Outline. Overview. Introduction. Well-Founded Monotone Algebras. Monotone algebras. Polynomial Interpretations. Dependency Pairs

Transcription

1 Overview Lecture 1: Introduction, Abstract Rewriting Lecture 2: Term Rewriting Lecture 3: Combinatory Logic Lecture 4: Termination Lecture 5: Matching, Unification Lecture 6: Equational Reasoning, Completion Lecture 7: Confluence Lecture 8: Modularity Lecture 9: Strategies Lecture 10: Decidability Lecture 11: Infinitary Rewriting Term Rewriting Systems - Lecture 4 1/71

2 Overview Outline Term Rewriting Subterm Systems Criterion - Lecture 4 2/71 Overview Introduction Well-Founded Monotone Algebras Monotone algebras Polynomial Interpretations Dependency Pairs Stepwise Termination Proofs Dependency Graphs

3 Overview Termination Term Rewriting Systems - Lecture 4 3/71

4 Introduction Termination, Example 1 Example A(x, s(y)) s(a(x, y)) A(x, 0) x A s s s s 0 s A A s s A R 0 A R 0 s R A R 0 s 0 s Looks terminating: second rule makes terms smaller first rule makes s move upwards Term Rewriting Systems - Lecture 4 4/71

5 Introduction Termination, Example 2 Example f(g(x)) g(f(x)) f(f(g(f(g(x))))) f(f(g(g(f(x))))) f(g(f(g(f(x))))) g(f(f(g(f(x))))) g(f(g(f(f(x))))) g(g(f(f(f(x))))) Looks terminating: f s move to the right g s move to the left Term Rewriting Systems - Lecture 4 5/71

6 Introduction Termination, Example 3 Example f(g(x)) g(g(f(f(x)))) Looks terminating: f s move to the right g s move to the left But we have an infinite rewrite sequence: f(g(g(x))) g(g(f(f(g(x))))) g(g(f(g(g(f(f(x)))))))... We need proofs of termination! Term Rewriting Systems - Lecture 4 6/71

7 Introduction Termination, Example 4 Example f(x, g(y)) f(x, x) Looks terminating: right side is smaller than the left side But we have an infinite rewrite sequence: f(g(x), g(x)) f(g(x), g(x))... We need proofs of termination! Term Rewriting Systems - Lecture 4 7/71

8 Introduction Definition (Termination SN(R)) A rewrite system R is terminating if there are no infinite rewrite sequences. Termination Methods Knuth-Bendix order, polynomial interpretations, multiset order, simple path order, lexicographic path order, semantic path order, recursive decomposition order, multiset path order, recursive path order, transformation order, elementary interpretations, type introduction, well-founded monotone algebras, general path order, semantic labeling, dummy elimination, dependency pairs, freezing, top-down labeling, monotonic semantic path order, context-dependent interpretations, match-bounds, size-change principle, matrix interpretations, predictive labeling, uncurrying, bounded increase, quasi-periodic interpretations, arctic interpretations, increasing interpretations, root-labeling,... Term Rewriting Systems - Lecture 4 8/71

9 Introduction Termination Research Termination Tools AProVE, Cariboo, CiME, Jambox, Termptation, Matchbox, MuTerm, NTI, Torpa, TPA, T T T 2, VMTL,... Term Rewriting Systems - Lecture 4 9/71

10 Well-Founded Monotone Algebras Outline Overview Introduction Well-Founded Monotone Algebras Monotone algebras Polynomial Interpretations Dependency Pairs Stepwise Termination Proofs Dependency Graphs Subterm Criterion Iterative Lexicographic Path Order Term Rewriting Systems - Lecture 4 10/71

11 Well-Founded Monotone Algebras Lemma TRS R is terminating well-founded order > on terms such that s > t whenever s R t Example TRS 0 + y y s(x) + y s(x + y) well-founded order > 1 if u = 0 ϕ(v) + 1 if u = s(v) s > t ϕ(s) > N ϕ(t) with ϕ(u) = 2ϕ(v) + ϕ(w) if u = v + w 0 otherwise Remark (very) inconvenient to check all rewrite steps Term Rewriting Systems - Lecture 4 11/71

12 Well-Founded Monotone Algebras Definitions A reduction order is well-founded order > on terms which is closed under contexts s > t = C[s] > C[t] closed under substitutions s > t = sσ > tσ Notation R > if l > r for all rules l r in R Theorem TRS R is terminating R > for reduction order > Term Rewriting Systems - Lecture 4 12/71

