definable incompleteness and friedberg splittings

Definable Incompleteness and Friedberg SplittingsAuthor(s): Russell MillerSource: The Journal of Symbolic Logic, Vol. 67, No. 2 (Jun., 2002), pp. 679-696Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2694945 .

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THE JOURNAL OF SYMBOLIC LOGIC

Volume 67, Number 2, June 2002

DEFINABLE INCOMPLETENESS AND FRIEDBERG SPLITTINGS

RUSSELL MILLER

Abstract. We define a property R (A0, A1) in the partial order g of computably enumerable sets under inclusion, and prove that R implies that A0 is noncomputable and incomplete. Moreover, the property is nonvacuous, and the AO and Al which we build satisfying R form a Friedberg splitting of their union

A, with Al prompt and A promptly simple. We conclude that A0 and Al lie in distinct orbits under

automorphisms of A, yielding a strong answer to a question previously explored by Downey, Stob, and Soare about whether halves of Friedberg splittings must lie in the same orbit.

?1. Introduction. The computably enumerable sets form an upper semi-lattice under Turing reducibility. Under set inclusion, they form a lattice A, as first noted by Myhill in [14], and the properties of a c. e. set as an element of g often help determine its properties under Turing reducibility. Even before Myhill, Post had suggested that there should be a nonvacuous property of c. e. sets, definable without reference to the Turing degrees, which would imply that the Turing degree of such a set must lie strictly between the computable degree 0 and the complete c. e. degree 0'.

Post's own attempts to find such a property failed. The properties he defined turned out to be extremely useful in computability theory, but each of them- simplicity, hypersimplicity, and hyperhypersimplicity actually does hold of some complete set. The existence of a Turing degree between 0 and 0' was first proven by completely different means, namely the finite injury constructions of Friedberg and Muchnik ([6], [13]).

The term "Post's Program" eventually came to denote the search for an 9- definable property implying incompleteness. Of the properties proposed by Post, all except hypersimplicity turned out to be definable in A, and other i-definable properties, such as maximality, were developed and studied in their own right. Nevertheless, Post's Program remained unfinished until 1991, when Harrington and Soare ([7]) found a property Q(A) definable in W such that every A satisfying Q must be both noncomputable and Turing-incomplete. We give their definition of Q (A):

Q(A): (0C)ACmC (VB C C) (:D C C) (VS)SC

(B n (S - A) = D n (S - A)

= (:T) [C c T &A n (S n T) = B n (S n T)]).

Received June 3, 2000; revised May 5, 2001. This article is the second chapter of a Ph. D. thesis at the University of Chicago under the supervision

of Robert I. Soare, to whom the author is grateful for extensive conversations and suggestions.

( 2002, Association for Symbolic Logic 0022-4812/02/6702-001 2/$2.80

679



680 RUSSELL MILLER

Here S E C abbreviates (3S) [S U S = C & S n = 0]. (All variables represent elements of t, namely c. e. sets.) A H B denotes the union of two disjoint sets A and B. Also, A Cm C abbreviates "A is a major subset of C," meaning that A c C with C - A infinite such that for every W, if C c W, then A - W is finite. Since the property of being finite is W-definable, the statement A Cm C is i-definable as well.

In this paper we generalize the property Q(A) to an i-definable property R(AO, Al) of two c. e. sets. The statement of R is as follows:

R(Ao, A1): AO n Al 0 & (3C) (VB C C) (]D C C) (VS C C) (3T)

AO U Al Cm C & [(B n (S - AO)) U Al = (D n (S - AO)) U Al

[C ciT & (AOSo nT) U A =(B n S nT) U A1]]].

This property can be read to say that AO satisfies the Q-property on Al. Indeed, the statement R(AO, 0) is equivalent to Q(Ao). In Section 2 we prove that just as with the Q-property, R(Ao, AI) implies that AO is not of prompt degree, and hence not Turing complete in Y2?. (A set which is not of prompt degree is said to be tardy, and since AO satisfies an i-definable property implying tardiness, we say that AO is "definably tardy." Since all tardy sets are incomplete, we also say that AO is "definably incomplete.")

