Logics Definition A logic ⊢ is a consequence relation over the set of formulas Fm of The computational complexity of the Leibniz an algebraic language,which is substitution invariant in the sense hierarchy that if Γ ⊢ ϕ , then σ ( Γ ) ⊢ σ ( ϕ ) for all substitutions σ : Fm → Fm . Tommaso Moraschini ◮ Logics are consequence relations (as opposed to sets of valid Institute of Computer Science of the Czech Academy of Sciences formulas). June 28, 2017 ◮ Example: IPC is the logic defined as follows: Γ ⊢ IPC ϕ ⇐ ⇒ for every Heyting algebra A and � a ∈ A , if Γ A ( � a ) = 1, then ϕ A ( � a ) = 1 . 1 / 14 2 / 14 Relative equational consequence Algebraizable logics Example: Consider Definition IPC = intuitionistic propositional logic Let K be a class of similar algebras. Given a set of equations HA = variety of Heyting algebras Θ ∪ { ϕ ≈ ψ } , we define ◮ Pick the translations between formulas and equations: Θ � K ϕ ≈ ψ ⇐ ⇒ for every A ∈ K and � a ∈ A , ϕ �− → ϕ ≈ 1 if ǫ A ( � a ) = δ A ( � a ) for all ǫ ≈ δ ∈ Θ , α ≈ β �− → { α ↔ β } . then ϕ A ( � a ) = ψ A ( � a ) . ◮ These translations allow to equi-interpret ⊢ IPC and ⊢ HA : The relation � K is the equational consequence relative to K. Γ ⊢ IPC ϕ ⇐ ⇒{ γ ≈ 1 : γ ∈ Γ } � HA ϕ ≈ 1 Θ � HA ϕ ≈ ψ ⇐ ⇒{ α ↔ β : α ≈ β ∈ Θ } ⊢ IPC { ϕ ↔ ψ } . ◮ Example: If K is the variety of Heyting algebras, then ◮ Moreover, the translations are one inverse to the other: ϕ ≈ 1 , ϕ → ψ ≈ 1 � K ψ ≈ 1 . ϕ ≈ ψ = || = HA ϕ ↔ ψ ≈ 1 and ϕ ⊣⊢ IPC ϕ ↔ 1 . ◮ Hence ⊢ IPC and � HA are essentially the same. 3 / 14 4 / 14
Algebraization Problem ◮ Intuitive idea: a logic ⊢ is algebraizable when it can be essentially identified with a relative equational consequence � K . ◮ We study the computational aspects of the following problem: Definition Algebraization Problem A logic ⊢ is algebraizable when there exists: Given a logic ⊢ , determine whether ⊢ is algebraizable or not. 1. A class of algebras K (of the same type as ⊢ ); 2. A set of equations τ ( x ) in one variable x ; ◮ Logic can be presented (at least) in two ways: 3. A set of formulas ρ ( x , y ) in two variables x and y such that τ and ρ equi-interpret ⊢ and � K : syntactically = by means of Hilbert calculi semantically = by means of collections of logical matrices . Γ ⊢ ϕ ⇐ ⇒ τ ( Γ ) � K τ ( ϕ ) Θ � K ϕ ≈ ψ ⇐ ⇒ ρ (Θ) ⊢ ρ ( ϕ, ψ ) Theorem (M. 2015) and the two interpretations are one inverse to the other: The Algebraization Problem for logics presented by finite consistent ϕ ≈ ψ = || = K τρ ( ϕ, ψ ) and ϕ ⊣⊢ ρτ ( ϕ ) . Hilbert calculi is undecidable. 5 / 14 6 / 14 Semantic Algebraization Problem A useful EXPTIME -complete problem Given a finite reduced logical matrix � A , F � of finite type, ◮ We want to prove that the Semantic Algebraization Problem is determine whether its induced logic is algebraizable or not. complete for EXPTIME . ◮ We need to construct a polynomial-time reduction to such a ◮ There is an easy decision procedure for this problem because: complete problem. Theorem The Problem Gen-Clo Let � A , F � be a finite reduced matrix and ⊢ its induced logic. ⊢ is Given a finite algebra A of finite type and a function h : A n → A , algebraizable iff there is a finite set of equations τ ( x ) and a finite determine whether h belongs to the clone of A or not. set of formulas ρ ( x , y ) such that ◮ Gen-Clo 1 3 is the same problem, restricted to the case where h a = b ⇐ ⇒ ρ ( a , b ) ⊆ F is unary and the operations of A are at most ternary. a ∈ F ⇐ ⇒ A � τ ( a ) . Theorem (Bergman, Juedes, and Slutzki) ◮ Since finitely generated free algebras over V ( A ) are finite, we Both Gen-Clo and Gen-Clo 1 3 are complete for EXPTIME . can just check the existence of the sets ρ ( x , y ) and τ ( x ) . ◮ Hence the Semantic Algebraization Problem is in EXPTIME . ◮ We will construct a polynomial reduction of Gen-Clo 1 3 to the Semantic Algebraization Problem. 7 / 14 8 / 14
