Model Checking for Modal Intuitionistic Dependence Logic

Transcription

1 1/71 Model Checking for Modal Intuitionistic Dependence Logic Fan Yang Department of Mathematics and Statistics University of Helsinki Logical Approaches to Barriers in Complexity II Cambridge, March, 2012 Joint work with Johannes Ebbing, Peter Lohmann, Institute for Theoretical Computer Science, Leibniz University Hannover

2 Outline 2/71 1 First-order Dependence Logic with Intuitionistic Implication First-order Dependence Logic First-order Intuitionistic Dependence Logic 2 Modal Intuitionistic Dependence Logic Introduction Some Properties of MID 3 Complexity of Model Checking Problem for fragments of MID MID-MC is PSPACE-complete

3 Characterizing dependence between variables 3/71 First Order Quantifiers x 1 y 1 x 2 y 2 φ Henkin Quantifiers (Henkin, 1961) ( x1 y 1 x 2 y 2 ) φ Independence Friendly Logic (Hintikka, Sandu, 1989) x 1 y 1 x 2 y 2 /{x 1 }φ Dependence Logic (Väänänen 2007) x 1 y 1 x 2 y 2 (=(x 2, y 2 ) φ)

4 Characterizing dependence between variables 3/71 First Order Quantifiers x 1 y 1 x 2 y 2 φ Henkin Quantifiers (Henkin, 1961) ( x1 y 1 x 2 y 2 ) φ Independence Friendly Logic (Hintikka, Sandu, 1989) x 1 y 1 x 2 y 2 /{x 1 }φ Dependence Logic (Väänänen 2007) x 1 y 1 x 2 y 2 (=(x 2, y 2 ) φ)

5 Characterizing dependence between variables 3/71 First Order Quantifiers x 1 y 1 x 2 y 2 φ Henkin Quantifiers (Henkin, 1961) ( x1 y 1 x 2 y 2 ) φ Independence Friendly Logic (Hintikka, Sandu, 1989) x 1 y 1 x 2 y 2 /{x 1 }φ Dependence Logic (Väänänen 2007) x 1 y 1 x 2 y 2 (=(x 2, y 2 ) φ)

6 Characterizing dependence between variables 3/71 First Order Quantifiers x 1 y 1 x 2 y 2 φ Henkin Quantifiers (Henkin, 1961) ( x1 y 1 x 2 y 2 ) φ Independence Friendly Logic (Hintikka, Sandu, 1989) x 1 y 1 x 2 y 2 /{x 1 }φ Dependence Logic (Väänänen 2007) x 1 y 1 x 2 y 2 (=(x 2, y 2 ) φ)

7 Syntax of First-order Dependence Logic (D) 4/71 D = FO + =(t 1,..., t n ) Well-formed formulas of D (in negation normal form) are given by the following grammar φ ::= α =(t 1,..., t n ) =(t 1,..., t n ) φ φ φ φ xφ xφ where α is a first order literal and t 1,..., t n are first-order terms.

8 Syntax of First-order Dependence Logic (D) 5/71 D = FO + =(t 1,..., t n ) Well-formed formulas of D (in negation normal form) are given by the following grammar φ ::= α =(t 1,..., t n ) =(t 1,..., t n ) φ φ φ φ xφ xφ where α is a first order literal and t 1,..., t n are first-order terms.

9 6/71 Semantics of dependence logic: Team semantics D adopts team semantics, originally introduced by W. Hodges (1997) for IF-logic. Important points of team semantics: 1 satisfaction is defined w.r.t. sets of assignments (teams) instead of single assignments (Tarskian semantics) 2 the semantics is compositional

10 7/71 Theorem (Enderton, Walkoe, Väänänen) D sentences have the same expressive power as sentences of the second order Σ 1 1 fragment.

