reflection ranks and proof theoretic ordinals
play

Reflection ranks and proof theoretic ordinals (based on joint work - PowerPoint PPT Presentation

Reflection ranks and proof theoretic ordinals (based on joint work with James Walsh) Fedor Pakhomov Steklov Mathematical Institute, Moscow pakhf@mi.ras.ru Logic Colloquium 2018, Udine 24 July 2018 Con -Order def U proves consistence


  1. Reflection ranks and proof theoretic ordinals (based on joint work with James Walsh) Fedor Pakhomov Steklov Mathematical Institute, Moscow pakhf@mi.ras.ru Logic Colloquium 2018, Udine 24 July 2018

  2. ≺ Con -Order def ⇒ U proves consistence of T . T ≺ Con U ⇐ Empirical fact: ≺ Con is a linear well-founded preorder on natural theories I Σ 1 ≺ Con . . . ≺ Con I Σ n ≺ Con PA ≡ Con ACA 0 ACA 0 ≺ Con Π 1 1 -CA 0 ≺ Con Π 1 2 -CA 0 ≺ Con . . . ≺ Con Π 1 ∞ -CA 0 = PA 2 PA 2 ≺ Con PA 3 ≺ Con . . . ≺ Con PA ∞ ≡ Con Z Z ≺ Con Z +∆ 0 -Coll ≺ Con Z +Π 1 -Coll ≺ Con . . . ≺ Con Z +Π ∞ -Coll = ZF ZF ≺ Con ZFC + ∃ κ κ is inaccessible ≺ Con . . . Although it is possible to construct artificial examples of descending chains consisting of true theories. T 0 ≻ Con T 1 ≻ Con T 2 ≻ Con . . .

  3. Π 1 1 soundndess and Π 1 1 reflection Let T be an r.e. extension of ACA 0 . ACA 0 = PA + second order axiom of induction + ∃ X ∀ x ( ϕ ( n ) ↔ x ∈ X ) , for all arithmetical ( Π 0 ∞ ) formulas ϕ ( x ) . The Π 1 1 ( T ) is Π 1 1 reflection principle RFN Π 1 1 sentence expressing T is Π 1 1 -sound, e.g. T proves only true Π 1 1 sentences. More formally RFN Π 1 1 ( T ) is given by the sentence ∀ ϕ ∈ Π 1 1 ( Prv ( T , ϕ ) → Tr Π 1 1 ( ϕ )) , 1 ( x ) is the partial truth definition for Π 1 where Tr Π 1 1 formulas.

  4. Well-foundedness in reflection order We put def T ≺ Π 1 ⇐ ⇒ U ⊢ RFN Π 1 1 ( T ) . 1 U Note that T ≺ Π 1 1 U ⇒ T ≺ Con U . Theorem 1 on Π 1 The restriction of ≺ Π 1 1 -sound extensions of ACA 0 is a well-founded relation.

  5. Proof of Well-Foundedness of ≺ Π 1 1 The negation of our theorem is the sentence DS 1 starting with Π 1 DS: “there is a descending chain in ≺ Π 1 1 -sound r.e. theory” We will show that ACA 0 + DS ⊢ Con ( ACA 0 + DS ) . Then by G¨ odel’s second incompleteness theorem ACA 0 + DS is inconsistent and hence ACA 0 ⊢ ¬ DS. Let us reason in ACA 0 + DS. We have sequence T 0 ≻ Π 1 1 T 1 ≻ Π 1 1 . . . , where T 0 is Π 1 1 -sound. Let S be the Σ 1 1 -sentence saying that “there is a descending sequence in ≺ Π 1 1 starting from T 1 .” Since S is true and T 0 is Π 1 1 -sound, there is a (countably coded) model M | = T 0 + S But since T 0 proves Π 1 1 -soundness of T 1 , M | = DS .

  6. The case of RCA 0 Over RCA 0 there are no truth definition for the class Π 1 1 but there are truth definitions for smaller classes Π 1 1 (Π 0 n ) , e.g. formulas of the form ∀ � X ϕ , where ϕ ∈ Π 0 n . And we have reflection principles RFN Π 1 n ) ( T ) . 1 (Π 0 Theorem 3 ) on Π 1 1 (Π 0 The restriction of ≺ Π 1 3 ) -sound extensions of RCA 0 is 1 (Π 0 a well-founded relation. Clarification : Note that we need partial truth definition for class of formulas Γ to make reflection principle RFN Γ a single sentence. Otherwise we put RFN Γ be the scheme ∀ � x ( Prv ( T , ϕ ( � x )) → ϕ ( � x )) , where ϕ ∈ Γ .

  7. Reflection in first-order arithmetic Over the system of first-order arithmetic EA we have partial truth definitions Tr Π 0 n ( x ) and reflection principles RFN Π 0 n ( T ) . Theorem (Friedman, Smorynski, Solovay) There are no recursive sequences of theories � T i | i ∈ N � such that T 0 is consistent and EA ⊢ ∀ x Prv ( T x , � Con ( T x + 1 ) � ) . Theorem There are no recursive sequences of theories � T i | i ∈ N � such that T 0 is Π 0 3 -sound and T 0 ≻ Π 0 3 T 1 ≻ Π 0 3 . . .

  8. Recursive descending chains Recursive descending chain in ≺ Π 0 2 : T 0 ≻ Π 0 2 T 1 ≻ Π 0 2 T 2 ≻ Π 0 2 . . . 2 ( PA ) or RFN p − n T n : I Σ 1 + “ either RFN Π 0 2 ( I Σ 1 ) , where p is G¨ odel Π 0 number of the first proof of false Σ 0 1 sentence in PA” Note that all T n are true arithmeical theories.

