How and when axioms can be transformed into good structural rules

Transcription

1 How and when axioms can be transformed into good structural rules Kazushige Terui National Institute of Informatics, Tokyo Laboratoire d Informatique de Paris Nord (Joint work with Agata Ciabattoni and Nikolaos Galatos) TUWien, 24/10/07 p.1/51

2 Introduction Our Setting: Full Lambek calculus [Ono 90] = Intuitionistic logic structural rules = Noncommutative intuitionistic linear logic (without exponentials) TUWien, 24/10/07 p.2/51

3 Introduction Our Setting: Full Lambek calculus [Ono 90] = Intuitionistic logic structural rules = Noncommutative intuitionistic linear logic (without exponentials) We know some axioms ff ffi 1 ff ffi ff Ω ff ff Ω fi ffi fi Ω ff correspond to structural rules ) Π ; ) Π (w) ;A; ) Π ;A;A; ) Π (c) ;A;B; ) Π ;B;A; ) Π (e) ;A; TUWien, 24/10/07 p.2/51

4 Introduction Our Setting: Full Lambek calculus [Ono 90] = Intuitionistic logic structural rules = Noncommutative intuitionistic linear logic (without exponentials) We know some axioms ff ffi 1 ff ffi ff Ω ff ff Ω fi ffi fi Ω ff correspond to structural rules ) Π ; ) Π (w) ;A; ) Π ;A;A; ) Π (c) ;A;B; ) Π ;B;A; ) Π (e) ;A; What about ffi Φ (ffi ffi? & flffi ffi) ffi fl Ω ffi? TUWien, 24/10/07 p.2/51

5 Introduction Q1: Which axioms correspond to structural rules? TUWien, 24/10/07 p.3/51

6 Introduction Q1: Which axioms correspond to structural rules? - [This talk] N 2 -axioms in the substructural hierarchy. TUWien, 24/10/07 p.3/51

7 Introduction Q1: Which axioms correspond to structural rules? - [This N talk] 2 -axioms in the substructural hierarchy. Q2: Why do you want to transform axioms into structural rules? TUWien, 24/10/07 p.3/51

8 Introduction Q1: Which axioms correspond to structural rules? - [This N talk] 2 -axioms in the substructural hierarchy. Q2: Why do you want to transform axioms into structural rules? - Cut-elimination. TUWien, 24/10/07 p.3/51

9 Introduction Q1: Which axioms correspond to structural rules? - [This N talk] 2 -axioms in the substructural hierarchy. Q2: Why do you want to transform axioms into structural rules? - Cut-elimination. Q3: Do all structural rules admit cut-elimination? TUWien, 24/10/07 p.3/51

10 Introduction Q1: Which axioms correspond to structural rules? - [This N talk] 2 -axioms in the substructural hierarchy. Q2: Why do you want to transform axioms into structural rules? - Cut-elimination. Q3: Do all structural rules admit cut-elimination? - No. ) Π ;A;A; ) Π (c) ; ±; ±; ) Π ;A; (seq c) ; ±; ) Π TUWien, 24/10/07 p.3/51

11 Introduction Q4: Which structural rules admit cut-elimination? TUWien, 24/10/07 p.4/51

12 Introduction Q4: Which structural rules admit cut-elimination? - Weakly substitutive (propagating) ones (exact as far as separated ones are concerned). [Terui 07, Ciabattoni-Terui 06] TUWien, 24/10/07 p.4/51

13 Introduction Q4: Which structural rules admit cut-elimination? - Weakly substitutive (propagating) ones (exact as far as separated ones are concerned). [Terui 07, Ciabattoni-Terui 06] Q5: Which structural rules can be transformed into those admitting cut-elimination? TUWien, 24/10/07 p.4/51

14 Introduction Q4: Which structural rules admit cut-elimination? - Weakly substitutive (propagating) ones (exact as far as separated ones are concerned). [Terui 07, Ciabattoni-Terui 06] Q5: Which structural rules can be transformed into those admitting cut-elimination? - All additive structural rules. [Terui 07] TUWien, 24/10/07 p.4/51

