Effective embeddings into strong degreestructures

Timothy H. McNicholl∗

July 27, 2007

AbstractWe show that any partial order with a Σ3 enumeration can be

effectively embedded into any partial order obtained by imposing astrong reducibility such as ≤tt on the c.e. sets. As a consequence,we obtain that the partial orders that result from imposing a strongreducibility on the sets in a level of the Ershov hiearchy below ω + 1are co-embeddable.

Primary classification number : 03.

Keywords : computability theory, strong reducibilities, Erhsov hierar-chy.

1 Introduction

We start with the following question of Odifreddi.

Is the partial order of the c.e. truth-table degrees isomorphic to the partialorder of the truth-table degrees below the truth-table degree of the halting

set?

By a result of Epstein et. al [5], the truth-table degrees below the truthtable degree ofK are precisely the truth-table degrees of the ω-c.e. sets. Thus,before investigating Nies’s question, it makes sense to look at the following.

∗This research was supported by a grant from the O’Hara Chemical Sciences Instituteof the University of Dallas.

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For n < ω, is the partial order of the n-c.e. truth-table degrees isomorphicto the partial order of the c.e. truth-table degrees?

There are other reasons for studying this question. For example, it iswell-known that the c.e. Turing degrees are dense [16]. However, Cooper et.al. have shown that the 2-c.e. degrees are not dense [3]. Hence, with regardsto Turing reducibility, more mind changes make a structural difference. Dothey make a difference with regards to strong reducibilities such as ≤tt,≤wtt?

Now, consider the case n = 2. Clearly, every c.e. set is 2-c.e.. Thus, anatural starting point when investigating the above question is the following.

Is there an embedding of the 2-c.e. truth-table (or other strong reducibility)degrees into the corresponding partial order of the c.e. degrees?

Such an embedding can be obtained from the observation that there isa computable partial order which is universal among all countable partialorders (see exercise VI.1.6 of [17]). However, we obtain a stronger answerto this question. Namely, that there is an effective embedding of the 2-c.e.degrees into the c.e. degrees. That is, there is an algorithm that, given anindex of a representative of an element in the domain of the embeddingproduces an index of a representative of the image of that element. Weobtain similar results about several other structures involving other strongreducibilities.

Our results fall into three catagories: complexity, universality, and co-embeddability. With regards to complexity, the fundamental notion is thefollwing.

Definition 1.1 Let P = (X,≤) be a poset. A Σ3 enumeration of P is asurjection ν : ω → X such that {(m,n) | ν(m) ≤ ν(n)} is Σ3.

This generalizes the notion of a constructivization of a model due to Er-shov et. al. (see e.g. [6] or [10]). For all a ≤ ω and ≤r∈ {≤btt,≤tt,≤wtt},we let Dr

a−c.e. be the partial order consisting of the r-degrees that containan a-c.e. set. We let Dm

c.e. be the partial order consisting of the m-degreesthat contain a c.e. set besides ω, ∅. (We exclude these sets in order to ensurethis structure has a 0.) We show that these partial orders all have Σ3 enu-merations. With regards to universality, we show that all partial orders withΣ3 enumerations can be effectively embedded into these structures. That is,given a partial order P with Σ3 enumeration ν, there is an algorithm that

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given a natural number n produces an index of a representative of the imageof ν(n). In this way, these degree structures behave like Σ3-complete sets.Finally, we use the universality and complexity results to show that thesestructures are all effectively co-embeddable. Along the way, we answer aquestion of Downey [4].

Unfortunately, we do not answer the question of Nies nor the proposedvariant of this question. The effectiveness of the embeddings allows us toexert some control over their images, and so hopefully will make it easier tosimultaneously construct an inverse map. In the conclusion, we discuss theobstacles to this approach and possible future directions.

Preliminary Notations, Definitions, and Propositions

Unless otherwise stated, all computability-theoretic terminology is as in Soare[17]. In particular, we make use of the following facts.

1. If φe,t(x) ↓, then φe,t(x) < t.

2. If {e}At (x) ↓, then {e}A

t (x) < t and u(A; e, x) ≤ t.

All model-theoretic terminology is as in Chang and Kiesler [2]. All ter-minology related to computable model theory is as in Ash and Knight [1].

The following definitions of ≤tt,≤btt are easily verified to be equivalentto any of the usual definitions (see, for example, [17] or [14]) .

Definition 1.2 Let A,B ⊆ ω. Then, A ≤btt B if and only if there existse, n such that A = {e}B and {e}σ(x) ↓ for all x and all σ ∈ 2n.

Definition 1.3 Let A,B ⊆ ω. Then, A ≤tt B if and only if there exists esuch that A = {e}B and such that for all x there exists n such that {e}σ(x) ↓for all σ ∈ 2n.

Definition 1.4 Let A be a set. A computable approximation of A is a uni-formly computable sequence of sets {At}t∈ω such that A0 = ∅ and A(x) =limt→∞At(x) for all x ∈ ω.

Thus, by the Limit Lemma, A ≤T K if and only if A has a computableapproximation.

We use the following definition of the n-c.e. sets for n ≤ ω.

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Definition 1.5 Let n ∈ ω, and let A ⊆ ω.

1. We say that A is n-c.e. if it has a computable approximation {At}t∈ω

such that for all x, |{t | At(x) 6= At+1(x)}| ≤ n.

2. We say that A is ω-c.e. if it has a computable approximation {At}t∈ω

for which there exists a computable function f : ω → ω such that forall x, |{t | At(x) 6= At+1(x)}| ≤ f(x).

Clearly, a set is c.e. if and only if it is 1-c.e.. A d.c.e set is a set that isthe difference of two c.e. sets. It well-known that the d.c.e. sets are preciselythe 2-c.e. sets.

Facts 1.6 1. A ≤tt K ⇔ A ≤wtt K.

2. A ≤tt K ⇔ A ω − c.e..

3. A ≤btt K ⇔ ∃n < ω A n− c.e..

4. A ≤m K ⇔ A c.e..

Fact 1 is Exercise III.8.14 of [14]. Fact 2 is proven in [5]. By TheoremIII.8.7 of [14], B ≤btt K if and only if B is in the Boolean algebra generatedby the c.e. sets. It is well known that these are precisely the sets that aren-c.e. for some n < ω. Thus, Fact 3. Fact 4 was proven by Post [15].