13 Well-Founded Monotone Algebras Theorem TRS R is terminating R > for reduction order > Proof. Let R be terminating. We define > =. Then: > is well-founded since R is terminating, > is closed under substitutions since is, and > is closed under contexts since is. Hence > is a reduction order. Moreover R. Term Rewriting Systems - Lecture 4 13/71

14 Well-Founded Monotone Algebras Theorem TRS R is terminating R > for reduction order > Proof. Let > be reduction order such that R >. Recall that is the smallest relation S such that: R S, S is closed under contexts, and S is closed under substitutions. Then > since > has these properties. Assume there exists an infinite rewrite sequence: t 0 t 1 t 2... Then also t 0 > t 1 > t 2... since > However, this contradicts well-foundedness of >. Term Rewriting Systems - Lecture 4 14/71

15 Monotone algebras Constructing Reduction Orders Idea: give semantics to terms by interpreting them into an algebra. Definition A Σ-algebra (A, [ ]) consists of: a non-empty set A, and for every f Σ an interpretation function [f ]: A ar(f ) A. Example (We use the Σ-algebra (N, [ ]) with) [0] = 1 [s](x) = x + 1 [A](x, y) = x + 2 y A s 5 A 4 s R s 10 A A 4 s R s 6 A s 2 R 0 1 s 5 s 4 A R s 3 s Term Rewriting Systems - Lecture 4 15/71

16 Monotone algebras Interpretation of terms Definition (Interpretation of Terms) Let α : X A be an interpretation of the variables. We define the evaluation of terms [, α] : T (Σ, X ) A inductively: [x, α] = α(x) [f (t 1,..., t n ), α] = [f ]([t 1, α],..., [t n, α]) if x X Example [0] = 1 [s](x) = x + 1 [A](x, y) = x + 2 y Let α(x) = 1, α(y) = 3, we calculate: [A(0, s(0)), α] = 5, [A(s(x), y), α] = 8. Term Rewriting Systems - Lecture 4 16/71

17 Monotone algebras Monotone Σ-algebras A function [f ] is monotone (w.r.t. >) if Definition a > b implies [f ](..., a,...) > [f ](..., b,...) A well-founded monotone Σ-algebra (A, [ ], >): a Σ-algebra (A, [ ]), a well-founded order > on A (no infinite chains a 1 > a 2 >...), such that for all f Σ the function [f ] A is monotone. Relation > A on terms: s > t if [s, α] > [t, α] for all assignments α. Lemma > A is a reduction order for every well-founded monotone Σ-algebra (A, [ ], >) Term Rewriting Systems - Lecture 4 17/71

18 Monotone algebras Lemma Let A well-founded monotone Σ-algebra. Then > A is closed under substitutions. Proof. Let s, t T (Σ, X ) such that t > A s. Let σ be a substitution. We show tσ > A sσ. That is: [tσ, α] > [sσ, α] for every α : X A. Define β : X A by β(x) = [σ(x), α]. Claim: [uσ, α] = [u, β] for all u. Proof of the claim by induction over the term structure of u: Hence [tσ, α] = [t, β] > [s, β] = [sσ, α]. [xσ, α] = [σ(x), α] = [x, β] [f (t 1,..., t n )σ, α] = [f ]([t 1 σ, α],..., [t n σ, α]) IH = [f ]([t 1, β],..., [t n, β]) = [f (t 1,..., t n ), β] Term Rewriting Systems - Lecture 4 18/71

19 Monotone algebras Lemma Let A well-founded monotone Σ-algebra. Then > A is closed under contexts. Proof. Let s, t T (Σ, X ) such that t > A s. We show C[s] > A C[t] for every C[ ]. That is: [C[t], α] > [C[s], α] for every α : X A. Let α : X A. We show [C[t], α] > [C[s], α] by induction over structure of C[ ]: C[ ] [ ] [C[t]] = [t] > [s] = [C[s]] C[ ] f (t 1,..., t n ) Let i such that t i contains the hole. Then C[u] = f (t 1,..., t i [u],..., t n ). Hence [C[s], α] = [f ]([t 1, α],..., [t i [s], α],..., [t n, α]). [C[t], α] = [f ]([t 1, α],..., [t i [t], α],..., [t n, α]). By IH [t i [t], α] > [t i [s], α]. By monotonicity [C[t], α] > [C[s], α]. Term Rewriting Systems - Lecture 4 19/71

20 Monotone algebras Lemma Let A well-founded monotone Σ-algebra. Then > A is well-founded. Proof. Assume there is an infinite sequence t 1 > A t 2 > A t 3 > A.... Let α : X A. Then by definition of > A : [t 1, α] > [t 2, α] > [t 3, α] >.... This infinite decreasing > chain contradicts well-foundedness of >. We have shown that: > A is closed under substitutions > A is closed under contexts > A is well-founded Hence we have proven that: Lemma > A is a reduction order for every well-founded monotone Σ-algebra (A, [ ], >) Term Rewriting Systems - Lecture 4 20/71