Alternatively, we can interpret R(AO, Al) in the lattice t/ I, where a is the principal ideal in g generated by A1. (See [15], p. 225.) In this lattice, C C? D is defined to mean C C D U AI, and C ~ D if C c i D and D C s C. Essentially, R(Ao, A1) says that Q(AO) holds in W Id, with containment and equality replaced by Ci and z The only differences are that we cannot state the properties AO n A1 = 0 or AI C C in 9/1, and that we have left the quantifier (VS E C) in R(AO, Al) just as in the original Q-property, rather than restating it to hold on AI. Choosing not to restate it makes the R-property slightly stronger, but the stronger version can still be satisfied.

In Section 3 we construct c. e. sets AO and Al satisfying R, to show that the R-property is non-vacuous. AO and Al will also be noncomputable. Thus, the following R2-definable formula is non-vacuous:

(3Al) [Ao >To 0&R(AO, AI)].

This formula guarantees that AO is noncomputable and incomplete, just as the property Q(A) does for A. (Recall that computability is equivalent to the property of having a complement in 9 .)

We then consider Friedberg splittings. Two disjoint c. e. sets Bo and B, form a Friedberg splitting of B = Bo L B1 if for every c. e. W:

W - B is not c. e. a==> neither W - Bo nor W - B, is c. e.

The sets Bo and B, are each said to be half of this Friedberg splitting. The sets Ao and AI which we construct will have the additional property of forming a Friedberg splitting of their union.

We use the R-property to show that AO and AI cannot lie in the same orbit under automorphisms of A. (In the argot of this topic, we say that AO and Al are not automorphic. Two sets are automorphic if they lie in the same orbit.) This will



DEFINABLE INCOMPLETENESS AND FRIEDBERG SPLITTINGS 681

follow because the AI we construct will be of prompt degree, hence automorphic to a complete set, by another result of Harrington and Soare in [7].

The orbits of halves of Friedberg splittings have been a subject of interest for some time, at least since the discovery of the hemimaximal sets. A set is hemimaximal if it is half of a nontrivial splitting of a maximal set. This is W-definable, and Downey and Stob proved that the hemimaximal sets form an orbit (see [3]).

Since the maximal sets themselves form an orbit, and since few orbits are known in A, this led to the conjecture that if & is any orbit in A, then the collection of "hemi-&" sets, i.e. halves of nontrivial splittings of sets in &, might also be an orbit. Alternatively, it was conjectured that halves of Friedberg splittings of sets in a might form an orbit. (For the orbit of maximal sets, these classes coincide, since any nontrivial splitting of a maximal set is automatically a Friedberg splitting.)

Downey and Stob refuted both conjectures in [5], by producing two Friedberg splittings Bo H B = Co Li Cl of the same set B, which were definably different in A. Hence Bo and Co satisfy different 1-types in the language of inclusion and cannot be automorphic.

The present result goes a step further. Since AO is definably tardy, every set in its orbit must also be tardy, and hence AI must lie in a different orbit. This is thus the first example of a single Friedberg splitting with the two halves known to lie in different orbits in A. It is also the first application of Harrington and Soare's Q-property to derive results about Friedberg splittings.

Our notation mostly follows that of [16]. The finite sets form an ideal ? c A, and we write W* for the lattice W/3r. (Computability is definable in X as the property of possessing a complement, and then finiteness is definable, since a set is finite if and only if all its subsets are computable.) We write A C* B if B - A is finite, and A =* B if A C* B and B C* A.

We use the standard enumeration { We }ec of the computably enumerable sets, with finite approximations { Wes t to each. For the c. e. sets which we construct ourselves, we will also give finite approximations, usually writing A- U,'cAs . If A and B are both enumerated this way, we write A"B = { x: (3s) [x C As-Bs] }, and A \ B = {x E AnB: (]s)[x E As'-Bs]}. ThuswhenanelementnotyetinBen- ters A, we put it into A . B, and if it later enters B, then we put it into A \ B as well.

?2. The R-property. In order to guarantee that the set AO is not automorphic to a complete set, we will force it to satisfy the lattice-definable property R defined in Section 1, and prove that this implies tardiness of AO. Tardiness itself does not guarantee that a set cannot be automorphic to a complete set, of course, but satisfaction of R does, since every other set automorphic to AO must also satisfy R and therefore must also be tardy, hence incomplete. (A tardy set must be half of a minimal pair under <T, as shown in [1 6], and therefore must be incomplete.) We restate the R-property here:

R(AO, A1): AO n Al = 0 & (]C) (VB C C) (]D C C) (VS E C) (ST)

[Ao U AI Cm C & [(B n (SAO)) U AI = (D n (S-AO)) U Al

i>4C[c T&(AoSon T)UAl =(B nsnT)UAI]].



