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Math. Log. Quart. 51, No. 5, 470 – 492 (2005) / DOI 10.1002/malq.200410044 / www.mlq-journal.org The minimal complementation property above 0 Andrew E. M. Lewis Department of Pure Mathematics, School of Mathematics, University of Leeds, Leeds, LS2 9JT, U. K. Received 30 April 2004, revised 11 January 2005, accepted 19 January 2005 Published online 1 August 2005 Key words Computability, minimal degrees, complementation, definability. MSC (2000) 03D25 Let us say that any (Turing) degree d > 0 satisfies the minimal complementation property (MCP) if for every degree 0 < a < d there exists a minimal degree b < d such that a b = d (and therefore a b = 0). We show that every degree d 0 satisfies MCP. c 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim 1 Introduction We say that a degree d > 0 has the complementation property if for every degree 0 < a < d there exists a degree b < d such that a b = d and a b = 0. Before it could be shown that 0 has the complementation property many partial results had to be obtained along the way. Robinson [14] was able to show that every degree in low 2 has a complement below 0 , Posner [12] showed that the same is true of every degree in high 1 and Epstein [4] showed that this result holds for the c. e. degrees. In 1981 Posner [13] showed that every degree not in low 2 has a complement below 0 , so that 0 satisfies the complementation property, as required. At this point in time, then, the only proof in existence was non-uniform in nature. Slaman and Steel [18] were able to improve upon this situation by giving a uniform proof that every non-trivial degree below 0 (i. e. those other than 0 or 0 ) has a 1-generic complement. In this paper we shall be concerned with the minimal complementation property. We say that d > 0 satisfies the minimal complementation property (MCP) if for every degree 0 < a < d there exists a minimal degree b < d such that a b = d (and therefore a b = 0). In [7] we extended ideas developed by Slaman and Seetapun in order to show that 0 satisfies MCP, answering a question of Posner [13]. The aim of this paper is to show that this result can be extended to include all degrees above 0 , that every degree d 0 satisfies MCP. This is a very strong result and requires difficult techniques. The following question remains open. Question 1.1 Is 0 the least degree such that all degrees above it satisfy MCP? A good knowledge of the proof detailed in [7] would certainly be helpful to the reader in understanding the proof that we shall present here, but is not essential. Many of the core ideas of that paper will be useful, but achieving the stronger result presents new problems. Where previously we were able to suppose given an approximation to a set A of the degree a < 0 for which we wished to find a minimal complement, that is clearly no longer possible. The convergent approximation was used to show that if splittings of a certain required variety cannot be found, then A is computable. Certainly, then, this key aspect of the construction (at least) requires new methods. In order to specify the minimal complements in question we shall describe a full approximation construction in which we shall work for all a and all d simultaneously, or more accurately for all A ω and all D ω simultaneously, to approximate a tree such that every set on the tree is of minimal degree. If D is of degree d 0 and A is of degree 0 < a < d, then we shall show that there is a set B A,D on this tree, of minimal degree b < d, such that b a = d. As is entirely standard in the construction of sets of minimal degree we shall be using the splitting tree method – for a clear account see [19]. Since we shall be using full approximation the e-mail: [email protected] c 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

The minimal complementation property above 0′

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Math. Log. Quart. 51, No. 5, 470 – 492 (2005) / DOI 10.1002/malq.200410044 / www.mlq-journal.org

The minimal complementation property above 0′

Andrew E. M. Lewis∗

Department of Pure Mathematics, School of Mathematics,University of Leeds, Leeds, LS2 9JT, U. K.

Received 30 April 2004, revised 11 January 2005, accepted 19 January 2005Published online 1 August 2005

Key words Computability, minimal degrees, complementation, definability.MSC (2000) 03D25

Let us say that any (Turing) degree d > 0 satisfies the minimal complementation property (MCP) if for everydegree 0 < a < d there exists a minimal degree b < d such that a ∨ b = d (and therefore a ∧ b = 0). Weshow that every degree d ≥ 0′ satisfies MCP.

c© 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

We say that a degree d > 0 has the complementation property if for every degree 0 < a < d there exists a degreeb < d such that a ∨ b = d and a ∧ b = 0. Before it could be shown that 0′ has the complementation propertymany partial results had to be obtained along the way. Robinson [14] was able to show that every degree in low2

has a complement below 0′, Posner [12] showed that the same is true of every degree in high1 and Epstein [4]showed that this result holds for the c. e. degrees. In 1981 Posner [13] showed that every degree not in low2 hasa complement below 0′, so that 0′ satisfies the complementation property, as required. At this point in time, then,the only proof in existence was non-uniform in nature. Slaman and Steel [18] were able to improve upon thissituation by giving a uniform proof that every non-trivial degree below 0′ (i. e. those other than 0 or 0′) has a1-generic complement.

In this paper we shall be concerned with the minimal complementation property. We say that d > 0 satisfiesthe minimal complementation property (MCP) if for every degree 0 < a < d there exists a minimal degree b < dsuch that a ∨ b = d (and therefore a ∧ b = 0). In [7] we extended ideas developed by Slaman and Seetapun inorder to show that 0′ satisfies MCP, answering a question of Posner [13]. The aim of this paper is to show thatthis result can be extended to include all degrees above 0′, that every degree d ≥ 0′ satisfies MCP. This is a verystrong result and requires difficult techniques. The following question remains open.

Question 1.1 Is 0′ the least degree such that all degrees above it satisfy MCP?

A good knowledge of the proof detailed in [7] would certainly be helpful to the reader in understandingthe proof that we shall present here, but is not essential. Many of the core ideas of that paper will be useful,but achieving the stronger result presents new problems. Where previously we were able to suppose given anapproximation to a set A of the degree a < 0′ for which we wished to find a minimal complement, that is clearlyno longer possible. The convergent approximation was used to show that if splittings of a certain required varietycannot be found, then A is computable. Certainly, then, this key aspect of the construction (at least) requires newmethods.

In order to specify the minimal complements in question we shall describe a full approximation constructionin which we shall work for all a and all d simultaneously, or more accurately for all A ⊆ ω and all D ⊆ ωsimultaneously, to approximate a tree such that every set on the tree is of minimal degree. If D is of degreed ≥ 0′ and A is of degree 0 < a < d, then we shall show that there is a set BA,D on this tree, of minimaldegree b < d, such that b ∨ a = d. As is entirely standard in the construction of sets of minimal degree we shallbe using the splitting tree method – for a clear account see [19]. Since we shall be using full approximation the

∗ e-mail: [email protected]

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Math. Log. Quart. 51, No. 5 (2005) / www.mlq-journal.org 471

required splitting trees shall be enumerated during the course of the construction, one splitting at a time. Theremay be many candidates for each of these trees – along the true path of the construction corresponding to anygiven A, D ⊆ ω there will be the usual finite injury as far as these splitting tree candidates are concerned. Theconstruction itself specifies their enumeration procedure.

Definition 1.2 Let Ψjj≥0 be some effective listing of the Turing functionals. For any j, n ≥ 0 and σ ∈ 2<ω

it will be convenient to assume that Ψσj (n)↓ only if Ψσ

j (n′)↓ for all n′ < n and only if the computation con-verges in ≤ |σ| steps. Given k ≥ 2 we shall refer to strings σ0, . . . , σk−1 as a k-fold Ψj splitting if the stringsΨσ0

j , . . . , Ψσk−1j are pairwise incompatible. Given l ≥ 0 we shall refer to the strings σ0, . . . , σk−1 as a k-fold

Ψj splitting of length at least l if it is also the case that the strings Ψσ0j , . . . , Ψσk−1

j are all of length at least l.

Let us begin, then, by considering a typical situation that arises in the course of any construction in which webuild a set of minimal degree by full approximation. Given σ ∈ 2<ω, which is an initial segment of our currentapproximation to B (the set of minimal degree under construction) and j ≥ 0 we search for a Ψj splitting aboveσ and lying on a c. e. tree, Tr say, in order that we can enumerate this splitting into a tree Tr′. Tr may be partial,but for the sake of simplicity let us suppose until further notice that it is not. If such a splitting σ0, σ1 ∈ 2<ω

is found, then, depending on whether this action is compatible with other demands of the construction, we maychoose one of these strings to be an initial segment of our next approximation to B. Now suppose that for eachA, D ⊆ ω we are to construct a set BA,D. In fact, we can use this choice in order to try and code some argumentof D in a very obvious way. Suppose that we wish to code D(n). Then we require that σ0 be an initial segment ofBA,D if D(n) = 0, and otherwise we require the same of σ1. Now suppose that, using oracles for A and BA,D

in order to compute D, we already know that after a certain point in the construction when σ (⊂ BA,D) was aninitial segment of our approximation to BA,D we searched for a Ψj splitting above σ and lying on Tr (and thatno other splitting search interferes with this action). Eventually we shall find the splitting and then, using theoracle for BA,D, we shall be able to determine D(n). Of course this simple coding strategy is no use as it standsfor the simple reason that the splitting in question may never be found. In this case we must be able to defineBA,D in such a fashion that, using oracles for A and BA,D, we can eventually determine that this splitting isnever found and then compute D(n) in this case also. Essentially we must ensure that the construction is codedinto A and BA,D.

The most primitive intuition behind the construction, then, is that for various A ⊆ ω certain strings on Trshould be designated ‘∆(A, Tr′) strings’; if σ′ ⊃ σ is such a string and σ′ ⊂ BA,D, then this will mean that theΨj splitting we were searching for above σ was never found and that there are no Ψj splittings above σ′ (or atleast that there is no k-fold Ψj splitting for some k ≥ 2) on Tr. By designating such strings in pairs we can, onceagain, use the choice as to which is to be an initial segment of BA,D in order to code an argument of D. So ourinitial task is to specify a computable procedure for

a) the designation of ∆(A, Tr′) strings (for A ranging over 2ω) and

b) the search for a Ψj splitting above σ on Tr,

with only the two following outcomes possible in the case that A is not computable. Either we are able to find asplitting above σ on Tr which is suitable for enumeration into Tr′ in the sense that sufficiently many of the stringsin this splitting could be used as an initial segment of BA,D without giving false information that the Ψj splittingsearch above σ (or indeed any other successful splitting search) was not successful, or there is a pair of stringsextending σ which are designated ∆(A, Tr′) strings, which we are able to choose between as an initial segmentof BA,D and above which there is no k-fold Ψj splitting on Tr for some k ≥ 2.

For the sake of simplicity we have already made the assumption that Tr is not partial. For the same reasonlet us now make the (unreasonable) assumption that the technique we are in the process of describing is notnecessary in the enumeration of the tree Tr, or indeed in the enumeration of any of the splitting trees other thanthe Ψj splitting tree we are presently attempting to enumerate, Tr′. Of course Tr itself will be a Ψj′ splittingtree for some j′, but we shall assume for the moment that any string enumerated into this tree can be used as aninitial segment of BA,D without coding false information about the construction. In this simplified case a suitablesplitting search procedure is not difficult to define, but in order to do so we shall need to review a basic techniqueused in [7].

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472 A. E. M. Lewis: The minimal complementation property above 0′

Lemma 1.3 Suppose that S0, . . . , Sk−1 are such that

1) for 0 ≤ i ≤ k − 1, Si contains at least k − i pairwise incompatible strings, and

2) if 0 ≤ i < j ≤ k − 1, then every string in Sj is longer than every string in Si.

Then we may choose from each Si a string σi so that σi is incompatible with σj if i = j.

The proof of this lemma is left to the reader. For now we shall restrict ourselves to the case where we arelooking for a 2-fold splitting and so the use we shall make of this lemma is only the following. Suppose thatwe find a 2-fold splitting above a string σ0 and then, having set l to be larger than any number yet mentionedduring the course of the construction, subsequently find a 2-fold splitting of length at least l above a string σ1

incompatible with σ0. Then we can form a 2-fold splitting with precisely one string extending each of σ0, σ1 bychoosing one string from each of the splittings found.

1.1 A basic splitting search procedure

Suppose that we have a fixed (computable) procedure which, given k ≥ 2 and any σ′ on Tr, produces k pair-wise incompatible strings on Tr extending σ′ any of which are suitable as an initial segment of BA,D (for anyA, D ⊆ ω) since they do not code false information regarding the construction. In order to search for a 2-foldΨj splitting extending σ on Tr begin with five pairwise incompatible strings on Tr extending σ. Label the stringsσ0,0

0 , σ0,01 , σ0,1

0 , σ0,11 and σ0

e . The string which we have labelled σ0e corresponds, in terms of the use that we

make of it, to those strings which we shall later refer to as ‘e strings’ and to those strings which we referred to as‘escape routes’ in [7]. Designate the strings σ0,0

0 and σ0,01 as ∆(A, Tr′) strings for all A ⊆ ω such that A(0) = 0.

Designate the strings σ0,10 and σ0,1

1 as ∆(A, Tr′) strings for all A ⊆ ω such that A(0) = 1. Now continue tosearch for a 2-fold Ψj splitting on Tr above any of the latter four strings using any fixed (exhaustive) searchprocedure. If no such splitting is found, then our approach has been successful. Suppose that we are wantingto code D(n). If A(0) = 0 and D(n) = 0, then we can insist that BA,D extend σ0,0

0 and this will code theinformation, recoverable using oracles for A and BA,D, that the Ψj splitting we were searching for was neverfound, that there does not exist a 2-fold Ψj splitting above σ0,0

0 on Tr, and that D(n) = 0. If A(0) = 1 andD(n) = 1, then we can insist that σ0,1

1 be an initial segment of BA,D and this will similarly code the requiredinformation, and so on.