It is well known that a set is (2n+1)-c.e. if and only if it is the union of ac.e. set and n d.c.e sets. It is also well known that a set is (2n+2)-c.e. if andonly if it is the union of (n+ 1) d.c.e. sets. It follows that for n < ω, there isan effective enumeration of the n-c.e. sets, {Wa,n}n∈ω. It follows from Fact 2that we obtain an effective enumeration of the ω-c.e. sets by letting

Wω,n,t = {z ≤ t | ∃n ≤ t ∀σ ∈ 2n {e}σt (z) ↓ ∧ {e}Kt

t (z) = 1}.

In the following propositions, we summarize the properties of these struc-tures with regards to 0, 1 elements. Some of these merely summarize well-known or trivial facts about these structures, in which case the proofs areomitted.

Proposition 1.7 [K]m is the 1 of Dmc.e.. The 0 of Dm

c.e. is the degree of anycomputable set. For a > 1, Dm

a−c.e. has a 1, but it is not the m-degree of K.

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Proof: The first part is well-known. Let a > 1. Then, as is shown in [5],there is an a-c.e. set C that is not c.e.. Hence, by Fact 4, C 6≤m K. However,if we take the m-degree of ⊕n∈ω Wa,n, then we obtain the 1 of this structure.

Proposition 1.8 For n < ω, the 0 of Dbttn−c.e. is [∅]btt, and the 1 of Dbtt

n−c.e.

is [K]btt. The structure Dbttω−c.e. has a 0 and 1, but [K]btt is not the 1 of this

structure.

Proof: The first part is well known. As is shown in [5], there is an ω-c.e.set C that is not n-c.e. for any n < ω. Hence, by Fact 3, C 6≤btt K. However,if we take the m-degree of ⊕n∈ωWa,n, we obtain the 1 of this structure.

Proposition 1.9 Let n ≤ ω. Let ≤r∈ {≤tt,≤wtt}. The 0 of Dr is [∅]r, andthe 1 of Dr is [K]r.

Proof: This follows from Facts (1) and (2).

With regards to universality and co-embeddability, we are only interestedin degree structures in which the degree of K is the 1. It is for this reasonthat we do not bother with the m-degrees of the a-c.e. sets for a > 1. Also,Dbtt

ω−c.e. is excluded from results from results.

Definition 1.10 Let P be a partial order with Σ3 enumeration ν. Let P0 ⊇P be a partial order with Σ3 enumeration µ. We write µ >> ν if there is acomputable function λ such that ν = µ ◦ λ.

Proposition 1.11 Let P be a poset with a Σ3 enumeration ν. Then, thereexists a poset P0 ⊇ P with 0, 1 and a Σ3 enumeration µ >> ν.

Proof: It is quite easy to see that there is a poset P0 with 0, 1 such thatP0 = P ∪ {0, 1}. Let ν be a Σ3 enumeration of P. For all n, let

µ(n) =

0 if n ≡ 0 mod 31 if n ≡ 1 mod 3

ν((n− 2)/3) if n ≡ 2 mod 3

It is then easily verified that µ is a Σ3 enumeration of ν and that µ >> ν.

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The following observation is crucial.

Theorem 1.12 Let L be a computable language with no function symbols.Then, the set of all consistent Π1 sentences of L is computable.

Proof: Let S be the set of all consistent Π1 sentences of L. Let φ be aΠ1 sentence of L. Let L(φ) be the language of φ (i.e. the set of all relationalsymbols, variables, and constants that appear in φ). Then, φ is consistent ifand only if there is a model for L(φ), A, such that A |= φ. Also, since φ isΠ1, for all models A,B for L(φ)

A |= φ ∧ B ⊂ A ⇒ B |= φ.

Since L has no function symbols, φ has a model if and only if φ has a model inwhich every domain element is the interpretation of a constant in L(φ). Letk be the number of constants in φ. It follows that φ has a model if and onlyif φ has a model with domain {1, . . . , k}. We can effectively search throughall such models to see if any satisfy φ. It follows that S is computable.

2 Degree structures with Σ3 enumerations

The goal of this section is to prove that for a ≤ ω and ≤r∈ {≤m,≤tt,≤wtt},Dr

a−c.e. has a Σ3 enumeration.

Definition 2.1 Let A,B ⊆ ω, and let e, t ∈ ω.

1. lm(e, A,B, t) =df max{y ≤ t | ∀x < y φe,t(x) ↓ ∧ A(x) = B(φe(x))}.

2. ltt(e, A,B, t) =df max{y ≤ t | ∀x < y ∃n ≤ t A(x) = {e}Bt (x) ∧ ∀σ ∈

2n {e}σt (x) ↓}.

Let e = 〈a, b〉.

3. lbtt(e, A,B, t) =df max{y ≤ t | ∀x < y A(x) = {a}Bt (x) ∧ ∀σ ∈

2b {e}σt (x) ↓}.

4. lwtt(e, A,B, t) =df max{y ≤ t | ∀x < y φb,t(x) ↓ ∧ A(x) = {a}Bt (x) ∧

u(B; a, x, t) < φb(x)}.

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Lemma 2.2 Let A,B ≤T K. Let {At}t∈ω, {Bt}t∈ω be computable approxi-mations to A,B respectively. Let ≤r∈ {≤m,≤btt,≤tt,≤wtt}. Then, the fol-lowing are equivalent.

1. limt→∞ lr(e, At, Bt, t) = ∞.

2. ∀N ∃t lr(e, At, Bt, t) ≥ N .

3. ∀N ∃∞t lr(e, At, Bt, t) ≥ N .

Proof: (1) ⇒ (2): Obvious.

(2) ⇒ (3): Suppose that for all N there exists t such that lr(e, At, Bt, t) ≥ N .Let N ∈ ω. It suffices to show that

∀t lr(e, At, Bt, t) ≥ N ⇒ ∃s > t lr(e, As, Bs, s) ≥ N.

Let t be such that lr(e, At, Bt, t) ≥ N . There exists s such that lr(e, As, Bs, s) ≥t + 1. By Definition 2.1, t ≥ lr(e, At, Bt, t). Hence, lr(e, As, Bs, s) ≥ N .Again, by Definition 2.1, s ≥ lr(e, As, Bs, s). Hence, s ≥ t+ 1 > t.