682 RUSSELL MILLER

THEOREM 2.1. If AO and A1 are two c. e. sets such that R(AO, A1) holds, then AO is not of prompt degree.

PROOF. The proof is similar to the corresponding result for the Q-property in [7]. Given AO and A1, we pick a set C as specified in R (AO, A1) and fix enumerations {A}s s, of AO and {Cs}sE,, of C such that AO C C \ AO.

To prove that a given (Oe is not a promptness function for AO, we need to find an infinite c. e. set Wi with standard enumeration { Wis s, satisfying the tardiness requirement 35e:

[(VS) WOe (S) S] (VX) (VS) [X E Wi - s-I == AO s x =

'e(s) r x].

We will prove independently for each e that Se holds. Having fixed e, we will assume for the rest of this section that (Pe is total with (Pe (S) > S for every s, since otherwise Se is automatically fulfilled. We will build a strong array { V(a,k),n }k,nEc;aEo Xo

of c. e. sets with enumerations { V$sa , s. The Slowdown Lemma then gives a computable function f such that for each (a, k) and each n, Wf ((azk) n) = V(a,k),n

and V(a,k),n \ Wf ((c,k),n) = V(a,k),n, so that no element of V(ak),n enters Wf ((O.k),n) until it has already entered V(ak),n . Periodically the strategy for a given (a, k) may be injured by a higher-priority strategy. If this happens while we are enumerating V(aek),n, then we give up on V(o,k),n and start enumerating V(a,k) ,n+ . There will exist an (a, k) which is only injured n times (with n < co), yet receives attention at infinitely many stages, and the corresponding V(ok),, will be infinite and will be the set which proves satisfaction of S7e.

We define the function n ((a, k), s) to keep track of which V(ak), nwe are enumerating at stage s. In particular, if the (a, k) -strategy receives attention at stage s + 1, then we may add an element to Vs+' To avoid national chaos, (ca,k),n((ca,k),s?I) T vi oainlcas however, we will write Vs+' in the construction and understand Vs+1 for it.

To ensure that one of these Wf((ak),n) will satisfy Te, we build a c. e. set B to which to apply the property R. When we want to preserve Ao 0 x from stage s until stage (0e(s) so as to satisfy Te, we do so by restraining all elements < x from entering B until stage e(s). The R-property then prohibits such elements from entering Ao, since if they did, we would then hold them out of B forever after, thereby contradicting R (Ao, A1).

To apply the R-property, we need to know which c. e. set Wi is the D specified by the property. Of course, we do not have this information, but our strategy is to use S to cover all the possibilities. Specifically, in the construction we will split C into the disjoint union of c. e. sets:

C =H Si. iCo

and apply the R-property to each Si, with Si in the role of S. (Clearly each Si z C.) We use each Si to handle the possibility that D = Wi.

Of course, the R-property states that the restraints we place on elements from entering B only affect Ao on S n T n A 1. Since R (Ao, A 1) also states that Ao n A 1 is empty, we do not need to worry about elements of A1, for they can never enter Ao. We are allowed to choose the S, since the matrix of R applies for all S, and indeed




we have already done so above (namely S = Si, for each i in turn). However, we can only guess at the set T.

To determine the index ] such that T = Wj corresponds to the set S which we choose, we use a 1f1? guessing procedure, since the conclusion in the matrix of R is a ri2 property. The j for which T = Wj will be the least ] which receives infinitely many guesses under this procedure. (We ensure that the hypothesis of the matrix holds, by periodically putting all elements of DS n (SS - A') into BS.) Moreover, in the construction, we will subdivide each Si into the disjoint union of c. e. sets Si ,:

Jan~~~~~~~~~~~~~~~ Si LI sJ.

Sinj is used to handle the possibility that T = Wj, so we pay attention to Sij each time j is named by the guessing procedure. Thus the Sij corresponding to the correct T will receive attention infinitely often.

