First-order definability on ordered structures
Lax485149.FirstOrderDefinability · concepts/Lax485149/FirstOrderDefinability.lean · lax-485149
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Definition
A decision problem over a relational vocabulary is FO() definable when there is a first-order sentence over such that, for every nonempty finite -structure and every linear order on , is a yes-instance of if and only if . The sentence may use the order; the problem does not depend on it.
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| 1 | import Mathlib.ModelTheory.Order |
| 2 | import Mathlib.ModelTheory.Semantics |
| 3 | import Lax904597.Problems |
| 4 | import Lax904597.Interpretations |
| 5 | |
| 6 | /-! |
| 7 | --- |
| 8 | title: First-order definability on ordered structures |
| 9 | type: definition |
| 10 | --- |
| 11 | A decision problem over a relational vocabulary is FO() |
| 12 | definable when there is a first-order sentence over |
| 13 | such that, for every nonempty finite -structure |
| 14 | and every linear order on , is a yes-instance of if and only if |
| 15 | . The sentence may use the order; the problem |
| 16 | does not depend on it. |
| 17 | -/ |
| 18 | |
| 19 | namespace Lax485149.FirstOrderDefinability |
| 20 | |
| 21 | open Lax904597.Problems |
| 22 | |
| 23 | open FirstOrder |
| 24 | |
| 25 | open Language |
| 26 | |
| 27 | variable {L : Language.{0, 0}} [L.IsRelational] |
| 28 | |
| 29 | /-- A decision problem is *FO(≤) definable* if a single sentence over the |
| 30 | ordered expansion of its vocabulary decides it on nonempty finite ordered |
| 31 | structures. The equivalence is required for |
| 32 | *every* linear order, so the notion is order-invariant: the sentence sees the |
| 33 | order, the problem does not. -/ |
| 34 | def FODefinable (P : DecisionProblem L) : Prop := |
| 35 | ∃ φ : (L.sum Language.order).Sentence, |
| 36 | ∀ (A : Type) [L.Structure A] [LinearOrder A] [Finite A] [Nonempty A], P A ↔ A ⊨ φ |
| 37 | |
| 38 | end Lax485149.FirstOrderDefinability |
| 39 |
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