21 Monotone algebras Theorem TRS R is terminating R > A for well-founded monotone algebra (A, >) Proof. Follows since we have shown that > A is a reduction order. As Σ-algebra we take (A, [ ], >) with A = T (Σ, X ), [f ](t 1,..., t n ) = f (t 1,..., t n ), > :=. Then > is well-founded since R is terminating. Monotonicity of > follows from closure of under contexts: s > t [f ](..., s,...) = f (..., s,...) > f (..., t,...) = [f ](..., t,...) R > A since for all l r R and α : X A: [l, α] = lα > rα = [r, α]. Term Rewriting Systems - Lecture 4 21/71

22 Monotone algebras Well-Founded Monotone Algebras used in termination proofs/tools : polynomial interpretations over N polynomial interpretations over Q and R matrix interpretations over N matrix interpretations over N { } Term Rewriting Systems - Lecture 4 22/71

23 Polynomial Interpretations Outline Overview Introduction Well-Founded Monotone Algebras Monotone algebras Polynomial Interpretations Dependency Pairs Stepwise Termination Proofs Dependency Graphs Subterm Criterion Iterative Lexicographic Path Order Term Rewriting Systems - Lecture 4 23/71

24 Polynomial Interpretations Polynomial interpretations Definition A polynomial interpretations over N consists of: Σ-algebra is (N, [ ]), > defined as usual on N, the interpretations [f ] are polynomials. The first two conditions of well-founded monotone Σ-algebras are fulfilled: (N, [ ]) is a Σ-algebra > is well-founded However, to prove termination we need to check: monotonicity of the polynoms [f ], and R >, that is, [l, α] > [r, α] for all l r R and α : X A Term Rewriting Systems - Lecture 4 24/71

25 Polynomial Interpretations Example 1 Example f(g(x)) f(f(x)) Find a polynomial interpretation over N which proves termination: [f](x) = x [g](x) = x + 1 Are the functions [f ] monotone? Yes, since whenever a > b, then [f](a) = a > b = [f](b), [g](a) = a + 1 > b + 1 = [g](b), Does [l, α] > [r, α] hold? Yes since, [f(g(x)), α] = [f]([g](α(x))) = α(x) + 1 > α(x) = [f(f(x)), α] Hence we have proven termination. Term Rewriting Systems - Lecture 4 25/71

26 Polynomial Interpretations Example 2 Example f(g(x)) g(f(x)) Find a polynomial interpretation over N which proves termination: [f](x) = 2 x [g](x) = x + 1 Are the functions [f ] monotone? Yes, since whenever a > b, then [f](a) = 2 a > 2 b = [f](b), [g](a) = a + 1 > b + 1 = [g](b), Does [l, α] > [r, α] hold? Yes since, [f(g(x)), α] = 2 (α(x) + 1) > 2 α(x) + 1 = [g(f(x)), α] Hence we have proven termination. Term Rewriting Systems - Lecture 4 26/71

27 Polynomial Interpretations Example 3 Example A(x, 0) x A(x, s(y)) s(a(x, y)) Find a polynomial interpretation over N which proves termination: [0] = 1 [s](x) = x + 1 [A](x, y) = x + 2 y Are the functions [f ] monotone? Yes, since whenever a > b, then [s](a) = a + 1 > b + 1 = [s](b), [A](a, y) = a + 2 y > b + 2 y = [A](b, y), and [A](x, a) = x + 2 a > x + 2 b = [A](x, b), Hence we have proven termination. Term Rewriting Systems - Lecture 4 27/71

28 Polynomial Interpretations Why do we need the conditions? Example (> not well-founded) Let R = { f(x) f(f(x)) } with the Σ-algebra (Z, [ ]) and [f](x) = x 1 and > as usual on Z. Then R is not terminating f(x) f(f(x)) f(f(f(x)))... but [f ] is monotone, and [f(x), α] = α(x) 1 > α(x) 2 = [f(f(x)), α]. Hence > needs to be well-founded! Term Rewriting Systems - Lecture 4 28/71

29 Polynomial Interpretations Why do we need the conditions? Example ([f] not monotone) Let R = { f(x) g(f(x)) } with the Σ-algebra (N, [ ]) and [f](x) = x + 1 [g](x) = 0 and > as usual on N. Then R is not terminating f(x) g(f(x)) g(g(f(x)))... but > is well-founded, and [f(x), α] = α(x) + 1 > 0 = [g(f(x)), α]. Hence the functions [f ] need to be monotone! Term Rewriting Systems - Lecture 4 29/71