To simplify the notation, we let the variable a (i, j) range over co x co, and define:

Dot= Wi

Sc = Sij

To = Wi.

We order the elements a of co x co by pulling back the usual order < on co to co x co via a standard pairing function. Thus each a has only finitely many predecessors under <.

For each a, let F (a) be the conjunction of the hypothesis and conclusion in the matrix of the R-property:

(1) F(a): (Bn(Sa -Ao))U AI= (Dogn (S. -AO)) U Al

(2) &[C c To, & (AO n So n ToJ U Al = (B n So n To) U Al].

Then F (a) is a T1- condition, uniformly in a, so there is a computable total function g such that F (a) holds just if g'- (a) is infinite. We enumerate the c. e. set Z, -

g'- 1 (a) by setting Z" { t < s : g (t) = a Now we narrow down each T,, to a c. e. subset U0, enumerated by:

-UO, = U,,,-' U { x c T, - Cs: x < Z,} I .

Thus, if T,, actually is the T corresponding to Si, then Uc, will contain all of T,, except certain elements of C. Hence F (a) will hold with Us, in place of T,. On the other hand, if F (a) fails, then Z,, and Us, are both finite.

If F (a) holds, then C C Uc, so AO C* U, U Al, because AO U AI Cm C. For the least a such that F(a) holds, our construction of S+1 will yield C -AO C* So, UA1, with Sp finite for all /? < a. Hence there will exist a k such that

(3) C-AO C So,UA1 U{O,1,...k-1}.

Line (3) is a 11f statement, uniformly in k and a, since our definition of So, will be uniform in a. Therefore, there exists a total function ho, such that (3) holds if and only if hj (k) is infinite. We define:

h(s) = hg(s)(n), where n t < s: g(t) = g(s)}j.



684 RUSSELL MILLER

We will enumerate sets V(a,k),n for each a, k and n. For the least a with Zc, infinite and the least k with hgj l(k) infinite, the set V(ak),n (for some n) will be the W1 required by 57fe. Elements of each V(ak),n (the "witness elements" for the requirement 5e) will be denoted VS Each VS k) will enter V(ak),n for at most one n.

The Slowdown Lemma (see [16], p. 284) then yields a computable function f such that, for every (a, k) and every n, V(ak),n = Wf ((c k),n) and at every stage s,

(VZ(ak),n - V (ak),) 0 Wf((cak),n),s 0-

When a witness element Vsk) enters V(a~k),n, we will find the stage tsk) > s at

which VKsk) enters Wf((a,k),n) and restrain (with priority (a, k)) elements < Vs

from entering AO until stage Oe (tQSk))* (Recall that ye, assumes (Oe to be total.)

Thus we will have A(k) -V(k) At ((auk)) rVSk). If we can achieve this for all

V(k) in the (infinite) set V(ak),n for some n, then the set Wf((ak),n) will be the set

required by Se to prove that (Oe is not a promptness function for Ao. At stage 0, for all (a, k), we set n ((a, k), O) = O and V? k) 0, with vok) 1 and

tok) '. Also, let every S- 0 and let BO - 0.

At stage s + 1, we first define each So,+. For each x C Cs+' - Cs, find the least a such that x E Us and put x into Ss+1. If there is no such a, put x into Ss+'. (The c. e. set Set simply collects elements which enter C without entering any So. Thus C = Lla<w, So.)

Set a = g(s), and define:

Bs+=Bs U { x: xE C- As_ & (/3f a) [x ED+1 0S+1

& (V8 < /?) (Vk < s) [t(,k)L \X > V(,5k)]] }

For each strategy which is injured at stage s + 1, we begin enumerating a new witness set. To this end, set n((y, k), s + 1) = n((y, k), s) + 1 and vs+l) T and t(s+)t

for each (y, k) satisfying any of the following conditions:

Y y>a.

* y = a and k>h(s). * There exists x < k with x C As+'- As

* There exists < y with Ss+' + Ss.

* There exists /? < y such that Up+1 contains an element > m, where m

min(Bs+l - Bs).

