XIV. Arithmetic Hierarchy Yuxi Fu BASICS, Shanghai Jiao Tong - - PowerPoint PPT Presentation

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XIV. Arithmetic Hierarchy Yuxi Fu BASICS, Shanghai Jiao Tong - - PowerPoint PPT Presentation

Oct 22, 2022 394 likes •745 views

XIV. Arithmetic Hierarchy Yuxi Fu BASICS, Shanghai Jiao Tong University We introduce a hierarchy of sets in terms of logical formula and prove its relationship to the hierarchy 0 , 0 , 0 , . . . of Turing degree. Computability Theory,

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  • XIV. Arithmetic Hierarchy

Yuxi Fu

BASICS, Shanghai Jiao Tong University

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We introduce a hierarchy of sets in terms of logical formula and prove its relationship to the hierarchy 0, 0′, 0′′, . . . of Turing degree.

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Synopsis

  • 1. Arithmetic Hierarchy
  • 2. Post Theorem
  • 3. Σn-Complete Set
  • 4. Relative Arithmetic Hierarchy

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Arithmetic Hierarchy

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Arithmetic Hierarchy

A set B is in Σ0 (Π0) if B is recursive. A set B is in Σn, where n ≥ 1, if there is a recursive relation R(x, y1, y2, . . . , yn) such that x ∈ B iff ∃y1.∀y2.∃y3. . . . Qnyn.R(x, y1, y2, . . . , yn). A set B is in Πn, where n ≥ 1, if there is a recursive relation R(x, y1, y2, . . . , yn) such that x ∈ B iff ∀y1.∃y2.∀y3. . . . Qnyn.R(x, y1, y2, . . . , yn). ∆n = Σn ∩ Πn.

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Arithmetic Set

B is arithmetical if B ∈

n∈ω(Σn ∪ Πn).

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Basic Property

  • Theorem. The following hold:

(i) A ∈ Σn iff A ∈ Πn. (ii) If A ∈ Σn (Πn) then ∀m>n.A ∈ Σm ∩ Πm. (iii) If A, B ∈ Σn (Πn) then A ∪ B, A ∩ B ∈ Σn (Πn). (iv) If R ∈ Σn ∧ n > 0 ∧ A = {x : (∃y)R(x, y)} then A ∈ Σn. (v) If B ≤m A ∧ A ∈ Σn then B ∈ Σn. (vi) If R ∈ Σn (Πn) and A, B are defined by x, y ∈ A ⇔ ∀z < y.R(x, y, z), x, y ∈ B ⇔ ∃z < y.R(x, y, z), then A, B ∈ Σn (Πn).

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Fin ∈ Σ2

  • Fact. Fin ∈ Σ2.

x ∈ Fin ⇔ Wx is finite ⇔ ∃s.∀t. (t ≤ s ∨ Wx,t = Wx,s) .

  • Fact. Inf ∈ Π2.

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Cof ∈ Σ3

  • Fact. Cof ∈ Σ3.

x ∈ Cof ⇔ Wx is finite ⇔ ∃y.∀z. (z ≤ y ∨ z ∈ Wx) ⇔ ∃y.∀z.∃s. (z ≤ y ∨ z ∈ Wx,s) .

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Tot ∈ Π2

  • Fact. {x, y | Wx ⊆ Wy} ∈ Π2.

Wx ⊆ Wy ⇔ ∀z. (z ∈ Wx ⇒ z ∈ Wy) ⇔ ∀z. (z / ∈ Wx ∨ z ∈ Wy) ⇔ ∀z. (∀s.z / ∈ Wx,s ∨ ∃t.z ∈ Wy,t) ⇔ ∀z.∀s.∃t. (z / ∈ Wx,s ∨ z ∈ Wy,t) ⇔ ∀w.∃t.

  • (w)0 /

∈ Wx,(w)1 ∨ (w)0 ∈ Wy,t

  • .
  • Fact. {x, y | Wx = Wy} ∈ Π2.
  • Fact. Tot = {x | Wx = ω} ∈ Π2.

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Rec ∈ Σ3

  • Fact. Rec ∈ Σ3.

x ∈ Rec ⇔ Wx is recursive ⇔ ∃y.

  • Wx = Wy
  • ⇔

∃y. (Wx ∩ Wy = ∅ ∧ Wx ∪ Wy = ω) ⇔ ∃y. ((∀s.Wx,s ∩ Wy,s = ∅) ∧ (∀z.∃s.z ∈ Wx,s ∪ Wy,s)) .

