MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
ANDREW LEWIS
Abstract. It is shown that for every (Turing) degree 0 < a < 00 there there is a minimal
degree m < 00 such that a ∨ m = 00 (and therefore a ∧ m = 0).
1. Introduction.
The aim of this paper is to prove the following theorem.
Theorem 1.1. Every (Turing) degree 0 < a < 00 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≥0 to a set A of degree 0 < a < 00
and an enumeration {Ks }s≥0 of K. For convenience we shall assume that for all s ≥ 0, As is a
binary string of length s. We will enumerate an approximation {Xs }s≥0 to a set X of minimal
degree and we will enumerate axioms for a Turing functional Γ such that:
ΓA,X = K.
The axioms that we shall enumerate for Γ will be of the form Γσ,τ (n) = c, where σ and τ are
finite binary strings and n ∈ ω, c ∈ {0, 1}. Throughout the construction we will have to ensure
that at every stage s ≥ 0 and for every n ∈ ω,
ΓAs ,Xs (n)[s] ↓ → ΓAs ,Xs (n)[s] = Ks (n)
and that ultimately, for every n ∈ ω, ΓA,X (n) ↓.
Let {Ψj }j≥0 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 Ψj , X lies on some c.e. tree T r which is either Ψj
splitting or which is such that if ΨX
j is total there is some initial segment τ of X lying on T r
for which no τ1 , τ2 extending τ and lying on T r are a Ψj splitting. To be more precise, for each
j ≥ 0 we will attempt to enumerate an infinite tree T rj such that X lies on T rj and which is Ψj
splitting. If we do not succeed and there is a greatest j 0 < j for which we do then T rj 0 will act
as the T r above demonstrating that if ΨX
j is total then it is computable. We note that there
may be many candidates for each T rj – the tree T rj,i should be regarded as the ith candidate
for T rj .
1
2
ANDREW LEWIS
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 0
then the strings that H 0 has been provided with will extend those that H has been provided
with. If H puts Xs through a string τ then any string τ 0 which H 0 puts Xs through will be
an extension of τ . 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, A, B, C and D strategies. We use
the variable Υ to range over the set {A, B, C, D}. If we refer to a strategy of type Υ as an
Υ(n) strategy this means that the strategy is concerned with ensuring that ΓA,X (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 σ, τ and for n ∈ ω we say that
σ ∈ γ(τ, n) if there are initial segments of σ and τ , σ 0 and τ 0 respectively, for which we have
0 0
already enumerated the axiom Γσ ,τ (n) = c for some c. Note that the set γ(τ, n) may change as
the construction progresses.
The construction will be such that for any n, n0 ∈ ω such that n > n0 we shall never enumerate
an axiom of any form Γσ,τ (n) = c at any stage s, for any σ ⊆ As , τ ⊆ Xs and c ∈ {0, 1}, unless
0 0
we have previously enumerated an axiom of some form Γσ ,τ (n0 ) = c0 such that σ 0 ⊂ σ, τ 0 ⊂ τ
and c0 ∈ {0, 1}. The following is an immediate consequence.
(†0 ) For any string τ and any n, n0 ∈ ω such that n > n0 it will always be the case that
γ(τ, n) ⊆ γ(τ, n0 ).
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 ensuring that ΓA,X (n) is
correctly defined. Suppose further that all of the following are true.
1) H ? is provided with four strings φ, τ1 , τ2 , τ3 such that τ1 , τ2 , τ3 are (pairwise) incompatible
extensions of φ.
2) For i 6= j, γ(τi , n) ∩ γ(τj , n) = γ(φ, n). For 1 ≤ i ≤ 3 define Pi = γ(τi , n) (as this set is defined
at stage s).
3) At stage s no axioms have been enumerated on proper extensions of τi for any i, i.e. for
1 ≤ i ≤ 3 we have not enumerated any axioms of any form Γσ,τ (n0 ) = c such that τ properly
extends τi .
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
3
4) The required ‘environment’ of the strategy H ? states that at any stage s0 , As0 ∈
/ γ(φ, 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 s0 such that As0 ∈
/ γ(φ, n).
Why should the required environment of the strategy H ? state that at any stage s0 , As0 ∈
/
γ(φ, n)? We have assumed that the strategy is concerned with ensuring that ΓA,X (n) is correctly
defined. Now suppose that at some stage s0 ≥ s the strategy is passed control and that As0 ∈
γ(φ, n) – let us suppose that we have enumerated some axiom Γσ,τ (n) = 0 for σ ⊆ As0 and
τ ⊆ φ, but that Ks0 (n) = 1. Whichever string τi the strategy H ? puts Xs0 through, the result
will be that ΓAs0 ,Xs0 (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 ΓA,X (n) is correctly defined? At any stage s0 at
which the strategy is passed control it can proceed as follows. By 2) there exists at most one
i ∈ {1, 2, 3} such that As0 ∈ Pi . Let us suppose that As0 ∈ P1 . Now if we put Xs0 through τj ,
j 6= 1, then the fact that As0 ∈
/ Pj and that H ? will only ever enumerate Γ(n) axioms (axioms
σ,τ
of the form Γ (n) = c for σ, τ ∈ 2<ω and c ∈ {0, 1}), means that by the observation (†0 )
made above, As0 ∈
/ γ(τj , n + 1). The strategy H ? can thus proceed to put Xs0 through τj and
pass control to another strategy which is concerned with ensuring that ΓA,X (n + 1) is correctly
defined (and which you might imagine being provided with three pairwise incompatible strings,
τ10 , τ20 , τ30 say, extending τj such that for i0 6= j 0 , γ(τi00 , n + 1) ∩ γ(τj0 0 , n + 1) = γ(τj , n + 1)) and
this strategy will be provided with the correct environment. Namely that As0 ∈
/ γ(τj , n + 1). So
at any stage s0 such that As0 ∈ P1 the strategy H ? is presented with two choices. It can put
Xs0 through τ2 or through τ3 . At stages s0 such that As0 ∈ P1 and n ∈
/ Ks0 the strategy can put
σ,τ
2
Xs0 through τ2 and enumerate the axiom Γ (n) = 0 where σ is the initial segment of As0 of
length τ2 . At stages s0 such that As0 ∈ P1 and n ∈ Ks0 the strategy can put Xs0 through τ3 and
enumerate the axiom Γσ,τ3 (n) = 1 where σ is the initial segment of As0 of length τ3 .
Similarly at stages s0 such that As0 ∈ P2 the strategy can put Xs0 through either τ1 or τ3 . At
0
0
0
stages s0 such
S3 that As ∈ P3 the strategy can put Xs through either τ1 or τ2 . At 0 stages s such
that As0 ∈
/ i=1 Pi the strategy can proceed just as if As0 ∈ P1 . At any stage s at which the
strategy puts Xs0 through the string τi it should enumerate the axiom Γσ,τi (n) = Ks0 (n) where
σ is the initial segment of As0 of length τi . If the strategy is passed control at all but a finite
number of stages then it is able to ensure that ΓA,X (n) is correctly defined and the fact that
we provide the strategy with three incompatible strings extending φ rather than just two means
that it is able to pass control to strategies concerned with ensuring that ΓA,X (n + 1) is correctly
defined and these strategies will be provided with the correct environment.
Definition 1.2. Given j ≥ 0 and τ ∈ 2<ω let n be the least such that Ψτj (n) ↑. We say that Ψτj
is of length n.
Definition 1.3. By a k-fold Ψj splitting of length at least l we mean k strings τ1 , .., τk whose
images Ψτj i (1 ≤ i ≤ k) are 1) of length at least l and 2) pairwise incompatible – for 1 ≤ i < i0 ≤ k
τ0
if the lengths of Ψτj i and Ψj i are n1 and n2 respectively then there exists n3 < n1 , n2 such that
τ0
Ψτj i (n3 ) 6= Ψj i (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, τ1 , .., τk say, above a string φ in order that we can enumerate
this splitting into a tree T rj,i , we shall insist that for i0 6= j 0 we have γ(τi0 , n) ∩ γ(τj 0 , n) = γ(φ, n)
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ANDREW LEWIS
(and where n is such that on this part of the tree of strategies we are concerned with ensuring
that ΓA,X (n) is correctly defined).
Now let us describe each of the strategy types in turn.
1.2. The A(n) strategy. To each strategy H, of any variety, we shall be associating for some
m a finite set of trees,
t(H) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im }
where j1 < .. < jm and where for any 1 ≤ r ≤ m, T rjr ,ir should be regarded as the ith
r candidate
for T rjr . T r−1,0 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 j0 = −1, i0 = 0. Assume for
now that m ≥ 1 and suppose that 0 ≤ r0 < 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 T rjr ,ir so as to
be a subtree of T rjr0 ,ir0 (actually we shall have to qualify this statement slightly in 1.4).
For some k, l ∈ ω it will be the task of an A(n) strategy H, with t(H) as above, to search
for a k-fold Ψjm splitting of length at least l lying on T rjm−1 ,im−1 . 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 ΓA,X (n) is correctly defined and will witness the fact that T rjm ,im
is finite. The strategy will proceed to achieve these specified tasks as follows.
The first stage s at which an A(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 φ, τ1 , τ2 , τ3 such that τ1 , τ2 , τ3 are (pairwise) incompatible extensions of φ and such
that for i 6= j, γ(τi , n) ∩ γ(τj , n) = γ(φ, n). The strings τ1 , τ2 , τ3 will be strings that we have
enumerated into T rjm−1 ,im−1 , while φ will extend a leaf of T rjm ,im . At stage s we shall have that
no axioms have been enumerated on proper extensions of τi for any i. Once again it will be part
of the required environment of the A(n) strategy that it is only passed control at stages s0 such
that As0 ∈
/ γ(φ, n). The A(n) strategy can therefore proceed to operate much as the strategy
H ? . It will choose which string to put Xs0 through at any stage s0 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 Ψjm splitting of length at least l lying on T rjm−1 ,im−1 and
such that there exists 1 ≤ i ≤ 3 every string in the splitting extends τi . It is worth emphasizing
that the A(n) strategy searches only for a splitting such that there exists some i, every string
in the splitting extends τi .