For all other (y, k), set n((y, k), s + 1) = n((y, k), s). We now define the witness sets at stage s + 1. For each (/?, k) < (a, h(s)) (in the

lexicographic order) which was not injured at stage s + 1:

1. If V<Sf/)t and (/3,k) 7 (a, h(s)), let vs<+' and tsg+) diverge also, with Vs+ =

V(flk) ,n- 2. If as ( letvS+ s + 1, with Vs+' = Us and ts+ (t ~~~~~(a,h (s )) (a,h (s ))' (s s) ),n (ash (s )), (ash (




3. If V(k)J. but t k) , let v = Vsk) and ask whether the following holds:

(4) (VY)k<y<v1 [y CA+' V y C As+' V y C (uS11 -

V y E (Cs+1 - Bs+') n Ss+ n 'S+'

If (4) holds, let V(+1s = /3 n U {ivs+1 } and (/3,k),n (/3,k),n (/3~k)

tS +k) =ju t [Vs +k IC Wf ( (f,Bk),n),t]-

(Such a t must exist, since Wf ((#,k),n) = V(#,k),n .) If (4) fails, then let 17`n

(pk),n and tI(flk)t

4. If VKs3k). and tJ/3k). and .pes(t(/3k)) .1< s, then let v(s+)t and t S +;)T, with

r(/, k) n V(3,k),n

5. If vs and tKpk)J. but either pes(ts/3k))J. > S or (Pe,s (ts/3k)) diverges, then let

V(fl#k),n r(/,k),n= '(fk) = V ,(fk) and tS+i - tk

This completes the construction. We now use the sets B and Sc, to prove that requirement ?e is satisfied.

LEMMA 2.2. If Zfl isfinite, then there exists a stage s, such that tKs3 k) Tfor all s > s and all k.

PROOF. Pick a stage so such that no s > so satisfies g(s) =,, and let k'

maxf h (s) : g (s) }. Then for all k > k', vs3 k)t for all s, and hence tKs/k)t for

all s. (The construction makes it clear that for any k and S, tKs/k) can converge only if V(s/k) converges.)

Now suppose k < k' and vs k) for all s > so. This means that we never execute

Step (4) in the construction after stage so, and that the (/B, k) strategy is never injured after stage so. But if tsk) ever converges after stage so, then eventually we must

reach Step (4), since we assumed (pe to be total. Hence tKs/k) must diverge for all s > so.

Finally, suppose k < k' and VSI/k T for some Sl ,k > so. Then vS will diverge for (/3~k) (fl,k)

all subsequent s, since it can only be newly defined at a stage s with g (s) =,. Thus tKs3k) will diverge for all subsequent s as well. Letting si = maXk<k/ s1,k completes the proof. -

LEMMA 2.3. F (a) holds for some a, andfor the least such a, there exists a k such that h-j1 (k) is infinite.

PROOF. First we claim that some Za, must be infinite. Suppose not, so Zc, is finite for all a, and F (a) fails for all a. However, the R-property holds, so there must be some a for which line (1) fails. Choose the least such a. Then

(B n (s, -AO)) U A1 (D, n (Sa -AO)) U A1.

Suppose x E B n (Sc - AO). Pick s such that x E Bs+- BS. Now to go into BS+l, x must have been in Ds+' n Ss+1 for some ,B. Since x c S,, we know x V S/

for all ,B a a. Hence x c Do, and so

(B n (S.-AO)) U A1 C (D, n (S.-AO)) U A1.



686 RUSSELL MILLER

Therefore, there must be some element x C AI n B n Dc, n (S. - AO). Assume x is the least such element. Now for every /3 < a, line (1) must hold and line (2) must fail, since we chose a to be minimal satisfying the R-property. Hence for all /3 < a,

(B n (Sl-Ao)) U Al = (DA n (Sl-Ao)) U Al.

Now since every Zf with 3 < Oa is finite, there is a stage so such that for all s > so, g(s) > a, and we may also assume that so is so large that x C Sso n Dso n Cso (Notice that x C S, forces x C C.)

Now use Lemma 2.2 to find a stage si > so such that:

(Vs > sl) (V# < a) (Vk) [tS,])].k)

Since (pe is total, there must be a stage s > si such that tsa k) i, and once we reach

this stage s, x must go into BS+l, contradicting our assumption that x V B. Thus, there must be some a such that Zc, is infinite. Let a be the least such.

