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R ? S ?? ?f ?x, y [x R y ? f(x) S f(y)].

Here, f is taken from a suitable class of effective functions. For us the relations will be on natural numbers, and f must be computable. We show that there is a ?01-complete equivalence relation, but no ?0k-complete for k ? 2. We show that ?0k preorders arising naturally in the above-mentioned areas are ?0k -complete. This includes polynomial time m-reducibility on exponential time sets, which is ?02, almost inclusion on r.e. sets, which is ?03, and Turing reducibility on r.e. sets, which is ?04.

§1. Introduction. Mathematicians devote much of their energy to the classification of mathematical structures. In this endeavor, the tool most commonly used is the notion of a reduction: a map reducing a seemingly more complicated question to a simpler question. As a simple example, one classifies the countable rational vector spaces (under the equivalence relation of isomorphism) using the notion of dimension, by proving that two such vector spaces are isomorphic if and only if they have the same dimension over Q. More generally, we have the following definition. x E y

f(x) F f(y).

Definition 1.1. If E and F are equivalence relations on the domains DE and DF , a reduction from E to F is a function f : DE ? DF with the property that, for all x, y ? DE ,

?? In our example, E is the isomorphism relation on the class DE of countable rational vector spaces (with domain , say, to avoid set-theoretical issues), and F is the equality relation on the set {0, 1, 2, . . . , } of possible dimensions of these spaces.

A reduction f of a complicated equivalence relation E to a simpler equivalence relation F is entirely possible if one allows f to be sufficiently complex. (For instance, every equivalence reduction E with exactly n classes can be reduced to the equality relation on {1, 2, . . . , n}, but this reduction has the same complexity as E.) Normally, however, the goal is for f to be a readily understandable function, Received February 4, 2013. 2010 Mathematics Subject Classification. Primary 03D78; Secondary 03D32.

Key words and phrases. computable reducibility, computability theory, equivalence relations, m-reducibility, recursion theory. c 2014, Association for Symbolic Logic 0022-4812/14/7903-0012/$3.40

DOI:10.1017/jsl.2013.33
so that we can actually learn something from the reduction. In our example, one
should ask how hard it is to compute the dimension of a rational vector space. It
is natural to restrict the question to computable vector spaces over Q (i.e., those
where the vector addition is given as a Turing-computable function on the domain
of the space). Yet even when its domain DE such vector spaces, computing the
function which maps each one to its dimension requires a 0 -oracle, hence is not
as simple as one might have hoped. (The reasons why 0 is required can be gleaned
from [

1.1. Effective reductions. Reductions are normally ranked by the ease of
computing them. In the context of Borel theory, for instance, a large body of research is
devoted to the study of Borel reductions (the standard book reference is [

A further body of research is devoted to the study of the same question for
equivalence relations E and F on , and reductions f : ? between them
which are computable. If such a reduction from E to F exists, we say that E is
computably reducible to F , and write E ?c F , or often just E ? F . These reductions
will be the focus of this paper. Computable reducibility on equivalence relations was
perhaps first studied by Ershov [

The main purpose of this paper is to investigate the complexity of equivalence
relations under these reducibilities. In certain cases we will generalize from
equivalence relations to preorders on . We restrict most of our discussion to relatively low
levels of the hierarchy, usually to ?0n and ?0n with n ? 4. One can focus more closely
on very low levels: Such articles as [

1.2. Completeness. Equivalence relations on are subsets of 2, and so those with arithmetical definitions can be ranked in the usual hierarchy as ?0n and/or ?0n for various n ? . The question of completeness then naturally arises. A ?0n equivalence relation F is complete (among equivalence relations under ?c ) if for every ?0n equivalence relation E, we have E ?c F . ?0n-completeness is defined similarly. Much of this paper is devoted to the study of complete equivalence relations at various levels of the arithmetical hierarchy, and we will often speak of their ?0 n completeness or ?0n -completeness without specifying the reducibility ?c . At certain times we will need to consider other reducibilities on subsets of , such as Turing reducibility ?T or 1-reducibility ?1, and these reducibilities give rise to their own (analogous) notions of completeness, but when referring to those notions we will always name them specifically.

The standard notion of an m-reduction from one set to another is relevant here, since computable reducibility on equivalence relations can be seen as one of two natural ways to extend m-reducibility from sets to binary relations. However, when comparing the complexity between equivalence relations, it is more natural to use computable reducibility. Following Definition 1.1, we define:

A ?c B ?? ? computable g ?x, y [ x, y ? A ? g(x), g(y) ? B].

(1)

The resulting reducibility has many interesting and, at times, surprising properties. It can be readily seen that (1) is a stronger property than m-reducibility: A ?c B via g implies A ?m B via the function f given by f(x, y) = g(x), g(y) , so a relation complete for a level of the arithmetic hierarchy in the sense of relations as in (1) is also complete in the sense of sets. The converse fails: in Subsection 3.2 we shall see that some levels do not have complete equivalence relations at all.

It is easy to see that for every n, there exists an equivalence relation E that is ?0n -complete under ?c : let (Ve)e? be an effective list of all symmetric ?0n relations, let (Se )e? be the effective list of their transitive closure, and let E = e Se be their effective sum. In Section 3 we demonstrate the existence of ?01-complete equivalence relations, which is not nearly as obvious. We give a natural example from complexity, namely equality of quadratic time computable functions. In contrast, we also show that for n ? 2, no equivalence relation is ?0n-complete.

1.3. Preorders. Equivalence relations were the first context in which reductions
arose, but the notion of a reduction can be applied equally well to any pair of binary
relations: The E and F in Definition 1.1 need not be required to be equivalence
relations. In fact, a reduction could be considered as a homomorphism of
structures in the language containing a single binary relation symbol, whether or not that
symbol defines an equivalence relation on either structure. Preorders (i.e., reflexive
transitive binary relations , with x y x allowed even when x = y) form a
natural collection of binary relations on which to study reductions, and in Section 4
we will turn to the question of computable reducibility on preorders, with some
interesting uses of effectively inseparable Boolean algebras. In this way we obtain
some examples of natural ?0n -complete preorders from computability theory. For
instance, almost inclusion ?? of recursively enumerable (r.e.) sets, and weak truth
table reducibility on r.e. sets, are ?03 complete preorders. To obtain
corresponding completeness results for equivalence relations, we use the easy Fact 2.1 that
completeness of a preorder P at some level of the arithmetical hierarchy implies
completeness of the corresponding symmetric fragment which is an equivalence
relation. Using this we obtain, for instance, the ?03 completeness of several of the
equivalence relations considered in [

Our results can be applied to obtain completeness results for preorders and
equivalence relations from effective algebra. For instance, Ianovski [

The articles already cited here are among the principal references for this subject.
For ordinary computability-theoretic questions, we suggest [

§2. Preliminaries. Let X [

Fact 2.1. Let P be a preorder, E = {{x, y} : x, y ? P and y, x ? P}, and n ? . If P is complete among ?0n [?0n ] preorders under ?c , then E is complete among ?0n [?0n ] equivalence relations under ?c .