(3) ⇒ (1): Suppose that for all N there exist infinitely many t such thatlr(e, At, Bt, t) ≥ N . Let N ∈ ω. We show that there exists t such that forall s ≥ t, lr(e, As, Bs, s) ≥ N . Let t0 be such that lr(e, At0 , Bt0 , t0) ≥ N . If≤r=≤m, then for all x ≤ N φe(x) ↓< t0. In this case, we let M = t0. If≤r=≤btt, then {a}σ

t0(x) ↓for all x ≤ N and all σ ∈ 2b. In this case, we let

M = b. If ≤r=≤tt, then for all x ≤ N there exists nx such that eσt0(x) ↓ for

all σ ∈ 2nx . In this case, we let M = max{nx | x ≤ N}. If ≤r=≤wtt, thenfor all x ≤ N , φb(x) ↓. In this case, we let M = max{φb(x) | x ≤ N}. Thereexists t1 ≥ t0 such that At1 � (N + 1) = A � (N + 1), Bt1 � M = A � M , andlr(e, At1 , Bt1 , t1) ≥ N .

Suppose ≤r = ≤m. Then, for all x ≤ N φe,t1(x) ↓ and A(x) = At1(x) =Bt1(φe(x)) = B(φe(x)). It follows that for all s ≥ t1, l

r(e, As, Bs, s) ≥ N .

Suppose that ≤r = ≤btt. Then, for all x ≤ N , A(x) = At1(x) = {e}Bt1t1 (x).

Also, for all x ≤ N , Bt1 � u(Bt1 ; e, x) = B � u(Bt1 ; e, x). It follows that forall s ≥ t1, l

r(e, As, Bs, s) ≥ N .Suppose that ≤r∈ {≤tt,≤wtt}. Then, for all x ≤ N , A(x) = At1(x) =

{a}Bt1t1 (x). Also, for all x ≤ N , Bt1 � u(Bt1 ; a, x) = B � u(Bt1 ; a, x). Thus,

for all s ≥ t1, lr(e, As, Bs, s) 6= N .

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Proposition 2.3 Let A,B ≤T K, and let {At}t∈ω, {Bt}t∈ω be computableapproximations of A,B respectively. Let ≤r∈ {≤m,≤btt,≤tt,≤wtt}.

1. λe,t.lr(e, At, Bt, t) is computable. Furthermore, an index for this func-

tion can be obtained from indices for the computable approximations ofA,B.

2. A ≤r B if and only if there exists e such that limt→∞ lr(e, At, Bt, t) =

∞.

Proof: The proof of (1) is clear. We prove (2).

Case: ≤r = ≤m.

(⇒): Suppose A ≤m B. Then, by definition, there exists e such that φe

is total and A(x) = B(φe(x)) for all x. Let N ∈ ω. There exists t0 suchthat for all x ≤ N φe,t0(x) ↓. There exists t1 ≥ t0 such that for all x ≤ N ,At1(x) = Bt1(φe(x)). Hence, lm(e, At1 , Bt1 , t1) ≥ N . By Lemma 2.2, thatlimt→∞ l

m(e, At, Bt, t) = ∞.

(⇐): Suppose that limt→∞ lm(e, At, Bt, t) = ∞. Let x ∈ ω. There exists t0

such that lm(e, At0 , Bt0 , t0) > x. Thus, φe,t0(x) ↓ < t0. There exists t1 ≥ t0such that A(x) = At1(x), B � t0 = Bt1 � t0, and lm(e, At1 , Bt1 , t1) > x. Hence,A(x) = At1(x) = Bt1(φe(x)) = B(φe(x)).

Case: ≤r = ≤btt.

(⇒): Suppose that A ≤btt B. Thus, by Definition 1.2, there exists e, n suchthat A = {e}B, and {e}σ(x) ↓ for all x and all σ ∈ 2n. Let e′ = 〈e, n〉. LetN ∈ ω. There exists t0 such that for all x ≤ N and all σ ∈ 2n {e}σ

t0(x) ↓.

There exists t1 ≥ t0 such that for all x ≤ N , At2(x) = {e}Bt2t2 (x). Hence,

lbtt(e′, At1 , Bt1 , t1) ≥ N . Thus, by Lemma 2.2, limt→∞ lbtt(e′, At, Bt, t) = ∞.

(⇐): Suppose limt→∞ lbtt(e′, At, Bt, t) = ∞. Let e, n be such that e′ = 〈e, n〉.

Let x ∈ ω. There exists t0 such that lbtt(e′, At0 , Bt0 , t0) > x. Hence, for allσ ∈ 2n, {e}σ

t0(x) ↓. There exists t1 ≥ t0 such that A � (x+1) = At1 � (x+1),

B � n = Bt1 � n, and lbtt(e′, At1 , Bt1 , t1) > x. Therefore, A(x) = At1(x) =

{e}Bt1t1 (x). Clearly, u(Bt1 ; e, x) ≤ n. It follows that {e}Bt1

t1 (x) = {e}B(x).

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Thus, A(x) = {e}B(x). Hence, A ≤btt B.

Case: ≤r = ≤tt.

(⇒): Suppose A ≤tt B. By Definition 1.3, there exists e such that A = {e}B

and for all x there exists n such that {e}σ(x) ↓ for all σ ∈ 2n. Let N ∈ ω.There exists t0 such that for all x ≤ N there exists n ≤ t0 such that{e}σ

t0(x) ↓ for all σ ∈ 2n. There exists t1 ≥ t0 such that for all x ≤ N

At1(x) = {e}Bt1t1 (x). Hence, ltt(e, At1 , Bt1 , t1) ≥ N . By Proposition 2.2,

limt→∞ ltt(e, At, Bt, t) = ∞.

(⇐): Suppose that limt→∞ ltt(e, At, Bt, t) = ∞. Let x ∈ ω. There exists t0

such that ltt(e, At0 , Bt0 , t0) > x. There exists n ≤ t0 such that for all σ ∈ 2n

{e}σt0(x) ↓. There exists t1 ≥ t0 such that At1 � (x + 1) = A � (x + 1), Bt1 �

n = B � n, and ltt(e, At1 , Bt1 , t1) > x. Thus, A(x) = At1(x) = {e}Bt1t1 (x).

