Minimal Complements for Degrees below 0´

Minimal Complements for Degrees below 0´Author(s): Andrew LewisSource: The Journal of Symbolic Logic, Vol. 69, No. 4 (Dec., 2004), pp. 937-966Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/30041773 .

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THE JOURNAL OF SYMBOLIC LOGIC

Volume 69, Number 4, Dec. 2004

MINIMAL COMPLEMENTS FOR DEGREES BELOW 0'

ANDREW LEWIS

Abstract. It is shown that for every (Turing) degree 0 < a < O' there is a minimal degree m < O0' such that a V m = O' (and therefore a Am = 0).

11. Introduction. The aim of this paper is to prove the following theorem. THEOREM 1.1. Every (Turing) degree 0 < a < 0' has a minimal complement. In 1992 Seetapun and Slaman wrote what they had hoped to be a proof sketch of

this result. In order to present a valid proof we shall use most elements of the proof sketch that they described, but new ideas will also be required.

Before we go on to give some account of the intuition it is helpful first to give a brief description of the 'architecture' of the construction in order that we may define terms with which to frame the subsequent discussion.

We shall suppose that we are given an approximation {As}s>o to a set A of degree 0 < a < 0' and an enumeration {K, }s)> of K. For convenience we shall assume that for all s > 0, A, is a binary string of length s. We will enumerate an approximation {Xs}s>o to a set X of minimal degree and we will enumerate axioms for a Turing functional F such that:

FAX = K.

The axioms that we shall enumerate for F will be of the form P," (n) = c, where o and z are finite binary strings and n ow, c E {0, 1 }. Throughout the construction we will have to ensure that at every stage s > 0 and for every n E co,

FrAs'Xs(n)[s ] _--_ rA, (n)[s] = Ks(n)

and that ultimately, for every n E co, FAx (n) 1. Let

{j }jj>o be an effective listing of the Turing functionals. We must also

construct X in such a way as to ensure that it is a set of minimal degree. This we accomplish (as is standard) by ensuring that for each Turing functional Kj, X lies on some c.e. tree Tr which is either YI splitting or which is such that if PY is total there is some initial segment z of X lying on Tr for which no Ti, z2 extending z and lying on Tr are a TY splitting. To be more precise, for each j > 0 we will attempt to enumerate an infinite tree Trj such that X lies on

Try and which is Ij splitting.

If we do not succeed and there is a greatest j' < j for which we do then Trj1 will act as the Tr above demonstrating that if TT is total then it is computable. We note

Received April 10, 2002; revised November 28, 2003.

Q 2004, Association for Symbolic Logic 0022-4812/04/6904-0001/$4.00

937



938 ANDREW LEWIS

that there may be many candidates for each Trj-the tree Tr1,i should be regarded as the ith candidate for Trj.

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'. We use the variable H to range over the set of strategies. At each stage s of the construction we will start at the bottom node of the tree of strategies and each strategy will, in turn, pass control to another strategy one level up on the tree until stage s activity ceases, with a finite number of strategies having been passed control. Each strategy may be provided with a finite set of binary strings and at every stage s at which the strategy is passed control it may 'put Xs through' one of these strings i.e., the strategy may insist that Xs be an extension of this string. If a strategy H passes control to a strategy H' then the strings that H' has been provided with will extend those that H has been provided with. If H puts Xs through a string z then any string r' which H' puts X, through will be an extension of T. When stage s activity ceases we define Xs to be the longest string that it was put through during stage s activity. For technical completeness, if Xs is not put through any strings during stage s then we define Xs to be the empty string (of course this must only happen a finite number of times). It is also worth noting that the outcome of every strategy will be finitary: for every strategy there will be an s such that after stage s either the strategy is never passed control or the strategy is always passed control and the outcome of the strategy is always the same. We may therefore define a 'true path' of the construction (those strategies passed control at an infinite number of stages) computable in K.

1.1. There will be four basic varieties of strategy involved, V, .', V and 9 strategies. We use the variable T to range over the set {I, T, 'F, 9 }. If we refer to a strategy of type T as an T(n) strategy this means that the strategy is concerned with ensuring that FA, (n) is correctly defined. Before we go on to describe each of the strategy types in some detail consider first the following definition and the subsequent observation concerning the description of a strategy H*, which is crucial to the intuition behind the construction.

DEFINITION 1.1. At any stage of the construction, for any strings a, z and for n E co we say that a E y(z, n) if there are initial segments of a and z, a' and z' respectively, for which we have already enumerated the axiom F","' (n) = c for some c. Note that the set y (r, n) may change as the construction progresses.

The construction will be such that for any n, n' c co such that n > n' we shall never enumerate an axiom of any form Fr"~ (n) = c at any stage s, for any 1r As, z C Xs and c E {0, 1 }, unless we have previously enumerated an axiom of some form

F1'1'/(n') = c' such that a' c a, z' c z and c' E {0, 1}. The following is an

immediate consequence.

(to) For any string z and any n, n' E co such that n > n' it will always be the case that 7 (, n) C y (z, n').

We satisfy ourselves, for now, with simply stating that this is the case-but the truth of this fact will soon become perfectly obvious.

So suppose that a strategy, let us call it H*, is passed control for the first time at stage s of the construction and that this is a strategy which is concerned with



MINIMAL COMPLEMENTS FOR DEGREES BELOW 0' 939

ensuring that FA,X (n) is correctly defined. Suppose further that all of the following are true.

(1) H* is provided with four strings 4, ri, T2, 13 such that TI, T2, T3 are (pairwise) incompatible extensions of 4.

(2) For i -1

j, y(ri, n) n y(z j, n) = y (0, n). For 1 < i < 3 define Pi = y(zi, n) (as this set is defined at stage s).

(3) At stage s no axioms have been enumerated on proper extensions of ri for any i, i.e., for 1 < i < 3 we have not enumerated any axioms of any form F11 (n') = c such that T properly extends zi.

(4) The required 'environment' of the strategy H* states that at any stage s', As, / y(4, n). So if the strategy H* is only passed control at stages such that there is the required environment then it will only be passed control at stages s' such that As, V 7(o, n).

Why should the required environment of the strategy H* state that at any stage s', As,I y(0, n)1 We have assumed that the strategy is concerned with ensuring that FAX (n) is correctly defined. Now suppose that at some stage s' > s the strategy is passed control and that As, E (0, n)-let us suppose that we have enumerated some axiom F"~'(n) = 0 for a C As, and r

C_ , but that Ks,(n) = 1. Whichever

string ri the strategy H* puts Xs, through, the result will be that FAs'/xv' (n) is not correctly defined.

Given that the strategy H* is only passed control at stages such that there is the required environment, how can the strategy ensure that FAX(n) is correctly defined1 At any stage s' at which the strategy is passed control it can proceed as follows. By (2) there exists at most one i E {1,2, 3} such that As, E Pi. Let us suppose that As, E P1. Now if we put Xs, through rj, j 4 1, then the fact that As, V Pj and that H* will only ever enumerate F(n) axioms (axioms of the form FT"~(n) = c for a, r e 2<w and c E {0, 1}), means that by the observation (to) made above, As, (zrj, n + 1). The strategy H* can thus proceed to put Xs, through zj and pass control to another strategy which is concerned with ensuring that FAX (n + 1) is correctly defined (and which you might imagine being provided with three pairwise incompatible strings, r', -r, T- say, extending rj such that for i' j', y (,, n + 1) y (,, n + 1) = y(zj, n + 1)) and this strategy will be provided with the correct environment. Namely that As, V 1y(r, n + 1). So at any stage s' such that As, E Pi the strategy H* is presented with two choices. It can put Xs, through T2 or through T3. At stages s' such that As E P1 and n V Ks, the strategy can put Xs, through Z2 and enumerate the axiom F,T2 (n) = 0 where a is the initial segment of As, of length 1|21. At stages s' such that As E P1 and n E Ks, the strategy can put Xs, through z3 and enumerate the axiom FT~3 (n) = 1 where a is the initial segment of As, of length 1-3 1.

Similarly at stages s' such that As, E P2 the strategy can put X,' through either ri or (3. At stages s' such that As, E P3 the strategy can put Xs, through either Ti or T2. At stages s' such that

As, 4

U-=1 Pi the strategy can proceed just as if As, E P1. At

any stage s' at which the strategy puts Xs, through the string xi it should enumerate the axiom F""' (n) = Ks, (n) where a is the initial segment of As, of length ri |. If the strategy is passed control at all but a finite number of stages then it is able to ensure that FAX (n) is correctly defined and the fact that we provide the strategy with three



940 ANDREW LEWIS

incompatible strings extending 0 rather than just two means that it is able to pass control to strategies concerned with ensuring that FAx (n + 1) is correctly defined and these strategies will be provided with the correct environment.

DEFINITION 1.2. Given j > 0 and z e 2<" let n be the least such that P .(n)

T. We say that T' is of length n.

DEFINITION 1.3. By a k-fold yPj splitting of length at least 1 we mean k strings Tl ..., k whose images

1.' (1 < i < k) are (1) of length at least / and (2) pairwise

incompatible-for 1 < i < i' < k if the lengths of P7.

and T'P' are ni and n2 respectively then there exists n3 < ni, n2 such that 17'(n3) 7i' (n3).

What we learn from this example is that when we search for splittings to enumerate into any of the splitting trees, we must search for splittings with a certain intersection property as regards the axioms that have been enumerated on the strings in the splitting-generally, if we are searching for a k-fold splitting, tl,...,Tk say, above a string 0 in order that we can enumerate this splitting into a tree Trj1,, we shall insist that for i' f j' we have y

(ri', n) n y

(pj,, n) = 7 (0, n) (and where n is such that

on this part of the tree of strategies we are concerned with ensuring that FA,x (n) is correctly defined).

Now let us describe each of the strategy types in turn. 1.2. The s' (n) strategy. To each strategy H, of any variety, we shall be associating

for some m a finite set of trees,

t(H) = { Tr-1,o, Trj,,i,,...., Trim im}

where jl < ... < j,, and where for any 1 < r < m, Trjir should be regarded as the i,th candidate for Trir. Tr-l,o

is a tree which we enumerate for the sake of technical convenience, but which is not required to be any kind of splitting tree. Let us agree that jo = -1, io = 0. Assume for now that m > 1 and suppose that 0 < r' < r < m. If, for some strategy H, t(H) is defined as above then during the course of the construction as a whole we shall construct

Trjr,, so as to be a subtree

of Trjr,,,i, (actually we shall have to qualify this statement slightly in 1.4). For some k, 1 E co it will be the task of an W (n) strategy H, with t(H) as above,

to search for a k-fold Y/j, splitting of length at least I lying on Trjm-_,im

,. We shall explain why it is necessary to search for splittings of at least a certain length subsequently. If the strategy ever finds such a splitting then it will be declared 'successful' and it will never be passed control again. If no such splitting exists and the strategy is passed control at an infinite number of stages then it will ensure that FA,X (n) is correctly defined and will witness the fact that Trjm,i~ is finite. The

strategy will proceed to achieve these specified tasks as follows. The first stage s at which an d (n) strategy is passed control it will be provided with

an environment like that which we described for the strategy H* in 1.1. It will be provided with four strings 0, Ti, T2, T3 such that Tl, z2, T3 are (pairwise) incompatible extensions ofq0 and such that for i = j, 7 (Ti, n)ny(zj, n) = y (0, n). The strings zl, z2, Z3

will be strings that we have enumerated into Trim ,1im-

, while 0 will extend a leaf

of Trjm,i~. At stage s we shall have that no axioms have been enumerated on proper extensions of ri for any i. Once again it will be part of the required environment of the W (n) strategy that it is only passed control at stages s' such that A1, V y (0, n).




