Selected Topics in the Theory of the Computably Enumerable Sets and Degrees

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1 Selected Topics in the Theory of the Computably Enumerable Sets and Degrees Klaus Ambos-Spies Ruprecht-Karls-Universität Heidelberg Institut für Informatik Handout 3-8 November Effective reducibilities and their degrees of unsolvability In this section we introduce some of the most common strong reducibilities and classify them according to their strength. We start with some general definitions on reducibilities (Section 4.1). We then introduce truth-table reducibility and bounded versions of it (Sections 4.2 and 4.3). In Section 4.4 we use the finite extension method in order to separate the reducibilities of truth-table type, and in Section 4.5 we show how this approach can be formalized in order to get generic separations of these reducibilities. In particular, there we define (weakly) 1-generic sets, a notion which is of independent interest. In Section 4.6 we show that the separations can be obtained on the c.e. sets too. For this sake we introduce the priority method which is the fundamental technique for analysing c.e. sets. We then turn to another type of strong reducibilities which are obtained by imposing by imposing bounds on the oracle queries (i.e., use function) in a Turing reduction (Section 4.7). More specifically we introduce weak truth table reducibility and the strongly bounded Turing reducibilities, ibt and cl, and we separate these reducibilities on the c.e. sets. Finally we separate wtt-reducibility from Turing reducibility (Section 4.8), compare the truth-table type reducibilities and the bounded Turing reducibilities (Section 4.9) and summarize the results of this section (Section 4.10). 4.1 General definitions A binary relation r on sets is a reducibility if it is reflexive (A r A for all A) and transitive (A r B and B r C implies A r C for all A, B, C). For any reducibility r, sets A and B are called r-equivalent (A = r B) if they are reducible to each other, A = r B A r B & B r A, and A and B are called r-incomparable (A r B) if neither is reducible to the other, i.e., A r B A r B & B r A. 1

2 For any reducibility r, r-equivalence is an equivalence relation. The r-degrees are the corresponding equivalence classes, and we let deg r (A) = {B : A = r B} be the r-degree of A. Then r induces a partial ordering on the r-degrees, a b A a B b (A r B). The partial ordering of the r-degrees is denoted by (D r, ). The partial ordering of the c.e. r-degrees is denoted by (R r, ), where an r-degree is computably enumerable (computable) if it contains a c.e. (computable) set. A reducibility r is stronger than a reducibility r (and r is weaker than r) if, for all sets A and B such that A r B, it holds that A r B. 1 Note that if r is stronger than r then the r-degree of any set A is contained in the r -degree of A, deg r (A) deg r (A). A reducibility r is an effective reducibility if r is stronger than Turing reducibility. So, for an effective reducibility r, A r B implies that A is (Turing) computable in B. Also note that any effective reducibility has the countable predecessor property, i.e., for any set A, the class {B : B r A} is countable. So, in particular, any r-degree is countable whence D r has the cardinality of the continuum. (On the other hand, for any reducibility r, R r is countable since there are only countably many c.e. sets and since r-degrees are nonempty and pairwise disjoint.) A reducibility r is closed under finite variants if for any sets A, B, Â, ˆB such that A r B and  and ˆB are finite variants of A and B (i.e.,  = A and ˆB = B),  r ˆB too; a reducibility r is computably invariant if for any computable permutation and any set A, A = r p(a); a reducibility r is sufficiently general if, for any sets A and B such that A is computable, A is r-reducible to B; and a reducibility r respects joins if, for any sets A and B, the r-degree of A B is the least upper bound (supremum, join) of deg r (A) and deg r (B) in the r-degrees, i.e., deg r (A B) = deg r (A) deg r (B). A set A is r-hard for a class C if C r A for all sets C C, and A is r-complete for C if A is r-hard for C and A C. A is r-hard (r-complete) if A is r-hard for E (r-complete for E). Note that if A is r-hard for C and r is stronger than r then A is r -hard for C (and similarly for complete in place of hard). Finally, we call a reducibility r regular if it is effective, closed under finite variants, computably invariant, sufficiently general, and respects joins, and we call r normal if it is regular and there are r-complete sets. 2 Moreover, we say that a reducibility has any of the above properties on a class if it has this property if we consider only sets from this class. For instance, Turing reducibility is normal whereas many-one reducibility is normal on 2 ω \ {, ω} but not normal (see Exercise 4.1). 1 We use stronger in the inclusive sense. I.e., any reducibility is stronger than itself. If we want to exclude this case we say strictly stronger. 2 The terms sufficiently general, respects joins, regular and normal introduced here are not standard and may be used with some different meanings in the literature. 2

3 Note that, for an effective reducibility r which is sufficiently general, the computable sets form an r-degree, denoted by {0}, and this is the least (c.e.) r-degree. If r respects joins then the partial orderings (D r, ) and (R r, ) are upper semilattices, and if there are r-complete sets then (R r, ) has a greatest element which we denote by 0. So, for a normal reducibility r, (R r, ) is a countable upper semilattice with least and greatest element (for the structure of (D r, ) see Exercise 4.4). Exercises. Exercise 4.1 (i) 1-reducibility is stronger than m-reducibility and m-reducibility is stronger than Turing reducibility. (ii) 1-reducibility is computably invariant. Moreover, any reducibility which is weaker than a computably invariant reducibility is computably invariant too. (iii) Turing reducibility is normal. Exercise 4.2 (i) Describe the partial ordering (C m, ) of the m-degrees of computable sets. (ii) Show that m-reducibility is regular on 2 ω \ {, ω} but not regular. Exercise 4.3 (i) Describe the partial ordering (C 1, ) of the 1-degrees of computable sets. (ii) Show that there are computable sets witnessing that 1-reducibility is not closed under finite variants, is not sufficiently general, and does not respect joins (though (C 1, ) is an upper semilattice). Exercise 4.4 For a regular reducibility r, the partial ordering (D r, ) of the r- degrees is an uncountable upper semilattice with the countable predecessor property which has a least element but does not have maximal elements (i.e., for any r-degree a there is an r-degree b such that a < b). 4.2 Truth-table reducibility While Turing reducibility captures the intuitive notion of relative computability, in many cases where A is computable in B this can be captured by a stronger reduction like a many-one-reduction or even a one-one-reduction. In this section we look at an effective reducibility of intermediate strength, truth-table reducibility, which was introduced by Post (1944). In a truth-table reduction all oracle queries and an evaluation function of the oracle answers to the queries are specified in advance. This has been formalized in various (equivalent) ways. Definition 4.5 A truth-table reduction (tt-reduction for short) is a pair (g, h) of computable functions g : ω ω (the selector function) and h : ω {0, 1} {0, 1} (the evaluation function). 3 A set A is tt-reducible to a set B via (f, g) if for all numbers x A(x) = h(x, B(y 0 ),..., B(y n )) where g(x) = (y 0,..., y n ). 3 Formally, these functions cannot be formally computable since they are from the wrong type. So, in a formal definition, we have to code finite sequences of numbers by single numbers thereby replacing ω by ω. 3