11 Intuitionistic Implication [Abramsky, Väänänen, 2009] 8/71 In a general context of W. Hodges team semantics, Abramsky, Väänänen introduced intuitionistic implication and Boolean disjunction., satisfy the Galois connection: φ ψ = χ φ = ψ χ,, satisfy axioms of intuitionistic propositional logic. first-order intuitionistic dependence logic: ID = D + +

12 Intuitionistic Implication [Abramsky, Väänänen, 2009] 8/71 In a general context of W. Hodges team semantics, Abramsky, Väänänen introduced intuitionistic implication and Boolean disjunction., satisfy the Galois connection: φ ψ = χ φ = ψ χ,, satisfy axioms of intuitionistic propositional logic. first-order intuitionistic dependence logic: ID = D + +

13 Intuitionistic Implication [Abramsky, Väänänen, 2009] 8/71 In a general context of W. Hodges team semantics, Abramsky, Väänänen introduced intuitionistic implication and Boolean disjunction., satisfy the Galois connection: φ ψ = χ φ = ψ χ,, satisfy axioms of intuitionistic propositional logic. first-order intuitionistic dependence logic: ID = D + +

14 9/71 First-order Intuitionistic Dependence Logic Theorem ([Abramsky, Väänänen 2009],[Y. 2010]) Sentences of ID have the same expressive power as sentences of the full second order logic.

15 PD and PID 10/71 The underlying propositional logic of D and ID are propositional dependence logic (PD) and propositional intuitionistic dependence logic (PID), respectively. Syntactically: PD = CPL+ =(p 1,..., p n ) PID = IPL+ =(p 1,..., p n ) PID is essentially equivalent to inquisitive logic [Ciardelli and Roelofsen, 2009], studied in the field of linguistics.

16 PD and PID 10/71 The underlying propositional logic of D and ID are propositional dependence logic (PD) and propositional intuitionistic dependence logic (PID), respectively. Syntactically: PD = CPL+ =(p 1,..., p n ) PID = IPL+ =(p 1,..., p n ) PID is essentially equivalent to inquisitive logic [Ciardelli and Roelofsen, 2009], studied in the field of linguistics.

17 11/71 Modal Intuitionistic Dependence Logic Modal Dependence Logic (Väänänen 2008) MD = Modal Logic (M)+ =(p 1,, p n ) Modal Intuitionistic Dependence Logic MID = MD + + Well-formed formulas of MID (in negation normal form) are defined by the following grammar : =(p 1,..., p n ) ϕ ::= p p =(p 1,, p n ) =(p 1,, p n ) ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

18 11/71 Modal Intuitionistic Dependence Logic Modal Dependence Logic (Väänänen 2008) MD = Modal Logic (M)+ =(p 1,, p n ) Modal Intuitionistic Dependence Logic MID = MD + + Well-formed formulas of MID (in negation normal form) are defined by the following grammar : =(p 1,..., p n ) ϕ ::= p p =(p 1,, p n ) =(p 1,, p n ) ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

19 11/71 Modal Intuitionistic Dependence Logic Modal Dependence Logic (Väänänen 2008) MD = Modal Logic (M)+ =(p 1,, p n ) Modal Intuitionistic Dependence Logic MID = MD + + Well-formed formulas of MID (in negation normal form) are defined by the following grammar : =(p 1,..., p n ) ϕ ::= p p =(p 1,, p n ) =(p 1,, p n ) ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

20 11/71 Modal Intuitionistic Dependence Logic Modal Dependence Logic (Väänänen 2008) MD = Modal Logic (M)+ =(p 1,, p n ) Modal Intuitionistic Dependence Logic MID = MD + + Well-formed formulas of MID (in negation normal form) are defined by the following grammar : =(p 1,..., p n ) ϕ ::= p p =(p 1,, p n ) =(p 1,, p n ) ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ

21 12/71 Modal Intuitionistic Dependence Logic Modal Dependence Logic (Väänänen 2008) MD = Modal Logic (M)+ =(p 1,, p n ) Modal Intuitionistic Dependence Logic MID = MD + + Well-formed formulas of MID (in negation normal form) are defined by the following grammar : ϕ ::= p p =(p 1,, p n ) =(p 1,, p n ) ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ ϕ MD = MID[, =( ),,,, ],

22 13/71 Team Semantics of MID Definition A Kripke model is a triple K = (S, R, π) consisting of a nonempty set S, a binary relation R S S, and a labeling function π : S (PROP). s 3 p 1 p 2 p 3 p 4 p 4

23 14/71 Teams Teams: sets of possible worlds s 3 p 1 p 1 p 2 p 3 s 5 p 4 p 5 p 4 team T = {, s 3, }