The W(n) strategy can therefore proceed to operate much as the strategy H*. It will choose which string to put Xs, through at any stage s' and will enumerate axioms in the manner described previously, except that at every stage at which it is passed control the strategy also performs one more step in some exhaustive search procedure (which clearly there is no need to specify) looking for a k-fold YIjm splitting of length at least I lying on Trjim,,i,_ and such that there exists 1 < i < 3 every string in the splitting extends ri. It is worth emphasizing that the W(n) strategy searches only for a splitting such that there exists some i, every string in the splitting extends zi.

At every stage s' at which the s(n) strategy puts Xs, through the same string it will pass control to the same 2 (n + 1) strategy, so that there are three different -9(n + 1) strategies that an W(n) strategy may pass control to at any stage at which it is passed control.

We shall refer to the V (n) strategy with parameters 0, rl, T2, T3, k, I and t as above, as the strategy H = W[t, k, 0, rl, t2, z3](n) where 1(H) 1= 1. We consider the argument I separately since it may be redefined a finite number of times during the course of the construction. We may also use the notation k(H) to denote the value k, aswell as the notation t(H) to denote the value t. The strings 0, zl, z2, z3 we refer to as the 'base strings' of the strategy.

1.3. The W (n) strategy. Suppose that we are searching for a splitting above 0 to enumerate into Tr jm,im and that the d(n) strategy, as described in 1.2, is eventually declared successful. Let us suppose that it finds a k-fold Y!j, splitting, r ..., 1, such that every string in the splitting extends Ti. This splitting is not suitable for enumeration into Trjm,im, Before we could enumerate such a splitting into this tree we would require that for i - j, y(T1,n) n y(r',n) = y(q, n), but common to the axioms that we have enumerated on each of the strings in this splitting are the axioms that the W (n) strategy has enumerated on Tz. If the

sW (n) strategy is passed control at an infinite number of stages then it will ensure that FA,X (n) is correctly defined and will witness that

Trj,,im is finite, but if the strategy is declared successful

then the splitting that it finds will not be suitable for enumeration.

LEMMA 1.1. Suppose that S1,..., Sk are such that, (1)for 1 < i < k, Si contains at least k - i + 1 pairwise incompatible strings, and (2) if 1

_ i < j < k then every

string in Sj is longer than every string in Si. Then we may choose from each Si a string ai so that ai is incompatible with aj if i 5 j.

The proof of this lemma is left to the reader. Now it will be the aim of a T.4(n) strategy H with t(H) =

{Tr-l,o, Trj,,i,,..., Trim,im}, m > 1, jl < "'"

< j,, to find (for some k e co) a k-fold 'Pj, splitting that we can actually enumerate into Trjm,im. A splitting, that is, with the required intersection properties as regards y. The basic intuition behind the strategy is that we should proceed as follows. Suppose that we wish to find such a splitting to enumerate into Trjm,im every string of which extends b. Then at the first stage s at which the strategy is passed control we should be provided with 3k (pairwise) incompatible strings extending 4, which have been enumerated into Trj_,im_, (let us agree once again that jo = -1, io = 0) and such that for distinct r, z' we have y(z, n) n y(r', n) = y(4, n). The required environment of the strategy states that no axioms should have been enumerated on proper extensions of the strings that we have been provided with and that H should only be passed control at stages s'

_ s such that As, ( ,

(4, n). We may then divide



942 ANDREW LEWIS

these strings into k triples, so that each triple provides a suitable environment for an .V(n) strategy. Let the variable ~ range over this set of triples and for each i let us agree that 4u = {z,,1, r ,2, Ty,3} and that, for any n', y(u, n') = Ui y(ruc, n'). At any stage at which the T(n) strategy is passed control we shall pass control to an

Ms(n) strategy, H' say, with base strings 0 1,z, 1, 9,,2, Z,,3

for some t and such that t(H') = t(H), k(H') = k(H). Let us call this the 'd(n) strategy for p'. Suppose that H is passed control at some stage s' such that none of the d'(n) strategies that we have passed control to have yet been declared successful. If As, E 7y(u, n) for some u then we will pass control to the V(n) strategy for p, H' say-just choose the required parameter 1(H') to be some fixed number. If As,, 7((u, n) for any p then we will just choose u and pass control to the V (n) strategy for Up.

Suppose that one of the V(n) strategies is eventually declared successful. Then for each of the ds(n) strategies H' redefine 1(H') to be larger than any number yet mentioned in the course of the construction. Let us call the string above which a splitting has been found 4r and suppose that 'H E l (generally, if z is the ith string above which a splitting is found by the V strategies passed control by a M strategy H we shall refer to z as rH). We 'discard' the other two strings in ktl and we shall never again pass control to the W'(n) strategy for pul. At any subsequent stage s' at which the W (n) strategy is passed control such that As, / 7(iH, n) we can now proceed basically as before-the idea is that if all k of the V (n) strategies are declared successful then in accordance with Lemma 1.1 we can form a splitting of the required variety by choosing one string from each of the splittings found. If

As, E y(1u, n) for some up :# up then we may pass control to the s(n) strategy for

p, and if not then we may choose a triple and proceed similarly. If another .(n) strategy should subsequently find a splitting above a string r2H E 42 then, so long as we have not passed control to the W(n) strategy for pu2 at any stage s' such that

As, E y(rH, n), we shall have that y(r, n) n y(H, n) = y(4, n). Since we shall not have enumerated any F(n) axioms on proper extensions of either TH or r2H it will be the case that any two strings, one chosen from each of the splittings found, will also satisfy this property. Each time that another W(n) strategy is declared successful we should redefine 1(H') for each of the V(n) strategies H' to be larger than any number yet mentioned in the course of the construction.

After the first splitting has been found, above the string 4TH, how should the M (n) strategy now proceed at stages s' such that As~ E y (4H, n)1 If it passes control to some sd(n) strategy which then finds a splitting above a string /2H it may be the case that we will not be able to choose a string from each of these splittings in order to form a splitting with the required intersection properties-since it might then be the case that y (rH, n) n y (H, n) H y (, n). We must insist, in fact, that the V(n) strategy is not passed control at stages s' such that A,, E C(T4, n) since the strategy is now unable to act at such stages. More generally, if k' (< k) of the W(n) strategies have been declared successful and have found splittings above strings t41 .

H. k then we cannot pass control to the 9(n) strategy at stages s'

such that As, E U/i1 4 , n)

DEFINITION 1.4. In the course of discussing a & or a ' strategy H every string T under consideration will extend one of the base strings for H. We let

/H (Z)

designate this string.




So in the event that the T(n) strategy becomes unusable after a certain stage of the construction we must devise some sort of contingency plan. More generally we must consider, in fact, T (n) strategies which given strings q1,..., 1, (which we refer to as the base strings for the strategy) such that r divides k search for a k-fold splitting zli...,..k

such that k/r of these strings extend each Oi, and such that if i - j then y (i, n) n y (j, n) = Y()H( Zi), n) n Y()H (zj), n). The required environment of the strategy will be such that it should only be passed control at stages s' such that As, V Ui=, y(4i, n). We shall refer to a T(n) strategy with base strings 01,..., r and with t and k as above as the strategy T[t, k,

141...,,r](n). We shall proceed (roughly) as follows. In the 'initial environment' of the 1M(n)

strategy we shall require a much larger number of strings. When k' of the V(n) strategies that a W(n) strategy has passed control to have been declared successful we shall regard the strategy as being in state k'. Whenever one of the V'(n) strategies is declared successful, so that the T(n) strategy enters a new state, we shall take a number of the triples above which we are yet to find splittings and choose one string from each of these triples, discarding the other two. We shall then take an extension of each of these strings in order to form what we shall call 'escape routes'. We shall not pass control to W (n) strategies for those u from which we have discarded strings. Each time that a T(n) strategy enters a new state k' (< k) we shall have to discard all of the previously defined escape routes and choose k' new sets of escape routes-one set for each of the splittings that has been found. Suppose that p = {zrp,1, Tp,2, ... } is a set of escape routes that we have defined for the W(n) strategy H upon being declared to be in state k'. Each Tp,i will satisfy the following property: for 1 < j < k', y(zrp,i,n) ny(rH,n) =

Y,(OH(Zp,i),,n)ny (H(rH'),fn). The importance of this is that, if at any stage s' at which H is in state k' there would be an environment appropriate in order to pass control to H except that As, y(r~ , n) for some j, then for each rp,i we have As, ~ y(zp,i, n). Other factors aside there will thus be an environment appropriate such that we may pass control to a M(n) strategy with base strings defined to be those in p-a T(n) strategy 'above p' let's say. We can now hope to make progress with this T(n) strategy instead. We shall leave it to the following sections to describe in detail the manner in which we must define escape routes, but in 1.4 we shall explain why it is that whenever a T(n) strategy enters a new state k' > 0 we must define k' sets of escape routes.

From the preceding discussion it is hopefully clear that the number of strategies any ~4(n) strategy may pass control to is fixed and determined by k. A 'T(n) strategy passes control to s'(n) strategies. If H is a T(n) strategy and H' is a T(n) strategy above p, a set of escape routes that we have defined for H, then control will be passed to H' by the V(n) strategy immediately below H on the tree of strategies. If a V(n) strategy ever finds the splitting that it is searching for then it will be declared successful and it will never be passed control again.

1.4. The '(n) strategy. This is the most complicated of the strategy types and we shall aim in this introduction only to give some impression of the manner in which these strategies will operate. For finite binary strings 4) ..., 1r (which we refer to as the base strings for the strategy), k e co and for some t = { Tr-l,o, Trj,i,, ...,

Trj,.,i, } with jl < '

< jm the strategy H = '[t, k, 41l .. 1,,](n) must achieve

the following. Either it must find some string above which there is no Ijm splitting on Trim

l,im_1 (presuming m > 1) or it must find a k-fold Ijm splitting, rl,..., rk



944 ANDREW LEWIS

say, suitable for enumeration into Trj.,im with the intersection properties described previously-that for i #: j, y (i, n)ny(zj, n) = ((H (zi), n)n y (H (-j), n). Where any individual W (n) strategy may fail-and where in fact every individual W(n) strategy may fail to find a splitting of the required magnitude-the F(n) strategy must organize a potentially unbounded number of W.(n) strategies in such a way that, should we not find any string above which there is no YTjm splitting on Trjm ,im_-I, then we can find amongst the splittings that these T.(n) strategies col-

lectively produce a splitting of the variety required. Should the W(n) strategy find the required splitting then it will be declared successful and it will never be passed control again. We shall have to use the non-computability of A in order to prove that any W(n) strategy only passes control to a finite number of different strategies and that should it be passed control at an infinite number of stages then there will be a stage after which it always passes control to the same strategy.

How (roughly speaking) should the strategy H = [t, k, 1,- . k.. ,r,](n) go about achieving its specified task1 In the case that t = { Tr-1,o}

the instructions are very simple. At the first stage s at which the strategy is passed control it will just choose k/r incompatible extensions of each qi and enumerate them into Tr 1,0 (we shall have that r divides k). The strategy will be declared successful and we shall terminate stage s activity for the construction. In fact, we shall have specified that H was intended to provide an environment for another strategy H'. Upon being declared successful H will declare H' to be in state 0 and will 'deliver' the strings that it enumerated into

Tr-1,o to H'. So suppose that m > 1. The required environment

of the strategy H will be such that it should only be passed control at stages s' such that As, Ur, y (i, n). The first stage at which H is passed control we shall have that k1,..., 1

extend leaves of each of the trees in t and that no axioms have been enumerated on proper extensions of these strings. The first course of action is to pass control to the strategy H1 =- [t, k, 4~1...,r](n), but first of all we must find the large number of strings required in the initial environment of this strategy. We therefore pass control to the strategy H1 = F[tl, ki,1,..... r](n), where tl = t - { Trj,im} and ki is the number of strings that are required in the initial environment of H1. We specify that H' is intended to provide an environment for H1. Should it be the case that H1 is never declared successful then at every stage at which H is passed control it will simply pass control to the strategy H1 which will witness that one of the trees in ti is finite. In this case the strategies above H' will ensure that FA,X (n) is correctly defined.

