A single minimal complement for the c.e. degrees

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1 A single minimal complement for the c.e. degrees Andrew Lewis Leeds University, April 2002 Abstract We show that there exists a single minimal (Turing) degree b < 0 s.t. for all c.e. degrees 0 < a < 0, 0 = a b. Since b is minimal this means that b complements all c.e. degrees 0 < a < 0. Since every n-c.e. degree bounds a (non-zero) c.e. degree, b complements every n-c.e. degree 0 < a < 0. 1 Introduction. In the paper Minimal complements for degrees below 0 [2002] we extended the ideas developed by Slaman and Seetapun [1992] in order to show that for every (Turing) degree 0 < a < 0 there exists a minimal b < 0 s.t. 0 = a b. Since b is minimal this means that b complements a. Using finite injury arguments it is also possible to show that there exists no finite minimal complement set, in other words that there does not exist a finite set of minimal degrees below 0 such that every non-zero degree below 0 is complemented by some degree in this set. In the paper Finite cupping sets [2002] we strengthened the latter result by showing that there does not exist a finite set of (any) degrees strictly below 0, b 1,.., b l say, such that for every degree 0 < a < 0 there exists 1 i l, 0 = a b i (that there does not exist a finite cupping set ). The aim of this paper is to show that there does exist a single minimal degree b < 0 s.t. for all c.e. degrees 0 < a < 0, 0 = a b. Since b is minimal this means that b complements all c.e. degrees 0 < a < 0. Since every n-c.e. degree bounds a (non-zero) c.e. degree this gives us the very strong result that b complements every n-c.e. degree 0 < a < 0. This significantly strengthens the result of Cooper, Seetapun and (independently) Li that there exists a degree 0 < b < 0 which cups every c.e. degree 0 < a < 0 to 0. The proof is quite difficult Mathematics Classification. Primary 03Dxx 1

2 Before we proceed to give some explanation of the intuition behind the proof it is helpful first to give a brief description of the architecture of the construction, so that we may define terms with which to frame our discussion. For each i ω, {W i,s } s 0 is an enumeration of W i. For convenience we shall assume that, given any i, s ω, W i,s is a finite binary string of length s. We assume given an enumeration {K s } s 0 of K and we shall enumerate an approximation {B s } s 0 to a set B of minimal degree b. Let {Ψ j } j 0 be an effective listing of the Turing functionals. In order to ensure that B is a set of minimal degree we shall use the familiar method of splitting trees. Thus for each j ω we shall attempt to construct an infinite Ψ j splitting tree T r j, such that B is on this tree. For each j ω there may be many candidates for T r j, the tree T r j,i should be regarded as the i th candidate for T r j. If for some j ω, there exists i such that T r j,i is infinite we will be able to conclude that Ψ B j is total and that B T Ψ B j. If on the other hand there exists no such i for some j, then we will have to work a little harder in order to show that if Ψ B j is total then it is computable. The construction itself will take place on a tree, so that each node of the tree is assigned one strategy. This tree we consider to grow upwards. At each stage s of the construction control is initially passed to that strategy which is assigned to the node at the bottom of the tree of strategies. This strategy may then pass control to another, one level up on the tree of strategies. This latter strategy may in turn pass control to another one level up again on the tree of strategies, and so on until stage s activity is terminated for the construction with control having been passed to a finite number of strategies. To each strategy we will be associating some base string, p 2 <ω we shall define this term more precisely later in the introduction. When any strategy is initially passed control it will be considered to be in state 0. At every stage at which a strategy, with base string p say, is passed control in state 0 and passes control to another it will always pass control to the same strategy one level up on the tree of strategies, also with base string p. In the subsequent discussion we shall describe the circumstances in which a strategy which is in state 0 may be declared to be in state 1. If a strategy is in state 1 then it will have, in its environment a set of four incompatible strings all extending the base string for this strategy these strings will be delivered to it (we shall define these terms formally in the context of each strategy as we describe them). At any stage s at which the strategy is passed control in state 1 it may put B s through one of these strings i.e. the strategy may choose some string in its environment and insist that B s be an extension of this string. At every stage s at which a strategy is passed control in state 1 and puts B s through the same string q before passing control to another strategy, it will pass control to the same 2

3 strategy with base string q, one level up on the tree of strategies. We shall use the variable H to range over the set of strategies. So if the strategy H is passed control in state 1 at stage s, puts B s through one of the strings in its environment, q say, and passes control to the strategy H with base string q, then if H is in state 1 the strings in the environment of H will be (pairwise) incompatible extensions of q. If H puts B s through q then any string q which H puts B s through will be an extension of q. When stage s activity ceases we define B s to be the longest string that it was put through during stage s activity. For technical completeness, if B s is not put through any strings at stage s then we define B s to be the empty string. Of course this only happens a finite number of times. So let us initially consider a simplified version of the construction. We shall remove most elements of the construction that are concerned with enumerating axioms for Turing functionals which ensure that for all c.e. degrees 0 < a < 0, b a = 0, and we shall concentrate on those aspects of the construction which are concerned with ensuring that B is a set of minimal degree. In the full construction there will be two basic varieties of strategy involved. We shall use strategies of the form D[t, p, r](m, n) and of the form C[t, p, r](m, n) D strategies and C strategies respectively. Here n, m, r ω, p is a finite binary string and t is a finite set of candidates for the T r j, t = {T r j1,i 1,.., T r jk,i k } say, with j 1 <.. < j k. The parameters n, m and r are used to specify the form of the axioms that should be enumerated by the strategy and therefore play no part in the present discussion. We use the variables Υ, Υ, Υ.. to range over the set {C, D}. Given any strategy H = Υ[t, p, r](m, n) we let t(h) denote the set t. It will be convenient to adopt the convention that for any σ 2 <ω and i, n ω, Ψ σ i (n) only if the computation converges in less than σ steps and Ψ σ i (n ) for all n < n. Definition 1.1 Given j 0 and a finite binary string σ let n be the least such that Ψ σ j (n). We say that Ψ σ j is of length n. Definition 1.2 By a 4-fold Ψ j splitting of length at least l we mean 4 strings σ 1,.., σ 4 whose images Ψ σ i j (1 i 4) are 1) of length at least l and 2) pairwise incompatible for 1 i < i 4 if the lengths of Ψ σ i j and Ψ σ i j are n 1 and n 2 respectively then there exists n 3 < n 1, n 2 such that Ψ σ i j (n 3 ) = Ψ σ i j (n 3 ). Definition 1.3 If a strategy H is of the form Υ[., p,.](.) for some finite binary string p, then we say p is the base string for H. Before we go on to a generalized discussion we shall consider a specific example. Suppose that control is passed to a strategy H = D[t 1, p,.](.,.) for the first time at stage s with t 1 = {T r 0,0, T r 1,0 }, so that on this part of the tree of strategies we are concerned with constructing the splitting trees T r 0,0 and T r 1,0. During the course of the construction we shall construct the tree T r 1,0 in such a way that it may be regarded as a sub-tree of 3