24 15/71 Operations on teams For any set T of states in a Kripke model K, we define R(T ) = {s K s T, s.t. s Rs}, s 3 p 1 p 1 p 2 s 5 p 3 p 4 p 4

25 16/71 Operations on teams For any set T of states in a Kripke model K, we define R(T ) = {s K s T, s.t. s Rs}, s 3 p 1 p 1 p 2 s 5 p 3 p 4 p 4

26 17/71 Operations on teams Let T be a team. Define T = {T T R(T ) and s T, R(s) T }. s 3 p 1 p 1 p 2 s 5 p 3 p 4 p 4

27 18/71 Operations on teams Let T be a team. Define T = {T T R(T ) and s T, R(s) T }. s 3 p 1 p 1 p 2 s 5 p 3 p 4 p 4 { } { }, s 5, s 3 Clearly, R(T ) T.

28 18/71 Operations on teams Let T be a team. Define T = {T T R(T ) and s T, R(s) T }. s 3 p 1 p 1 p 2 s 5 p 3 p 4 p 4 { } { }, s 5, s 3 Clearly, R(T ) T.

29 19/71 Team Semantics of MID Let K = (S, R, π) be a Kripke model, T S a team. Define K, T = ϕ iff K, R(T ) = ϕ K, T = ϕ iff there exists nonempty T T such that K, T = ϕ

30 20/71 Team Semantics of MID Let K = (S, R, π) be a Kripke model, T S a team. Define K, T = ϕ iff K, R(T ) = ϕ K, T = ϕ iff there exists nonempty T T such that K, T = ϕ K, T = p iff p π(s) for all s T K, T = p iff p / π(s) for all s T r p r T = {, } K, T = r K, T = p

31 Team Semantics of MID Let K = (S, R, π) be a Kripke model, T S a team. Define K, T = ϕ iff K, R(T ) = ϕ K, T = ϕ iff there exists nonempty T T such that K, T = ϕ K, T = p iff p π(s) for all s T K, T = p iff p / π(s) for all s T K, T = =(p 1,, p n, q) iff for any, T such that π( ) {p 1,, p n } = π( ) {p 1,, p n }, we have that π( ) {q} = π( ) {q} K, T = =(p 1,, p n, q) iff T = r p T = {, } K, T = =(r, p) r K, T = =(r) 21/71

32 Team Semantics of MID Let K = (S, R, π) be a Kripke model, T S a team. Define K, T = ϕ iff K, R(T ) = ϕ K, T = ϕ iff there exists nonempty T T such that K, T = ϕ K, T = p iff p π(s) for all s T K, T = p iff p / π(s) for all s T K, T = =(p 1,, p n, q) iff for any, T such that π( ) {p 1,, p n } = π( ) {p 1,, p n }, we have that π( ) {q} = π( ) {q} K, T = =(p 1,, p n, q) iff T = K, T = ϕ ψ iff K, T = ϕ and K, T = ψ K, T = ϕ ψ iff there are teams T 1, T 2 with T = T 1 T 2 such that K, T 1 = ϕ and K, T 2 = ψ p r T = {, } K, T = p r K, T = =(p) =(p) 22/71

33 Team Semantics of MID 23/71 Let K = (S, R, π) be a Kripke model, T S a team. Define K, T = ϕ iff K, R(T ) = ϕ K, T = ϕ iff there exists nonempty T T such that K, T = ϕ K, T = p iff p π(s) for all s T K, T = p iff p / π(s) for all s T K, T = =(p 1,, p n, q) iff for any, T such that π( ) {p 1,, p n } = π( ) {p 1,, p n }, we have that π( ) {q} = π( ) {q} K, T = =(p 1,, p n, q) iff T = K, T = ϕ ψ iff K, T = ϕ and K, T = ψ K, T = ϕ ψ iff there are teams T 1, T 2 with T = T 1 T 2 such that K, T 1 = ϕ and K, T 2 = ψ K, T = ϕ ψ iff K, T = ϕ or K, T = ψ K, T = ϕ ψ iff for any subteam T T if K, T = ϕ then K, T = ψ.

