAG of A on word α Σ ω is the directed cyclic grph G = (V, E) where V = (Q i {i}) where Q 0 = I nd Q i+1 = {q } i 0 E = {((q, i), (q, i + 1)) i 0, (q, α(i), q ) T } q Q i, (q,α(i),q ) T A pth in run DAG is ccepting iff it visits F N infinitely often. The utomton ccepts word α Σ ω iff some pth in the run DAG on α is ccepting. Exmple 5.1. Consider the following nondeterministic Büchi utomton A. The lnguge of A consists of ll infinite words over {, b} with infinitely mny bs, i.e., ( b) ω :, b, b, b p q r b The run DAG of A on α = bb... is shown below. Note tht strting with level 1, every level of the grph includes the ccepting stte q, indicting tht for every number n 1, there exists run tht visits q in the nth position. However, since α is not in the lnguge of A, there is no run tht visits q infinitely often. b b p, 0 p, 1 p, 2 p, 3 p, 4 p, 5 p, 6 p, 7... q, 1 q, 2 q, 3 q, 4 q, 5 q, 6 q, 7... r, 3 r, 4 r, 5 r, 6 r,

24 To prove tht none of the pths in given run DAG is ccepting, we use n nnottion of the vertices of the run DAG with nturl numbers, clled rnking. The numbers ssigned by the rnking come from finite set, whose size is derived from the number of sttes of the utomton. We require tht vertices tht re lbeled with odd numbers do not contin ccepting sttes nd tht the rnk never increses long pths in the run DAG. If furthermore ll pths eventully sty in some odd rnk, then we cll the rnking odd. Definition 5.2. A rnking for run DAG G is function f : V {0,..., 2 Q } such tht for ll (q, i) V where f(q, i) is odd, we hve tht q / F, nd for ll ((q, i), (q, i )) E, we hve tht f(q, i ) f(q, i). A rnking is odd iff for ll pths (q 0, i 0 )(q 1, i 1 )(q 2, i 2 )... in G, there is position n N such tht f(q n, i n ) is odd nd, for ll j 0, f(q n+j, i n+j ) = f(q n, i n ). Exmple 5.2. Continuing our exmple, we nnotte the run DAG with n odd rnking. The rnks re displyed with the following colors: rnk 1 - rnk 2 - rnk 3 - rnk 4. b b p, 0 p, 1 p, 2 p, 3 p, 4 p, 5 p, 6 p, 7... q, 1 q, 2 q, 3 q, 4 q, 5 q, 6 q, 7... r, 3 r, 4 r, 5 r, 6 r, 7... Clerly, the existence of n odd rnking implies tht there is no ccepting run. Lemm 5.1. If there exists n odd rnking for the run DAG G of nondeterministic Büchi utomton A on word α Σ ω, then A does not ccept α. Proof. Consider tht by the definition of n odd rnking, every pth eventully gets trpped in some odd rnk, nd if f(q, i) is odd, then q / F. We conclude tht every pth of G visits F only finitely often. The more complicted rgument is to show tht if there is no ccepting run, then there exists n odd rnking. We begin by introducing some helpful terminology. Let G be subgrph of G. We cll vertex (q, i) sfe in G, if for ll vertices (q, i ) rechble from (q, i), we hve tht q / F, nd endngered in G, if only finitely mny vertices re rechble from (q, i). We now define sequence G 0 G 1 G 2... of subgrphs of G inductively s follows: G 0 = G 20