30 Polynomial Interpretations Question How to find suitable polynomials? Modern Approach (a) choose abstract polynomial interpretations (linear, quadratic,... ) (b) (c) (d) (e) transform rewrite rules into polynomial ordering constraints add monotonicity and well-definedness constraints eliminate universally quantified variables translate resulting constraints to SAT or SMT problem Term Rewriting Systems - Lecture 4 30/71

31 Polynomial Interpretations Example rewrite system interpretations 0 + y y s(x) + y s(x + y) 0 A = 0 s A (x) = x A (x, y) = 2x + y + 1 diophantine constraints x, y N e 1 0 da + f > 0 e be 0 dc + f bf c > 0 a 0 b 1 c 0 d 1 e 1 f 0 possible solution a = 0 b = 1 c = 1 d = 2 e = 1 f = 1 Term Rewriting Systems - Lecture 4 31/71

32 Polynomial Interpretations Example TRS 0 + y y 0 y 0 s(x) + y s(x + y) s(x) y y + (x y) interpretations in N 0 A = 1 + A (x, y) = 2x + y s A (x) = x + 1 A (x, y) = 2xy + x + y + 1 constraints x, y N 2 > 0 3y + 1 > 0 1 > 0 1 > 0 s(0) s(s(0)) s(s(0)) + (0 s(s(0))) s(s(0)) + 0 s(s(0) + 0) 18 > 17 > 7 > 6 s(s(0 + 0)) s(s(0)) > 5 > 3 Term Rewriting Systems - Lecture 4 32/71

33 Polynomial Interpretations Example TRS (x + y) (x) + (y) (α) = 1 (x y) (x) (y) (β) = 0 (x y) ( (x) y) + (x (y)) (x y) (( (x) y) (x (y))) (y y) interpretations in N α A = β A = 0 A = 1 A = 1 + A (x, y) = A (x, y) = A (x, y) = A (x, y) = x + y + 3 A (x) = x 2 + 6x + 6 constraints x, y N x 2 + y 2 + 2xy + 12x + 12y + 33 > x 2 + y 2 + 6x + 6y > 1 x 2 + y 2 + 2xy + 12x + 12y + 33 > x 2 + y 2 + 6x + 6y > 1 x 2 + y 2 + 2xy + 12x + 12y + 33 > x 2 + y 2 + 7x + 7y + 21 x 2 + y 2 + 2xy + 12x + 12y + 33 > x 2 + y 2 + 7x + 9y + 27 Term Rewriting Systems - Lecture 4 33/71

34 Polynomial Interpretations Remark Numerous terminating TRSs are not polynomially terminating. polynomial interpretations terminating TRSs Term Rewriting Systems - Lecture 4 34/71

35 Polynomial Interpretations A non-simply terminating TRS: f (f (x)) f (g(f (x))) Example (S = {f (f (x)) f (g(f (x)))}) There exists no monotone Σ-algebra A with A = N proving SN(S). Assume contrary it would exist. Let α : X A, then: [f (f (x)), α] > [f (g(f (x))), α]. But then also [f (x), α] > [g(f (x)), α] since otherwise: [f (x), α] = [g(f (x)), α] implies [f (f (x)), α] = [f (g(f (x))), α], [f (x), α] < [g(f (x)), α] implies [f (f (x)), α] < [f (g(f (x))), α] by monotonicity. But then A would also prove termination of f (x) g(f (x)). Thus we need another Σ-algebra, for example A = N 2. Term Rewriting Systems - Lecture 4 35/71

36 Polynomial Interpretations A non-simply terminating TRS: f (f (x)) f (g(f (x))) Example (S = {f (f (x)) f (g(f (x)))}) We choose the Σ-algebra (N 2, [ ]) with: ( ) ( ) [f]( x) = 0 0 x + [g]( x) = where > on N 2 is defined as follows: ( ) ( ) x1 y1 x > 2 y x 1 > y 1 and x 2 y 2 2 Let α: X A be arbitrary; write x = α(x). We obtain [f (f (x))] = Hence S is terminating. ( ) ( ) x + > ( ) ( ) x + ( ) ( ) x + = [f (g(f (x)))] Term Rewriting Systems - Lecture 4 36/71