Suppose that H' is eventually declared successful. Then at every subsequent stage s' at which H is passed control there will be an environment appropriate such that it can pass control to H1 unless this strategy is in state k' > 0 and As, e y(Cr', n) for some 1 < j < k' (here the superfix is used as a counter rather than to indicate exponentiation). Let us suppose that this latter situation applies. We shall have defined kI different sets of (non-discarded) escape routes for H1 at this stage, and we associate each of these sets of escape routes with a different one of the strings z' 1 < i < k' . We would now like to pass control to a .'(n) strategy, H2 say, above

p which is the non-discarded set of escape routes for H1 that is associated with zr

such that As, E y ,(C',

n), but first of all we must provide an initial environment for this strategy. We therefore pass control to a '(n) strategy H2 such that t(H2) = tl




as above, with base strings defined to be those strings in p and which searches for a k2-fold splitting where k2 is the required number of strings in the initial environment of the strategy H2. We specify that H2 is intended to provide an environment for H2. Should it have been the case that As, e

,(H , ,n)

for somej' j then we would have passed control to the V (n) strategy intended to provide an environment for a T(n) strategy above p' associated with H',-we would have 'used' the set of

escape routes p' instead. Suppose that H is passed control at some stage s' after which H2 has been

declared successful and that once again we cannot pass control to H1 because it is in state k1 and As, e y (zr', n). Then we shall pass control to H2 unless it is in state

k2 > 0 (say) and As, E ,(172, n) for some 1 < j' < k2. In this latter case we should then like to pass control to a W(n) strategy above the non-discarded set of escape routes for H2 which are associated with zr~, but first of all we must pass control to a ~(n) strategy in order to provide the strings required in the initial environment, and so on.

The reason that, when a W(n) strategy enters a new state, we define a different set of escape routes for each splitting that this strategy has found can be understood as follows. Let H, H1, H2 and rHI be as above-so that H2 is the 2'(n) strategy above the set of escape routes defined for H1 upon being declared to be in state k1 which is associated with z Le. Let r $=Ar be a string above which the V (n) strategies that Hi passes control to have found a splitting when H1 is declared to be in state k' and let T' be any string above which the V (n) strategies which H2 passes control to find a splitting. Since we shall not pass control to the strategy H2 at any stage s' such that As, E y (z, n)-we would use a different set of escape routes at any such stage-it will be the case that y(-, n) n y(z', n) = Y(OH(Z), n) n Y(OH(r'), n), as required. Although there may be a stage after which the strategy H cannot pass control to H1, at any such stage s' all but one of the splittings that H1 has found (which one this is depends on the specific value of As,) remains useful to H. Thus, while HI is in state k1, the V strategies above non-discarded escape routes for H1 need only search for a (k - k' + 1)-fold splitting.

H is declared successful when any of the T(n) strategies that it has passed control to are declared successful, since then we are able to find a splitting of the required variety in amongst the splittings that these strategies have collectively produced.

How (roughly speaking) can we use the non-computability of A in order to show that any W(n) strategy is finitary1 Let H and H1 be as above. Suppose that H is passed control at an infinite number of stages (so that H is never declared successful) and that there are an infinite number of different strategies to which H passes control. Then we may consider an infinite 'chain' ofV(n) strategies, defined in the following way. At any stage of the construction at which H is passed control it first checks to see whether it can pass control to the strategy H1. The V(n) strategy which is intended to provide an environment for H1 must eventually be declared successful, otherwise at every stage at which H is passed control it would simply pass control to this '(n) strategy. Let k1 be the greatest state that H1 is declared to be in. Then k' > 0 otherwise there would be a stage, after which, whenever H is passed control it would pass control to H1. Clearly k' < k otherwise H would be declared successful. Define M1 = k - 1. Since the given approximation to A



946 ANDREW LEWIS

converges there will be a set of escape routes defined for H1 such that after a certain stage whenever H is passed control it checks to see whether it can pass control to

H1, finds that there is not an environment appropriate, and elects to 'use' this set of escape routes. Let H2 be the W(n) strategy above this set of escape routes. The '(n) strategy which is intended to provide an environment for H2 must eventually

be declared successful, otherwise there would be a stage after which whenever H is passed control it passes control to this W(n) strategy. If k2 is the greatest state that H2 is ever declared to be in then k2 > 0, otherwise there would be a stage after which whenever H is passed control it passes control to H2. Since all but one of the splittings that H1 has found are still apparently usable at any stage at which H2 is passed control (as explained above), the strategy H2 must search for a (k - Mi)- fold splitting. If it was ever successful in this task then we would declare H to be successful. Thus k2 < k - M1. Define M2 = M1 + k2 - 1, so that M2 is just the total number of splittings found by the strategies H1 and H2 which are apparently usable at any stage at which, rather than passing control to H2, H elects to use one of the sets of escape routes that we have defined for H2. Since our approximation to A converges there must be a set of escape routes that are defined for H2 such that after a certain stage whenever H is passed control it checks to see whether it can pass control to H2, finds that there is not an environment appropriate, and elects to use this set of escape routes. Let H3 be the .(n) strategy above this set of escape routes. The 9'(n) strategy which is intended to provide an environment for H3 must eventually be declared successful. H3 searches for a (k - M2)-fold splitting. Let k3 be the greatest state that H3 is declared to be in. Then 0 < k3 < k - M2. Define M3 = M2 + k3 - 1, and so on.

Now { Mi }i>l is a nondecreasing sequence of natural numbers such that, for all i > 1, k - Mi > 2. Thus limi,,(k - Mi) J> 2. This means that there exists j E w, Vi > j(k' = 1). For i > j consider the strategy Hi. Let Ji be the set of all those strings a such that, before Hi is declared to be in state 1, we have enumerated some axiom F', (n) = c such that z is compatible with rjH but properly extends

OHi (TH') (and for some c). For all i > j it must be the case that A is compatible with one of the strings in the set Ji, but as i --, oo this allows us to decide longer and longer initial segments of A. Thus A would be computable, which gives us the required contradiction.

DEFINITION 1.5. We say that 'z 2<w is compatible with a tree Tr if z has an extension on Tr, or if r extends a leaf of Tr.

Throughout the course of the construction, the following condition must clearly be satisfied.

(t I) When we enumerate a splitting into a tree Tr, every string in the splitting must extend (what was prior to this enumeration) a leaf of Tr.

If the '(n) strategy H is eventually declared successful then we are able to form a k-fold splitting by choosing one string from each of k different splittings found by strategies above it. In order that (tl) should be satisfied in future stages of the construction we must then find an extension of each string in this splitting, long enough that it extends a leaf of each of the trees in t (and long enough that no axioms have been enumerated on proper extensions of the string). We then enumerate the extensions that we have found into Tr,,im, declare the strategy which




H was intended to provide an environment for to be in state 0, 'deliver' the strings in the splitting we have just enumerated into Trjm,im to this strategy, and then terminate the present stage of the construction. In 1.1 we suggested that for r > r',

Trj,,ir would be constructed so as to be a subtree of Trjr,, ir,. The action that we have just described means that, in fact, this is not quite true. When we enumerate a string into

Trj,,ir we shall ensure that it is compatible with Trjr,,ir,. So long as there is

some 'active' part of the tree of strategies where we are still interested in building Trjirir we shall ensure that this string remains compatible with

Trjl,,i,,. So the tree

Trjr ,r will be 'thinner' than Trjr,,i1,,

but possibly taller at any given stage. 1.5. The 9(n) strategy. When the strategy H = 9[t, 0](n), with b a finite binary

string and t = { Tr ,o, Tri, .... Trim,im, }, j < ... < jm, is first passed control we shall have that 0 extends a leaf of each of the trees in t and that no axioms have been enumerated on strings properly extending 4. The strategy is initially considered to be in state -1. At every stage at which the strategy is passed control in state -1 it simply passes control to the strategy H' = C[t, 3, 0](n), performing no other instructions. We specify that H' is intended to provide an environment for H. If H' is never declared successful then at every stage at which H is passed control it will do nothing more than pass control to H' which will witness that one of the trees in t is finite. In this case the strategies above H' will ensure that FA, (n) is correctly defined. Control will eventually always be passed to the same V (n) strategy which will eventually always pass control to the same 9 (n + 1) strategy.

If H' is ever declared successful then it will declare H to be in state 0. The strategy H can now proceed to operate just as the strategy H* described in 1.1. If it is passed control at an infinite number of stages then H can ensure that FA X (n) is correctly defined. Every stage s at which H puts Xs through the same string it will pass control to the same 9 (n + 1) strategy, so there will be three different 9 (n + 1) strategies that H may pass control to at any stage at which it is passed control.

1.6. Now that we have described each of the strategy types in some detail let us consider a simple example with the aim of illustrating how these strategies may be combined in order to approximate a set of minimal degree.

So suppose that at stage s of the construction control is passed to the strategy H = 9[t, 0](n) for the first time, where t = {Trl,0, Tro,o, Trl,2}. At stage s we shall have that 0 extends a leaf of each of the trees in t and that no axioms have been enumerated on proper extensions of 4. At stage s, and at all subsequent stages at which H is passed control until it is declared to be in state 0, H will do nothing more than pass control to the strategy H' = F[t, 3, 0](n).

At stage s the strategy H' will pass control to the strategy H" = F'[t', k, 0](n) where t' = { Tr-1,o, Tro,o} and k is the number of strings required in the initial environment of the strategy T[t, 3, 0](n). The strategy H" will pass control to the strategy H"' = 2[t", k', q](n) where k' is the number of strings required in the initial environment of the strategy T[t', k, 0](n) and t" = { Tr-1,o}. The strategy H'" will then choose k' incompatible extensions of A, enumerate these strings into

Tr-l,o, declare 4[t', k, k](n) to be in state 0 and 'deliver' these strings to that

strategy. We shall declare H'" to be successful and will terminate stage s activity.

DEFINITION 1.6. We let z be a (computable) injection from the strategies that are passed control during the course of the construction into cw.



948 ANDREW LEWIS

At all subsequent stages at which H is passed control, until H" is declared successful, it will pass control to H' which will pass control to H". If H" is never declared successful and is passed control at an infinite number of stages then (using the non-computability of A) we shall be able to show that there is a stage after which it always passes control to the same T(n) strategy which eventually always passes control to the same W(n) strategy H1 which, in turn, always eventually puts Xs, through the same string at any stage s' at which it is passed control and thus passes control to the same 9(n + 1) strategy Hi*. H1 will ensure that FAx (n) is correctly defined. How should we define t (H*)1 If the strategy H1 ever finds the Po splitting for which it is searching then it will be declared successful and H1 and H* will never be passed control again. Thus at any stage at which H* is passed control this fact in itself ensures that such a splitting has not been found. This strategy therefore assumes that Tro,o will be finite. As we have said before, however, we must continue to construct Trl,o in such a way that it may be regarded as a 'subtree' of Tro,o so at this point on the tree of strategies we must introduce a new candidate for TrI, a tree for which we do not need to satisfy this condition. We therefore define t(Hr) = {Tr-,o, Trl,z(H*)}. The role of the function z, then, is just to ensure that above Hj we have a different candidate for a T1 splitting tree than anywhere else on the tree of strategies.