Proof. We give the proof for ?0 . Note that E is clearly ?0n. Let F be an arbitrary n ?0n equivalence relation. In particular, F can be seen as a preorder and so there is a computable function f such that for every x, y, {x, y} ? F iff f(x), f(y) ? P. Then, f witnesses that F ?c E.

In parts of the Subsections 3.1 and 4.3 we will assume familiarity with the basic
notions of computational complexity theory; see for instance [

Recall that a function h from numbers to strings with h(n) ? n is time constructible if h(n) can be computed in O(h(n)) steps on a multitape TM. Suppose c ? is a constant. If a function h : ? is time constructible, then we can equip any given Turing machine with a ?clock? on an extra tape. This is a counter initialized at c · h(n) and decremented each step of the given machine. It stops the given machine on reaching 0. Since h is time constructible, the number of steps the clocked machine needs is still within O(h). Numbering all Turing machines clocked in this way yields an effective listing of the class DTIME(h).

§3. ?0n -completeness for equivalence relations. In this section we show that there is a ?01-complete equivalence relation, but no ?0n-complete equivalence relation for n ? 2.

3.1. ?01 equivalence relations. For each function f : fx denote the function n ? f(x, n). If f is computable, × ? and x ? , let Ef = { x, y : fx = fy } (2) (3)

Proposition 3.1. For each ?01 equivalence relation E, there is a computable binary function f such that E = Ef , namely, for each x, y, we have

x E y ? fx = fy . appears to be a natural example of a ?01 equivalence relation. We show that indeed every ?01 equivalence relation is obtained in this way. A slight extension of the argument yields a function f such that the corresponding equivalence relation Ef is ?01-complete. We show that f can in fact be chosen polynomial time computable.

For each ?01 equivalence relation E, there is a computable sequence of cofinite
relations (Et )t? contained in [

be as in (2). We define f(x, n) by recursion on x. Let f(x, n) = f(r, n), if r < x for the least r such that r Emax(x,n) x, x,

otherwise.

Note that for each z, for sufficiently large k, we have f(z, k) = min[z]E . Here, [z]E is the equivalence class of z with respect to E. We verify that (3) is satisfied.

For the implication ???, we show the following by induction on x.

Claim. For each y < x, for each n, if y Emax(x,n) x then fy (n) = fx(n). To see this, let r < x be as in the definition of f(x, n). Then r ? y < x. By definition of f we have fx (n) = fr (n). By transitivity in (2), we have r Emax(x,n) y, and hence r Emax(y,n) y. Then, by inductive hypothesis, fr (n) = fy (n).

For the implication ???, suppose that y ¬E x. Then, fy (k) = fx (k) for sufficiently large k by the remark after the definition of f above.

As a corollary we obtain a presentation of ?01 equivalence relations as the uniform intersection of recursive ones.

Corollary 3.2. For each ?01 equivalence relation E, there is a uniformly recursive sequence (Fn)n? of equivalence relations such that Fn ? Fn+1 and E = n Fn.

Proof. Let E = Ef for a computable function f. Let

Fn = { x, y : fx n= fy n}, where g n denotes the tuple g(0), . . . , g(n ? 1) .

We obtain a ?01-complete equivalence relation by exposing a uniformity in the proof of Proposition 3.1.

Theorem 3.3. There is a computable binary function g such that the equivalence relation Eg = { x, y : gx = gy } is ?01-complete.

Proof. Let Ei = [

We can in fact obtain in Theorem 3.3 a function on binary strings that is quadratic time computable (see Section 2). We thank Moritz M u¨ller at the Kurt G o¨del Research Institute in Vienna for suggesting a simplified proof that we provide at the end of this subsection.

Lemma 3.4. For each computable binary function g, there is a quadratic time computable binary function G defined on strings over {0, 1}, and a computable unary function p mapping numbers to strings, such that g(x, n) = G (p(x), p(n)), and G (w, v) = 0 for any string v not of the form p(n).

Unary quadratic time functions over the alphabet {0, 1} are given by indices for multitape Turing machines (TM) that are equipped with a counter forcing them to stop in time O(n2), where n is the input length.

Theorem 3.5. (i) There is a quadratic time computable binary function G such that EG is ?01-complete. (ii) Equality of unary quadratic time computable functions is a ?01-complete equivalence relation.

Proof of Theorem. (i) Let G be the function obtained from the function g of Theorem 3.3 according to the lemma. Let p be as in the lemma. Then x Eg y ? Gp(x) = Gp(y). Since Eg is complete, so is EG . (ii) Since all the Turing machines considered stop on every input in a quadratic time bound, the equivalence relations is ?0. The completeness follows from 1 (i) since from a string w we can compute an index for a quadratic time TM computing Gw .

Proof of Lemma 3.4. Let g(x, y) (with x, y represented in unary) be computable in time t(x, y) where t is a computable function that is increasing in both x and y. We want a time constructible h satisfying t(n, k) ? h(n) + h(k). This can be achieved by having h(n) run g(n, n) and counting the number of steps. This satisfies t(n, k) ? h(n) + h(k) as for n ? k, t(n, k) ? t(k, k) = h(k), and h(n) = m can be calculated in time quadratic in m: Place a binary counter at the start of the tape and simulate g to the right of it. In the worst case scenario for every step of f the machine would iterate over m + log m cells, and whenever the counter grows in size the entire tape would need to be shifted right which in total would take m log m, coming to m log m + m(m + log m) or O(m2).

To compute G (a, b), verify whether a = 1x01h(x) and b = 1n01h(n). If so output g(x, n), else output 0. Observe that G is quadratic time: We can verify whether a = 1x 01z is of the form 1x 01h(x) by beginning to compute g(x, x), stopping whenever the number of steps exceeds z. If the input is of the right form we can compute g(x, n) in time t(x, n) ? h(x) + h(n) = O(|a| + |b|).

Note that for every x, n,

G (1x 01h(x), 1n01h(n)) = g(x, n), and for all other values the function is 0. Thus the function p(y) = 1y 01h(y) is as required.

In work independent of ours [10, Proposition 3.1] the authors offer a characterisation of ?01 preorders: Every such preorder is computably isomorphic to the inclusion relation on a uniformly computable family of sets. As an equivalence relation is merely a symmetric preorder, a ?01 equivalence relation is then isomorphic to equality on such a family. Since sets can be viewed as 0/1 valued functions, this yields another proof of Proposition 3.1. Methods similar to the ones employed in the proof of Theorem 3.5 above now yield a version of that Theorem for the inclusion of quadratic time computable languages.

Theorem 3.6 ([

For a proof see [19, Theorem 5.5]. Using that result one can also obtain a stronger form of Corollary. 3.2 where the Fn have at most 2n equivalence classes.

3.2. For n ? 2 there is no ?0n -complete equivalence relation. ?0n equivalence
relations often occur naturally. For instance, equality of r.e. sets as a relation on indices
is ?02. See [

Here we show that there is no ?0n-complete equivalence relation for n ? 2. Intuitively speaking, one would then expect that the computable reducibility degrees of ?0n equivalence relations are complicated.