37 Dependency Pairs Termination via Dependency Pairs Term Rewriting Systems - Lecture 4 37/71

38 Dependency Pairs Minimal Terms We call a term is minimal if all its strict subterms are terminating. Theorem Let R be a non-terminating TRS. Then there exists a minimal term which is non-terminating. Proof. Since R is non-terminating, there exists a non-terminating term t. We prove: every non-terminating term t contains a minimal subterm. We use induction on the term structure of t: Base case: all strict subterms of t are terminating. Then t is minimal itself. Induction step: t has a strict subterm t which is not terminating. Then by IH the term t contains a minimal, non-terminating subterm t. The term t is a minimal non-terminating subterm of t. Term Rewriting Systems - Lecture 4 38/71

39 Dependency Pairs Steps below the root do not change minimality: Lemma Let t be minimal and t s a rewrite step below the root. Then s is minimal. Proof. We have t = f (t 1,..., t n ) and s = f (s 1,..., s n ) and i N such that: t i s i, and t j = s j for all j i. Then since t is minimal, it follows that all t k are terminating. Since t k s k, we obtain s k is terminating for all 1 k n. Hence s is minimal. Term Rewriting Systems - Lecture 4 39/71

40 Dependency Pairs Dependency Pairs, Introduction Assume R is non-terminating. There exists a minimal, non-terminating term t 0. t 0 R t 1 R t 2 R... At some point there must be a root step t i t i+1. Proof. Assume all steps would be below the root. Then: all t i are minimal, all rewrite steps are in terminating terminating. Notation: we use top R to denote root rewrite steps. Term Rewriting Systems - Lecture 4 40/71

41 Dependency Pairs Dependency Pairs, Introduction Let t 0 be a minimal and t 0 R t 1 R t 2 R... an infinite rewrite sequence. We consider the first root rewrite step t i t i+1 : t i lσ l t i+1 rσ r SN SN & minimal SN σ s σ SN The temr t i+1 contains a minimal, non-terminating subterm s. The root of s lies in the pattern of r. Hence there exists a non-variable subterm r of r such that s = r σ. Idea: add a rule l r then t i s. Term Rewriting Systems - Lecture 4 41/71

42 Dependency Pairs Dependency Pairs For every f Σ let f # be a fresh symbol with the same arity as f. By t # we denote f # (t 1,..., t n ) for t = f (t 1,..., t n ) T (Σ, X ). Definition (Dependency Pairs) DP(R) = {l # r # l r R, r r with r X } Example R = { f(x) g(f(x)) } DP(R) = { f # (x) g # (f(x)), f # (x) f # (x) } Term Rewriting Systems - Lecture 4 42/71

43 Dependency Pairs Dependency Pairs Let t 0 be a minimal and t 0 R t 1 R t 2 R... an infinite rewrite sequence. We consider the first root rewrite step t i t i+1 : t i lσ l t i+1 rσ r SN SN & minimal SN σ s σ SN Then t 0,# R... R t i,# Repeating the construction with s yields: t 0,# R top DP(R) R top DP(R) s # top DP(R) R top DP(R)... an infinite rewrite sequence containing infinitely many DP(R) steps. Term Rewriting Systems - Lecture 4 43/71

44 Dependency Pairs Dependency Pairs Lemma Let R be a non-terminating TRS. Then there exists a rewrite sequence: such that: t 0,# R top DP(R) R top DP(R) R the sequence contains infinitely many DP(R) steps, all R steps are below the root, and all DP(R) steps are at the root. top DP(R)... Term Rewriting Systems - Lecture 4 44/71

45 Dependency Pairs Dependency Pairs, Examples Example R is non-terminating: R = { f(g(x)) g(g(f(f(x)))) } DP(R) = { f # (g(x)) g # (g(f(f(x)))), f # (g(x)) g # (f(f(x))), f # (g(x)) f # (f(x)), f # (g(x)) f # (x) } f(g(g(x))) g(g(f(f(g(x))))) g(g(f(g(g(f(f(x)))))))... How does an infinite R top DP(R) rewrite sequence look like? f # (g(g(x))) top DP(R) f # (f(g(x))) R f # (g(g(f(f(x))))... Term Rewriting Systems - Lecture 4 45/71

46 Dependency Pairs Dependency Pairs, Examples Example R = { A(x, s(y)) s(a(x, y)), A(x, 0) x } DP(R) = { A # (x, s(y)) s # (A(x, y)), A # (x, s(y)) A # (x, y) } Term Rewriting Systems - Lecture 4 46/71

47 Dependency Pairs Dependency Pairs, Main Theorem Definition (Relative Termination) A relation 1 is called terminating relative to 2, denoted SN( 1 / 2 ), if every 1 2 rewrite sequence contains only finitely many 1 steps. Lemma SN( 1 / 2 ) SN( 2 1 2) The main theorem from dependency pairs is: Theorem SN(R) SN(DP(R) top /R) That is, a TRS R is terminating if and only if top DP(R) terminates relative to R. Term Rewriting Systems - Lecture 4 47/71