Then every Up with fl < a is finite. Since F (a) holds, we have C C TO,, so by our construction, C C Ua, and by the major subset property, AO C* Ua U A1.

For this a, we claim that C - AO C* So, U Al. Suppose x e C - AO. All but finitely many such x lie in Up, U AI, as noted above. If x C AI, we are done. For each sufficiently large x c C-A A- Al, there exists s such that x C Uas - Ua -1. By definition of Uas, we must have x V Cs. But x E C, so x E Ct+l - Ct for some t > s. Hence x E Sjt+- by definition of St+1, unless there exists /3 < a with x C Up. But all Up with /3 < a are finite, by our choice of a, so all but finitely many of these x lie in S,. Therefore, line (3) holds for some k, and hj (k) is infinite. -

Use Lemma 2.3 to take the lexicographically least (a, k) such that F (a) holds and hj (k) is infinite. Then there are infinitely many stages s for which g (s) = a and h (s) = k, but only finitely many for which (g(s), h(s)) precedes (a, k) in the lexicographic ordering. Let so be the least stage with (g(so), h (so)) = (a, k) such that:

* Aso k =Aork, and * Bs0 rm B rm, where m = max Up<, Upl, and * for all s > so, (g (s), h (s)) > (a, k) lexicographically, and * Ss? = S# for all < a.

The final condition is possible since each S# C Up, which is finite for every /3 < a. We also let so < si < s2 < ... be all the stages s > so with (g(s), h(s)) = (a, k).

Now the (a, k)-strategy is never injured after stage so, so for every s > so, n ((a, k),so) = n ((a, k), s), and we write n = n ((a, k),so). (Thus n is the number of times the (a, k)-strategy was injured during the construction.) Moreover, minimality of so implies that this strategy was injured at some stage s < so such that there is no sq with s < s-I < so and (g(sq),h(sq)) (a,k). Therefore,

VUs~n = Vso is empty We claim that the subset V(a,k),n satisfies requirement 3Se. For this we need:

LEMMA 2.4. For this (a, k), and for each y > k, there exists an s such that the matrix of line (4) holds of y, (a, k), and s.

PROOF. Let y > k. If y C AO U AI, we are done. If y C C, then y C Tc, since F (a) holds. But Zc, is infinite, so To, - C C Uc,, and y is in Uc, - C, hence in some U0S+1 _ Cs+I.




So suppose y E C - Ao - Al. Now since hl l(k) is infinite and y > k, we know by line (3) that y E Sa. But Sa C Ua C Ta, by definition of Ss+1. Since y V (B nSa n Ta) U Al by line (2), we know y V B. Thus there is an s with y E (CS+l - Bs+l) n Ss+' n Uas+1. This proves the Lemma. -

Now V(ak),n = Wf ((aok),n), and if s' is the stage at which Vs enters V(a,k),n, then (a~k)

tuck) J. > S' by our choice of f from the Slowdown Lemma. Let s" ,Oe (tsk))

Then s' < s", since we assumed (We to be increasing.

LEMMA 2.5. V(ak),n is infinite. Moreover, for any element Vsk) of V(a,k) n, with s' and s" as above, we have:

BS rV(,k) = BS [rV(,k) and As rVck) =Av rVtk)

PROOF. For each k) with s > so, Lemma 2.4 guarantees that there will be a

stage at which Step (3) of the construction applies. The first such stage will be s', since at that stage vak) V( k) will enter V(a,k),n and t ' will be defined. But

VS~k (a~k) since We is total, we will eventually reach the stage s" > s' at which Step (4) applies, leaving vs" +1 undefined. Then at the next Sn > s", we will define vs-+1 - Sin + 1, levn (a~k) ,wV(a~k)-Si which is not yet in Via, )fn. Thus, V(a,,k),n must be infinite.

Now pick v E V(aOk)nI with s' and s" as above. Since Vs'k)n is empty, (ca,k) akn

we know that s' > so. If s is any stage with s' < s < s", then we see from the definition of BS+l that an element y can only enter BS+l on behalf of some y such that y E Ss+'. But then y US+l. Since we chose so to let Bs' [m = B [m, we

must have y > oa. But tsak) d, soy ? V ~alc) Vsk) by definition of Bs+'. Hence

Bsl [VS/k BS11 [VS/k) BSp(a k) = BSp(a k) .