So let us suppose that at least one splitting is found – above σ0,00 say. Then we let l be larger than any number

yet mentioned during the course of the construction and continue searching for a 2-fold Ψj splitting of length atleast l on Tr above either of σ0,1

0 or σ0,11 . We also take five pairwise incompatible strings on Tr extending σ0

e .Label these strings σ1,0

0 , σ1,01 , σ1,1

0 , σ1,11 and σ1

e . Designate the strings σ1,00 and σ1,0

1 as ∆(A, Tr′) strings for all A

such that A(0) = 0 and A(1) = 0. Designate the strings σ1,10 and σ1,1

1 as ∆(A, Tr′) strings for all A such thatA(0) = 0 and A(1) = 1. We continue to search for a 2-fold Ψj splitting lying on Tr above any one of the latterfour strings as well as searching for a splitting above one of σ0,1

0 or σ0,11 as described previously. Now suppose

that a 2-fold Ψj splitting of length at least l is found above either one of σ0,10 or σ0,1

1 , the former say. Then wemay form a 2-fold Ψj splitting by choosing one string from the splitting found above this string and another fromthe splitting found above σ0,0

0 . Actually this splitting is not quite suitable for enumeration into the Ψj splittingtree under construction since for any particular A, D we can only ensure that there is at least one string in thesplitting which is suitable as an initial segment of BA,D (since it will not code false information concerning theconstruction) when we actually need two in order to be able to code the relevant argument of D, but this is a detailwhich we shall not worry about for the moment. For now we shall regard this outcome as successful. So supposethat no 2-fold Ψj splitting of length at least l is found above either one of σ0,1

0 or σ0,11 . Then our approach has

still been successful, unless A(0) = 0, since if A(0) = 1, then there is a pair of strings on Tr which we havedesignated ∆(A, Tr′) strings and above which there is no k′-fold Ψj splitting on Tr for some k′ ≥ 2. The crucialpoint, then, is that failure of the approach that we have detailed so far is enough to decide A(0). But the sameargument that we have detailed regarding the first set of five strings that we considered can now be applied to thesecond set. We can assume that A(0) = 0. If no splitting is found above any one of σ1,0

0 , σ1,01 , σ1,1

0 , σ1,11 , then our

approach has been successful. If we find a splitting above one of the second two strings and then subsequentlyfind a splitting above one of the first two, then we may form a splitting in the same manner as before. Otherwise

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assume we find a splitting above a string in only one of these two sets before beginning work above σ1e . If the

approach that we have detailed is not successful, then this allows us to determine A(1), and so on.

1.2 An improved splitting search procedure

We now drop the assumption made previously, that there is a computable procedure which, given k ≥ 2 andany σ′ on Tr, produces k incompatible strings extending σ′ on Tr any one of which can be used as an initialsegment of BA,D (whichever A and indeed whichever D we consider) without giving false information about theconstruction, and replace it with a weaker one. Suppose that in the course of searching for a Ψj′′ splitting (forsome j′′ ≥ 0) above a string σ0, suitable for enumeration into a tree Tr′′, we designate a string σ1 ⊃ σ0 as a∆(A, Tr′′) string for some A ⊆ ω. If the splitting that we are searching for is subsequently found and enumeratedinto Tr′′, then we shall enumerate A into γ(σ2) for all σ2 ⊇ σ1. Given A, D ⊆ ω and σ2 ∈ 2<ω, if we find atany point of the construction that A ∈ γ(σ2) this simply means, then, that we cannot define σ2 to be an initialsegment of BA,D without coding false information about the construction. The weaker assumption that we shallmake is that there is a computable procedure which, given k ≥ 2 and any σ′ on Tr, produces k incompatiblestrings extending σ′ on Tr, σ0, . . . , σk−1 say, such that for any A not already in γ(σ′) there is at most one0 ≤ i ≤ k − 1 with A ∈ γ(σi). Ultimately we must describe a procedure which given any k ≥ 2 searches fora k-fold Ψj splitting on Tr above σ (without altering γ(σ)) and with only the two following outcomes possiblein the case that A is not computable and A /∈ γ(σ). Either we find a k-fold splitting σ0, . . . , σk−1 above σ suchthat for any A′ /∈ γ(σ) there is at most one 0 ≤ i ≤ k − 1 such that A′ ∈ γ(σi) (after the appropriate values ofγ have been adjusted subsequent to enumeration of this splitting in Tr′), or there is a pair of strings extending σwhich are designated ∆(A, Tr′) strings and above which there is no k′-fold Ψj splitting on Tr, for some k′ ≥ 2.First we shall describe such a procedure for the case that k = 2.

Consider the solution that we gave to the simplified situation in 1.1. Suppose that we find a 2-fold splittingabove σ0,1

0 and then, having set l to be larger than any number yet mentioned during the course of the construction,subsequently find a 2-fold splitting of length at least l above σ0,0

1 . If we now form a 2-fold splitting σ0, σ1

by choosing one string from each of the splittings found and enumerate this splitting into Tr′, then for anyA ∈ γ(σ0,1

0 ) such that A(0) = 0 we shall have that A ∈ γ(σ0)∩ γ(σ1). For any A ∈ γ(σ0,01 ) such that A(0) = 1

we shall also have that A ∈ γ(σ0) ∩ γ(σ1). Let us forget, momentarily, about the need to define those stringswhich we refer to as e strings – those strings which will perform a function analogous to that of the strings σi

e

in 1.1. We shall just assume that when we have need of such strings they are available to us. We might thendescribe a new solution along the following lines.

As we describe the procedure we shall establish some notation which is not strictly necessary but which shouldmake for an easier discussion. In place of the pair σ0,0

0 , σ0,01 we have a number of pairs, m say. Let us call this

set of m pairs α0. In place of the pair σ0,10 , σ0,1

1 we also have m pairs, let us call this set of pairs β0. Let thevariable µ range over this set of 2m pairs. For each pair µ ∈ β0 we must choose a pair µ′ in α0 which weshall call ∆0(µ). For each µ ∈ α0 we must choose a pair µ′ in β0 which we shall call ∆1(µ). Suppose thatµ = σ0, σ1. If µ ∈ β0, then for each A ∈ γ(µ) = γ(σ0) ∪ γ(σ1) such that A(0) = 0 we designate each stringσ2 ∈ ∆0(µ) as a ∆(A, Tr′) string. If µ ∈ α0, then for each A ∈ γ(µ) such that A(0) = 1 we designate eachstring σ2 ∈ ∆1(µ) as a ∆(A, Tr′) string. We shall only be searching for a Ψj splitting above σ on Tr in orderto define BA,D for specific values of A, for A ∈ Π say. We may assume that Π ∩ γ(σ) = ∅. For A ∈ Π whichsatisfy also the condition that there is not yet any pair of strings µ which have been designated ∆(A, Tr′) stringsand such that A(0) = 0 just choose µ ∈ α0 and designate each string σ2 ∈ µ as a ∆(A, Tr′) string, ensuring thatif A ∈ γ(µ′), then µ = µ′. If these conditions are satisfied but A(0) = 1, then just choose µ ∈ β0 and designateeach string σ2 ∈ µ as a ∆(A, Tr′) string. Again, if A ∈ γ(µ′), we must ensure that µ = µ′. Of course we do allof this in a computable and finitely describable fashion. We satisfy also the following conditions:

(†1) If µ, µ′ ∈ α0 and µ = µ′, then ∆1(µ) = ∆1(µ′). If µ, µ′ ∈ β0 and µ = µ′, then ∆0(µ) = ∆0(µ′).

(†2) If µ ∈ α0 and ∆1(µ) = µ′, then µ = ∆0(µ′). (If µ ∈ β0 and ∆0(µ) = µ′, then µ = ∆1(µ′).)

Before going into some detail, let us describe the procedure in outline. Initially we are in state 0 and we proceedto search for a 2-fold splitting above any one of the strings. If we find a splitting above a string in a pair µ, thenwe do not continue to search for a splitting above the other string in µ – we continue this search only for those

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474 A. E. M. Lewis: The minimal complementation property above 0′

pairs µ′ above which we are yet to find a splitting. As we find splittings we may have to ‘discard’ pairs from α0

or β0. If µ, in α0 say, is discarded this means that we no longer search for a splitting above the strings in µ, andthat we no longer consider µ to be in α0. We may also have to redefine ∆0(µ) or ∆1(µ) for various µ ∈ α0 ∪β0.When we redefine ∆i(µ) it should be understood that we automatically designate each string in (the new valueof) ∆i(µ) as a ∆(A, Tr′) string, for each A ∈ γ(µ) such that A(0) = i. So long as we are in state 0 we shallensure that (†2) is satisfied. The same is not true of (†1) but, so long as we are in state 0, we shall also ensurethat the three following conditions are satisfied:

(†3) If we have found a splitting above a string in µ ∈ α0 and µ′ = µ is in α0, then, for any µ′′ ∈ β0, ifit has ever been the case that ∆1(µ) = µ′′, then it has never been the case that ∆1(µ′) = µ′′. If wehave found a splitting above a string in µ ∈ β0 and µ′ = µ is in β0, then, for any µ′′ ∈ α0, if it hasever been the case that ∆0(µ) = µ′′, then it has never been the case that ∆0(µ′) = µ′′.

(†4) For any µ ∈ α0 ∪ β0 and for any i ∈ 0, 1, we shall not redefine ∆i(µ) unless µ′ which is thepresent value ∆i(µ) is discarded or a splitting is found above a string in µ′. If a splitting is foundabove a string in µ ∈ α0, then we shall discard all pairs µ′ from β0 such that ∆0(µ′) = µ, other thanthose above which we have already found splittings. If a splitting is found above a string in µ ∈ β0,then we shall discard all pairs µ′ from α0 such that ∆1(µ′) = µ, other than those above which wehave already found splittings.

(†5) If µ ∈ α0, then ∆1(µ)↓ and is equal to some pair in β0 above which we are still searching for asplitting. If µ ∈ β0, then ∆0(µ)↓ and is equal to some pair in α0 above which we are still searchingfor a splitting.

While we are in state 0 we ensure that for every A ∈ Π there is precisely one pair µ ( = µ′ if A ∈ γ(µ′)), in α0

if A(0) = 0 and in β0 otherwise, above which we are still searching for a splitting and such that the strings in µhave been designated ∆(A, Tr′) strings. We declare the splitting search procedure to be in state 1 the momentthat we find two splittings above strings in α0, or two splittings above strings in β0. Let us suppose that theformer case applies. We now ensure that for every A ∈ Π such that A(0) = 1 there is precisely one pair µ( = µ′ if A ∈ γ(µ′)) in β0 above which we are still searching for a splitting and such that the strings µ havebeen designated ∆(A, Tr′) strings. Setting l to be larger than any number yet mentioned during the course of theconstruction, we now search for splittings of length at least l, but only above the remaining pairs in β0. Let µ0

and µ1 be those pairs above which we have found splittings in α0. We ensure that ∆1(µ0) ↓ = ∆1(µ1) ↓ andthat both of these values are pairs in β0 above which we are still searching for splittings. We also ensure that(†4) is still satisfied. For those A ∈ Π such that A(0) = 0 we continue to work above some suitably definede strings – the procedure so far seems to have failed for such values of A but we have been able to decide A(0).Now suppose that a splitting is subsequently found above a string in one of the remaining pairs µ2 ∈ β0. By (†4)it cannot be the case that (ever) ∆0(µ2) = µ0 or ∆0(µ2) = µ1. By (†3) (and (†4)), and by the action that wetook upon entering state 1, there can be at most one i ∈ 0, 1 such that (ever) ∆1(µi) = µ2. Let i ∈ 0, 1be such that it has never been the case ∆1(µi) = µ2. We can now form a splitting of the required variety bychoosing one string from the splitting found above µ2 and one string from the splitting found above µi. Of coursethere is an assumption that we have made here. Let the string above which we found a splitting in µ2 be calledσ0 and let the string above which we found a splitting in µi be called σ1. Let the strings that we have taken fromthese splittings in order to form the 2-fold splitting required be σ′

0 and σ′1 respectively. We have assumed that

γ(σ′0) ⊆ γ(σ0) ∪ Π0, where Π0 is the set of all those A for which σ0 was designated a ∆(A, Tr′) string and that

γ(σ′1) ⊆ γ(σ1) ∪ Π1, where Π1 is the set of all those A for which σ1 was designated a ∆(A, Tr′) string. Unlike

some of those assumptions that we have made previously, however, this is a reasonable assumption to make.We shall now describe the splitting search procedure in more detail. Initially we search for a 2-fold splitting

above any one of the strings. If no such splitting is found, then it is easy to see that our procedure has beensuccessful. So suppose that such a splitting is found, above one of the strings in µ0 ∈ α0 say. We must discardthat pair µ0 ∈ β0 such that ∆0(µ0) = µ0. There is one pair µ1 = µ0 in α0 such that ∆1(µ1) = µ0. Choose any(remaining) pair µ2 in β0 such that

a) µ2 = ∆1(µ0) and

b) ∆0(µ2) = µ1.

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Math. Log. Quart. 51, No. 5 (2005) / www.mlq-journal.org 475

Redefine ∆1(µ1) = µ2. So far, then, we have not required any more than four pairs in each of α0, β0. If A ∈ Πand there is presently no pair in (α0 ∪ β0) − µ0 which are designated ∆(A, Tr′) strings, then just choose one( = µ′ if A ∈ γ(µ′)), in α0 − µ0 if A(0) = 0 and in β0 otherwise, to designate as such. We now proceed tosearch for splittings above all of the remaining strings in (α0 ∪ β0)−µ0. If no such splitting is found, then theprocedure has been successful. We may assume, then, that such a splitting is found.

Suppose first that this splitting is found above another pair, let us call it µ1, in α0. Then we immediatelydeclare our splitting search procedure to be in state 1. We must discard that pair, let us call it µ0, in β0 such that∆0(µ0) = µ1. If µ0 = ∆1(µ0), then choose some remaining pair in β0 other than ∆1(µ1) to redefine as ∆1(µ0).If A ∈ Π, A(0) = 1 and there is presently no pair in β0 which are designated ∆(A, Tr′) strings, then just chooseone ( = µ′ if A ∈ γ(µ′)) to designate as such. The crucial point is this. We now set l to be larger than anynumber yet mentioned during the course of the construction. From now on we shall search for 2-fold splittingsof length at least l and only above the remaining pairs in β0. If such a splitting is found above a string in β0, astring in a pair µ2 say, then we shall be able to form a splitting of the required variety by choosing one stringfrom the splitting found above a string in µ2 and then one string from the splitting found above either µ0 or µ1.This follows since it cannot be the case that ∆0(µ2) has ever been defined to be equal to either of µ0 or µ1 andfor at most one i ∈ 0, 1 is it the case that ∆1(µi) = µ2. If such a splitting is found, then the procedure we aredescribing has been successful. If no such splitting is found this is still the case, unless A(0) = 0. For such A wemay now continue to work above suitably defined e strings – proceeding to define α1, β1 and so on. So far wehave not required that there initially be any more than four pairs in each of α0, β0.