At every stage s0 at which the A(n) strategy puts Xs0 through the same string it will pass
control to the same D(n + 1) strategy, so that there are three different D(n + 1) strategies that
an A(n) strategy may pass control to at any stage at which it is passed control.
We shall refer to the A(n) strategy with parameters φ, τ1 , τ2 , τ3 , k, l and t as above, as the
strategy H = A[t, k, φ, τ1 , τ2 , τ3 ](n) where l(H) ↓= l. We consider the argument l 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 φ, τ1 , τ2 , τ3 we refer to as the ‘base strings’ of the strategy.
1.3. The B(n) strategy. Suppose that we are searching for a splitting above φ to enumerate
into T rjm ,im and that the A(n) strategy, as described in 1.2, is eventually declared successful.
Let us suppose that it finds a k-fold Ψjm splitting, τ10 , .., τk0 , such that every string in the splitting
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
5
extends τ1 . This splitting is not suitable for enumeration into T rjm ,im . Before we could enumerate such a splitting into this tree we would require that for i 6= j, γ(τi0 , n) ∩ γ(τj0 , n) = γ(φ, n),
but common to the axioms that we have enumerated on each of the strings in this splitting are
the axioms that the A(n) strategy has enumerated on τ1 . If the A(n) strategy is passed control
at an infinite number of stages then it will ensure that ΓA,X (n) is correctly defined and will
witness that T rjm ,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 σi so that σi is incompatible with
σj if i 6= j.
The proof of this lemma is left to the reader. Now it will be the aim of a B(n) strategy
H with t(H) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im }, m ≥ 1, j1 < .. < jm , to find (for some k ∈ ω) a
k-fold Ψjm splitting that we can actually enumerate into T rjm ,im . A splitting, that is, with the
required intersection properties as regards γ. 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
T rjm ,im every string of which extends φ. Then at the first stage s at which the strategy is passed
control we should be provided with 3k (pairwise) incompatible strings extending φ, which have
been enumerated into T rjm−1 ,im−1 (let us agree once again that j0 = −1, i0 = 0) and such that
for distinct τ, τ 0 we have γ(τ, n) ∩ γ(τ 0 , n) = γ(φ, 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 s0 ≥ s such that
As 0 ∈
/ γ(φ, n). We may then divide these strings into k triples, so that each triple provides a
suitable environment for an A(n) strategy. Let the variable µ range over this set
S of triples and
for each µ let us agree that µ = {τµ,1 , τµ,2 , τµ,3 } and that, for any n0 , γ(µ, n0 ) = i γ(τµ,i , n0 ). At
any stage at which the B(n) strategy is passed control we shall pass control to an A(n) strategy,
H 0 say, with base strings φ, τµ,1 , τµ,2 , τµ,3 for some µ and such that t(H 0 ) = t(H), k(H 0 ) = k(H).
Let us call this the ‘A(n) strategy for µ’. Suppose that H is passed control at some stage s0
such that none of the A(n) strategies that we have passed control to have yet been declared
successful. If As0 ∈ γ(µ, n) for some µ then we will pass control to the A(n) strategy for µ, H 0
say – just choose the required parameter l(H 0 ) to be some fixed number. If As0 ∈
/ γ(µ, n) for
any µ then we will just choose µ and pass control to the A(n) strategy for µ.
Suppose that one of the A(n) strategies is eventually declared successful. Then for each of the
A(n) strategies H 0 redefine l(H 0 ) 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 τ1H and suppose
that τ1H ∈ µ1 (generally, if τ is the ith string above which a splitting is found by the A strategies
passed control by a B strategy H we shall refer to τ as τiH ). We ‘discard’ the other two strings
in µ1 and we shall never again pass control to the A(n) strategy for µ1 . At any subsequent stage
s0 at which the B(n) strategy is passed control such that As0 ∈
/ γ(τ1H , n) we can now proceed
basically as before – the idea is that if all k of the A(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 As0 ∈ γ(µ, n) for some µ 6= µ1 then we may pass control to
the A(n) strategy for µ, and if not then we may choose a triple and proceed similarly. If another
A(n) strategy should subsequently find a splitting above a string τ2H ∈ µ2 then, so long as we
have not passed control to the A(n) strategy for µ2 at any stage s0 such that As0 ∈ γ(τ1H , n),
we shall have that γ(τ1H , n) ∩ γ(τ2H , n) = γ(φ, n). Since we shall not have enumerated any Γ(n)
axioms on proper extensions of either τ1H or τ2H it will be the case that any two strings, one
6
ANDREW LEWIS
chosen from each of the splittings found, will also satisfy this property. Each time that another
A(n) strategy is declared successful we should redefine l(H 0 ) for each of the A(n) strategies H 0
to be larger than any number yet mentioned in the course of the construction.
After the first splitting has been found, above the string τ1H , how should the B(n) strategy now
proceed at stages s0 such that As0 ∈ γ(τ1H , n)? If it passes control to some A(n) strategy which
then finds a splitting above a string τ2H 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 γ(τ1H , n) ∩ γ(τ2H , n) 6= γ(φ, n). We must insist,
in fact, that the B(n) strategy is not passed control at stages s0 such that As0 ∈ γ(τ1H , n) since
the strategy is now unable to act at such stages. More generally, if k 0 (< k) of the A(n) strategies
have been declared successful and have found splittings above strings τ1H , .., τkH0 then we cannot
S 0
pass control to the B(n) strategy at stages s0 such that As0 ∈ ki=1 γ(τiH , n).
Definition 1.4. In the course of discussing a B or a C strategy H every string τ under consideration will extend one of the base strings for H. We let φH (τ ) designate this string.
So in the event that the B(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, B(n) strategies which given strings φ1 , .., φr (which we refer to as the base strings for the
strategy) such that r divides k search for a k-fold splitting τ1 , .., τk such that k/r of these strings
extend each φi , and such that if i 6= j then γ(τi , n) ∩ γ(τj , n) = γ(φH (τi ), n) ∩ γ(φH (τj ), n). The
required environment
S of the strategy will be such that it should only be passed control at stages
s0 such that As0 ∈
/ ri=1 γ(φi , n). We shall refer to a B(n) strategy with base strings φ1 , .., φr
and with t and k as above as the strategy B[t, k, φ1 , .., φr ](n). We shall proceed (roughly) as
follows. In the ‘initial environment’ of the B(n) strategy we shall require a much larger number
of strings. When k 0 of the A(n) strategies that a B(n) strategy has passed control to have been
declared successful we shall regard the strategy as being in state k 0 . Whenever one of the A(n)
strategies is declared successful, so that the B(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 A(n) strategies
for those µ from which we have discarded strings. Each time that a B(n) strategy enters a new
state k 0 (< k) we shall have to discard all of the previously defined escape routes and choose k 0
new sets of escape routes – one set for each of the splittings that has been found. Suppose that
ρ = {τρ,1 , τρ,2 , ..} is a set of escape routes that we have defined for the B(n) strategy H upon
being declared to be in state k 0 . Each τρ,i will satisfy the following property: for 1 ≤ j ≤ k 0 ,
γ(τρ,i , n) ∩ γ(τjH , n) = γ(φH (τρ,i ), n) ∩ γ(φH (τjH ), n). The importance of this is that, if at any
stage s0 at which H is in state k 0 there would be an environment appropriate in order to pass
control to H except that As0 ∈ γ(τjH , n) for some j, then for each τρ,i we have As0 ∈
/ γ(τρ,i , n).
Other factors aside there will thus be an environment appropriate such that we may pass control
to a B(n) strategy with base strings defined to be those in ρ – a B(n) strategy ‘above ρ’ let’s
say. We can now hope to make progress with this B(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 B(n) strategy enters a new state k 0 > 0 we
must define k 0 sets of escape routes.
From the preceding discussion it is hopefully clear that the number of strategies any B(n)
strategy may pass control to is fixed and determined by k. A B(n) strategy passes control to
A(n) strategies. If H is a B(n) strategy and H 0 is a B(n) strategy above ρ, a set of escape routes
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
7
that we have defined for H, then control will be passed to H 0 by the C(n) strategy immediately
below H on the tree of strategies. If a B(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 C(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 φ1 , .., φr (which we refer to as the base strings for the
strategy), k ∈ ω and for some t = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } with j1 < .. < jm the strategy
H = C[t, k, φ1 , .., φr ](n) must achieve the following. Either it must find some string above
which there is no Ψjm splitting on T rjm−1 ,im−1 (presuming m ≥ 1) or it must find a k-fold Ψjm
splitting, τ1 , .., τk say, suitable for enumeration into T rjm ,im with the intersection properties
described previously – that for i 6= j, γ(τi , n) ∩ γ(τj , n) = γ(φH (τi ), n) ∩ γ(φH (τj ), n). Where
any individual B(n) strategy may fail – and where in fact every individual B(n) strategy may
fail to find a splitting of the required magnitude – the C(n) strategy must organize a potentially
unbounded number of B(n) strategies in such a way that, should we not find any string above
which there is no Ψjm splitting on T rjm−1 ,im−1 , then we can find amongst the splittings that
these B(n) strategies collectively produce a splitting of the variety required. Should the C(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 C(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 = C[t, k, φ1 , .., φr ](n) go about achieving its
specified task? In the case that t = {T r−1,0 } 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 φi and enumerate them into T r−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 0 . Upon
being declared successful H will declare H 0 to be in state 0 and will ‘deliver’ the strings that it
enumerated into T r−1,0 to H 0 . So suppose that m ≥ 1. The required environment ofSthe strategy
H will be such that it should only be passed control at stages s0 such that As0 ∈
/ ri=1 γ(φi , n).
