Bar recursion over finite partial functions Thomas Powell (joint - PowerPoint PPT Presentation

Bar recursion over finite partial functions Thomas Powell (joint work with Paulo Oliva) University of Innsbruck CCC 2015 Kochel am See, Germany 15 September 2015 Thomas Powell (Innsbruck) Bar recursion over partial functions 1 / 32

Backward recursion in the continuous functionals Outline 1 Backward recursion in the continuous functionals 2 The computational interpretation of countable choice 3 Backward recursion as a learning realizer Thomas Powell (Innsbruck) Bar recursion over partial functions 2 / 32

Backward recursion in the continuous functionals Spector’s bar recursion BR g,h,ϕ : ρ ∗ → σ is defined by � g ( s ) if ϕ (ˆ s ) < len( s ) BR g,h,ϕ ( s ) = σ h s ( λx . BR g,h,ϕ ( s ∗ x )) otherwise Thomas Powell (Innsbruck) Bar recursion over partial functions 3 / 32

Backward recursion in the continuous functionals Spector’s bar recursion BR g,h,ϕ : ρ ∗ → σ is defined by � g ( s ) if ϕ (ˆ s ) < len( s ) BR g,h,ϕ ( s ) = σ h s ( λx . BR g,h,ϕ ( s ∗ x )) otherwise Recursion input s : X ∗ a finite sequence; recursive calls made over all one-element extensions s ∗ x of s ; g assigns base values to the recursion. Thomas Powell (Innsbruck) Bar recursion over partial functions 3 / 32

Backward recursion in the continuous functionals Spector’s bar recursion BR g,h,ϕ : ρ ∗ → σ is defined by � g ( s ) if ϕ (ˆ s ) < len( s ) BR g,h,ϕ ( s ) = σ h s ( λx . BR g,h,ϕ ( s ∗ x )) otherwise Recursion input s : X ∗ a finite sequence; recursive calls made over all one-element extensions s ∗ x of s ; g assigns base values to the recursion. Termination � s ( k ) if k < | s | otherwise i.e. a canonical embedding of s into ρ N ; s := λk. ˆ 0 ϕ : ρ N → N controls the recursion, terminating it when Spector’s point ϕ (ˆ s ) is less than the length of the input i.e. ϕ (ˆ s ) < len( s ) holds. Thomas Powell (Innsbruck) Bar recursion over partial functions 3 / 32

Backward recursion in the continuous functionals Bar recursion exists in continuous models by a standard argument: Thomas Powell (Innsbruck) Bar recursion over partial functions 4 / 32

Backward recursion in the continuous functionals Bar recursion exists in continuous models by a standard argument: 1. Extend. If BR ( s 0 ) = ⊥ there exists an infinite sequence s 0 ≺ s 1 ≺ s 2 ≺ . . . satisfying ϕ (ˆ s i ) ≥ len( s i ) and s i +1 = s i ∗ x i and BR ( s i +1 ) = ⊥ Thomas Powell (Innsbruck) Bar recursion over partial functions 4 / 32

Backward recursion in the continuous functionals Bar recursion exists in continuous models by a standard argument: 1. Extend. If BR ( s 0 ) = ⊥ there exists an infinite sequence s 0 ≺ s 1 ≺ s 2 ≺ . . . satisfying ϕ (ˆ s i ) ≥ len( s i ) and s i +1 = s i ∗ x i and BR ( s i +1 ) = ⊥ 2. Limit. Let α : ρ N be the domain-theoretic limit of the s i i.e. � α := s i = λk . s k +1 ( k ) i ∈ N Thomas Powell (Innsbruck) Bar recursion over partial functions 4 / 32