4 And A is tt-reducible to B (A tt B) if A is tt-reducible to B via some ttreduction (g, h). Alternatively, we can describe truth-table reducibility in terms of total Turing functionals where a Turing functional Ψ is total if Ψ X (x) for all sets X and numbers x. (This characterization has been independently found by Trakhtenbrot (1955) and Nerode (1957).) Lemma 4.6 (Characterization Lemma for tt-reducibility) For any sets A and B the following are equivalent. (i) A tt B. (ii) There is a total Turing functional Ψ such that A T B via Ψ, i.e., A = Ψ B. Proof. (i) (ii). Fix (g, h) such that A tt B via (g, h). Then this reduction can be simulated by a total oracle machine M as follows. M X on input x first computes g(x) = (y 0,..., y n ) (without using oracle X), then, by querying the oracle about y 0,..., y n, M X produces X(y 0 ),..., X(y n ), and, finally, M X computes h(x, X(y 0 ),..., X(y n )) (without using the oracle). (ii) (i). Assume that A = Ψ B where the Turing functional Ψ is total. Then, for any x, the computation tree CT Ψ (x) of Ψ on input x is finite. So the depth f(x) = depth(ct Ψ (x)) of CT Ψ (x) is finite and computable from x. Moreover, f(x) is a bound on the length of the computations Ψ X (x) for any oracle X, hence a bound on the oracle queries in any of the computations Ψ X (x) (for X ω). So if we let g(x) = (0, 1, 2,..., f(x)) and h(x, i 0,..., i f(x) ) = Ψ {y f(x):iy=1} (x) then g and h are computable and A tt B via (g, h). Exercises. Exercise 4.7 Truth-table reducibility is normal. 4.3 Bounded truth-table reducibilities In a truth-table reduction the number of queries may grow with the input. In a bounded truth-table reduction the number of queries is fixed and does not depend on the input. This number is called the norm of the reduction. Definition 4.8 Let k 1. A truth-table reduction (g, h) is a truth-table reduction of norm k or a k-tt-reduction for short if the selection function is of type g : ω ω k. A is k-tt-reducible to B if there is a k-tt-reduction (g, h) such that A is tt-reducible to B via (g, h). 4 (ii) A truth-table reduction (g, h) is a bounded truth-table (btt) reduction if (g, h) is a k-tt-reduction for some k 1. A is btt-reducible to B if A is k-tt-reducible to B for some k 1. 4 Note that for a k-tt-reduction (g, h) we may assume that the evaluator h is of type h : ω {0, 1} k {0, 1} since the other values of h are not used in the definition of the reduction. In the following we tacitly make this assumption. 4

5 As one can easily check, 1-tt and btt are normal reducibilities (whereas, for k 2, k-tt-reducibility is not transitive hence not a reducibility in our sense; see Exercise 4.21 below). Also note that a many-one reduction f may be viewed as a 1-tt reduction (f, h) where f is the selector function and the evaluation is positive, i.e., h(x, i) = i. Exercises. Exercise tt-reducibility and bounded truth-table reducibility are normal. 4.4 Separating the reducibilities of truth-table type on the noncomputable sets The following implications among the reducibilities of truth-table type (including 1 and m ) are immediate by definition. A 1 B A m B A 1 tt B A btt B A tt B (1) Here we will show that these implications are strict even if we consider only noncomputable sets. (We ignore the computable sets in order to avoid trivialities when dealing with the first two implications. Note that all computable sets are 1-tt-equivalent whereas by Exercises 4.2 and 4.3 there are 3 m-degrees and infinitely many 1-degrees of computable sets. So, on the computable sets, 1- reducibility is strictly stronger than m-reducibility and m-reducibility is strictly stronger than 1-tt-reducibility whereas 1-tt-reducibility, btt-reducibility and ttreducibility are sufficiently weak hence collapse all computable sets in a single degree.) For a reducibility r and the next weaker reducibility r in (1) we separate r from r as follows. First, for any set A we define a set A r such that A r r A for all sets A where the canonical reductions use some features provided by r- reductions but not by r -reductions. Then, by a diagonal argument, we define a specific (noncomputable) set A such that all r -reductions fail to reduce A r to A. We start with separating 1-reducibility from m-reducibility. The property A m which we consider here is the self-join A A of A. Lemma 4.10 For any set A, A A m A. Proof. Obviously, A A m A via f where f(2x) = f(2x + 1) = x. Note that the canonical reduction function f above is not one-to-one. We exploit this fact for (not effectively) constructing a set A such that A A 1 A. For this sake we use a special form of a diagonalization argument, called a finite extension argument. In a finite extension argument, a set A with some desired property (or properties) is constructed. In a first step this property (or these properties) are split into a list {R e } e 0 of finitary requirements such that meeting these requirements will guarantee the required property/ies. Then, inductively, longer and longer initial segments of (the characteristic sequence of) A, A l(s), are defined where 5