48 Dependency Pairs Dependency Pairs, Termination Proofs Definition A well-founded weakly monotone Σ-algebra (A, [ ], >, ) consists of: a Σ-algebra (A, [ ]) with relations >, on A > is well-founded, > > (compatibility), for all f Σ the function [f ] is monotone w.r.t. Theorem SN(DP(R) top /R) if there exists a weakly monotone Σ-algebra s.t. DP(R) > that is, [l, α] > [r, α] α, l r DP(R) R that is, [l, α] [r, α] α, l r R Advantages: no monotonicity for >, and not well-founded. Term Rewriting Systems - Lecture 4 48/71

49 Dependency Pairs Frequently used are polynomial interpretations over N: > as usual on N and := the interpretations [f ] are polynomials We will see some examples... Term Rewriting Systems - Lecture 4 49/71

50 Dependency Pairs Example: f(f(x)) f(g(f(x))) Example DP(R) = { f # (f(x)) f # (g(f(x))), f # (f(x)) g # (f(x)), f # (f(x)) f # (x) } [f](x) = x + 1 [f # ](x) = x + 1 [g](x) = 0 [g # ](x) = 0 Are the functions [f ] monotone w.r.t.? Yes, since whenever a b, then [f](a) = a b = [f](b), [g](a) = 0 0 = [g](b), Hence we have proven termination. Term Rewriting Systems - Lecture 4 50/71

51 Stepwise Termination Proofs Stepwise Termination Proofs Stepwise termination proofs with monotone Σ-algebras: Theorem If there exists a monotone Σ-algebra (A, [ ], >) s.t. R, and R >, and where := > =. Then SN(R) = SN(R R ) This theorem allows us to stepwise remove rules until none are left. Remark Instead of we can more generally use a monotone relation with > > Term Rewriting Systems - Lecture 4 51/71

52 Stepwise Termination Proofs Example Example f(f(x)) g(x) f(g(x)) g(f(x)) We use the following interpretation: [f](x) = x + 1 [g](x) = x + 1 We get the following interpretation of rules: [f(g(x)), α] = α(x) + 2 > α(x) + 1 = [g(x), α] [f(g(x)), α] = α(x) + 2 α(x) + 2 = [g(f(x)), α] The first rule is strictly decreasing, hence we can remove it. Thus for termination it sufficies to show SN(f(g(x)) g(f(x))). We have already shown this a few slides ago. Hence we have proven termination. Term Rewriting Systems - Lecture 4 52/71

53 Stepwise Termination Proofs Stepwise Termination Proofs Stepwise termination proofs with dependency pairs: Theorem If there exists a weakly monotone Σ-algebra s.t. T 1 > T 2 R Then SN(T 2,top /R) = SN((T 1 T 2 ) top /R) That is, we may remove the strictly decreasing top-rules. Typically, T 1, T 2 DP(R). But we can also tackle other top-termination problems. We are not allowed to remove strictly decreasing rules in R! (for removing from R we need monotonic interpretations) Term Rewriting Systems - Lecture 4 53/71

54 Stepwise Termination Proofs Example minus(x, 0) x minus(s(x), s(y)) minus(x, y) quot(0, s(y)) 0 quot(s(x), s(y)) s(quot(minus(x, y), s(y))) DP(R) = { minus # (s(x), s(y)) minus # (x, y) We use the interpretation: quot # (0, s(y)) 0 # quot # (s(x), s(y)) s # (quot(minus(x, y), s(y))) quot # (s(x), s(y)) quot # (minus(x, y), s(y)) quot # (s(x), s(y)) minus # (x, y) quot # (s(x), s(y)) s # (y) } [minus # ](x, y) = 1 [quot # ](x, y) = 1 [minus](x, y) = x [s](x) = x [f]( x) = 0 for all other symbols f Term The Rewriting following Systems - Lecture interpretation 4 removes the remaining DP rules and proves: 54/71

55 Dependency Graphs Dependency Graphs Example minus(x, 0) x minus(s(x), s(y)) minus(x, y) quot(0, s(y)) 0 quot(s(x), s(y)) s(quot(minus(x, y), s(y))) Dependency graph: analysis which DP-rules may follow each other (1) minus # (s(x), s(y)) minus # (x, y) (2) quot # (s(x), s(y)) minus # (x, y) (3) quot # (s(x), s(y)) quot # (minus(x, y), s(y)) Idea: consider only strongly connected components SN({1} top /R), SN({3} top /R). Term Rewriting Systems - Lecture 4 55/71