Theorem 3.7. For each ?02 equivalence relation E there is a ?02 equivalence relation L such that L ?c E.

Let ? be a computable ordinal. Relativizing to ?(?) yields an immediate corollary. We say that E ?0(?) F if Definition 1.1 holds for E and F for some f ?T ?(?).

Corollary 3.8. Let ? be a computable ordinal. No ?0?+2 equivalence relation can be hard for all the ?0?+2 equivalence relations under the reducibility ?0(?) , let alone under computable reducibility ?c.

In contrast, there is a ?11-complete equivalence relation E. Similar to the case of ?0n relations, the class of symmetric ?11 relations is uniformly closed under taking the transitive closure. Let (Ve )e? be an effective list of all symmetric ?11 relations, let (Se )e? be the effective list of their transitive closure, and let E = e Se .

Proof of Theorem 3.7. E be any ?02 equivalence relation on . Since E is ?0,? 1 we may fix a ? -computable function f such that x, y ? E iff f(x, y, t) = 1 for every t. Now define L as follows.

For e0 = e1, { e0, s0 , e1, s1 } ? L for any s0, s1. For each e, we distinguish two cases.

Case 1. There is a number t, chosen least, such that

?e( e, 0 ) ?= x, ?e ( e, 1 ) ?= y, ?e ( e, t + 2 ) ?= z, and f(x, y, t) = 0. Let v be least such that f(x, z, v) = 0 or f(y, z, v) = 0.

? If f(x, z, v) = 0 we set { e, 0 , e, t + 2 } ? L.

? If f(y, z, v) = 0 we set { e, 1 , e, t + 2 } ? L.

Declare { e, s0 , e, s1 } ? L for every other pair s0 = s1.

Case 2. There is no such number t. Then we set { e, s0 , e, s1 } ? L for every s0 = s1.

To see that L is ?02, it suffices to note that for each e, if Case 1 applies then by the transitivity of E the number v exists. Transitivity of L follows from the fact that we relate each element to at most one other element.

Now suppose that L ?c E via the total function ?e. Since { e, 0 , e, 1 } ? L it follows that the least number t described above exists. By construction we ensure that either { e, 0 , e, t + 2 } ? L and {x, z} ? E, or we have { e, 1 , e, t + 2 } ? L and {y, z} ? E, a contradiction.

Remark 3.9. Observe that given any two ?0n+2 equivalence relations R, S, the disjoint union R S = {{2n, 2m} | n R m} ? {{2n + 1, 2m + 1} | n S m} is also a ?0n+2 equivalence relation and we have R, S ?c R S. Hence Theorem 3.7 shows that amongst ?0n+2 equivalence relations there can be no maximal element under ?c.

The following is an immediate consequence of Theorem 3.7; for preorders we also apply Fact 2.1.

Corollary 3.10. For each n ? 2, there is no ?0n-complete preorder, and no ?0n -complete equivalence relation, or preorder.

There also is no ?0 analog of Corollary. 3.2. Let us say a ?02 equivalence relation E 2 is effectively ?02 if there is a uniformly r.e. sequence (Fn)n? of equivalence relations such that E = n Fn. (We may also require Fn ? Fn+1 without loss of generality.) If S is an r.e. nonrecursive set then the ?02 equivalence relation with the two classes S and ? S is not effectively ?02. Note that we can effectively list the effectively ?20 equivalence relations. By putting together all the effectively ?02 equivalence relations in the obvious way we see that there is a complete one under computable reducibility.

§4. ?0k -complete preorders. The concept of an effectively inseparable r.e. Boolean
algebra was introduced by Pour-El and Kripke [28, Lemmas 1,2] when they studied
the complexity of logical derivability for recursively axiomatizable theories. The
method was developed in more generality by Montagna and Sorbi [

We are interested mostly in semantic preorders such as reducibilities, which are usually at higher levels of the arithmetical hierarchy. So we extend the methods discussed above in order to show the completeness of a ?0k preorder, using effectively inseparable ?0k Boolean algebras. Thereafter we apply it to preorders in the subrecursive setting, which are ?02, and to preorders on the r.e. sets arising from computability theory, which are ?0. Each time, we naturally embed an effectively 3 inseparable ?0k Boolean algebra for the appropriate k into the preorder.

For sets X, Y , we write X ?? Y (X is almost contained in Y ) if X \ Y is finite. We write X =? Y if X ?? Y ?? X . The preorders we will consider in this section for the cases k = 2, 3 are given on certain natural classes of sets, either by almost containment, or by reducibilities.

4.1. The method of effectively inseparable ?0k Boolean algebras. Our proofs rest on the following notion, which is an analog of creativity for disjoint pairs of sets. For details see [30, Exercise II.4.13]. We relativize r.e. sets to ?(k?1), thereby obtaining ?0k sets. Note, however, that the analog g of the productive function remains computable.

Definition 4.1 (see [30, Exercise II.4.13]). We say that disjoint ?0k sets A, B ? are effectively inseparable (e.i.) if there is a computable binary function g (called a productive function) such that for each p, q ? , if A ? Wp?(k?1) , B ? Wq?(k?1) and Wp?(k?1) ? Wq?(k?1) = ?,

then g(p, q) ? Wp?(k?1) ? Wq?(k?1) .

Remark 4.2. (a) If we enlarge both components of the disjoint pair to disjoint ?0 sets, the function g is still a productive function.

k (b) We say that a disjoint pair A, B of ?0k sets is m-complete for disjoint pairs if for each disjoint pair U, V of ?0 sets, there is a computable map ? such that k x ? U ? ?(x) ? A, and x ? V ? ?(x) ? B.

It is a well-known result going back to Smullyan [

We consider Boolean algebras in the language with partial order ?, meet ?, join ?, and complementation . A ?0k Boolean algebra B is represented by a model ( , , ?, ?, ) such that is a ?0k preorder, ?, ? are computable binary functions, is a computable unary function, and the quotient structure ( , , ?, ?, )/? is isomorphic to B. Here, ? is the equivalence relation corresponding to , and we assume the functions are compatible with ?. We may also assume that 0 ? denotes the least element of the Boolean algebra, and 1 ? denotes the greatest element. If necessary, we write B etc. to indicate the Boolean algebra a relation on belongs to.

The usual effectiveness notions defined for sets or functions on the natural numbers can be transferred to B. For instance, we say that an ideal I of B is ?0k if I is ?0k when viewed as a subset of .

Let F be a computable Boolean algebra that is freely generated by a computable
sequence (pn)n? . For instance, we can take as F the finite unions of intervals [x, y)
in [0, 1)Q. We fix an effective encoding of F by natural numbers. Then, equivalent
to the definition above, a ?0k Boolean algebra is given in the form F /I, where I is
a ?0k ideal of F . We rely on this view for coding a ?0k preorder into a ?0k Boolean
algebra. We slightly extend concepts and results of [

Lemma 4.3. For any ?0k preorder , there is a ?0k ideal I of F such that n k ? pn ? pk ? I.