So suppose instead that the second splitting is found above a string in one of the pairs in β0, µ1 say (where µ0 isas defined previously). If µ1 = ∆1(µ0): we must redefine ∆1(µ0). Let µ0 = ∆0(µ1). Choose any µ1 ∈ β0 otherthan ∆1(µ0) (and µ1) and other than that such that there are two µ2, ∆1(µ2) = µ1 and redefine ∆1(µ0) = µ1.There is a single pair in α0 other than µ0, µ2 say, such that ∆1(µ2) = µ1. Discard this pair. For the single pairµ3 ∈ β0 such that ∆0(µ3) = µ2 redefine ∆0(µ3) to be any pair in α0 other than µ0, ∆0(µ1) and other thanany pair µ4 such that ∆1(µ4) = µ3. If A ∈ Π and there is presently no pair in (α0 ∪ β0) − µ0, µ1 whichare designated ∆(A, Tr′) strings, then just choose one ( = µ′ if A ∈ γ(µ′)), in α0 − µ0 if A(0) = 0 and inβ0 − µ1 otherwise, to designate as such. So far we have not required that there should initially be any morethan six pairs in each of α0, β0.

Otherwise (if µ1 = ∆1(µ0)): we must discard from α0 those µ such that ∆1(µ) = µ1. There are a maximumof two such. We must then redefine ∆0(µ0) for those µ0 ∈ β0 such that ∆0(µ0) is a pair we have just discarded.Just choose any µ1 in α0 other than µ0, ∆0(µ1), and other than any µ2 such that ∆1(µ2) = µ0. If A ∈ Π andthere is presently no pair in (α0 ∪ β0) − µ0, µ1 which are designated ∆(A, Tr′) strings, then just choose one( = µ′ if A ∈ γ(µ′)), in α0 − µ0 if A(0) = 0 and in β0 − µ1 otherwise, to designate as such. Once again,we have not required that there should initially be any more than six pairs in each of α0, β0.

We proceed to search for a 2-fold Ψj splitting on Tr above any of the strings in the pairs in (α0∪β0)−µ0, µ1.If no such splitting is found, then clearly the procedure has been successful. So assume that such a splitting isfound, above a string in the pair µ2 say. Then we declare the splitting search procedure to be in state 1. If µ2 ∈ α0

we must remove all of those µ from β0 such that ∆0(µ) = µ2. If ∆1(µ0) is amongst these pairs, then we mustredefine it to be µ0 ∈ β0 other than µ1 and ∆1(µ2). If A ∈ Π, A(0) = 1 and there is presently no pair inβ0−µ1 which are designated ∆(A, Tr′) strings, then just choose one ( = µ′ if A ∈ γ(µ′)) to designate as such.For all A ∈ Π such that A(0) = 0 we continue to work above a set of suitably defined e strings, defining α1, β1

and so on.If µ2 ∈ β0 we must remove all of those µ from α0 such that ∆1(µ) = µ2. If ∆0(µ1) is amongst these

pairs, then we must redefine ∆0(µ1) to be µ0 ∈ α0 other than µ0 and ∆0(µ2). If A ∈ Π, A(0) = 0 and thereis presently no pair in α0 which are designated ∆(A, Tr′) strings, then just choose one ( = µ′ if A ∈ γ(µ′))to designate as such. For all A ∈ Π such that A(0) = 1 we continue to work above a set of suitably definede strings, defining α1, β1 and so on.

Choose l to be larger than any number yet mentioned during the course of the construction. If a splitting oflength at least l is ever subsequently found above a string in a pair µ ∈ α0 if µ2 ∈ β0, or above a string in a pairµ ∈ β0 if µ2 ∈ α0, then the splitting search procedure terminates – we have found a splitting of the requiredvariety. Consider the number of µ initially in α0 above which we have found a splitting or which have beendiscarded, by the point of the procedure at which we have found a splitting above a pair in α0 and a splittingabove a pair in β0 (presuming that this does indeed occur). This number is no more than three, and the same

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476 A. E. M. Lewis: The minimal complementation property above 0′

is true of β0. At this latter point of the procedure, if distinct µ1, . . . , µi ∈ α0 satisfy the condition that, for1 ≤ i1, i2 ≤ i, ∆1(µi1 ) = ∆1(µi2 ), then i ≤ 3. Similarly if distinct µ1, . . . , µi ∈ β0 satisfy the condition that,for 1 ≤ i1, i2 ≤ i, ∆0(µi1) = ∆0(µi2), then i ≤ 3. Clearly, then, if α0 and β0 each consist of eight pairs, thenthis will suffice. The same will be true of all αi, βi.

1.3 Defining e strings

We must describe how to alter this procedure in order that we can properly define suitable e strings. The firstobservation to be made is that, whereas in 1.1 it sufficed to use one e string σi

e for each i ≥ 0, we shall now needat least two. Suppose that we were to find a splitting extending each of σ0, σ1 ⊃ σ0

e . If we were to form a 2-foldsplitting σ′

0, σ′1 by choosing one string from each of these splittings and enumerate this splitting into Tr′, then for

any A ∈ γ(σ0e) we would have that A ∈ γ(σ′

0)∩γ(σ′1). So while searching for a 2-fold splitting we shall consider

e strings in sets of two. If = σ,0, σ,1 is a set of e strings, then we agree that γ() = γ(σ,0) ∪ γ(σ,1)and insist that we do not use for those A ∈ γ(). What this means, more precisely, is that for such values ofA we shall not designate strings extending those in as ∆(A, Tr′) strings – for such values of A we shall haveto designate ∆(A, Tr′) strings elsewhere on Tr. The solution, then, is that we shall initially require two sets of estrings, 0

0 and 01. We shall replace the set of pairs α0 with two such sets α(0

0) and α(01). We replace the set

of pairs β0 with two such sets β(00) and β(0

1). Initially we are in state 0. We proceed to operate with the twosets of pairs α(0

0) and β(00) in the same way that we described in 1.2 for α0 and β0, proceeding similarly with

α(01) and β(0

1). We do not search for splittings above the e strings. For i ∈ 0, 1 and A ∈ γ(0i ): if A(0) = 0

we ensure that those strings we have designated ∆(A, Tr′) strings are in α(0i ) and if A(0) = 1 we ensure that

those strings we have designated ∆(A, Tr′) strings are in β(0i ). We enter state 1 as soon as splittings have been

found above two pairs in (any single) one of α(00), α(0

1), β(00), β(0

1). Let us suppose that we enter state 1upon finding splittings above two pairs in α(0

0). Let Π be as described earlier. Then we act in order to ensurethat for all A ∈ Π such that A(0) = 1 there is a pair of strings ( = µ′ if A ∈ γ(µ′)) in β(0

0) or β(01) which have

been designated ∆(A, Tr′) strings and above which we are still searching for a splitting – in a manner almostidentical to that which we described in 1.2. For all A ∈ γ(0

1), however, (so not just those such that A(0) = 1)we insist that there be a pair of strings in β(0

1) which have been designated ∆(A, Tr′) strings and above whichwe are still searching for a splitting. For all A ∈ Π such that A(0) = 0, except those A ∈ γ(0

1), we shall nowwork above the set of e strings 0

1.

Let 01 = σ0

1,0, σ01,1. When working above this set of e strings we shall operate in exactly the same

fashion. We must insist, though, that all strings in α(10) or α(1

1) extend σ01,0 and that all strings in β(1

0) orβ(1

1) extend σ01,1. For each set of e strings 1

i we insist that there is one string in the set extending each ofσ0

1,0, σ01,1. Previously we made the assumption that there is a computable procedure which, given k ≥ 2 and

any σ′ on Tr, produces k incompatible strings extending σ′ on Tr, σ0, . . . , σk−1 say, such that for any A notalready in γ(σ′) there is at most one 0 ≤ i ≤ k − 1 with A ∈ γ(σi). We now need the stronger assumption thatthere is a computable procedure which, given k ≥ 2 and p incompatible strings ϕ0, . . . , ϕp−1 on Tr where p|k,produces k incompatible strings on Tr with k

p extending each ϕi, σ0, . . . , σk−1 say, such that for any A notalready in

⋃i γ(ϕi) there is at most one 0 ≤ i ≤ k − 1 with A ∈ γ(σi). Now suppose that another splitting is

eventually found above one of the remaining pairs µ ∈ β(00)∪β(0

1). If µ ∈ β(00), then, by the same reasoning

that was applicable previously, we are now able to form the 2-fold splitting required. But if µ ∈ β(01) this is

also the case, since then if A ∈ γ(µ) or A is such that we have designated the strings in µ as ∆(A, Tr′) strings,we have not designated any strings in α(0

0) as ∆(A, Tr′) strings. If A ∈ γ(µ′) for µ′ ∈ α(00), then we have not

designated the strings in µ as ∆(A, Tr′) strings.

The following tasks are still outstanding. It must be shown how to search for a k-fold splitting for any k ≥ 2above strings ϕ0, . . . , ϕp−1. In order to do so we must revise a technique developed in [7]. In the procedureoutlined above for the case that k = 2 suppose that there is a separate strategy responsible for the ‘management’of each four sets of pairs α(i

0), β(i0), α(i

1), β(i1). When any strategy enters state 1 it has just one set of

e strings available for ‘use’. In the case that k > 2 each strategy must be responsible for more than four setsof pairs and may enter states > 1 before a splitting of the required variety is found. When such a strategy is instate k′ ≥ 1 we must ensure that there are k′ sets of e strings available for use. This will all be explained in the

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discussion that follows. We must also drop the assumption that Tr is not partial (we now do so) and show how tocombine these splitting search procedures in order to construct the sets of minimal degree in question.

1.4 The strategy B[t, u, l, k, ϕ0, . . . , ϕp−1](na, n) (in outline)

Here u ∈ 0, 1 and is just used in order to indicate the kind of position the strategy occupies on the tree ofstrategies; l ∈ ω and is used in specifying the length of splittings that the strategy must search for; k ∈ ω andrecords the fact that this strategy must search for a k-fold splitting with an equal number of strings extending eachof ϕ0, . . . , ϕp−1 ∈ 2<ω (we shall have that p|k); n ∈ ω and indicates that we are concerned with coding D(n);na ∈ ω and indicates that ‘failure’ of this strategy is enough to decide A(na) and t is a finite set of splitting treeswhich are presently under construction. Let us suppose that t = Tr−1,0, Trj1,i1 , . . . , Trjm,im for some m ∈ ωand for j1 < · · · < jm. In this notation Trj,i should be regarded as the ith candidate for a Ψj splitting tree. Thetrees in t are those which we are concerned with constructing on this part of the tree of strategies. The tree Tr−1,0

is just a tree which we enumerate for the sake of technical convenience, but which is not required to be any kindof splitting tree. During the course of the construction as a whole, the tree Trj1,i1 will be constructed so as to be asubtree of Tr−1,0. The tree Trj2,i2 will be constructed so as to be a subtree of Trj1,i1 , and so on. Let us agree thatj0 = −1 and i0 = 0 and assume for now that m ≥ 1. The variables H, H ′, H ′′, . . . will be used to range over theset of strategies, let H = B[t, u, l, k, ϕ0, . . . , ϕp−1](na, n). Every stage at which H is passed control we will havedefined a value Π(H) which is the set of all A with which the strategy must be concerned. In fact the value Π(H)may be redefined a finite number of times during the course of the construction to be a superset of its previousvalue. In the context of discussing the strategy H every string σ under consideration will extend one of the ϕi, welet ϕH(σ) designate this string. The aim of the strategy H , then, is to find a k-fold Ψjm splitting, σ0, . . . , σk−1

say, on Trjm−1,im−1 with kp strings extending each ϕi and such that this splitting is suitable for enumeration into

Trjm,im since we shall require that then, for 0 ≤ r < r′ < k, γ(σr) ∩ γ(σr′) = γ(ϕH(σr)) ∩ γ(ϕH(σr′)). Forthose who have read [7] this intersection property will be familiar, for others the following explanation shouldbe helpful. At no stage in the construction shall we work above any string σ for those A ∈ γ(σ). Certainly,then, if A ∈ Π(H) (at any stage at which H is passed control) we shall have that A /∈ ⋃p−1

r=0 γ(ϕr). This morecomplicated intersection property therefore amounts to little more than the intersection property we strived forin 1.2 and 1.3 – for 0 ≤ r < r′ < k we shall insist that, once the splitting has been enumerated into Trjm,im

and the appropriate values of γ adjusted, γ(σr) ∩ γ(σr′) ∩ Π(H) = ∅, γ(σr) − γ(ϕH(σr)) ⊆ Π(H), andγ(σr′)− γ(ϕH(σr′)) ⊆ Π(H). We cannot insist that γ(σr)∩ γ(σr′) = ∅ since all of the strings ϕi, for example,may extend the same string σ such that γ(σ) = ∅. The reader may also require clarification as to why it is nowthe case that p strings, in particular ϕ0, . . . , ϕp−1, have replaced the single string σ above which we consideredsearching for a splitting previously. Recall that in searching for a 2-fold splitting we required sets consisting oftwo e strings. In the case that k > 2 we will require such sets consisting of more than two e strings and H maybe a strategy which works above such a set of e strings. Initially the strategy is in state −1.