25 G 2j+1 = G 2j \ {(q, i) (q, i) is endngered in G 2i } G 2j+2 = G 2j+1 \ {(q, i) (q, i) is sfe in G 2i+1 } Exmple 5.3. Consider gin the run DAG from our previous exmples with the coloring from Exmple 5.2. G 0 is the entire grph. There re no endngered vertices in G 0, hence G 1 = G 0. The red vertices re sfe in G 1 ; hence, G 2 consists of ll vertices except the red vertices. In G 2, the green vertices re endngered; hence, G 3 consists of ll vertices except the red nd green vertices. In G 3, the blue vertices re sfe; leving only the non-colored vertices for G 4. G 4 is finite, hence ll remining vertices re endngered. G 5 nd ll further subgrphs thus re empty. In the exmple, G 5 nd ll further subgrphs re empty. In generl, if A hs n sttes, then G 2n+1 is empty. We estblish this fct with the inductive rgument of the following lemm: Lemm 5.2. If A does not ccept α, then the following holds: For every n 2 Q there exists n l n N such tht for ll i l n we hve t most Q n vertices of the form (_, i) in G 2n. Proof. We prove the lemm by induction on n N. The cse of n = 0 is strightforwrd, s every subgrph of G hs t most Q vertices of the from (_, i). So, let n be greter thn 0. First ssume tht G 2(n 1) is finite. Then every vertex in G 2(n 1) is endngered nd correspondingly, both G 2n 1 nd G n re empty. Next, ssume tht G 2(n 1) is infinite. Then there must exist sfe vertex (q, m) in G 2n 1, s otherwise, we cn construct pth within G contining infinitely mny visits to F. We choose l n = m nd clim tht for ll j l n, there re t most Q n mny vertices of the form (_, j) in G 2n. Our clim is proven s follows: Since (q, l n ) G 2n 1, we know tht it is not endngered in G 2(n 1). Hence, there re infinitely mny vertices rechble from (q, l n ) in G 2(n 1). It follows by König s Lemm, tht there exists n infinite pth p = (q, l n )(q 1, l n + 1)(q 0, l n + 2)... in G 2(n 1). Given tht the pth is infinite, clerly no vertex of p is endngered in G 2(n 1). Thus, p is lso pth of G 2n 1. Since (q, l n ) is sfe in G 2n 1, ll vertices on p re sfe in G 2n 1 s well. Hence, by the construction of G 2n, we cn conclude tht none of the vertices of p re in G 2n. Thus, for ll j l n, the number of vertices of the form (_, j) in G 2n is strictly smller thn in G 2(n 1), concluding the proof. With these preprtions, we re now redy to show tht the existence of n odd rnking is not only sufficient for the bsence of n ccepting run, but lso necessry. Lemm 5.3. If A does not ccept α, then there exists n odd rnking for the run DAG of A on α. Proof. We define the function f by f(q, i) = 2n if the vertex (q, i) is endngered in G 2n nd by f(q, i) = 2n + 1 if (q, i) is sfe in G 2n+1. We clim tht f is n odd rnking: By Lemm 5.2, G j is empty for j > 2 Q. Hence, f : V {0,..., 2 Q }. Furthermore: For ll successors (q, i ) of vertex (q, i) it holds tht f(q, i ) f(q, i): Let j = f(q, i). Then if j is even, the vertex (q, i) is endngered in G j. Hence, either (q, i ) is not in G j, nd therefore f(q, i ) j, or (q, i ) is in G j, nd thus endngered, 21

26 i.e., f(q, i ) = j. Anlogously, if j is odd, then the vertex (q, i) is sfe in G j such tht either (q, i ) is not in G j, nd therefore f(q, i) < j, or (q, i ) is in G j, nd thus sfe, i.e. f(q, i) = j. f is odd: For every pth (q 0, i 0 )(q 1, i 1 )(q 2,, i 2 )... in G there exists n n N such tht for ll j n we hve tht f(q j, i j ) = f(q n, i n ). Now, suppose tht k = f(q n, i n ) is even. Then, (q n, i n ) is endngered in G k. Furthermore, ll (q j, i j ) re in G k, since f(q j, i j ) = k for ll j N. However, this contrdicts tht (q n, i n ) is endngered in G k such tht k must be odd. In our complementtion construction, we build n utomton tht constructs the run DAG level by level nd nondeterministiclly guesses the odd rnking. A level rnking is function g : Q {0,..., 2 Q } { } such tht if g(q) is odd, then q / F. The specil element indictes tht the stte is not present in the level of the run DAG. Let R be the set of ll level rnkings. We sy tht level rnking g covers level rnking g if for ll q, q Q, where g(q) nd (q, σ, q ) T, it holds tht g (q ) g(q). Construction 5.1. For given Büchi utomton A = (Σ, Q, I, T, büchi(f )) we construct Büchi utomton A = (Σ, Q, I, T, büchi(f )) with L(A ) = Σ ω \ L(A) s follows. Q = R 2 Q I = {(g 0, ) g 0 R, g 0 (q) = iff q / I} T = {((g, ), σ, (g, P )) g covers g, P = {q Q g (q ) is even}} {((g, P ), σ, (g, P )) P, g covers g, F = R { } P = {q Q (q, σ, q ) T, q P, g (q ) is even}} Intuitively, the first component of ech stte identifies the level rnking, nd the second component trcks the sttes whose corresponding vertices in the run DAG hve even rnks. Pths tht trverse such vertices should eventully rech vertex with n odd rnk nd the cceptnce condition ensures tht ll pths visit vertex with odd rnk infinitely often. Theorem 5.1. For every Büchi utomton A there exists Büchi utomton A such tht L(A ) = Σ ω \ L(A). Proof. L(A ) Σ ω L(A): Let α L(A ) nd let r = (g 0, P 0 )(g 1, P 1 )... be n ccepting run of A on α. Further, let G = (V, E) be the run DAG of A on α. Then the function f : (q, i) g i (q) for q Q i nd i N is rnking for G, becuse if g i (s) is odd then s / F nd for every ((q, i), (q, i + 1)) E it holds tht g i+1 (q ) g i (q). We clim tht f is n odd rnking for G. By wy of contrdiction ssume tht it is not. Then there is pth (q 0, i 0 )(q 1, i 1 )(q 2, i 2 )... in G such tht for infinitely mny n N the rnk f(q n, i n ) is even. Hence, there exists n index m N, such tht f(q m, i m ) is even nd for ll j m we hve tht f(q j, i j ) = f(q n, i n ). Now, since r is ccepting, it follows tht P k = for infinitely mny k. Let k be the smllest such index greter or equl thn n. However, then lso P j for ll j > k, yielding the desired contrdiction. We conclude tht α / L(A), s there is n odd rnking. 22

