Finitary proof systems for Kozen s µ

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1 Finitary proof systems for Kozen s µ Bahareh Afshari Graham Leigh TU Wien University of Gothenburg homc & cdps 16, Singapore 1 / 17

2 Modal µ-calculus Syntax: p p φ ψ φ ψ φ φ x µx φ νx φ Semantics: For Kripke structure K = (W,, λ) and valuation V : Var 2 W p V K = {u W p λ(u)} φ ψ K V = φ K V ψ K V φ V K = {u W v(u v v φ K V )} x K V = V (x) and similarly for p, and µx φ V K = least fixpoint of the function X φ K V [x X ] { = X W φ K V [x X ] X } νx φ V K = greateast fixpoint of the function X φ K V [x X ] { } = X W X φ K V [x X ] Examples: µx( x p); νx( x p); νxµy( y (p x)) Duality: Define φ as the De Morgan dual of φ: µxφ(x) = νxφ(x) φ V K = W \ φ K y V (y) 2 / 17

3 Validity and proofs Let φ be a closed formula Define K = φ iff φ K = W where K = (W,, λ), = φ (φ is valid) iff K = φ for every K Theorem (Kozen 1983; Walukiewicz 2000) For every closed φ, = φ iff Koz φ Soundness: Proved by Kozen (1983) Completeness: Kozen: aconjuntive fragment; Walukiewicz: full µ-calculus: 1 Completeness of disjunctive fragment: tableaux 2 Provable equivalence between disjunctive and µ-formulæ: tableaux, games/automata 3 / 17

4 almost always implies infinitely often Consider the formula ψ = µxνyφ(x, y) ν yµxφ(x, y) ψ is valid easy semantic argument Koz ψ non-trivial Questions: 1 Is there a more direct/constructive proof of completeness? 2 Is cut necessary? 3 Are there other natural sound and complete finitary proof systems? 4 / 17

5 Tableaux proofs A tableau is a Fix-tree in which every infinite path contains a ν-thread Theorem (Niwinski, Walukiewicz 1996; Studer 2008; Friedman 2013) For every closed guardedguarded formula φ, = φ iff there exists a tableau for φ ( ) Y (νxy ), X (νyx ) νx νxy, X (νyx ) Y (νxy ),νyx ( ) φ(νxy,y (νxy )), φ(x (νyx ),νyx ) µy + µ x Y (νxy ), X (νyx ) ( ) νy Y (νxy ),νyx ( ) νx νxµyφ,νyµxφ where Y (x) = µyφ(x, y) and X (y) = µxφ(x, y) 5 / 17

6 Stirling s tableaux proofs with names Fix a set of names for each variable: N x = {x 0, x 1, } and N = x:var N x An annotated sequent is an expression a 0 φ a 1 1,, φa n n st a 0,, a n N Definition A Stirling proof is a finite tree built from rules Fix N + reset x + ν x + exp st 1 For every sequent a φ a 0 0,, φa k and every i k, a k i a; 2 Every non-axiom leaf has the configuration axb axb Π resetx axb Π axb where: 21 x appears in every sequent between repetition 22 reset x occurs between the two nodes We write Stir Γ if there exists a Stirling proof with root ε {φ ε φ Γ} Theorem (Stirling 2014) Tableaux can be effectively transformed into Stirling proofs 6 / 17

7 Stirling proofs: interpreting tableaux [xy Y (νxy ) x, X (νyx ) y ] reset x [xy Y (νxy ) x, X (νyx ) y ] reset y xyx Y (νxy ) xx, X (νyx ) y ν x xy νxy x, X (νyx ) y xyy Y (νxy ) x, X (νyx ) yy ν y xy Y (νxy ) x,νyx y xy φ(νxy,y (νxy )) x, φ(x (νyx ),νyx ) y µ x + µ y xy Y (νxy ) x, X (νyx ) y ( ) νy x Y (νxy ) x,νyx ε ν x ε νxy ε,νyx ε Pros: 1 Fintary proofs 2 Proofs constructed semantically Cons: 1 Non-locality 2 Guessing resets 7 / 17

8 Circular proofs with ν-closure Consider the rule Definition [ Γ,νxφ(x) ax ] Γ, φ(νxφ) ax where a x, x is not in Γ or a clo Γ,νxφ a Clo Γ iff there exists a finite tree built from rules Fix N + ν x + clo satisfying 1 Every sequent has the form ε φ a 1 1,, φa k k ; 2 Every leaf is either an axiom or discharged by an application of clo: Theorem Stir Γ implies Clo Γ Proof 1 Unravel Stirling proofs; 2 Delete resets; 3 Search for clo 8 / 17

9 Example: Stirling to circular proofs (I) [xy Y (νxy ) x, X (νyx ) y ] reset x [xy Y (νxy ) x, X (νyx ) y ] reset y xyx Y (νxy ) xx, X (νyx ) y ν x xy νxy x, X (νyx ) y xyy Y (νxy ) x, X (νyx ) yy ν y xy Y (νxy ) x,νyx y xy φ(νxy,y (νxy )) x, φ(x (νyx ),νyx ) y µ x + µ y xy Y (νxy ) x, X (νyx ) y ( ) νy x Y (νxy ) x,νyx ε ν x ε νxy ε,νyx ε 9 / 17