4 T r 0,0. Given that this is the first stage at which H has been passed control, p will extend (not necessarily properly) a leaf of each of the trees in t 1. When it is passed control at stage s this strategy will be in state 0. At this stage and at every subsequent stage at which the strategy H is passed control in state 0 it will simply pass control to a strategy H 1 = C[t 1, p,.](.,.), performing no other instructions. It will be the role of the C strategies to search for the splittings. So once it has entered state 1 the strategy H 1 will search for a 4-fold Ψ 1 splitting lying on T r 0,0, every string of which extends p. If it is eventually successful in this task then it will declare H to be in state 1, it will deliver the four strings in the splitting that it finds to H and enumerate this splitting into T r 1,0. Let us briefly discuss the meaning of this outcome. The four strings that are delivered to H, in this case, are a splitting that has been enumerated into T r 1,0 and lying on T r 0,0 so that these strings are also a Ψ 0 splitting. When H is first declared to be in state 1 each of the four strings will extend leaves of these trees. At each stage s at which H is passed control in state 1 it can choose one of the four strings it has been provided with, q say, put B s through q and pass control to a strategy H = D[t, q,.](.,.). Since we have been successful in the task of enumerating splittings into T r 0,0 and T r 1,0 extending p we shall have that T r 0,0, T r 1,0 t we are still interested in constructing these splitting trees since we do not have evidence yet that it is not possible to do so but we shall add to t also, another candidate for some other splitting tree. The strategy H, since it is a D strategy, will then proceed to operate much as the strategy H. The strategies above H will search for splittings to enumerate into the trees in t extending q. Initially, however, the strategy H 1 will be passed control in state 0. At every stage at which the strategy H 1 is passed control in state 0 it simply passes control to a strategy H 2 = C[t 2, p,.](.,.) where t 2 = {T r 0,0 }, performing no other instructions. So, since H 1 is a C strategy in state 0 passing control to another, in order to form t 2 we have taken the set of trees t 1 and removed one tree that tree T r j,i such that j is greatest. Once it has entered state 1 the strategy H 2 will search for a 4-fold Ψ 0 splitting every string of which extends p and lying on the non-splitting tree, T r 1,0, which is not required to be any kind of splitting tree and so is a tree that we can enumerate just as is convenient (the use of this non-splitting tree is just a technical convenience). If it is eventually successful in this task then it will declare H 1 to be in state 1, it will deliver the four strings in the splitting that it finds to H 1 and enumerate this splitting into the tree T r 0,0. Initially, however, it will be passed control in state 0 and so the strategy H 2 will pass control to the strategy H 3 = C[t 3, p,.](.,.). where t 3 =. Now the aim of this strategy is to find four incompatible strings that we can enumerate into T r 1,0 but since this tree is not required to be a splitting tree the strategy H 3 can proceed simply as follows. The first time that it is passed control it will just choose four incompatible extensions of p, enumerate them into T r 1,0, deliver them to H 2 and declare H 2 to be in state 1 before terminating stage s activity for the construction. So at the first stage s at which H is passed control it passes control to the strategy H 1 4

5 which passes control to H 2 which passes control to H 3. The strategy H 3 then declares H 2 to be in state 1 and delivers four incompatible extensions of p to this strategy. Now at every subsequent stage at which H is passed control, until H 2 is successful in the task of finding a 4-fold Ψ 0 splitting lying on the tree T r 1,0 and declares H 1 to be in state 1, H will still be in state 0 and will therefore pass control to H 1 which will still be in state 0 and will therefore pass control to H 2. In this simplified version of the construction, in which no axioms for Turing functionals are being enumerated, the strategy H 2 can then always proceed to put B s through the same string in its environment at any stage s at which it is passed control in state 1 in order to be consistent with later terminology let us call this the axiom enumeration string. The strategy H 2 will, in fact, search for a 4-fold splitting such that every string in the splitting extends its axiom enumeration string. At any stage s at which it is passed control in state 1 the strategy H 2 will put B s through its axiom enumeration string, q say, and will pass control to a strategy H2 = D[t 2, q,.](.,.). How should we define t 2? If the strategy H 2 ever finds the splitting that it is searching for then it will declare H 1 to be in state 1 and the strategies H 2 and H2 will never be passed control again. Thus at any stage at which H2 is passed control, this fact in itself ensures that H 2 has not yet found the required splitting. The strategy H2 therefore assumes that T r 0,0 will be finite. We must, however, continue to construct the tree T r 1,0 in such a way that we search for splittings lying on T r 0,0 the tree T r 1,0 may be regarded as assuming that the tree T r 0,0 will be infinite. The first time that control is passed to the strategy H2 we might define t 2 as follows. Let i be the least such that we have not yet passed control to any strategy H with T r 1,i t(h ). Define t 2 = {T r 1,i }. This is a new candidate for the Ψ 1 splitting tree and the strategies above H2 will search for splittings to enumerate into this tree lying on T r 1,0. The strategy H2, being a D strategy, will proceed to operate much as the strategy H except, of course, that in this example t 2 has one less tree in it than t 1. The correct interpretation of this outcome, then, is that the tree T r 0,0 is finite and that we must therefore introduce a new candidate for T r 1 on this part of the tree of strategies, for which we do not have to search for splittings lying on T r 0,0. Now suppose that the strategy H 2 eventually finds a 4-fold Ψ 0 splitting lying on the tree T r 1,0. Then the strategy H 2 will declare H 1 to be in state 1, it will enumerate these strings into T r 0,0 and it will deliver them to H 1 before terminating activity for this stage of the construction. At any subsequent stage at which H is passed control, until H 1 has declared it to be in state 1, H will pass control to H 1. This strategy we now suppose to have had four strings delivered to it, which have been enumerated into T r 0,0. In this simplified version of the construction the strategy H 1 can then always proceed to put B s through its axiom enumeration string, q say, at any stage s at which it is passed control in state 1. The strategy H 1 will, in fact, search for a 4-fold splitting such that every string in the splitting extends q. At any stage s at which it is passed control in state 1 the strategy H 1 will put B s through q and will pass control to a strategy H 1 = D[t 1, q,.](.,.). How should we define t 1? At any stage at which H 1 is passed control, this fact in itself 5

