Arithmetical Hierarchy Peter Mayr Computability Theory, February 26, 2021 DTM vs functions onN For a partial functionf write � f(x) ifx is in the domain off; ↓ � f(x) ifx is not in the domain off. ↑ ϕe (x):N p N is computed by the DTM with Goedel numbere → Facts � A N k is computably enumerable iffA is the domain of some ⊆ partial recursive function. � A N k is computable iff the characteristic function ofA is ⊆ recursive. � The Diagonal Halting Problem K := x N:ϕ x (x) { ∈ ↓} is c.e. but not computable. Properties of recursive functions are not computable Rice’s Theorem LetC be a class ofk-ary recursive functions. Then e N:ϕ e C is computable iffC= orC is the class of all { ∈ ∈ } ∅ k-ary recursive functions. Example None of the following are computable: � K := x N:ϕ x (x) { ∈ ↓} � F := x N:ϕ x hasfinite domain { ∈ } � T := x N:ϕ x is total { ∈ } The arithmetical hierarchy of subsets ofN Idea: Classify problems that are not computable by the complexity of formulas that describe them. Example (Diagonal halting problemK) x Kiffϕ (x) ∈ x ↓ iff y(config(x,x,y)) =t ∃ 0 computable predicate �stating computation�� � halts after y steps Definition LetP(¯x)beak-ary predicate onN,n N: ∈ 0 � P isΣ n if there is a computable predicateR: P(¯x) y y y ... / y :R(¯x,¯y) ≡ ∃ 1∀ 2∃ 3 ∃ ∀ n n alternating quantifiers starting with ∃ 0 � �� � � P isΠ n if there is a computable predicateR: P(X) y y y ... / y :R(¯x,¯y) ≡ ∀ 1∃ 2∀ 3 ∃ ∀ n n alternating quantifiers starting with ∀ 0 0 � �� � � Σ0 =Π 0 = computable predicates � Δ0 :=Σ 0 Π 0 n n ∩ n Note The superscript 0 denotes quantification over type-0-objects (elements inN). Example 0 1. K isΣ 1 2. T= e N:ϕ e is total { ∈ } e Tiff xϕ (x) ∈ ∀ e ↓ iff x y(config(e,x,y)) =t ∀ ∃ 0 0 HenceT isΠ 2. 3. F= e N:ϕ e hasfinite domain { ∈ } e Fiff z y x(config(e,x,y)) =t x z ∈ ∃ ∀ ∀ 0 ⇒ ≤ 0 HenceF isΣ 2. Arithmetical hierarchy 0 0 n N Σn = n N Πn (sets defined infirst order arithmetic, ∈ ∈ hence called arithmetical) � � . 0 0 Σ2 Π2 0 Δ2 0 0 Σ1 Π1 0 Δ1 Question Are all subsets ofN arithmetical? Closure properties For proving the previous picture we need some preparation. Lemma Letn 1. ≥ 0 0 1. Σn is closed under existential quantification,Π n is closed under universal quantification. 2. Σ0,Π 0 are both closed under , , bounded quantifiers n n ∧ ∨ x<y, x<y, and substitution of total computable ∀ ∃ functions. 3. Σ0 =Π 0, Π 0 =Σ 0 ¬ n n ¬ n n Proof sketch. LetR be computable,P(x,z)= y y ...R(x,z,y ,...) beΣ 0. ∃ 1 ∀ 2 1 n 1. Claim:Q(x) := z P(x,z) isΣ 0 ∃ n Q(x) z y y ...R(x,z,y ,y ,...,y ) ≡∃ ∃ 1 ∀ 2 1 2 n u y ...R(x,(u) ,(u) ,y ...,y ) ≡ ∃ ∀ 2 0 1 2 n Dual argument for andΠ 0. ∀ n 2. Substitution: Letf(x) total, computable. 0 Claim:Q(x) :=P(x,f(x)) isΣ n Q(x) y y ...R(x,f(x),y ,y ,...) ≡∃ 1 ∀ 2 1 2 computable sinceR is Conjunction: Induct� onn (HW).�� � 3. Negation: immediate. 0 Σ1 is computably enumerable Normal Form Theorem for c.e. sets 0 P is c.e. iffP isΣ 1. Proof : LetP N k be c.e. ⇒ ⊆ (k) � ThenP= domainϕ e for somee (HW). � x Piffϕ e (x) ∈ ↓ iff n(config(e,x,n)) =t ∃ 0 =:ϕe,n (x) ↓ 0 � P isΣ because the predicateϕ e,n (x) (“M e computes 1 � �� � ↓ ϕe (x) inn steps”) is computable. : LetP(x) y R(x,y) forR computable. ⇐ ≡∃ � Thenψ(x) :=µy R(x,y) is recursive. � P= domainψ is c.e. 0 DuallyΠ 1 is co-c.e. 0 UniversalΣ n predicates 0 0 Idea: Enumeratek-aryΣ n predicates by a singlek + 1-aryΣ n predicate. Definition 0 0 Ak + 1-aryΣ n predicateU(e,x) is universalΣ n fork-ary predicates if 0 1. U(e,x) isΣ n 0 2. for everyk-aryΣ n predicateP(x) there is somee such that P(x) U(e,x) ≡ 0 UniversalΠ n-predicates are defined similarly. Example From the last proofU(e,x) := nϕ (x) is universalΣ 0. ∃ e,n ↓ 1 Enumeration Theorem For allk,n 1, universalΣ 0- andΠ 0-predicates exist. ≥ n n Proof by induction onn andk: Base case:U(e,x) := nϕ (x) is universalΣ 0 by the Normal ∃ e,n ↓ 1 Form Theorem for c.e. predicates. Note: IfU(e,x) is universalΣ 0, then U(e,x) is universalΠ 0 n ¬ n (and conversely). 0 Induction step: LetU(e,y,¯x) be universalΣ n fork + 1-ary predicates. Then y U(e,y,¯x) is universalΠ 0 fork-ary predicates since ∀ n+1 0 1. it is inΠ n+1 and 0 2. for everyk-aryΠ n+1-predicateP(¯x) there exists ak + 1-ary 0 Σn-predicateQ(y,¯x) such that P(¯x)= y Q(y,¯x) ∀ The arithmetical hierarchy does not collapse Corollary For eachn 1 there existΣ 0-predicates that are notΠ 0 (and ≥ n n conversely). Proof. 0 � LetU n(e,x) a unary universalΣ n-predicate. 0 � ThenK n(x) :=U n(x,x) isΣ n. 0 0 � Seeking a contradiction, supposeK n isΠ . Then K isΣ . n ¬ n � Hence K n(x) U n(e,x) for somee. ¬ ≡ � Contradiction forx=e..