  9. Reflection Rank For an r.e. extension T of ACA 0 we put | T | ACA 0 = α if T is in well-founded part of ≺ Π 1 1 and α is it’s well-founded rank | T | ACA 0 = ∞ , otherwise More standard measure is Π 1 1 proof-theoretic ordinal: | T | WO = sup {| α | | α is recursive linear order and T ⊢ WO ( α ) } . Reflection ranks and proof-theoretic ordinals of some theories: | · | ACA 0 | · | WO ACA 0 0 ε 0 ACA 0 + Con ( ACA 0 ) 0 ε 0 ACA 0 + RFN Π 1 1 ( ACA 0 ) 1 ε 1 ACA ′ ω ε ω 0 ACA ε 0 ε ε 0 ACA + ϕ ( 2 , 0 ) ϕ ( 2 , 0 ) 0 ATR 0 Γ 0 Γ 0

  10. Iterations of reflection principles For recursive ordinal notations α we could define iterations RFN α Γ ( T ) : ◮ RFN 0 Γ ( T ) = T ◮ RFN α + 1 ( T ) = T + RFN Γ ( RFN α Γ ( T )) Γ ◮ RFN λ RFN α Γ ( T ) = � Γ ( T ) , λ ∈ Lim . α<λ Theorem (Turing) For each true Π 1 sentence F there is recursive ordinal notation α Con α ( PA ) ⊢ F . Theorem (Feferman) For each true Π 0 ∞ sentence F there is recursive ordinal notation α RFN α ∞ ( PA ) ⊢ F . Π 0

  11. Iterations of Π 1 1 -reflection Theorem RFN α 3 ) RFN ε α 1 ( ACA 0 ) ≡ Π 1 3 ) ( RCA 0 ) 1 (Π 0 Π 1 Π 1 1 (Π 0 Proposition | RFN β 3 ) ( RCA 0 ) | RCA 0 = | β | Π 1 1 (Π 0 Proposition ACA 0 ⊢ ∀ α ( WO ( α ) ↔ RFN α + 1 3 ) ( RCA 0 )) Π 1 1 (Π 0 Corollary | RFN α 1 ( ACA 0 ) | WO = | ε α | . Π 1

  12. Proving RFN α 3 ) RFN ε α 1 ( ACA 0 ) ≡ Π 1 3 ) ( RCA 0 ) 1 (Π 0 Π 1 Π 1 1 (Π 0 Let us consider pseudo- Π 1 1 language Π 0 ∞ , i.e. arithmetical formulas ϕ ( X ) with free unary predicate X . We have embedding of pseudo- Π 1 1 language into second-order arithmetic ϕ ( X ) �− → ∀ X ϕ ( X ) . RFN α ∞ RFN α 1 ( ACA 0 ) ≡ Π 0 ∞ ( PA ( X )) , Π 1 Π 0 RFN α 3 RFN α 3 ) ( RCA 0 ) ≡ Π 0 3 ( I Σ 1 ( X )) . Π 1 1 (Π 0 Π 0 Schmerl-style formula for uniform pseudo- Π 1 1 reflection RFN α 3 RFN ε α ∞ ( PA ( X )) ≡ Π 0 3 ( I Σ 1 ) Π 0 Π 0 Thus RFN α 1 ( ACA 0 ) ≡ Π 0 ∞ RFN α ∞ ( PA ( X )) ≡ Π 0 3 RFN ε α 3 ( I Σ 1 ) ≡ Π 0 3 RFN ε α 3 ) ( RCA 0 ) Π 1 Π 0 Π 0 Π 1 1 (Π 0

  13. Calculus RC 0 Beklemishev approach to proof of Schmerl formula employs ordinal notation system based on reflection principles. Reflection calculus RC: Formulas: F ::= ⊤ | F ∧ F | ✸ n F , where n ranges over N . Sequents: A ⊢ B , for RC-formulas A and B . 1. A ⊢ A ; A ⊢ ⊤ ; if A ⊢ B and B ⊢ C then A ⊢ C ; 2. A ∧ B ⊢ A ; A ∧ B ⊢ B ; if A ⊢ B and A ⊢ C then A ⊢ B ∧ C ; 3. if A ⊢ B then ✸ n A ⊢ ✸ n B , for all n ∈ N ; 4. ✸ n ✸ n A ⊢ ✸ n A , for every n ∈ N ; 5. ✸ n A ⊢ ✸ m A , for all n > m ; 6. ✸ n A ∧ ✸ m B ⊢ ✸ n ( A ∧ ✸ m B ) , for all n > m .

  14. Beklemishev’s Ordinal Notation System def A < 0 B ⇐ ⇒ B ⊢ ✸ 0 A def A ∼ B ⇐ ⇒ A ⊢ B and B ⊢ A Theorem (Beklemishev) ( RC 0 / ∼ , < 0 ) is a well-ordering with order type ε 0 . It were done by Beklemishev by embedding this system in Cantor ordinal notation system for ε 0 .

  15. Well-Foundedness Proof Let us interpret RC-formulas by L 2 -theories. We interpret ⊤ as ⊤ ⋆ = ACA 0 . And we interpret ✸ n A as ( ✸ n A ) ⋆ = RFN Π 1 n + 1 ( A ⋆ ) . It is easy to see that A ⊢ B implies A ⋆ ⊢ B ⋆ . Hence A < 0 B implies A ⋆ < Π 1 1 B ⋆ . Thus < 0 is a well-founded relation on the set of RC 0 formulas.

  16. Thank You!

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend


More recommend