15 Introduction Q4: Which structural rules admit cut-elimination? - Weakly substitutive (propagating) ones (exact as far as separated ones are concerned). [Terui 07, Ciabattoni-Terui 06] Q5: Which structural rules can be transformed into those admitting cut-elimination? - All additive structural rules. [Terui 07] Q6: Is there anything else? TUWien, 24/10/07 p.4/51

16 Introduction Q4: Which structural rules admit cut-elimination? - Weakly substitutive (propagating) ones (exact as far as separated ones are concerned). [Terui 07, Ciabattoni-Terui 06] Q5: Which structural rules can be transformed into those admitting cut-elimination? - All additive structural rules. [Terui 07] Q6: Is there anything else? - [This talk] Those equivalent to acyclic rules. TUWien, 24/10/07 p.4/51

17 Outline 1. Introduction 2. Substructural hierarchy and structuralization 3. Acyclic structural rules and simplification 4. Semantic cut-elimination: an introduction 5. Simple rules admit conservativity and cut-elimination 6. Conservativity implies acyclicity 7. Conclusion TUWien, 24/10/07 p.5/51

18 A ) A Identity Syntax of Full Lambek Calculus Formulas: A & B, A Φ B, A Ω B, A ffi B, Bffi A, >,?, 0, 1. Sequents: ) Π ( : sequence of formulas, Π: stoup with at most one formula) Inference rules: ) A 1;A; 2 ) Π Cut 1; ; 2 ) Π 1 ;A;B; 2 ) Π 1 ;AΩ B; 2 ) Π Ωl ) A ) B ; ) A Ω B Ωr ) A 1;B; 2 ) Π ;A ffi B; 2 ) Π ffil A; ) B ) A ffi B ffir 1; ) A 1;B; 2 ) Π ; 2 ) Π ffi l ;A) B ) Bffi A ffi r 1;Bffi A; TUWien, 24/10/07 p.6/51

19 ) A ) A Φ B Φr 1 1 ; 0; 2 ) Π 0l ) > >r ) B ) A Φ B Φ ) A ) B ) A & B Syntax of Full Lambek Calculus 1 ;A; 2 ) Π 1 ;B; 2 ) Π Φl 1 ;AΦ B; 2 ) Π 1 ;A; 2 ) Π 1 ;A& B; 2 ) Π &l 1 1 ;B; 2 ) Π 1 ;A& B; 2 ) Π &l 2 1 ; 2 ) Π 1 ; 1; 2 ) Π 1l ) 1 1r )?l ) )??r? TUWien, 24/10/07 p.7/51

20 What is a structural rule? Examples: ) Π ;A;A; ) Π (c) ;A;A; ) ;A; ;A; ) (wc) ±; ±; ) Π ; ±; ) Π (seq c) ; ± 1 ; ) Π ; ± 2 ; ) Π ; ; ± 1 ; ± 2 ) Π (min) Ingredients: Metavariables for A; B; C; : : : formulas: Metavariables for ; ; ±;::: sequences: Metavariables for Π;::: stoups: TUWien, 24/10/07 p.8/51

21 1 ) Ψ 1 ::: n ) Ψ n 0 ) Ψ 0 What is a structural rule? A structural rule is where 0 ;:::; n : sequences of A s and s. Ψ 0 ;:::; Ψ n : one A or Π or empty. TUWien, 24/10/07 p.9/51

22 Substructural hierarchy The P sets ; N n n of formulas defined by: P (0) = N 0 = 0 the set of atomic formulas N (P1) P n+1 n A; B 2 P (P2) =) A Φ B; A Ω B;1; n+1 2 P 0 n+1 P (N1) N n+1 n A; B 2 N (N2) =) A & B;?; > 2 N n+1 n+1 A 2 P (N3) ;B 2 N n+1 =) A ffi B; Bffi A 2 N n+1 n+1 TUWien, 24/10/07 p.10/51