Suppose that H" is eventually declared successful. Then at every subsequent stage at which it is passed control until it is declared to be in state 0, H will pass control to H'. If H' is never declared successful and is passed control at an infinite number of stages then we shall be able to show that there is a stage after which control is always passed to the same dW(n) strategy above H', H2 say. H2 will eventually always put Xs, through the same string and thus pass control to the same 0 (n + 1) strategy H2 at any stage s' at which it is passed control. H2 will ensure that FA,X (n) is correctly defined. How should we define t(H)1 If H2 is searching for a 0 splitting then we should define t(H2) in a similar way to that in which we defined t (Hr) previously. So suppose that H2 is searching for a P1 splitting. If H2 ever finds such a splitting then it will be declared successful and H2 and Hf will never be passed control again. At any stage at which Hf is passed control this fact in itself therefore ensures that such a splitting has not been found and this strategy therefore assumes that Trl,o will be finite. We therefore define t(H*) = { Tr-1,o, Tro,o, Tr2,z(H2)}.

So finally, then, suppose that H' is eventually declared successful. At any subsequent stage at which H is passed control it will pass control to 9(n + 1) strategies. For any 2 (n + 1) strategy H3 that H passes control to we define t(H3) = {Tr-1,o, Tro,o, Trl,2, Tr2,z(H3)}. To sum up let us briefly consider the general case of which the previous discussion was an example. 9(n) strategies in state -1 will pass control to F(n) strategies which will pass control to F'(n) and 1.P(n) strategies until declared successful. W(n) strategies pass control to M'(n) strategies until declared successful while d' (n) strategies pass control to 9 (n + 1) strategies until declared successful. Once a ', ., or

M strategy is declared successful it is never passed control again. Once a .(n) strategy is declared to be in state 0 it passes control to (n + 1) strategies. Any

'chain' of (n) strategies such that each passes control to the next will be finite in length. Each '(n) strategy will be 'finitary'. Thus each (n) strategy H and the ', 9, and ' strategies lying above it and below (n + 1) strategies should be




viewed as a (finite) 'cluster', which we shall refer to as the 'H cluster of strategies'. If H is a D(n) strategy then the strategies in the H cluster will enumerate strings into the trees in t(H) and if H is passed control at an infinite number of stages then the strategies in the H cluster will ensure that FA, (n) is correctly defined and will eventually always pass control to the same 9 (n + 1) strategy.

There are two distinguishable ways in which this may happen. It may be the case that there is a stage after which control is always passed to the same Vs(n) strategy and after which this strategy always puts X, through the same string at any stage s. When any d(n) strategy H' is passed control at a stage s and puts X, through a string r it will pass control to a strategy H" = 9[t, r](n + 1) where we define t as follows. Let t(H') = {Trl,o, Trj,i,,..., Trim,im} with jl < ... < im. Then t = (t(H') - {Trjm,im)) U

{Trjm+1,z(H")}. The second possibility is that H may be declared to be in state 0. At any subsequent stage s at which H puts Xs through a string r, it will pass control to a strategy H' = [t, z](n + 1) where we define t as follows. Let t(H) =

{Tr-l,o, Tri,i,-...., Trim,im} with ji < .. < jm. Then t = t(H) U { Trj,+I,z(H,) . At this point it is probably best that we begin to articulate the specifics of the

construction. From now on it will be convenient to adopt the convention that if Tr is empty then a = 0 is a leaf of Tr.

12. The construction. In order that this section should be self contained we shall restate definitions when required.

DEFINITION 2.1. At any stage of the construction, for any strings r, z and for n E co we say that a E y(z, n) if there are initial segments of a and z, a' and r'

respectively, for which we have already enumerated the axiom F1'O"' (n) = c for

some c.

DEFINITION 2.2. If a strategy H 'puts Xs through' a string z then we insist that X, be an extension of T. When stage s activity is terminated we define Xs to be the longest string that it was put through during stage s. If X, is not put through any strings during stage s then we define Xs = 0.

DEFINITION 2.3. Given j > 0 and z E 2<" let n be the least such that I' (n) T. We say that T is of length n.

DEFINITION 2.4. By a k-fold Ij splitting of length at least I we mean k strings

rl ...7.. k whose images ' (1 < i < k) are (1) of length at least I and (2) pairwise incompatible-for 1 < i < i' < k if the lengths of P;' and Y'i are nl and n2 respectively then there exists n3 < ni, n2 such that I'1 (n3)

- YT" (n3).

DEFINITION 2.5. We let z be a computable injection from the set of strategies that are passed control during the course of the construction into co.

2.1. The W[t,k, , l, ti,2, 3](n) strategy. Here k > 1, n > 0, 4, 'l, "2, "3

are binary strings such that b c

'i (for 1 < i < 3) and the

"i are pairwise incompatible

while t is a finite set oftrees, t = {Tr-1,o, Tr1,i, ,..., Tr,,i m} say, withjl < 1

. < jm,

m > 1. Let H = d [t, k, 4, Tzl, 2, T3](n). At every stage at which H is passed control we will have defined the value 1(H) E co. We call 4, ri, r2, Z3 the 'base strings' for H.



950 ANDREW LEWIS

The strategy searches for a k-fold YTjm splitting 'above' Zl, T2 or z3 (i.e., such that there exists 1 < i < 3 every string in the splitting extends zi) of length at least 1(H) and lying on Trjim ,im_ .

It also enumerates axioms of the form FT"'i (n) = c (1 < i < 3). As we describe each of the strategies we shall specify initial and active environments. These specifications do not play any active role in the construction- our intention in stating them is simply to aid the reader's intuition.

2.1.1. The initial environment. Suppose the strategy is first passed control at stage s. Define Pi = y(zi, n) for each 1 < i < 3. The following conditions will be satisfied.

(1) As V 7(4, n). (2) For i 7 j, Pi n Pj = 7(, n).. (3) No axioms have been enumerated on proper extensions of ri for any i, i.e., for

1 < i < 3 we have not enumerated any axioms of the form FT'(n') = c such that r properly extends zi (and for any a E 2<w, n' E w, cG {0, 1}).

(4) 0 will extend a leaf of Trji,iM, 1' say. (5) Suppose Tr E t - { Trjimi1}. Each zi extends a different leaf of Tr, zi(Tr) say.

Each zi is a string that we have enumerated into Trim j,im_- (let us agree that

jo = - 1, io = 0). 2.1.2. The active environment. Every stage s at which the strategy is passed con-

trol the following conditions will be satisfied.

(1) As Vy (40,n). (2) No F(n) axioms have been enumerated on strings compatible with any zi

since the strategy was first passed control except by this strategy, i.e., no other strategy has since enumerated an axiom of the form F", (n) = c for r

compatible with some ri (and some ar E 2<1, c E {0, 1}).

(3) b', as defined as in (4) of 2.1.1 above, is still a leaf of Trim m.

(4) Suppose Tr e t - {Trji,im} and for 1 < i < 3 let zi(Tr) be defined as in (5) of 2.1.1. If any string r has been enumerated into Tr extending Ti(Tr) then T extends zi and was enumerated into this tree by a strategy above H on the tree of strategies.

2.1.3. The instructions. Suppose the strategy is passed control at stage s of the construction. We wish to enumerate axioms of the form F"'i (n) = c where a is of the same length as zi. So if I is the length of the longest of the ri and s < 1 then terminate stage s activity for the construction (recall that As is of length s). Otherwise we perform the following steps: (1) If via some exhaustive procedure (which clearly there is no need to specify)

we have not yet found a k-fold I]jm splitting such that there exists 1 < i < 3

every string in the splitting extends zi, of length at least 1(H) (noting that this value may have changed since the last stage at which H was passed control) and lying on Trjm_,i~_,, then proceed immediately to the next step. Otherwise we declare the strategy to be successful and terminate stage s activity for the construction, performing no more instructions at this stage.

(2) Perform another step in the 'exhaustive procedure'. (3) If Ks (n) = 0 then if:

(a) A~ C P1 put Xs through T2.

(b) As E P2 put Xs through z3.




(c) As G P3 then put Xs through T2. (d) A, Ui Pi then put X, through Z2.

(4) If Ks (n) = 1 then if: (a) As E P1 put X, through Z3. (b) As E P2 U P3 then put X, through 1l. (c) A, V Ui Pi then put X, through Z3.

(5) Suppose that we have put X, through ri. Then if FAst1'(n) is not already defined enumerate the axiom F1'j (n) = Ks (n) where a is the restriction of As to the length of zi.

(6) Suppose that we have put Xs through ri. H passes control to the strategy H' = 9[t', zi](n + 1) where t' = (t - {Trjm,im}) U {

Trj,+l,z(H'). DEFINITION 2.6. In the course of discussing a , F or a 9 strategy H every

string z under consideration will extend one of the base strings for H. We let OH (z) designate this string.

2.2. The W[t, k, 1l,...

r](n) strategy. Here k, r > 1, n > 0, q1,....r

r 2<` and t is a finite set of trees, t = { Tr_ 1,o,

Trjj1,i ..., Tryj,im } say, with jl < . < jm,

m > 1. We shall also have that r divides k. Let H = W[t, k, 1l,...,~9r](n).

We call 01, ... br the 'base strings' for H. The strategy aims to produce a k-fold

]jm splitting,

zli...,.Tk say, that we can enumerate into Trjm,im and such that for i # j,

S(Zri, n) n y ( i, n) = (H (Ti), n) n Y(H(Zj), n).

2.2.1. The initial environment. If H is first passed control at stage s the following conditions will be satisfied.

(1) H will be provided with k groups of (3Ei11 i) + 3 strings with all the strings in k/r of these groups extending each i (so we have an equal number of incompatible strings extending each ji) and such that for distinct z, r', y (z, n)n y(Z', n) = Y (OH(Z), n) 7y (OH(z'), n). These strings will have been 'delivered' to H.

(2) As 4 Ur= (di,n). (3) No axioms have been enumerated on proper extensions of any strings delivered

to H. (4) For 1 < i < r, 4i extends a leaf of Triji, q say. (5) Suppose Tr E t - { Trim,im }. Each of the strings that has been delivered to H

is a string that has been enumerated into Trjmi ,im _, (we agree that jo = -1, io = 0). Each such string r extends a different leaf of Tr, r(Tr) say.

2.2.2. The active environment. Every stage s at which the strategy is passed control the following conditions will be satisfied.

(1) Except in the case of strings which are extended by escape routes for H no F(n) axioms have been enumerated on strings compatible with those delivered to H since H was first passed control, except by the d' strategies which H has passed control to. In the case of the strings which are escape routes such axioms have only been enumerated on proper extensions of the string.

(2) As Ui=

1(i, n). (3) If the strategy is in state k' > 1 then As 4 C(H, k') (to be defined subse-

quently). (4) For each 1 a i < r, as defined in (4) of 2.2.1 is still a leaf of TrJm,im.



952 ANDREW LEWIS

(5) Suppose Tr E t - {Trjm,im}. For each string z delivered to H let z(Tr) be defined as in (5) of 2.2.1. If any strings have been enumerated into Tr extending r(Tr) these strings extend z.

2.2.3. The initial instructions. Divide the set of strings delivered to H into triples so that every string in a triple extends the same ji. Then divide these into k groups of (k-l1 i) + 1 triples so that every triple in the same group extends the same 0i. Let the variable ,u range over this set of triples and for each u let us agree that 1 = {Zf,,1,Z p,2, Tp,3}

and that, for any n', y(u, n') = U y (ry,i, n'). Call the groups of triples fiH ... itkH'. Proceed immediately to carry out the instructions for the strategy in state 0 (H will already have been declared to be in state 0 by another strategy) .