6 l(0) = 0 and where l(s+1) > l(s) and the finite extension A l(s+1) of A l(s) are chosen so that the requirement R s is met (no matter how A will look like on the numbers l(s + 1)). Usually this inductive definition is given as a construction in stages s (s 0). Stage 0 is vacuous, i.e., l(0) = 0 and A l(0) = λ. Then, given l(s) and A l(s), at stage s + 1 the new length l(s + 1) of the extended initial segment of A and the extension A l(s + 1) of A l(s) are defined. So, in the following, when we give a finite extension argument then we only list the requirements {R e } e 0, describe stage s + 1 of the construction, and explain why the finite extension A l(s + 1) of the previously given initial segment A l(s) guarantees that requirement R s is met. A further observation will be helpful. Note that the definition of the new length l(s + 1) and of the extended initial segment A l(s + 1) depends on l(s) and A l(s). But the requirements are of a form such that no matter what l(s) and A l(s) are, we can find the appropriate extension A l(s + 1) guaranteeing that requirement R s is met. This observation allows us to merge two finite extension arguments in one: if the required properties P 0 and P 1 can be guaranteed by the requirements {Re} 0 e 0 and {Re} 1 e 0, respectively, then we can define one set A with property P 0 and property P 1 by meeting requirement Rs 0 at (the odd) stage 2s + 1 and requirement Rs 1 at (the even) stage 2s + 2 (i.e., by working with the requirements {R e } e 0 where R 2e = Re 0 and R 2e+1 = Re). 1 In the following we exploit this. When we have to construct a noncomputable set A with a certain property P then we may ignore the task of making A noncomputable. Namely, noncomputability can be enforced by a direct diagonalization and such a diagonalization can be performed by a simple finite extension argument. To be more precise, in order to make A noncomputable, it suffices to meet the requirements R e : A ϕ e for e 0, and given l(s) and A l(s), A l(s+1) will guarantee that requirement R s is met if we let l(s + 1) = l(s) + 1 and let A(l(s)) = 0 if ϕ s (l(s)) = 1 and A(l(s)) = 1 otherwise. Lemma 4.11 There is a noncomputable set A, such that A A 1 A. Proof. As pointed out above, it suffices to construct a set A by a finite extension argument such that A A 1 A. For this sake it suffices to meet the requirements R e : If ϕ e is total and one-to-one then A(x) A(ϕ e (2x + j)) for some x 0 and j 1. for e 0. (Note that R e ensures that A A is not 1-reducible to A via ϕ e. Namely, if ϕ e is total and one-to-one and A(x) A(ϕ e (2x + j)) then, by A(x) = (A A)(2x) = (A A)(2x + 1), (A A)(2x + j) A(ϕ e (2x + j)).) Given l(s) and A l(s) we can guarantee that R s is met by defining l(s + 1) and A l(s + 1) as follows. Let x = l(s). If ϕ s (2x) or ϕ s (2x + 1) or ϕ s (2x) = ϕ s (2x + 1) then R s is trivially met (since either ϕ s is not total or ϕ s 6

7 is not one-to-one) and we may let l(s + 1) = l(s) + 1 and A l(s + 1) = A l(s) (i.e., A(l(s)) = 0). So we may fix y 0 and y 1 such that ϕ s (2x + j) = y j (j = 0, 1) where y 0 y 1. Now, it suffices to ensure that A(x) A(y j ) for some j 1. If y j < l(s) for some j (whence A(y j ) has been specified already) then this can be ensured by letting l(s + 1) = l(s) + 1 = x + 1 and A(x) = 1 A(y j ). Finally, if y 0, y 1 l(s), then for j such that y j 1 < y j, x < y j. So we can ensure A(x) A(y j ) by letting l(s + 1) = y j + 1 and by letting A(x) = 0 and A(y) = 1 for all y such that x < y < l(s + 1). Theorem 4.12 There are noncomputable sets A and B such A m A 1 B. B and Proof. By Lemmas 4.10 and For separating m-reducibility from 1-tt-reducibilty we let A 1-tt be the complement of A. Lemma 4.13 For any set A, A 1-tt A. Proof. Obviously, A 1-tt A via (g, h) where g(x) = x and h(x, i) = 1 i. Note that the canonical 1-tt-reduction above is not positive hence not an m-reduction. Lemma 4.14 There is a noncomputable set A, such that A m A. Proof. The requirements R e : If ϕ e is total then A(x) = A(ϕ e (x)) for some x. for e 0 ensure that A m A. Given l(s) and A l(s) we can guarantee that R s is met by defining l(s+1) and A l(s+1) as follows. Let x = l(s). If ϕ s (x) then R s is trivially met and we may let l(s + 1) = l(s) + 1 and A l(s + 1) = A l(s) (i.e., A(l(s)) = 0). So w.l.o.g. we may fix y such that ϕ s (x) = y. It suffices to ensure that A(x) = A(y). If y < l(s) (whence A(y) has been specifies already) then this can be ensured by letting l(s + 1) = l(s) + 1 = x + 1 and A(x) = A(y). Finally, if y l(s) then we can ensure A(x) = A(y) by letting l(s + 1) = y + 1 and by letting A(z) = 0 for all z such that x z < l(s + 1). Theorem 4.15 There are noncomputable sets A and B such that A 1-tt B and A m B. Proof. By Lemmas 4.13 and (For an alternative, structural proof of Theorem 4.15, see Exercise 4.18 below.) For the remaining two separations we only define the sets used for the separations but leave the proofs which follow the same pattern as an exercise. In the first case we do not only consider the separation of 1-tt from btt but also the stronger separations of k-tt from (k + 1)-tt for all k 1. For any set A let A k = {x : {x, x + 1,..., x + k} A } (k 1) 7

8 and A = {x : {x, x + 1,..., 2x} A }. Note that the canonical reduction of A k to A is a (k +1)-tt-reduction (but not a k-tt-reduction) and the canonical reduction of A to a A is a tt-reduction (but not a btt-reduction). Theorem 4.16 For any k 1 and for any set A, A k (k+1)-tt A. On the other hand, for any k 1 there is a (noncomputable) set A such that A k k-tt A. Hence, for any k 1, there are (noncomputable) sets A and B such that A (k+1)-tt B and A k-tt B. So, in particular, 1-tt-reducibility is strictly stronger than btt-reducibility. Theorem 4.17 For any set A, A tt A. On the other hand, there is a (noncomputable) set A such that A btt A. Hence btt-reducibility is strictly stronger than tt-reducibility. Note that the constructions above are effective relative to the (general) halting problem. So the above separations can be obtained by noncomputable sets which are K-computable. In order to get the separations (where possible) on the c.e. sets, we need some more sophisticated diagonalization arguments. Before we introduce these techniques in Section 4.6 below, we will have a closer look at the finite extensions method and show that it provides generic separations of the reducibilities of truth-table type. Exercises. Exercise 4.18 (i) Show that the class E of the c.e. sets is closed downwards under m. (ii) Conclude from (i) that, for any noncomputable c.e. set A, the set A and its complement A are m-incomparable, i.e., A m A and A m A. So, in particular, K m K and K m K. (iii) Conclude from (ii) that, for any noncomputable c.e. set A there is a set B such that A and B are 1-tt-equivalent and m-incomparable. Exercise 4.19 Prove Theorem In particular, by a finite extension argument construct a set A such that A k k-tt A. Exercise 4.20 Prove Theorem In particular, by a finite extension argument construct a set A such that A btt A. Exercise 4.21 Show that, for k 2, k-tt-reducibility is not transitive. (Hint. Apply Theorem 4.16.) 4.5 Separating the reducibilities of truth-table type by generic sets In the preceding section where we used the finite extension method in order to separate the reducibilities of truth-table type on the noncomputable sets, we already observed that two finite extension arguments can be combined in one. This can be easily extended to countably many properties. So if we have countably many properties which can be forced by a finite extension argument 8