The first stage at which H is passed control we shall have that φ1 , .., φr 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 = B[t, k, φ1 , .., φr ](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 H 1 = C[t1 , k1 , φ1 , .., φr ](n), where t1 = t − {T rjm ,im }
and k1 is the number of strings that are required in the initial environment of H1 . We specify
that H 1 is intended to provide an environment for H1 . Should it be the case that H 1 is never
declared successful then at every stage at which H is passed control it will simply pass control to
the strategy H 1 which will witness that one of the trees in t1 is finite. In this case the strategies
above H 1 will ensure that ΓA,X (n) is correctly defined.
Suppose that H 1 is eventually declared successful. Then at every subsequent stage s0 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 1 > 0 and As0 ∈ γ(τjH1 , n) for some 1 ≤ j ≤ k 1 (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 k 1 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 τiH1 , 1 ≤ i ≤ k 1 . We would now like to pass control to a B(n) strategy, H2 say,
8
ANDREW LEWIS
above ρ which is the non-discarded set of escape routes for H1 that is associated with τjH such
that As0 ∈ γ(τjH1 , n), but first of all we must provide an initial environment for this strategy.
We therefore pass control to a C(n) strategy H 2 such that t(H 2 ) = t1 as above, with base
strings defined to be those strings in ρ 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 H 2 is
intended to provide an environment for H2 . Should it have been the case that As0 ∈ γ(τjH0 1 , n)
for some j 0 6= j then we would have passed control to the C(n) strategy intended to provide an
environment for a B(n) strategy above ρ0 associated with τjH0 1 – we would have ‘used’ the set of
escape routes ρ0 instead.
Suppose that H is passed control at some stage s0 after which H 2 has been declared successful
and that once again we cannot pass control to H1 because it is in state k 1 and As0 ∈ γ(τjH1 , n).
Then we shall pass control to H2 unless it is in state k 2 > 0 (say) and As0 ∈ γ(τjH0 2 , n) for some
1 ≤ j 0 ≤ k 2 . In this latter case we should then like to pass control to a B(n) strategy above
the non-discarded set of escape routes for H2 which are associated with τjH0 2 , but first of all
we must pass control to a C(n) strategy in order to provide the strings required in the initial
environment, and so on.
The reason that, when a B(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 τjH1 be as above – so that H2 is the B(n) strategy above the set of escape routes defined
for H1 upon being declared to be in state k 1 which is associated with τjH . Let τ 6= τjH1 be a
string above which the A(n) strategies that H1 passes control to have found a splitting when
H1 is declared to be in state k 1 and let τ 0 be any string above which the A(n) strategies which
H2 passes control to find a splitting. Since we shall not pass control to the strategy H2 at any
stage s0 such that As0 ∈ γ(τ, n) – we would use a different set of escape routes at any such stage
– it will be the case that γ(τ, n) ∩ γ(τ 0 , n) = γ(φH (τ ), n) ∩ γ(φH (τ 0 ), n), as required. Although
there may be a stage after which the strategy H cannot pass control to H1 , at any such stage s0
all but one of the splittings that H1 has found (which one this is depends on the specific value
of As0 ) remains useful to H. Thus, while H1 is in state k 1 , the B strategies above non-discarded
escape routes for H1 need only search for a (k − k 1 + 1)-fold splitting.
H is declared successful when any of the B(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 C(n)
strategy is finitary? 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’ of B(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 C(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 C(n)
strategy. Let k 1 be the greatest state that H1 is declared to be in. Then k 1 > 0 otherwise
there would be a stage, after which, whenever H is passed control it would pass control to H1 .
Clearly k 1 < k otherwise H would be declared successful. Define M1 = k 1 − 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 ,
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
9
finds that there is not an environment appropriate, and elects to ‘use’ this set of escape routes.
Let H2 be the B(n) strategy above this set of escape routes. The C(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 C(n) strategy. If
k 2 is the greatest state that H2 is ever declared to be in then k 2 > 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 − M1 )-fold splitting. If it was
ever successful in this task then we would declare H to be successful. Thus k 2 < k − M1 . Define
M2 = M1 + k 2 − 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 B(n) strategy above this set of escape routes. The C(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 k 3 be the greatest state that H3 is declared to be in.
Then 0 < k 3 < k − M2 . Define M3 = M2 + k 3 − 1, and so on.
Now {Mi }i≥1 is a nondecreasing sequence of natural numbers such that, for all i ≥ 1, k −Mi ≥
2. Thus limi→∞ (k − Mi ) ↓≥ 2. This means that there exists j ∈ ω, ∀i ≥ j(k i = 1). For i ≥ j
consider the strategy Hi . Let Ji be the set of all those strings σ such that, before Hi is declared
to be in state 1, we have enumerated some axiom Γσ,τ (n) = c such that τ is compatible with
τ1Hi but properly extends φHi (τ1Hi ) (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 → ∞ 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 τ ∈ 2<ω is compatible with a tree T r if τ has an extension on T r,
or if τ extends a leaf of T r.
Throughout the course of the construction, the following condition must clearly be satisfied.
(†1 ) When we enumerate a splitting into a tree T r, every string in the splitting must extend
(what was prior to this enumeration) a leaf of T r.
If the C(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 (†1 ) 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 T rjm ,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 T rjm ,im to this strategy, and then terminate the present
stage of the construction. In 1.1 we suggested that for r > r0 , T rjr ,ir would be constructed so
as to be a subtree of T rjr0 ,ir0 . The action that we have just described means that, in fact, this
is not quite true. When we enumerate a string into T rjr ,ir we shall ensure that it is compatible
with T rjr0 ,ir0 . So long as there is some ‘active’ part of the tree of strategies where we are still
interested in building T rjr ,ir we shall ensure that this string remains compatible with T rjr0 ,ir0 .
So the tree T rjr ,ir will be ‘thinner’ than T rjr0 ,ir0 , but possibly taller at any given stage.
10
ANDREW LEWIS
1.5. The D(n) strategy. When the strategy H = D[t, φ](n), with φ a finite binary string and
t = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im }, j1 < .. < jm , is first passed control we shall have that φ extends
a leaf of each of the trees in t and that no axioms have been enumerated on strings properly
extending φ. 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 0 = C[t, 3, φ](n),
performing no other instructions. We specify that H 0 is intended to provide an environment for
H. If H 0 is never declared successful then at every stage at which H is passed control it will
do nothing more than pass control to H 0 which will witness that one of the trees in t is finite.
In this case the strategies above H 0 will ensure that ΓA,X (n) is correctly defined. Control will
eventually always be passed to the same A(n) strategy which will eventually always pass control
to the same D(n + 1) strategy.
If H 0 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 ΓA,X (n) is correctly defined. Every stage s at
which H puts Xs through the same string it will pass control to the same D(n + 1) strategy,
so there will be three different D(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 = D[t, φ](n)
for the first time, where t = {T r−1,0 , T r0,0 , T r1,2 }. At stage s we shall have that φ extends a
leaf of each of the trees in t and that no axioms have been enumerated on proper extensions of
φ. 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 0 = C[t, 3, φ](n).
At stage s the strategy H 0 will pass control to the strategy H 00 = C[t0 , k, φ](n) where t0 =
{T r−1,0 , T r0,0 } and k is the number of strings required in the initial environment of the strategy
B[t, 3, φ](n). The strategy H 00 will pass control to the strategy H 000 = C[t00 , k 0 , φ](n) where k 0
is the number of strings required in the initial environment of the strategy B[t0 , k, φ](n) and
t00 = {T r−1,0 }. The strategy H 000 will then choose k 0 incompatible extensions of φ, enumerate
these strings into T r−1,0 , declare B[t0 , k, φ](n) to be in state 0 and ‘deliver’ these strings to that
strategy. We shall declare H 000 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 ω.
At all subsequent stages at which H is passed control, until H 00 is declared successful, it will
pass control to H 0 which will pass control to H 00 . If H 00 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 B(n) strategy
which eventually always passes control to the same A(n) strategy H1 which, in turn, always
eventually puts Xs0 through the same string at any stage s0 at which it is passed control and
thus passes control to the same D(n + 1) strategy H1? . H1 will ensure that ΓA,X (n) is correctly
defined. How should we define t(H1? )? If the strategy H1 ever finds the Ψ0 splitting for which
it is searching then it will be declared successful and H1 and H1? will never be passed control
again. Thus at any stage at which H1? is passed control this fact in itself ensures that such a
splitting has not been found. This strategy therefore assumes that T r0,0 will be finite. As we
have said before, however, we must continue to construct T r1,0 in such a way that it may be
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
11
regarded as a ‘subtree’ of T r0,0 so at this point on the tree of strategies we must introduce a new
candidate for T r1 , a tree for which we do not need to satisfy this condition. We therefore define
t(H1? ) = {T r−1,0 , T r1,z(H1? ) }. The role of the function z, then, is just to ensure that above H1?
we have a different candidate for a Ψ1 splitting tree than anywhere else on the tree of strategies.
Suppose that H 00 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 0 . If H 0 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 A(n) strategy
above H 0 , H2 say. H2 will eventually always put Xs0 through the same string and thus pass
control to the same D(n + 1) strategy H2? at any stage s0 at which it is passed control. H2 will
ensure that ΓA,X (n) is correctly defined. How should we define t(H2? )? If H2 is searching for
a Ψ0 splitting then we should define t(H2? ) in a similar way to that in which we defined t(H1? )
previously. So suppose that H2 is searching for a Ψ1 splitting. If H2 ever finds such a splitting
then it will be declared successful and H2 and H2? will never be passed control again. At any
stage at which H2? is passed control this fact in itself therefore ensures that such a splitting has
not been found and this strategy therefore assumes that T r1,0 will be finite. We therefore define
t(H2? ) = {T r−1,0 , T r0,0 , T r2,z(H2? ) }.