Clearly, u(Bt1 ; e, x) ≤ n. Hence, {e}Bt1t1 (x) = {e}B. Thus, A ≤tt B.

Case: ≤r = ≤wtt.

(⇒): By definition, there exists a, b such that A = {a}B, φb is total, andu(B; a, x) < φb(x) for all x. Let e = 〈a, b〉. Let N ∈ ω. There existst0 such that for all x ≤ N , φb,t0(x) ↓. There exists t1 ≥ t0 such that

for all x ≤ N , At1(x) = {a}Bt1t1 (x) and u(Bt1 ; a, x, t1) < φb(x). Thus,

lwtt(e, At1 , Bt1 , t1) ≥ N . By Lemma 2.2, limt→∞ lwtt(e, At, Bt, t) = ∞.

(⇐): Suppose that limt→∞ lwtt(e, At, Bt, t) = ∞. Let a, b such that e = 〈a, b〉.

Let x ∈ ω. Let t0 be such that lwtt(e, At0 , Bt0 , t0) > x. Thus, φe,t0(x) ↓ < t0.There exists t1 ≥ t0 such that A � (x + 1) = At1 � (x + 1), B � t0 =

Bt1 � t0, and lwtt(e, At1 , Bt1 , t1) > x. Hence, A(x) = At1(x) = {a}Bt1t1 (x) and

u(Bt1 ; a, x, t1) < φb(x). Thus, {a}Bt1t1 (x) = {a}B(x) and u(B; a, x) < φb(x).

It follows that A ≤wtt B.

Proposition 2.4 Let ≤r∈ {≤m,≤btt,≤tt,≤wtt}. Let a ≤ ω.

1. λe,m,n,t.lr(e,Wa,m,t,Wa,n,t, t) is computable.

2. {(m,n) | Wa,m ≤r Wa,n} is Σ3.

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Proof: Part (1) is clear. We have

Wa,m ≤r Wa,n ⇔ ∃e ∀N ∃t lr(e,Wa,m,t,Wa,n,t, t) ≥ N.

Part (2) follows.

Proposition 2.5 The following have Σ3 enumerations.

1. Any final segment of Dc.e.m .

2. Any final segment of Dbtta−c.e. for all a < ω.

3. Any final segment of Dra−c.e. for all a ≤ ω and ≤r∈ {≤tt,≤wtt}.

Proof: Let C 6= ∅, ω be c.e.. For all n, let ν(n) = [Wn ⊕ C]m. Then, ν isa Σ3 enumeration of the final segment of Dm

a−c.e. above [C]m. The proofs of(2),(3) are similar.

3 Main Theorem

All of our results about embeddings will be obtained from the theorem inthis section. The proof of this theorem is a tree construction in the style ofSoare [17].

Definition 3.1 For all A,B ⊆ ω, we write A ≤∗1 B if there is a one-to-onecomputable function f such that for all sufficiently large n, A(n) = B(f(n)).

Notation: For all A, A =df ∅ ⊕ A⊕ ω.

Proposition 3.2 Let A,B ⊆ ω.

1. K ≤∗1 B ⇒ K ≤1 B.

2. A ≡btt A.

3. A ≤∗1 B ⇒ A ≤1 B.

4. A ≤T ∅ ⇒ A ≤1 B.

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Proof: Obvious.

Theorem 3.3 Let P = (X,≤, 0, 1) be a poset with Σ3 enumeration ν. Let≤r∈ {≤m,≤btt,≤tt,≤wtt}. Let C <r K be such that C 6= ∅, ω. Then, thereexists uniformly c.e. {An}n∈ω such that the following hold.

1. ∀n [ν(n) = 1 ⇒ K ≤1 An].

2. ∀n [ν(n) = 0 ⇒ An ≤T ∅ ∧ An 6= ∅, ω].

3. ∀m,n [ν(m) ≤ ν(n) ⇒ Am ≤1 An].

4. ∀m,n [ν(m) 6≤ ν(n) ⇒ Am 6≤r An ⊕ C].

Proof: We construct a uniformly c.e. family {Bn}n∈ω such that the fol-lowing hold.

(1’) ∀n [ν(n) = 1 ⇒ K ≤∗1 Bn].

(2’) ∀n [ν(n) = 0 ⇒ Bn ≤T ∅].

(3’) ∀m,n [ν(m) 6= 0 ∧ ν(m) ≤ ν(n) ⇒ Bm ≤∗1 Bn].

(4’) ∀m,n [ν(m) 6≤ ν(n) ⇒ Bm 6≤wtt Bn ⊕ C].

We then let An = Bn for all n.We begin by giving a sketch of the construction. Let T = 2<ω. Let

R ⊆ ω5 be computable such that for all m,n

ν(m) ≤ ν(n) ⇔ ∃e ∀x ∃y R(e, x, y,m, n).

For all e,m, n, letRe(m,n) hold if for all x there exists y such thatR(e, x, y,m, n).We assign statements of the form Re(m,n) to nodes of T . We also assignstatements of the form Bm ≤r Bn ⊕ Bnvia φe to nodes of T . No node isassigned more than one statement. Since these statements are all Π2, we canassociate τ∧(0) with the outcome that the statement assigned to τ is trueand τ∧(1) with the outcome that the statement assigned to τ is false. If forsome e a node τ witnesses that Re(m,n) holds, then we may assign to τmodule that codes Bm into Bn in a standard way. On the other hand, if τwitnesses a reduction of Bm to Bn ⊕ C but does not witness a value of e forwhich Re(m,n) holds, then we assign a module to τ which ensures τ is not

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on the true path. This module works by coding K into Bm while imposingan increasing restraint on Bn. Hence, if τ is on the true path, then C ≤r Bm

and Bn is computable. Thus, C ≤r K- a contradiction.There are at least two potential problems with this module’s strategy.