9 then there will be one set with all of these properties. In particular, if we consider only properties which can be forced by finite extension constructions where the potential extensions are uniformly computably enumerable (of which there are only countably many!) then there are sets A which have all of these properties. Here we define Σ 1 -generic (or 1-generic for short) sets which have these properties, and we will show that there are 1-generic sets computable in K. Using this fact, we get simplified modular proofs of the results in the preceding section. We start with some auxiliary definitions used in the definition of the 1- generic sets. We call a set S {0, 1} <ω of strings a condition set or, shortly, a condition, and we say that S is a c.e. condition if the set Ŝ = { σ : σ S} of the codes of the strings in S is c.e. Definition 4.22 Let S be a condition set and let A be (the characteristic sequence of) a set of natural numbers. (i) S is dense if σ τ σ [τ S]. (ii) S is dense along A if n τ A n [τ S]. (iii) A meets S if (iv) A avoids S if n [A n S]. n τ A n [τ S]. Note that a dense condition set S is dense along all sets A ω. Also note that a set A avoids a condition S if and only if S is not dense along A. Definition 4.23 (Hinman, Jockusch) (a) A set A ω is Σ 1 -generic 1-generic for short) if (or S {0, 1} <ω c.e. [A meets S or A avoids S]. (2) (b) A set A is weakly 1-generic if A meets any c.e. condition which is dense. Note that (2) is equivalent to S {0, 1} <ω c.e. [S dense along A A meets S]. (3) So, in particular, a 1-generic set meets all dense c.e. conditions. So 1-generic sets are weakly 1-generic. Now, the finite extension arguments considered in the previous subsection can be captured by the above notions as follows. There, we wanted to get a set A with a certain property P. In the first step we split this property in a sequence {R e } e 0 of requirements such that any set A which meets all of these 9

10 requirements has property P. Then we inductively defined longer and longer initial segments A l(s) of A such that the finite extension A l(s + 1) of A l(s) ensured that A meets requirement R s. In order to be able to find such an extension, requirement R s must have the following property. No matter what string σ is given (σ = A l(s)) there is a string τ σ such that any set A with τ A meets the requirement R s (so we may let τ = A l(s + 1)). So if we say that τ forces R s if the latter holds and let S Rs be the set of all strings τ forcing R s then, by the former (extendibility) property, the condition set S Rs is dense while, by the latter (forcing) property, any set A which meets S Rs meets requirement R s. So in order to show that any weakly 1-generic set (hence any 1-generic set) A has property P it suffices to show that the conditions S Rs are c.e. The advantage of this genericity approach is that we get one set A (namely any (weakly) 1-generic set) with all of the desired properties and that, in order to show that A has a desired property, it suffices to analyse a single corresponding requirement (which gives a simplified modular approach ). We illustrate this by the following theorem. Theorem 4.24 Let A be (weakly) 1-generic. Then the following hold. (i) A is not computable. (ii) A A 1 A. (iii) A m A. (iv) A k k-tt A (for k 1). (v) A btt A. Proof. We prove the first two parts of the theorem and leave the other parts as an exercise. Since any 1-generic set is weakly 1-generic, w.l.o.g. we may assume that A is weakly 1-generic (hence meets any c.e. condition which is dense). (i). Given a total computable function f, we have to show that A f. Since A is weakly 1-generic, it suffices to define a condition S such that S is c.e. and dense and such that, for any set X which meets S, X f. Such a condition S is defined as follows. For any string σ let σ be the string of length 1 defined by σ = (1 f( σ )). Then S = {σσ : σ {0, 1} <ω }. (4) Obviously, S is dense. Moreover, since σ can be computed from σ and since τ S iff there is an initial segment σ of τ such that τ = σσ, S is computable (hence c.e.). Finally, if X (n + 1) S then X(n) f(n). So S has the required properties. (ii) Given a computable one-to-one function f, we have to show that A A is not 1-reducible to A via f, i.e., that there is a number n such that A(n) A(f(2n)) or A(n) A(f(2n+1)) (note that A(n) = (A A)(2n) = (A A)(2n+ 1)). Since A is weakly 1-generic, it suffices to define a condition S such that S is c.e. and dense and such that, for any set X which meets S, X(n) X(f(2n+i)) for some n 0 and i 1. Such a condition S is obtained by defining S by (4) 10

11 where, for any string σ, the string σ is defined as follows. Let n be the length of σ and fix j 1 such that f(2n + j) > f(2n + (1 j)). Then { σ (1 σ(f(2n + (1 j)))) if f(2n + (1 j)) < n = 01 f(2n+j) n otherwise. (Note that, in the first case of the definition of σ, σσ (n) σσ (f(2n+(1 j))) while, in the second case, n < f(2n + j) < σσ and σσ (n) σσ (f(2n + j)).) Obviously S is dense and, since σ can be computed from σ, S is computable (hence c.e.). Finally, assume that X (m + 1) S. Fix σ such that X (m + 1) = σσ, let n = σ and fix j such that f(2n + j) > f(2n + (1 j)). Then, by definition of σ, X(n) X(f(2n + (1 j))) if f(2n + (1 j)) < n and X(n) X(f(2n + (1 j))) otherwise. So S has the required properties. In order to show that Theorem 4.24 gives a separation of the reducibilities of truth-table type on the noncomputable sets ( T K), it remains to show that there is a (weakly) 1-generic set A ( T K). 5 Theorem 4.25 (Jockusch (?)) There is a 1-generic set G T K. Proof. It suffices to construct a set G T K meeting the requirements R s : If W s is dense along G then G meets W s. for all s 0 (where, by identifying a condition with its set of codes, we view the c.e. set W s as a condition). Such a set G is constructed by a finite extension argument where stage s + 1 is as follows. Given l(s) and γ s = G l(s), l(s + 1) > l(s) and γ s+1 = G l(s + 1) are defined as follows: If there is a string τ W s such that τ γ s then, for the least such τ let l(s + 1) = τ and γ s+1 = τ. Otherwise let l(s + 1) = l(s) + 1 and γ s+1 = γ s 0. 5 The finite extension method is closely related to the theory of Baire Category in topology. By identifying reals with (characteristic sequences of) sets of natural numbers, the basic notions of this theory are as follows. A class C of sets is open dense if there is a dense condition S such that a set A is in C iff it extends a string σ in S; and a class C is comeager if it contains the intersection of countably many open dense classes as a subclass. Then Baire s Theorem says that any comeager class is nonempty (in fact not countable). Note that the class of the sets sharing a property P which can be forced by a finite extension argument is comeager, hence by Baire s Theorem nonempty. Namely, the condition S Re corresponding to a requirement R e is dense whence the class of sets meeting requirement R e is open dense. Since there are only countable many requirements, it follows that the class of sets with property P is comeager. Similarly, since there are only countably many c.e. conditions (hence only countable many dense c.e. conditions) we may argue that the class of weakly 1-generic sets is comeager. So, by Baire s Theorem, weakly 1-generic sets exist. But Baire s Theorem does not give us the stronger result, namely that there are weakly 1-generic sets computable in K. (Namely, the class of sets computable in K is countable hence might be in the complement of a comeager class.) Finally, we remark that Baire s Theorem also yields the existence of 1- generic sets. For this sake it suffices to note that there is a countable class of dense conditions such that any set which meets these conditions is 1-generic (see Exercise 4.30). 11