So finally, then, suppose that H 0 is eventually declared successful. At any subsequent stage at
which H is passed control it will pass control to D(n + 1) strategies. For any D(n + 1) strategy
H3 that H passes control to we define t(H3 ) = {T r−1,0 , T r0,0 , T r1,2 , T r2,z(H3 ) }.
To sum up let us briefly consider the general case of which the previous discussion was an
example. D(n) strategies in state -1 will pass control to C(n) strategies which will pass control to
C(n) and B(n) strategies until declared successful. B(n) strategies pass control to A(n) strategies
until declared successful while A(n) strategies pass control to D(n + 1) strategies until declared
successful. Once a C, B, or A strategy is declared successful it is never passed control again.
Once a D(n) strategy is declared to be in state 0 it passes control to D(n + 1) strategies. Any
‘chain’ of C(n) strategies such that each passes control to the next will be finite in length. Each
C(n) strategy will be ‘finitary’. Thus each D(n) strategy H and the C, B, and A strategies lying
above it and below D(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 ΓA,X (n) is correctly defined and will
eventually always pass control to the same D(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 A(n) strategy and after which this
strategy always puts Xs through the same string at any stage s. When any A(n) strategy H 0
is passed control at a stage s and puts Xs through a string τ it will pass control to a strategy
H 00 = D[t, τ ](n + 1) where we define t as follows. Let t(H 0 ) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } with
j1 < .. < jm . Then t = (t(H 0 ) − {T rjm ,im }) ∪ {T rjm +1,z(H 00 ) }.
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 τ , it will pass control to a strategy H 0 = D[t, τ ](n + 1)
where we define t as follows. Let t(H) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } with j1 < .. < jm . Then
t = t(H) ∪ {T rjm +1,z(H 0 ) }.
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 T r is empty then σ = ∅ is a
leaf of T r.
12
ANDREW LEWIS
2. 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 σ, τ and for n ∈ ω we say that
σ ∈ γ(τ, n) if there are initial segments of σ and τ , σ 0 and τ 0 respectively, for which we have
0 0
already enumerated the axiom Γσ ,τ (n) = c for some c.
Definition 2.2. If a strategy H ‘puts Xs through’ a string τ then we insist that Xs be an
extension of τ . When stage s activity is terminated we define Xs to be the longest string that
it was put through during stage s. If Xs is not put through any strings during stage s then we
define Xs = ∅.
Definition 2.3. Given j ≥ 0 and τ ∈ 2<ω let n be the least such that Ψτj (n) ↑. We say that Ψτj
is of length n.
Definition 2.4. By a k-fold Ψj splitting of length at least l we mean k strings τ1 , .., τk whose
images Ψτj i (1 ≤ i ≤ k) are 1) of length at least l and 2) pairwise incompatible – for 1 ≤ i < i0 ≤ k
τ0
if the lengths of Ψτj i and Ψj i are n1 and n2 respectively then there exists n3 < n1 , n2 such that
τ0
Ψτj i (n3 ) 6= Ψj i (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 ω.
2.1. The A[t, k, φ, τ1 , τ2 , τ3 ](n) strategy. Here k ≥ 1, n ≥ 0, φ, τ1 , τ2 , τ3 are binary strings such
that φ ⊂ τi (for 1 ≤ i ≤ 3) and the τi are pairwise incompatible while t is a finite set of trees,
t = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } say, with j1 < .. < jm , m ≥ 1. Let H = A[t, k, φ, τ1 , τ2 , τ3 ](n).
At every stage at which H is passed control we will have defined the value l(H) ∈ ω. We call
φ, τ1 , τ2 , τ3 the ‘base strings’ for H.
The strategy searches for a k-fold Ψjm splitting ‘above’ τ1 , τ2 or τ3 (i.e. such that there
exists 1 ≤ i ≤ 3 every string in the splitting extends τi ) of length at least l(H) and lying on
T rjm−1 ,.im−1 . It also enumerates axioms of the form Γσ,τi (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 = γ(τi , n) for each 1 ≤ i ≤ 3. The following conditions will be satisfied.
1) As ∈
/ γ(φ, n).
2) For i 6= j, Pi ∩ Pj = γ(φ, n).
3) No axioms have been enumerated on proper extensions of τi for any i, i.e. for 1 ≤ i ≤ 3 we
have not enumerated any axioms of the form Γσ,τ (n0 ) = c such that τ properly extends τi (and
for any σ ∈ 2<ω , n0 ∈ ω, c ∈ {0, 1}).
4) φ will extend a leaf of T rjm ,im , φ0 say.
5) Suppose T r ∈ t − {T rjm ,im }. Each τi extends a different leaf of T r, τi (T r) say. Each τi is a
string that we have enumerated into T rjm−1 ,im−1 (let us agree that j0 = −1, i0 = 0).
2.1.2. The active environment. Every stage s at which the strategy is passed control the following conditions will be satisfied.
1) As ∈
/ γ(φ, n).
2) No Γ(n) axioms have been enumerated on strings compatible with any τi since the strategy
was first passed control except by this strategy i.e. no other strategy has since enumerated an
axiom of the form Γσ,τ (n) = c for τ compatible with some τi (and some σ ∈ 2<ω , c ∈ {0, 1}).
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
13
3) φ0 , as defined as in 4) of 2.1.1 above, is still a leaf of T rjm ,im .
4) Suppose T r ∈ t − {T rjm ,im } and for 1 ≤ i ≤ 3 let τi (T r) be defined as in 5) of 2.1.1. If any
string τ has been enumerated into T r extending τi (T r) then τ extends τi 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 Γσ,τi (n) = c where σ is of the same length as τi .
So if l is the length of the longest of the τi and s < l 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 Ψjm splitting such that there exists 1 ≤ i ≤ 3 every string in the splitting
extends τi , of length at least l(H) (noting that this value may have changed since the last stage
at which H was passed control) and lying on T rjm−1 ,im−1 , 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) As ∈ P1 put Xs through τ2 .
b) As ∈ P2 put Xs through τ3 .
c) As ∈ P
S3 then put Xs through τ2 .
d) As ∈
/ i Pi then put Xs through τ2 .
4) If Ks (n) = 1 then if:
a) As ∈ P1 put Xs through τ3 .
b) As ∈ S
P2 ∪ P3 then put Xs through τ1 .
c) As ∈
/ i Pi then put Xs through τ3 .
5) Suppose that we have put Xs through τi . Then if ΓAs ,τi (n) is not already defined enumerate
the axiom Γσ,τi (n) = Ks (n) where σ is the restriction of As to the length of τi .
6) Suppose that we have put Xs through τi . H passes control to the strategy H 0 = D[t0 , τi ](n+1)
where t0 = (t − {T rjm ,im }) ∪ {T rjm +1,z(H 0 ) }.
Definition 2.6. In the course of discussing a B, C or a D strategy H every string τ under
consideration will extend one of the base strings for H. We let φH (τ ) designate this string.
2.2. The B[t, k, φ1 , .., φr ](n) strategy. Here k, r ≥ 1, n ≥ 0, φ1 , .., φr ∈ 2<ω and t is a finite set
of trees, t = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } say, with j1 < .. < jm , m ≥ 1. We shall also have that
r divides k. Let H = B[t, k, φ1 , .., φr ](n). We call φ1 , .., φr the ‘base strings’ for H. The strategy
aims to produce a k-fold Ψjm splitting, τ1 , .., τk say, that we can enumerate into T rjm ,im and
such that for i 6= j, γ(τi , n) ∩ γ(τj , n) = γ(φH (τi ), n) ∩ γ(φH (τj ), 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 (3Σk−1
i=1 i) + 3 strings with all the strings in k/r of these
groups extending each φi (so we have an equal number of incompatible strings extending each
φi ) and such that for distinct τ, τ 0 , γ(τ, n) ∩ γ(τ 0 , n) = γ(φH (τ ), n) ∩ γ(φH (τ 0 ), n). These strings
will haveSbeen ‘delivered’ to H.
2) As ∈
/ ri=1 γ(φi , n).
3) No axioms have been enumerated on proper extensions of any strings delivered to H.
4) For 1 ≤ i ≤ r, φi extends a leaf of T rjm ,im , φ0i say.
14
ANDREW LEWIS
5) Suppose T r ∈ t − {T rjm ,im }. Each of the strings that has been delivered to H is a string
that has been enumerated into T rjm−1 ,im−1 (we agree that j0 = −1, i0 = 0). Each such string τ
extends a different leaf of T r, τ (T r) 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 Γ(n) axioms
have been enumerated on strings compatible with those delivered to H since H was first passed
control, except by the A 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. S
2) As ∈
/ ri=1 γ(φi , n).
3) If the strategy is in state k 0 ≥ 1 then As ∈
/ C(H, k 0 ) (to be defined subsequently).
0
4) For each 1 ≤ i ≤ r, φi as defined in 4) of 2.2.1 is still a leaf of T rjm ,im .
5) Suppose T r ∈ t − {T rjm ,im }. For each string τ delivered to H let τ (T r) be defined as in 5)
of 2.2.1. If any strings have been enumerated into T r extending τ (T r) these strings extend τ .
