first order theories
play

First-order theories Gabriele Puppis LaBRI / CNRS Definition Fix a - PowerPoint PPT Presentation

First-order theories Gabriele Puppis LaBRI / CNRS Definition Fix a class C of structures (e.g. graphs) and a logic L (e.g. FO). The L L L -theory of C C C is the set of all formulas in L that can be satisfied by some structure in C . The theory


  1. First-order theories Gabriele Puppis LaBRI / CNRS

  2. Definition Fix a class C of structures (e.g. graphs) and a logic L (e.g. FO). The L L L -theory of C C C is the set of all formulas in L that can be satisfied by some structure in C . The theory is decidable if there is an algorithm that receives formulas as input and tells whether they are in the theory or not.

  3. Definition Fix a class C of structures (e.g. graphs) and a logic L (e.g. FO). The L L L -theory of C C C is the set of all formulas in L that can be satisfied by some structure in C . The theory is decidable if there is an algorithm that receives formulas as input and tells whether they are in the theory or not. Examples first-order theory of the class of all graphs monadic theory of the class of all linear orders monadic theory of N monadic theory of the grid

  4. Undecidability of first-order theory One cannot decide whether a given formula of FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] is satisfied over some labelled grid labelled grid labelled grid.

  5. Undecidability of first-order theory One cannot decide whether a given formula of FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] is satisfied over some labelled grid labelled grid labelled grid. Given a Turing machine M , construct ψ M ψ M defining its halting runs : ψ M ● ● ● ● ⋯ ● ● ● ● ⋯ ● ● ● ● ⋯ ● ● ● ● ⋯ ⋮ ⋮ ⋮ ⋮ ⋱ q halt q halt q halt

  6. Undecidability of first-order theory One cannot decide whether a given formula of FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] is satisfied over some labelled grid labelled grid labelled grid. Given a Turing machine M , construct ψ M ψ M ψ M defining its halting runs : ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 0 q 0 q 0 encode initial configuration by top row 1 ● ● ● ● ⋯ ● ● ● ● ⋯ ● ● ● ● ⋯ ⋮ ⋮ ⋮ ⋮ ⋱ q halt q halt q halt

  7. Undecidability of first-order theory One cannot decide whether a given formula of FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] is satisfied over some labelled grid labelled grid labelled grid. Given a Turing machine M , construct ψ M ψ M ψ M defining its halting runs : ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 0 q 0 q 0 encode initial configuration by top row 1 ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 1 q 1 q 1 a a a encode next configurations by next rows 2 ● ● ● ● ⋯ q 2 ⊔ ⊔ ⊔ a q 2 q 2 a a b b b ● ● ● ● ⋯ ⊔ ⊔ ⊔ q 3 q 3 q 3 a a a b b b ⋮ ⋮ ⋮ ⋮ ⋱ q halt q halt q halt

  8. Undecidability of first-order theory One cannot decide whether a given formula of FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] is satisfied over some labelled grid labelled grid labelled grid. Given a Turing machine M , construct ψ M ψ M ψ M defining its halting runs : ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 0 q 0 q 0 encode initial configuration by top row 1 ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 1 q 1 q 1 a a a encode next configurations by next rows 2 ● ● ● ● ⋯ q 2 ⊔ ⊔ ⊔ a q 2 q 2 find a row with halting configuration a a b b b 3 ● ● ● ● ⋯ ⊔ ⊔ ⊔ q 3 q 3 q 3 a a a b b b ⋮ ⋮ ⋮ ⋮ ⋱ q halt q halt q halt

  9. Undecidability of first-order theory One cannot decide whether a given formula of FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] is satisfied over some labelled grid labelled grid labelled grid. (and equally for MSO [ E 1 , E 2 ] MSO [ E 1 , E 2 ] over the grid N × N MSO [ E 1 , E 2 ] N × N N × N ) Given a Turing machine M , construct ψ M ψ M ψ M defining its halting runs : ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 0 q 0 q 0 encode initial configuration by top row 1 ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 1 q 1 q 1 a a a encode next configurations by next rows 2 ● ● ● ● ⋯ q 2 ⊔ ⊔ ⊔ a q 2 q 2 find a row with halting configuration a a b b b 3 ● ● ● ● ⋯ ⊔ ⊔ ⊔ q 3 q 3 q 3 a a a b b b MSO can even guess the labelling! ⋮ ⋮ ⋮ ⋮ ⋱ q halt q halt q halt

  10. Undecidability of first-order theory One cannot decide whether a given formula of FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] FO [ Σ , E 1 , E 2 ] is satisfied over some labelled grid labelled grid labelled grid. (and equally for MSO [ E 1 , E 2 ] MSO [ E 1 , E 2 ] over the grid N × N MSO [ E 1 , E 2 ] N × N N × N ) Given a Turing machine M , construct ψ M ψ M ψ M defining its halting runs : ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 0 q 0 q 0 encode initial configuration by top row 1 ● ● ● ● ⋯ ⊔ ⊔ ⊔ ⊔ ⊔ ⊔ q 1 q 1 q 1 a a a encode next configurations by next rows 2 ● ● ● ● ⋯ q 2 ⊔ ⊔ ⊔ a q 2 q 2 find a row with halting configuration a a b b b 3 ● ● ● ● ⋯ ⊔ ⊔ ⊔ q 3 q 3 q 3 a a a b b b MSO can even guess the labelling! ⋮ ⋮ ⋮ ⋮ ⋱ q halt q halt q halt

  11. Consequences (Church ’36, Turing ’37, Trakhtenbrot ’50, ...) The FO theory of the class of all finite structures is undecidable (provided that signature contains a binary predicate besides = ).