S t a t e −1. In the procedures described previously we began with a number of strings on Tr. The rolethat was played by Tr is now played by Trjm−1,im−1 . In fact we must first pass control to another strategy inorder that these strings should be provided. In searching for a 2-fold splitting we initially required four lots ofeight pairs as well as two sets of two e strings, a total of sixty eight strings. In general H initially demandsx0 = 17

2 (k3−k2 +4k) strings. We shall explain this number below. Putting k = 2 in this expression we find thatx0 = 68. In fact the case k = 2 is special in terms of the number of strings required, but since having extra stringswill never cause a problem it suffices always just to demand the number of strings specified by the expressionabove. While in state −1 the strategy H passes control to the strategy H ′ = B[t′, 1, 0, x0, ϕ0, . . . , ϕp−1](0, n)where t′ = t−Trjm,im. We define Π(H ′) = Π(H). The argument u here takes the value 1 in order to indicatethat H ′ is the first strategy (or rather the lowest on the tree of strategies which we consider to ‘grow upwards’)of a set of strategies whose purpose is to search for the required splitting – in that sense H ′ is analogous to thestrategy which would have been responsible for 0

0, 01, α(0

0), α(01), β(0

0) and β(01) in 1.3. H will remain in

state −1 until declared to be in state 0 by another strategy.So while in state −1 the strategy H passes control to the same (single) strategy at any stage at which it is passed

control. More generally, however, the fact that we are working for all A ⊆ ω and all D ⊆ ω simultaneously meansthat at any given stage a strategy H ′′ may pass control to several others one ‘level’ up on the tree of strategies. Itshould be understood that these strategies act simultaneously. If a strategy H ′′′ terminates stage s activity, then

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478 A. E. M. Lewis: The minimal complementation property above 0′

all other strategies on the same level as H ′′′ (which have been passed control at stage s) are allowed to completetheir instructions for the stage, but are not allowed to pass control to another.

S t a t e 0. We are provided with 172 (k3 − k2 + 4k) strings on Trjm−1,im−1 . The first time that the strategy is

passed control in state 0 we divide these strings up in the following way. The fact that we are searching for a k-fold splitting will mean that the maximum state that the strategy can be declared to be in (before being terminatedbecause a splitting of the required variety has been found) is state k − 1. Each set of e strings must consist ofk strings, with k

p extending each ϕi. If = σ,0, . . . , σ,k−1 is a set of e strings, we shall say that this setof e strings is used by H when this strategy passes control to another of the form B[. . . , σ,0, . . . , σ,k−1](·, ·).While in state 1 we use one set of e strings. One set will already be useless by this stage (as was the case in 1.3).While in state 2, H will use two (new) sets, and generally while in state k′′ we shall require k′′ (new) sets ofe strings. Upon entering state 2 it may be the case that another set of e strings is made useless, as was the caseupon entering state 1. In total, then, we require k(1

2k(k − 1) + 2) e strings. Let these sets of e strings be calledH0 , H

1 , . . . . Apart from the e strings we also require k groups of strings, with all the strings in each groupextending the same ϕi and with an equal number of strings in each group. Choose any linear ordering of thesek groups, it will be convenient to be able to refer to the 0th group, the 1st group, the 2nd group and so on. Callthe 0th group αH and the 1st group βH . Divide αH into sets of eight pairs, one for each set of e strings. Labelthese α(H

0 ), α(H1 ), . . . . Similarly divide βH into sets of eight pairs and label these β(H

0 ), β(H1 ), . . . . We

divide up the remaining groups of strings in the same way, but since we shall be using these strings differently itseems reasonable to use different notation. For r ≥ 2 we call the sets of eight pairs in the rth group of strings(H

0 , r), (H1 , r), . . . . Where it is convenient we may also use this notation for r < 2. If = σ,0, . . . , σ,k−1

is a set of e strings, then we associate every string in this set with a different one of the k groups, in such away that each σ,i is associated with a group every string of which extends ϕH(σ,i). In total, then, this gives16k(1

2k(k − 1) + 2) strings and 16k(12k(k − 1) + 2) + k(1

2k(k − 1) + 2) = 172 (k3 − k2 + 4k).

Every stage at which the strategy is passed control in state 0 we proceed just as we did in 1.2 for each pairα(H

i ) and β(Hi ). So we search only for splittings above the strings in αH and βH . While we are in state 0 we

insist (as previously) that for all A ∈ Π(H) there is always some pair of strings µ ( = µ′ if A ∈ γ(µ′)), in αH

if A(na) = 0 and in βH otherwise, above which we are still searching for a splitting and which are designated∆(A, Trjm,im) strings. For A ∈ γ(H

i ) such that A(na) = 0 we ensure that those strings which we havedesignated ∆(A, Trjm,im) strings are in α(H

i ). For A ∈ γ(Hi ) such that A(na) = 1 we ensure that those strings

which we have designated ∆(A, Trjm,im) strings are in β(Hi ). For any i, r ≥ 0 let γ(H

i , r) =⋃

µ∈(Hi ,r) γ(µ).

For any i ≥ 0 and r ≥ 2, if A ∈ γ(Hi , r), then we ensure that the pair of strings which we have designated

∆(A, Trjm,im) strings are in α(Hi ) if A(na) = 0 and are in β(H

i ) otherwise. The strategy is declared to be instate 1 as soon as two splittings are found above strings in any (single one) of α(H

0 ), β(H0 ), α(H

1 ), β(H1 ), . . . .

S t a t e 1. Let us suppose that two splittings have been found above strings in β(H1 ). Then the first time

that the strategy is passed control in state 1 we perform the same instructions for α(H1 ) and β(H

1 ) as werespecified in 1.2. For all A ∈ Π(H) such that A(na) = 0 we ensure that there is some pair of strings µ ( = µ′

if A ∈ γ(µ′)) in αH above which we are still searching for a splitting and which are designated ∆(A, Trjm,im)strings. We enumerate 1 into E(H) – this set records all those i such that H

i is now useless as a set of e strings.For A ∈ γ(H

i ) such that A(na) = 0 we ensure that those strings which we have designated ∆(A, Trjm,im)strings are in α(H

i ). For any i ≥ 0 and r ≥ 2, if A ∈ γ(Hi , r) and A(na) = 0, then we ensure that the pair

of strings which we have designated ∆(A, Trjm,im) strings are in α(Hi ). We also choose some set of e strings

Hi for i /∈ E(H), H

0 say. For all A ∈ γ(H0 ) (so not just those such that A(na) = 0) we ensure that there

is a pair of strings in α(H0 ) above which we are still searching for a splitting and which have been designated

∆(A, Trjm,im) strings. Every stage at which the strategy is passed control in state 1, for all A ∈ Π(H) such thatA(na) = 1 except those in γ(H

0 ), we shall now use the set of e strings H0 . We consider this set of e strings to

be ‘active’ and, choosing l1 larger than any number yet mentioned during the course of the construction, continueto search for splittings of length at least l1 above all nondiscarded pairs in αH other than those above which wehave already found splittings. The strategy is declared to be in state 2 if another splitting is found above a stringin αH , or terminates in this case if k = 2.

S t a t e 2. Let us suppose that the strategy is declared to be in state 2 because a splitting is found above astring in µ ∈ α(H

4 ). The first time that the strategy is declared to be in state 2 we enumerate 4 into E(H). Wealso enumerate 0 into this set since any sets of e strings which were active while in state 1 are now useless – and

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Math. Log. Quart. 51, No. 5 (2005) / www.mlq-journal.org 479

are no longer considered to be active. Choose a set of e strings Hi such that i /∈ E(H), H

2 say. Discard thestring from this set which is associated with αH , so that this set of e strings now consists of k − 1 strings. Thisset of e strings is now active. Choose another set of e strings H

i′ which is not active and such that i′ /∈ E(H),H3 say. Discard the string from this set which is associated with βH . This set of e strings is now active. For all

A ∈ γ(H2 ) ensure that there is a pair of strings µ ∈ (H

2 , 2) which are designated ∆(A, Trjm,im) strings. Forall A ∈ γ(H

3 ) ensure that there is a pair of strings µ ∈ (H3 , 2) which are designated ∆(A, Trjm,im) strings.

Choose l2 larger than any number yet mentioned during the course of the construction. Every stage at which thestrategy is passed control in state 2 we continue to search for a splitting of length at least l2 above any string ina pair µ ∈ (H

2 , 2) ∪ (H3 , 2). For all A ∈ Π(H) except those A ∈ γ(H

2 ) ∪ γ(H3 ) we use one of these sets

of e strings. The reason that we define these two sets of active e strings should be understood as follows. Wehave now found two splittings, above σ0 ∈ αH and σ1 ∈ βH say, such that we can choose one string from eachof these splittings in order to form a 2-fold splitting with the required intersection properties as regards γ. Forall A ∈ γ(σ0) and all those A such that σ0 has been designated a ∆(A, Trjm,im) string we shall use the set ofe strings H

3 . For such A our approach has not yet been successful but as we continue to search for splittingsabove this set of e strings the splitting that we have found above σ1 remains useful. We shall not proceed todesignate strings extending those in H

3 as ∆(A′, Trjm,im) strings for A′ ∈ γ(σ1) and for those A′ such that σ1

has been designated a ∆(A′, Trjm,im) string since for these A′ we shall use the other set of e strings H2 . Above

the strings in H3 , then, we need only search for a (k − 1)-fold splitting and the same is true above the strings

in H2 . The strategy is declared to be in state 3 if a splitting is found above a string in µ ∈ (H

2 , 2) ∪ (H3 , 2), or

terminates in this case if k = 3.

S t a t e k′ > 2. Suppose that the strategy is declared to be in state k′ because a splitting is found above a stringin µ ∈ (H

6 , k′ − 1). We have now found k′ splittings, above the strings σ0, . . . , σk′−1 in the 0th, . . . , (k′ − 1)th

group respectively say, such that we can choose one string from each in order to form a splitting with the requiredintersection properties. The first time that the strategy is passed control in state k′ we enumerate into E(H) allof those i (including i = 6) such that H

i was active while the strategy was in state k′ − 1. For each 0 ≤ i < k′ inturn, perform the following. Choose a set of e strings H

i′ which is not active and such that i′ /∈ E(H). Discardall strings from this set which are associated with the first k′ groups (those groups above which we have founda splitting) except that which is associated with the ith group. The set of e strings H

i′ is now active and will beused by all A ∈ Π(H) such that A ∈ γ(σi) or such that σi has been designated a ∆(A, Trjm,im) string. For allA ∈ γ(H

i′ ) ensure that there is a pair of strings µ ∈ (Hi′ , k′) which are designated ∆(A, Trjm,im) strings.

Let E′ be the set consisting of those i such that Hi is active. Choose lk′ to be larger than any number yet

mentioned during the course of the construction. Every stage at which the strategy is passed control in state k′

we continue to search for splittings of length at least lk′ above the strings in any pair µ ∈ ⋃i∈E′(H

i , k′). For allA ∈ Π(H) except those A ∈ ⋃

i∈E′ γ(Hi ) we use precisely one of the sets of e strings which are active. The

strategy is declared to be in state k′ + 1 when another splitting is found, or terminates in this case if k = k′ + 1.

Let us conclude our discussion of the strategy H by considering the possible outcomes. Assume that His passed control at an infinite number of stages. If H is never declared to be in state 0, then at every stageat which it is passed control it passes control to the strategy H ′ = B[t′, 1, 0, x0, ϕ0, . . . , ϕp−1](0, n) wheret′ = t − Trjm,im and where we define Π(H ′) = Π(H). In this case one of the trees in t′ is partial. Since|t′| < |t|, both these sets are finite, and since Tr−1,0 is not partial, we may regard this as a successful outcome.So suppose that H is eventually declared to be in state 0. Let Π(H) take its final value. In this case, if H is neverdeclared to be in state 1, then our approach may be regarded as successful since then for all A ∈ Π(H) thereis a pair of strings on Trjm−1,im−1 which are designated ∆(A, Trjm,im) strings by H and above which there isno k′-fold splitting on Trjm−1,im−1 for some k′ ≥ 2. Suppose that H is eventually declared to be in state 1. IfH is never declared to be in state 2, then, if A ∈ Π(H) is such that there is no pair of strings designated by Has ∆(A, Trjm,im) strings and above which there is no k′-fold splitting on Trjm−1,im−1 for some k′ ≥ 2, we areable to determine A(na). So suppose that H is eventually declared to be in state 2 and let k′′ be the greateststate that H is ever declared to be in. If k′′ = k, then H terminates and we have found a splitting suitable forenumeration into Trjm,im . Otherwise for all A ∈ Π(H) it is either the case that there is a pair of strings onTrjm−1,im−1 which are designated ∆(A, Trjm,im) strings by H and above which there is no k′-fold splitting onTrjm−1,im−1 for some k′ ≥ 2, or that k′′ − 1 of the splittings found by H remain useful when working above the

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480 A. E. M. Lewis: The minimal complementation property above 0′

set of e strings used for this value of A while in state k′′ – so that above this set of e strings we need only searchfor a (k − k′′ + 1)-fold splitting. This outcome may therefore also be regarded as successul.

1.5 The strategy D[t, ϕ](n) (in outline)

Let H = D[t, ϕ](n). Here ϕ ∈ 2<ω, n ∈ ω and indicates that we are concerned with coding D(n), t is a finite setof splitting trees and we will have defined Π(H) to be the set of all A ⊆ ω with which H must be concerned. Thestrategy is very simple, its use is really just in organizing the tree of strategies. Initially we regard the strategyas being in state −1. While in this state it passes control to the strategy H ′ = B[t, 1, 0, 3, ϕ](0, n) where wedefine Π(H ′) = Π(H). If it is ever declared to be in state 0 (by one of those B strategies above it on the treeof strategies), then we are provided with three strings and for each A ∈ Π(H) and D ⊆ ω we can choose oneof these strings as an initial segment of BA,D without coding false information about the construction and whilesuccessfully coding D(n).

Definition 1.4 We say that σ ∈ 2<ω is compatible with a tree Tr if σ has an extension on Tr or if σ extends aleaf of Tr.