2.2.3. The initial instructions. Divide the set of strings delivered to H into triples so that every
k−1
string in a triple extends the same φi . Then divide these into k groups of (Σi=1
i) + 1 triples
so that every triple in the same group extends the same φi . Let the variable µ range over
this set of Striples and for each µ let us agree that µ = {τµ,1 , τµ,2 , τµ,3 } and that, for any n0 ,
γ(µ, n0 ) = i γ(τµ,i , n0 ). Call the groups of triples β1H , .., βkH . 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 A 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 τ1H and suppose that τ1H ∈ µ. Relabel if necessary so that µ ∈ β1H . We
wish to form the set of escape routes ρH
1,1 . 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’ τ1H . Choose µ1 ∈ β1H other than µ 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 − {T rjm ,im } and long enough such that no axioms have been enumerated
H
on proper extensions of the string, to give us our first string in ρH
1,1 . Choose µi from each βi ,
i > 1, discard two strings from this triple and find an escape route extension of the remaining
string to give us another string in ρH
1,1 , so that this set consists of k escape routes. From each
0
H
µ ∈ β1 other than µ1 , choose one string and discard the other two – ensuring that the string
τ1H is not discarded. From the set of 21 k(k − 1) non-discarded strings from triples in β1H other
than µ1 , take τ1H and any k − 2 others, discarding the rest. Let this set of k − 1 strings be called
R(H, 1). Define C(H, 1) = {σ : ∃τ ∈ R(H, 1), σ ∈ γ(τ, 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 B strategy enters a new state. The
point is that we cannot predict the order in which the A 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 As ∈ γ(µ, n) we transfer control to the strategy H 0 = A[t, k, φH (τµ,1 ), τµ,1 , τµ,2 ,
τµ,3 ](n) – which we call the ‘A(n) strategy for µ’. If l(H 0 ) has not yet been defined then choose
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
15
it to be larger than any number previously mentioned in the construction. If As ∈
/ γ(µ, n) for
any triple µ then just choose one and proceed similarly.
2.2.5. The instructions for the strategy in state 0 < k 0 < 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 µ we shall refer to µ as a ‘remaining triple’.
1) For 1 ≤ i ≤ k 0 we will have defined a string τiH and a set R(H, i).
H
2) We will have defined k 0 sets of non-discarded escape routes, ρH
k0 ,1 , ..., ρk0 ,k0 . Each set contains
(k − k 0 + 1) strings.
S 0
3) For distinct τ, τ 0 each of which is either in ki=1 R(H, i) or is a non-discarded escape route,
γ(τ, n) ∩ γ(τ 0 , n) = γ(φH (τ ), n) ∩ γ(φH (τ 0 ), n). The same is true for distinct τ, τ 0 each of which
S 0
is either in ki=1 R(H, i) or is in remaining µ, unless ∃µ(τ, τ 0 ∈ µ).
We perform the following instructions.
1) If one of the A strategies which H has passed control to has been declared successful since H
was declared to be in state k 0 then proceed as follows. Declare the strategy to be in state k 0 + 1.
If k 0 + 1 = k then label the string above which we have just found a splitting as τkH , declare H
to be successful and terminate stage s activity for the construction. Otherwise (given that one
of the A strategies has been successful since H was declared to be in state k 0 ):
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 splitting τkH0 +1 and suppose that τkH0 +1 ∈ µ.
Relabel if necessary so that µ ∈ βkH0 +1 (and so that each τiH , 1 ≤ i ≤ k 0 is from a triple in βiH ).
Choose remaining µk0 +1 ∈ βkH0 +1 other than µ, and discard two strings from this triple. Find an
‘escape route extension’ of the remaining string to give us the first escape route in ρH
k0 +1,k0 +1 .
H
0
Choose some remaining µi from each βi , 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 ρH
k0 +1,k0 +1
so that this set consists of k − k 0 escape routes. From each remaining µ0 ∈ βkH0 +1 choose one
string and discard the other two – ensuring that the string τkH0 +1 is not discarded. From the set
of non-discarded strings from triples in βkH0 +1 other than µk0 +1 , take τkH0 +1 and any (k − k 0 − 2)
others, discarding the rest. Let this set of (k − k 0 − 1) strings be called R(H, k 0 + 1).
c) For each 1 ≤ i ≤ k 0 , in turn, define a set of escape routes ρH
k0 +1,i as follows. Remove one
H
string from R(H, i) other than τi and find an escape route extension to give us the first escape
0
0
route in ρH
k0 +1,i . For each i > k + 1 choose some remaining µ and from that triple choose one
string, discarding the other two. Find an escape route extension of the remaining string to give
0
us another escape route in ρH
k0 +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).
S 0 +1
R(H, i), σ ∈ γ(τ, n)}. Terminate stage s activity for
d) Define C(H, k 0 + 1) = {σ : ∃τ ∈ ki=1
the construction.
2) Otherwise if it is the case that As ∈ γ(µ, n) for some remaining µ then pass control to the
A(n) strategy for µ, H 0 . If l(H 0 ) has not been redefined since H was declared to be in state
k 0 then redefine it to be larger than any number previously mentioned in the construction. If
As ∈
/ γ(µ, n) for any remaining µ then just choose remaining µ and proceed similarly.
2.3. The C[t, k, φ1 , .., φr ](n) strategy. Here k, r ≥ 1, n ≥ 0, φ1 , .., φr ∈ 2<ω and t is a finite set
of trees, t = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } say, with j1 < .. < jm . We also have that r divides k.
Let H = C[t, k, φ1 , .., φr ](n). We call φ1 , .., φr the ‘base strings’ for H.
16
ANDREW LEWIS
2.3.1. The initial environment. If the strategy is first passed control at stage s the following
conditions
S will be satisfied.
1) As ∈
/ ri=1 γ(φi , n).
2) No axioms have been enumerated on proper extensions of φi for any i.
3) Suppose T r ∈ t. Each φi extends a different leaf of T r, φi (T r) say.
2.3.2. The active environment. Suppose m ≥ 1. Every stage s at which H is passed control the
followingSconditions will be satisfied.
1) As ∈
/ ri=1 γ(φi , n).
2) No Γ(n) axioms have been enumerated on strings compatible with φ1 , .., φr since the strategy
was first passed control except by strategies above H.
3) For each T r ∈ t and for each 1 ≤ i ≤ r let φi (T r) be defined as in 2.3.1. Then φi (T rjm ,im ) is
still a leaf of T rjm ,im . If T r ∈ t − {T rjm ,im } and strings have been enumerated into T r extending
φi (T r) then these strings extend φi and have been enumerated into T r by strategies above H.
The C strategies with t = {T r−1,0 } are very simple. The first stage at which H is passed
control it will choose k/r incompatible extensions of each φi , enumerate these strings into T r−1,0 ,
deliver them to the strategy H 0 that we have specified H is intended to provide an environment
for, declare H 0 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 B 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 = B[t, k, φ1 , .. , φr ](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 H 1 = C[t0 , k1 , φ1 , .., φr ](n) where t0 = t − {T rjm ,im } and
k1 is the number of strings required in the initial environment of H1 i.e. k1 = 3k( 21 k(k − 1) + 1).
We specify that H 1 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
k 1 > 0 (here, and everywhere in the description of the C strategy, the superfix is used as a
counter rather than to indicate exponentiation). If As ∈
/ C(H1 , k 1 ) then pass control to H1 .
Sk 1
Otherwise ∃τ ∈ i=1 R(H1 , i), (As ∈ γ(τ, n)). There can only be one such τ (of course we shall
have to prove that this is the case). Suppose this τ is in R(H1 , j 1 ). This decides the set of
1
escape routes that we shall use. Let r1 = (k − k 1 + 1) and let ρH
= {τ11 , .., τr11 }. Define
k1 ,j 1
H2 = B[t, r1 , τ11 , ..., τr11 ](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 H 2 = C[t0 , k2 , τ11 , ..., τr11 ](n) where t0 , τ11 , .., τr11 are as in
step 1 and k2 is the number of strings required in the initial environment of H2 . We specify that
H 2 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
S 2
2
k > 0. If As ∈
/ C(H2 , k 2 ) then pass control to H2 . Otherwise ∃τ ∈ ki=1 R(H2 , i)(As ∈ γ(τ, n)).
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
17
There can only be one such τ . Suppose this τ is in R(H2 , j 2 ). Let r2 = (r1 − k 2 + 1) and let
2
= {τ12 , .., τr22 }. Define H3 = B[t, r2 , τ12 , ..., τr22 ](n). Proceed to the next step.
ρH
k2 ,j 2
Step y) At step y − 1, having found that we could not pass control to the strategy Hy−1 we will
have defined a strategy Hy = B[t, ry−1 , τ1y−1 , ..., τry−1
y−1 ](n). Check to see whether Hy has been
declared to be in state 0.
0
If NOT. Pass control to the strategy H y = C[t0 , ky , τ1y−1 , ..., τry−1
y−1 ](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 H y
is intended to provide an environment for Hy .
If SO. If Hy is in state 0 then pass control to it. Otherwise suppose that it is in state k y > 0.
S y
If As ∈
/ C(Hy , k y ) then pass control to Hy . Otherwise ∃τ ∈ ki=1 R(Hy , i)(As ∈ γ(τ, n)). There
can only be one such τ . Suppose this τ is in R(Hy , j y ). Let ry = (ry−1 − k y + 1) and let
H
ρkyy,j y = {τ1y , ..., τryy }. Define Hy+1 = B[t, ry , τ1y , ..., τryy ](n). Proceed to the next step.
So at each step in the iteration we check to see whether a certain B strategy has previously
been declared to be in state 0. If not then we pass control to a C strategy in order to provide
the strings required in the initial environment of this B 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 B 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 B 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 B 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 B 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 B strategies H0 that H has
passed control to has been declared successful. First we shall enumerate a set of splittings Ω.
Step 0. The first strategy that we shall consider is H0 . Take each of the splittings that have
been found by the A strategies that H0 passes control to and enumerate these splittings into Ω.