12 Obviously, this guarantees that requirement R s is met. (Namely, in the first case of the definition of τ, G meets W s and, in the second case of the definition of τ, G avoids W s.) Moreover, the oracle K can tell whether a string τ as above exists and, if so, K can compute the least such τ. So the construction is effective in K hence G T K. Do the 1-generic sets provide separations of the reducibilities of truth-table type on the computably enumerable sets too? For this sake it would suffice to show that there is a computably enumerable (weakly) 1-generic set. But this is not the case. Lemma 4.26 Let A be (weakly) 1-generic. Then A is not c.e. Proof. Since, by Theorem 4.24 (i), (weakly) 1-generic sets are not computable hence infinite, given an infinite c.e. set B it suffices to show that A B. To do so, consider the condition S = {σ0 : σ B}. Then S is dense (since B is infinite) and S is c.e. (since B is c.e.). So A meets S, say A n+1 S. But, by definition of S, this implies A(n) = 0 1 = B(n). In fact, for any 1-generic set A, the Turing degree of A is not c.e. By noncomputability of 1-generic sets, this follows from the following even stronger result. Theorem 4.27 (Reference?) Let A and B be sets such that A is 1-generic, B T A and B is c.e. Then B is computable. Proof. It suffices to show that the complement B of B is c.e. Fix a Turing functional Φ such that B = Φ A, and define the condition set S by S = {σ : x (x B & Φ σ (x) = 0)} Then S is c.e. (since B is c.e.) and A does not meet S (since B = Φ A ). Since A is 1-generic it follows that A avoids S. So there is a string γ A such that σ γ x [Φ σ (x) = 0 B(x) = 0] (5) On the other hand, by B = Φ A, But (5) and (6) imply x [B(x) = 0 σ γ (Φ σ (x) = 0)] (6) x B σ γ [Φ σ (x) = 0] So B is computably enumerable. Corollary 4.28 (Shoenfield) There is a set A T K such that deg T (A) is not c.e. 12

13 Proof. By Theorem 4.25 let A be a 1-generic set such that A T K. By Theorem 4.27, A is not Turing equivalent to any c.e. set. Exercises. Exercise 4.29 Show that the complement of a (weakly) 1-generic set is (weakly) 1-generic too. Exercise 4.30 Call a set A weakly 2-generic if A meets all dense Σ 2 -conditions where a set B is Σ 2 if there is a computable relation R such that x B iff y 0... y m z 0... z n R(x, y 0,..., y m, z 0,..., z). Show that any weakly 2- generic set is 1-generic (Kurtz). (Hint: Given a c.e. condition C consider the following Σ 2 -condition Ĉ = {σ : σ C τ σ (τ C)}.) Exercise 4.31 Prove parts (iii) - (v) of Theorem Exercise 4.32 For a 1-generic set, the even part {x : 2x A} and the odd part {x : 2x + 1 A} are Turing incomparable. 4.6 Separating the reducibilities of truth-table type on the noncomputable c.e. sets If we consider only noncomputable c.e sets then the implications in (1) are strict with exception of the implication A m B A 1 tt B. (For the equivalence of m-reducibility and 1-tt-reducibility on the noncomputable c.e. sets see Exercise 4.37.) Here the separations require a more sophisticated technique, however, namely a so-called priority argument or finite injury argument. Theorem 4.33 (i) There are noncomputable c.e. sets A and B such that A m B and A 1 B. (ii) For any k 1 there are (noncomputable) c.e. sets A and B such that A (k+1)-tt B and A k-tt B. (Hence, in particular, there are c.e. sets A and B such that A btt B and A 1-tt B.) (iii) There are c.e. sets A and B such that A tt B and A btt B. Since, for any c.e. set A, the sets A A, A k and A (defined in Section 4.4) are c.e. too 6, it suffices to prove the following lemmas (compare with Lemma 4.11, Theorem 4.16 and Theorem 4.17 in Section 4.4). Lemma 4.34 There is a noncomputable c.e. set A such that A A 1 A. Lemma 4.35 For any k 1 there is a c.e. set A such that A k k-tt A. Lemma 4.36 There is a c.e. set A such that A btt A. 6 Note that the separation of m-reducibility and 1-tt-reducibility on the noncomputable sets (which fails on the noncomputable c.e. sets) was obtained by considering the complement A of A and that, for noncomputable c.e. A, A is not c.e. 13