Since we are about to describe the manner in which these strategies should be combined in order to constructthe sets of minimal degree in question, the following point is worth making now. Clearly, when enumeratingany splitting into a tree, we must ensure that every string in the splitting extends what was (until this point) aleaf of the tree. Now suppose that t = Tr−1,0, Trj1,i1 , . . . , Trjm,im with j1 < · · · < jm and that the strategyH = B[t, u, l, k, ϕ0, . . . , ϕp−1](na, n) terminates because it finds the k-fold Ψjm splitting on Trjm−1,im−1 thatis is searching for, σ0, . . . , σk−1 say. In order to form a splitting that is suitable for enumeration into Trjm,im

(and indeed a set of strings suitable for that strategy which H now declares to be in state 0 to work with) wemust actually take an extension of each σi long enough that it extends a leaf of each of the trees in t. If thereason behind this is not clear now, then it will become so later. We said previously that, during the course of theconstruction, the tree Trj1,i1 would be constructed so as to be a subtree of Tr−1,0, while the tree Trj2,i2 wouldbe constructed so as to be a subtree of Trj1,i1 , and so on. In fact this is not quite true. For r ≥ 1, when weenumerate a splitting into Trjr ,ir we shall have that every string σ in this splitting extends a different leaf σ′ ofTrjr−1,ir−1 (j0 = −1, i0 = 0) and thereafter remains compatible with the latter tree since any string subsequentlyenumerated into Trjr−1,ir−1 extending σ′ will extend σ.

1.6 The tree of strategies (an example)

At the base of the tree of strategies is the strategy H = D[t, ∅](0) where t = Tr−1,0, Tr0,0 and where wedefine Π(H) = 2ω. While in state −1, H passes control to the strategy H1 = B[t, 1, 0, 3, ∅](0, 0). We defineΠ(H1) = Π(H). Suppose that H is eventually declared to be in state 0. Every stage at which H is passed controlin state 0 it passes control to three strategies H2 = D[t2, ϕ2](1), H3 = D[t3, ϕ3](1) and H4 = D[t4, ϕ4](1),where ϕ2, ϕ3 and ϕ4 are the three strings that H is provided with upon being declared to be in state 0.

Definition 1.5 Let z be some computable injection from the set of strategies which are passed control duringthe course of the construction into ω.

We assume that z(H) = 0. Since we do not yet have any reason to believe that Tr0,0 will be partialwe shall continue building this tree above each Hi, i ∈ 2, 3, 4. For i ∈ 2, 3, 4 we therefore defineti = Tr−1,0, Tr0,0, Tr1,z(Hi). The point of the function z, then, is just that each of these strategies should have adifferent candidate for a Ψ1 splitting tree. Let Π = 2ω−⋃

i∈2,3,4 γ(ϕi). We define Π(H2) = γ(ϕ3)∪γ(ϕ4)∪Π,Π(H3) = γ(ϕ2) ∪ γ(ϕ4) ∪ Π and Π(H4) = γ(ϕ2) ∪ γ(ϕ3), so that for all A, D ⊆ ω there are precisely twoof these strings which we can use as an initial segment of BA,D without coding false information as regardsthe construction, and so precisely one which we can use as an initial segment of BA,D in order to correctlycode D(0). Fix A, D ⊆ ω and suppose that, in order to correctly code D(0), we specify that ϕ2 ⊂ BA,D (sothat necessarily A ∈ Π(H2)). Suppose that H2 is never declared to be in state 0. Then at every stage at whichH2 is passed control it passes control to the strategy H5 = B[t2, 1, 0, 3, ϕ2](0, 1), which we shall suppose isdeclared to be in state 0 but is never declared to be in state 1 and where we define Π(H5) = Π(H2). Every stageat which H5 is passed control in state 0 it passes control to a different strategy for each string above which it isstill searching for a splitting. There is a single pair of strings which H5 designates as ∆(A, Tr1,z(H2)) strings and

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above which we never cease searching for a splitting. D(1) dictates which of these two strings we should defineto be an initial segment of BA,D, σ say. Every stage at which it is passed control in state 0, H5 passes control tothe strategy H6 = D[t6, σ](2) (amongst others) where Π(H6) is defined to be the set of all those A′ ∈ Π(H2)for which H5 designates σ a ∆(A′, Tr1,z(H2)) string. Since H5 failed to find a Ψ1 splitting on Tr0,0 above σ, H6

assumes that Tr1,z(H2) is partial. We therefore define t6 = Tr−1,0, Tr0,0, Tr2,z(H6). Let us suppose that H6

is never declared to be in state 0 so that every stage at which it is passed control it passes control to the strategyH7 = B[t6, 1, 0, 3, σ](0, 2), which we shall suppose is declared to be in state 2 but never terminates and wherewe define Π(H7) = Π(H6). H7 finds a splitting above a string in αH7 and a splitting above a string in βH7 , letus say σ0 and σ1 respectively, such that we can choose one string from each of these splittings in order to form asplitting with the required intersection properties as regards γ. For i ∈ 0, 1 let Πi be the set of all those A′ inΠ(H7) such that A′ ∈ γ(σi) or such that σi is designated a ∆(A′, Tr2,z(H6)) string. Let the two sets of e stringswhich are active while H7 is in state 2 be H7

0 and H71 . Let Π2 = Π(H7) − Π0 ∪ Π1 ∪ γ(H7

0 ) ∪ γ(H71 )

and suppose that for i ∈ 0, 1 and A′ ∈ Πi we use the set of e strings H7i while H7 is in state 2. Among the

strategies that H7 passes control to while in state 2 are those above the two sets of active e strings,H8 = B[t6, 0, l, 2, σ

H70 ,0

, σ

H70 ,1

](0, 2) and H9 = B[t6, 0, l, 2, σ

H71 ,0

, σ

H71 ,1

](0, 2), where l is chosen to be

larger than any number yet mentioned during the course of the construction when H7 enters state 2 and wherewe define Π(H8) = Π0 ∪ Π2, Π(H9) = Π1. Suppose that A ∈ Π(H8) (as this set is eventually defined), thatH8 is declared to be in state 1 but never terminates and that there is no pair of strings which this strategy des-ignates as ∆(A, Tr2,z(H6)) strings above which it never ceases to search for a splitting. Let the set of e stringsthat are active while H8 is in state 1 be H8

0 . Every stage at which it is passed control in state 1 the strategy H8

passes control to the strategy H10 = B[t6, 0, l, 2, σ

H80 ,0

, σ

H80 ,1

](1, 2) (amongst others) which we shall supposeis never declared to be in state 0. Every stage at which it is passed control H10 passes control to the strategyH11 = B[t11, 1, 0, 102, σ

H80 ,0

, σ

H80 ,1

](0, 2) where t11 = t6 −Tr2,z(H6). Since Tr−1,0 is not partial H11 must

enter state 0, but let us suppose that it is never declared to be in state 1. Then there is a pair of strings whichH11 designates as ∆(A, Tr0,0) strings and above which it never ceases to search for a splitting. D(2) dictateswhich of these strings, σ′ say, should be an initial segment of BA,D. At every stage at which it is passed controlH11 passes control to the strategy H12 = D[t12, σ′](3) (amongst others). Since H11 has failed to find a Ψ0 split-ting above σ′ on Tr−1,0, H12 does not continue to build Tr0,0. Since we searched for splittings to enumerateinto Tr1,z(H2) lying on Tr0,0 we must introduce a new candidate for a Ψ1 splitting tree. We therefore definet12 = Tr−1,0, Tr1,z(H12).

2 The construction

2.1 The strategy B[t, u, l, k, ϕ0, . . . , ϕp−1](na, n)

Here u ∈ 0, 1, l, k, na, n ∈ ω, ϕ0, . . . , ϕp−1 ∈ 2<ω and t is a finite set of splitting trees under con-struction, t = Tr−1,0, Trj1,i1 , . . . , Trjm,im say, with j1 < · · · < jm. We shall also have that p|k. LetH = B[t, u, l, k, ϕ0, . . . , ϕp−1](na, n). At any stage at which H is passed control we shall have defined aset Π(H) ⊆ 2ω. Π(H) may be redefined a finite number of times during the course of the construction to be asuperset of its previous value. If m = 0, so that t = Tr−1,0, the instructions at stage s are very simple. Justchoose k

p incompatible extensions of each ϕi, enumerate these strings into Tr−1,0, ‘deliver’ them to that strategyH1 which passes control to H and declare H1 to be in state 0 before terminating stage s activity. So supposethat m ≥ 1. Initially the strategy is in state −1.

2.1.1 The instructions while in state −1

Every stage at which the strategy is passed control in state −1 it passes control to the strategy

H ′ = B[t′, 1, 0, x, ϕ0, . . . , ϕp−1](0, n),

where t′ = t − Trjm,im and x = 172 (k3 − k2 + 4k). We define Π(H ′) = Π(H), so that if Π(H) is redefined,

then Π(H ′) will be redefined also.

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482 A. E. M. Lewis: The minimal complementation property above 0′

2.1.2 The instructions while in state 0

Every stage s at which H is passed control in state 0 we perform the instructions below.1) If this is the first stage at which H has been passed control in state 0: the strategy will have been provided

with 172 (k3 − k2 + 4k) incompatible strings on Trjm−1,im−1 (we agree that j0 = −1 and i0 = 0). These strings

will have been ‘delivered’ to it. We divide these strings up as follows. We require 12k(k− 1)+2 sets of e strings,

each set consisting of k strings and with kp of these strings extending each ϕi. Let these sets of e strings be called

H0 , H

1 , . . . . For each 0 ≤ i < 12k(k−1)+2 let H

i = σHi ,0, . . . , σH

i ,k−1. If we subsequently remove strings

from Hi , so that this set now consists of k′ strings, relabel so that H

i = σHi ,0, . . . , σH

i ,k′−1. Apart from thee strings we also require k groups of strings, with all the strings in each group extending the same ϕi and with anequal number of strings in each group. Choose any linear ordering of these k groups, it will be conveneint to beable to refer to the 0th group, the 1st group, the 2nd group and so on. Call the 0th group αH and the 1st group βH .Divide αH into sets of eight pairs, one for each set of e strings. Label these α(H

0 ), α(H1 ), . . . . Divide βH into

sets of eight pairs and label these β(H0 ), β(H

1 ), . . . . We divide up the remaining groups of strings in the sameway – for r ≥ 2 we call the sets of eight pairs in the rth group of strings (H

0 , r), (H1 , r), . . . . Where it is

convenient we shall also use this notation for r < 2. For 0 ≤ i < 12k(k − 1) + 2 we associate every string in the

set of e strings Hi with a different one of the k groups, in such a way that each σH

i ,i′ is associated with a groupevery string of which extends ϕH(σH

i ,i′).2) We assume given some fixed (exhaustive) search procedure SH which searches for k-fold Ψjm splittings

such that:a) The splitting is of length at least l, or in fact of length at least l′ > l if so specified during the course of the

construction.

b) Each string in the splitting lies on Trjm−1,im−1 .

c) There is a nondiscarded pair µ ∈ αH ∪ βH , such that SH has not yet declared a splitting above µ andsuch that, for some σ ∈ µ, every string in the splitting extends σ. During the course of the construction we mayspecify that SH should search for splittings above strings other than those in pairs µ ∈ αH ∪ βH , but it will notdo so until so notified.

When SH finds a splitting of the kind that it is searching for, then it ‘declares’ this splitting. The reason that wedeclare such splittings rather than just observing that they have been found is so that when referring to a declaredsplitting it is clear that, in fact, this is a splitting of a specific (required) variety. If the splitting has been declaredabove a string σ ∈ µ, then we also say that a splitting has been declared above µ. SH may only declare onesplitting at each stage of the construction.

The strategy checks to whether two splittings have been declared above any (single one) of the setsα(H

0 ), β(H0 ), α(H

1 ), β(H1 ), . . . . If so, then the strategy is declared to be in state 1. H then proceeds

immediately to carry out the instructions for the strategy while in state 1 (at stage s). Otherwise proceed tostep 3).

Definition 2.1 If r ∈ 0, 1, then r = |1 − r|.3) For each pair α(H

i ), β(Hi ) we carry out the following instructions. Initially (the first stage at which H is

passed control in state 0) each such pair is in state a).The instructions for the pair α(H

i ), β(Hi ) while in state a):

(i) If this is the first stage at which H has been passed control in state 0: for each pair µ ∈ β(Hi ) we choose

a pair µ′ ∈ α(Hi ) and define ∆0(µ) = µ′. For each µ ∈ α(H

i ) choose µ′ ∈ β(Hi ) and define ∆1(µ) = µ′.

Suppose µ = σ0, σ1. When we (re)define ∆i(µ) (for i ∈ 0, 1), for each A ∈ γ(µ) such that A(na) = i, wedesignate each string σ2 ∈ ∆i(µ) as a ∆(A, Trjm,im) string. We do this so as to satisfy (†1) and (†2) below:

(†1) If µ, µ′ ∈ α(Hi ) and µ = µ′, then ∆1(µ) = ∆1(µ′). If µ, µ′ ∈ β(H

i ) and µ = µ′,then ∆0(µ) = ∆0(µ′).

(†2) If µ ∈ α(Hi ) and ∆1(µ) = µ′, then µ = ∆0(µ′).

(ii) The pair α(Hi ), β(H

i ) is declared to be in state b) if a splitting is declared by SH above anyµ ∈ α(H

i ) ∪ β(Hi ). In this case we then proceed immediately to carry out the instructions for this pair in

state b) at stage s.

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Math. Log. Quart. 51, No. 5 (2005) / www.mlq-journal.org 483

The instructions for the pair α(Hi ), β(H

i ) while in state b):

(i) If the pair α(Hi ), β(H

i ) was declared to be in state b) at (the present) stage s: suppose that a splitting hasbeen declared above µ0 ∈ (H

i , r) (r ∈ 0, 1). We discard that pair µ0 ∈ (Hi , r) such that ∆r(µ0) = µ0 (so

that this pair is no longer considered to be in (Hi , r)). There is one pair µ1 ∈ (H

i , r) such that ∆r(µ1) = µ0.Choose any (nondiscarded) pair µ2 ∈ (H

i , r) such that µ2 = ∆r(µ0) and ∆r(µ2) = µ1. Redefine ∆r(µ1) = µ2.

(ii) The pair α(Hi ), β(H

i ) is declared to be in state c) if a splitting is declared by SH above any µ ∈ (Hi , r)

(with r as in (i) above). In this case we then proceed immediately to carry out the instructions for this pair instate c) at stage s.