Backward recursion in the continuous functionals Bar recursion exists in continuous models by a standard argument: 1. Extend. If BR ( s 0 ) = ⊥ there exists an infinite sequence s 0 ≺ s 1 ≺ s 2 ≺ . . . satisfying ϕ (ˆ s i ) ≥ len( s i ) and s i +1 = s i ∗ x i and BR ( s i +1 ) = ⊥ 2. Limit. Let α : ρ N be the domain-theoretic limit of the s i i.e. � α := s i = λk . s k +1 ( k ) i ∈ N 3. Continuity. The value of ϕ ( α ) depends only on some finite initial segment [ α (0) , . . . , α ( N − 1)] of its argument. Take any M ≥ N, ϕ ( α ) + 1. Then ϕ ( α ) < ϕ ( α ) + 1 ≤ M ≤ len( s M ) ϕ (ˆ s M ) = ���� continuity: N ≤ M Thomas Powell (Innsbruck) Bar recursion over partial functions 4 / 32

Backward recursion in the continuous functionals What is so useful about bar recursion? One answer: self-reference . Suppose Q ( s, i ) is some predicate on ρ ∗ × N , and that whenever ∀ k < len( s ) Q ( s, k ) we can compute an extension a s : ρ such that ∀ k < len( s ∗ a s ) Q ( s ∗ a s , k ) . Then bar recursion allows us to compute a chain [] ≺ s 1 ≺ s 2 ≺ . . . ≺ s M with ∀ k < len( s i ) Q ( s i , k ) for each i , and moreover s M is a leaf with ϕ (ˆ s M ) < len( s M ) therefore we have Q ( s M , ϕ (ˆ s M )) We will see why this in important in Part 2! Thomas Powell (Innsbruck) Bar recursion over partial functions 5 / 32

Backward recursion in the continuous functionals Symmetric bar recursion sBR g,h,ϕ : ρ † → σ is defined by  g ( u ) if ϕ (ˆ u ) ∈ dom( u )  sBR g,h,ϕ ( u ) = σ h s ( λx . sBR φ,b,ϕ ( u ⊕ ( ϕ (ˆ u ) , x ) )) otherwise  Thomas Powell (Innsbruck) Bar recursion over partial functions 6 / 32

Backward recursion in the continuous functionals Symmetric bar recursion sBR g,h,ϕ : ρ † → σ is defined by � u ) ∈ dom( u ) g ( u ) if ϕ (ˆ sBR g,h,ϕ ( u ) = σ h s ( λx . sBR φ,b,ϕ ( u ⊕ ( ϕ (ˆ u ) , x ))) otherwise Recursion input u : ρ † a finite partial function; recursive calls made over one-element domain-theoretic extensions u ⊕ ( ϕ (ˆ u ) , x ) of u ; g assigns base values to the recursion. Thomas Powell (Innsbruck) Bar recursion over partial functions 6 / 32

Backward recursion in the continuous functionals Symmetric bar recursion sBR g,h,ϕ : ρ † → σ is defined by � u ) ∈ dom( u ) g ( u ) if ϕ (ˆ sBR g,h,ϕ ( u ) = σ h s ( λx . sBR φ,b,ϕ ( u ⊕ ( ϕ (ˆ u ) , x ))) otherwise Recursion input u : ρ † a finite partial function; recursive calls made over one-element domain-theoretic extensions u ⊕ ( ϕ (ˆ u ) , x ) of u ; g assigns base values to the recursion. Termination � u ( k ) if k ∈ dom( u ) , a canonical embedding of u into ρ N ; u := λk. ˆ 0 otherwise ϕ : ρ N → N controls recursion, terminating when Spector’s point is in the u ) ∈ dom( u ) holds. domain of u i.e. ϕ (ˆ Thomas Powell (Innsbruck) Bar recursion over partial functions 6 / 32

Backward recursion in the continuous functionals Need to adapt standard argument: Thomas Powell (Innsbruck) Bar recursion over partial functions 7 / 32