First, suppose that at a stage t, we set τ ’s restraint on Bn equal to t + 1.Nodes below and to the right of τ are then bound not to enumerate anynumber below t+ 1 into Bm at any future stage. However, it does not followthat “up to finite injury” Bn � (t + 1) has settled at this stage. For, it maythe case that τ ’s module enumerates into Bm at this stage. Then, at somefuture stage, a module assigned to a node ρ1 above τ and charged with thetask of ensuring Bm is reducible to some set Bm1 might enumerate into Bm1

to record this event. It could then happen that a module assigned to a nodeρ2 above τ and charged with the task of ensuring that Bm1 is reducible to Bn

will enumerate a number into Bn to record this event. Futhermore, it couldbe that this number is less than t+1. Hence, the module assigned to τ mightinadvertantly injure itself! The resolution of this difficulty has to do withthe assignment of modules. Although the node τ may not have observedan e such that Re(m,n) holds, the information it has observed implies thatsuch an e exists. Thus, when assigning modules we must take into accountnot only the relations which have been explicitly recorded by a node, butall relations which follows from them. Forturnately, since the axioms for thetheory of partial orders are Π1, we can do this in an effective manner.

But even when modules have been more carefully assigned to nodes, otherproblems of a less serious nature remain. For, in the above scenario, it maystill not be the case that “up to finite injury” Bn � (t+1) settles at stage t+1.For, it may be that some node below τ enumerates into some set Bm1 at staget+1. It may then be the case that as a result of this event, a coding moduleassigned to a node above τ enumerates into a set Bm2 at a future stage. Itmay then be the case that as a result of this enumeration into Bm2 , a codingmodule assigned to a node above τ enumerates into Bn. The resolution ofthis difficulty is obtained by showing that even though Bn � (t+ 1) may notsettle at this stage, we can, “up to finite injury”, compute a future stage atwhich it will settle.

We now proceed to describe the construction in detail. Let {Ct}t∈ω be acomputable approximation of C.

Let ρ ∈ T . Let u, e,m, n be such that lh(ρ) = 〈u, e,m, n〉. If u is even,then for all t we let l(ρ, t) = lr(e, Bm,t, Bn,t ⊕ Ct, t).

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If u is odd, then for all t we let

l(ρ, t) = max{z ≤ t | ∀x < z ∃y ≤ t R(e, x, y,m, n)}.

Thus,Re(m,n) ⇔ lim

t→∞l(ρ, t) = ∞.

It follows from Proposition 2.4 that λρ,t.l(ρ, t) is computable.Let δ0 = ∅. Let t ∈ ω. Assume δ0, . . . , δt have been defined. For all η, let

vη(t) = max({0} ∪ {s < t | δs+1 ⊇ η}).

Let δt+1,0 = ∅. Let s ≤ t. Assume δt+1,0, . . . , δt+1,s have been defined. Letη = δt+1,s. Let

δt+1,s+1 =

{η∧(0) if l(η, t) > l(η, vη(t))η∧(1) otherwise.

Let δt+1 = δt+1,t+1.Let f = lim inft δt. It follows that for all m,n, Bm ≤r Bn if and only if

there exists u, e,m, n, ρ such that ρ∧(0) ⊂ f and lh(ρ) = 〈2u, e,m, n〉. Also,for all e,m, n,

Re(m,n) ⇔ ∃ρ, u lh(ρ) = 〈2u+ 1, e,m, n〉 ∧ ρ∧(0) ⊂ f.

Let Lpo be the language of posets with 0, 1. Let Lusl be the language ofupper semi-lattices.

Let ν0, ν1, . . . be distinct new constants effectively indexed. Let Ldom =Lpo ∪ {ν0, ν1, . . .}.

Let c, b0, b1, . . . be distinct new constants effectively indexed. Let Lran =Lusl ∪ {c, b0, b1, . . .}.

Fix numbers n0, n1 such that ν(n0) = 0 and ν(n1) = 1. Let T be theset of axioms for posets with 0, 1 together with the statements νn0 = 0 andνn1 = 1. Let τ ∈ T . Let T dom

τ = T ∪ {νm ≤ νn | ∃v, e, ρ ρ∧(0) ⊆ τ ∧ lh(ρ) =〈2v, e,m, n〉}.

Let φ be a Σ0 sentence of Ldom. We say that τ believes φ if T domτ ` φ.

Hence, for all m,n,

ν(m) ≤ ν(n) ⇔ ∃τ ⊂ f τ believes νm ≤ νn.

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Also, for all n,

ν(n) ≤ 0 ⇔ ∃τ ⊂ f τ believes νn ≤ 0, and1 ≤ ν(n) ⇔ ∃τ ⊂ f τ believes 1 ≤ νn.

We say that τ believes bm ≤ bn ∧ c if and only if there exists v, e, ρ suchthat ρ∧(0) ⊆ τ and lh(ρ) = 〈2v + 1, e,m, n〉. Since the axioms in T are Π1,it follows from Theorem 1.12 that {(τ, φ) | τ believes φ} is computable.

We say that τ ∈ T is diseased if there exists m,n ≤ lh(τ) such that τbelieves bm ≤ bn ∧ c but τ does not believe νm ≤ νn. It follows that the setof diseased nodes is computable. The construction will ensure that there areno diseased nodes on f . This ensures (4’). In order to ensure (1’),(2’),(3’),we assign requirements to non-diseased nodes. These requirements are asfollows.

Pn : ν(n) = 1 ⇒ K ≤∗1 Bn

Qn : ν(n) = 0 ⇒ Bn ≤T ∅Rm,n : ν(m) ≤ ν(n) ∧ ν(m) 6= 0 ⇒ Bm ≤∗1 Bn

We define an effective assignment of these requirements to nodes as follows.Let τ ∈ T . Assume that for all ρ ⊂ τ it has been determined which require-ments are assigned to ρ. If lh(τ) ≡ 0 mod 3, then for all n, Pn is assigned toτ if n ≤ lh(τ), τ believes 1 ≤ νn, and Pn is not assigned to any ancestor of τ .If lh(τ) ≡ 1 mod 3, then for all n, Qn is assigned to τ if n ≤ lh(τ), τ believesνn ≤ 0, and Qn is not assigned to any ancestor of τ . If lh(τ) ≡ 2 mod 3,then for all m,n, Rm,n is assigned to τ if m,n ≤ lh(τ), τ believes νm ≤ νn,τ does not believe νm ≤ 0, and Rm,n is not assigned to any ancestor of τ . Itis easily shown that the following hold for all m,n.

(i) If ν(n) = 1, then there exists ρ ⊂ f such that Pn is assigned to ρ.

(ii) If ν(n) = 0, then there exists ρ ⊂ f such that Qn is assigned to ρ.

(iii) If ν(m) ≤ ν(n) and ν(m) 6= 0, then there exists ρ ⊂ f such that Rm,n

is assigned to ρ.