34 Team Semantics of MID 24/71 Let K = (S, R, π) be a Kripke model, T S a team. Define K, T = ϕ iff K, R(T ) = ϕ K, T = ϕ iff there exists nonempty T T such that K, T = ϕ K, T = p iff p π(s) for all s T K, T = p iff p / π(s) for all s T K, T = =(p 1,, p n, q) iff for any, T such that π( ) {p 1,, p n } = π( ) {p 1,, p n }, we have that π( ) {q} = π( ) {q} K, T = =(p 1,, p n, q) iff T = K, T = ϕ ψ iff K, T = ϕ and K, T = ψ K, T = ϕ ψ iff there are teams T 1, T 2 with T = T 1 T 2 such that K, T 1 = ϕ and K, T 2 = ψ K, T = ϕ ψ iff K, T = ϕ or K, T = ψ K, T = ϕ ψ iff for any subteam T T if K, T = ϕ then K, T = ψ.

35 25/71 Lemma (Empty team property) Empty team satisfies all formulas φ of MID in any Kripke model K, namely K, = φ.

36 Example 26/71 r 2 p 2 t 1 r 2 T = { t 1 } K, T = (r 2 =(p 2 ))

37 Example 27/71 r 2 p 2 t 1 r 2 T = { } t 1 K, T = (r 2 =(p 2 )) T 0 = { } K, T 0 = r 2 =(p 2 )

38 Simple properties 28/71 =(p 1,, p n, q) ( =(p 1 ) =(p n ) ) =(q) =(p) p p p p ; =(p 1,, p n ) =(p 1,, p n ) ;

39 29/71 Important Properties of MID Theorem (Downwards Closure) For any formula φ of MID, if K, T = φ and T T, then K, T = φ. Definition (Flatness) We say that φ is flat if for all Kripke models K and teams T K, T = φ (K, {s} = φ for all s T ). Theorem Formulas without any occurrences of =( ) and are flat, namely MID[,,,, ] formulas are flat.

40 29/71 Important Properties of MID Theorem (Downwards Closure) For any formula φ of MID, if K, T = φ and T T, then K, T = φ. Definition (Flatness) We say that φ is flat if for all Kripke models K and teams T K, T = φ (K, {s} = φ for all s T ). Theorem Formulas without any occurrences of =( ) and are flat, namely MID[,,,, ] formulas are flat.

41 29/71 Important Properties of MID Theorem (Downwards Closure) For any formula φ of MID, if K, T = φ and T T, then K, T = φ. Definition (Flatness) We say that φ is flat if for all Kripke models K and teams T K, T = φ (K, {s} = φ for all s T ). Theorem Formulas without any occurrences of =( ) and are flat, namely MID[,,,, ] formulas are flat.

42 30/71 Satisfaction Invariance Theorem Theorem Satisfaction of MID is invariant under generated submodels, p-morphic images, disjoint unions.

43 31/71 Satisfaction Invariance Theorem Theorem Satisfaction of MID is invariant under generated submodels, p-morphic images, disjoint unions.

44 32/71 Important Properties of MD [Sevenster 2009]: Theorem On singleton teams, MD = M. There is a translation from MD into the usual modal logic. But the translation causes an exponential blow-up in the size of the formulas. Theorem On arbitrary teams, MD > M.

45 32/71 Important Properties of MD [Sevenster 2009]: Theorem On singleton teams, MD = M. There is a translation from MD into the usual modal logic. But the translation causes an exponential blow-up in the size of the formulas. Theorem On arbitrary teams, MD > M.

46 33/71 Complexity results of MD Theorem (Sevenster 2009) Satisfiability problem for MD is NEXPTIME-complete. Theorem (Ebbing, Lohmann 2011) Model checking problem for MD is NP-complete.

47 34/71 MID Model Checking Problem Definition Let L be a sublogic (fragment) of MID. The model checking problem for L is defined as the decision problem of the set L MC := { K, T, ϕ K = (S, R, π) is a Kripke model, T S, ϕ L and K, T = ϕ }

48 35/71 Complexity results for fragments of MD-MC (Ebbing, Lohmann 2010) Operators Complexity Ref. =( ) NP [E& L 2011] NP [E& L 2011] + + NP [E& L 2011] in P [E& L 2011] in P [CES 1986] + : operator present : operator absent : complexity independent of operator

49 36/71 Computational Complexity of MID Model Checking Theorem Model checking problem for MID is PSPACE-complete.