2.2.4. The instructions for the strategy in state 0. Suppose that we are stage s.

(1) If one of the V strategies that H has passed control to has been declared successful then proceed as follows. Declare the strategy to be in state 1. Label the string above which a splitting was found 4r and suppose that l' E /t. Relabel if necessary so that yG fl H. We wish to form the set of escape routes

ptfi The first subfix indicates that this is a set of escape routes defined for H while in state 1. The second subfix indicates that it is the set of escape routes 'associated with' zH . Choose l e llH other than p and discard two strings from this triple. Find an 'escape route extension' of the remaining string i.e., a string extending a leaf in each of the trees in t - { Trjm,im } and long enough such that no axioms have been enumerated on proper extensions of the string, to give us our first string in p/1 . Choose pi from each fl', i > 1, discard two strings from this triple and find an escape route extension of the remaining string to give us another string in

pH1,1, so that this set consists of k escape

routes. From each p' E fl' other than pi, choose one string and discard the other two--ensuring that the string rt H is not discarded. From the set of

lk(k - 1) non-discarded strings from triples in flH other than 1 i, take 4 and any k - 2 others, discarding the rest. Let this set of k - 1 strings be called R(H, 1). Define C (H, 1) = { f : 3r E R(H, 1), a E (z, n) }. Terminate stage s activity for the construction.

It may initially seem strange that having required so many strings in the initial environment we then proceed to discard a large number of them when the M strategy enters a new state. The point is that we cannot predict the order in which the .s strategies will find splittings. If a given group of triples is the first above which we find a splitting then most of the strings in this group of triples will have proved to be unnecessary, but not so if it is the last.

(2) Otherwise, if A E y(u, n) we transfer control to the strategy H' = ds[t, k, ckH (Tfl1), ,ip,1,

1p,2, z2,3](n)-which we call the '.W(n) strategy for y'. If 1(H')

has not yet been defined then choose it to be larger than any number previously mentioned in the construction. If A Is y (,a, n) for any triple u then just choose one and proceed similarly.

2.2.5. The instructions for the strategy in state 0 < k' < k. Suppose that we are stage s. The following conditions will be satisfied, together with those in 2.2.2. If we have not yet discarded strings from some ,z we shall refer to / as a 'remaining triple'.




(1) For 1 < i < k' we will have defined a string H

and a set R(H, i). (2) We will have defined k' sets of non-discarded escape routes, pkl, . Pk'H,k".

Each set contains (k - k' + 1) strings. (3) For distinct z, v' each of which is either in

Ui-1 R(H, i) or is a non-discarded

escape route, y7(, n) n y(-r', n) = y (OH(Z), n) 7 (H(Tr'), n). The same is true for distinct z, z' each of which is either in U-1 R(H, i) or is in remaining u, unless 3u (z, z' e cu).

We perform the following instructions.

(1) If one of the d strategies which H has passed control to has been declared successful since H was declared to be in state k' then proceed as follows. Declare the strategy to be in state k' + 1. If k' + 1 - k then label the string above which we have just found a splitting as 4kH, declare H to be successful and terminate stage s activity for the construction. Otherwise (given that one of the d' strategies has been successful since H was declared to be in state k'): (a) Discard all escape routes, and the strings delivered to H that they extend. (b) Label the string above which we have just found a splittingTI ,H and

suppose that TkH+l E . Relabel if necessary so that p E Pf,+1 (and so that each ZiH 1 < i < k' is from a triple in Pfl). Choose remaining

Uk'+l E fik'+l other than jp, and discard two strings from this triple. Find an 'escape route extension' of the remaining string to give us the first escape route in Pk'+l,k'+1. Choose some remaining pi from each PiH, i > k' + 1, discard two strings from this triple and find an escape route extension of the remaining string to give us another escape route in pH'+1,k'+ so that this set consists of k - k' escape routes. From each remaining /' E fk+1 choose one string and discard the other two--ensuring that the string

k,+1 is not discarded. From the set of non-discarded strings from triples in k'+l other than Uk'+l, take rTk+ and any (k-k' -2) others, discarding

the rest. Let this set of (k - k' - 1) strings be called R(H, k' + 1). (c) For each 1 < i < k', in turn, define a set of escape routes 1H i as

follows. Remove one string from R(H, i) other than TH and find an escape route extension to give us the first escape route in pkH+1,i. For each i' > k' + 1 choose some remaining y and from that triple choose one string, discarding the other two. Find an escape route extension of the remaining string to give us another escape route in

p+1,i so that this set

consists of k - k' strings. (So we define a different set of escape routes for every splitting that we have found).

(d) Define C (H, k' + 1) = {I :3- e Ul + R(H, i), a E y(-, n) }. Terminate stage s activity for the construction.

(2) Otherwise if it is the case that A e ~ (y(,, n) for some remaining u then pass control to the Y(n) strategy for u, H'. If 1(H') has not been redefined since H was declared to be in state k' then redefine it to be larger than any number previously mentioned in the construction. If As ( 7 (/,

n) for any remaining

/u then just choose remaining u and proceed similarly.

2.3. The F'[t, k, q1,..., ,](n) strategy. Here k, r > 1, n 2 0, b1,..., 1r E 2<" and t is a finite set of trees, t = { Tr-q,o,

Trj,, ..., Trjm,i } say, with jl <

" < jm.



954 ANDREW LEWIS

We also have that r divides k. Let H = -[t, k, q1,.... 1r](n). We call 0 1,..., 1r the 'base strings' for H.

2.3.1. The initial environment. If the strategy is first passed control at stage s the following conditions will be satisfied.

(1) As i U =1 y(, n). (2) No axioms have been enumerated on proper extensions of qi for any i. (3) Suppose Tr E t. Each 4i extends a different leaf of Tr, qi(Tr) say. 2.3.2. The active environment. Suppose m > 1. Every stage s at which H is

passed control the following conditions will be satisfied.

(1) As V U=,l I(ij, n). (2) No F(n) axioms have been enumerated on strings compatible with 1, ..., ,

since the strategy was first passed control except by strategies above H. (3) For each Tr E t and for each 1 < i < r let qi(Tr) be defined as in 2.3.1. Then

4i(Trj,,im,) is still a leaf of Trj,imM. If Tr E t - { TrIjmM} and strings have been

enumerated into Tr extending Oi(Tr) then these strings extend q; and have been enumerated into Tr by strategies above H.

The F strategies with t = { Tr-1,o} are very simple. The first stage at which H is passed control it will choose k/r incompatible extensions of each ji, enumerate these strings into Tr_1,0, deliver them to the strategy H' that we have specified H is intended to provide an environment for, declare H' to be in state 0 and terminate stage s activity for the construction. So suppose that m > 1.

2.3.3. The instructions. Suppose we are at stage s. If any of the w strategies that H has passed control to has been declared successful then we declare H successful and proceed immediately to carry out the 'instructions upon being declared successful' at stage s.

Otherwise, if no such strategy has been declared successful, we perform the following iteration in order to decide which strategy to pass control to. As soon as control is passed to another strategy stage s activity for H is terminated.

STEP 1. Check to see whether H1 = 6,[t, k, 1, ..., 1](n) has been declared to be in state 0-we say that H1 is 'checked for use' at stage s.

If not. Pass control to the strategy H1 = -'[t',kl,1l,...,1r](n)

where t' = t - { Trimi, } and ki is the number of strings required in the initial environment of

H1, i.e., ki = 3k(1k(k - 1) + 1). We specify that H1 is intended to provide an environment for H1.

If so. If H1 is in state 0 then pass control to it. Otherwise suppose that H1 is in state k1 > 0 (here, and everywhere in the description of the '2 strategy, the superfix is used as a counter rather than to indicate exponentiation). If As V C(HI, k') then pass control to H1. Otherwise r E Uki=l R(H1, i), (As E y(,r, n)). There can only be one such z (of course we shall have to prove that this is the case). Suppose this - is in R(HI, j'). This decides the set of escape routes that we shall use. Let r'= (k-k'+11)andletpL '1= { .z.r, }. DefineH2 =~[t, r', z... rl](n).

Proceed to step two. STEP 2. Check to see whether H2 has been declared to be in state 0. If not. Pass control to the strategy H2 - '[t',k2, l...,b,](n) where t',

,...,..rl, are as in Step 1 and k2 is the number of strings required in the initial




environment of H2. We specify that H2 is intended to provide an environment for H2

If so. If H2 is in state 0 then pass control to it. Otherwise suppose that it is in state k2 > 0. If A, C(H2, k2) then pass control to H2. Otherwise E3 E

Uk=l R(H2, i)(As E y(r, n)). There can only be one such T. Suppose this r

is in R(H2, j2). Let r2 (r1 - k2 + 1) and let pHk2 -2 }. Define H3 =

_ _ p kz,j2

, r

. .W[t, r2, Z ... r2](n). Proceed to the next step. STEP y. At step y - 1, having found that we could not pass control to the strategy

H,_1 we will have defined a strategy Hy = W[t, ry -, z 1,... ](n). Check to-1 see whether Hy has been declared to be in state 0.

If not. Pass control to the strategy Hy = FY[t', ky, Z11 ...... r,](n), where t' is as in Step 1 and ky is the number of strings required in the initial environment of Hy. We specify that Hy is intended to provide an environment for H,.

If so. If Hy is in state 0 then pass control to it. Otherwise suppose that it is in state ky > 0. If As V C(Hy, ky) then pass control to Hy. Otherwise 3T E U1k

R(Hy, i) (As E 7y(, n)). There can only be one such

r. Suppose this r is

in R(H,,jy). Let ry (rY - ky + 1) and let p, {. }. Define

Hy+i - [t, ry, z-r ...,1ZYy](n). Proceed to the next step. So at each step in the iteration we check to see whether a certain T strategy has

previously been declared to be in state 0. If not then we pass control to a F strategy in order to provide the strings required in the initial environment of this ' strategy. If so, then we check to see whether there is an environment such that we can pass control to it at the present stage. If we cannot then we decide which set of escape routes to use. The T strategy above these escape routes will look for the same number of splittings as the number of strings in this set of escape routes. Given that at any stage of the construction H can only have passed control to a finite number of different T strategies it is clear that this iteration must come to an end.

Although every strategy is different the main difference between this proof and the proof sketch previously put forward is as follows. In the proof sketch control is passed to a succession of what are the equivalent of the M strategies and it is hoped that one such strategy will eventually produce the k-fold splitting required. Because, however we choose them, the escape routes already have axioms defined on them this will not necessarily happen. The key to a successful proof is to find a method of organizing the T strategies in such a way that if enough splittings are found then we can choose from amongst the splittings that they have collectively produced, a splitting of the type required.

2.3.4. The instructions upon being declared successful. One of the M strategies Ho that H has passed control to has been declared successful. First we shall enumerate a set of splittings Q.

STEP 0. The first strategy that we shall consider is H0. Take each of the splittings that have been found by the at strategies that Ho passes control to and enumerate these splittings into Q. If Ho was the first 9 strategy ever passed control by H then our enumeration of Q is completed. Otherwise proceed to the next step.

STEP y. Suppose that Hy_1 was a strategy with Pk,k2 as its set of base strings, for some 9 strategy H' that H has passed control to and for some k', k2 e co. Define



956 ANDREW LEWIS

H, = H'. If k' = 1 then proceed to the next step, unless Hy is the first W strategy that H ever passed control to in which case (presuming k' = 1) our enumeration of Q is completed. Otherwise we have defined strings zy

..... 'kH.