6 ensures that H 1 has not yet found the splitting that it is searching for and the strategy H 1 therefore assumes that T r 1,0 will be finite. The first time that control is passed to the strategy H 1 we might define t 1 as follows. Let i be the least such that we have not yet passed control to any strategy H with T r 2,i t(h ). Define t 1 = {T r 0,0, T r 2,i }. This is a new candidate for the Ψ 2 splitting tree and we shall search for splittings to enumerate into this tree lying on T r 0,0. The correct interpretation of this outcome is that the tree T r 1,0 is finite, while it still seems as if the tree T r 0,0 will be infinite, so that on this part of the tree of strategies we should introduce a new candidate for another splitting tree and search for splittings to enumerate into this tree lying on T r 0,0. Finally then, let us suppose that the strategy H 1 is eventually successful in the task of finding a 4-fold Ψ 1 splitting on the tree T r 0,0. Then, as discussed previously, it will enumerate this splitting into T r 1,0, declare H to be in state 1 and deliver the strings in the splitting that it has found to H before terminating activity for this stage of the construction. We shall choose the strings in the splitting so that they extend leaves of T r 0,0. In this simplified version of the construction the strategy H can then always proceed to put B s through its axiom enumeration string at any stage s at which it is passed control in state 1, q say. At every stage at which it does so it will pass control to a strategy H = D[t, q,.](.,.). The first time that we pass control to H we might define t as follows. Let i be the least such that we have not yet passed control to any strategy H with T r 2,i t(h ). Define t = {T r 0,0, T r 1,0, T r 2,i }. Now let us briefly consider the general situation of which the previous discussion is an example, in the context of the full construction. As we do so we shall introduce some terminology that will be useful in the subsequent discussion concerning methods of axiom enumeration. Definition 1.4 If H is a strategy D[t,..] then define t (H) = t. If H is a strategy C[t,..] then if t = define t (H) =, and otherwise if t = {T r j1,i 1,.., T r jk,i k }, j 1 <.. < j k, then define t (H) = t {T r jk,i k }. When a strategy H = Υ[t, p, r](m, n) is passed control in state 0 it simply passes control to the strategy H = C[t (H), p, r](m, n) carrying out no other instructions. Unless, that is, Υ = C and t = in which case it will choose four incompatible extensions of its base string before delivering these strings to the strategy which passed control to it, declaring that strategy to be in state 1, and terminating activity for this stage of the construction. Suppose that the latter case does not apply. Let t (H) = {T r j1,i 1,.., T r jk,i k } such that j 1 <.. < j k. The strategy H, initially passed control in state 0, will then proceed in a similar fashion. If t (H) = then at the first stage at which it is passed control H will choose four incompatible extensions of p before delivering these strings to H, declaring H to be in state 1, and terminating activity for this stage of the construction. Otherwise, should H enter state 1, the fact 6

7 that it is a C strategy will mean that it then searches for a 4-fold Ψ jk splitting on the tree T r jk 1,i k 1 (or T r 1,0 if k = 1), each string of which extends p, in order that it can enumerate this splitting into the tree T r jk,i k. If it is successful in this task then it will deliver these four strings to the strategy H and will declare the strategy H to be in state 1. Suppose that t(h) = {T r j1,i 1,.., T r jk,i k }. If H is a C strategy then once it has been declared to be in state 1 it will proceed to search for a Ψ jk splitting lying on T r jk 1,i k 1 (or T r 1,0 if k = 1). Now in the context of discussing the full construction, whether H is a C strategy or not, there may be reasons (discussed subsequently) why the strategy H, passed control in state 1 at stage s, may not be allowed to put B s through its axiom enumeration string. We consider the four strings p 1,.., p 4 which are delivered to the strategy H to be in the environment of the strategy and we label them as follows: 1) p 1 we label the axiomatic release 2) p 2 we label the axiom enumeration string 3) p 3 we label the split escape 4) p 4 we label the marked escape. axiom enumeration string split escape axiomatic release marked escape p 2 p 3 p 1 p 4 p Fig 1.1 At any stage s at which H is passed control in state 1 it will put B s through one of the strings p 2, p 3, or p 4. The significance of the axiom enumeration string, as might be expected, is that this is the string on which the strategy may enumerate axioms. If H is a C strategy then it searches for splittings such that each string in the splitting extends the axiom enumeration string. The significance of the other labelled strings will be explained shortly. If a D strategy, H, is passed control in state 1 at stage s then it does not search for splittings. If it puts B s through its axiom enumeration string at stage s then it may 7

8 enumerate axioms on this string. Whichever string, p say, the strategy puts B s through it will pass control to another D strategy, H, with base string p and such that t(h ) t(h). In the example given previously we have hopefully established some understanding of the manner in which t(h ) should be defined. We have already stated that when any strategy H is passed control in state 0 it immediately passes control to a C strategy H such that t(h ) t(h), unless H is a C strategy and t(h) = in which case H does not pass control to any other strategy. When a C strategy, H, is passed control in state 1 at stage s and puts B s through its axiom enumeration string, p say, it will pass control to a D strategy, H, with base string p such that t(h ) t(h). It is hoped, once again, that the example described previously provides some understanding of the way in which we should define t(h ). If the C strategy H is passed control in state 1 at stage s and puts B s through a string p in its environment which is not its axiom enumeration string, what variety of strategy should H then pass control to? The fact that H searches for a splitting above its axiom enumeration string, which is incompatible with p, means that the strategy which H passes control to at stage s cannot assume that the tree into which this splitting would be enumerated is finite. H must therefore pass control to another C strategy, H, with base string p and such that t(h ) = t(h). Any D strategy H and all of those C strategies above it and below any strategies H strictly above H which are D strategies, should thus be regarded as a (finite) D cluster of strategies and the strategies in this D cluster will enumerate strings into the trees in t(h). Of course, any such D cluster of strategies must be finite (and not just finite at any given stage of the construction) otherwise there would only be enumerated a finite number of infinite splitting trees. When any strategy H in state 1 passes control to another strategy H for the first time we are yet to specify exactly the manner in which we should define t(h ). We leave it until the following sections to describe precisely how this is done complications introduced due to the methods of axiom enumeration mean that we must deviate just slightly from the intuition of the example described previously. This covers most aspects of the construction which are concerned with ensuring that B is a set of minimal degree, but we must also ensure that for all c.e. degrees 0 < a < 0, b a = 0. Definition 1.5 For l ω, U l is the set of all finite sets of ordered pairs of the form {(y 0, z 0 ),.., (y g, z g )} with g ω, 0 y g l and z g ω for 0 g g and such that if (y, z), (y, z ) are distinct elements of u for u U l then y y. For any l ω, then, any u U l is just a partial function defined on some arguments l. Each u U l should be regarded as a guess as to which trees T r j,i such that j l will be infinite the guess is that T r j,i will be infinite precisely if (j, i) u. In order that b should complement all c.e. degrees 0 < a < 0 we shall satisfy the following. For each i ω, if W i is non-computable, there will exist u U i such that we enumerate axioms for a functional Γ i,u so that for all but finitely many n ω: 8