(iv) If Pn is assigned to ρ, then ρ believes 1 ≤ νn and n ≤ lh(ρ).

(v) If Qn is assigned to ρ, then ρ believes νn ≤ 0 and n ≤ lh(ρ).

(vi) If Rm,n is assigned to ρ, then ρ believes νm ≤ νn, ρ does not believeνm ≤ 0, and m,n ≤ lh(ρ).

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(vii) For all ρ, there is at most one requirement assigned to ρ.

Let y : T×ω2 → ω be an effective bijection. We write yτ,n,x for y(τ, n, x).We choose this bijection so that x < yτ,n,x for all τ, n, x. For all τ, n, τ mayonly enumerate numbers of the form yτ,n,x into Bn. We let Bn,t,τ be theset of all numbers enumerated into Bn at τ by the end of stage t. That is,Bn,t,τ = {z | ∃x z = yτ,n,x ∧ z ∈ Bn,t}. Thus, for all n, t, Bn,t = ∪τBn,t,τ .We let Bn,τ = ∪tBn,t,τ .

For all n, η, t, rn(η, t) is the restraint placed on Bn by node η at stage t.We now describe the contruction.

Stage 0: For all η, n, rn(η, 0) =df −1. For all m, τ , Bm,0,τ =df ∅. Thus,for all m, Bm,0 = ∅.

Stage t + 1: Let δt+1 be the maximum non-diseased ancestor of δt+1. Foreach s ≤ t+ 1, do the following. Let τ = δt+1 � s.

Suppose τ ⊆ δt+1. If no requirement is assigned to τ , then do nothing.Suppose there is a requirement assigned to τ . If Qn is assigned to τ , then forall n′ such that τ believes νn′ ≤ νn let rn′(τ, t + 1) = t. If Pn is assigned toτ , then let

Bn,t+1,τ = Bn,t,τ ∪ {yτ,n,x | x ∈ Kt ∧ ∀τ ′ < τ yτ,n,x > rn(τ ′, t)}.

If Rm,n is assigned to τ , then let

Bn,t+1,τ = Bn,t,τ ∪ {yτ,n,x | x ∈ Bm,t ∧ ∀τ ′ < τ yτ,n,x > rn(τ ′, t)}.

Now, suppose τ ⊃ δt+1. If lh(τ) > lh(δt+1)+1, then do nothing. Supposelh(τ) = lh(δt+1) + 1. Thus, τ is diseased, but no ancestor of τ is. Letη = τ−. Let u, e,m, n be such that lh(η) = 〈u, e,m, n〉. Hence, u is even andτ believes bm ≤ bn ∧ c. Let rn′(τ, t + 1) = t for all n′ such that τ believesνn′ ≤ νn. Let

Bm,t+1,τ = Bm,t,τ ∪ {yτ,m,x | x ∈ Kt ∧ ∀τ ′ < τ yτ,m,x > rm(τ ′, t)}.

This completes the construction. We now turn to its verification. Let∆τ,t = max{z | ∀ρ ⊂ τ ∀n Bn,ρ,t+1 � z = Bn,ρ,t � z}. It follows that∆τ,t = max{z | ∀ρ ⊂ τ ∀n ≤ lh(ρ) Bn,ρ,t+1 � z = Bn,ρ,t � z}. Hence, λτ,t.∆τ,t

is computable. We let rn(η, t) = min{rn′(η, t) | η belives νn′ ≤ νn}.

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Lemma 3.4 Let τ ⊂ f be such that no ancestor of τ is diseased. Supposethat τ does not believe 1 ≤ νn, and that Bn′,τ = ∅ for all n′ such that τbelieves νn′ ≤ νn. Then,

∀∞t ∀z [z < min{rn(τ, t),∆τ,t} ⇒ Bn � z = Bn,t � z].

Proof: Let t0 be such that for all t ≥ t0, δt+1 ≥L τ . Let t1 be such thatt1 ≥ t0 and δt1+1 ⊇ τ . Let z be such that z < rn(τ, t1 + 1),∆τ,t1 . By way ofcontradiction, suppose that Bn � z 6= Bn,t1 � z. Let y0 be such that y0 < zand y0 ∈ Bn − Bn,t1 . There exists unique ρ0 such that y0 ∈ Bn,ρ0 . Sincern(τ, t1 + 1) > z, ρ0 ≤ τ . By the choice of t0, ρ0 ⊆ τ . Since Bn,τ = ∅,ρ0 ⊂ τ . Let n0 = n. Since ρ0 is not diseased and τ does not believe 1 ≤ νn,there exists m0 such that Rm0,n0 is assigned to ρ0. There exists y1 such thaty0 = yρ0,n0,y1 . Hence, y1 < y0 < z < ∆τ,t1 . Thus, y1 6∈ Bm0,t1+1 −Bm0,t1 .

There exists s0 such that y1 ∈ Bm0,s0+1 − Bm0,s0 . It follows that s0 ≥ t1.For, suppose that s0 < t1. Thus, y1 ∈ Bm0,t1 . Since δt1+1 ⊇ τ and y0 ∈ Bn,it follows from the construction that y0 ∈ Bn,t1+1,ρ −Bn,t1,ρ. But, y0 < ∆τ,t1 ;a contradiction. Thus, s0 ≥ t1.

There exists unique ρ1 such that y1 ∈ Bm0,ρ1 . Since y1 < rn(τ, t1), andsince τ believes νm0 ≤ νn, it follows that ρ1 ≤ τ . By the choice of t0, ρ1 ⊆ τ .Since τ believes νm0 ≤ νn, it follows that ρ1 ⊂ τ . There exists n1, y2 such thaty1 = yρ1,n1,y2 . Therefore, n1 = m0. It follows that there exists m1 such thatRm1,n1 is assigned to ρ1. It follows that τ believes νm1 ≤ νm0=n1 ≤ νn0=n.Also, y2 < y1 < z. Thus, y2 6∈ Bm1,t1+1,ρ1 − Bm1,t1 . There exists s1 suchthat y2 ∈ Bm1,s1+1 − Bm1,s1 . It follows that s1 ≥ t1. There exists uniqueρ2 such that y2 ∈ Bm1,ρ2 . It again follows that ρ2 ⊂ τ . Continuing in thisway we obtain an infinite descending sequence y0 > y1 > y2 > . . .. Thus,Bn � z = Bn,t1 � z.