Proof. Let I be the ideal of F generated by {pn ? pk : n k}. The implication ??? follows from the definition. For the implication ???, let B be the Boolean algebra generated by the subsets of of the form i? = {r : r i }. The map pi ? i? extends to a Boolean algebra homomorphism g : F ? B that sends I to 0. If n k, then n? ? k? , and hence pn ? pk ? I.

We will frequently use the following criterion to show that a ?0k Boolean algebra B is effectively inseparable.

Then, B is effectively inseparable.

Proof. Let A = {x : g(x) ? 0} and B = {x : g(x) ? 1}. Then A, B are ?0k sets, A ? B = ?, U ? A, and V ? B. By Remark 4.2 this implies that A, B are effectively inseparable as ?0k sets, whence B is effectively inseparable as a ?0k Boolean algebra.

We slightly extend a result [24, Proposition. 3.1]. As mentioned already, that work
goes back to results of Pour-El and Kripke [28, Lemmas 1,2] who only worked in
the setting of axiomatizable theories. It is worth including a proof of the extension,
because we only partially relativize the setting of [

Theorem 4.6. Suppose a Boolean algebra B is ?0k-effectively inseparable. Then, B is a ?0k -complete preorder.

Proof. We say that a ?0k Boolean algebra is ?0k -complete if each ?0k Boolean algebra C is computably embedded into it. We first show that the given Boolean algebra B is ?0k -complete. For an element z of a Boolean algebra define z(0) = z and z(1) = z .

Given a ?0k Boolean algebra C, we construct the desired computable embedding h of C into B recursively. Recall that the domain of a presentation of C is by our definition. Suppose yi = h(i ) has been defined for i < n. For each n-bit string let p = i ( i ) and

y = i<n i<n yi( i ).

In particular, if n = 0, we let p? = 1C and y? = 1B.

We define h in such a way that for each n and each string of length n, we have p ?C 0 ? y ?B 0. (?) This is clearly the case for n = 0. Inductively assume (?) holds for n. To define h(n) we use the following.

Claim. From a string of length n we can effectively determine an element z B y such that n ? p ?C 0 ? z ?B 0, and n

C p ? z ?B y .

A = {r : r ? y ?B 0}, B = {r : r is disjoint. By Remark 4.2(a) this pair is e.i. with the same productive function g as the one for the e.i. pair of sets {r : r ?B 0}, {r : r ?B 1B}.

Since y ?B 0, by (?) for n we have p ?C 0. Hence, the pair of ?0k sets U = {k : k ? p ?C 0}, V = {k : k

C p } is also disjoint. By Remark 4.2(b), we are uniformly given an m-reduction ? from U, V to A, B. We let z = y ? ?(n). Then, the claim is satisfied in case y ?B 0.

If y ?B 0, then, by (?) for n, we have p ?C 0. None of the pairs of sets is disjoint now, but (based on the productive function g) we still have a computable index for a function ?, and hence a definition of z such that z ?B 0, which vacuously satisfies the claim.

Now let h(n) = | |=n z . It is clear that (?) holds for n + 1. Then, by induction, the map h is a computable embedding of C into B, as required.

To conclude the proof of the theorem, let P be any ?0k preorder. By Lemma 4.3 there is a ?0k ideal I of the free Boolean algebra F such that n k ? pn ? pk ? I. Since B is ?0k -complete, there is a computable embedding g of F /I into B. Thus, n P k ? g(pn) B g(pk ).

A forth-and-back version of the argument above shows that any two e.i. ?0
k
Boolean algebras are effectively isomorphic. This fact for k = 1 was already noted
in [

4.2. Derivability in first-order logic. Let T be a recursively axiomatized
sufficiently strong theory in the language of arithmetic, such as Robinson arithmetic Q.
Note that such a theory T can be finitely axiomatizable. The following was first
observed in [24, Section 4], building on Pour-El and Kripke [

Theorem 4.7. For sentences ?, in the language of T , let ? if T ? ? . Then, is a ?01-complete preorder.

Proof. Clearly is ?01. It was first observed by Smullyan [

In [

4.3. ?02 preorders in the resource-bounded setting. For relevant notation from complexity theory see Section 2. In particular, there is an effective listing of the class DTIME(h), which we fix below without further mention. We use a technical tool:

Lemma 4.8. Given a disjoint pair of ?02 sets U, V , we can effectively in x determine a linear time computable language Lx such that

x ? U ? Lx is finite, and x ? V ? Lx is cofinite. (Sx )x? , (Tx )x?

Proof. It follows from [30, p. 66] that there is a uniformly r.e. pair of sequences such that x ? U ? Sx is finite, and x ? V ? Tx is finite.

To define Lx , at stage n, in linear time we determine whether w ? Lx for each string w of length n. Let t ? n be largest such that t > 0 and in n steps one can verify that

B = ({A ? L : L is linear time}, ??)/ =? is ?02 effectively inseparable. To with the canonical representation given by (Gi )i? 02 effectively inseparable sets. If see this, in Lemma 4.8 let U, V be any pair of ? x ? U , then A ? Lx =? ?. If x ? V , then A ? Lx =? A. Now we apply Fact 4.5.

The preorder of B on indices for linear time computable languages is ?02-complete by Theorem 4.6. From i we can compute an index for A ? Gi within our effective listing of the quadratic time sets. Hence, we can effectively reduce the preorder on B to almost inclusion ?? among quadratic time computable languages.

In the following, ?rp will denote a polynomial time reducibility in between polynomial time many-one (m) and Turing (T) reducibility. Clearly this reducibility is ?02 on DTIME(h).

Theorem 4.10. Suppose the function h is time constructible and dominates all polynomials. Then ?rp on sets in DTIME(h) is ?02-complete preorder.

For instance, we can let h(n) = 2n, or h(n) = nlog log n. The proof relies on the
following notion of Ambos-Spies [

Definition 4.11. Let f : ? be a strictly increasing, time constructible function. We say that a language A ? {0}? is super sparse via f if A ? {0f(k) : k ? } and ?0f(k) ? A ?? can be determined in time O(f(k + 1)).

Super sparse sets exist in the time classes we are interested in.

Lemma 4.12 ( [

Sketch of Proof. Let f(n) = h(n)(0). Since h eventually dominates all polynomials, we can construct A ? {0f(k) : k ? } such that A ? DTIME(h), but still diagonalize against all polynomial time machines.

We will keep this set A fixed in what follows. Given a reducibility ?rp, we write a = degrp(A) for the degree of A. We write [0, a] for the initial segment of the degrees below a, viewed as a partial order.

Proof of Theorem 4.10. Recall the e.i. ?02 Boolean algebra B from the proof of Theorem 4.9. It is represented via an effective listing (Gi ) of linear time computable languages. We will embed a ?02 quotient of B into [0, a].

Since A is super sparse, by [26, Lemma 6.2.9] for polynomial time sets U, V we have

A ? U ?rp A ? V ? A ? (U ? V ) ? P.

