Finite Injury Priority Method

Qiang Yin

@BASICS, SJTU

August 15, 2014

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Post’s Problem

Post’s Problem: is there is a r.e. degree a such that 0 < a < 0′.

It was solved by Friedberg[1957] and Muchnik[1956] independently.

The method used to solve the problem was named finite injurypriority method.

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Finite Injury Priority Method

Recursively enumerate a set A =⋃

sAs to meet certainrequirements {Rn}n∈ω.

1 If n < m, then Rn is given priority over Rm.

2 Actions take for Rm at some stage s may at a later staget > s be undone when action is taken for for Rn, n < m.(Injury)

3 Finite Injury Property. Each requirement is injured at mostfinitely often.

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Overview

1 Low Simple SetConstruct a simple set A s.t. A′ ≡T ∅′

2 The Original Friedberg-Muchnik TheoremConstruct two r.e. set A and B such that A|TB

3 Sacks Splitting Theorems

Every nonrecursive r.e set can be split as a disjoint union two

incomparable r.e. subsets.

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Notation

I {e}As (x) = y if x, y, e < s, s > 0, φAe (x) = y in < s steps andonly numbers z < s are used the computation.

I The use function u(A; e, x, s) is 1+ the maximum numberused the computation if {e}As (y) ↓ and 0 otherwise.

LemmaLet {As}s∈ω be an enumeration such that A =

⋃sAs, then

I If {e}A(x) = y, then ∃s ∀t ≥ s {e}Att (x) = y.

I {e}Ass (x) = y and let r = u(As, e, x, s). If As�r = A�r then

{e}A(x) = y.

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Low Simple Set

TheoremThere is a simple set A such that A′ ≡T ∅′.Remark. The jump operator is not 1:1.

Corollary (Friedberg-Muchnik)

There is a nonrecursive incomplete r.e. degree a (i.e., 0 < a < 0′).

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Requirements

It’s sufficient to recursively enumerate a coinfinite r.e. setA =

⋃sAs to meet for all e the requirements:

(simplicity) Pe: We is infinite =⇒ We ∩A 6= ∅.

(lowness) Ne: (∃∞s) [{e}Ase (e) ↓] =⇒ {e}A(e) ↓.

Fact. {Ne}e∈ω implies A′ ≤T ∅′

g(e, s) =

{1 if {e}As

s (e) ↓;0 otherwise.

{Ne}e∈ω implies g(e) = lims g(e, s) exists for all e. But g ≤T ∅′,and hence A′ ≤T ∅′.

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Strategy to Meet Ne

Ne : (∃∞s) [{e}Ase (e) ↓] =⇒ {e}A(e) ↓

Given As define for all e

(restraint function) r(e, s) = u(As, e, e, s).

Observation. If {e}Ass (e) ↓ and Ne succeeds in preventing any

x ≤ r(e, s) from entering A, then {e}A(e) ↓.

Strategy. Restrain with priority Ne any elements x ≤ r(e, s) fromentering As+1.

Such elements can only enter A for the sake of some Pi withstronger priority.

N0 > P0 > N1 > P1 > N2 > P2 > . . .

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Construction

Stage s = 0. Let A0 = ∅.Stage s+ 1. Given As we have r(e, s) for all e. Choose the leasti ≤ s such that

Wi,s ∩As = ∅; (1)

and(∃x)[x ∈Wi,s ∧ x > 2i ∧ (∀e ≤ i)[r(e, x) < x]] (2)

I If i exists, chooses the least x satisfying (2), letAs+1 = As ∪ {x} and say that Pi receives attention. HenceWi ∩As+1 6= ∅ and Pi is satisfied for all stage > s+ 1.

I If i dose not exists, let As+1 = As.

Let A =⋃

sAs.

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Finite Injury Property

x injury Ne at stage s+ 1 if x ∈ As+1 −As and x ≤ r(e, s).Defined the injury set for Ne as

Ie = {x : (∃s)[x ∈ As+1 −As ∧ x ≤ r(e, s)]}

Lemma (FIP)

(∀e) [Ie is finite ].

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Correctness

LemmaFor every e requirement Ne is meet and r(e) = lims r(e, s) exists.