Backward recursion in the continuous functionals Need to adapt standard argument: 1. Extend. If sBR ( u 0 ) = ⊥ there exists an infinite sequence u 0 ⊏ u 1 ⊏ u 2 ⊏ . . . satisfying n i := ϕ (ˆ u i ) / ∈ dom( u i ) and u i +1 = u i ⊕ ( n i , x i ) and sBR ( u i +1 ) = ⊥ Thomas Powell (Innsbruck) Bar recursion over partial functions 7 / 32

Backward recursion in the continuous functionals Need to adapt standard argument: 1. Extend. If sBR ( u 0 ) = ⊥ there exists an infinite sequence u 0 ⊏ u 1 ⊏ u 2 ⊏ . . . satisfying n i := ϕ (ˆ u i ) / ∈ dom( u i ) and u i +1 = u i ⊕ ( n i , x i ) and sBR ( u i +1 ) = ⊥ 2. Limit. Let α : N → ρ ⊥ be the domain-theoretic limit of the u i i.e. � u i ( k ) ( k ) where i ( k ) least s.t. k ∈ dom( u i ( k ) ) � α := u i = λk . undefined if no such index exists . i ∈ N α : ρ N denote the canonical extenion: Let ˆ � u i ( k ) ( k ) where i ( k ) least s.t. k ∈ dom( u i ( k ) ) α = λk . ˆ 0 ρ if no such index exists . Thomas Powell (Innsbruck) Bar recursion over partial functions 7 / 32

Backward recursion in the continuous functionals 3. Continuity. The value of ϕ (ˆ α ) depends only on some finite initial segment [ˆ α (0) , . . . , ˆ α ( N − 1)] of its argument. α ) + 1. Since α = � u i there exists some I such that Take any M ≥ N, ϕ (ˆ ∀ i < M ( u I ( i ) = α ( i )) , or equivalently, ∀ i < M (ˆ u I ( i ) = ˆ α ( i )) which implies that n I := ϕ (ˆ u I ) = ϕ (ˆ α ) < ϕ (ˆ α ) + 1 ≤ M. ���� continuity: N ≤ M ∈ dom( u I ) and n I < M we have n I / ∈ dom( α ). But Since n I / u I +1 = u I ⊕ ( n I , x I ) , and since u I +1 ⊏ α we have n I ∈ dom( α ), a contradiction. Therefore n I = ϕ (ˆ u I ) ∈ dom( u I ). Thomas Powell (Innsbruck) Bar recursion over partial functions 8 / 32

Backward recursion in the continuous functionals Summary: Two ways of achieving self-reference Spector’s bar recursion � g ( s ) if ϕ (ˆ s ) < len( s ) BR g,h,ϕ ( s ) = σ h s ( λx . BR g,h,ϕ ( s ∗ x )) otherwise makes recursive calls over the tree s 0 ≺ s 1 ≺ s 2 ≺ . . . until it reaches a leaf s M such that ϕ ( ˆ s M ) < len( s M ). This tree is well-founded in continuous models. Symmetric bar recursion generalises this idea: � g ( u ) if ϕ (ˆ u ) ∈ dom( u ) BR g,h,ϕ ( u ) = σ h s ( λx . BR φ,b,ϕ ( u ⊕ ( ϕ (ˆ u ) , x ))) otherwise making recursive calls over the tree u 0 ⊏ u 1 ⊏ u 2 ⊏ . . . until it reaches a leaf u M such that ϕ (ˆ u M ) ∈ dom( u M ). This tree is well-founded in continuous models. Thomas Powell (Innsbruck) Bar recursion over partial functions 9 / 32

Backward recursion in the continuous functionals Some facts about symmetric bar recursion, taken from (Oliva/P. 2015). Thomas Powell (Innsbruck) Bar recursion over partial functions 10 / 32

Backward recursion in the continuous functionals Some facts about symmetric bar recursion, taken from (Oliva/P. 2015). Theorem 1. BR is primitive recursively definable from sBR , provably in E - HA ω (extensional Heyting arithmetic in all finite types). Proof. Straightforward. Thomas Powell (Innsbruck) Bar recursion over partial functions 10 / 32