So we may well-define a map ? : B ? [0, a] as follows. For a linear time set L = Gi , let

?(A ? L/ =?) = degrp(A ? L).

Let I be the ?02 ideal of B consisting of all the equivalence classes A ? L/ =? such that A ? L ? P. Then, C = B/I is in a natural way a ?02 Boolean algebra, and C is effectively embedded into [0, a]. Since B is e.i. and the map ? is effective, it follows that C is e.i. Hence, by Theorem 4.6, the ?02 preorder { e, i : A ? Ge ?rp A ? Gi } is complete. Via a computable transformation of indices, this preorder is effectively reducible to ?rp on DTIME(h).

4.4. Almost inclusion of r.e. sets.

Theorem 4.13. The preordering { e, i : We ?? Wi } of almost inclusion among r.e. sets is ?03-complete.

Proof. All sets in this proof will be r.e. Fix a nonrecursive r.e. set A. By X Y = A we denote that r.e. sets X, Y are a splitting of A. That is, X ? Y = ? and X ? Y = A. We also write X A to denote that X is part of a splitting of A.

The following technical tool parallels Lemma 4.8.

Lemma 4.14. Given a disjoint pair of ?03 sets U, V , we can effectively in x determine an r.e. splitting A = Ex Fx such that

x ? U ? Ex is computable, andx ? V ? Fx is computable. (Px,n )x,n?

Proof of lemma. By [30, p. 66] we may choose a uniformly r.e. double sequence of initial segments of such that x ? U ? ?n [Px,2n = ] and x ? V ? ?n [Px,2n+1 = ].

We fix x. Let g(k, s ) be the greatest t ? s such that t = 0, or Px,k,t?1 = Px,k,t . We enumerate sets Ex , Fx with Ex Fx = A as follows. At stage s > 0, if a ? As ? As?1, see if there is a k ? s be least such that a < g(k, s ). If so, choose k least. If k is even, enumerate a into Fx ; otherwise, or if there is no such k, enumerate a into Ex .

First suppose that x ? U . Then, the least k such that Px,k = is even. We have lims g(k, s ) = ?. If k > 0, then also r = maxi<k lims g(i, s ) < ?. For any v ? r, we have v ? Ex ? v ? Ex,s , where s is least such that g(k, s ) > v. Therefore, Ex is computable.

If x ? V , then the least k such that Px,k = is odd. A similar argument shows that Fx is computable.

Since A is nonrecursive, there is a major subset D ?m A (see [30, Section X.4]). Thus, if R ? A is computable, then R ?? D. We now define an e.i. ?03 Boolean algebra B, which is a subalgebra of the complemented elements in [D, A]/ =?. Suppose we are given a pair of e.i. ?03 sets U, V . Let (Ex )x? , (Fx )x? be the u.r.e. sequences obtained through the foregoing lemma. We let B be the Boolean algebra generated by all the elements of the form (Ex ? D)? for x ? . Since uniformly in x we can recursively enumerate both Ex and its complement Fx = A ? Ex , it is clear that B is a ?03 Boolean algebra. Furthermore, by Lemma 4.14 and Fact 4.5, B is effectively inseparable. So by Theorem 4.6 the preorder of B, namely, { x, y : Ex ?? Ey ? D}, is ?03-complete. Since the sequence (Ex )x? is uniformly r.e., this establishes the theorem.

Note that we have in fact shown that almost inclusion of sets of the form (We ?
D) ? A is ?03-complete. It is worth noting that by a result of Maass and Stob [

Corollary 4.15. The equivalence relation { e, i : We =? Wi } is ?03-complete. Proof. This follows immediately from Fact 2.1.

The relation =? on r.e. indices is the same relation as that called E0ce in [11,
Theorem 3.4]. In the spirit of that article, we consider this and several other equivalence
relations carried over from the Borel theory. The relations E0, E1, E2, and E3 are
all standard Borel equivalence relations, whose definitions can be found in [

i E0ce j i E1ce j i E2ce j i E3ce j ?? ?? ?? ?? |Wi " Wj | < ? ??n (Wi )n = (Wj )n 1 i < ? i?Wi Wj ?n |(Wi )n " (Wj )n| < ? .

( ??

Wi =? Wj ) Now it is not hard to see that E3ce is ?04-complete as a set (since =? is ?03-complete as a set), and, therefore, not computably reducible to any of the others, since the others are all ?03. In the Borel theory, E0 <B E1, E0 <B E2, and E0 <B E3, with no other reductions holding among these relations. The proofs of the Borel reducibilities can be adapted to give proofs that E0ce ?c E1ce , E0ce ?c E2ce , and E0ce ?c E3ce . Therefore, Corollary 4.15 gives a further result, providing the relation left open in [11, Figure. 3].

Corollary 4.16. E0ce , E1ce , and E2ce are all ?03-complete equivalence relations, and hence are bireducible with each other under ?c.

In contrast, Corollary 3.8 showed that the relation E3ce cannot be complete under ?c at any level of the arithmetical hierarchy, as it is ?04 but not ?04 by virtue of its ?04-completeness as a set of pairs.

4.5. Weak truth-table reducibility on r.e. sets. It is well-known that weak
truthtable reducibility ?wtt on r.e. sets is a ?03 preorder with computable supremum
operation. It determines the distributive upper semilattice of r.e. weak truth-table
degrees; see, e.g., [

Theorem 4.17. The preorder { e, i : We ?wtt Wi } is ?03-complete.

Proof. Ambos-Spies [

Note that the degrees {degwtt (X ) : X A} form a Boolean algebra of complemented elements in [0, a] with the usual degree ordering. We establish an e.i. ?03 subalgebra C of this Boolean algebra. We are given a pair of e.i. ?03 sets U, V . Let (Ex ), (Fx ) be uniformly r.e. sequences as in Lemma 4.14. Thus, Ex A for each x. Let C be the Boolean algebra generated by the degwtt (Ex ), x ? . Since A is antimitotic, intersection, union, and complementation for splits of A correspond to the operations of meet, join, and complementation in [0, a]. Since the former operations are effective, C is indeed a ?03 Boolean algebra.

If x ? U , then Ex is computable, so deg(Ex ) = 0. If x ? V , then Fx is computable, so deg(Ex ) = a. This implies that C is effectively inseparable. So the preordering of C, namely, { x, y : Ex ?wtt Ey }, is ?03-complete by Theorem 4.6. Since the sequence is uniformly r.e., this establishes the theorem. §5. Turing reducibility on r.e. sets is a ?04-complete preorder.

Theorem 5.1. The preorder { i, j | Wi ?T Wj } is ?04-complete. Hence, the equivalence relation { i, j | Wi ?T Wj } is also ?04-complete.

Proof. We fix a ?04 preorder R. We give a construction with the overall aim to bNuiielsd[a25u]nwifeorfimx aseuqnuiefonrcmeolyf rr..ee.. sseetqsu(eVnic)ei?Xei,,pj of initial segments of , such that such that i R j iff Vi ?T Vj . Following i R j ? For almost all e and p, |Xei,,pj| < ?, ¬ i R j ? ?e?p Xei,,pj = .