56 Subterm Criterion Subterm Criterion Theorem (Subterm Criterion) Let R be a TRS, T 1, T 2 DP(R), and π : Σ # N such that: s π(f# ) t π(g# ) for every rule f # (s 1,..., s n ) g # (t 1,..., t m ) T 1 s π(f# ) = t π(g# ) for every rule f # (s 1,..., s n ) g # (t 1,..., t m ) T 2 Then: SN(T 2,top /R) = SN((T 1 T 2 ) top /R) Proof. After the dependency pairs transformation, we consider only minimal terms. We can only finitely often make a terminating term smaller ( ). Term Rewriting Systems - Lecture 4 56/71

57 Subterm Criterion Example: Ackermann function Example We use the interpretation: Ack(0, y) s(y) Ack(s(x), 0) Ack(x, s(0)) Ack(s(x), s(y)) Ack(x, Ack(s(x), y)) DP(R) = { Ack # (0, y) s # (y) Ack # (s(x), 0) Ack # (x, s(0)) Ack # (s(x), 0) s # (0) Ack # (s(x), 0) 0 # Ack # (s(x), s(y)) Ack # (x, Ack(s(x), y)) Ack # (s(x), s(y)) Ack # (s(x), y) Ack # (s(x), s(y)) s # (x) } [Ack # ](x, y) = 1 [Ack](x, y) = 0 [s](x) = 0 [0] = 0 s # (x) = 0 We use the subterm criterion: π(ack # ) = 1 Term Rewriting Systems - Lecture 4 57/71

58 Iterative Lexicographic Path Order Iterative Lexicographic Path Order (ILPO) Term Rewriting Systems - Lecture 4 58/71

59 Iterative Lexicographic Path Order ILPO... Historical overview Kamin and Lévy [1980] (lexicographic path order, LPO): Kruskal s Tree Theorem was used in the original proofs Buchholz [1995] simplified the proof: Kruskal not needed Bergsta and Klop [1985]: Iterative version of RPO: star method Operational definition of reduction order via an auxiliary term rewriting system (with stars) Klop, van Oostrom and de Vrijer [2005]: Extension of the star method to LPO Iterative lexicographic path order (ILPO) Term Rewriting Systems - Lecture 4 59/71

60 Iterative Lexicographic Path Order ILPO... The star TRS Lex Given a terminating relation on signature Σ, define TRS Lex Signature: Σ Σ, where Σ = {f f Σ} f is fresh and has the same arity as f Reduction rules (four types, for arbitrary f, g Σ): f ( x) put f ( x) f ( x) select x i f ( x) copy g(f ( x),..., f ( x)) f ( x, g( y), z) lex f ( x, g ( y), l,..., l) if f g where l = f ( x, g( y), z) Definition (ILPO) ilpo is the restriction of + Lex to terms over Σ, i.e. t ilpo s t + Lex s t, s T (Σ V ) Term Rewriting Systems - Lecture 4 60/71

61 Iterative Lexicographic Path Order ILPO Claim ilpo is a reduction order, that is: ilpo is well-founded, and closed under substitution and contexts The closure under substitutions and contexts is immediate since + Lex is. Only required: proof of termination of ilpo. Corollary A TRS R is terminating if R ilpo. Proof. ilpo is a reduction order with R ilpo. Term Rewriting Systems - Lecture 4 61/71

62 Iterative Lexicographic Path Order Example, Addition and multiplication A(x, 0) x A(x, S(y)) S(A(x, y)) M(x, 0) 0 M(x, S(y)) A(x, M(x, y)) Use relation R given by M A and A S. For each reduction rule a corresponding Lex-reductions: A(x, 0) put A (x, 0) select x A(x, S(y)) put A (x, S(y)) copy S(A (x, S(y))) lex S(A(x, S (y))) select S(A(x, y)) M(x, 0) put M (x, 0) select 0 M(x, S(y)) put M (x, S(y)) copy A(M (x, S(y)), M (x, S(y))) select A(x, M (x, S(y))) lex A(x, M(x, S (y))) select A(x, M(x, y)) Term Rewriting Systems - Lecture 4 62/71

63 Iterative Lexicographic Path Order Termination for ilpo via termination of Lex ω But Lex is in general not terminating, e.g. if A > S, then A(x, y) put A (x, y) copy S(A (x, y)) copy S(S(A (x, y)))... Starred symbol A is used infinitely often. This is essential in any infinite reduction! Idea: use numbers instead of stars, the numbers fix how often a symbol can be used. Yields a terminating TRS Lex ω. Term Rewriting Systems - Lecture 4 63/71