50 37/71 Complexity results for fragments of MID-MC Operators Complexity Method/Ref. =( ) PSPACE reduct. from TQBF PSPACE see above PSPACE see above conp reduct. from TAUT in P [CES 1986] NP [E& L 2011] NP [E& L 2011] + + NP [E& L 2011] in P [E& L 2011] in P [E& L 2011] + : operator present : operator absent : complexity independent of operator

51 38/71 In the rest of the talk, we will prove: Theorem MID-MC is PSPACE complete. MID-MC is in PSPACE. MID-MC is PSPACE-hard.

52 39/71 Theorem MID-MC is in PSPACE.

53 PSPACE algorithm of MID-MC 40/71 check(k = (S, R, π), ϕ, T ): case ϕ when ϕ = p foreach s T i f not p π(s) then return f a l s e return t r u e when ϕ = p foreach s T i f p π(s) then return f a l s e return t r u e when ϕ = =(p 1,..., p n, q) foreach (s, s ) T T i f π(s) {p 1,..., p n} = π(s ) {p 1,..., p n} then i f ( q π(s) and not q π(s ) ) or ( not q π(s) and q π(s ) ) then return f a l s e return t r u e when ϕ = =(p 1,..., p n) i f S = return t r u e return f a l s e when ϕ = ψ χ e x i s t e n t i a l l y guess two sets of s t a t e s T 1, T 2 S i f not T 1 T 2 = T then return f a l s e return ( check ( K, T 1, ψ ) and check ( K, T 2, χ ) )

54 PSPACE algorithm of MID-MC (cont.) 41/71 when ϕ = ψ χ return ( check ( K, T, ψ ) or check ( K, T, ψ ) ) when ϕ = ψ χ return ( check ( K, T, ψ ) and check ( K, T, χ ) ) when ϕ = ψ T := foreach s S foreach s T i f (s, s ) R then T := T {s } return check ( K, T, ψ ) when ϕ = ψ e x i s t e n t i a l l y guess a set of s t a t e s T S foreach s T i f there i s no s T with (s, s ) R then return f a l s e return check ( K, T, ψ ) when ϕ = ψ χ u n i v e r s a l l y guess a set of s t a t e s T T i f ( not check ( K, ψ, T ) or check ( K, χ, T ) ) return t r u e return f a l s e

55 42/71 Next, we show Theorem MID-MC is PSPACE-hard. Proof. Reduction from TQBF.

56 43/71 Kripke model K ϕ Definition Let ϕ(p 1,, p n ) be a formula of CPL. We define a Kripke model K ϕ = (S ϕ, R ϕ, π ϕ ) by letting S ϕ := {,..., s n,,..., s n }, R ϕ :=, π ϕ (s i ) := {r i, p i }, π ϕ (s i ) := {r i }. r 1 p 1 s n r 2 r n p 2 p n r 1 r 2 r n s n

57 44/71 σ vs T ϕ,σ Let σ : Prop {, } be a valuation of CPL. The team T ϕ,σ of K ϕ induced by σ is defined as Example: T ϕ,σ = {s i S ϕ σ(p i ) = } {s i S ϕ σ(p i ) = }. σ: p 1 p 2... p n T ϕ,σ = {,,..., s n } r 1 p 1 s n r 2 r n p 2 p n r 1 r 2 r n s n

58 45/71 ϕ Definition Let r be a fixed propositional variable. For every formula ϕ of CPL in negation normal form without any occurrence of r, we inductively define a formula ϕ of MID as follows: p := r p, ( p) := r p, (ϕ ψ) := ϕ ψ, (ϕ ψ) := ϕ ψ.