The instructions for the pair α(Hi ), β(H

i ) while in state c):If the pair α(H

i ), β(Hi ) was declared to be in state c) at (the present) stage s: suppose that SH has declared

a splitting above µ1 ∈ (Hi , r) at stage s and that it previously declared a splitting above µ0 ∈ (H

i , r).If µ1 = ∆r(µ0): we must redefine ∆r(µ0). Let µ0 = ∆r(µ1). Choose any µ1 ∈ (H

i , r) other than∆r(µ0), µ1 and other than that such that there are two µ2, ∆r(µ2) = µ1. Redefine ∆r(µ0) = µ1. There isa single pair in (H

i , r) other than µ0, µ2 say, such that ∆r(µ2) = µ1. Discard this pair. For the single pairµ3 ∈ (H

i , r) such that ∆r(µ3) = µ2 redefine ∆r(µ3) to be any pair in (Hi , r) other than µ0, ∆r(µ1) and other

than any pair µ4 such that ∆r(µ4) = µ3.If µ1 = ∆r(µ0): we discard from (H

i , r) those µ such that ∆r(µ) = µ1. We must then redefine ∆r(µ0) forthose µ0 ∈ (H

i , r) such that ∆r(µ0) is a pair we have just discarded. Just choose any µ1 ∈ (Hi , r) other than

µ0, ∆r(µ1) and other than any µ2 such that ∆r(µ2) = µ0.

4) We consider the pairs in each (Hi , r) to be linearly ordered. For A ∈ γ(H

i ) such that A(na) = 0 wedesignate those strings, in the first nondiscarded pair µ ∈ α(H

i ) above which SH is yet to declare a splitting,as ∆(A, Trjm,im) strings. For A ∈ γ(H

i ) such that A(na) = 1 we designate those strings, in the first nondis-carded pair µ ∈ β(H

i ) above which SH is yet to declare a splitting, as ∆(A, Trjm,im) strings. For r < 2 ifA ∈ γ(H

i , r), say A ∈ γ(µ), and A(na) = r, then we designate those strings in the first nondiscarded pairµ′ = µ above which SH is yet to declare a splitting in (H

i , r) as ∆(A, Trjm,im) strings. For r ≥ 2, ifA ∈ γ(H

i , r), then we designate those strings in the first nondiscarded pair µ above which SH is yet to declarea splitting, in α(H

i ) if A(na) = 0 or in β(Hi ) otherwise, as ∆(A, Trjm,im) strings. Let

Π = Π(H) − ⋃r,i(γ(H

i , r) ∪ γ(Hi )).

If A ∈ Π, then we designate those strings in the first nondiscarded pair µ above which SH is yet to declare asplitting, in α(H

0 ) if A(na) = 0 or in β(H0 ) otherwise, as ∆(A, Trjm,im) strings.

5) The strategies that H passes control to while in state 0:For every string in a nondiscarded pair µ ∈ αH ∪ βH above which SH is yet to declare a splitting, H passes

control to a different strategy. Suppose that σ is a string in such a pair. The strategy which H passes control to,corresponding to σ, is the strategy H ′ = D[t′, σ](n+1) where t′ = (t−Trjm,im)∪Trjm+1,z(H′) and wherewe define Π(H ′) to be all those A ∈ Π(H) such that σ has been designated a ∆(A, Trjm,im) string.

2.1.3 The instructions while in state 1

Suppose that we are at stage s.

1) If H was declared to be in state 1 at stage s: suppose that SH has declared a splitting above µ1 ∈ (Hi , r)

at stage s and that SH has previously declared a splitting above µ0 in this set of pairs. We discard all pairs µ0

from (Hi , r) such that ∆r(µ0) = µ1. For any µ1 ∈ (H

i , r) such that ∆r(µ1) is amongst the pairs we havejust discarded choose one of the remaining pairs µ2 in (H

i , r) other than ∆r(µ1) and above which SH is yet todeclare a splitting. Redefine ∆r(µ1) = µ2. Choosing l1 to be larger than any number yet mentioned during thecourse of the construction, we insist that SH should now only search for splittings above strings in nondiscardedpairs, in αH if r = 1 or in βH otherwise, above which it is yet to declare a splitting and of length at least l1.Enumerate i into E(H). For A ∈ γ(H

i ) such that A(na) = r designate the strings in the first nondiscardedpair in (H

i , r) above which SH is yet to declare a splitting as ∆(A, Trjm,im) strings. If A ∈ γ(Hi , r), say

A ∈ γ(µ), and A(na) = r designate the strings in the first nondiscarded pair µ′ = µ above which SH is yetto declare a splitting in (H

i , r) as ∆(A, Trjm,im) strings. For r′ ≥ 2 if A ∈ γ(Hi , r′) and A(na) = r, then

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484 A. E. M. Lewis: The minimal complementation property above 0′

designate those strings in the first nondiscarded pair in (Hi , r) above which SH is yet to declare a splitting as

∆(A, Trjm,im) strings. Choose i′ = i such that 0 ≤ i′ ≤ 12k(k−1)+2. For all A ∈ γ(H

i′ ) designate those stringsin the first nondiscarded pair in (H

i′ , r) above which SH is yet to declare a splitting as ∆(A, Trjm,im) strings.The set of e strings H

i′ is now considered to be ‘active’.

2) The strategy checks to whether SH has declared a splitting subsequent to the stage at which it was declaredto be in state 1. If so, then the strategy is declared to be in state 2, or terminates if k = 2. If k = 2, then Hproceeds immediately to carry out the ‘instructions upon termination’ at stage s. If k > 2 then H then proceedsimmediately to carry out the instructions for the strategy while in state 2 at stage s. Otherwise proceed to step 3).

3) Let Π = Π(H) − ⋃r,i(γ(H

i , r) ∪ γ(Hi )). Let r = 0 if SH now searches for splittings above strings in

pairs µ ∈ αH . Let r = 1 otherwise. For all A ∈ Π such that A(na) = r designate those strings in the firstnondiscarded pair µ ∈ (H

0 , r) above which SH is yet to declare a splitting as ∆(A, Trjm,im) strings.

4) The strategies that H passes control to while in state 1:For every string above which SH is searching for a splitting H passes control to a different strategy. Sup-

pose that σ is such a string. The strategy which H passes control to, corresponding to σ, is the strategyH ′ = D[t′, σ](n + 1) where t′ = (t − Trjm,im) ∪ Trjm+1,z(H′) and where we define Π(H ′) to be all thoseA ∈ Π(H) such that σ has been designated a ∆(A, Trjm,im) string. Suppose that the set of e strings H

i′ is active.H also passes control to the strategy H ′′ = B[t, 0, l, k, σH

i′ ,0, . . . , σH

i′ ,k−1](na + 1, n) where we define Π(H ′′)to be all those A ∈ Π(H) such that A(na) = r and A /∈ γ(H

i′ ) (where r is as in 3) above).

2.1.4 The instructions while in state k′ ≥ 2

Suppose that we are at stage s.

1) If H was declared to be in state k′ at stage s: suppose that SH has declared a splitting above µ ∈ (Hi , r)

at stage s. Enumerate i and all i′ such that Hi′ was active while H was in state k′ − 1 into E(H). If i′ ∈ E(H),

then we no longer consider the set of e strings Hi′ to be active. If k′ > 2, then let the string above which

SH has declared a splitting at stage s be called σHk′−1 – in this case we shall already have specified strings σH

k′′

for 0 ≤ k′′ < k′ − 1. If k′ = 2, then we must still do so. In this case let the string above which SH hasdeclared a splitting at stage s be called σH

r . Suppose that H was declared to be in state 1 because SH declaredtwo splittings above strings in (H

i′ , r), above µ0 and µ1 say. Choose r′ ∈ 0, 1 such that it has never been thecase ∆r(µr′) = µ. We call the string in µr′ above which SH declared a splitting σH

r .For each 0 ≤ i1 < k′ in turn, perform the following. Choose a set of e strings H

i2 which is not active andsuch that i2 /∈ E(H). Discard all strings from this set which are associated with the first k′ groups (those whichwe refer to as the 0th group, . . . , the (k′ − 1)th group) except that which is associated with the (i1)th group ofstrings. This set of e strings is now active and will be associated with the string σH

i1 . For all A ∈ γ(Hi2 ) designate

those strings in the first pair in (Hi2 , k′) as ∆(A, Trjm,im) strings. Choosing lk′ to be larger than any number yet

mentioned during the course of the construction, we now insist that SH should only search for splittings of lengthat least lk′ above the first pair in each (H

i3 , k′) such that the set of e strings Hi3 is active.

2) The strategy checks to whether SH has declared a splitting subsequent to the stage at which it was declaredto be in state k′. If so, then the strategy is declared to be in state k′ +1, or terminates if k = k′ +1. If k = k′ +1,then H proceeds immediately to carry out the ‘instructions upon termination’ at stage s. If k > k′ + 1 then Hthen proceeds immediately to carry out the instructions for the strategy while in state k′ +1 at stage s. Otherwiseproceed to step 3).

3) The strategies that H passes control to while in state k′ ≥ 2:For every string above which SH is searching for a splitting H passes control to a different strategy. Sup-

pose that σ is such a string. The strategy which H passes control to, corresponding to σ, is the strategyH ′ = D[t′, σ](n + 1) where t′ = (t − Trjm,im) ∪ Trjm+1,z(H′) and where we define Π(H ′) to be all thoseA ∈ Π(H) such that σ has been designated a ∆(A, Trjm,im) string. Let E′ be the set of all those i′ such that H

i′

is active. For each i′ ∈ E′ it is also the case that H passes control to a different strategy. The strategy which Hpasses control to, corresponding to i′ ∈ E′, is the strategy H ′′ = B[t, 0, lk′ , k−k′+1, σH

i′ ,0, . . . , σHi′ ,k−k′ ](0, n)

where lk′ is as in 1) above. Let Π′ be the set of all those A such that one of the following applies:

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a) A ∈ γ(σHi ) for some 0 ≤ i < k′.

b) There exists 0 ≤ i < k′ such that σHi has been designated a ∆(A, Trjm,im) string.

c) A ∈ γ(Hi′′) for some i′′ ∈ E′.

We define Π(H ′′) to be the set of all A such that A ∈ γ(σHi ) or such that σH

i has been designated a∆(A, Trjm,im) string, where H

i′ is associated with σHi , unless i = 0 in which case we define Π(H ′′) to be

this set together with all those A ∈ Π(H) − Π′.Definition 2.2 We use the variable Υ to range over the set B,D. If H ′ is a strategy Υ[t′, u′, . . .](n′

a, ·), thent(H ′) = t′, u(H ′) = u′ and na(H ′) = n′

a (where such arguments exist).

Definition 2.3 Suppose that H0, H1 are B strategies, i. e. strategies of the form B[. . .](·, ·). We say that H1 isin the same ‘splitting search cluster’ as H0 if there exists a B strategy H2 and two sequences of B strategies,H0, . . . , Hg and H ′

0, . . . , H′g′ for some g, g′ ≥ 0, such that:

a) H2 = H0, H2 = H ′0, H0 = Hg, H1 = H ′

g′ .

b) For all 0 ≤ i ≤ g, t(Hi) = t(H0).c) For all 0 ≤ i ≤ g′, t(H ′

i) = t(H0).d) For all 0 ≤ i < g, Hi passes control to Hi+1.

e) For all 0 ≤ i < g′, H ′i passes control to H ′

i+1.

Definition 2.4 At the base of the tree of strategies is the strategy D[t, ∅](0). We consider this strategy to beon level 0 of the tree of strategies. If H ′ is on level i of the tree of strategies and passes control to H ′′, then thelatter strategy is on level i + 1.

Definition 2.5 If H ′ is a D strategy, then we define t(H ′) = t. If H ′ is a B strategy, then suppose thatt(H ′) = Tr−1,0, Trj1,i1 , . . . , Trjm,imwith j1 < · · · < jm. If m ≥ 1, then define t(H ′) = t(H ′)−Trjm,im.

2.1.5 The instructions upon termination

If there is another strategy H ′ on the same level of the tree of strategies and in the same splitting search clusteras H , which terminates at stage s and such that z(H ′) < z(H), then H carries no more instructions at stage sother than to terminate stage s activity for the construction. Otherwise we proceed as follows. First we shallenumerate a set of splittings, Ω say.

S t e p 0. The first strategy that we shall consider taking splittings from is H0 = H . Take each of the splittingsthat have been declared by SH above σH

0 , . . . , σHk−1 and enumerate these splittings into Ω. If u(H) = 1, then our

enumeration of Ω is completed. Otherwise proceed to step 1.S t e p r. Suppose that Hr−1 was a strategy B[. . . , ϕ′

0, . . . , ϕ′p′−1](·, ·). Let Hr be the strategy which passes

control to Hr−1 and suppose that Hr

i′ = ϕ′0, . . . , ϕ

′p′−1. Let k′ ≥ 1 be the state which Hr is in when it passes

control to Hr−1. If k′ = 1, then proceed to the next step, unless u(Hr) = 1 in which case (presuming k′ = 1)our enumeration of Ω is completed. Otherwise we have defined strings σHr

0 , . . . , σHr

k′−1. Take the splittings that

have been declared by SHr above each of these strings except that which SHr has declared above σHr

i , whereHr

i′ is associated with σHr

i , and enumerate these splittings into Ω. If u(Hr) = 1, then we have completed ourenumeration of Ω, otherwise proceed to step r + 1.

Definition 2.6 Suppose that H ′ is a strategy of the form Υ[. . . , ϕ′0, . . . , ϕ

′p′−1](·, ·). We call ϕ′

0, . . . , ϕ′p′−1

the base strings for H ′.If |Ω| = k′, then label the splittings in Ω as ω0, . . . , ωk′−1 according to the order in which they were declared,

so that ωk′−1 was the first and ω0 the last. Take a string σ0 from ω0 and another string σ1 from ω1 such that Ψσ0jm

and Ψσ1jm

are incompatible. Take a string σ2 from ω2 such that Ψσ2jm

, Ψσ1jm

and Ψσ0jm

are pairwise incompatible, andso on. For each 0 ≤ i < k′ take an extension σ′

i of σi compatible with all of the trees in t(H), longer than anystrings in the trees in t(H), and such that no strings properly extending σ′

i have been designated ∆(A, Tr) stringsfor any A ⊆ ω and any Tr. Enumerate σ′

0, . . . , σ′k′−1 into Trjm,im , deliver them to that strategy H ′ which passes

control to Hr where we completed our enumeration of Ω at step r in the iteration above, and declare H ′ to be instate 0. If σ is a string which properly extends one of the base strings for H ′ and which has been designated a∆(A, Trjm,im) string for some A ⊆ ω, then enumerate A into γ(σ′) for all σ′ ⊇ σ. Terminate stage s activityfor the construction.