23 P 2 N 6 P 1 N 6 P 0 N 0 ff ffi 1; ff ffi ff Ω ff; ff Ω fi ffi fi Ω ff 2 N 2 (ff ffi fi) Φ (fi ffi ff) 2 P 2 ((ff ffi fi) ffi fi) ffi (fi ffi ff) ffi ff 2 N 3 ff? Φ ff?? 2 P 3 Substructural Hierarchy p p p p6 p p p p p p p p p6 p p p p p Alternation of positive (1; 0; Φ; layers Ω) and negative (?; >; &; ffi; ones ffi ) TUWien, 24/10/07 p.11/51

24 1 ) A 1 ::: ff n ) A n ff 1 ;:::;ff n ) B ff ) fi B 1 ;:::;A n ) fi A From axioms to structural rules Theorem: Any axiom N in 2 corresponds to a set of structural rules. Key Lemma: An axiom A 1 ;:::;A n ) B is equivalent to and also to with ff 1 ;:::;ff n ;fi fresh. TUWien, 24/10/07 p.12/51

25 ffi Φ (ffi ffi? & flffi ffi) ffi fl Ω ffi 2 N 2 Ω ffi ) fi fl ) fi ffi ) fi fl;ffi ) fi ffi Ω ffi ) fi fl ffi? & flffi ffi ) fi ffi ) fi fl;ffi ffi? & flffi ffi ) fi ffi ) ffi ffi? & flffi ffi ff ) fi ff From axioms to structural rules ffi ffi fl Ω ffi (ffi ffi? & flffi ffi) ffi fl Ω ffi ffi ) fl Ω ffi (ffi ffi? & flffi ffi) ) fl Ω ffi KeyLemma fl;ffi ) fi KeyLemma OK TUWien, 24/10/07 p.13/51

26 ) ffi ffi? & flffi ffi ff ) fi ff ) fi ff ) ffi ffi? ff ) flffi ffi fl;ffi ) fi ff D ) B C; ) B D ) fi ffi;ff) ff; ffi ) fl fl;ffi ) fi ff D ) B D; A ) A; D ) C C; ) B A From axioms to structural rules fl;ffi ) fi ffi Φ (ffi ffi? & flffi ffi) ffi fl Ω ffi is equivalent to TUWien, 24/10/07 p.14/51

27 N n+2 From axioms to structural rules Theorem: Any axiom N in 2 corresponds to a set of structural rules. More generally, any axiom N in n+2 corresponds to a (non-structural) rule in which N only n formulas appear: structural rules N n =) Partial converse result: Any weakly acyclic structural rule corresponds to an axiom N in 2. Question: P Does 3, to which the linearity axiom belongs, correspond to structural rules in hypersequent calculus? What N about 3? TUWien, 24/10/07 p.15/51

28 Outline 1. Introduction 2. Substructural hierarchy and structuralization 3. Acyclic structural rules and simplification 4. Semantic cut-elimination: an introduction 5. Simple rules admit conservativity and cut-elimination 6. Conservativity implies acyclicity 7. Conclusion TUWien, 24/10/07 p.16/51

29 1 ) Ψ 1 ::: n ) Ψ n Acyclicity Given a structural rule ) Ψ draw edges between metavariables occurring in the premises (upper sequents): ff! fi if ff 2 i and fi 2 Ψ i for some 1» i» n identify two occurrences of the same metavariable A structural rule is acyclic if the resulting DAG is. In other words, if ;ff; r ) ff l never belongs to the cut closure of the premises. TUWien, 24/10/07 p.17/51