59 46/71 In team T ϕ,σ of the Kripke model K ϕ, MID formula ϕ behaves like the CPL formula ϕ under valuation σ. Lemma For any formula ϕ and any valuation σ of CPL, σ(ϕ) = K ϕ, T ϕ,σ = ϕ. Proof. By induction on ϕ. Case ϕ = p: We have that S p = {s, s} and that σ(p) = T p,σ = {s} K p, T p,σ = r p. r p r s s

60 47/71 Complexity of MID Model Checking Theorem MID-MC is PSPACE-hard. Proof. We give a polynomial-time reduction from TQBF to MID-MC. Let ψ = x 1 x 2... x n 1 x n ϕ with ϕ quantifier-free (thus in CPL) be a QBF instance. The corresponding MID-MC instance is defined as (K, T 0, f (ψ)) where

61 47/71 Complexity of MID Model Checking Theorem MID-MC is PSPACE-hard. Proof. We give a polynomial-time reduction from TQBF to MID-MC. Let ψ = x 1 x 2... x n 1 x n ϕ with ϕ quantifier-free (thus in CPL) be a QBF instance. The corresponding MID-MC instance is defined as (K, T 0, f (ψ)) where

62 47/71 Complexity of MID Model Checking Theorem MID-MC is PSPACE-hard. Proof. We give a polynomial-time reduction from TQBF to MID-MC. Let ψ = x 1 x 2... x n 1 x n ϕ with ϕ quantifier-free (thus in CPL) be a QBF instance. The corresponding MID-MC instance is defined as (K, T 0, f (ψ)) where

63 48/71 K = (S, R, π), where S = 1 i n S i, R = 1 i n R i, for 1 i n/2 S 2i 1 = {i 1, i 1 } S 2i = {i, i } {t i } {t i1,, t i(i 1) } R 2i 1 = {(i 1, i 1 ), (i 1, i 1 )} R 2i = {(t i, t i1 ), (t i1, t i2 ),, (t i(i 2), t i(i 1) )} {(t i(i 1), i ), (t i(i 1), i )} {(i, i ), (i, i )} π(s j ) = {r j, p j }, for 1 j n π(t) =, for t / {s j, s j 1 j n}; T 0 = {s i, s i 1 i n, i odd} {t i 1 i n/2};

64 49/71 f : QBF MID is the reduction function defined by f (ψ) = ( (r1 =(p 1 ) ) ( (r3 =(p 3 ) ) ( (rn 1 =(p n 1 ) ) ϕ ))... )).

65 Model K 50/71 r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3.. s n 1 s n 1 r n 1 p n 1 r n 1 t n/2... s n s n r n p n r n

66 Model K 51/71 r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3.. s n 1 s n 1 r n 1 p n 1 r n 1 t n/2... s n s n r n p n r n

67 Model K 52/71 r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3.. s n 1 s n 1 r n 1 p n 1 r n 1 t n/2... s n s n r n p n r n

68 53/71 It suffices to show: for any QBF formula ψ = x 1 x 2... x n 1 x n ϕ, ψ TQBF K, T 0 = f (ψ).

69 An example 54/71 Let ψ = x 1 x 2 x 3 x 4 ϕ be a QBF formula with ϕ quantifier-free. Then f (ψ) = (r 1 =(p 1 )) ( (r 3 =(p 3 )) ϕ ) Claim: ψ TQBF K, T 0 = f (ψ). r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3

70 Proof of Claim = 55/71 = : Suppose x 1 x 2 x 3 x 4 ϕ TQBF, i.e. x 1 x 2 x 3 x 4 ϕ. We will show that K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 0 = { blue points }

71 Proof of Claim = 55/71 = : Suppose x 1 x 2 x 3 x 4 ϕ TQBF, i.e. x 1 x 2 x 3 x 4 ϕ. We will show that K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 0 = { blue points }

72 Proof of Claim = 56/71 = : Suppose x 1 x 2 x 3 x 4 ϕ TQBF, i.e. x 1 x 2 x 3 x 4 ϕ. Claim: K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 1 = { red points } K, T 1 = r 1 =(p 1 )

73 Proof of Claim = 57/71 = : Suppose x 1 x 2 x 3 x 4 ϕ. Claim: K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 1 = { red points } K, T 1 = r 1 =(p 1 ) σ: x 1

74 Proof of Claim = 58/71 = : Suppose x 1 x 2 x 3 x 4 ϕ. Claim: K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 1 = { red points } K, T 1 = r 1 =(p 1 ) σ: x 1, x 2