27 Σ ω L(A) L(A ): Let α / L(A) nd let G = (V, E) be the run DAG of A on α. Then there exists n odd rnking f on G nd there is run r = (g 0, P 0 )(g 1, P 1 )... of A on α, where { f(q, i) if q Q i g i (q) = otherwise, P 0 =, nd { {q Q g i+1 (q) is even} if P i = P i+1 = {q Q q Q i P i. ((q, i), (q, i + 1)) E, g i+1 (q ) is even} otherwise. The run r is ccepting, s otherwise there would be n index n such tht P j for ll j n, yielding pth in G tht visits only finitely often n odd rnk. Hence, we conclude tht α L(A ). 23

28 24

29 6 McNughton s Theorem We lredy estblished tht while the lnguges tht cn be recognized with nondeterministic Büchi utomt re exctly the ω-regulr lnguges, the lnguges tht cn be recognized with deterministic Büchi utomt re strictly smller set. We now repir this deficiency with more expressive cceptnce condition, the Muller condition. McNughton s theorem sttes tht the set of lnguges recognizble by deterministic Muller utomt re gin exctly the ω-regulr lnguges. We will see lter tht it is very useful to hve deterministic utomton for given ω-lnguge, for exmple in synthesis, where we construct the gme between the system nd the environment from deterministic utomton tht recognizes the winning plys for the system plyer. Since the complementtion of deterministic Muller utomt is very simple opertion, McNughton s theorem lso provides n lterntive proof for the result of the previous section tht the ω-regulr lnguges re closed under complementtion. 6.1 The Muller Acceptnce Condition Definition 6.1. The Muller condition muller(f) on s set of sets of sttes F 2 Q is the set muller(f) = {α Q ω Inf(α) F} An utomton A = (Σ, Q, I, T, Acc) with Acc = muller(f) is clled Muller utomton. The set F is clled the set of ccepting subsets (or the tble) of A. Exmple 6.1. Consider the deterministic utomton over the lphbet Σ = {, b} shown below. For the tble F = {{q}} we obtin the Muller utomton A recognizing the lnguge L(A) = ( + b) b ω, for F = {{q}, {p, q}} we obtin the Muller utomton A recognizing L(A ) = ( b) ω. b p b q As first exercise, we trnslte Büchi utomt into Muller utomt: Construction 6.1. For (deterministic) Büchi utomton A = (Σ, Q, I, T, büchi(f )) we define the (deterministic) Muller utomton A = (Σ, Q, I, T, muller(f)) using F = {S Q S F } Since the construction does not modify the trnsitions, the Muller utomton is gin deterministic if the Büchi utomton is deterministic. It is strightforwrd to see tht the utomt recognize the sme lnguge. Theorem 6.1. For every (deterministic) Büchi utomton A, there is (deterministic) Muller utomton A such tht L(A) = L(A ). Proof. The utomton A of Construction 6.1 complies with our requirements, since büchi(f ) = {α Q ω Inf(α) F } = {α Q ω Inf(α) F} = muller(f) 25