Having seen that no y < vsak) can enter B between stages s' and s", we prove that no such y can enter A0 at those stages either. First, we know that As' [k = A0 [k by choice of so. So suppose k < y < V(s'). Now since v(sak) entered V(a,k),n at

stage s', we know by line (4) that

y c AS V y As Vyc(US- C') VY (C' B ) sj0 US'

If y E As', then As'(y) = A" (y), and if y E AI, then y V A0 at all. Therefore,

we will assume that y V As' U AI and prove that y As If the final clause holds, then y E (Ci - Ba) n S. n Us. Hence y V Bs", by the

first half of the lemma. If y E As", then y V B, since no element that has entered A0 can later enter B. But then

(A0O n S n T.) U Al + (B n Sa n T,) U Al

since y is on the left side and not on the right side. (Notice that y E U, implies y E T.,) This contradicts line (2), which we knows holds because F(av) holds. Therefore y V As".

So suppose the third clause holds, i.e., y a (us - Cs ). Then y f Bs since Bs C Cs, and so y V Bs". If y E As", then we must have y E Cs"-1 since we chose enumerations such that A0 C C \ AO. Picks such that y E Cs - Cs-1; then s' < s < s" and y V As. Now y E Us' C T.', and by definition of Ss we will have



688 RUSSELL MILLER

y E S'. (Recall that so was chosen so large that SI = Sp for all /3 < oa.) But now

y V AtO, since otherwise

(Ao n so, n T.) U A1 :& (B n so n T,) U Al

just as in the preceding paragraph. H

Hence V(c,.k),n = Wf ((cak),n) is an infinite c. e. set which satisfies the tardiness requirement Ye. This completes the proof of Theorem 2.1. H

?3. Satisfaction of R. We now prove that the R-property defined in Section 2 is nontrivial. The theorem establishes several other properties of the sets AO and Al as well, in order to yield the corollaries.

THEOREM 3.1. There exists a c. e. set A with Friedberg splitting A = LO H AI such that all of the following hold:

1. A is promptly simple of high degree. 2. Al has prompt degree. 3. R(Ao,Al).

COROLLARY 3.2. The formula in one free variable Ao:

(:AI) [Ao >T 0&R(AOA1)]

is definable in W and non-vacuous, and implies that AO is a noncomputable incomplete set.

PROOF OF COROLLARY. The statement AO >T 0 is equivalent to the statement that AO does not have a complement in A, hence is I-definable. The AO and Al constructed in Theorem 3.1 satisfy the matrix, since halves of a Friedberg splitting must be noncomputable. Finally, Theorem 2.1 shows that AO is tardy, hence incomplete. H

COROLLARY 3.3. There exists a Friedberg splitting A = AO L AI such that AO and A1 are not automorphic in the lattice of c. e. sets.

PROOF OF COROLLARY. Take the splitting given by Theorem 3.1. If an automorphism D of g satisfied D(A0) = A 1, then R (A 1, D(A 1)) would have to hold. By Theorem 2.1, then, AI would be tardy, contradicting the promptness of Al. -1

PROOF OF THEOREM. Let C be any promptly simple set, with computable enumeration C = {Cs}lsc. Then C is also of prompt degree, so let v and w be the prompt-simplicity and promptness functions for this enumeration of C, satisfying for every i:

Wi infinite (I?s) (x E Wis - Wi,s- ) [x E Cv (s)]

Wi infinite (3j s) (3x E Wis - Wi,s-) [CW(s) [x :& Cs [x].

We construct disjoint sets AO and Al and auxiliary sets Di and T1,j, and set A = AO H Al. The approximations to A, AO, and Al at stage s will be written As, As, and As, and will be defined so that As = As U As C Cs for all s. The construction will satisfy the following requirements for all i and j:




r(i~j) (matrix of R-property):

[W, CC&W) C C&C- Wic.e. & (wi n (Wj -AO)) U Al = (Di n (Wj -o)) U A1]

: (]T)[C C T & (AO n Wj n T) U Al =* (Wi n Wj n T) U A1]

X(i (major subset requirement):

C C W, A C* W1.