9 Γ W i,b i,u (n) = K(n). The axioms that we enumerate for Γ i,u (for any i ω and u U i ) will be of the form Γ σ,τ i,u (n) = c for n ω, c {0, 1} and finite binary strings σ and τ. Definition 1.6 We let the function f be defined as follows. For any i, j ω, f( i, j ) = i, where.,. is just the standard pairing function. We may now explain the use of the parameters m, n and r when discussing a strategy H = Υ[t, p, r](m, n). If H is passed control at any stage s and puts B s through its axiom enumeration string q then when there are conditions appropriate it may enumerate axioms of the form Γ σ,q f(m),u (n) = c where σ is the prefix of W f(m),s of length r, u U f(m) and c {0, 1} so once again the use of these parameters is that they specify what kind of axioms the strategy is to enumerate. The use of the function f is that, intuitively speaking, as we proceed up the tree of strategies we should take it in turn to enumerate axioms for the functionals Γ i,u associated with each W i. Thus, depending on precisely how we choose the pairing function.,. to be defined, if we suppose that f(0) = 0 then the strategy at the bottom of the tree of strategies might be a strategy of the form Υ[..](0,.) and thus be concerned with ensuring that functionals of the form Γ 0,u, such that u U 0, are correctly defined on argument 0. If we suppose that f(1) = 1 then a strategy one level up on the tree of strategies might be of the form Υ [..](1,.) and thus be concerned with ensuring that functionals of the form Γ 1,u, such that u U 1, are correctly defined on argument 0. If we suppose that f(2) = 0 then any strategy one level up on the tree of strategies again might be of the form Υ [..](2,.) and thus be concerned with ensuring that functionals of the form Γ 0,u such that u U 0 are correctly defined, either on argument 1 if it looks on this part of the tree of strategies as if the bottom strategy has already successfully ensured that they are correctly defined on argument 0, or otherwise on argument 0 and so on. Definition 1.7 Suppose that at some stage s of the construction we have enumerated an axiom Γ σ,τ i,u (n) = c for some i, n ω, u U i, c {0, 1} and finite binary strings σ and τ. At a stage s > s of the construction we say that this axiom is expired if the prefix of W i,s of length σ is not equal to σ. Now before we enumerate any splitting into a tree we must clearly consider the axioms that we have already enumerated on the strings in that splitting. It is in order that given any specific tree we need only consider those axioms that we have enumerated of any form Γ σ,τ i,u (n) = c for a finite set of i ω that we consider functionals of the form Γ i,u for u U i. As previously stated, each u U i may be regarded as a guess as to which of the trees T r j,i such that j i will be infinite. 9

10 ( 0 ) No strategy H with base string p will enumerate axioms for a functional Γ i,u if there is a C strategy below H which is still searching for a splitting, above a string compatible with p, to enumerate into the tree T r j,i and (j, i) u. The guess associated with u, in that case, indicates that such a strategy H will not be on the true path of the construction. So before enumerating a splitting p 1,.., p 4 into a tree, T r j,i say, such that each of p 1,.., p 4 extend a string p which was (until this point) a leaf of T r j,i, we will insist that all axioms that we have enumerated of any form Γ σ,τ i,u (n) = c for i < i + j, u U i, n ω, c {0, 1} and finite binary strings σ and τ such that p τ p i for some 1 i 4, are expired or rather we shall insist that all such axioms are either expired or have already been labelled hidden axioms, a term which we will now explain. After we have enumerated p 1,.., p 4 into T r j,i all axioms that we have enumerated of any form Γ σ,τ i,u (n) = c for any i, n ω (not just those i < i + j), u U i, c {0, 1} and finite binary strings σ and τ such that p τ p i for some 1 i 4 and which are not expired axioms we shall label hidden axioms if they are not already labelled as such. This is the only way in which axioms that we have enumerated may be labelled as hidden and, as such, may be taken as the definition of what we mean by a hidden axiom. So the intuition behind the labelling of hidden axioms, initially anyway, is just that it allows us to simplify the way in which we view the construction. Once an axiom has been labelled as hidden the construction will proceed just as if this axiom had never been enumerated. Apart from the fact, that is, that we must still ensure that the axioms we enumerate for all of the various functionals are consistent. When a splitting is enumerated into any tree then, apart form this consistency requirement, we shall be able to proceed just as if there are no axioms enumerated on the strings in that splitting which need be taken into consideration. Subsequently we shall see that the use of hidden axioms is also vital to the construction in other ways. For any i ω, let u (i ) be the set of all those pairs (j, i) such that j i and T r j,i is infinite. Then u (i ) U i and by ( 0 ) axioms that we enumerate for Γ i,u (i ) will only be labelled hidden axioms when strings are enumerated into trees T r j,i such that j + i i and (j, i) / u (i ). But there are a finite number of such trees and each such tree is finite. Thus there will be a finite number of axioms that we enumerate for this functional which are labelled hidden axioms. If W i is non-computable then it will be for this functional Γ i,u (i ) that it is the case, for all but finitely many n, Γ W i,b i,u (i ) (n) = K(n). For such i we shall be able to guarantee, for each n ω, that there exists some strategy H on the true path of the construction which will achieve the following. H will successfully ensure that Γ W i,b i,u (i )(n) = K(n) unless there are hidden axioms which (for the sake of the consistency requirement) restrict the enumeration of the axioms which would be required in order to ensure that this is the case. The real difficulty that we face in this proof is in engineering a situation whereby in searching for a splitting to enumerate into the tree T r j,i, say, we are either able to demon- 10