Proof.Fix e. By FIP, choose se such that Ne is not injured at any stages > se. If {e}As

s (e) converges for s > se then by induction ont ≥ s, r(e, t) = r(e, s) and {e}At

t (e) = {e}Ass (e) for all t ≥ s.

LemmaFor every i, requirement Pi is meet.

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The Original Friedberg-Muchnik Theorem

Recursively enumerate A and B to meet for all e the requirements:

R2e : A 6= {e}B,R2e+1 : B 6= {e}A.

Theorem (Friedberg 1957, Muchnik 1956)

There exsit r.e. set A and B such that A|TB, and hence∅ <T A,B <T ∅′.

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Strategy to Meet a Requirement

Attach to R2e a witness x not yet enumerate in A and look for astage s+ 1 such that

eBss (x) ↓= 0.

If no such stage exists then we do nothing.

If s+ 1 exists, we say R2e receive attention at stage s+ 1.

I Enumerate x into As+1;

I Rest the wall r(2e, s+ 1) = u(Bs; e, x, s);

I Restrain with priority R2e any numbers y ≤ r(2e, s+ 1) fromlater entering B.

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Construction of A and B

Stage s = 0.Set A0 = B0 = ∅, x0e = 〈0, e〉 and r(e, 0) = −1 for all e.

Stage s+ 1. Requirement R2e requires attention if

{e}Bss (xs2e) ↓= 0 and r(2e, s) = −1 (3)

and R2e+1 require attention if

{e}Ass (xs2e+1) ↓= 0 and r(2e+ 1, s) = −1 (4)

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Construction of A and B (cont’d)

Choose the least i ≤ s such that Ri requires attention and we sayRi receive attention by doing the follows. Suppose that i = 2e.

I Enumerate xs2e into As+1 and set xs+12e = xs2e;

I Reset the wall r(2e, s+ 1) = u(Bs, e, xs2e, s);

I For j < 2e, set r(j, s+ 1) = r(j, s) and xs+1j = xsj ;

I For j > 2e, set r(j, s+ 1) = −1 (Rj may require attention inthe future) and let xs+1

j to be

min{y ∈ ωj : y 6∈ As+1 ∪Bs+1 ∧ y > xsj∧ y > max{r(k, s+ 1) : k ≤ 2e}}

If i fails to exist then do nothing.

Remark. Ri can be injured at most 2i − 1 times.

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Correctness

LemmaFor every i, Ri receives attention at most finitely often and iseventually satisfied.

Proof.Fix i and assume by induction that the Lemma holds for all j < i.Choose s minimal so that no Rj , j < i received attention at astage t ≥ s.

1 ∀t ≥ s, xti = xsi = xi, and xi 6∈ As ∪Bs.

2 If Ri receive attention at some stage t ≥ s, then byassumption Ri will not injured thereafter, and hence Ri issatisfied.

3 If Ri never receive attention after stage s, then A(xi) = 0 andit is not the case that {e}D(xi) ↓= 0. (D = B if i is even andD = A if i is odd.)

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Sacks Theorem

Given a nonrecursive r.e. set C, recursively enumerate a coinfiniteset A =

⋃sAs that meets the requirements

Ne : C 6= {e}A;Pe : We is finite =⇒We ∩A 6= ∅.

Theorem (Sacks)

For every nonrecursive r.e. set C there is a simple set A such thatC 6≤T A ( and hence ∅ <T A <T ∅′).

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Strategy to Meet Ne

Sacks’s method for constructing A to meet Ne of the formC 6= {e}A is to preserve agreement between Cs and {e}As

s ratherthan disagreement as in the Friedberg strategy.

Sufficient preservation of agreement will guarantee that ifC = {e}A then C is recursive.

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Sacks’s Construction

Stage s = 0. Set A0 = ∅.Stage s+ 1. Given As, defined for all elength function:

l(e, s) = max{x : (∀y < x)[{e}Ass (y) ↓= Cs(y)]}

restraint function:

r(e, s) = max{u(As; e, x, s) : x ≤ l(e, s)}.