Here, |X | denotes the cardinality of the set X . The facts ?Xei,,pj = ? and ?Xei,,pj < ?? are (uniformly) ?02 and ?02, respectively, so each such statement can be measured at a single node on a priority tree measuring infinitary or finitary behavior. That is, |¬Xiei,,Rpj| jhawsoturulde binefienqituairvyaloeunttctoomteh?e. fSaicmt itlharalty,?i?fei?Rp jsutchhenth?aftorthaelmnoosdteamll eea,spu,ritnhge node measuring |Xei,,pj| has true finitary outcome?. Hence, the ordered pairs in R will determine the true path of the construction, i.e., which nodes are visited infinitely often and which are not. We will arrange the strategies on the construction tree to align our actions with the true path.

In this proof we use Y k to mean the first k + 1 bits of Y . We say ?Y changes below k? to mean a change in Y k.

5.1. Requirements and a high-level description. The requirements to be met are Si,j for i, j ? :

Si,j : i R j ? Vi ?T Vj .

The way in which each Si,j is satisfied can only be answered by a 0 oracle, and so it is difficult to split Si,j explicitly into subrequirements in a meaningful way. We will instead describe the local action of each node on the priority tree and then describe how the priority tree ensures that each requirement Si,j is met globally.

i?, j?, e?, p?, X? for i, j, e, p, Xei,,pj, respectively.

Outcome ? stands for X? = , while outcome 0 stands for X? finite.

We now describe the local action of each node ?. During the construction ? has a dual role. During the nonexpansionary stages (i.e., ? ? 0 stages) ? builds a Turing reduction aiming to make Vi ?T Vj. Loosely speaking, this strategy monitors for changes in Vi and responds with a change in Vj each time it sees an element entering Vi . The other (conflicting) strategy of ? acts at each expansionary stage (? ? ? stage). At each expansionary stage ? checks to see if it can make Vi = ?eV?j by enumerating some number in Vi and restraining Vj (henceforth, we will write ?? instead of ?e? for convenience). These two strategies are clearly conflicting, and we always initialize the Turing reduction whenever an expansionary stage occurs for ?.

We now describe how Si,j can be met globally. Suppose ¬ i R j. Then, for each e there is some node ? along the true path with i? = i, j? = j, e? = e and having true outcome ?. This ? will ensure that ?eVj = Vi , and so overall we get Vi ?T Vj. In this case there could be infinitely many nodes along the true path also working for i, j, but which have true outcome 0. The local strategy for will attempt to demonstrate Vi ?T Vj, which cannot possibly be achieved. Indeed each such node along the true path will be injured infinitely often by another node ? further along the true path aiming to diagonalize ?eVj = Vi . This is the key difference between this construction and a typical ? tree argument. In the latter case we usually only allow a finite amount of injury to each requirement along the true path, whereas in our case it is necessary that some node gets injured infinitely often along the true path.

Now suppose that i R j. Then, almost every node along the true path working for the pair i, j must have true outcome 0. In particular, there is a shortest node ? along the true path with true outcome 0 (we will later call all possible candidates for ? a top node). In this case there could be finitely many top nodes ? ? ? with true outcome 0. However, the Turing reducibilities built by these ? will all be destroyed (by nodes between ? and ? which are attempting to diagonalize), and Vi ?T Vj will be demonstrated by the final top node ? along the true path. It is important to note that each top node ? cannot possibly know anything about the nodes on the true path which are longer than ? , and so the situation described above cannot be avoided.

We conclude this section with some technical definitions. We say that ? is a top node if there is no ? ? such that i? = i and j? = j , or if ?(| |) = ? for the maximal such . Given a node ? we define the top of ? to be the maximal top node ? such that i? = i and j? = j . We say that ? is a child of if ? has top .

We let Z?(?) = { ? ? | ? ? ? and is not a top node}, which are all the nodes extended by ? running the ? strategy. Similarly, we let Z0(?) = { ? ? | ? 0 ? and is a top node}.

5.2. Ensuring Vi ?T Vj. Each top node ? will build a Turing reduction ?? to ensure that Vi? ?T Vj? in the case that ? ? 0 is along the true path. The Turing reduction is built indirectly by specifying the coding markers { ?(x)[s ]}x,s? . Each ? (x)[s ] is not in Vj? [s ] and obeys the following standard marker rules: For all x and s , we ensure (i) ?(x)[s ] < ?(x + 1)[s ]. (ii) If ?(x)[s ] = ?(x)[s + 1] or if ?(x)[s ] ? and ? (x)[s + 1] ?, then Vj?,s?1 ( ? (x)[s ]) = Vj?,s ( ? (x)[s ]). (iii) If x is enumerated in Vi? at stage s , and if ?(x)[s ] is defined then Vj? must change below ?(x)[s ] at the same stage or later.

(iv) lims ?(x)[s ] exists (if ? ? 0 is along the true path).

It is easy to see that if these rules are ensured, and if ? ? 0 is visited infinitely often, then Vi? ?T Vj? : To compute if x is in Vi? we find recursively in Vj? a stage s where Vj? [s ] is correct up to ? (x)[s ]. This stage exists because of (iv). At all stages t > s we have ? (x)[s ] = ?(x)[t] because of (ii). By (iii) we must have x ? Vi? iff x ? Vi? [s ].

5.3. Informal description of the strategy. We now describe the strategy and the method of overcoming the key difficulties. Each top node ? will have a dual role. At ? ? 0 stages it must extend the Turing reducibility ??Vj by picking a fresh value for ? (x). It must also ensure the correctness of ??Vj (x) by enumerating ?(x) if x had been enumerated into Vi recently. At each ? ? ? stage ? will attempt to diagonalize by enumerating (if possible) some number l into Vi and preserving Vj . Clearly these two strategies are conflicting and we give the diagonalization strategy higher priority; Hence at each ? ? ? stage we reset ?? and all ? markers. If ? ? 0 is along the true path, then we ensure Vi = ??Vj , and if ? ? ? is along the true path, then we ensure ??Vj = Vi . In the latter case, the reduction ?? fails. Analyzing the overall outcome of the construction we note that if ¬ i R j then for every e we ensure ?eVj = Vi at some node along the true path, and if i R j then there is a final top node along the true path where Vi = ?Vj is successfully maintained. In this case there can be finitely many top nodes ? where ? ? ? for which ??Vj is initialized infinitely often.