11 strate that Ψ B j is not total or is computable, or we are able to find a splitting of the kind described above in which all of the axioms we have enumerated of a specific variety are either expired axioms or have already been labelled hidden axioms. To this end we use a technique of marking strategies, the product of which is (intuitively speaking), for those i such that W i is computable, that we may push all but finitely many of those strategies which appear to be on the true path of the construction and which are successfully enumerating axioms for functionals Γ i,u, further and further up the tree of strategies. More specifically, then, we are presented with the problem that a C strategy may find an infinite number of splittings, but that it may be the case none of these splittings satisfy the conditions that we require as regards the axioms that have been enumerated on the strings in this splitting. We therefore proceed (roughly) as follows. Whenever a C strategy, H, finds a splitting which is not yet suitable for enumeration into the relevant tree because of the axioms that have been enumerated on the strings in that splitting, we may label those strings splitting strings w.r.t. all strategies above H. Definition 1.8 Given a strategy H with base string p and a string q with p q in the environment of H, suppose that H is passed control at stage s. We say that q is split at stage s if there is some string q with p q, q is compatible with q and q is labelled a splitting string w.r.t. H at stage s. When any strategy H = Υ[.., r](m, n), say, above H is passed control at a stage s in state 1 and finds that its axiom enumeration string is split at stage s, it will not be allowed to enumerate axioms at this stage. Let us call the axiom enumeration string for H, q. If it is the case that this strategy has already enumerated axioms Γ σ,q f(m),u (n) = K s(n) for σ W f(m),s (and for those functionals Γ f(m),u which are appropriate for axiom enumeration on this part of the tree of strategies) then H may proceed to put B s through q at stage s, since in this case there is no need to enumerate any more axioms at this stage. The strategies above that which H passes control to at any stage s at which it puts B s through q may be seen as assuming that H will successfully ensure that Γ W f(m),b f(m),u (n) = K(n) for all of those u U f(m) such that Γ f(m),u is appropriate for axiom enumeration on this part of the tree of strategies. Otherwise, if such axioms have not been enumerated, H may put B s through the string which we labelled the split escape for this strategy, q say. It will then pass control to a strategy, H say, of the form Υ[., q, r](m, n). In this case also, we shall therefore have that H does not enumerate axioms at stage s. The strategy H is therefore able to wait and see whether all of the (relevant) axioms that have been enumerated on the strings in this splitting become expired axioms at some subsequent stage, without having to consider the possibility that further axioms may be enumerated on these strings. Now the fact that the strategies H and H have the same parameters r, m and n means that, intuitively speaking, when the strategy H is forced, at any stage s, to put B s through its split escape and pass control to H, W f(m) has not lost its turn to ensure that 11

12 the appropriate functionals Γ f(m),u are defined correctly on the argument n. The action that we have taken might, nonetheless, be regarded as a form of injury. If it is the case that, at all but a finite number of stages s, H is passed control and is forced to put B s through its split escape before passing control to the strategy H, clearly we must ensure that the same is not true of an infinite chain of strategies H, H, H..., each defined from its predecessor in this fashion. Definition 1.9 For any r, m, n ω we define a maximal (r, m, n) set to be any set of strategies M which is the smallest set satisfying the following criteria for some strategy H of the form Υ[.., r](m, n) (for some Υ): 1) H is in M. 2) If a strategy H is of the form Υ [.., r](m, n) and there is a stage at which it passes control to, or has control passed to it by, a strategy in M then H is in M. More specifically, then, we shall have to show that, for any r, m, n ω, every maximal (r, m, n) set is finite. For this reason we cannot actually proceed to label all of the strings in the splittings that H finds as splitting strings. We make sure to label enough of these strings as splitting strings in order to ensure the following. If at any stage and for any one of the splittings that H has found it is the case that all of the axioms of the relevant variety, which had been enumerated prior to the point at which H found this splitting, have become hidden or expired axioms then we are able to find a splitting which is suitable for enumeration into the relevant tree. Suppose that H (the same C strategy) has base string p and that t(h) = {T r j1,i 1,.., T r jk,i k }, j 1 <.. < j k. At each stage s at which it is passed control H takes each of the splittings that it has found and which are not yet suitable for enumeration into T r jk,i k (unless H has already found a splitting of the required variety) and for each of these 4-fold splittings, q 1,,.q 4 say, it chooses an index i < j k + i k to blame for that splitting so this index i will be such that prior to the stage of the construction at which we found this splitting we have enumerated an axiom of some form Γ σ,τ i,u (n) = c for τ such that there exists 1 i 4, p τ q i, which is not expired and which has not been labelled a hidden axiom. Once it has decided to blame an index i for a given splitting, H then marks at stage s each of those strategies above it, of any form Υ[..](m, n ) such that f(m ) = i which have enumerated the offending axioms. If any of these marked strategies, Υ[..](m, n ) say, is passed control at stage s then they will put B s through the marked escape in the environment of that strategy, p say, and will pass control to a strategy Υ[..](m + 1, n ) with base string p. In that case, intuitively speaking, W i loses its turn to ensure that the appropriate functionals Γ i,u are defined correctly on the relevant argument. Thus for each splitting that is found by the strategy H and which never becomes suitable for enumeration into T r jk,i k there exists i < j k +i k such that we have enumerated 12

13 an axiom for some Γ i,u on one of the strings in the splitting which never becomes expired an axiom of some form Γ σ,τ i,u (n) = c such that σ W i. Suppose that H finds an infinite number of splittings but that none of these splittings ever becomes suitable for enumeration into T r jk,i k. Let Y be the set of all those i < j k + i k such that there exist an infinite number of these splittings such that there exists a stage s, i is blamed for that splitting at stage s. We leave it until later in this introduction to describe roughly the manner in which the blame procedure will decide which i < j k + i k should be blamed for any given splitting at any given stage, but we shall be able to show that if i Y then W i is computable. This we shall achieve basically by ensuring that there is no upper bound on the length of those strings σ obtained from each of the splittings as above. Of course it is not the case that if W i is computable then necessarily we have i Y. Now clearly there can only be a finite number of splittings found such that all of the axioms that have been enumerated on the strings in the splitting of the relevant variety, for Turing functionals of the form Γ i,u such that i Y and u U i, become hidden or expired. But our technique of marking strategies will mean that above H, those strategies which appear to be on the true path of the construction and which are successfully enumerating axioms for functionals Γ i,u such that i Y, will be found higher and higher up on the tree of strategies as the construction progresses. We shall thus be able to show that if Ψ B j k is total and non-computable there do exist an infinite number of (individual) strings in the splittings found (in fact an infinite number of initial segments of B) such that all axioms of the relevant variety which have been enumerated on these strings for functionals Γ i,u such that i Y become hidden or expired. Let this set of strings be Λ. Now consider the set of strings Λ = {Ψ p j k : p Λ}. By the above observations, although clearly further argument is required, we shall be able to show that there exists l ω such that: 1) there are not l incompatible strings in Λ, and 2) for every n ω, there exists σ Λ extending the initial segment of Ψ B j k of length n. This gives us the required contradiction. We shall leave it to the following sections to describe the blame procedure precisely, but it is probably appropriate that we should give here a good impression of the manner in which this procedure will operate. The blame procedure for any strategy H stores the splittings which are found by that strategy in the order in which they are found, so that if an infinite number of splittings are found then for any j ω we may refer to the j th splitting. At each stage s at which H is passed control, such that H has not yet found a splitting which is suitable for enumeration into the relevant tree (T r jk,i k say) at stage s, the blame procedure will consider each of the splittings in turn and choose some index i < j k + i k to blame for each. If i is blamed for the j th splitting then we shall say that i is blamed for j at stage s. At each stage s at which the blame procedure is run, and for each j such that H has found at least j splittings by stage s, we shall define g(h, j, s) to be the set of all those i < j k +i k such that there are axioms (of the relevant variety) which we enumerated, prior to the stage at which this splitting was found, for functionals Γ i,u 13