Pi requires attention at stage s+ 1 if i ≤ s, Wi,s ∩As = ∅, and

(∃x)[x ∈Wi,s ∧ x > 2i ∧ (∀e ≤ i)[r(e, s) < x]]

We enumerate in As+1 the least such x for each Pi requiresattention at s+ 1.

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Finite Injury Property

Injury Set:

Ie = {x : (∃s)[x ∈ As+1 −As ∧ x ≤ r(e, s)]}

Lemma (FIP)

Ie is finite for all e. (Indeed |Ie| < e.)

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Correctness

Lemma(∀e)[C 6= {e}A].

Proof.Assume that C = {e}A. Then lims l(e, s)=∞. By FIP we canchoose s′ such that Ne is never injured after stage s′. We shallrecursively compute C contrary to hypothesis.

To compute C(p) for some p ∈ ω find the least s > s′ such thatl(e, s) > p. It follows by induction on t ≥ s that

(∀t ≥ s)[l(e, t) > p ∧ r(e, t) > max{u(As; e, x, s) : x ≤ p}]

and hence C(p) = {e}AP = {e}Ass (p).

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Correctness (cont’d)

Lemma(∀e)[lims r(e, s) exists and is finite ].

Proof.Choose p = (µx)[C(x) 6= {e}A(x)]. Choose s′ sufficient large suchthat for all s ≥ s′

1 Ne is not injured at stage s;

2 (∀x ≤ p)[{e}Ass (x) ↓= {e}A(x)]; and

3 (∀x ≤ p)[Cs(x) = C(x)].

If {e}Att (p) ↓ for some t ≥ s′, then lims r(e, s) = r(e, t), otherwise

lims r(e, s) = r(e, s′).

Lemma(∀e)[We is infinite =⇒We ∩A 6= ∅].

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Sacks Splitting Theorem

Theorem (Sacks 1963)

Let B and C be r.e. sets such that C is nonrecursive. Then thereexist r.e. sets A0 and A1 such that

1 A0 ∪A1 = B and A0 ∩A1 = ∅;2 C 6≤T Ai for i = 0, 1; and

3 Ai ≤T ∅′ for i = 0, 1.

Corollary

Let C be a nonrecursive r.e. set. Then there exist low r.e. sets A0

and A1 such that A0|TA1, C = A0 ∪A1 and A0 ∩A1 = ∅.

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Requirements

Let {Bs}s∈ω and {Cs}s∈ω be recursive enumeration of B and Csuch that B0 = ∅ and |Bs+1 −Bs| = 1 for all s.

We give recursive enumerations {Ai,s}s∈ω, i = 0, 1 satisfying therequirements

P : x ∈ Bs+1 −Bs =⇒ [x ∈ A0,s+1 ∨ x ∈ Ai,s+1]

N〈e,i〉 : C 6= {e}Ai (i = 0, 1)

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Sacks’s Construction

Stage s = 0. Define Ai,0 = ∅, i = 0, 1.Stage s+ 1. Given i, s define the recursive function li(e, s) andri(e, s) as follows

li(e, s) = max{x : (∀y < x)[{e}Ais

s (y) ↓= Cs(y)]}

ri(e, s) = max{u(Ais; e, x, s) : x ≤ l(e, s)}.

Let x ∈ Bs+1 −Bs. Choose the 〈e′, i′〉 to be the least 〈e, i〉 suchthat x ≤ ri(e, s), and enumerate x in A1−i′,s+1.

If 〈e′, i′〉 fails to exist, enumerate x into A0.

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Correctness

LemmaFor all i = 0, 1 and all e

I Iie is finite.

I C 6= {e}Ai ; and

I ri(e) = lims ri(e, s) exists and is finite.

LemmaAi ≤T ∅′ for i = 0, 1.

Proof.Define a recursive function g as follows

g(e)Ai(y) =

{C(0) if y = 0 and {e}Ai(e) ↓,↑ otherwise.

e ∈ A′i ⇐⇒ {e}Ai(e) ↓ ⇐⇒ {g(e)}Ai(0) = C(0) ⇐⇒ lim

sli(g(e), s) > 0

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Thanks.August 15, 2014