30 fi ) fl fl ) ffi ffi ) ff ff; fi; fl ) fl ff; ) ffi fi;fl fi; fl ) fl ff;? Ψ?? - Acyclicity Given a structural rule, we draw a DAG on the metavariables occurring in the premises. A structural rule is acyclic if the DAG is. The following structural rule is not acyclic: ff Ψ while the following is acyclic: fl ffi ff fi ff; fi ) fl fl ffi TUWien, 24/10/07 p.18/51

31 Simplification procedure Given an acyclic rule, we apply: 1. Conclusion separation 2. Premise separation 3. Removing redundant premises 4. Sequencing 5. Contexing 6. Linearization The resulting rule is simple. Always admits cut-elimination. TUWien, 24/10/07 p.19/51

32 ) B; ) B (r) B; ) B? ) B B ffi? ) B; B; B ffi? ) B (r) A & B; B ffi? ) B??? Simp (1): conclusion separation A rule with shared variables in the conclusion is problematic: B ) B A & B ) B (cut) A & B; B ffi? ) B 6+ ) B? ) B B ffi? ) B; & B; B ffi? ) A TUWien, 24/10/07 p.20/51

33 ) B; ) B (r) =) B; ) B ) C B; ) C B; ) B? ) B B ffi? ) B; Simp (1): conclusion separation Conclusion separation: (r 0 ) with C fresh. Now the cut can be removed. A & B; B ffi? ) ) B B & B ) B A (r 0 ) A & B; B ffi? ) B TUWien, 24/10/07 p.21/51

34 Simp (2): premise separation A rule with shared variable in the premises ) A ) A A; ±; A ) Π ; ; ± ) Π is also problematic. It can be transformed into ; ±; ) Π ; ±; ) Π ; ±; ) Π ; ±; ) Π ; ; ± ) Π Acyclicity is crucial. The procedure does not apply to a premise A; ) like A. As a result, variables are separated: LHS-variables and RHS-variables are disjoint. TUWien, 24/10/07 p.22/51

35 Simp (3): removing redundant premises TUWien, 24/10/07 p.23/51

36 Simp (4): sequencing Contraction alone (without Exchange) does not admit cut elimination: A Ω B ) A Ω B A Ω B ) A Ω B A Ω B; A Ω B ) (A Ω B) Ω (A Ω B) A ) A B ) B Cntr A; B ) A Ω B A Ω B ) (A Ω B) Ω (A Ω B) Cut A; B ) (A Ω B) Ω (A Ω B) 6+ A ) A B ) B A ) A B ) B A; B ) A Ω B A; B ) A Ω B A; B; A; B ) (A Ω B) Ω (A Ω B)??? A; B ) (A Ω B) Ω (A Ω B) TUWien, 24/10/07 p.24/51

37 Simp (4): sequencing Sequencing [Terui 07]: A; A; ) Π ; A; ) Π (c) =) ; ±; ±; ) Π ; ±; ) Π ; (seq c) Now the cut can be eliminated: A ) A B ) B A ) A B ) B A; B ) A Ω B A; B ) A Ω B A; B; A; B ) (A Ω B) Ω (A Ω B) (seq c) A; B ) (A Ω B) Ω (A Ω B) TUWien, 24/10/07 p.25/51

38 A ffi B ) B A;?;A ffi B ) B A; ffi B ) B A?;A ffi B ) B (mix) A; Simp (5): contexing The following version of mix is problematic: ) ) Π ) Π (mix) ; For example, ) ffi B ) A ffi B?. A ) A ffi B (mix) B?;A ffi (cut).??.?? 6+ A;? ) TUWien, 24/10/07 p.26/51

39 ) A; A ffi B ) B??;A ffi B ) B (mix0 ) A; Simp (5): contexing Contexing: ) ) Π ) Π (mix) =) ; ) ± l ; ; ± r ) Π l ; ; ; ± r ) Π (mix0 ) ± Now the cut can be eliminated: TUWien, 24/10/07 p.27/51