14 Proof (of Lemma 4.34). We enumerate a c.e. set A with the desired properties in stages, i.e., we define a computable enumeration {A s } s 0 of A where A 0 = and A s+1 is defined at stage s + 1 of the construction. Just as in the proof of Lemma 4.11, it suffices to meet the requirements R e : If ϕ e is total and one-to-one then A(x) A(ϕ e (2x + j)) for some x 0 and j 1. for e 0. (We will argue below that the construction automatically ensures that A is not computable.) In contrast to the (nonconstructive) proof of Lemma 4.11, however, here we cannot meet one requirement after the other but we have to work on more and more requirements simultaneously. The reason is that we cannot effectively tell whether, for a potential diagonalization candidate x, ϕ e (2x) and ϕ e (2x + 1) are defined and differ. (Note that the requirement R e is trivially met if this not the case, whereas if this is the case then we can meet R e by appropriately defining A on x (and ϕ e (2x) and ϕ e (2x + 1)).) The strategy for meeting a single requirement R e is as follows. 1. Pick a number x = x e which may be enumerated into A only for the sake of requirement R e. This number is called a follower of R e. 2. Wait for a stage s such that ϕ e,s (2x), ϕ e,s (2x + 1) and ϕ e,s (2x) ϕ e,s (2x + 1). (Note that if there is no such stage s then the hypothesis of R e fails hence R e is trivially met.) Then, at stage s + 1, do the following. Fix j 1 minimal such that ϕ e (2x + j) x. If ϕ e (2x + j) A s then do nothing (hence, by keeping x out of A forever, A(x) A(ϕ e (2x + j)); and if ϕ e (2x + j) A s then put x into A at stage s + 1 and keep ϕ e (2x + j) out of A forever (hence, again, A(x) A(ϕ e (2x + j))). In either case declare R e to be satisfied at stage s + 1. Note that this strategy may negatively affect the strategy for meeting some other requirement R e. This happens if we have to keep ϕ e (2x + j) out of A and x e = ϕ e (2x + j) is the follower of R e. Then, it may happen, that the R e -strategy wants to put x e into A at a later stage whereas the R e -strategy wants to keep x e out of A. In order to resolve this conflict, we give R e higher priority than R e iff e < e. So if R e wants to keep ϕ e (2x + j) out of A then it cancels all followers of lower priority at stage s+1 and requests that any follower appointed later is greater than ϕ e (2x + j). (Note that, by our convention, this can be ensured by making followers greater than or equal to the stage they are appointed at.) Since a strategy will cancel lower priority followers only once (unless its own follower is cancelled later), it follows by induction that eventually any requirement has a permanent follower and that this follower witnesses that the requirement is met. The above discussion leads to the following construction of A where initializing a requirement means that its follower is cancelled (if there is any) and the requirement is declared to be not satisfied. Moreover, if x e is the follower of R e (or R e is satisfied) at the end of stage s then x e is the follower of R e (or R e is satisfied) at the end of stage s + 1 unless R e is initialized at stage s + 1. Similarly, A s+1 = A s unless we explicitly say that a number x is enumerated into A at stage s

15 Stage 0. Let A 0 = and initialize all requirements. Stage s + 1. Requirement R e requires attention at stage s + 1 if either (i) R e has no follower at the end of stage s or (ii) R e has follower x at the end of stage s, R e is not satisfied at the end of stage s, and ϕ e,s (2x) = ϕ e,s (2x + 1). Fix the least e such that R e requires attention. If (i) holds appoint x = s + 1 as R e -follower. If (ii) holds then declare R e to be satisfied; fix j 1 minimal such that ϕ e (2x + j) x; and put x into A iff ϕ e (2x + j) A s. In any case declare that R e receives attention or is active at stage s + 1. Moreover, for any e > e, initialize R e and declare R e to be injured. Obviously the construction is effective hence {A s } s 0 is a computable enumeration of A. So A is c.e. In order to show that the requirements R e are met we prove a more general claim by induction on e. Claim 1. For e 0, R e has a permanent follower x, requires attention at most finitely often and is met. Proof. Fix e and, by inductive hypothesis, fix s minimal such that no requirement R e with e < e requires attention after stage s e. Then R e is initalized at stage s e. Moreover, whenever R e requires attention after stage s e then it receives attention, and R e is not initialized after stage s e. So R e requires and receives attention via clause (i) at stage s e + 1, hence the R e -follower x = s e + 1 is appointed at stage s e + 1 and this follower is permanent. Hence, if it is not the case that ϕ e (2x) ϕ e (2x + 1), then R e does not require attention after stage s e + 1 and R e is trivially met since ϕ e is not total or not one-to-one. So for the remainder of the proof we may assume that ϕ e (2x) ϕ e (2x + 1). Hence we may pick j 1 minimal such that ϕ e (2x + j) x and we may fix s s e + 1 minimal such that ϕ e,s (2x) ϕ e,s (2x + 1). Then R e receives attention via clause (ii) at stage s + 1. Since R e is not initialized later, R e is satisfied at all stages s + 1 hence R e does not require attention after stage s + 1. Moreover, by construction, A s+1 (x) A s+1 (ϕ e (2x + j)). Since, neither R e nor a higher priority requirement R e (e < e) acts after stage s + 1 and since all lower priority requirements R e (e > e) are initialized at stage s + 1, it follows that no number s+1 enters A after stage s+1. Since x, ϕ e (2x+j) s it follows that A(x) A(ϕ e (2x + j)). So R e is met. (This completes the proof of Claim 1.) In order to complete the proof of Lemma 4.34 it only remains to show that A is noncomputable. Since, for any infinite and co-infinite computable set A, A A 1 A (see Exercise 4.2), it suffices to show that A and A are infinite. Since, for e < e, the permanent follower x e of R e is less than the permanent follower x e of R e this can be done as follows. For the former it suffices to note that there are infinitely many e such that ϕ e (x) = x for all x and that, for such e and for the permanent R e -follower x e, at least one of the numbers x e, 2x e, 2x e + 1 is in A. For the latter it suffices to note that there are infinitely many e such that ϕ e is nowhere defined and that, for such e and for the permanent R e -follower x e, x e A. 15