30 A slightly more difficult construction is to trnslte the Muller utomton bck into Büchi utomton. Construction 6.2. Let A = (Σ, Q, I, T, muller({f 1,..., F n })) be Muller utomton nd < some rbitrry totl order on Q. We construct the Büchi utomton A = (Σ, Q, I, T, büchi(f )) with L(A ) = L(A) s follows: Q = Q n ({i} F i F i ) I = I i=1 T = T {(q, σ, (i, q, q )) 1 i n, (q, σ, q ) T, q F i } {((i, q, p), σ, (i, q, p )) 1 i n, (q, σ, q ) T, p if q p p = min(f i ) if q = p = mx(f i ) min(f i {r r p}) if q = p < mx(f i ), q, p, q F i } F = n ({i} {min(f i )} {min(f i )}) i=1 A run of the Büchi utomton first simply simultes (while in sttes Q) the Muller utomton nd then guesses the ccepting subset of the Muller utomton. The purpose of the order < on the sttes is tht we cn step through the sttes of the ccepting subset in order to mke sure tht ll sttes in the ccepting subset ctully occur infinitely often. In sttes {i} F i F i, the first component records the index i of the ccepting subset, the second component the currently visited stte of the Muller utomton, nd the third component the next stte (ccording to the order on the sttes) we need to see in order to mke progress towrds ccepting the input word. The Büchi utomton ccepts if we step through the sttes of the ccepting subset infinitely often. Recll tht we used similr trick in the construction of the Büchi utomton for the intersection of two Büchi-recognizble lnguges in Construction 3.2. Theorem 6.2. For every Muller utomton A there is Buchi utomton A such tht L(A) = L(A ). Proof. We show tht the Büchi utomton A of Construction 6.2 nd A re equivlent. L(A) L(A ): Let α L(A) nd r = q 0 q 1 q 2... be n ccepting run of A on α. As r is ccepting, we hve tht Inf(r) F, i.e., there is some 1 i n such tht Inf(r) = F i. Let m the first position such tht q j Inf(r) for ll j m nd consider some run r = q 0 q 1... q m 1 (i, q m, p 0 )(i, q m+1, p 1 )(i, q m+2, p 2 )... of A on α, which nondeterministiclly switches to (i, q m, p 0 ) t position m. For the ske of contrdiction ssume tht r is not ccepting. Then there is position k 0 such tht p j = p k for ll j k. Then lso q m+j p j for ll j k. However, this contrdicts tht p k F i. L(A ) L(A): Let α L(A ) nd r = q 0 q 1... q m 1 (i, q m, p 0 )(i, q m+1, p 1 )(i, q m+2, p 2 )... be some ccepting run of A on α, which hs to switch to some (i, q m, p 0 ) t some position m, s otherwise it would not be ccepting. By construction q j F i for ll j m nd for ech 26

31 p F i there re infinitely mny positions k such tht q k = p k = p. Thus, Inf(pr 2 (r)) = F i, yielding n ccepting run r = q 0 q 1 q 2... of A on α. We now show tht deterministic Muller utomt re closed, like nondeterministic Büchi utomt, under the Boolen opertions (complementtion, union, nd intersection). Construction 6.3. For deterministic Muller utomton A = (Σ, Q, I, T, muller(f)) we construct the deterministic Muller utomton A C = (Σ, Q, I, T, muller(2 Q F)) with L(A C ) = Σ ω L(A). We use the function pr n for n N to project to the (n 1)-th component of tuple of rbitrry length. If the tuple hs no such component, pr n is undefined. The first component is ccessed using pr 0, nd for sets we hve tht pr n (S) = s S pr n(s). Construction 6.4. For Muller utomt A 1 = (Σ, Q 1, I 1, T 1, muller(f 1 )) nd A 2 = (Σ, Q 2, I 2, T 2, muller(f 2 )) over the sme lphbet Σ, we construct the Muller utomton A = (Σ, Q 1 Q 2, I 1 I 2, T, muller(f )) with L(A ) = L(A 1 ) L(A 2 ) nd where A is deterministic if A 1 nd A 2 re deterministic, s follows: T = {((q 1, q 2 ), σ, (q 1, q 2)) (q 1, σ, q 1) T 1, (q 2, σ, q 2) T 2 } F = {P Q 1 Q 2 pr 0 (P ) F 1, pr 1 (P ) F 2 } Theorem 6.3. The lnguges recognizble by deterministic Muller utomt re closed under Boolen opertions (complementtion, union, intersection). Proof. Deterministic Muller utomt re closed under complementtion: For deterministic Muller utomton A, the utomton A of Construction 6.3 recognizes the complement lnguge, becuse ny set F / F hs to be in the complement, i.e., F 2 Q F. Deterministic Muller utomt re closed under intersection: For deterministic Muller utomt A 1 nd A 2, the utomton A of Construction 6.4 recognizes the intersection. Let r 1 = q 1 0q nd r 2 = q 2 0q be ccepting runs of A 1 nd A 2 on some α. Then r = (r 1 0, r 2 0)(r 1 1, r 2 1)... is n ccepting run of A on α nd vice vers. Deterministic Muller utomt re closed under union: They re closed under complement nd intersection, which suffices by De Morgn s lw. Theorem 6.4. A lnguge L is recognizble by deterministic Muller utomton if nd only if L is Boolen combintion of lnguges W where W Σ is regulr. Proof. : If W is regulr, then W is recognizble by deterministic Büchi utomton. Hence, W is recognizble by deterministic Muller utomton. Thus, the boolen combintion L is recognizble by deterministic Muller utomton. : A deterministic Muller utomton A ccepts some word α with unique run r if for some F F we hve tht Inf(r) = F. Thus, there is some F F such tht for ll q F we hve tht α W q nd for ll q / F we hve tht α / W q, where W q = L(A q ) for the finite-word utomton A q = (Σ, Q, I, T, {q}). Hence, α ( W q (Σ ω ) W q ). F F q F q / F 27