14 on the strings in the j th splitting which (at stage s) are not yet hidden or expired. From the previous discussion concerning splitting strings it is hopefully clear why we shall only have to consider those axioms enumerated prior to the stage at which the splitting was found. The set g(h, j, s), then, contains precisely those i < j k + i k which are eligible for blame as regards the j th splitting at stage s. Clearly for any j 1 and any s > s, if it is the case that both g(h, j, s) and g(h, j, s ) are defined, then g(h, j, s ) g(h, j, s). Definition 1.10 Given any C strategy H and j 1 we define g (H, j) = {i 1,.., i l } if there exist an infinite number of stages s such that g(h, j, s) = {i 1,.., i l }. Definition 1.11 Given any C strategy H we say that any set Y = {i 1,.., i l } is an infinitely occurring index set for H if there exist an infinite number of j such that g (H, j) = {i 1,.., i l }. We say that Y is a minimal infinitely occurring index set for H if Y is an infinitely occurring index set for H and for no other infinitely occurring index set for H, Y say, is it the case that Y Y. Let us suppose that H finds an infinite number of splittings (and that none of these splittings ever becomes suitable for enumeration into T r jk,i k ). If, for some i, there exist an infinite number of j such that there exists s, i is blamed for j at stage s, then the first thing that we shall do (with the ultimate aim of proving that W i is computable) is to ensure that i Y for some minimal infinitely occurring index set Y. This we shall achieve as follows. Suppose that we wish to decide which index i < j k + i k to blame at stage s for the j th splitting found by H. First of all we check to see whether all of the following are satisfied: 1) H has run the blame procedure at a previous stage. If so then let s be the last stage at which it was run. 2) At stage s, H had already found at least j splittings. If so then let i be such that we blamed i for j at stage s. 3) i g(h, j, s). If all of these conditions are satisfied then let i be as above and blame i for j at stage s. Otherwise proceed as follows. Test to see whether there exists j 1 < j such that g(h, j 1, s) g(h, j, s). If not then just choose some i g(h, j, s) and blame i for j at stage s. Otherwise let j 1 be the greatest such. Then test to see whether there exists j 2 < j 1 such that g(h, j 2, s) g(h, j 1, s). If not then find i such that we blamed i for j 1 at stage s and blame i for j at stage s. Otherwise let j 2 be the greatest such. Test to see whether there exists j 3 < j 2 such that g(h, j 3, s) g(h, j 2, s). If not then find i such that we blamed i for j 2 at stage s and blame i for j at stage s, and so on. Why do we first check whether the statements 1),2) and 3) apply? We wish to ensure that for any strategy H it is either the case that H is marked by H at all but a finite number of stages, or that H is only marked by H at a finite number of stages. Since any 14

15 strategy can only be marked by the finite number of strategies below it this will ensure that for any strategy H it is either the case that H is marked (by some strategy) at all but a finite number of stages or that, at all but a finite number of stages H is not marked by any strategies at all. Amongst other things this is clearly necessary in showing that we can define a true path for the construction. In order to show that this procedure achieves what it is supposed to, choose j1 large enough such that for no j j1 is it the case that g (H, j) is not an infinitely occurring index set. Choose j2 large enough such that for every Y which is a minimal infinitely occurring index set for H, there exists j with j1 j j2 such that g (H, j) = Y. Choose s large enough such that for all 1 j j2, g(h, j, s) = g (H, j). Now suppose that for j > j2, s > s and some i < j k + i k, we blame i for j for the first time at stage s. Consider the iteration that took place. If g(h, j, s ) is a minimal infinitely occurring index set then clearly we may conclude that i is in a minimal infinitely occurring index set. Otherwise there exists j 1 < j such that g(h, j 1, s ) g(h, j, s ) and g(h, j 1, s ) is a minimal infinitely occurring index set. Let j 1 be the greatest number (of any kind) < j such that g(h, j 1, s ) g(h, j, s ). If g(h, j 1, s ) is a minimal infinitely occurring index set then clearly we may conclude, since i g(h, j 1, s ), that i is in a minimal infinitely occurring index set. Otherwise there exists j 2 < j 1 such that g(h, j 2, s ) g(h, j 1, s ) and g(h, j 2, s ) is a minimal infinitely occurring index set. Let j 2 be the greatest < j 1 (of any kind) such that g(h, j 2, s ) g(h, j 1, s ), and so on. At some point in this iteration we must find j u such that g(h, j u, s ) is a minimal infinitely occurring index set, and it must then be the case that i g(h, j u, s ). Let i be such that there exist an infinite number of j such that there exists s, i is blamed for j at stage s. Then, by the argument above, we may choose Y which is a minimal infinitely occurring index set for H and such that i Y. Let j1 be as above. Suppose that for j j1 we find at some stage s that g(h, j, s) = Y. Clearly ( 1 ) it cannot be the case at any subsequent stage s that g(h, j, s ) g(h, j, s). In order to show how we can use this fact in order to decide an initial segment of W i we shall consider a simple example. Suppose that, at the stage at which we found the j th splitting, the only axioms of the relevant variety which we had enumerated on the strings in this splitting, for functionals Γ i,u such that u U i, and which were not then hidden or expired axioms were the two axioms Γ σ,τ i,u (n) = c and Γσ,τ i,u (n ) = c. Clearly σ and σ must be compatible. Let us suppose that σ < σ. Then by ( 1 ) the first of these two axioms cannot become an expired axiom, so that at stage s we may conclude that σ W i. Suppose that at some subsequent stage the first of these two axioms becomes hidden. At this stage we may conclude that σ W i. Let j 1 < j 2 <.. be all those j j 1 such that g (H, j) = Y. For each j n let σ n be the longest string which we are able to deduce is an initial segment of W i, by considering those axioms that have been enumerated on the strings in the j th n splitting in the manner described above. For each n 1 there is a strategy of some form H n = Υ n [.., r n ](m n,.), say, such that f(m n ) = i, r n = σ n and which enumerates an axiom 15