16 We leave the proofs of Lemmas 4.35 and 4.36 as exercises. Exercises. Exercise 4.37 Let A and B be c.e. sets such that B, ω and A 1 tt B. Show that A m B. (Hint: Show that the negative queries can be answered effectively using the fact that A and B are c.e.) Exercise 4.38 (i) Prove Lemma (ii) Prove Lemma Bounded Turing reducibilities An alternative way to strengthen Turing reducibility is to impose some bounds on the use functions of the reductions, i.e., on the size of the oracle queries. Definition Let f be any function. A set A is f-bounded Turing reducible to a set B if A is Turing reducible to B via a reduction where the use function is bounded by f, i.e., if there is a number e such that A = Φ B e and ϕ B e (x) f(x) + 1 for all x A set A is bounded-turing (bt-) reducible or weak-truth-table (wtt-) reducible to a set B (A bt B or A wtt B) if A is f-bounded Turing reducible to B for some computable function f. 3. A set A is identity-bounded-turing (ibt-) reducible to a set B (A ibt B) if A is f-bounded Turing reducible to B for the identity function f(x) = x. 4. A set A is (i+k)-bt-reducible to a set B (A (i+k)bt B) if A is f-bounded Turing reducible to B for f(x) = x + k. 5. A set A is computable Lipschitz (cl-) reducible to a set B (A cl B) if A is (i + k)-bt-reducible to B for some k 0. We call a Turing functional Φ a bounded Turing (bt) functional or a wttfunctional if there is a total computable function f such that, for any set X and any number x such that Φ X (x), ϕ X (x) f(x) + 1. cl- and ibt-functionals are defined correspondingly. Note that in the definition of A wtt B we only require that there is a Turing functional Φ and a computable function f such that A = Φ B and the use function ϕ B with oracle B (but not necessarily the use function ϕ X for other oracles X) is bounded by f. Still we can convert the Turing functional Φ reducing A to B into a wtt-functional ˆΦ reducing A to B. Namely, ˆΦ X (x) simulates Φ X (x) step by step unless Φ X (x) asks an oracle query y > f(x). In this case the oracle query is supressed and the computation is aborted. Obviously, ˆΦ is a wtt-functional (via the bound f) and, since the use function of Φ B is bounded by f, Φ B and ˆΦ B agree. So A bt B via the wtt-functional ˆΦ. Similarly, any cl-reduction and any ibt-reduction is witnessed by a cl-functional and ibt-functional, respectively. In the following we will refer to identity-bounded Turing reducibility and computable Lipschitz reducibility as the strongly bounded Turing (sbt) reducibilities. Obviously, A ibt B A cl B A wtt B A T B (7) 16

17 holds. In fact these implications are strict, even if we restrict ourselves to c.e. sets. Strictness of the first two implications are obtained by simple shift arguments. (Strictness of the third implication is shown in Section 4.8 below.) Theorem 4.40 (Downey et al. [5], Barmpalias and Lewis [3]) Let A be a noncomputable set. (i) For A + 1 = {x + 1 : x A}, A + 1 < ibt A and A + 1 = cl A. (ii) For 2A = {2x : x A}, 2A < cl A and 2A = wtt A. (Note that, for c.e. A, the sets A + 1 and 2A are c.e. again.) Proof. We give the proof of (i). The proof of (ii) is similar. Obviously, A + 1 = cl A and A + 1 ibt A. So it suffices to show that A ibt A + 1. For a contradiction assume that A ibt A + 1 and fix an ibt-functional Φ such that A = Φ A+1. We claim that A is self-reducible, i.e., there is a reduction A = ˆΦ A such that on input x all queries of ˆΦ A (x) (if any) are less than x. Since self-reducible sets are computable this will complete the proof. The functional ˆΦ is obtained from Φ as follows. For x > 0, ˆΦ A (x) simulates Φ A+1 (x) replacing any oracle query y > 0 by the query y 1 (note that y x, hence y 1 < x) and by answering the query y = 0 negatively. Corollary 4.41 For any noncomputable c.e. set A there are noncomputable c.e. sets B and ˆB such that (i) A cl B but A ibt B and (ii) A wtt ˆB but A cl ˆB. Proof. By Theorem 4.40, it suffices to let B = A + 1 and ˆB = 2A. The preceding theorem also show that the strongly bounded Turing reducibilities are not weaker than 1-reducibility. Corollary 4.42 For any noncomputable c.e. set A there is a noncomputable c.e. set B such that A 1 B but A cl B. Proof. Let B = 2A. Then A 1 B via f(x) = 2x. So the claim follows by Theorem 4.40 (ii). Exercises. Exercise 4.43 (i) Show that ibt-reducibility (hence cl-reducibility) is sufficiently weak and closed under finite variants. (ii) Show that cl-reducibility (hence ibt-reducibility) is not computably invariant. (iii) Show that wtt-reducibility is a normal reducibility. 17

18 Exercise 4.44 (i) Are there ibt-complete sets? (ii) Show that there is an ibt-hard set H such that H T K. Hint (Merkle): Effectively split ω into intervals I n of length 2(n + 1). For any (sufficiently large) n effectively code all information about the sets W 0 I n+1,..., W n I n+1 into H I n. (Note that this guarantees W e ibt H for all e.) For this sake it suffices to code the e n such that W e is the last of the sets W 0,..., W n which changes on I n+1 and to code the number x I n which is enumerated into W e last (we may assume that, at any stage s + 1, there is at most one e and at most one x such that x W e,s+1 \ W e,s and that W 0 = ω; so e and x are unique and well defined). Moreover, e n and x can be coded by its position ˆx inside I n+1 where ˆx I n+1 hence ˆx 2(n + 2). Since any number < 2 n+1 can be coded by a binary string of length n + 1, for sufficiently large n there are strings σ e and σˆx of length n + 1 coding e and ˆx (hence x), and it suffices to let the (local) characteristic string of H I n be σ e σˆx. Finally note that K can compute e and x whence H T K. There is an alternative proof of this fact using Kolmogorov complexity and algorithmic randomness: any left-c.e. 1-random real is ibt-hard (see Downey and Hirschfeldt [4], Theorem ). Exercise 4.45 (i) Call f an unbounded shift if f is strictly increasing and f(x) x is unbounded, and let A f = {f(x) : x A} be the f-shift of A. Extend Theorem 4.40 (ii) by showing that, for any unbounded computable shift f, A f < cl A. (ii) (Ambos-Spies et al. [2]) Show that for any noncomputable c.e. set A there is a c.e. set B and an unbounded computable shift f such that A = ibt B f. (Hint: Let D be an infinite computable subset of A, let f be the function enumerating the complement D of D in order of magnitude, and define B by letting B(n) = A(f(n)).) (iii) (Barmpalias) Show that for any c.e. set A there is a c.e. set B such that A < cl B. Conclude that there are no cl-complete sets. 4.8 Separating wtt and Turing on the c.e. sets For separating weak truth-table reducibility and Turing reducibility on the c.e. sets, first for any c.e. set A we define a c.e. set  such that A T  but the canonical reduction from A T  is not necessarily a wtt-reduction. Then, by a priority argument, we enumerate a set A such that A wtt Â. Actually, the set  which we consider here does not only depend on A but also on a given computable enumeration {A s } s 0 of A. It is defined in terms of the following standard marker of A w.r.t. {A s } s 0. Definition 4.46 Let A be a c.e. set and let {A s } s 0 be a computable enumeration of A. The standard marker γ of A w.r.t. {A s } s 0 is defined by γ(x, 0) = x, 0 { x, s + 1 if A s+1 x + 1 A s x + 1 γ(x, s + 1) = γ(x, s) otherwise. The function γ should be viewed as a movable marker where γ(x, s) denotes the position of the marker γ(x) for x at stage s. Note that γ is computable 18