Lemma 3.5 There are no diseased nodes on f .

Proof: By way of contradiction, suppose otherwise. Let τ ⊂ f be the leastsuch. Let η = τ−. There exists v, e,m, n such that lh(η) = 〈2v, e,m, n〉.Also, η∧(0) = τ . Clearly, τ does not believe νm ≤ νn, and τ believesbm ≤ bn ∧ c. Thus, Qn is not assigned to any ρ ⊂ τ . It follows thatlimt→∞ max{rm(τ ′, t) | τ ′ < τ} exists. Let M be this limit. It follows fromthe construction that for all x > M ,

x ∈ K ⇔ yτ,x,m ∈ Bm.

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Hence, K ≤∗1 Bm. Since τ ⊂ f , Bm ≤r Bn ⊕ C.We claim that Bn is computable. For, by construction, limt→∞ rn(τ, t) =

∞. Since τ does not believe νm ≤ νn, it follows that τ does not believe1 ≤ νn. By construction, for all n′ 6= m, Bn′,τ = ∅. Thus, for all ν ′ such thatτ believes νn′ ≤ νn, Bn′,τ = ∅. It now follows from Lemma 3.4 that Bn iscomputable.

Now, if ≤r 6=≤m, it follows that Bm ≤r C; a contradiction. So, supposethat ≤r=≤m. We have

K ≤∗1 Bm ≤m Bn ⊕ C.

If C 6∈ {∅, ω}, then it follows that K ≤m C; a contradiction. So, supposethat C ∈ {∅, ω}. Then, Bn ⊕ C is computable. It then follows that K iscomputable; again, a contradiction.

Hence, there are no diseased nodes on f .

Lemma 3.6 Let m,n ∈ ω. If ν(m) ≤ ν(n), and if ν(m) 6= 0, then Bm ≤∗1Bn.

Proof: Suppose ν(m) ≤ ν(n), and that ν(m) 6= 0. Then, there ex-ists τ ⊂ f such that Rm,n is assigned to τ . Thus, τ believes νm ≤ νn.There are no diseased ρ ⊂ τ . Furthermore, for all ρ ⊂ τ , Qn is not as-signed to ρ. For, otherwise τ believes νm ≤ 0 which contradicts the factthat ν(m) 6= 0. Since there are no diseased ancestors of τ , it follows thatlimt→∞ max{rm(τ ′, t) | τ ′ < τ} exists. Let M be this limit. By construction,for all x > M

x ∈ Bm ⇔ yτ,x,n ∈ Bn.

Lemma 3.7 For all n, if ν(n) = 0, then Bn is computable.

Proof: Suppose that ν(n) ≤ 0. There exists τ ⊂ f such that Qn isassigned to τ . It then follows from the contruction that limt→∞ rn(τ, t) = ∞.It follows from Lemma 3.4 that Bn is computable.

Lemma 3.8 For all n, if 1 = ν(n), then K ≤∗1 Bn.

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Proof: Suppose that 1 ≤ ν(n). There exists τ ⊂ f such that Pn isassigned to τ . Therefore, τ believes 1 ≤ νn. Hence, τ does not believeνn ≤ 0. Thus, limt→∞ max{rn(τ ′, t) | τ ′ < τ} exists. Let M be this limit. Itfollows from the construction that for all x > M

x ∈ K ⇔ yτ,x,n ∈ Bn.

It now follows that the construction is successful.

4 Universality results

Definition 4.1 Let P be a poset with Σ3 enumeration ν. Let a ≤ ω, andlet ≤r∈ {≤m,≤btt,≤tt,≤wtt}. A map φ : P → Dr

a−c.e. is called effective withrespect to ν if there is a computable function g such that φ(ν(n)) = [Wg(n)]rfor all n.

Definition 4.2 Let P be a poset with a Σ3 enumeration. Let I be a finalsegment of Dr

a−c.e. where ≤r∈ {≤m,≤btt,≤tt,≤wtt} and a ≤ ω. We say thatP effectively embeds into I if for each Σ3 enumeration of P, ν, there is anembedding φ : P → Dr

a−c.e. that is effective with respect to ν. If P has a 0,then we also require φ to preserve 0. If P has a 1, then we also require φ topreserve 1.

Theorem 4.3 Let ≤r∈ {≤btt,≤tt,≤wtt}. If ≤r 6= ≤btt, then let a ≤ ω.Otherwise, let a < ω. Then, every partial order with Σ3 enumeration can beeffectively embedded into every non-trivial final segment of Dr

a−c.e..

Proof: Let P be a partial order with Σ3 enumeration ν. Let P0 ⊇ P bea partial order with 0, 1 and Σ3 enumeration µ such that µ >> ν. Let C bean a− c.e. set such that C <r K. Let {An}n∈ω be as given by Theorem 3.3.For all n, let

ψ(µ(n)) = [An ⊕ C]r.

If ν(n) ≤ ν(n′), then An ≤1 An′ . Hence, if µ(n) = µ(n′), then An ≡1 An′ .It follows that ψ is a well-defined homomorphism. If µ(n) 6≤ µ(n′), then

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An 6≤r An′ ⊕ C. It follows that ψ is a monomorphism. If µ(n) = 0, thenψ(µ(n)) = [C]r. Hence, ψ preserves 0, If µ(1) = 1, then ψ(µ(n)) = [K]r.Hence, ψ preserves 1. We then let φ = ψ � dom(P). Since µ >> ν, it followsthat φ is effective with respect to ν.

Corollary 4.4 Let ≤r∈ {≤btt,≤tt,≤wtt}. If ≤r = ≤btt, then let a < ω. Oth-erwise, let a ≤ ω. Then every partial order with Σ3 enumeration effectivelyembeds into Dr

a−c.e..

Theorem 4.5 Every partial order with Σ3 enumeration effectively embedsinto every non-trivial final segment of Dm

c.e..