16 (of the relevant variety) on one of the strings in the jn th hidden or expired. splitting, which never becomes Suppose that there exists an upper bound on the length of the strings σ n. As stated previously, we shall be able to show that for all r, m, n ω every maximal (r, m, n) set is finite. It will also be an obvious consequence of the way in which we define the construction, that if the first strategy passed control in any maximal (r, m, n) set is passed control at stage s, then r = s when passing control for the first time at some stage s to any strategy which is the first to be passed control in the maximal (r, m, n) set to which it belongs, we shall just define r = s. Since a finite number of strategies are passed control at any stage of the construction (and there is an upper bound on the length of the strings σ n ) the set of strategies {H n } n 1 must therefore be finite. Now in the course of searching for splittings for any strategy H we shall use a procedure, which we shall call the splitting search procedure, and which will output the splittings that it finds (and which are not necessarily appropriate yet for enumeration into the relevant tree). The splitting search procedure will ensure that for any strategy H which enumerates an axiom which never becomes hidden or expired, and with axiom enumeration string q say, there are a finite number of splittings outputted such that both: 1) At least one string in the splitting extends q. 2) Not every string in the splitting extends q. Since the set of strategies {H n } n 1 is finite it must be the case, for at least one of the strategies H n, that there are an infinite number of splittings outputted by the splitting search procedure for H such that at least one string in the splitting extends the axiom enumeration string for H n. Let H n be such a strategy and let the axiom enumeration string for this strategy be q. But then, by the above, it must be the case for the strategy H n that every string in an infinite number of the splittings outputted by H extends q. This means that q B so that, in fact, for all but a finite number of the splittings found by H it must be the case that at least one string in the splitting extends q. But there exist an infinite number of j such that there exists s, i is blamed for j. Thus H n would be marked at an infinite number of stages. We have stated previously that this will mean H n is marked at all but a finite number of stages which gives us the required contradiction since in that case q B. Definition 1.12 We define the function z as follows. For any j ω, z(j) = {(i, n) : i, n j}. From that which we have already said it follows that for any i ω, if W i is noncomputable, then any individual C strategy will only mark a finite number of strategies of any form Υ[..](m, n) such that f(m) = i. The danger remains that for fixed n ω an infinite number of strategies on the true path will mark strategies on the true path of some form Υ[..](m, n ) such that f(m ) = i. We shall use the function z, as defined 16

17 above, precisely in order to remove this possibility. Thus any strategy which is searching for Ψ j splittings, say, will not be allowed to mark a strategy of any form Υ[..](m, n ) such that f(m ) = i and (i, n ) z(j) so we shall restrict the strings amongst which it is searching for splittings to those on which we have not enumerated axioms of any form Γ σ,τ i,u (n ) = c such that (i, n ) z(j) and which are not hidden or expired. We stated previously that in defining t(h) the first time we pass control to any strategy H, we must deviate just slightly from the intuition of the example described. It is the following point that we were then referring to. Above a strategy of the form Υ[..](m, n ) such that f(m ) = i and such that (i, n ) z(j), which puts B s through its axiom enumeration string at stage s, we may have to introduce at stage s a new candidate for T r j. We shall be able to show (and hopefully it is clear anyway) that, for any j ω, there exist a finite number of strategies on the true path of the construction which put B s through their axiom enumeration strings at all but a finite number of stages, and which search for Ψ j splittings. Our use of the function z therefore achieves what we required of it. We are yet to explain the use of those strings which are labelled the axiomatic release. Suppose that a strategy C[t, p, r](m, n) with t = {T r j1,i 1,.., T r jk,i k }, j 1 <.. < j k, finds a splitting such that all of the axioms we have enumerated of the relevant variety are either expired or have already been labelled hidden axioms. We must find an extension of each of the strings in this splitting that extends a leaf of each tree in t in order that we may enumerate these extensions into T r jk,i k. It is in order that we can perform this operation without having to consider axioms that have been enumerated on extensions of those strings in the splitting that we provide every strategy in state 1 with an axiomatic release. The use of the axiomatic release is really just a technical convenience. When any strategy H is passed control at a stage s it will not put B s through the string which we have labelled its axiomatic release. In what follows it will be convenient to adopt the convention that if T r is empty then σ = is a leaf of T r. At this point it is probably best that we begin to articulate the specifics of the construction. 2 The D[t, p, r](m, n) strategy. Here n, m, r ω, p is a finite binary string and t is a finite set of candidates for the T r j, t = {T r j1,i 1,.., T r jk,i k } say, with j 1 <.. < j k. Definition 2.1 Given a strategy H we say that u U l for some l ω is appropriate for axiom enumeration at H if there are no strategies C[t..] below and including H on the tree of strategies with t = {T r j 1,i,.., T r 1 j k,i }, j k 1 <.. < j k and (j k, i k ) u. (Explanation: such a C strategy is searching for a splitting to enumerate into T r j k,i ). k 17