If H0 was the first B strategy ever passed control by H then our enumeration of Ω is completed.
Otherwise proceed to the next step.
0
Step y. Suppose that Hy−1 was a strategy with ρH
k1 ,k2 as its set of base strings, for some B
0
strategy H that H has passed control to and for some k 1 , k 2 ∈ ω. Define Hy = H 0 . If k 1 = 1
then proceed to the next step, unless Hy is the first B strategy that H ever passed control to in
which case (presuming k 1 = 1) our enumeration of Ω is completed. Otherwise we have defined
H
H
strings τ1 y , .., τk1y . Take the splittings that have been found by the A strategies passed control
H
by Hy above each of these strings, except that which was found above τk2y , and enumerate these
splittings into Ω. If Hy was the first B strategy passed control by H then our enumeration of Ω
is completed, otherwise proceed to the next step.
18
ANDREW LEWIS
Label the splittings in Ω as ω1 , .., ωk according to the order in which they were found, so that
ωk was the first and ω1 the last. Take a string τ1 from ω1 and another string τ2 from ω2 such
1
2
1
2
3
that Ψτjm
and Ψτjm
are incompatible. Take a string τ3 from ω3 such that Ψτjm
, Ψτjm
and Ψτjm
are
0
pairwise incompatible, and so on. For each 1 ≤ i ≤ k take an extension τi of τi long enough
that it extends a leaf in each of the trees in t − {T rjm ,im }, 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 τ10 , .., τk0 into T rjm ,im , deliver these strings to H 0 which is the strategy which we have
specified H is intended to provide an environment for, declare H 0 to be in state 0 and terminate
stage s activity for the construction.
2.4. The D[t, φ](n) strategy. Here n ≥ 0, φ is a finite binary string and t is a finite set of trees,
t = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } say with j1 < .. < jm . Let H = D[t, φ](n). We call φ 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 ∈
/ γ(φ, n).
2) No axioms have been enumerated on proper extensions of φ.
3) Suppose T r ∈ t. Then φ extends a leaf of T r, φ(T r) say.
2.4.2. The active environment. Every stage s at which H is passed control the following conditions will be satisfied.
1) As ∈
/ γ(φ, n).
2) No Γ(n) axioms have been enumerated on strings compatible with φ since the strategy was
first passed control except by strategies above H.
3) For each T r ∈ t let φ(T r) be defined as in 2.4.1. If T r ∈ t and strings have been enumerated
into T r extending φ(T r) then these strings extend φ and have been enumerated into T r 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 C[t, 3, φ](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 τ1 , τ2 , τ3 – these strings will have been delivered to
H. At this stage define Pi = γ(τi , n) (so that Pi , unlike γ(τi , n) is fixed and will not change at
subsequent stages). Let l be the length of the longest τi .
At any stage s we now proceed as follows:
1) If l > s then terminate stage s activity for the construction.
2) If Ks (n) = 0 then if:
a) As ∈ P1 put Xs through τ2 .
b) As ∈ P2 put Xs through τ3 .
c) As ∈ P
S3 put Xs through τ2 .
d) As ∈
/ i Pi put Xs through τ2 .
3) If Ks (n) = 1 then if:
a) As ∈ P1 put Xs through τ3 .
b) As ∈ S
P2 ∪ P3 put Xs through τ1 .
c) As ∈
/ i Pi put Xs through τ3 .
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
19
4) Suppose that we have put Xs through τi . If ΓAs ,τi (n) is not yet defined then enumerate the
axiom Γσ,τi (n) = Ks (n) where σ is the restriction of As to the length of τi .
5) Suppose that we have put Xs through τi . Pass control to the strategy H 0 = D[t0 , τi ](n + 1)
where t0 = t ∪ {T rjm +1,z(H 0 ) }.
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 = D[t, ∅](0), where t = {T r−1,0 , T r0,z(H) }. At the beginning of each stage s, control is first
passed to this strategy.
3. The verification.
When an A 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 H0 enumerates the splitting
into T r, since the construction would have permanently terminated otherwise.
Definition 3.1. If H is a strategy Υ[t0 , ..](.) then t(H) = t0 . Suppose that for some m ≥ 1
and j1 < .. < jm , t0 = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im }. If Υ = D then t? (H) = t0 . Otherwise
t? (H) = t0 − {T rjm ,im }.
Lemma 3.1. At each stage s a finite 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 C strategy H such that
t(H) = {T r−1,0 }, 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 C strategy H0 checks to see whether a B strategy H1 has yet been
declared to be in state 0 we say that H1 is ‘checked for use’, or more specifically that ‘H0 checks
H1 for use’.
Definition 3.3. In the context of discussing a B strategy H0 suppose that we have defined a set
of escape routes ρ. We refer to the B strategy H1 with base strings defined to be those in ρ, such
that t(H1 ) = t(H0 ) and which searches for a k-fold splitting, where k is the number of strings in
ρ, as the B strategy ‘above’ ρ.
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 ‘α 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 B
strategy H was responsible for passing control to any B strategy above a set of escape routes
defined for H, and to any C strategy intended to provide an environment for H.
20
ANDREW LEWIS
Definition 3.4. We define the α ordering by recursion. Suppose H0 is a B strategy and let H1
be the C strategy which passes control to H0 . We say that H2 is α-below H0 if either:
a) H2 = H0 ,
b) H2 is α-below H1 , or
c) H2 is one of the B strategies that H1 checks for use before H0 at any stage at which H0 is
passed control.
Suppose H0 is a C strategy and let H1 be the strategy that H0 is intended to provide an environment for. We say that H2 is α-below H0 if either H2 = H0 or H2 is α-below H1 . Finally
suppose that H0 is a D or an A strategy and let H1 be the strategy which passes control to H0 .
We say that H2 is α-below H0 if H2 = H0 or H2 is α-below H1 .
Definition 3.5. We say that H0 is α-above H1 if H1 is α-below H0 . We say that H0 is strictly
α-below/ α-above H1 if H0 is α-below/ α-above H1 and H0 6= H1 . We say that H0 is the α
predecessor of H1 if H0 is strictly α-below H1 and for all strategies H2 strictly α-below H1 , H0
is α-above H2 .
It is convenient to assume that A strategies are in state 0 until declared successful, that B
strategies are in state -1 until declared to be in state 0, and that C strategies are in state -1 until
declared successful.
Definition 3.6. We say that H0 is an α successor of H1 if H1 is the α predecessor of H0 – if
H1 is an A strategy or a C strategy we also say that H0 is an α successor of H1 while in state
k, for any k ∈ ω, in this case. Suppose H1 is a D strategy. We say that H0 is an α successor
of H1 while in state k ∈ ω if H1 is declared to be in state k, H0 is an α successor of H1 and H0
is a C strategy if k = −1, H0 is a D strategy if k = 0. Finally suppose H1 is a B strategy. We
say that H0 is an α successor of H1 while in state k ∈ ω if H0 is an α successor of H1 , H1 is
declared to be in state k, and while H1 is in this state either:
a) k ≥ 0 and H0 is an A strategy, none of the base strings for which have been discarded by H1 .
b) k ≥ 0 and H0 is a B strategy above a set of escape routes we have defined for H1 and which
are not discarded.
c) k = −1 and H0 is a C strategy.
Definition 3.7. We define a strategy H0 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 α-below H0 , H1 say, is
declared to be in a new state k such that none of the α successors of H1 while in state k are
α-below H0 .
Definition 3.8. If H is a B, C, or D strategy then the α base strings for H are just the base
strings for H. Suppose H is an A strategy and that the base strings for H are φ, τ1 , τ2 , τ3 , so
that the τi are pairwise incompatible extensions of φ. The α base strings for H are τ1 , τ2 , τ3 .
Lemma 3.2. Suppose t(H0 ) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im }, m ≥ 1, j1 < .. < jm , j0 = −1, i0 = 0
and that 0 ≤ r0 < r. If H1 enumerates a splitting into T rjr ,ir then each string in the splitting
extends a different leaf of T rjr0 ,ir0 .
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 T rjr0 ,ir0 . Let t(H1 ) =
0 so that j = j 0 and i = i0 . For fixed r 0 the
{T r−1,0 , T rj10 ,i01 , .., T rj 0 0 ,i0 0 } with j10 < ... < jm
0
r
r
m0
m0
m m
0
0
result follows immediately by induction on r since r = m , jr−1 = jm0 −1 and ir−1 = i0m0 −1 .
Lemma 3.3. Suppose that H0 is active when H1 which is not α-above or α-below H0 is passed
control and that T r ∈ t? (H0 ) ∩ t(H1 ). Let φ1 , .., φr be the α base strings for H0 and let φ01 , .., φ0r0
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
21
be the α base strings for H1 . For 1 ≤ i ≤ r and 1 ≤ i0 ≤ r0 , φi and φ0i0 extend incompatible
strings on T r.
Proof. Let H2 be the α-highest of all the strategies α-below H0 and H1 . Since T r ∈
t? (H0 ) ∩ t(H1 ) it must be the case that T r ∈ t? (H2 ). Suppose that for j1 < .. < jm ,
t? (H2 ) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im }. If 1 ≤ i ≤ r and 1 ≤ i0 ≤ r0 , φi and φ0i0 extend incompatible strings on T rjm ,im which H2 was provided with upon being declared to be in state
0 (let us suppose that an A 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 T r.
We restate the convention that if T r is empty then ∅ is regarded as a leaf of T r.
Lemma 3.4. When a strategy H0 is first passed control or checked for use:
a) the α base strings for H0 will extend leaves of each of the trees T r ∈ t(H0 ).
b) no axioms will have been enumerated on strings properly extending the α base strings for H0 .