We now describe the main issue which will arise in implementing the above strategy. Suppose ? ? are both working for the same i, j, and ? has true outcome 0, while has true outcome ?. We may as well assume that ? is the top of . Now ? and will have conflicting actions. At each ? ? stage, we may want to diagonalize by putting some number l in Vi . If ?(l ) has already been defined, then ? will later want to correct ?? by putting ? (l ) into Vj , which can destroy the diagonalization previously obtained by . We overcome this by delaying from diagonalizing. In the meantime, at each ? ? stage we lift ? (k) for some fixed number k by enumerating ?(k) into Vj and later picking a fresh value for ?. Since there are infinitely many ? ? stages, ?(k) goes to infinity. Hence, destroys ?? but will now be able to meet ?Vj = Vi (via diagonalization or an undefined computation). ??Vj is now destroyed. However, this is fine, because witnesses the fact that ? is not the final top node along the true path. In this case the reduction ?Vj = Vi will be built further down the true path (if i R j) or may not be built at all (if ¬ i R j).

In the general situation (when considering nodes devoted to other pairs i , j ) we consider the nodes 0, 1 ? ? where 0, 1 have true outcome 0, ? has true outcome ?, i = i? = i 1 , j 1 = i 0 , and j 0 = j? = j. Again if ? wants to diagonalize at some ? ? ? stage it must be careful; A number l enumerated in Vi might cause 1 to later correct ?Vj 1 by changing Vj 1 , which might in turn cause 0 to correct ?Vj by 1 0 changing Vj, destroying the previous diagonalization attempt of ?. In this case we observe that along the true path, we must have one of the following three situations hold: (i) There is some child 0 of 0 where 0 ? ? is along the true path. (ii) There is some child 1 of 1 where 1 ? ? is along the true path. (iii) Almost every working for the same pair i, j as ? along the true path has true 0 outcome.

One of these three situations must hold since R is transitive. If (iii) holds, then it is fine for ? to fail in diagonalizing even though it has true outcome ?. If (i) or (ii) holds, then v (for some v < 2) witnesses that v is not the final top node along the true path; In this case v will destroy ? v by pushing some v (k) to infinity. In this case we delay ? from diagonalizing until we find that ??Vj (l ) has converged on a number l with use u safely. That is, the combined response of 0 and 1 to the entry of l into Vi will not change Vj below u.

5.4. Markers for capricious destruction. Each node ?, which is not a top node, is given a parameter k?. At each ?-expansionary stage, ? will enumerate (k?) into Vj and lift this marker. Hence, if ? ? ? is along the true path, then (k?) goes to infinity. This is fine because is not the maximal top node along the true path; in this case there may be a top node further down the true path, or there may be none along the true path, and so the Turing reducibility witnessing Vi ?T Vj need not be built at . The important thing here is to ensure that for each top node where ? 0 is along the true path and is not the maximal top node along the true path, we have to destroy the markers and create an interval (i, ?) where every point is eventually cleared of markers. This allows a conflicting diagonalization strategy of lower priority to succeed.

If is the final top node along the true path, then each marker is lifted finitely often this way, and so the Turing reducibility witnessing Vi ?T Vj will be built at .

5.5. Believable computations. A computation ??Vj? (l )[s] with use u is said to be ?-believable at stage s if the following two conditions are met.

? For every ? Z0(?) there is no marker below u which is pending. A marker (x) is said to be pending at t if x was enumerated into Vi at some s < t where (x)[s] is defined and Vj has not changed below (x)[s]. ? There does not exist a sequence of distinct nodes 0, . . . , n ? Z0(?) such that i n = i?, j 0 = j?, and for every m < n, j m+1 = i m , and m (l )[s] ?? u. The first item says that a computation is ?-believable if there is no marker below the use that will be enumerated by some ? ? due to coding. The second item ensures that if l were to be enumerated by ? into Vi? for the sake of diagonalization then the resulting sequence of correction actions will not cause the computation to be later destroyed. 5.6. Formal construction. At stage 0 we initialize every node. This means to reset ? and k? to be undefined. At stage s > 0 we define TPs of length s , where TPs is the stage s approximation to the true path of construction. Assume that ? ? TPs has been defined. If |X?| has increased since the last ?-stage, then we play outcome ? for ?, otherwise play outcome 0 for ?. Initialize every node to the right of ? ? o where o is the outcome played. We now act for ?. There are four cases. (i) ? is not a top node and outcome 0 is played. Do nothing. (ii) ? is a top node and outcome ? is played. Initialize ?. Check if it is possible to diagonalize, and if so, do it. This means to check if there exists some l < s such that ??Vj? (l )[s ] ? is ?-believable, l ? [?] ? Vi? [s ], and for every l ? l we have Vi? (l ) = ??Vj? (l )[s ]. We also require l to be larger than the previous stage where ? was initialized by another node. If such l is found enumerate l into Vi? and initialize every node extending ?. (iii) ? is not a top node and outcome ? is played. Let be the top of ?. If k? ? we pick a fresh value for it. Otherwise if (k?) ? enumerate it into Vj and make it (and all larger marker) undefined. Check if it is possible to diagonalize, and if so, do it. (iv) ? is a top node and outcome 0 is played. We will correct and extend the Turing reduction ?? . For correction, we check to see if there is any number l enumerated into Vi? since the last visit to ?. Let l be the least. If ? (l ) ? we enumerate it in Vj? and make ?(l ) ?. To extend ?? we let k < s be the least such that ? (k) ?. We pick a fresh value x ? [?] and set ? (k) ?= x. 5.7. Verification. Let TP = lim infs TPs be the true path of the construction, which clearly exists as the construction tree is finitely branching. We first prove the following key lemma.

Lemma 5.2. Suppose ? is a node on the true path. Then, ? makes finitely many diagonalization attempts.

Proof. Assume we are at a stage of the construction where each ? ? makes no more diagonalization attempt, and we never move left of ?. Suppose ? makes infinitely many diagonalization attempts after this. Let l be the smallest number put in by ? due to a diagonalization attempt, at some stage t. At t we have ??Vj? (l )[t] ?= 0 with use u, and is an ?-believable computation. We claim that at t, Vj? is stable below u, which contradicts the assumption that infinitely many attempts are made.

By convention u < t, and so the only nodes which can enumerate a number less than u into any set after stage t, are nodes of the form ?. In fact we cannot have = ?, because if ? is a top node then it is initialized at t and any further diagonalization action by ? involves a number l > l , and if ? is not a top node then k? or (k? ) is undefined before the diagonalization at t (here is the top of ?). Similarly, we cannot have ? ? ?, because if is a top node then it is initialized at t and never again performs diagonalization, and if is not a top node then k or (k ) is undefined when ? was visited at t. Hence, if is a node which enumerates a number smaller than u into any set after t, it must satisfy 0 ? Z0(?).

Let 0 ? Z0(?) be the first node after t to change Vj? below u, say at t0 > t. At t0, 0 is correcting ? 0 and responding to the action of 1 at t1, t1 < t0. Let l0 be the number which was enumerated by 1 into Vi 0 at t1. We cannot have t1 < t because of ?-believability at t. Suppose t1 > t. Since l < (l ) for every l, , we have 1 ? Z0(?). Then, we have correction at t1 and so j 1 = i 0 . We can then repeat to get a sequence 0, . . . , n ? Z0(?) and t0 > · · · > tn > tn+1 = t, where n+1 = ?, j 0 = j? , i n = i? , and for every m < n, j m+1 = i m . Furthermore for every m ? n, lm < m (lm) = lm?1, and ln = l . Since m (lm)[tm] ?< u we have m (lm)[t] ?< u (since markers are picked fresh), and so m (l )[t] ?< u. By removing any cycles in the sequence, we may assume that the nodes 0, . . . , n are distinct. This contradicts the second requirement for ?-believability at t.