40 ) B B ) A Φ B B ) A A ) A Φ B A ) A A ) A Φ B A Φ B ) A Φ B A Φ B; A Φ B ) A Φ B Exp A ) B B ) A Φ B B Simp (6): linearization A rule with multiple occ. of the same metavariable in conclusion is problematic: ±; ) Π ; ±; ±; ) Π (exp) ; Cut A; A Φ B ) A Φ B Cut A; B ) A Φ B 6+??? A; B ) A Φ B TUWien, 24/10/07 p.28/51

41 ±; ) C ; ; ) C ; ±; ; ) C ; Simp (6): linearization Linearization [Terui 07]: ±; ) Π ; ±; ±; ) Π (exp) =) ; (min) Now the cut can be eliminated: ) A A ) A Φ B A ) B B ) A Φ B B (min) A; B ) A Φ B TUWien, 24/10/07 p.29/51

42 1 ) ::: m ) 0 ) Simple structural rules Theorem: Every acyclic structural rule is equivalent to a simple structural rule of the form: 1 ) ::: m ) ± l ; m+1 ; ± r ) Π :::± l ; n ; ± r ) Π ± l ; 0 ; ± r ) Π or where 0 consists only of metavariables for sequences (not formulas) and each occurs at most once in 0. TUWien, 24/10/07 p.30/51

43 Outline 1. Introduction 2. Substructural hierarchy and structuralization 3. Acyclic structural rules and simplification 4. Semantic cut-elimination: an introduction 5. Simple rules admit conservativity and cut-elimination 6. Conservativity implies acyclicity 7. Conclusion TUWien, 24/10/07 p.31/51

44 Conservativity and cut-elimination Given a logic L, one can consider its extension L! with infinitary V, W. To L show is a conservative extension of L, one proves: Any L-algebra can be completed, i.e., embedded into! a complete L-algebra. In fact, given a formula A without V, W, `L! A () 8 B: complete L-alg. B j= A () 8 B: L-alg. B j= A () `L A: TUWien, 24/10/07 p.32/51

45 Conservativity and cut-elimination For intuitionistic and modal logics, completion (Stone, Jonsson-Tarski) can be most effectively done via Kripke frames. Given a Heyting algebra B, B define = (W;R) + (the dual Kripke frame) by W = the set of prime filters of B urv = u v: Let (B) + + = the complete Heyting algebra associated to (W;R). There is an embedding B! (B) + +. TUWien, 24/10/07 p.33/51

46 Conservativity and cut-elimination Cut-elimination is stronger than conservativity. To show cut-elimination, one proves a stronger completion: Any intransitive (cut-free) L-structure can be quasi-embedded into an L-algebra. [Okada 96, Belardinelli-Jipsen-Ono 04, Galatos-Jipsen 07] From IL to FL: IL Heyting algebras Kripke frames S-J-T completion FL Residuated lattices Residuated frames (GJ07) Dedekind-MacNeille completion TUWien, 24/10/07 p.34/51

47 Residuated lattices A (pointed) residuated lattice is = hp; ^; _; Ω; ffi; ffi ; 1;bi P hp; ^; _i 1. is a lattice. 2. hp; Ω; 1i is a monoid. 3. For any x; y; z 2 P, x Ω y» z () x» zffi y () y» x ffi z: 4. b 2 P. A valuation is a homomorphism f : Fm! P (Fm: the absolutely free algebra of formulas). A is true under f () 1» f (A). TUWien, 24/10/07 p.35/51

48 (z l ; _;z r ) u) () (z l x; _;z r ) u) Residuated frames Given a set P of atoms, define P Λ = the free monoid generated by P (sequences) P con = P Λ P Λ (P [ f"g) (contexts) (x l ;x r ;u) 2 P con is written (x l ; _;x r ) u). ("; "; a) written a. A (simple) residuated frame W = is a binary relation between P Λ and P con such that () (z l ; _;yz r ) u) for any x; y 2 P Λ and (z l ; _;z r ) u) 2 P con. TUWien, 24/10/07 p.36/51