Proof. The proof is by induction on the point of the construction at which H0 is first passed
control or checked for use. Suppose that H1 is the α predecessor of H0 and that H0 is first passed
control or checked for use at stage s. If H1 is in state -1 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. T r ∈ t? (H1 ). Suppose first that H1 is an A or a D strategy, or that H1 is a B strategy
and H0 is an A strategy. When H1 was declared to be in state 0, b) above and also a) above as
regards T r were satisfied. If H1 is a B or a D strategy this follows since we are working under
the assumption that the construction would permanently terminate otherwise. If H1 is an A
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 T r only)
are those α-above H1 . Suppose H2 is α-above H1 and is passed control or checked for use before
H0 , after the point of the construction when H1 was declared to be in state 0. Let φ1 , .., φr be
the α base strings for H0 and let φ01 , .., φ0r0 be the α base strings for H2 . For 1 ≤ i ≤ r and
1 ≤ i0 ≤ r0 , φi and φ0i0 extend incompatible strings which H1 was provided with upon being
declared to be in state 0. So suppose that H1 is a B strategy and H0 is a B strategy also, and
suppose that H1 is in state k when H0 is first checked for use. Repeat the relevant parts of the
argument above, replacing ‘ H1 was declared to be in state 0 ’ with ‘H1 was declared to be in
state k’. Clearly we need not consider b) in the cases that follow, since certainly T r−1,0 ∈ t? (H1 ).
Case 2. T r ∈ t? (H0 ) − t? (H1 ). Then a), as regards T r, follows immediately since the only
strategies that can enumerate strings into T r are those α-above H0 .
Case 3. T r ∈
/ t? (H0 ) and T r ∈
/ t? (H1 ). Let H2 be the strategy α-highest of all those α-below H0
?
such that T r ∈ t (H2 ). When H2 was first passed control or checked for use, the base strings for
this strategy extended leaves of T r, by the induction hypothesis. By lemma 3.3 it follows that
the only strategies which could subsequently have violated this condition are those α-above H2 .
But the fact that H0 is passed control or checked for use at stage s means that such strategies
have not yet enumerated any strings into T r.
Lemma 3.5. Suppose that H0 enumerates a splitting into T r. Each string in the splitting
extends what was (prior to this enumeration) a leaf of T r.
Proof. Let H1 be the strategy which H0 declares to be in state 0 upon being declared
successful. When H1 was first passed control or checked for use, the α base strings for this
strategy each extended a leaf of T r. By lemma 3.3 it follows that the only strategies which
22
ANDREW LEWIS
could have subsequently violated this condition (prior to the enumeration in question) are those
α-above H1 . No such strategy has enumerated strings into T r prior to stage s, otherwise H0
would not be passed control at this stage.
Lemma 3.6. Let t(H0 ) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } with j1 < .. < jm , m ≥ 1, and suppose
that at stage s0 , H0 enumerates a splitting τ1 , .., τk into T rjm ,im . Suppose given r < m. By
lemma 3.2 each string τi extends a leaf τi0 of T rjr ,ir , such that for 1 ≤ i < i0 ≤ k we have
τi0 6= τi00 , when this enumeration takes place. Suppose further that at stage s1 > s0 a strategy H1
enumerates a string τ into T rjr ,ir extending τi0 and that, subsequent to this enumeration, there
is at least one active strategy H2 with T rjm ,im ∈ t(H2 ). Then τ extends τi .
Proof. For 1 ≤ i < i0 ≤ k we have that τi0 and τi00 are incompatible. It therefore follows
by lemma 3.5 that if, at any point of the construction after H0 enumerates the splitting into
T rjm ,im , a string compatible with τi0 extends a leaf of T rjm ,im the string extends τi . Let H2 be
the α-highest of all the strategies α-below H0 such that T rjm ,im ∈ t(H2 ) and which is active
after H1 enumerates τ into T rjr ,ir . Then T rjm ,im ∈ t? (H2 ). By lemma 3.3 H1 must be α-above
H2 . Let H3 be the α-highest of all the strategies α-below H0 and H1 . Then T rjm ,im ∈ t? (H3 ).
Suppose first that H3 is declared to be in state 0 when, or at some point after, H0 enumerated
the splitting into T rjm ,im . But then τ extends one of the strings which H3 is provided with
upon being declared to be in state 0. Since T rjm ,im ∈ t? (H3 ), τ extends a leaf of T rjm ,im when
H3 is declared to be in state 0 and therefore extends τi . So suppose that H3 had been declared
to be in state 0 before H0 enumerated the splitting. By the definition of H3 it follows that
if H4 is the α successor of H3 α-below H0 and H5 is the α successor of H3 α-below H1 then
H4 6= H5 . If H5 is a D or an A strategy, or if H5 and H4 are both B strategies we obtain an
immediate contradiction, by lemma 3.2. So suppose H5 is a B strategy and that H4 is an A
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 τ extends an escape route which we defined for H3 after H0
enumerated the splitting into T rjm ,im and which, since it must be compatible with τi0 , therefore
extends τi .
Lemma 3.7. If H0 = Υ[t, k, φ1 , .., φr ](n) is passed control or checked for use and Υ ∈ {B, C}
then rk.
Proof. The proof is by induction on the α ordering. Suppose that H1 is the α precessor of
H0 . If H1 is a D strategy then r = 1, k = 3. If H1 is a B strategy which is in state k 0 > 0
at any stage at which H0 is passed control or checked for use then k = r. If H1 is a C or
B strategy which is in state -1 whenever H0 is passed control or checked for use then suppose
H1 = Υ[t1 , k1 , φ1 , .., φr ](n). We have, by the induction hypothesis, that rk1 . Since k1 k the result
follows.
Lemma 3.8. Suppose a B 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 B strategy which searches for a k-fold splitting we
specify that there should be k groups of 3( 21 k (k − 1) + 1) strings. We divide these into k groups
of 12 k(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 + .. + k − 1 = 12 k(k − 1). Suppose
that a group is the k 0th group above which we find a splitting (k 0 ≥ 1). Before finding this
splitting we will already have removed 1 + 2 + .. + k 0 − 1 triples from this group. If k 0 < k then
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
23
upon finding the splitting we need to be sure that we have k − k 0 + 1 triples remaining. But
then 12 k(k − 1) − (1 + 2 + .. + k 0 − 1) = k 0 + (k 0 + 1) + .. + (k − 1) ≥ k − k 0 .
Consider the three following statements.
(†2 ) Suppose that the strategy H is passed control or checked for use at stage s. If H is a
B(n), C(n) or D(n) strategy then As ∈
/ γ(φ, n) for any string φ which is a base string for H. If
H is an A(n) strategy then As ∈
/ γ(φ, n) for that string φ which is a base string but not an α
base string for H.
(†3 ) Suppose that a B(n) strategy H is passed control or checked for use at stage s and is in
S
state k > 0. For distinct τ, τ 0 each of which is either in ki=1 R(H, i) or is a non-discarded escape
route, γ(τ, n) ∩ γ(τ 0 , n) = γ(φH (τ ), n) ∩ γ(φH (τ 0 ), n). The same is true for distinct τ, τ 0 each of
S
which is either in ki=1 R(H, i) or is in remaining µ, unless ∃µ(τ, τ 0 ∈ µ).
(†4 ) Suppose that a C(n) strategy H0 is declared successful and enumerates a splitting τ1 , .., τk
into a tree T r. Let H1 be the strategy which H0 declares to be in state 0 and let φ1 , .., φr
be the base strings for H1 . Then k/r of the τi extend each φi0 and, for 1 ≤ i < i0 ≤ k,
γ(τi , n) ∩ γ(τi0 , n) = γ(φH1 (τi ), n) ∩ γ(φH1 (τi0 ), n).
Lemma 3.9. Statements (†2 ), (†3 ) and (†4 ) are correct.
Proof. The proof is by induction on the point of the construction in question. First we prove
(†2 ) for the induction step. Let H1 be the α predecessor of H. If H1 is a C strategy, or a B or
D 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 D(n − 1) strategy or a A(n − 1)
strategy which is in state 0 at any stage at which H is passed control. If H1 is a D(n − 1)
strategy let φ0 be the base string for this strategy and if H1 is an A(n) strategy then let φ0 be
the string which is a base string but not an α base string for this strategy. We have by the
induction hypothesis that when H1 was passed control at stage s, As ∈
/ γ(φ0 , n − 1). Since for
any string τ , γ(τ, n) ⊆ γ(τ, n − 1) it follows by the induction hypothesis on (†4 ) and by lemma
3.3 that H1 chooses τ to put Xs through such that As ∈
/ γ(τ, n). So suppose that H1 is a B(n)
strategy which is state k ≥ 0 at any stage at which H is passed control or checked for use. If H
is an A(n) strategy then the result follows immediately from the induction hypothesis. Assume,
then, that H is a B(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 (†2 ) and (†3 ).
We prove (†3 ) for the induction step. By the induction hypothesis on (†4 ), when H was
declared to be in state 0 the strings that were delivered to H satisfied the property that for
distinct τ, τ 0 , γ(τ, n) ∩ γ(τ 0 , n) = γ(φH (τ ), n) ∩ γ(φH (τ 0 ), n). By lemma 3.3 the only strategies
which could have subsequently violated this condition are those α-above H. Suppose first that
τ ∈ R(H, i) and that either τ 0 ∈ R(H, i0 ) for i0 ≥ i or τ 0 is in remaining µ0 . Then the strategies
strictly α-above H have only enumerated Γ(n0 ) axioms on strings compatible with τ, τ 0 for n0 > n.