We devote the remainder of the proof to showing that each requirement Si,j is met. Fix i and j such that iRj. Let ? be the shortest node along TP such that i? = i , j? = j and for every node such that i = i and j = j where ? ? TP, we have ? 0 TP. Clearly ? is a top node. We argue that Vi = ??Vj (by a symmetric argument we can then conclude Vi ?T Vj ). Since ? has true 0 outcome, there are finitely many self-initializations, and by Lemma 5.2, there are only finitely many initializations to ?. The marker rules (i)?(iii) are clearly met for ? . We argue (iv) holds. Since ? is visited infinitely often it suffices to check that each marker is made undefined finitely often. Each ? (k) can only be made undefined by some child of ?, and only when k = k?. A child of ? to the left of the true path only does this finitely often, while a child to the right of the true path has k? = k at finitely many stages. A child on the true path must have true 0 outcome and acts at only finitely many stages.

Now fix i and j such that ¬ i R j. Fix an e, and let ? be the node on the true path such that i? = i, j? = j, e? = e and ? ? ? ? TP. We argue that ??Vj = Vi . Suppose that ??Vj = Vi .

Lemma 5.3. For almost every l , the true computation ??Vj (l ) is eventually ?-believable.

Proof. Call a sequence of distinct nodes 0, . . . , n ? Z0(?) such that i n = i?, j 0 = j? , and for every m < n, j m+1 = i m , a bad sequence. There are only finitely many bad sequences, and for each bad sequence there exists a node which is a child of some m such that ? ? ? TP. This follows by the transitivity of R. Since is not a top node it is initialized finitely often and so achieves a stable value for k . We consider l larger than every k and argue that ??Vj (l ) is eventually ?-believable.

Fix l and fix a large stage s such that ? is visited and ??Vj (l )[s ] ? with the correct use u. The first requirement for ?-believability must be met because any pending marker from any node in Z0(?) must be enumerated in Vj below u before the next visit to ?. For the second requirement, we can assume that s is large enough such that for each bad sequence and associated , we have m (k )[s ] > u. This is possible, because at each visit to ? ? we make m (k ) undefined.

Let us assume the second requirement for ?-believability fails. Fix a bad sequence witnessing this, and let be the associated node. By the failure of believability we have m (l )[s ] ? u; On the other hand, by the choice of s we have m (l )[s ] > m (k )[s ] > u, a contradiction.

By Lemma 5.2, there are only finitely many initializations to ? initiated by a different node; Let s be large enough so that there are no more initializations to ? of this type. By Lemma 5.3 pick l large enough, and wait for ??Vj (l ) to become ?-believable, and for ??Vj to agree with Vi below l . Hence, ? will make another diagonalization attempt after s . Since this holds for any large s , we have a contradiction to Lemma 5.2. This ends the proof of Theorem 5.1.

We also obtain as a corollary, examples of complete equivalence relations at the ?0n+4 level for each n ? :

Corollary 5.4. For each n ? , the preorder { i, j | W ? i (n) ?T Wj? (n) } is

Proof. The proof of Theorem 5.1 produces a computable function f such that for each i, j, we have i R j iff Wf(i) ?T Wf(j). Relativizing this to ?(n), for each ?n+4 0 preorder R, we get a function f ? ?(n) such that i R j iff Wf?((ni)) ?T Wf?((nj)) ? ?(n).

For each oracle X and each function g ?T X there is a computable function h such that WgX(i) = WhX(i) for every i . To see this, observe that the set {(n, i ) : n ? WgX(i)} is ?01,X and is hence equal to the domain of ?X for some functional ?. The function h is obtained by applying the s-m-n Theorem.

Now applying this fact we may assume that f is computable. Let f? be a computable function such that WfX(i) ? X = WfX?(i) for all X and i . Then, f? witnesses that R is reducible to the preorder { i, j | Wi?(n) ?T Wj?(n) }.

complete.

, the preorder { i, j | Wi(n) ?T Wj(n)} is ?0n+4?(nP) r?oTofW.Tj?h(n)e?pr?o(no)f}oisf cCoomropllleatreyam5.o4nsghothwes?t0nh+a4t ptrheeorpdreerosr.dBeyr t{hei, ujnif|orWmi?it(ny) o?f Sacks? Jump Inversion Theorem (see [30, Corollary VIII.3.6.]), for each n there is a computable function qn such that for all x, we have Wq(nn()x) ?T Wx?(n) ? ?(n).

the relation { i, j : Wi?(n+1) ?1 Wj?(n+1) } is ?0n+4-complete.

§6. Discussion and open questions. Many effective equivalence relations from algebra can be considered under the aspect of relative complexity. For instance:

Question 6.1. Is the equivalence relation E of isomorphism between finite presentations of groups ?01-complete?

Adyan and Rabin independently showed in 1958 that the triviality problem, whether a finite presentation describes the trivial group, is m-complete. See Lyndon and Schupp [21, Theorem IV.4.1], letting the given group H there have a word problem m-equivalent to the halting problem. In fact, by the discussion following the proof there, every single equivalence class of E is m-complete: Being isomorphic to a particular finitely presented group P is incompatible with free products in their sense, because P ? A has higher rank than P for any nontrivial group A.

Question 6.2. Is the equivalence relation E of isomorphism between automatic equivalence relations ?01-complete?

Kuske, Liu, and Lohrey [

Computable isomorphism of computable Boolean algebras is ?0 complete by [

§7. Acknowledgment. Miller was partially supported by Grant # DMS ? 1001306 from the National Science Foundation, and by the European Science Foundation. Nies is supported by the Marsden Fund of New Zealand. Miller and Nies were also supported by the Isaac Newton Institute as visiting fellows.

DEPARTMENT OF COMPUTER SCIENCE

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DEPARTMENT OF MATHEMATICS, QUEENS COLLEGE 65-30 KISSENA BLVD., FLUSHING

NY 11367 USA; & PH.D. PROGRAMS IN MATHEMATICS & COMPUTER SCIENCE

CUNY GRADUATE CENTER, 365 FIFTH AVENUE, NEW YORK, NY 10016, USA E-mail: russell.miller@qc.cuny.edu DIVISION OF MATHEMATICAL SCIENCES

SCHOOL OF PHYSICAL & MATHEMATICAL SCIENCES

NANYANG TECHNOLOGICAL UNIVERSITY

21 NANYANG LINK, SINGAPORE E-mail: kmng@ntu.edu.sg DEPARTMENT OF COMPUTER SCIENCE, UNIVERSITY OF AUCKLAND

PRIVATE BAG 92019, AUCKLAND, NEW ZEALAND E-mail: andre@cs.auckland.ac.nz