64 Iterative Lexicographic Path Order Auxiliary TRS with numerical control symbols Given a terminating relation on signature Σ, define TRS Lex ω Signature: Σ Σ ω, where Σ ω = {f n f Σ, n N} f n is fresh and has same arity as f Reduction rules: f ( x) put f n ( x) f n ( x) select x i f n+1 ( x) copy g(f n ( x),..., f n ( x)) f n+1 ( x, g( y), z) lex f ( x, g n ( y), l,..., l) if f g where l = f n ( x, g( y), z) Term Rewriting Systems - Lecture 4 64/71

65 Iterative Lexicographic Path Order Back and forth between Lex and Lex ω From Lex ω to Lex : every reduction can be transformed by replacing f n by f. From Lex to Lex ω: every finite reduction can be lifted, in particular every reduction between two starless terms can be lifted. For example: M(x, S(y)) put M (x, S(y)) copy A(M (x, S(y)), M (x, S(y))) becomes: select A(x, M (x, S(y))) lex A(x, M(x, S (y))) select A(x, M(x, y)) M(x, S(y)) put M 2 (x, S(y)) copy A(M 1 (x, S(y)), M 1 (x, S(y))) select A(x, M 1 (x, S(y))) lex A(x, M(x, S 0 (y))) select A(x, M(x, y)) Term Rewriting Systems - Lecture 4 65/71

66 Iterative Lexicographic Path Order Lex and Lex ω Theorem + Lex and + Lex ω coincide on T (Σ V ) Note that the infinite reduction cannot be lifted: A(x, y) put A (x, y) copy S(A (x, y)) copy S(S(A (x, y)))... A(x, y) put A? (x, y) copy S(A? 1 (x, y)) copy S(S(A? 2 (x, y)))... Term Rewriting Systems - Lecture 4 66/71

67 Iterative Lexicographic Path Order Termination of Lex ω à la Buchholz Prove the implication t 1,..., t n are terminating = f l (t 1,..., t n ) is terminating by induction on triple f, t, l in ordering, ( Lex ω ) n, >. Here ( Lex ω ) n is the lexicographic order on n-tuples No label l counts as with > n. In general a term is SN if all one-step reducts are SN. We check all one step reducts of f l (t 1,..., t n ). Term Rewriting Systems - Lecture 4 67/71

68 Iterative Lexicographic Path Order Termination of Lex ω à la Buchholz Prove the implication t 1,..., t n are terminating = f l (t 1,..., t n ) is terminating by induction on triple f, t, l in ordering, ( Lex ω ) n, >. Case 1. Internal step f l (..., t i,... ) f l (..., t i,... ). The triple decreases in the second component. Case 2. f (t 1,..., t n ) put f n (t 1,..., t n ). We have a decrease in the third component. Case 3. f (t 1,..., t n ) select t i. By assumption t i is SN. Term Rewriting Systems - Lecture 4 68/71

69 Iterative Lexicographic Path Order Termination of Lex and Lex ω Hence we have proven: Theorem Lex ω is terminating Corollary + Lex is terminating on T (Σ V ) Proof. + Lex and + Lexω coincide on T (Σ V ) Term Rewriting Systems - Lecture 4 69/71

70 Iterative Lexicographic Path Order Example, Ackermann Function Example Ack(0, y) s(y) Ack(s(x), 0) Ack(x, s(0)) Ack(s(x), s(y)) Ack(x, Ack(s(x), y)) Find an order on Σ which proves termination. Ack s We get the following derivations: Ack(0, y) put Ack (0, y) copy s(ack (0, y)) select s(y) Ack(s(x), 0) put Ack (s(x), 0) lex Ack(s (x), Ack (s(x), 0)) select Ack(x, Ack (s(x), 0)) copy Ack(x, s(ack (s(x), 0))) select Ack(x, s(0)) Hence we have proven termination. Term Rewriting Systems - Lecture 4 70/71

71 Iterative Lexicographic Path Order Recursive definition of LPO Let be a strict order on signature Σ Define lpo on T (Σ, V ) by: s lpo t iff (LPO1) (LPO2) t Var(s) and s t, or s = f (s 1,..., s m ), t = g(t 1,..., t n ), and (LPO2a) (LPO2b) (LPO2c) 1 i m, with s i = t or s i lpo t, or f g and s lpo t j for all 1 j n, or f = g, and s lpo t j for all 1 j n, and there exists 1 i m, s.t. s 1 = t 1,..., s i 1 = t i 1 and s i lpo t i. Theorem ilpo is equivalent with lpo Term Rewriting Systems - Lecture 4 71/71