  12. Consequences (Church ’36, Turing ’37, Trakhtenbrot ’50, ...) The FO theory of the class of all finite structures is undecidable (provided that signature contains a binary predicate besides = ). The MSO theory of any class of graphs with unbounded grids as minors unbounded grids as minors unbounded grids as minors is undecidable. . . . . . . . . . . . . n ⋮ ⋮ ⋮ ⋮ ⋮ . . . . . . . . . . . . ⋮ ⋮ ⋮ ⋮ ⋮ . . . . . . . . . . . . ⋮ ⋮ ⋮ ⋮ ⋮ . . . . . . . . . . . . 4 n + 20

  13. Consequences (Church ’36, Turing ’37, Trakhtenbrot ’50, ...) The FO theory of the class of all finite structures is undecidable (provided that signature contains a binary predicate besides = ). The MSO theory of any class of graphs with unbounded grids as minors unbounded grids as minors unbounded grids as minors is undecidable. The MSO theory of ( N , +) ( N , +) ( N , +) is undecidable. . . . . . . . . . . . . n + 1 n + 2 n + 3 n + 4 n ⋮ ⋮ ⋮ ⋮ ⋮ . . . . . . . . . . . . 2 n + 2 2 n + 4 2 n + 6 2 n + 8 2 n ⋮ ⋮ ⋮ ⋮ ⋮ . . . . . . . . . . . . 3 n + 3 3 n + 6 3 n + 9 3 n + 12 3 n ⋮ ⋮ ⋮ ⋮ ⋮ . . . . . . . . . . . . 4 n + 4 4 n + 8 4 n + 16 4 n + 20 4 n

  14. Definition Presburger arithmetic is the first-order theory of ( N , + ) ( N , + ) ( N , + ) Examples of Presburger formulas ∃ x . ∀ y . ( x + y = y ) = ψ 0 ϕ ≤ ( x , y ) ∃ z . ( y = x + z ) = ϕ 2 × ( x , y ) ( x + x = y ) = ∀ x . ∃ y . ( x ≤ y ∧ ¬ x = y ) = ψ ω

  15. Decidability of Presburger arithmetic (Presburger ’29) One can decide if a Presburger sentence ψ holds over ( N , + ) . Originally proved by quantifier elimination . Here we use automata!

  16. Decidability of Presburger arithmetic (Presburger ’29) One can decide if a Presburger sentence ψ holds over ( N , + ) . Originally proved by quantifier elimination . Here we use automata! Encode numbers x ∈ N by reverse binary expansions [ x ] ∈ B ⋆ [ x ] ∈ B ⋆ [ x ] ∈ B ⋆ 1 e.g. [ 4 ] = 001, [ 0 ] = ε , . . .

  17. Decidability of Presburger arithmetic (Presburger ’29) One can decide if a Presburger sentence ψ holds over ( N , + ) . Originally proved by quantifier elimination . Here we use automata! Encode numbers x ∈ N by reverse binary expansions [ x ] ∈ B ⋆ [ x ] ∈ B ⋆ [ x ] ∈ B ⋆ 1 e.g. [ 4 ] = 001, [ 0 ] = ε , . . . Encode sum relation + ⊆ N × N × N by language L + ⊆ ( B × B × B ) ⋆ L + ⊆ ( B × B × B ) ⋆ L + ⊆ ( B × B × B ) ⋆ 2 1 1 0 e.g. [ + ( 3 , 1 , 4 ) ] [ 3 ] ⊗ [ 1 ] ⊗ [ 4 ] ( 0 ) ( 0 ) ( 1 ) = = 1 0 0

  18. Decidability of Presburger arithmetic (Presburger ’29) One can decide if a Presburger sentence ψ holds over ( N , + ) . Originally proved by quantifier elimination . Here we use automata! Encode numbers x ∈ N by reverse binary expansions [ x ] ∈ B ⋆ [ x ] ∈ B ⋆ [ x ] ∈ B ⋆ 1 e.g. [ 4 ] = 001, [ 0 ] = ε , . . . Encode sum relation + ⊆ N × N × N by language L + ⊆ ( B × B × B ) ⋆ L + ⊆ ( B × B × B ) ⋆ L + ⊆ ( B × B × B ) ⋆ 2 1 1 0 e.g. [ + ( 3 , 1 , 4 ) ] [ 3 ] ⊗ [ 1 ] ⊗ [ 4 ] ( 0 ) ( 0 ) ( 1 ) = = 1 0 0 0 ( 1 ) 0 A + ∶ p q 1 ( 0 ) 1 1 0 1 ( 0 ) , ( 0 1 1 ) , ( 1 ) 0 ( 0 ) , ( 0 ) , ( 1 ) 0 1 1 0 0 1

  19. Decidability of Presburger arithmetic (Presburger ’29) One can decide if a Presburger sentence ψ holds over ( N , + ) . We inductively translate every Presburger formula ϕ ( x 1 , ..., x m ) into a finite automaton A ϕ over Σ m = B m such that L ( A ϕ ) = { [ x 1 ] ⊗ ⋅ ⋅ ⋅ ⊗ [ x m ] ∈ Σ ⋆ L ( A ϕ ) = { [ x 1 ] ⊗ ⋅ ⋅ ⋅ ⊗ [ x m ] ∈ Σ ⋆ L ( A ϕ ) = { [ x 1 ] ⊗ ⋅ ⋅ ⋅ ⊗ [ x m ] ∈ Σ ⋆ m ∣ ( N , + ) ⊧ ϕ ( x 1 , ..., x m ) } m ∣ ( N , + ) ⊧ ϕ ( x 1 , ..., x m ) } m ∣ ( N , + ) ⊧ ϕ ( x 1 , ..., x m ) } so as to reduce satisfiability of ϕ to emptiness of L ( A ϕ ) .

Recommend


More recommend