Proof: Let ν be a Σ3 enumeration of P. Let P0 ⊇ P be a partial orderwith 0, 1 and Σ3 enumeration µ such that µ >> ν. Let C be a c.e. set suchthat C <m K and C 6= ∅, ω. Let {An}n∈ω be as given by Theorem 3.3. For alln, let ψ(µ(n)) = [An⊕C]m. It follows that ψ is a well-defined homomorphismof P0 into Dm

c.e.. If µ(n) 6≤ µ(n′), then An 6≤m An′ ⊕C. It then follows that ψis a monomorphism. If µ(n) = 1, then ψ(µ(n)) = [K⊕C]m = [K]m. Thus, ψpreserves 1. Since C 6= ∅, ω, An ⊕ C ≡m C if µ(n) = 0. Hence, ψ preserves0. Let φ = ψ � dom(P). Since µ >> ν, it follows that φ is effective withrespect to ν.

5 Co-embeddability results

Theorem 5.1 Let ≤r,≤s∈ {≤btt,≤tt,≤wtt}. If ≤r 6= ≤btt, then let a ≤ ω.Otherwise, let a < ω. If ≤r 6= ≤btt, then let b ≤ ω. Otherwise, let b < ω.Then, any non-trivial final segment of Dr

a−c.e. effectively embeds into anynon-trivial final segment of Ds

b−c.e..

Proof: Let I be a non-trivial final segment of Dra−c.e., and let J be a non-

trivial final segment of Dsb−c.e.. By Theorem 2.5, I has a Σ3 enumeration. By

Theorem 4.3, I effectively embeds into J .

The following answers a question of Downey [4].

Corollary 5.2 Let ≤r,≤s∈ {≤btt,≤tt,≤wtt}. If ≤r 6= ≤btt, then let a ≤ ω.Otherwise, let a < ω. If ≤r 6= ≤btt, then let b ≤ ω. Otherwise, let b < ω.Then, Dr

a−c.e. effectively embeds into Dsb−c.e..

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Theorem 5.3 Let ≤r∈ {≤btt,≤tt,≤wtt}. If ≤r 6= ≤btt, then let a ≤ ω. Oth-erwise, let a < ω. Then, any non-trvial final segment of Dr

a−c.e. is effectivelyco-embeddable with any non-trivial final segment of Dm

c.e..

Proof: Let I be a non-trivial final segment of Dra−c.e.. Let J be a non-

trivial final segment of Dmc.e.. Both I, J have Σ3 enumerations. By Theorems

4.3 and 4.5, these orderings are effectively co-embeddable.

Conclusion

We return to the following question which was posed in the introduction: isDtt

2−c.e. isomorphic to Dttc.e.? We have shown that Dtt

2−c.e. can be effectivelyembedded into Dtt

c.e.. In doing so, for each 2-c.e. set W2,n, we have constructedan “image” An. In order to make this map invertible, we must construct, foreach c.e. set We a “pre-image” Be. In order to make the resulting map andits inverse homomorphisms, we would have to ensure that for all e, n

Wa,n ≤tt Be ⇒ An ≤tt We.

Unfortunately, there are no general techiniques for constructing a non-computablec.e. set that is truth-table reducible to a given non-computable c.e. set. Thisis evidenced by the fact that the c.e. truth-table degrees contain minimalelements [11]. This is the main obstacle to using the techniques in this paperto solve this problem. Perhaps more sophisticated techniques could be em-ployed to make guesses at the nature of the truth-table degrees below a givendegree with the resulting isomorphism culled from the degrees constructedaccording to the correct guesses.

A negative solution might also be investigated. Naturally, one would wantto find a sentence satisfied by one structure, but not the other. By the resultsin this paper, Dtt

2−c.e. and Dttc.e. satisfy the same Σ1 sentences. Namely, since

every finite partial order embeds into these structures, they satisfy all Σ1

sentences true in some partial order. Thus, it would seem natural to beginby looking at the Σ2 sentences. Among such sentences are those that assertthat a certain lattice can be embedded as a finite initial segment. Muchis known about which lattices can be embedded as finite initial segmentsof Dtt

c.e. (see, for example [11]). Unfortunately, not much is known aboutwhich lattices can not be so embedded into Dtt

c.e.. Harrington and Haughtclaim, in an unpublished manuscript, to have shown that diamond can not

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be embedded as an initial segment of Dttc.e.. However, this seems to be the

only result of its kind at the time of this writing. In any case, resolving thequestion in this way would involve constructing a finite initial segment ofDtt

c.e. which would contain a non-c.e. 2-c.e. degree. The following questionmight then be a starting point for this investigation.

Is there a minimal 2-c.e. degree which is not c.e.?

For their own sake, both lines of investigation are worthy of pursuit.

Acknowledgements I thank the O’Hara Chemical Sciences Institute for itsfinancial support of this project. I also thank the mathematics department ofThe George Washington University for the use of their computer equipmentwhile preparing an earlier draft of this paper.

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[4] R. Downey, review Undecidability and initial segments of the (r.e.) tt-degrees by C. Haught and R. Shore. Mathematical Reviews, August1991, pp. 4139 - 4140.

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[8] Y. Ershov, A hierarchy of sets, II, Algebra and Logic, vol. 7(1968),pp. 47 - 74.

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[9] Y. Ershov, A hierarchy of sets, III, Algebra and Logic, vol. 9(1970),pp. 34 - 51.

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[11] C. Haught, R. Shore, Undecidability and initial segments of the (r.e.)tt-degrees, Journal of Symbolic Logic, vol. 55 (1990), no. 3, 987 -1006.

[12] G. LaForte, Phenomena in the n-r.e. and n-REA degrees, Ph.D. disser-tation, University of Michigan, v+107 pp., 1995.

[13] A. Nies, R. Shore, Interpreting true arithmetic in the theory of the r.e.truth-table degrees, Annals of Pure and Applied Logic, vol. 75(1995)no. 3, 269 - 311.

[14] P.G. Odifreddi, Classical Recursion Theory. (Elsevier Science B.V.,1989).

[15] E.L. Post, Recursively enumerable sets of positive integers and their de-cision problems, Bulletin of the American Mathematical Society, vol. 50(1944), pp. 284-316

[16] G. Sacks, The recursively enumerable degrees are dense, Annals ofMathematics (2), vol. 80, pp. 300-312.

[17] R.I. Soare, Recursively Enumerable Sets and Degrees, 1st ed.(Springer-Verlag, Berlin, Heidelberg, 1987).

Department of MathematicsUniversity of DallasIrving, Texas 75062 USAemail: tmcnicho@acad.udallas.edu

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