Suppose τ ∈ µ. Before τ was enumerated into R(H, i), control was passed to the A(n) strategy
for µ every stage s0 at which H was passed control and As0 ∈ γ(µ, n). Control was passed to
the A(n) strategy for µ0 every stage s0 at which H was passed control and As0 ∈ γ(µ0 , n). From
the point at which τ was enumerated into R(H, i), control was never passed to H at any stage
s0 at which As0 ∈ γ(τ, n). Almost precisely the same argument suffices for the remaining cases.
If τ or τ 0 is a non-discarded escape route then it is clear that the strategies strictly α-above H
have only enumerated Γ(n) axioms on proper extensions of the string.
We prove (†4 ) for the induction step. Let H2 be the α successor of H0 and let H3 be the B
strategy which H0 has passed control to and which has been declared successful. For each of
24
ANDREW LEWIS
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 α-above H2 and α-below H3 upon being declared to be in state 0, that
precisely one of the two situations below applies.
a) There is precisely one τi , 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 τi which extends an escape route defined for H2 extending a string in
this group.
It follows immediately that k/r of the τi extend each φi0 by considering the induction hypothesis
as regards the strings that H2 was provided with upon being declared to be in state 0.
By lemma 3.4 we know that when H0 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
α-above H0 . We shall suppress mention of any use of lemma 3.3 in what follows. Suppose given
1 ≤ i < i0 ≤ k and (without loss of generality) suppose that all of the following are true.
a) τi extends a string in a splitting found by an A(n) strategy passed control by H i and τi0
0
extends a string in a splitting found by an A(n) strategy passed control by H i .
b) τi extends τi0 , a string which H i was provided with upon being declared to be in state 0, and
0
τi0 extends τi00 , a string which H i was provided with upon being declared to be in state 0.
c) The splitting above τi00 was found before that above τi0 .
We show, first of all, that γ(τi0 , n) ∩ γ(τi00 , n) = γ(φH i0 (τi0 ), n) ∩ γ(φH i0 (τi00 ), n).
0
Case 1. H i = H i . The result follows immediately by the induction hypothesis on (†3 ).
0
0
Case 2. H i 6= H i . Let τ be the non-discarded escape route for H i which τi0 extends. Since it follows from the induction hypothesis on (†3 ) that γ(τi00 , n)∩γ(τ, n) = γ(φH i0 (τi00 ), n)∩γ(φH i0 (τ ), n)
0
we are left to show that the strategies strictly α-above H i do not enumerate axioms of the form
0
0
0
Γσ ,τ (n) = c for τ 0 extending τ and σ 0 ∈ γ(τi00 , n). But this is clear – suppose H i is in state
i0
k 0 at any stage at which H i is passed control or checked for use and that τi00 = τjH . Suppose
i0
0
0
i
0
τ ∈ ρH
k0 ,j 0 . Then j 6= j and at any stage s such that H is checked for use while in state k and
i0
As ∈ γ(τi00 , n) we would proceed to check the B strategy above the set of escape routes ρH
k0 ,j for
use.
So we have shown that γ(τi0 , n) ∩ γ(τi00 , n) = γ(φH i0 (τi0 ), n) ∩ γ(φH i0 (τi00 ), n), but since no Γ(n)
axioms have been enumerated on proper extensions of τi0 or τi00 this means that γ(τi , n)∩γ(τi0 , n) =
γ(φH i0 (τi ), n) ∩ γ(φH i0 (τi0 ), n). It follows by repeated applications of (†3 ) that γ(φH i0 (τi ), n) ∩
γ(φH i0 (τi0 ), n) = γ(φH1 (φH i0 (τi )), n) ∩ γ(φH1 (φH i0 (τi0 )), n) = γ(φH1 (τi ), n) ∩ γ(φH1 (τi0 ), n). Thus
γ(τi , n) ∩ γ(τi0 , n) = γ(φH1 (τi ), n) ∩ γ(φH1 (τi0 ), 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.
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
25
Lemma 3.10. The axioms enumerated for Γ are consistent.
Proof. Suppose an A or a D strategy enumerates an axiom Γσ,τ (n) = c at stage s of the
construction. Then σ ⊆ As , prior to the enumeration of this axiom As ∈
/ γ(τ, n), and no Γ(n)
axioms have been enumerated on proper extension of τ .
Lemma 3.11. Every strategy is finitary 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 A , B and D strategies are finitary, so we are left to prove the result
for C strategies.
So suppose that the strategy H = C[t, k, φ1 , .., φr ](n) is passed control at an infinite number
of stages and that there is no B or C strategy H1 for which there exists s, after stage s H always
passes control to H1 . It must be the case that H is never declared successful.
Let H1 = B[t, k, φ1 , .., φr ](n). We may consider an infinite ‘chain’ of B(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 C(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 C(n) strategy. Let k 1 be
the greatest state that H1 is declared to be in (here the superfix is used as a counter rather
than to indicate exponentiation). Then k 1 > 0 otherwise there would be a stage, after which,
whenever H is passed control it would pass control to H1 . Clearly k 1 < k otherwise H would be
declared successful. Define M1 = k 1 − 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 B(n) strategy above
this set of escape routes. The C(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 C(n) strategy. If k 2 is the greatest state that H2 is
ever declared to be in (and where the superfix is used as a counter) then k 2 > 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 k 2 < k − M1 . Define M2 = M1 + k 2 − 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 B(n) strategy above this set of escape routes. The C(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 k 3 be the greatest state that H3 is declared to be in. Then 0 < k 3 < k − M2 .
Define M3 = M2 + k 3 − 1, and so on.
Now {Mi }i≥1 is a nondecreasing sequence of natural numbers such that, for all i ≥ 1, k −Mi ≥
2. Thus limi→∞ (k − Mi ) ↓≥ 2. This means that there exists j ∈ ω, ∀i ≥ j(k i = 1). For i ≥ j
consider the strategy Hi . Let Ji be the set of all those strings σ such that, before Hi is declared
to be in state 1, we have enumerated some axiom Γσ,τ (n) = c such that τ is compatible with
one of the strings in R(Hi , 1) but properly extends φHi (τ1Hi ) (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
l(i) be the greatest such that all strings in Ji agree on (and are defined on) arguments ≤ l(i).
Since the approximation to A must converge, it follows that as i → ∞ so does l(i). This allows
26
ANDREW LEWIS
us to decide longer and longer initial segments of A. Thus A would be computable, which gives
us the required contradiction.
Definition 3.9. Suppose that H is a D(n) strategy. We define the ‘H cluster of strategies’ to
be H and all the C(n), B(n) and A(n) strategies above H on the tree of strategies.
Lemma 3.12. If there is a stage s1 after which control is always passed to a strategy H =
D[t, φ](n) then there is a stage s2 , a string τ ⊃ φ and a D(n + 1) strategy, H 0 , such that ∀s ≥ s2
the strategies in the H cluster do not terminate stage s activity, put Xs through τ and pass
control to H 0 . Furthermore, the strategies in the H cluster ensure that ΓA,τ (n) = K(n).
Proof. Let H0 = 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+1 ↓ are both C strategies then t(Hi+1 ) ⊂ t(Hi ). Thus there
exists a least i0 such that Hi0 ↑. Each strategy Hi , i < i0 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 Hi0 −1 always puts Xs through the same string τ and passes control to the same D(n + 1)
strategy at any stage s. Hi0 −1 ensures that ΓA,τ (n) = K(n).
Lemma 3.13. X is defined and ΓA,X = K.
Proof. Let H0 be the strategy at the bottom of the tree of strategies and for each i > 0 let
Hi be the D 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 H0 cluster always put
Xs through the same (non-empty) string τ0 and pass control to H1 . The strategies in the H0
cluster ensure that ΓA,τ0 (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 τi ⊃ τi−1 and pass control to Hi+1 . The
strategies in the Hi cluster ensure that ΓA,τi (i) = K(i).
Lemma 3.14. X is a set of minimal degree.
Proof. Let H0 be the strategy at the base of the tree of strategies and for all i > 0 let Hi be that
D 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 i0 , i1 ∈ ω (where we use the superfix as a counter) such that for all i ≥ i0 ,
T rj,i1 ∈ t(Hi ). For each i ≥ i0 , the strategies in the Hi cluster enumerate strings into T rj,i1
extending φi ⊂ X, the base string for Hi . Thus X ≤T ΨX
j .
0
b) There exists i which is the greatest such that there exists i1 , T rj,i1 ∈ t(Hi0 ). Suppose
t(Hi0 ) = {T r−1,0 , T rj1 ,i1 , .., T rjm ,im } and that jr = j. Then there is a strategy in the Hi0 cluster
which is passed control at an infinite number of stages and which never ceases searching for a
Ψj splitting above the base string for Hi0 +1 on T rjr−1 ,ir−1 . Also T rjr−1 ,ir−1 ∈ t(Hi ) for all i > i0
so that for all i > i0 the strategies in the Hi cluster enumerate strings into T rjr−1 ,ir−1 extending
the base string for Hi . Thus, if it is total ΨX
j is computable.
Remark: In the paper ‘A single minimal complement for the c.e. degrees’ we show that
there exists a minimal degree below 00 which complements every c.e. degree 0 < a < 00 – 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 ≤ 00 and any uniformly ∆2 sequence of
degrees b0 , b1 , b2 , .. such that ∀i(bi 6≥ c), there exists 0 < a < 00 such that for all i ≥ 0, a ∨ bi 6≥ c.
If c is c.e. and b0 , b1 , b2 , .. are uniformly (strictly) below c then there exists such an a below c. In
MINIMAL COMPLEMENTS FOR DEGREES BELOW 00
27
‘The minimal complementation property above 00 ’ we show that every degree above 00 satisfies
the minimal complementation property that we have proved for 00 in this paper.
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Andrew Lewis: Department of Pure Mathematics, School of Mathematics, University of Leeds,
Leeds, LS2 9JT, U. K. Email: [email protected]