19 hence the marker moves are effective. If a marker position is moved then it is moved to a higher (greater) position which is at least as great as the current stage, and so are the positions of the markers for all greater numbers. At any stage s the marker positions are in order (i.e., if x < x then γ(x, s) < γ(x, s)), and for different numbers different positions are used (namely, by γ(x, s) ω [x], γ(x, s) γ(x, s ) for all x x and all stages s and s ). Note that the marker for x is moved only if A changes on the initial segment of length x + 1. So the marker for x can be moved at most x + 1 times, hence eventually reaches a final position γ (x), and the moves are controlled by A. In particular, A can tell whether a position is final or not, i.e., γ T A. Now the set  is just the set of the nonfinal marker positions of the standard marker for A. Lemma 4.47 Let A be a c.e. set and let {A s } s 0 be a computable enumeration of A, let γ be the standard marker of A w.r.t. {A s } s 0, and let  = {γ(x, s) : x, s 0 & s > s (γ(x, s ) γ(x, s))}. (8) Then  is c.e., A T  and  wtt A. Proof. Obviously,  is Σ 1 hence c.e. For a proof of A T  it suffices to note that x A iff x A s for the least s such that γ(x, s)  (namely, by the observations following Definition 4.46, γ(x, s) reaches a final position γ (x), this final position is the only position of γ(x) which is not in  and A does not change below x + 1 once this final position is reached). Finally, for a proof of  wtt A it suffices to note that x, s  iff γ(x, s) = x, s and A s x + 1 A x + 1 (whence the marker position γ(x, s) is not final). Theorem 4.48 There is a c.e. set A and a computable enumeration {A s } s 0 of A such that, for the corresponding c.e. set  defined by (8), A wtt Â. Proof. By a priority argument we define a computable enumeration {A s } s 0 of a c.e. set A such that the corresponding set  has the required properties. In order to ensure that A wtt  it suffices to meet the requirements R e : If ϕ e0 is total then A(x) Φ ϕe 0 (x) e 1 (x) for some x 0. for all e 0 where we assume that e = e 0, e 1. The basic idea how to meet requirement R e is as follows. We pick a follower x for which we want to ensure A(x) Φ ϕe 0 (x) e 1 (x) (provided that ϕ e0 (x) is defined; otherwise R e is met trivially). For this sake it suffices to enumerate x into A iff Φ ϕe 0 (x) e 1 (x) = 0. Now since the construction has to be effective, we have to work with approximations of the computation Φ ϕe 0 (x) e 1 (x). For this sake we first define a computable enumeration {Âs} s 0 of  by  s = {γ(x, t) : x, t < s & γ(x, t) γ(x, s)}. 19

20 (Note that Âs is given by the end of stage s of the construction since it only depends on A 0,..., A s.) Then, by the Use Lemma, ϕ e0 (x) and Φ ϕe 0 (x) e 1 (x) = 0 iff for all sufficiently large ϕ e0,s(x) & ΦÂs ϕe 0,s(x) e 1,s (x) = 0. (9) Hence if we do not see such a stage s then either ϕ e0 is not total or Φ ϕe 0 (x) e 1 (x) 0. So, in this case, R e will be met, provided that we keep x out of A. On the other hand, for any s such that (9) holds, it follows by the Use Lemma that we can preserve the computation ΦÂs ϕe 0 e,s(x) 1,s (x) thereby ensuring that (x) = 0 if we restrain all numbers < ϕ e0 (x) from  after stage s. So if we achieve the latter and at the same time put x into A at stage s + 1 then R e will be met. There is one problem with this, however: in order to restrain numbers < ϕ e0 (x) from  after stage s, we must not enumerate any number y into A after stage s such that γ(y, s) < ϕ e0 (x). So, since we have to put x into A at stage s + 1 which causes γ(y, s) to enter  at stage s + 1 for all y x, we have to be sure that γ(x, s) ϕ e0 (x)! This requires some extra action: when we appoint the follower x then we simultaneously appoint x 1 as clearer. Then, before we do the above attack, we wait for a stage s > x such that ϕ e0,s (x) s. Then, by enumerating the clearer x 1 into A at stage s + 1, for all y x 1, the marker position γ(y, s + 1) of y is moved Φ ϕe 0 (x) e 1 above ϕ e0 (x) at stage s + 1. So if we put x into A at a later stage s + 1 then γ(x, s) > ϕ e0 (x) whence the enumeration of γ(y, s) for y x into  does not injure the computation ΦÂs ϕe 0 e,s(x) 1,s (x). Summarizing, the basic strategy for meeting requirement R e is as follows. 1. Pick numbers x 1 and x which may be enumerated into A only for the sake of requirement R e. Call x 1 the clearer and x the follower of R e. 2. Wait for a stage s such that ϕ e0,s (x) < s. (Note that if there is no such stage s then ϕ e0 is not total and R e is trivially met.) Enumerate x 1 into A s +1. Note that this guarantees that γ(x, s) > ϕ e0 (x) for x x 1 and s > s. 3. Wait for a stage s > s such that ΦÂs ϕe 0 (x) e 1,s (x) = 0. (Note that if there is no such stage s then by the Use Principle Φ ϕe 0 (x) e 1 (x) 0 hence A(x) Φ ϕe 0 (x) e 1 (x) since x will not be enumerated into A.) Enumerate x into A s+1 and restrain all numbers y < x from A. (Note that the latter ensures that Âs ϕ e0 (x) =  ϕ e 0 (x). Namely if a marker position γ(y, t) will be enumerated into  after stage s then either x y s and t s + 1 or s y. In either case this implies that γ(y, t) ϕ e0 (x).) We leave the actual construction and verification as an exercise. Corollary 4.49 There are noncomputable c.e. sets A and B such that A T B and A wtt B. Proof. By Lemma 4.47 and Theorem Exercises. 20