10 Example: Stirling to circular proofs (II) Y (νxy ) x, X (νyx ) y reset x Y (νxy ) x, X (νyx ) y reset y Y (νxy ) xx, X (νyx ) y ν x νxy x, X (νyx ) y Y (νxy ) x, X (νyx ) yy ν y Y (νxy ) x,νyx y φ(νxy,y (νxy )) x, φ(x (νyx ),νyx ) y µ Y (νxy ) x, X (νyx ) y reset x Y (νxy ) x, X (νyx ) y reset y Y (νxy ) xx, X (νyx ) y ν x νxy x, X (νyx ) y Y (νxy ) x, X (νyx ) yy ν y Y (νxy ) x,νyx y φ(νxy,y (νxy )) x, φ(x (νyx ),νyx ) y µ x + µ y Y (νxy ) x, X (νyx ) y ν y Y (νxy ) x,νyx ε ν x νxy ε,νyx ε 10 / 17

11 Example: Stirling to circular proofs (III) Y (νxy ) xx x, X (νyx ) y reset x Y (νxy ) xx, X (νyx ) yy reset y Y (νxy ) xx x, X (νyx ) y clo x νxy xx, X (νyx ) y Y (νxy ) xx, X (νyx ) yy clo y Y (νxy ) xx,νyx y φ(νxy,y (νxy )) xx, φ(x (νyx ),νyx ) y µ Y (νxy ) xx, X (νyx ) y reset x Y (νxy ) x, X (νyx ) yy reset y Y (νxy ) xx, X (νyx ) y clo x νxy x, X (νyx ) y Y (νxy ) x, X (νyx ) yy clo y Y (νxy ) x,νyx y φ(νxy,y (νxy )) x, φ(x (νyx ),νyx ) y µ Y (νxy ) x, X (νyx ) y clo y Y (νxy ) x,νyx ε clo x νxy ε,νyx ε 11 / 17

12 Example: Stirling to circular proofs (IV) Y (νxy ) x, X (νyx ) y reset x Y (νxy ) xx, X (νyx ) y reset y Y (νxy ) xx, X (νyx ) y clo x [νxy xx, X (νyx ) y ] Y (νxy ) xx, X (νyx ) yy clo y [Y (νxy ) xx,νyx y ] φ(νxy,y (νxy )) xx, φ(x (νyx ),νyx ) y µ Y (νxy ) xx, X (νyx ) y reset x Y (νxy ) xx, X (νyx ) y clo x νxy x, X (νyx ) y Y (νxy ) x, X (νyx ) y reset y Y (νxy ) x, X (νyx ) yy clo y [Y (νxy ) x,νyx y ] φ(νxy,y (νxy )) x, φ(x (νyx ),νyx ) y µ Y (νxy ) x, X (νyx ) y clo y Y (νxy ) x,νyx ε clo x νxy ε,νyx ε 12 / 17

13 Example: Stirling to circular proofs (V) A proof in Clo: [νxy xx, X (νyx ) y ] [Y (νxy ) x, X (νyx ) y ] exp Y (νxy ) xx,νyx y φ(νxy,y (νxy )) xx, φ(x (νyx ),νyx ) y µ Y (νxy ) xx, X (νyx ) y clo νxy x, X (νyx ) y [Y (νxy ) x,νyx y ] φ(νxy,y (νxy )) x, φ(x (νyx ),νyx ) y µ Y (νxy ) x, X (νyx ) y clo Y (νxy ) x,νyx ε clo νxy ε,νyx ε 13 / 17

14 Taking stock We have shown φ valid φ has a tableau Stir φ Clo φ Corollary Circular proofs with ν-closure are complete for the modal µ-calculus Can we get closer to a sequent calculus? no discharge rules no annotations 14 / 17

15 Eliminating assumptions Let Z = νzψ (z) [Γ 1, Z az 1 ] [Γ k, Z az 1 z k ] (νzψ ) az 1 z i νz(γ i Γ 1 ψ a ) Theorem Γ k,ψ (Z ) az 1 z k clozk Γk, Z az 1 z k 1 Γ i,ψ (Z ) az 1 z i clozi Γ i, Z az 1 z i 1 Γ 1,ψ (Z ) az 1 cloz1 Γ1, Z a Clo Γ Koz +con+ d Γ Γ i,ψ a (νz Γ i Γ 1 ψ a ) d Γ i,ψ a (Γ i νz Γ i Γ 1 ψ a ) + νz Γ i,νz Γ i 1 Γ 1 ψ a (Γ i z)) ind Γ i,νyνz Γ i 1 Γ 1 ψ a (y z)) con Γ i,νz Γ i 1 Γ 1 ψ a Theorem Koz + con + d Γ implies Koz Γ 15 / 17

16 Summary Clo Stir Valid Koz + d + con Koz 1 We introduce two sound and complete cut-free proof systems; 2 Doing so we provide a new proof of completeness for Koz yielding a procedure for obtaining proofs from tableaux; 3 Koz is complete iff d and con are admissible 16 / 17

17 Proving almost always implies infinitely often νxy, X (νyx ) Y (νxy ),νyx d Y (νxy ),νyx φ(νxy,y (νxy )), φ(x (νyx ),νyx ) µx + µ y Y (νxy ), X (νyx ) (ν x + d + ind + con) νxy, X (νyx ) Y (νxy ),νyx where φ(νxy,y (νxy )), φ(x (νyx ),νyx ) µx + µ y Y (νxy ), X (νyx ) (ν y + d + ind + con) Y (νxy ),νyx νx νxµyφ,νyµxφ Y (x) = µyφ(x, y) X (y) = µxφ(x, y) Y = X (µyx ) Y X = Y (µxy ) X 17 / 17