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Ext ∈ Σ3

  • Fact. Ext ∈ Σ3.

x ∈ Ext ⇔ ∃y. (φx ⊆ φy ∧ Wy = ω) ⇔ ∃y.∀z.∃s.∃t. ((z / ∈ Wx,s ∨ φx,s(z) = φy,s(z)) ∧ z ∈ Wy,t) .

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Crt ∈ Σ3

  • Fact. Crt = {x | Wx is creative} ∈ Σ3.

x ∈ Crt ⇔ Wx is productive ⇔ ∃y.∀z.

  • Wz ⊆ Wx ⇒ (φy(z) ↓ ∧ φy(z) ∈ Wx \ Wz)
  • ⇔

∃y.∀z. (Wz ∩ Wx = ∅ ⇒ (φy(z) ↓ ∧ φy(z) / ∈ Wx ∪ Wz)) ⇔ ∃y.∀z. (Wz ∩ Wx = ∅ ∨ (φy(z) ↓ ∧ φy(z) / ∈ Wx ∪ Wz)) Now Wz ∩ Wx = ∅ iff ∃s.Wz,s ∩ Wx,s = ∅, and φy(z) ↓ ∧ φy(z) / ∈ Wx ∪ Wz iff ∃s.z ∈ Wy,s ∧ ∀s.(z / ∈ Wy,s ∨ φy,s(z) / ∈ Wx,s ∪ Wz,s).

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Let PTM be {x | Px runs in polynomial time}. x ∈ PTM ⇔ ∃c.∀z. (Px(z) terminates in czc) Hence PTM ∈ Σ2.

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Post Theorem

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Completeness

A set A ∈ Σn is Σn-complete if B ≤1 A for every B ∈ Σn. A set A ∈ Πn is Πn-complete if B ≤1 A for every B ∈ Πn.

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Post Theorem

(i) B ∈ Σn+1 iff B is r.e. in a Πn set iff B is r.e. in a Σn set.

Proof.

If B ∈ Σn+1, then x ∈ B iff ∃y.R(x, y) for some R ∈ Πn. So B is r.e. in {x, y | R} ∈ Πn. Suppose B is r.e. in some C ∈ Πn. Then for some e, x ∈ B iff x ∈ W C

e iff ∃s.∃σ.(σ ⊂ C ∧ x ∈ W σ e,s).

Now x ∈ W σ

e,s is recursive, and σ ⊂ C is C-recursive since

σ ⊂ C iff ∀y < |σ|.(σ(y) = 1 ∧ y ∈ C ∨ σ(y) = 0 ∧ y / ∈ C). Hence B ∈ Σn+1.

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Post Theorem

(ii) ∅(n) is Σn-complete for all n > 0.

Proof.

∅′ = K is Σ1-complete. Now assume ∅(n) is Σn-complete. Then B ∈ Σn+1 iff B is r.e. in some Σn set iff B is r.e. in ∅(n) iff B ≤1 ∅(n+1). Hence ∅(n+1) is Σn+1-complete.

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Post Theorem

(iii) B ∈ Σn+1 iff B is r.e. in ∅(n). (iv) B ∈ ∆n+1 iff B ≤T ∅(n).

Proof.

We have the following equivalence: B ∈ ∆n+1 iff B, B ∈ Σn+1 iff B, B are r.e. in ∅(n) iff B ≤T ∅(n).

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Hierarchy Theorem. ∀n > 0.∆n ⊂ Σn ∧ ∆n ⊂ Πn.

Proof.

∅(n) ∈ Σn \ Πn and ∅(n) ∈ Πn \ Σn.

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A Comment on Completeness

B ≤m ∅(n) ⇒ B ∈ Σn ⇒ B is r.e. in ∅(n−1) ⇒ B ≤1 ∅(n) ⇒ B ≤m ∅(n). The following is the relativized version of “K ≤m A iff K ≤1 A”: ∅(n) ≤m A iff ∅(n) ≤1 A.

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Σn-Complete Set

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Let (A1, A2) and (B1, B2) be two pairs of sets such that A1 ∩ A2 = ∅ and B1 ∩ B2 = ∅. Then (A1, A2) ≤m (B1, B2) if there is a recursive function f such that f (A1) ⊆ B1, f (A2) ⊆ B2 and f (A1 ∪ A2) ⊆ B1 ∪ B2. We write (A1, A2) ≤1 (B1, B2) if f is one-one. For n > 0 we write (Σn, Πn) ≤m (C, D) if (A, A) ≤m (C, D) for some Σn-complete set A. The notation (Σn, Πn) ≤1 (C, D) is defined similarly.