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486 A. E. M. Lewis: The minimal complementation property above 0′

2.2 The strategy D[t, ϕ](n)

Let H = D[t, ϕ](n). Here ϕ ∈ 2<ω, n ∈ ω, t is a finite set of splitting trees and we will have defined Π(H)to be the set of all A ⊆ ω with which H must be concerned. Let t = Tr−1,0, Trj1,i1 , . . . , Trjm,im withj1 < · · · < jm. Initially we regard the strategy as being in state −1.

2.2.1 The instructions while in state −1

While in state −1, H does nothing more than to pass control to the strategy H ′ = B[t, 1, 0, 3, ϕ](0, n) where wedefine Π(H ′) = Π(H).

2.2.2 The instructions for the strategy in state 0

We are provided with three strings, ϕ0, ϕ1, ϕ2 say. These strings will have been delivered to H . While in state 0,H passes control to three strategies H0 = D[t0, ϕ0](n + 1), H1 = D[t1, ϕ1](n + 1) and H2 = D[t2, ϕ2](n + 1).For i ∈ 0, 1, 2 we define ti = t ∪ Trjm+1,z(Hi). Let Π = Π(H) − ⋃

i∈0,1,2 γ(ϕi). We define

Π(H0) = γ(ϕ1) ∪ γ(ϕ2) ∪ Π,

Π(H1) = γ(ϕ0) ∪ γ(ϕ2) ∪ Π,

Π(H2) = γ(ϕ0) ∪ γ(ϕ1).

2.3 The tree of strategies

In order to complete our description of the tree of strategies we need only add the following. At the base of thetree of strategies is the strategy H = D[t, ∅](0), where we define Π(H) = 2ω, t = Tr−1,0, Tr0,0. At thebeginning of each stage s ≥ 0 of the construction control is initially passed to this strategy. All strategies onthe same level of the tree perform their instructions simultaneously at stage s (if passed control). If a strategyH ′ terminates stage s activity for the construction, then all strategies on the same level as H ′ complete theirinstructions for stage s, but do not pass control to other strategies at this stage.

Definition 2.7 Given a strategy H = D[. . .](n) we define the H cluster of strategies to be H and all theB strategies above H and below any strategy of the form D[. . .](n + 1).

2.4 Defining BA,D

Suppose given A, D ⊆ ω. For each i ≥ 0 we shall define a string σi, so that ultimately we may defineBA,D =

⋃i σi. For every strategy H let Π(H) take its final value. Let H0 be the strategy at the base of the

tree of strategies. At step 0 we consider the H0 cluster of strategies.S t e p r. If Hr is eventually declared to be in state 0: there are two strategies H ′ which Hr passes control to

while in state 0 and such that A ∈ Π(H ′), with base strings σ and σ′ respectively say. If D(r) = 0, then let σr

be the leftmost of σ, σ′ and otherwise define σr to be the rightmost of these two strings. At step r + 1 we shallconsider the Hr+1 cluster of strategies, where Hr+1 is the strategy with base string σr that Hr passes control towhile in state 0.

If Hr is not declared to be in state 0: let H ′ be the unique strategy in the Hr cluster of strategies which satisfiesboth of the following:

a) H ′ is passed control at an infinite number of stages.

b) H ′ designates a pair of strings, σ and σ′ say, as ∆(A, Tr) strings for some Tr ∈ t(Hr) and never ceases tosearch for a splitting above these strings.

Let σ and σ′ be as in b) above. If D(r) = 0, then let σr be the leftmost of σ, σ′, and otherwise define σr to bethe rightmost of these two strings. At step r + 1 we shall consider the Hr+1 cluster of strategies, where Hr+1 isthat strategy which H ′ passes control to with base string σr.

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3 The verification

When a B strategy H terminates it is instructed to enumerate a set of splittings Ω, to form a splitting σ0, . . . , σk−1

by choosing a string from each of those in Ω and then to find an extension σ′i of each σi compatible will all of the

trees in t(H), longer than any strings in the trees in t(H), and such that no strings properly extending σ′i have

been designated ∆(A, Tr) strings for any A ⊆ ω and any Tr. Until we know that this last instruction is alwayspossible we cannot know that the construction is well defined. There are other occasions too, where we mustprove that the instructions as stated can actually be carried out. When a B strategy is declared to be in state 0,for example, we must know that it is provided with the correct number of strings extending each of its basestrings, in order that it should be clear that the prescribed division of strings can take place. Strictly speaking,we cannot prove anything about stage s activity until we know that the instructions for the construction priorto this stage are properly specified. In order that we may avoid the task of a massive induction, it is thereforeconvenient to assume that in the event of receiving improperly defined or impossible instructions the constructionsimply resorts to permanent termination. Of course, it will turn out that such a provision is actually unnecessary,but proceeding in this way means that we are able to state and prove the required lemmas one at a time. Whileproving Lemma 3.6, for example, we are able to tacitly assume that the result of Lemma 3.8 has always held priorto the point at which H0 enumerates the splitting in question, or at least that if it has not, then this has not ledto imprecise or impossible instructions for the construction prior to this point, since otherwise the constructionwould already have terminated permanently so that the enumeration in question could not take place.

Lemma 3.1 At each stage s a finite number of strategies are passed control.

P r o o f. It follows immediately by induction on s that, at any given stage s, only a finite number of strategiesmay be passed control before control is passed to a B strategy H such that t(H) = Tr−1,0, whereupon stage sactivity is terminated. Of course other strategies may also terminate stage s activity.

It is also clear that the instructions for each strategy at any given stage are finite.

Definition 3.2 We define a strategy H0 to be active from the point at which it is first passed control until thepoint at which a strategy strictly below H0, H1 say, is declared to be in a new state such that none of the strategieswhich H1 passes control to while in this state are below or equal to H0.

Lemma 3.3 Suppose t(H0) = Tr−1,0, Trj1,i1 , . . . , Trjm,im with j1 < · · · < jm, j0 = −1, i0 = 0, and that0 ≤ r′ < r. If H1 enumerates a splitting into Trjr ,ir , then each string in the splitting extends a different leafof Trjr′ ,ir′ .

P r o o f. Since the construction will permanently terminate if we are not allowed to proceed as such, we mayassume that each string in the splitting extends a leaf of Trjr′ ,ir′ . Let t(H1) = Tr−1,0, Trj′1,i′1 , . . . , Trj′

m′ ,i′m′ so that jr = j′m′ , ir = i′m′ . For fixed r′ the result follows immediately by induction on r since r = m′,jr−1 = j′m′−1 and ir−1 = i′m′−1.

Lemma 3.4 Suppose that H0 is active when H1, which is not above or below H0 on the tree of strategies, ispassed control and that Tr ∈ t(H0)∩ t(H1). Let ϕ0, . . . , ϕp−1 be the base strings for H0 and let ϕ′

0, . . . , ϕ′p′−1

be the base strings for H1. For 0 ≤ i ≤ p− 1 and 0 ≤ i′ ≤ p′ − 1, ϕi and ϕ′i′ extend incompatible strings on Tr.

P r o o f. Let H2 be the strategy highest on the tree of strategies below H0 and H1. Since Tr ∈ t(H0)∩ t(H1)it must be the case that Tr ∈ t(H2). Suppose that for j1 < · · · < jm, t(H2) = Tr−1,0, Trj1,i1 , . . . , Trjm,im.If 0 ≤ i ≤ p− 1 and 0 ≤ i′ ≤ p′ − 1, ϕi and ϕ′

i′ extend incompatible strings on Trjm,im which H2 was providedwith upon being declared to be in state 0, and by Lemma 3.3 therefore extend incompatible strings on Tr.

It will be convenient to adopt the convention that if Tr is empty, then ∅ is a leaf of Tr.

Lemma 3.5 When a strategy H0 is first passed control:

a) The base strings for H0 will extend leaves of each of the trees Tr ∈ t(H0).b) No strings properly extending the base strings for H0 will have been designated ∆(A, Tr′) strings, for any

tree Tr′ and any A ⊆ ω.

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488 A. E. M. Lewis: The minimal complementation property above 0′

P r o o f. The proof is by induction on the point of the construction at which H0 is first passed control. SupposeH0 is first passed control at stage s by a strategy H1. If H1 is in state −1, then the result follows immediately,by the induction hypothesis and by Lemma 3.4. So suppose that H1 is in state k ≥ 0. There are three cases toconsider.

C a s e 1. Tr ∈ t(H1). When H1 was declared to be in state 0, b) above and also a) above as regards Tr,were satisfied since we are working under the assumption that the construction would permanently terminateotherwise. Since that point in the construction it follows by Lemma 3.4 that the only strategies which could haveviolated these conditions (where we consider a) as regards Tr only) are those above H1. Suppose H2 is aboveH1 and is passed control before H0 after the point of the construction at which H1 is declared to be in state 0.Let ϕ0, . . . , ϕp−1 be the base strings for H0 and let ϕ′

0, . . . , ϕ′p′−1 be the base strings for H2. For 0 ≤ i ≤ p − 1

and 0 ≤ i′ ≤ p′ − 1, ϕi and ϕ′i′ extend incompatible strings which H1 was provided with upon being declared to

be in state 0. The result follows by Lemma 3.3. Clearly we need not consider b) in the cases that follow.C a s e 2. Tr ∈ t(H0) − t(H1). Then a), as regards Tr, follows immediately since the only strategies that

can enumerate strings into Tr are those above H0.C a s e 3. Tr /∈ t(H0) and Tr /∈ t(H1). Let H2 be the strategy highest below H0 such that Tr ∈ t(H2).

When H2 was first passed control the base strings for this strategy extended leaves of Tr, by the induction hy-pothesis. By Lemma 3.4 it follows that the only strategies which could have subsequently violated this conditionare those above H2. But the fact that H0 is passed control at stage s means that such strategies have not yetenumerated any strings into Tr.

Lemma 3.6 Suppose that H0 enumerates a splitting σ0, . . . , σk−1 into Tr at stage s. Each string in thesplitting extends what was (prior to this enumeration) a leaf of Tr.

P r o o f. Let H1 be the strategy which H0 declares to be in state 0 upon termination. When H1 was firstpassed control the base strings for this strategy each extended a leaf of Tr. By Lemma 3.4 it follows that the onlystrategies which could have subsequently violated this condition (prior to the enumeration in question) are thosein the same splitting search cluster as H0. No such strategy has terminated prior to stage s, otherwise H0 wouldnot be passed control at this stage. If any such strategy H2 ( = H0) terminates at stage s, then z(H2) > z(H0),so that H2 does not enumerate any strings into Tr upon termination.

Lemma 3.7 Let t(H0) = Tr−1,0, Trj1,i1 , . . . , Trjm,im with j1 < · · · < jm and suppose that, at stage s0,H0 enumerates a splitting σ0, . . . , σk−1 into Trjm,im . Suppose given r < m. By Lemma 3.3 each σi extends aleaf σ′

i of Trjr ,ir , such that for 0 ≤ i < i′ < k we have σ′i = σ′

i′ , when this enumeration takes place. Supposefurther that at stage s1 > s0 a strategy H1 enumerates a string σ into Trjr ,ir extending σ′

i and that, subsequentto this enumeration, there is at least one active strategy H2 with Trjm,im ∈ t(H2). Then σ extends σi.

P r o o f. For 0 ≤ i < i′ < k we have that σ′i and σ′

i′ are incompatible. It therefore follows by Lemma 3.6 thatif, at any point of the construction after H0 enumerates the splitting into Trjm,im , a string compatible with σ′

i ex-tends a leaf of Trjm,im , the string extends σi. Let H2 be the highest strategy below H0 such that Trjm,im ∈ t(H2)and which is active after H1 enumerates σ into Trjr ,ir . Then Trjm,im ∈ t(H2). By Lemma 3.4, H1 must beabove H2. Let H3 be the highest strategy below H0 and H1. Then Trjm,im ∈ t(H3). Suppose first that H3 isdeclared to be in state 0 when, or at some point after, H0 enumerated the splitting into Trjm,im . But then σ ex-tends one of the strings which H3 is provided with upon being declared to be in state 0. Since Trjm,im ∈ t(H3),σ extends a leaf of Trjm,im when H3 is declared to be in state 0 and therefore extends σi. So suppose that H3

had been declared to be in state 0 before H0 enumerated the splitting. By the definition of H3, it follows that ifH4 is the strategy which H3 passed control to at any stage at which H0 was passed control and H5 is the strategywhich H3 passes control to at any stage at which H1 is passed control, then H4 = H5. By Lemma 3.3 we obtainan immediate contradiction.

Lemma 3.8 If H0 = B[t, u, l, k, ϕ0, . . . , ϕp−1](na, n) is passed control, then p|k.

P r o o f. The proof is by induction on the tree of strategies. Suppose that H0 is passed control by H1. IfH1 is a D strategy, then p = 1 and k = 3. If H1 is a B strategy, which is in a state k′ > 0 when it passescontrol to H0, then k = p. If H1 is a B strategy which is in state −1 when it passes control to H0, then supposeH1 = B[t1, u1, l1, k1, ϕ0, . . . , ϕp−1](n′

a, n). We have, by the induction hypothesis, that p|k1 and k1|k.

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Lemma 3.9 Suppose that a strategy H0 terminates and enumerates a splitting σ0, . . . , σk−1 into a tree Tr.Let H1 be the strategy which H0 declares to be in state 0 and let ϕ0, . . . , ϕp−1 be the base strings for H1. Thenkp of the σi extend each ϕi′ and, for 0 ≤ i < i′ < k, γ(σi) ∩ γ(σi′ ) = γ(ϕH1(σi)) ∩ γ(ϕH1(σi′ )).