3. Finally we add to A ♭ the following operation ♥ : Reduction if a m = c k and h ( a ) 5 = b n a 1 3 . We define a new algebra A ♭ as: Pick an input � A , h � for Gen-Clo 1 and m ∈ { 1 , 3 , 4 } ◮ The universe of A ♭ is eight disjoint copies A 1 , . . . , A 8 of A : if a m = c k a 2 An arbitrary finite set of elements in A ♭ can be denote as and h ( a ) 5 = b n and m ∈ { 2 , 5 , 6 , 7 , 8 } ♥ ( a m , b n , c k ) := a 4 { a m 1 1 , . . . , a m n if m , k ∈ { 1 , 3 , 4 } n } and (either a m � = c k or h ( a ) 5 � = b n ) for some a 1 , . . . , a n ∈ A and m 1 , . . . , m n ≤ 8. a 7 if { m , k } ∩ { 2 , 5 , 6 , 7 , 8 } � = ∅ and ◮ The basic operation of A ♭ are as follows: (either a m � = c k or h ( a ) 5 � = b n ). f on A ♭ as 1. For every n -ary basic f of A , we add an operation ˆ ◮ Then define F ⊆ A ♭ as follows: F := A 1 ∪ A 2 . ˆ f ( a m 1 . . . , a m n n ) := f A ( a 1 , . . . , a n ) 5 . ◮ The pair � A ♭ , F � is a finite reduced matrix of finite type, and 1 thus an input for the Semantic Algebraization Problem! 2. Then we add to A ♭ the following operation ✷ : a m Remark if m = 1 or m = 2 ✷ ( a m ) := a m − 1 if m is even and m ≥ 3 Since the arity of the operations of A is bounded by 3, the matrix a m + 1 if m is odd and m ≥ 3. � A ♭ , F � can be constructed in polynomial time. 9 / 14 10 / 14 → � A ♭ , F � can be used to Hardness result ◮ Variants of the construction A �− show that Theorem Theorem The problem of determining whether the logic of a finite reduced There is a polynomial-time reduction of Gen-Clo 1 3 to the Semantic matrix of finite type belongs to any of the following classes Algebraization Problem, i.e. given a finite algebra A of finite type, algebraizable logics whose basic operations are at most ternary, and a unary map protoalgebraic logics h : A → A , TFAE: equivalential logics 1. h belongs to the clone of A . truth-equational logics 2. The logic induced by the matrix � A ♭ , F � is algebraizable. order algebraizable logics, is hard for EXPTIME . Corollary The Semantic Algebraization Problem is complete for EXPTIME . ◮ For all the above classes of logics (except the one of truth-equational logics), the problem is complete for EXPTIME . 11 / 14 12 / 14
Further questions Finally... ◮ A similar situation appears in the study of Malsetv conditions: Theorem (Freese and Valeriote) The problem of determining whether a finite algebra A of finite type generates a congruence distributive (resp. modular) variety is complete for EXPTIME . ...thank you for coming! ◮ However, the above problems become tractable when A is idempotent, i.e when for every operation f of A and a ∈ A f A ( a , . . . , a ) = a Open Problem Find tractability conditions for Semantic Algebraization Problem. ◮ Remark: idempotency will not work here, since no idempotent non-trivial matrix determines an algebraizable logic. 13 / 14 14 / 14
Recommend
More recommend