49 Residuated frames Example 1 (the dual frame): P = (P; ^; _; Ω; ffi; ffi ; 1;b) If is a pointed residuated lattice, P then = (P;@) + is a simple residuated frame, where (z l ; _;z r ) u) () z l x z r»p u Example 2 (cut-free sequent calculus): (FL cf ) + = (Fm;@) is a simple residuated frame, (± l ; _; ± r ) Π) () ± l ; ; ± r ) Π is cut-free provable in FL TUWien, 24/10/07 p.37/51

50 From frames to algebras Let W = be a simple residuated frame. For any X W Λ and U W con, X B = fu 2 W con j8x 2 u)g (upperbounds of X) U C = fx 2 W Λ j8u 2 U u)g (lowerbounds of U) Eg. in (FL cf ) +, A C = fag C = f : ) A is cut-free provable g C(W) denotes the set of all closed subsets X = X BC of W. TUWien, 24/10/07 p.38/51

51 X Φ Y = (X [ Y ) BC X Ω Y = fxy : x 2 X; y 2 Y g BC? = f"g C TUWien, 24/10/07 p.39/51 From frames to algebras IfW = Theorem: is a simple frame, then the dual algebra + = (C(W); &; Φ; Ω; ffi; ffi ; 1;?) W is a complete pointed residuated lattice, where X & Y = X Y X ffi Y = fy : 8x 2 Xxy 2 Y g ffi X = fy : 8x 2 Xyx 2 Y g Y = f"g 1 BC

52 Conservativity Theorem [GJ07]: If P is a pointed residuated lattice, then (a) a is an embedding P! (P) + + f = C (Dedekind-MacNeille Completion). Corollary: FL! is conservative over FL. TUWien, 24/10/07 p.40/51

53 Cut-elimination Given two P; Q algebras in the language of FL, a quasi-homomorphism is a P! P(Q) function such that cq 2 F (cp) for c 2 fb; 1g; F (a)?q F (b) F (a?p b) for? 2 fω; ffi; ffi ; ^; _g, where X?Q Y = fx?q yjx 2 X; y 2 Y g. Theorem [GJ07]: Consider the frame (FL cf ) +. The function (A) fx = X BC : A 2 X A C g is a quasi-homomorphism F =! FL (FL. ) + + cf cf is analogous to F Schütte s semi-valuation Girard s reducibility candidates TUWien, 24/10/07 p.41/51

54 f"g BC f (A) Cut-elimination Recall that A C = f : ) A is cut-free provable in FLg. being quasi-homomorphism, there f is F a valuation on (FL such that (A), i.e.,. f (A) 2 F A 2 f (A) AC + )+ cf R If A is true under f, 1» f (A) " 2 f (A) " 2 A C ) A is cut-free provable in FL. Validity in (FL cf )+ + R implies cut-free provability. TUWien, 24/10/07 p.42/51

55 Cut-elimination Proof of cut-elimination `FL A sound =) (FL cf ) + + j= A (FLcf ) + + j= A complete =) `cf FL A A =) `FL FL A `cf TUWien, 24/10/07 p.43/51

56 Cut-elimination Proof of cut-elimination `FL A sound =) (FL cf ) + + j= A (FLcf ) + + j= A complete =) `cf FL A A =) `FL FL A `cf We have just replaced object cuts with a big META-CUT. TUWien, 24/10/07 p.43/51

57 Cut-elimination Proof of cut-elimination `FL A sound =) (FL cf ) + + j= A (FLcf ) + + j= A complete =) `cf FL A A =) `FL FL A `cf We have just replaced object cuts with a big META-CUT. Another criticism: it does not give a cut-elimination procedure. TUWien, 24/10/07 p.43/51

58 Cut-elimination Proof of cut-elimination `FL A sound =) (FL cf ) + + j= A (FLcf ) + + j= A complete =) `cf FL A A =) `FL FL A `cf We have just replaced object cuts with a big META-CUT. Another criticism: it does not give a cut-elimination procedure. Conjecture: If we eliminate the meta-cut, a concrete (object) cut-elimination procedure emerges. TUWien, 24/10/07 p.43/51