75 Proof of Claim = 59/71 = : Suppose x 1 x 2 x 3 x 4 ϕ. Claim: K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 2 = { blue points } σ: x 1, x 2

76 Proof of Claim = 60/71 = : Suppose x 1 x 2 x 3 x 4 ϕ. Claim: K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 3 = { red points } σ: x 1, x 2, x 3

77 Proof of Claim = = : Suppose x 1 x 2 x 3 x 4 ϕ. Claim: K, T 0 = (r 1 =(p 1 )) ((r 3 =(p 3 )) ϕ ) r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 4 = { blue points } σ: x 1, x 2, x 3, x 4 σ(ϕ) = It suffices to show K, T 4 = ϕ. 61/71

78 Proof of Claim = 62/71 But now, K r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3

79 Proof of Claim = 63/71 K r 1 p 1 r 1 r 2 p 2 r 2 s 3 r 3 p 3 r 4 p 4 r 3 r 4 s 3

80 Proof of Claim = 64/71 K r 1 p 1 r 1 r 2 p 2 r 2 s 3 r 3 p 3 r 4 p 4 r 3 r 4 s 3

81 Proof of Claim = 65/71 K ϕ r 1 p 1 r 1 r 2 p 2 r 2 s 3 r 3 p 3 r 4 p 4 r 3 r 4 s 3 T 4 = T ϕ,σ = { blue points } with σ: x 1, x 2, x 3, x 4 σ(ϕ) =

82 Proof of Claim = 65/71 K ϕ r 1 p 1 r 1 r 2 p 2 r 2 s 3 r 3 p 3 r 4 p 4 r 3 r 4 s 3 T 4 = T ϕ,σ = { blue points } with σ: x 1, x 2, x 3, x 4 σ(ϕ) =

83 Proof of Claim = 65/71 K ϕ r 1 p 1 r 1 r 2 p 2 r 2 s 3 r 3 p 3 r 4 p 4 r 3 r 4 s 3 T 4 = T ϕ,σ = { blue points } with σ: x 1, x 2, x 3, x 4 σ(ϕ) =

84 Proof of Claim = K ϕ r 1 p 1 r 1 r 2 p 2 r 2 s 3 r 3 p 3 r 4 p 4 r 3 r 4 s 3 T 4 = T ϕ,σ = { blue points } with σ: x 1, x 2, x 3, x 4 Hence σ(ϕ) = = K ϕ, T ϕ,σ = ϕ = K, T 4 = ϕ, since ϕ is modality-free = K, T 4 = ϕ since K is a generated submodel of K 66/71

85 Proof of Claim = K r 1 p 1 r 1 r 2 p 2 r 2 s 3 r 3 p 3 r 4 p 4 r 3 r 4 s 3 T 4 = T ϕ,σ = { blue points } with σ: x 1, x 2, x 3, x 4 Hence σ(ϕ) = = K ϕ, T ϕ,σ = ϕ = K, T 4 = ϕ, since ϕ is modality-free = K, T 4 = ϕ since K is a generated submodel of K 67/71

86 Proof of Claim = K r 1 p 1 r 1 t 1 r 2 p 2 r 2 s 3 r 3 p 3 r 3 t 2 r 4 p 4 r 4 s 3 T 4 = T ϕ,σ = { blue points } with σ: x 1, x 2, x 3, x 4 Hence σ(ϕ) = = K ϕ, T ϕ,σ = ϕ = K, T 4 = ϕ, since ϕ is modality-free = K, T 4 = ϕ, since K is a generated submodel of K 68/71

87 69/71 The other direction = is proved symmetrically. Hence Theorem MID-MC is PSPACE-hard. Theorem MID-MC is PSPACE-complete.

88 69/71 The other direction = is proved symmetrically. Hence Theorem MID-MC is PSPACE-hard. Theorem MID-MC is PSPACE-complete.

89 70/71 Complexity results for fragments of MID-MC Operators Complexity Method/Ref. =( ) PSPACE reduct. from TQBF PSPACE see above PSPACE see above conp reduct. from TAUT in P [CES 1986] NP [E& L 2011] NP [E& L 2011] + + NP [E& L 2011] in P [E& L 2011] in P [E& L 2011] + : operator present : operator absent : complexity independent of operator

90 71/71 That s all. Thank you!