Take the splittings that have been found by the A strategies passed control by Hy above each of these

strings, except that which was found above zk', and enumerate these splittings into D. If Hy was the first . strategy passed control by H then our enumeration of f2 is completed, otherwise proceed to the next step.

Label the splittings in Q as wi,..., 0k according to the order in which they were found, so that wk was the first and wi) the last. Take a string zl from wi1 and another string

"2 from 0)2 such that T and VP2 are incompatible. Take a string T3 from

Im m

W03 such that T'11, V.2

and VP3 are pairwise incompatible, and so on. For each 1 < i < k take an extension ' of -ri long enough that it it extends a leaf in each of the trees in t - { Trjm,i }, longer than any strings in the trees in t and long enough that no axioms have been enumerated on proper extensions of the string. Enumerate

T, .. ., T' into Trj,,im, deliver these strings to H' which is the strategy which we have specified H is intended to provide an environment for, declare H' to be in state 0 and terminate stage s activity for the construction.

2.4. The 9[t, q](n) strategy. Here n > 0, 4 is a finite binary string and t is a finite set of trees, t = {Trl,0, TrJ, ,,..., Trim,im} say with jl < ... < j,. Let H = 9[t, 0](n). We call 4 the 'base string' for H.

2.4.1. The initial environment. If the strategy is first passed control at stage s the following conditions will be satisfied.

(1) As V 7y(q, n). (2) No axioms have been enumerated on proper extensions of 4. (3) Suppose Tr E t. Then 0 extends a leaf of Tr, q(Tr) say. 2.4.2. The active environment. Every stage s at which H is passed control the

following conditions will be satisfied.

(1) Asy y(1(,n).

(2) No F(n) axioms have been enumerated on strings compatible with 4 since the strategy was first passed control except by strategies above H.

(3) For each Tr E t let 0$(Tr) be defined as in 2.4.1. If Tr E t and strings have been enumerated into Tr extending 0 (Tr) then these strings extend 0 and have been enumerated into Tr by strategies above H.

The strategy is initially considered to be in state -1. 2.4.3. The instructions for the strategy in state -1. If H is passed control in state

-1 at stage s it passes control to the strategy F [t, 3, 4](n), which we specify is intended to provide an environment for H.

2.4.4. The instructions for the strategy in state 0. The first stage at which H is

passed control in state 0 it is provided with three strings TI, z2, z3-these strings will have been delivered to H. At this stage define Pi = y (ri, n) (so that Pi, unlike y (zi, n) is fixed and will not change at subsequent stages). Let I be the length of the longest ri.

At any stage s we now proceed as follows:

(1) If I > s then terminate stage s activity for the construction. (2) If Ks (n) = 0 then if:




(a) As E Pi put X, through T2. (b) As E P2 put X, through T3. (c) As E P3 put X, through Z2. (d) As V U, Pi put Xs through T2.

(3) If Ks (n) = 1 then if: (a) As c P1 put Xs through T3. (b) As E P2 U P3 put Xs through rl. (c) As V Ui Pi put Xs through T3.

(4) Suppose that we have put X, through ri. If FAs,Tj (n) is not yet defined then enumerate the axiom F""' (n) = Ks (n) where a is the restriction of As to the length of Ti.

(5) Suppose that we have put X, through z,. Pass control to the strategy H' = 9[t', Ti](n + 1) where t' = t U

{Trjm+l,z(H')}. 2.5. The tree of strategies. In order to complete our description of the tree of strategies we need only add the following. To the bottom node of the tree of strategies we assign the strategy H = - [t, 0](0), where t = { Tr-l,o, Tro,z(H)}. At the beginning of each stage s, control is first passed to this strategy.

13. The verification. When an d' strategy is passed control at a stage s and does not terminate stage s activity it is instructed to put Xs through one of the base strings for the strategy. That the instructions specified are sufficient to isolate a single string to put Xs through at any stage s, however, requires proof. There are many other occassions, too, where it must be proved that the construction as specified is actually well defined. Strictly speaking, we cannot prove anything about some stage s of the construction until we know that the instructions prior to this point are well defined. In order that we may avoid the task of a massive induction, it is therefore convenient to assume that in the event of receiving improperly defined or impossible instructions the construction simply resorts to permanent termination. Of course, it will turn out that such a provision is unnecessary but proceeding in this manner means that we can state and prove the required lemmas one at a time. While proving lemma 3.5, for example, we are able to tacitly assume that the result of Lemma 3.7 has always held prior to the point at which Ho enumerates the splitting into Tr, since the construction would have permanently terminated otherwise.

DEFINITION 3.1. If H is a strategy T[t', ... ](.) then t(H) = t'. Suppose that for some m > 1 and ji < . < jm, t' =

{Tr-,0o, Trj1,i~,..., Trj,im }. If T = 0 then

t*(H) = t'. Otherwise t*(H) = t' - { Trj,imI}. LEMMA 3.1. At each stage s afinite number of strategies are passed control. PROOF. It follows immediately by induction on s that, at any given stage s, only

a finite number of strategies may be passed control before control is passed to a '

strategy H such that t(H) = { Tr_1,o}, whereupon stage s activity is terminated. Of course other strategies may also terminate stage s activity. -]

The following terminology was established during the description of the construction and the introduction, but is worth isolating now.

DEFINITION 3.2. When a & strategy Ho checks to see whether a . strategy H1 has yet been declared to be in state 0 we say that H1 is 'checked for use', or more specifically that 'Ho checks H1 for use'.



958 ANDREW LEWIS

DEFINITION 3.3. In the context of discussing a T strategy Ho suppose that we have defined a set of escape routes p. We refer to the V strategy H1 with base strings defined to be those in p, such that t(Hi) = t(Ho) and which searches for a k-fold splitting, where k is the number of strings in p, as the &. strategy 'above' p.

For much of the verification it is convenient to consider an ordering on the strategies which is different from that given by the tree of strategies. The 'a ordering' which we shall define below is basically that which would be given by the tree of strategies if it were is the case that a W strategy H was responsible for passing control to any O strategy above a set of escape routes defined for H, and to any V strategy intended to provide an environment for H.

DEFINITION 3.4. We define the a ordering by recursion. Suppose Ho is a . strategy and let H1 be the F strategy which passes control to Ho. We say that H2 is a-below Ho if either:

(a) H2 = Ho, (b) H2 is a-below H1, or (c) H2 is one of the T strategies that H1 checks for use before Ho at any stage at

which Ho is passed control.

Suppose Ho is a F strategy and let H1 be the strategy that Ho is intended to provide an environment for. We say that H2 is a-below Ho if either H2 = HO or H2 is a-below H1. Finally suppose that Ho is a 9 or an W strategy and let H1 be the strategy which passes control to Ho. We say that H2 is a-below Ho if H2 = Ho or H2 is a-below H1.

DEFINITION 3.5. We say that Ho is a-above H1 if H1 is a-below Ho. We say that Ho is strictly a-below/ a-above H1 if Ho is a-below/a-above H1 and Ho : H1. We say that Ho is the a predecessor of H1 if Ho is strictly a-below H1 and for all strategies H2 strictly a-below H1, Ho is a-above H2.

It is convenient to assume that QV strategies are in state 0 until declared successful, that M strategies are in state -1 I until declared to be in state 0, and that F strategies are in state -1 until declared successful.

DEFINITION 3.6. We say that Ho is an a successor of H1 if H1 is the a predecessor of H--if H1 is an d strategy or a F strategy we also say that Ho is an a successor of H1 while in state k, for any k E wo, in this case. Suppose H1 is a 9 strategy. We say that Ho is an a successor of H1 while in state k E co if H1 is declared to be in state k, Ho is an a successor of HI and Ho is a F strategy if k = -1, Ho is a 9 strategy if k = 0. Finally suppose H1 is a M strategy. We say that Ho is an a successor of HI while in state k E o if Ho is an a successor of H1, H1 is declared to be in state k, and while H1 is in this state either:

(a) k > 0 and Ho is an d strategy, none of the base strings for which have been discarded by H1.

(b) k > 0 and Ho is a 9 strategy above a set of escape routes we have defined for H1 and which are not discarded.

(c) k = -1 and Ho is a ' strategy.

DEFINITION 3.7. We define a strategy Ho to be active from the point at which it is first passed control or checked for use until the point at which a strategy strictly




a-below Ho, H1 say, is declared to be in a new state k such that none of the a successors of H1 while in state k are a-below Ho0.

DEFINITION 3.8. If H is a M, V, or 9 strategy then the a base strings for H are just the base strings for H. Suppose H is an V' strategy and that the base strings for H are 0, 71, z2, T3, so that the ri are pairwise incompatible extensions of 5. The a base strings for H are Ti, T2, z3.

LEMMA 3.2. Suppose t(Ho) = { Tr-l,o, Trj,,i,..., Trimim M 1, jl < J i < < jm, jo = -1, io = 0 and that 0 < r' < r. If H1 enumerates a splitting into Trjirir then each string in the splitting extends a different leaf of Trj ,,,i r

PROOF. Since the construction will permanently terminate if we are not allowed to proceed as such, we may assume that each string in the splitting extends a leaf of Trj,,ir,,. Let t(H) = {Tr-,0o, Trji,...,Tr ,i,

} with j < < , so that

jr = j', and i, = in,.

For fixed r' the result follows immediately by induction on r since r = m', jir-1- andir-1

= i-1.

LEMMA 3.3. Suppose that Ho is active when H1 which is not a-above or a-below Ho is passed control and that Tr c t*(Ho) n t (HI). Let 01,...,kr be the a base strings for Ho and let ....., be the a base stringsfor H1. For 1 < i < r and 1 < i' < r', 1i and 0', extend incompatible strings on Tr.

PROOF. Let H2 be the a-highest of all the strategies a-below Ho and H1. Since Tr E t*(Ho) n t(H1) it must be the case that Tr E t*(H2). Suppose that for jl < ... < jm, t*(H2) -= {Tr-1,, Tr-j,il..., Trj,,im}. If 1 < i < r and 1 < i' < r', qi and 0', extend incompatible strings on Trj,,i, which H2 was provided with upon being declared to be in state 0 (let us suppose that an V strategy is declared to be in state 0 upon being passed control for the first time), and by Lemma 3.2 therefore extend incompatible strings on Tr.

We restate the convention that if Tr is empty then 0 is regarded as a leaf of Tr. LEMMA 3.4. When a strategy Ho is first passed control or checked for use:

(a) the a base strings for Ho will extend leaves of each of the trees Tr E t(Ho). (b) no axioms will have been enumerated on strings properly extending the a base

strings for Ho. PROOF. The proof is by induction on the point of the construction at which Ho is

first passed control or checked for use. Suppose that H1 is the a predecessor of Ho and that Ho is first passed control or checked for use at stage s. If H1 is in state -1 I then the result follows immediately, by the induction hypothesis. So suppose that H1 is in state k > 0. There are three cases to consider.