59 1 ) Ψ 1 ::: n ) Ψ n Structural rules in residuated frames A simple residuated frame (P;@) satisfies (r) if ) Ψ Ψ 1 ::: Ψ n holds, where in the latter A stands for a 2 P, for x 2 P Λ Ψ Π for u 2 (P [ f"g). Example 1: If P satisfies a set R of structural rules, then P + satisfies R. Example 2: (FL cf ) + satisfies R. R (r) TUWien, 24/10/07 p.44/51

60 Structural rules in residuated frames =) P + =) P + + P R preserves??? FL cf =) (FL cf R ) + =) (FL cf R )+ + R Theorem: If a rule (r) is simple, then it is preserved by W 7! W +. Corollary: Let R be a set of simple structural rules. 1. R is preserved by Dedekind-MacNeille completion. 2. FL! R is a conservative extension of FL R. 3. FL! R enjoys cut-elimination. TUWien, 24/10/07 p.45/51

61 Outline 1. Introduction 2. Substructural hierarchy and structuralization 3. Acyclic structural rules and simplification 4. Semantic cut-elimination: an introduction 5. Simple rules admit conservativity and cut-elimination 6. Conservativity implies acyclicity 7. Conclusion TUWien, 24/10/07 p.46/51

62 fi ) fi ff; ) fi (we) fi;ff ) fi fl ) fi fl;ff Conservativity implies acyclicity Consider a cyclic rule and suppose that FL with R = R 0 [ f(we)g is conservative R over! FL R. We (we) show that is equivalent to an acyclic rule in FL R0. is equivalent to (we) ff; fi ) fi (we 0 ) TUWien, 24/10/07 p.47/51

63 Conservativity implies acyclicity Let fi be a solution of ff Ω fi» fi smaller than fi: ^ fi = ff n ffi fi: 0»n Then in FL! R:. ff; fi ) fi k ;fl ) fi : 0» kg fff. fl ) fi. (we 0 ) fl; ff ) fi fi ) fi (Cut) fl; ff ) fi Since FL! R is conservative over FL R,wehaveinFL R : k ;fl ) fi : 0» kg fff. TUWien, 24/10/07 p.48/51 fl; ff ) fi

64 ) fi ff; fl ) fi ff 2 ;fl ) fi ::: ff n ;fl ) fi fl ) fi fl;ff Conservativity implies acyclicity Since FL R is finitary, there is n such that ) fi ff; fl ) fi ff 2 ;fl ) fi ::: ff n ;fl ) fi fl. fl;ff ) fi R = R 0 [ f(we)g is equivalent to R 0 [ f(we 00 )g: (we 00 ) 00 is acyclic. (we ) Theorem: If FL is conservative over FL R, then R is R equivalent to a set of acyclic structural rules.! TUWien, 24/10/07 p.49/51

65 Main results Any N axiom in 2 can be transformed into a set of structural rules. For R any set of structural rules, the following are equivalent. 1. is equivalent to a set of acyclic rules. R 2. is equivalent to a set of simple rules. R 3. is equivalent to a R set of rules such that FL enjoys! R R (a stronger form of) 0 cut-elimination. 4. is preserved by Dedekind-MacNeille completion. R 5. FL is a conservative extension of FL R. R! TUWien, 24/10/07 p.50/51

66 6 Conclusion p p p p6 p p p p p p6 p p Structural rules in single-conclusion (RHS) calculi N capture 2. p p p p p p Acyclicity = Conservativity P 2 N 2 P 1 N 1 P 0 N 0 To P conquer ; N 3 3, to which more interesting axioms belong, one would need more sophisticated calculi (eg. hypersequent calculi). Question: How high can we go up? TUWien, 24/10/07 p.51/51