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Fin is Σ2-Complete, Tot is Π2-Complete

  • Theorem. (Σ2, Π2) ≤1 (Fin, Tot).

Proof.

Fin ∈ Σ2 and Tot ∈ Π2. Let A be in Σ2. There is a recursive relation R such that x ∈ A iff ∀y.∃z.R(x, y, z). By S-m-n Theorem there is a one-one recursive function s s.t. φs(x)(u) = 0, if ∀y ≤ u.∃z.R(x, y, z), ↑,

  • therwise.

Now x ∈ A ⇒ Ws(x) = ω ⇒ s(x) ∈ Tot and x ∈ A ⇒ Ws(x) is finite ⇒ s(x) ∈ Fin.

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Cof and Rec are Σ3-Complete

Let Cmp be {x | Wx ≡T K}, the set of Turing complete r.e. sets.

  • Theorem. (Σ3, Π3) ≤1 (Cof , Cmp) ≤1 (Rec, Cmp).
  • Corollary. Cof is Σ3-complete.
  • Corollary. (Rogers) Rec is Σ3-complete.

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(Σ3, Π3) ≤1 (Cof , Cmp)

Fix an A ∈ Σ3. Then some R ∈ Π2 exists such that x ∈ A iff ∃y.R(x, y). Since Inf is Π2-complete, a one-one recursive function g exists s.t. R(x, y) iff Wg(x,y) is infinite. We will construct an r.e. set Wf (x) =

s∈ω W s f (x) in stages s.t.

x ∈ A iff Wf (x) is cofinite.

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(Σ3, Π3) ≤1 (Cof , Cmp)

Let the elements of the cofinite set W s

f (x) be denoted by

bs

x,0 < bs x,1 < bs x,2 < . . . < bs x,k < . . . .

Let W 0

f (x) := ∅.

Let W s+1

f (x) := W s f (x). Additionally put bs x,k in W s+1 f (x) if k ≤ s and

Wg(x,k),s = Wg(x,k),s+1 ∨ k ∈ Ks+1 \ Ks. So we have constructed some programme Pf (x) that enumerates Wf (x), from which we can calculate f (x).

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(Σ3, Π3) ≤1 (Cof , Cmp)

If x ∈ A, then Wg(x,k) is infinite for some k; and |Wf (x)| ≤ k. If x / ∈ A, then Wg(x,k) is finite for all k. There is a stage when the first k + 1 elements bx,0 < bx,1 < bx,2 < . . . < bx,k of Wf (x) have all been fixed. So Wf (x) is infinite. Conclude that A ≤1 Cof . To prove A ≤1 Cmp, we show that if x / ∈ A then K ≤T Wf (x). For each k we can Wf (x)-recursively calculate a stage s(k) such that bs(k)

x,k = bx,k. Notice that k ∈ K iff k ∈ Ks(k).

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Relative Arithmetic Hierarchy

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Relative Post Theorem

Relative Post Theorem. For every n ≥ 0, the following hold: (i) A(n+1) is ΣA

n+1-complete.

(ii) B ∈ ΣA

n+1 iff B is r.e. in A(n).

(iii) B ≤T A(n) iff B ∈ ∆A

n+1.

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Low Degree and High Degree

A degree a ≤ 0′ is low if a′ = 0′. A degree a ≤ 0′ is high if a′ = 0′′.

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Low Degree

  • Theorem. For A ≤T ∅′, the following are equivalent:

(i) A is low. (ii) ΣA

1 ⊆ Π2.

(iii) A′ ≤1 ∅(2).

Proof.

The following equivalences hold: A is low iff A′ ≤T ∅′ iff A′ ∈ ∆2, Post Theorem, iff ΣA

1 ⊆ ∆2, A′ is ΣA 1 complete,

iff ΣA

1 ⊆ Π2, ΣA 1 ⊆ Σ∅′ 1 = Σ2,

iff A′ ≤1 ∅(2), ∅(2) is Π2 complete.

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High Degree

  • Theorem. For A ≤T ∅′, the following are equivalent:

(i) A is high. (ii) Σ2 ⊆ ΠA

2 .

(iii) ∅(2) ≤1 A(2).

Proof.

The following equivalences hold: A is high iff ∅′′ ≤T A′ iff ∅′′ ∈ ∆A

2 ,

iff Σ2 ⊆ ∆A

2 , ∅′′ is Σ2 complete,

iff Σ2 ⊆ ΠA

2 , Σ2 ⊆ ΣA 2 ,

iff ∅(2) ≤1 A(2), A(2) is ΠA

2 complete.

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