CASE 1. Tr E t*(HI). Suppose first that H1 is an V or a 9 strategy, or that H1 is a M strategy and Ho is an s strategy. When H1 was declared to be in state 0, b) above and also a) above as regards Tr were satisfied. If H1 is a I or a 9 strategy this follows since we are working under the assumption that the construction would permanently terminate otherwise. If H1 is an -

strategy then this follows by the induction hypothesis, since (as in the proof of Lemma 3.3) we assume that the stage at which H1 was declared to be in state 0 was the stage at which it was first passed control. Since that point in the construction it follows by Lemma 3.3. that the only strategies which could have violated these conditions (where we consider a) as regards Tr only) are those a-above H1. Suppose H2 is a-above H1 and is passed



960 ANDREW LEWIS

control or checked for use before Ho, after the point of the construction when H1 was declared to be in state 0. Let 1, ..., be the a base strings for Ho and let

,..., a, be the a base strings for H2. For 1 < i < r and 1 < i' < r', qi and

jb, extend incompatible strings which H1 was provided with upon being declared to be in state 0. So suppose that H1 is a 97 strategy and Ho is a M strategy also, and suppose that H1 is in state k when Ho is first checked for use. Repeat the relevant parts of the argument above, replacing 'H1 was declared to be in state 0' with 'HI was declared to be in state k'. Clearly we need not consider (b) in the cases that follow, since certainly Tr-1,0 E t*(Hi).

CASE 2. Tr E t*(Ho) - t*(Hi). Then (a), as regards Tr, follows immediately since the only strategies that can enumerate strings into Tr are those a-above Ho0.

CASE 3. Tr V t*(Ho) and Tr V t*(Hi). Let H2 be the strategy a-highest of all those a-below Ho such that Tr E t*(H2). When H2 was first passed control or checked for use, the base strings for this strategy extended leaves of Tr, by the induction hypothesis. By Lemma 3.3 it follows that the only strategies which could subsequently have violated this condition are those a-above H2. But the fact that Ho is passed control or checked for use at stage s means that such strategies have not yet enumerated any strings into Tr.

LEMMA 3.5. Suppose that Ho enumerates a splitting into Tr. Each string in the

splitting extends what was (prior to this enumeration) a leaf of Tr.

PROOF. Let H1 be the strategy which Ho declares to be in state 0 upon being declared successful. When H1 was first passed control or checked for use, the a base strings for this strategy each extended a leaf of Tr. By Lemma 3.3 it follows that the only strategies which could have subsequently violated this condition (prior to the enumeration in question) are those a-above H1. No such strategy has enumerated strings into Tr prior to stage s, otherwise Ho would not be passed control at this stage.

LEMMA 3.6. Let t(Ho) = {FTr-,o, Trj,,i, ...., Trjm,im} with ji < . < jm, m M> 1, andsuppose that at stage so, Ho enumerates a splitting zl

...., zk into Trjm,im. Suppose

given r < m. By Lemma 3.2 each string zi extends a leaf -i of Tr Jr ,i such that for 1 < i < i' < k we have z f z,, when this enumeration takes place. Suppose further that at stage sl > so a strategy H1 enumerates a string z into Trjr,ir extending z! and that, subsequent to this enumeration, there is at least one active strategy H2 with

Tr,imi. E t(H2). Then z extends zi. PROOF. For 1 < i < i' < k we have that

1i and r, are incompatible. It therefore

follows by Lemma 3.5 that if, at any point of the construction after Ho enumerates the splitting into Trjiim, a string compatible with Ti extends a leaf of Trimim the string extends zi. Let H2 be the a-highest of all the strategies a-below Ho such that Trjim, im t(H2) and which is active after H1 enumerates z into Trjrir. Then

Trjm,im E t*(H2). By Lemma 3.3 H1 must be a-above H2. Let H3 be the a-highest of all the strategies a-below Ho and H1. Then Trjm,im E t*(H3). Suppose first that H3 is declared to be in state 0 when, or at some point after, Ho enumerated the

splitting into Trji,im. But then z extends one of the strings which H3 is provided with upon being declared to be in state 0. Since TrT,,im E t*(H3), Z extends a leaf of Trj,im when H3 is declared to be in state 0 and therefore extends zi. So suppose that H3 had been declared to be in state 0 before Ho enumerated the splitting. By




the definition of H3 it follows that if H4 is the a successor of H3 a-below Ho and H5 is the a successor of H3 a-below H1 then H4

- H5. If H5 is a 9 or an V strategy,

or if H5 and H4 are both T strategies we obtain an immediate contradiction, by Lemma 3.2. So suppose H5 is a W strategy and that H4 is an s strategy. Suppose that H3 is in state k at any stage at which H5 is passed control or checked for use. Either the base strings for H4 and H5 pairwise extend incompatible strings which H3 was provided with upon being declared to be in state 0, which would give us an immediate contradiction by Lemma 3.2, or r extends an escape route which we defined for H3 after Ho enumerated the splitting into

Trjm,im and which, since it

must be compatible with c , therefore extends ti. LEMMA 3.7. If Ho - T[t, k, 1, ..r.

r,](n) is passed control or checked for use and

T E {c , } then r k.

PROOF. The proof is by induction on the a ordering. Suppose that H1 is the a precessor of Ho. If H1 is a 9 strategy then r = 1, k = 3. If H1 is a T strategy which is in state k' > 0 at any stage at which Ho is passed control or checked for use then k = r. If H1 is a V or R strategy which is in state -1 whenever Ho is passed control or checked for use then suppose H1 = T[tl, kl, 1,..... r](n). We have, by the induction hypothesis, that rlkl. Since kl Ik the result follows.

LEMMA 3.8. Suppose a W strategy is provided with the number of strings specified in the initial environment upon being declared to be in state 0. This number is sufficient that it can carry out the given instructions.

PROOF. In the initial environment of a R strategy which searches for a k-fold splitting we specify that there should be k groups of 3(1k (k - 1) + 1) strings. We divide these into k groups of 1k(k - 1) + 1 triples. If we never find a splitting above a group of triples then the maximum number of triples that we can remove from this group is 1 + 2 + 2 + + k - 1 = (k - 1). Suppose that a group is the k'th group above which we find a splitting (k' > 1). Before finding this splitting we will already have removed 1 + 2 +... + k' - 1 triples from this group. If k' < k then upon finding the splitting we need to be sure that we have k - k' + 1 triples remaining. But then

k(k- 1) - (1 + 2 + +k'- 1)= k' +(k'+ 1) +-.

+ (k - 1) > k - k'. Consider the three following statements.

(t2) Suppose that the strategy H is passed control or checked for use at stage s. If H is a W(n), F(n) or 9(n) strategy then As, y7(, n) for any string 0 which is a base string for H. If H is an &W(n) strategy then A, V y( , n) for that string 0 which is a base string but not an a base string for H.

(t3) Suppose that a W'(n) strategy H is passed control or checked for use at stage s and is in state k > 0. For distinct z, z' each of which is either in Uk=l R(H, i) or is a non-discarded escape route,

y,(r, n)ny(z', n) = y (OH(z ), (n)ny(H(z'), n). The same is true for distinct r, -' each of which is either in U=1 R(H, i) or is in remaining u, unless 3],(z, z' E C).

(t4) Suppose that a '(n) strategy Ho is declared successful and enumerates a splitting zl,..., zk into a tree Tr. Let H1 be the strategy which Ho declares to be in state 0 and let q1, ... ,r be the base strings for H1. Then k/r of the zi extend each #ib and, for 1 < i < i'

_ k, y(Ti,n) n y(Ti,,n) r

(1H,(Zi), n) n

'(1H, (ZLi' ), n).



962 ANDREW LEWIS

LEMMA 3.9. Statements (t2), (t3) and (t4) are correct.

PROOF. The proof is by induction on the point of the construction in question. First we prove (t2) for the induction step. Let H1 be the a predecessor of H. If H1 is a F strategy, or a M or 9 strategy which is in state -1 at any stage at which H is passed control then the result follows immediately by the induction hypothesis. Suppose that H1 is a 9(n - 1) strategy or a dW(n - 1) strategy which is in state 0 at

any stage at which H is passed control. If H1 is a - (n - 1) strategy let 0' be the base string for this strategy and if H1 is an d (n) strategy then let 0' be the string which is a base string but not an a base string for this strategy. We have by the induction hypothesis that when H1 was passed control at stage s, A , y (0', n - 1). Since for any string z, y (z, n) C y (r, n - 1) it follows by the induction hypothesis on (t4) and by Lemma 3.3 that H1 chooses z to put Xs through such that As V 7(y, n). So suppose that H1 is a W (n) strategy which is state k > 0 at any stage at which H is passed control or checked for use. If H is an d(n) strategy then the result follows immediately from the induction hypothesis. Assume, then, that H is a M (n) strategy above a non-discarded set of escape routes which we have defined for H1. The result follows immediately by the induction hypothesis on (t2) and (t3).

We prove (t3) for the induction step. By the induction hypothesis on (t4), when H was declared to be in state 0 the strings that were delivered to H satisfied the property that for distinct z, z', y7(, n) n y(z', n) = y

(/H(r), n) y( H(rZ'), n). By

Lemma 3.3 the only strategies which could have subsequently violated this condition are those a-above H. Suppose first that r E R(H, i) and that either -' e R(H, i') for i' > i or z' is in remaining u'. Then the strategies strictly a-above H have only enumerated F(n') axioms on strings compatible with z, z' for n' > n. Suppose z E u. Before z was enumerated into R(H, i), control was passed to the W1(n) strategy for ,u every stage s' at which H was passed control and As, E y (u, n). Control was passed to the V(n) strategy for u' every stage s' at which H was passed control and As, E y7(', n). From the point at which v was enumerated into R(H, i), control was never passed to H at any stage s' at which As, E (r, n). Almost precisely the same argument suffices for the remaining cases. If z or r' is a non-discarded escape route then it is clear that the strategies strictly a-above H have only enumerated F(n) axioms on proper extensions of the string.

We prove (t4) for the induction step. Let H2 be the a successor of Ho and let H3 be the W strategy which Ho has passed control to and which has been declared successful. 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 with regard to those strings delivered to each of the strategies strictly a-above H2 and a-below H3 upon being declared to be in state 0, that precisely one of the two situations below applies.

(a) There is precisely one zi, which is a string in a splitting found by one of the A.(n) strategies that H2 has passed control to, extending a string in this group.

(b) There is precisely one zi which extends an escape route defined for H2 extending a string in this group.

It follows immediately that k/r of the ti extend each 4/

by considering the induction hypothesis as regards the strings that H2 was provided with upon being declared to be in state O.




By Lemma 3.4 we know that when Ho was first passed control no axioms had been enumerated on strings properly extending the base strings for this strategy. By Lemma 3.3 we know that subsequent to this point the only strategies which may have violated this condition are those a-above Ho. We shall suppress mention of any use of Lemma 3.3 in what follows. Suppose given 1 < i < i' k and (without loss of generality) suppose that all of the following are true.

(a) ri extends a string in a splitting found by an V (n) strategy passed control by Hi and -i, extends a string in a splitting found by an sl(n) strategy passed control by Hi'.

(b) ri extends ri, a string which H' was provided with upon being declared to be in state 0, and ri, extends r4,, a string which Hi' was provided with upon being declared to be in state 0.

(c) The splitting above -c, was found before that above -r'. We show, first of all, that 7(rT, n) n (,, n) = 7Y(H' (), n) /' ( ), ). CASE 1. Hi = Hi'. The result follows immediately by the induction hypothesis

on (t3). CASE 2. Hi f Hi'. Let - be the non-discarded escape route for Hi' which

Iri extends. Since it follows from the induction hypothesis on (t3) that y 7(,, n) n y7(, n) =

Y(/Hi'1 (), n)NY (Hi/' (), n) we are left to show that the strategies strictly

a-above Hi' do not enumerate axioms of the form F"'' (n) = c for T' extending T

and a' E y(r:,, n). But this is clear-suppose Hi' is in state k' at any stage at which H' is passed control or checked for use and that -r, Zj Suppose - pH" Then j = j' and at any stage s such that Hi' is checked for use while in state k' and As E 7(' ,, n) we would proceed to check the W strategy above the set of escape routes Pk',j for use.