P r o o f. The proof is by induction on the stage of the construction at which the splitting is enumerated. LetH2 be the strategy which H1 passes control to while in state −1. For each of the k groups of strings which H2

was provided with upon being declared to be in state 0, it follows by considering the induction hypothesis withregard to those strings delivered to each of the strategies strictly above H2 and below or equal to H0 upon beingdeclared to be in state 0, that precisely one of the two situations below applies:

a) There is precisely one σi extending a string in this group.

b) There is precisely one σi extending an e string for H2 which is associated with this group.

It follows immediately that kp of the σi extend each ϕi′ , by considering the induction hypothesis as regards the

strings that H2 was provided with upon being declared to be in state 0.By Lemma 3.5 we know that when H1 was first passed control, no strings properly extending the base strings

for H1 had been designated ∆(A, Tr′) strings for any Tr and any A ⊆ ω. By Lemma 3.4 we know that subsequentto this point the only strategies which may have done so are those above H1. Suppose given 0 ≤ i < i′ < k and(without loss of generality) suppose that all of the following are true:

a) σi extends a string in a splitting found by the strategy Hi and σi′ extends a string in a splitting found by thestrategy Hi′ .

b) σi extends σ′i, a string which Hi was provided with upon being declared to be in state 0, and σi′ extends σ′

i′ ,a string which Hi′ was provided with upon being declared to be in state 0.

c) The splitting above σ′i′ was declared before that above σ′

i.

For clarity we shall use a non-standard notation for the duration of the proof of this lemma. For any σ ∈ 2<ω

we shall agree that γ1(σ) is γ(σ) as defined subsequent to the enumeration of the splitting in Tr, while γ0(σ) willdenote the value γ(σ) just prior to this enumeration. We show, first of all, that

γ1(σ′i) ∩ γ1(σ′

i′ ) = γ1(ϕHi′ (σ′i)) ∩ γ1(ϕHi′ (σ′

i′ )).

C a s e 1. Hi = Hi′ . By the induction hypothesis γ0(σ′i) ∩ γ0(σ′

i′ ) = γ1(ϕHi (σ′i)) ∩ γ1(ϕHi(σ′

i′ )). Supposethat σ′

i is in µ ∈ (Hi

ji, ri) and σ′

i′ is in µ′ ∈ (Hi

ji′ , ri′).If ji = ji′ , then for all A such that Hi has designated σ′

i a ∆(A, Tr) string, A(na(Hi)) = ri and for all Asuch that Hi has designated σ′

i′ a ∆(A, Tr) string, A(na(Hi)) = ri′ . Since it has never been the case that∆ri(µ′) = µ we may conclude that if A ∈ γ0(σ′

i′ ), then Hi has not designated σ′i as a ∆(A, Tr) string. Since

it has never been the case that ∆ri′ (µ) = µ′ we may similarly conclude that if A ∈ γ0(σ′i), then Hi has not

designated σ′i′ as a ∆(A, Tr) string.

If ji = ji′ : if A ∈ γ0(σ′i), then Hi has not designated σ′

i′ as a ∆(A, Tr) string, since for such A any stringswhich Hi has designated as ∆(A, Tr) strings in the rth

i′ group of strings that it was provided with upon be-ing declared to be in state 0 are in (Hi

ji, ri′ ). If A ∈ γ0(σ′

i′ ), then similarly Hi has not designated σ′i as a

∆(A, Tr) string. Those A for which Hi has designated σ′i a ∆(A, Tr) string are such that either A(na(Hi)) = ri

or A ∈ γ(Hi

ji). Those A for which Hi has designated σ′

i′ a ∆(A, Tr) string are such that either A(na(Hi)) = ri′

or A ∈ γ(Hi

ji′).

C a s e 2. Hi = Hi′ . Let σ be the e string for Hi′ which σ′i extends. Since it is clear that

γ1(σ) ∩ γ1(σ′i′) = γ1(ϕHi′ (σ′

i)) ∩ γ1(ϕHi′ (σ′i′)),

we are left to show that the strategies above Hi′ do not designate extensions of σ as ∆(A, Tr) stringsfor A ∈ γ1(σ′

i′). But this is clear – if H is that strategy which Hi′ passes control to, with base strings defined tobe the set of e strings for Hi′ of which σ is a member, we ensure Π(H) ∩ γ1(σ′

i′ ) = ∅.

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So we have shown that γ1(σ′i) ∩ γ1(σ′

i′ ) = γ1(ϕHi′ (σ′i)) ∩ γ1(ϕHi′ (σ′

i′ )). If H is that D strategy which Hi

passes control to with base string σ′i, then we define Π(H) ⊆ γ1(σ′

i). If H ′ is that D strategy which Hi′ passescontrol to with base string σ′

i′ , then we define Π(H ′) ⊆ γ1(σ′i′ ). Therefore

γ1(σi) ∩ γ1(σi′ ) = γ1(ϕHi′ (σi)) ∩ γ1(ϕHi′ (σi′ )).

But then, since it follows by repeated applications of the induction hypothesis applied to those strings deliveredto the strategies below Hi′ and above H1 upon being declared to be in state 0 that

γ1(ϕHi′ (σ′i)) ∩ γ1(ϕHi′ (σ′

i′ )) = γ1(ϕH1 (ϕHi′ (σi))) ∩ γ1(ϕH1 (ϕHi′ (σi′ )))= γ1(ϕH1 (σi)) ∩ γ1(ϕH1 (σi′ )),

we have that γ1(σi) ∩ γ1(σi′ ) = γ1(ϕH1 (σi)) ∩ γ1(ϕH1 (σi′ )), as required.

Lemma 3.10 Suppose that H is a B strategy. It is enough that there should be eight pairs in each set (Hi , r).

P r o o f. This is just a simple counting argument and is left to the reader.

The lemmas we have proved thus far suffice to show that the construction is well defined (and so neverpermanently terminates), although we are yet to show that when A is not computable BA,D is well defined.

Lemma 3.11 Given any strategy H0, suppose that Π(H0) is redefined. Then it is redefined to be a superset ofits previous value.

P r o o f. The proof is by induction on the tree of strategies. Suppose that H0 is passed control by H1. IfΠ(H0) is redefined this is either because Π(H1) is redefined or/and because H1 is a B strategy searching for asplitting to enumerate into a tree Tr, H0 is a D strategy and H1 has increased the set of those A such that the basestring for H0 has been designated a ∆(A, Tr) string.

Lemma 3.12 Suppose given A, D ⊆ ω such that A is not computable. Then BA,D is well defined.

P r o o f. Suppose that H0 is a D strategy which is passed control at an infinite number of stages and such that(letting Π(H) take its final value) A ∈ Π(H). We must show that either H0 is declared to be in state 0, or thereis a unique strategy in the H0 cluster which satisfies property ():

() The strategy is passed control at an infinite number of stages and designates a pair of strings abovewhich it never ceases searching for a splitting as ∆(A, Tr) strings for some Tr ∈ t(H0).

So suppose H0 is never declared to be in state 0 and that, for all i > 0, Hi is as specified below.If Hi−1 designates a pair of strings above which it never ceases to search for a splitting as ∆(A, Tr) strings,

for some Tr, then Hi ↑. Otherwise let Hi be the (unique) B strategy which is passed control at an infinite numberof stages by Hi−1 and such that (letting Π(Hi) take its final value) A ∈ Π(Hi).

Suppose there exists a least i0 such that Hi0 ↑. Then Hi0−1 satisfies property (). For those i < i0, it is clearthat Hi does not satisfy property (). But for each i < i0, it follows by Lemma 3.11 that Hi+1 is the uniqueB strategy H which Hi passes control to at an infinite number of stages and such that ever A ∈ Π(H). Thus ifanother strategy H = Hi0 in the H0 cluster is to satisfy property () it must be above Hi0 . There is no strategyH ′ in the H0 cluster which Hi0 passes control to at an infinite number of stages and such that ever A ∈ Π(H ′).So suppose towards a contradiction that for all i ≥ 0, Hi is defined. If i > i′, then t(Hi) ⊆ t(Hi′ ). Let i0 belarge enough such that, for all i ≥ i0, t(Hi) = t(Hi0). For all i ≥ i0, Hi must be declared to be in state 1. Butif i ≥ i0, Hi searches for a k-fold splitting and is declared to be in state k′ > 1 (necessarily k′ < k since Hi

cannot terminate), then Hi+1 searches for only a (k − k′ + 1)-fold splitting. We may therefore choose i1 > i0large enough such that for all i ≥ i1, Hi is declared to be in state 1 but is never declared to be in state 2. For alli ≥ i1 we are able to determine A(na(Hi)) (in a computable fashion). Therefore A is computable, which givesus the required contradiction.

It is clear that if A is not computable, D ≥T ∅′ and A ≤T D, then BA,D ≤T D. A strategy is passed controlso long as it is active. Whether or not it is passed control at an infinite number of stages is therefore computablein ∅′.

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Lemma 3.13 Suppose given A, D ⊆ ω such that A is not computable. Then BA,D is a set of minimal degree.

P r o o f. Let H0 be the strategy at the base of the tree of strategies and for all i > 0 let Hi be that D strategywhich is passed control at an infinite number of stages by a strategy in the Hi−1 cluster and which has base stringcompatible with BA,D. Since there is only finite injury as regards splitting tree candidates along this ‘chain’ ofstrategies, it is clear that for each j ≥ 0 one of the following two situations applies:

a) There exist i0, i1 ∈ ω such that for all i ≥ i0, Trj,i1 ∈ t(Hi). For each i ≥ i0, the strategies in the Hi cluster

enumerate strings into Trj,i1 extending σi ⊂ BA,D, the base string for Hi. Thus BA,D ≤T ΨBA,D

j .

b) There exists i0 which is the greatest such that there exists i1 such that Trj,i1 ∈ t(Hi0 ). Supposet(Hi0 ) = Tr−1,0, Trj1,i1 , . . . , Trjm,im and that jr = j. Then there is a strategy in the Hi0 cluster whichis passed control at an infinite number of stages and which never ceases searching for a Ψj splitting above thebase string for Hi0+1 on Trjr−1,ir−1 . Also Trjr−1,ir−1 ∈ t(Hi) for all i > i0 so that for all i > i0 the strategies

in the Hi cluster enumerate strings into Trjr−1,ir−1 extending the base string for Hi. Thus, if it is total ΨBA,D

j iscomputable.

Lemma 3.14 Suppose given A, D ⊆ ω such that A is not computable. If σ ⊂ BA,D, then at no stage is Aenumerated into γ(σ).

P r o o f. Given A, D as in the statement of the lemma we may define a ‘true path’ for the construction asfollows. Let H0 be the strategy at the base of the tree of strategies. For each i > 0 let Hi be the unique strategywhich Hi−1 passes control to at an infinite number of stages and such that BA,D is compatible with one of thebase strings for Hi. It follows by a straightforward induction on the strategies in this ‘true path’ that if σ is a basestring for some Hi, then A is never enumerated into γ(σ).

Lemma 3.15 Suppose that a strategy H0 designates σ a ∆(A, Tr) string for some A ⊆ ω, σ ∈ 2<ω andsome Tr. If a strategy H1 below H0 is subsequently (or simultaneously) declared to be in state 0, then A isenumerated into γ(σ).

P r o o f. Suppose that the strings delivered to H1 are a splitting that has been enumerated into Tr′. If Tr = Tr′,then the result follows immediately. Otherwise σ extends a string which has been designated a ∆(A′, Tr′) stringfor all A′ ∈ Π(H0).

Lemma 3.16 Suppose given A, D ⊆ ω such that A is not computable. Then D ≤T A ⊕ BA,D.

P r o o f. Given oracles for A and BA,D we may compute D as follows. We let H0 be the strategy at the baseof the tree of strategies.

S t e p r ≥ 0 (we compute D(r)). Run the construction until either Hr is declared to be in state 0 or a strategyin the Hr cluster = Hr designates a string σr ⊂ BA,D as a ∆(A, Tr) string for some Tr ∈ t(Hr).

If Hr is declared to be in state 0: there are two strategies H ′ which Hr passes control to while in state 0and such that A ∈ Π(H ′), with base strings σ and σ′ respectively say. If BA,D is compatible with the leftmostof σ, σ′, then D(r) = 0 and otherwise D(r) = 1. Define Hr+1 to be the strategy that Hr passes control to whilein state 0 such that the base string for Hr+1 is compatible with BA,D.

If a strategy, H = Hr say, in the Hr cluster designates a string σr ⊂ BA,D as a ∆(A, Tr) string forsome Tr ∈ t(Hr). We know that H remains active for the remaining duration of the construction since otherwiseone of the two following possibilities must occur:

a) A strategy below H is declared to be in state 0. Then by Lemma 3.15, A would be enumerated into γ(σr).By Lemma 3.14, σr ⊂ BA,D, giving us the required contradiction.

b) A strategy H ′ below H is declared to be in state k > 0 such that none of the strategies which H ′ passescontrol to while in state k are below H (and this was not the case before H ′ was declared to be in state k).Then all of the strategies which H ′ passes control to while in state k have base strings (every one of which is)incompatible with σr. Thus BA,D could not be compatible with σr, unless a strategy below H ′ is subsequentlydeclared to be in state 0. But then a) applies.

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492 A. E. M. Lewis: The minimal complementation property above 0′

We may conclude that H never ceases searching for a splitting above σr. For otherwise we would either defineBA,D to be compatible with another string which H designates as a ∆(A, Tr) string (and therefore incompatiblewith σr) or we would define BA,D to be compatible with one of the e strings for H . There are two strings whichH designates as ∆(A, Tr) strings and above which it never ceases searching for a splitting (those in µ of which σis an member). If σr is the leftmost, then D(r) = 0 and otherwise D(r) = 1. Define Hr+1 to be the D strategythat H passes control to such that the base string for Hr+1 is compatible with BA,D.

Acknowledgements The author was supported by EPSRC grant No. GR /S28730/01 and partially supported by theNSFC Grand International Joint Project, No. 60310213, New Directions in the Theory and Applications of Models ofComputation.

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