18 Definition 2.2 At any point in the construction and for any i, n ω, u U i and any finite binary string τ we define the set γ(τ, i, u, n) to be all of those strings σ such that we have enumerated an axiom of the form Γ σ,τ i,u (n) = c for some c {0, 1} and finite binary strings σ, τ such that σ σ and τ τ. Note that this set may change as we carry out the construction. The strategy is initially considered to be in state The instructions for the strategy in state 0. If the strategy is passed control in state 0 at stage s it simply passes control to the strategy C[t, p, r](m, n). 2.2 The instructions for the strategy in state 1. If the strategy is in state 1 it will have been provided with four incompatible strings, p 1,.., p 4 say, all extending p (these strings will have been delivered to it). We say that these strings are in the environment of this strategy. We label these strings as follows: 1) p 1 we label the axiomatic release 2) p 2 we label the axiom enumeration string 3) p 3 we label the split escape 4) p 4 we label the marked escape. If the strategy is passed control in state 1 at stage s we carry out the following instructions. a) If D[t, p, r](m, n) is not marked at stage s, K s (n) = 0 and p 2 is not split at stage s then put B s though p 2 and for any u U f(m) which is appropriate for axiom enumeration at D[t, p, r](m, n) and which is such that W f(m),s / γ(p 2, f(m), u, n) enumerate the axiom Γ σ,p 2 f(m),u (n) = 0 where σ is the prefix of W f(m),s of length r. b) If D[t, p, r](m, n) is not marked at stage s, K s (n) = 0 and p 2 is split at stage s then check to see if this strategy has ever been passed control in state 1 at a stage s < s and put B s through p 2. If not then put B s though p 3. Otherwise let s be the greatest stage < s at which this strategy was passed control and put B s through p 2. If the prefix of W f(m),s of length r is equal to the prefix of W f(m),s of length r then put B s through p 2. Otherwise put B s through p 3. c) If D[t, p, r](m, n) is not marked at stage s, K s (n) = 1 and p 2 is not split at stage s then let s be the least such that K s (n) = 1. Check to see whether this strategy was ever passed control before stage s. If not then put B s through p 2 and for any u U f(m) which is appropriate for axiom enumeration at D[t, p, r](m, n) and which is such that W f(m),s / γ(p 2, f(m), u, n) enumerate the axiom Γ σ,p 2 f(m),u (n) = 1 where σ is the prefix of W f(m),s of length r. Otherwise put B s through p 3. 18

19 d) If D[t, p, r](m, n) is not marked at stage s, K s (n) = 1 and p 2 is split at stage s then let s be the least such that K s (n) = 1. Check to see whether this strategy was ever passed control before stage s. If so then put B s though p 3. Otherwise check to see whether this strategy has ever been passed control in state 1 at a previous stage s and put B s through p 2. If not then put B s though p 3. Otherwise let s be the greatest stage less than s at which this happened. If the prefix of W f(m),s of length r is equal to the prefix of W f(m),s of length r then put B s though p 2 and otherwise put B s through p 3. e) If D[t, p, r](m, n) is marked at stage s then put B s through p 4. Suppose that D[t, p, r](m, n) has put B s through p i. If there is a previous stage s at which this strategy has put B s through p i then transfer control to the same strategy as the last stage at which it did so. Otherwise run the procedure TRANSFER with inputs D[t, p, r](m, n) and p i in order to decide which strategy to transfer control to. 3 The TRANSFER procedure. In order to understand the intuition behind 6) below we refer the reader to the section of the introduction in which we considered a simplified version of the contruction concentrating only on those aspects of the construction concerned with ensuring that B is a set of minimal degree, and also to the section of the introduction in which we discussed the use of the function z. Recall that any strategy H passed control at a stage s will not put B s through the string which we have labelled its axiomatic release. Suppose that we are given inputs Υ[t, p, r](m, n) and (the finite binary string) p at stage s. Then we transfer control to the strategy Υ [t, p, r ](m, n ) where Υ, t, r, m and n are defined as follows. 1) If Υ = D then Υ = D. 2) If Υ = C then, if p is the split or the marked escape for the strategy Υ[t, p, r](m, n), Υ = C and if p is the axiom enumeration string for this strategy then Υ = D. 3) If Υ = C and p is the split or the marked escape for the strategy Υ[t, p, r](m, n) then t = t. 4) If p is the split escape for Υ[t, p, r](m, n) then r = r, m = m and n = n. 5) If p is either the marked escape or the axiom enumeration string for the strategy Υ[t, p, r](m, n) then m = m + 1 and r = s. Find the number of strategies below and including Υ[t, p, r](m, n) on the tree of strategies of any form Υ [..](m, n ) such that f(m ) = f(m + 1) and for any n, Υ and which put B s through the axiom enumeration string for that strategy at stage s. We define n to be that number. 6) If Υ = D, or Υ = C and p is the axiom enumeration string for the strategy C[t, p, r] (m, n), then let t be defined as follows. Let us call the strategy Υ[t, p, r](m, n), H. When we say that one strategy is below / above another we do not mean that it must be properly below /above this strategy. Then find the least j such that i (T r j,i t (H)) and either: 19

20 a) there is no (as yet unsuccessful) strategy C[t..] below the strategy H with t = {T r j 1,i,.., T r 1 j k,i }, j k 1 <.. < j k (say), and such that j k = j. b) there is an (as yet unsuccessful) strategy C[t..] below the strategy H with t = {T r j 1,i,.., T r 1 j k,i }, j k 1 <.. < j k (say) and such that j k = j but if C[t,..] is the highest of these on the tree of strategies below H then k 1 0 and T r j / k 1,i t (H). k 1 c) there is an (as yet unsuccessful) strategy C[t..] below the strategy H with t = {T r j 1,i,.., T r 1 j k,i }, j k 1 <.. < j k (say) and such that j k = j. Furthermore there exists a strategy Υ [..](m, n ) above this C strategy and below H such that (f(m ), n ) z(j), which put B s through its axiom enumeration string at stage s and such that there does not exist a strategy Υ [t,..] strictly above Υ [..](m, n ) and below H such that j, T r j,j t. Let i be the least such that there is no strategy Υ [t,..] that we have yet passed control to with T r j,i t (so we choose a new candidate for T r j ). We define: t = {T r j,i } {T r j,i : T r j,i t (H), j < j}. 4 The C[t, p, r](m, n) strategy. Here n, m, r ω, p is a finite binary string and t is a finite set of candidates for the T r j, t = {T r j1,i 1,.., T r jk,i k } say, with j 1 <.. < j k. The instructions for the strategy at stage s in the case that t = are very simple: choose four incompatible extensions of p, p 1,.., p 4 say, and enumerate these into the nonsplitting tree, T r 1,0. Declare the strategy which passed control to it to be in state 1, deliver p 1,.., p 4 to it and terminate stage s activity for the entire construction. So suppose that t. The strategy (if t ) is initially considered to be in state The instructions for the strategy in state 0. If the strategy is passed control at stage s in state 0 then it immediately passes control to the strategy C[t, p, r](m, n) where t = t {T r jk,i k }. 4.2 The instructions for the strategy in state 1. If the strategy is in state 1 it will have been provided with four incompatible strings, p 1,.., p 4 say, all extending p (these strings will have been delivered to it). We say that these strings are in the environment of C[t, p, r](m, n). We label these strings as follows: 1) p 1 we label the axiomatic release 20

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