So we have shown that y(r, n) n n (T',, n) -= y (4Hi' (, n) y (O/Hi' (r), n), but since no F(n) axioms have been enumerated on proper extensions of zr or ri, this means that y (Ti, n)ny(Tt, n) = Y

(/Hi' (Zi), n)N (41H' (ri',),

n). It follows by repeated applications of (t3) that y(OHi,'(Zi), n) n Y7(Hi(ri'),n) -=

Y(Hi(O i/ (Z-i)), n)n Y (HI (Hi' i' )),) l= Y (HI (i), n)

n 1 (H, (Ti,), n). Thus y(T i, n) ny (Ti,, n) =

Y (OH (Zi), n) n 7 (OH, (iTi'), n) as required.

The lemmas we have proved thus far suffice to show that the construction is well defined, so that it does not permanently terminate. Since the environments specified for each of the strategies do not play an active role in the construction (the point of their being specified was just as an aid to the reader) there is no need to state separately and prove a lemma that every strategy receives the correct environment every stage at which it is passed control-but the truth of this fact follows immediately from the lemmas already proved.

LEMMA 3.10. The axioms enumerated for F are consistent.

PROOF. Suppose an & or a 9 strategy enumerates an axiom T"1 (n) = c at stage s of the construction. Then a c As, prior to the enumeration of this axiom

As y 7(z,n), and no F(n) axioms have been enumerated on proper extension ofT.



964 ANDREW LEWIS

LEMMA 3.11. Every strategy isfinitary, i.e., for any strategy there exists s such that if the strategy is passed control after stage s then the outcome of the strategy is always the same.

PROOF. It is clear that the V, T and 9 strategies are finitary, so we are left to prove the result for F strategies.

So suppose that the strategy H = [t, k, 1,..., 1](n) is passed control at an

infinite number of stages and that there is no W' or F strategy H1 for which there exists s, after stage s H always passes control to HI. It must be the case that H is never declared successful.

Let H1 = = [t, k, q1,..., 1r](n). We may consider an infinite 'chain' of W(n)

strategies, defined in the following way. At any stage of the construction at which H is passed control it first checks to see whether it can pass control to the strategy H1. The W(n) strategy which is intended to provide an environment for H1 must eventually be declared successful, otherwise at every stage at which H is passed control it would simply pass control to this W(n) strategy. Let k1 be the greatest state that HI is declared to be in (here the superfix is used as a counter rather than to indicate exponentiation). Then k' > 0 otherwise there would be a stage, after which, whenever H is passed control it would pass control to H1. Clearly k' < k otherwise H would be declared successful. Define M1 = k' - 1. Since the given approximation to A converges there will be a set of escape routes defined for H1 such that after a certain stage whenever H is passed control it checks to see whether it can pass control to H1, finds that there is not an environment appropriate, and elects to 'use' this set of escape routes. Let H2 be the W(n) strategy above this set of escape routes. The W(n) strategy which is intended to provide an environment for H2 must eventually be declared successful, otherwise there would be a stage after which whenever H is passed control it passes control to this ' (n) strategy. If k2 is the greatest state that H2 is ever declared to be in (and where the superfix is used as a counter) then k2 > 0, otherwise there would be a stage after which whenever H is passed control it passes control to H2. H2 searches for a (k - M1)-fold splitting. If it was ever successful in this task then we would declare H to be successful. Thus k2 < k - M1. Define M2 = M1 + k2 - 1. Since our approximation to A

converges there must be a set of escape routes that are defined for H2 such that after a certain stage whenever H is passed control it checks to see whether it can pass control to H2, finds that there is not an environment appropriate, and elects to use this set of escape routes. Let H3 be the W (n) strategy above this set of escape routes. The F(n) strategy which is intended to provide an environment for H3 must eventually be declared successful. H3 searches for a (k - M2)-fold splitting. Let k3 be the greatest state that H3 is declared to be in. Then 0 < k3 < k - M2. Define M3 = M2 + k3 - 1, and so on.

Now {Mi }i__l

is a nondecreasing sequence of natural numbers such that, for all i > 1, k - Mi > 2. Thus

limi_.(k - Mi) I> 2. This means that there exists

j E o, Vi 2 j(k' = 1). For i _

j consider the strategy Hi. Let Ji be the set of all those strings a such that, before Hi is declared to be in state 1, we have enumerated some axiom Fr"' (n) = c such that z is compatible with one of the strings in R(Hi, 1) but properly extends #H/T (rf) (and for some c). For all i

> j it must be the case

that A is compatible with one of the strings in the set Ji. For each i > j let 1(i) be the greatest such that all strings in Ji agree on (and are defined on) arguments




< 1(i). Since the approximation to A must converge, it follows that as i -- 00o so does 1(i). This allows us to decide longer and longer initial segments of A. Thus A would be computable, which gives us the required contradiction. -1

DEFINITION 3.9. Suppose that H is a 9(n) strategy. We define the 'H cluster of strategies' to be H and all the V'(n), (n) and ds(n) strategies above H on the tree of strategies.

LEMMA 3.12. If there is a stage sl after which control is always passed to a strategy H = 9[t, b](n) then there is a stage s2, a string Z D q and a 9 (n + 1) strategy, H', such that Vs > s2 the strategies in the H cluster do not terminate stage s activity, put X, through z and pass control to H'. Furthermore, the strategies in the H cluster ensure that FA',(n) = K(n).

PROOF. Let Ho = H and for each i > 0 let Hi be the unique strategy in the H cluster which is passed control at all but a finite number of stages by Hi-1 if there exists such, leaving Hi undefined otherwise. If Hi , and Hi+Il I are both T strategies then t (Hi+,) C t (Hi). Thus there exists a least io such that Hio, . Each strategy Hi, i < io can only terminate stage s activity at a finite number of stages s. Since the approximation to A converges there exists a stage after which Hi,,-1 always puts Xs through the same string z and passes control to the same 91(n + 1) strategy at any stage s.

Hio-1 ensures that FA,r (n) = K(n). -I

LEMMA 3.13. X is defined and FAX = K. PROOF. Let Ho be the strategy at the bottom of the tree of strategies and for each

i > 0 let Hi be the 9 strategy which is passed control at all but a finite number of stages by a strategy in the Hi_1 cluster. There is a stage after which the strategies in the Ho cluster always put X, through the same (non-empty) string to and pass control to H1. The strategies in the Ho cluster ensure that FA,,o(0) = K(0). For each i > 0 there is a stage after which the strategies in the Hi cluster always put Xs through the same string zi D Ti-1 and pass control to Hi+l. The strategies in the Hi cluster ensure that TA,zr (i) = K (i). -H

LEMMA 3.14. X is a set of minimal degree. PROOF. Let Ho be the strategy at the base of the tree of strategies and for all i > 0

let Hi be that 9 strategy which is passed control at all but a finite number of stages by a strategy in the Hi_1 cluster. Since there is only finite injury as regards splitting tree candidates along this 'chain' of strategies, it is clear that for each j > 0 one of the following two situations applies. (a) There exists io, i1 E co (where we use the superfix as a counter) such that for

all i > io, Trj,i EC t(Hi). For each i > io, the strategies in the Hi cluster enumerate strings into Trj,it extending Oi c X, the base string for Hi. Thus X <T TrX

(b) There exists io which is the greatest such that there exists i1, Try,i, E t(Hio). Suppose t(Hio) = {Tr_l,o, Trir, ,..., Trim,~ } and that j, = j. Then there is a strategy in the Hio cluster which is passed control at an infinite number of stages and which never ceases searching for a Ij splitting above the base string for Hio,1 on Trj,_,,i,_r. Also

Trjr_,i,_, E t(Hi) for all i > io so that

for all i > io the strategies in the Hi cluster enumerate strings into Trjrl ,ir_ extending the base string for Hi. Thus, if it is total 'IX is computable. -



966 ANDREW LEWIS

REMARK. In the paper 'A single minimal complement for the c.e. degrees' we show that there exists a minimal degree below 0' which complements every c.e. degree 0 < a < 0'-and therefore every such n-c.e. degree. In the paper 'Finite cupping sets' we prove the strong result in the negative direction that given any degree 0 < c < 0' and any uniformly A2 sequence of degrees bo, bl, b2, ... such that Vi (bi ; c), there exists 0 < a < O0' such that for all i > 0, a V bi 1 c. If c is c.e. and bo, bl, b2, ... are uniformly (strictly) below c then there exists such an a below c. In 'The minimal complementation property above 0" we show that every degree above 0' satisfies the minimal complementation property that we have proved for 0' in this paper.

REFERENCES

[1] C. T. CHONG, Generic sets and minimal a-degrees, Transactions of the American Mathematical

Society, vol. 254 (1979), pp. 157-169.

[2] S. B. COOPER, Minimal degrees and the jump operator, this JOURNAL, vol. 38 (1973), pp. 249-271.

[3] S. B. COOPER, A. E. M. LEWIS, and YUE YANG, Properly 12 minimal degrees and 0" complementation, to appear.

[4] R. L. EPSTEIN, Minimal degrees of unsolvability and the full approximation construction, Memoirs of the American Mathematical Society, vol. 163, 1975.

[5] M. FUKUYAMA, A remark on minimal degrees, Science Reports of Tokyo Daig., vol. 9 (1968), pp. 255-256.

[6] C. G. JOCKUSCH and D. POSNER, Double-jumps of minimal degrees, this JOURNAL, vol. 43 (1978), pp. 715-724.

[7] A. E. M. LEWIS, Finite cupping sets, to appear in Applied Mathematics Letters.

[8] , The minimal complementation property above 0', to appear. [9] - , A single minimal complement for the c. e. degrees, to appear. [10] A. LI and X. YI, Cupping the recursively enumerable degrees by d-r. e. degrees, Proceedings of the

London Mathematical Society, vol. 78 (1999), pp. 1-21.

[11] P. G. ODIFREDDI, Classical recursion theory, vol. 2, Elsevier, 1999.

[12] D. POSNER, High degrees, Ph. D. Thesis, University of California, Berkeley, 1977.

[13] - , The uppersemilattice of degrees below 0' is complemented, this JOURNAL, vol. 46 (1981), pp. 705-713.

[14] D. POSNER and R. W ROBINSON, Degrees joining to 0', this JOURNAL, vol. 46 (1981), pp. 714-722. [15] G. E. SACKS, Degrees ofunsolvability, 2nd ed., Princeton University Press, 1963.

[16] D. SEETAPUN and T. A. SLAMAN, Minimal complements, unpublished, 1992.

[17] J. R. SHOENFIELD, A theorem on minimal degrees, this JOURNAL, vol. 31 (1966), pp. 539-544.

[18] T. A. SLAMAN and J. STEEL, Complementation in the Turing degrees, this JOURNAL, vol. 54 (1989), pp. 160-176.

[19] R. I. SOARE, Recursively enumerable sets and degrees, Springer-Verlag, New York, 1987.

[20] C. SPECTOR, On degrees of recursive unsolvability, Annals ofMathematics, vol. 64 (1956), pp. 581- 592.

[21] C. E. M. YATES, Recursively enumerable degrees and the degrees less than 0', Models and recursion

theory (Crossley et al., editors), North Holland, 1967, pp. 264-271.

[22] , Initial segments of the degrees of unsolvability, part 2: minimal degrees, this JOURNAL, vol. 35 (1970), pp. 243-266.

DEPARTMENT OF PURE MATHEMATICS SCHOOL OF MATHEMATICS

UNIVERSITY OF LEEDS LEEDS, LS2 9JT, UK

E-mail: [email protected]



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Minimal Complements for Degrees below 0´

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