9 (prompt simplicity of A):

Wi infinite=i (]s) (]x E Wi - Wi, - 1) [x E A'(')].

di (promptness of A1):

Wi infinite= (]s) (3x E Wi - Wi,s- 1) [Aw (s) [x :& As [x].

Si (Friedberg requirementfor AO):

Wi \ A infinite Wi n Ao 0.

as (Friedberg requirement for A1):

Wi \Ainfinite Wi n AI 0.

In the requirement X(ij), of course, Wi plays the role of B and Wj the role of S in the matrix of the R-property. We will construct c. e. sets Ts,1 for each i and j, and then refine them to form the T demanded by each X(i ,) Once again we order co x co in order type co and write o = (i, j), this time with:

B- Wi DOG Di

SE,-WJ" where

Thus A,, says:

[BO C C & SOu H & = C & (BOG n (sc-AO)) U A1 = (Do n (S.-Ao)) U A1]

>. (T)[C C T & (Ao n S, n T)uA1 =* (Bo n S, n T) U A1].

A., is a negative requirement, trying to keep elements from entering AO until they can do so without harming the R-property (if ever). All the other requirements are positive ones, trying to put elements into AO or A1. There are no negative restraints on elements of C entering Al, except that they cannot already be in AO.

Each element which we try to put into AO to satisfy some F, or Ide must receive permission to enter AO from each A,, with oa < e. The restraint function q (x, s) will give the greatest oa < e which has not yet given this permission as of stage s. The priority function p(x, s) keeps track of which requirement 3e or Ide wanted x to enter AO. This can change from stage to stage, for several reasons. If a higher-priority requirement decides at stage s + 1 that it needs x to enter AO, then p (x, s + 1) < p(x, s). Alternatively, an 3e could find itself satisfied by another x' E As+' and no longer need to put x into AO, although in this case we leave



690 RUSSELL MILLER

p(x, s + 1) p(x, s) so as not to disrupt the flow of elements into AO. Finally, a higher-priority requirement could make x enter A'l+, in which case we define p (x, s + 1) 1, removing x from the flow of elements into AO since we need Ao nA 1 0.

We use the Recursion Theorem on our construction of AO, C, and D, to define the following FIn statement F (a) for each oa:

(B., n (S., - AO)) U Al = (D,, n (S., - AO)) U Al & B., C C & Sc, L Sc, - C.

Since F (a) is no, there is a computable function g: o -> (o x (o such that F(ae) holds if and only if the set Z_, = g'1(oa) is infinite. We let Z' = g'(ce) o {O, 1,... s - 1 }. Monitoring I Zc I will help us determine for which oa the hypothesis in the matrix of the R-property is satisfied. For those oa for which the hypothesis fails, IZc, I is finite, and A,, will only restrain finitely many elements from entering AO, since we need not satisfy the conclusion of the R-property for such an oa.

At stage s = 0, we set AO = AO 0. Also, let all p(x, 0) and q(x, 0) diverge. At stage s + 1, we first define T,?+l for each oa:

Ta+ 1 = Tc U{ x E Cs+1 :x < IZc'+l}

Next we determine which elements of Ci+' to add to A' to create A'+'. For this, we need movable markers for elements currently in C - A. Write

C -+1 - As - {do"', dv+l ... dS+l }

preserving the order of the markers from the preceding stage. (That is, if d[= di81+1

and dj8 = dj7,+l, then i < j if and only if i' < j'; and if df+1 E CC and dj'+l C5,

then i < j.) For the sake of Ie, we define

Ve + = Ve U { X E We,s+l - Ci" (Vy < X) [y E We,s+l U C ] }-

(For each e, the sets Ve1 enumerate a c. e. set Ve. If C 0 We, then Ve will be finite, but if C C We, then C C Ve C We.)

For each e < s, define the e-state of each d,8?l at stage s + 1 to be:

u(e, d;, s + 1) {i < e : ds+l E Vjs+ }.

We order the different possible e-states by viewing them as binary strings. Find the least i < s such that there exist e and j with e < i < j < s and

u(e, dfS', s + 1) = u(e, dJ?', s + 1) and df+' V Ve]+1 and dJe' E V]s'. Forth least such e and the least corresponding j, we say that "'e wants to put into AO all the elements df+', d+1. dj8+1, so as to give the marker di a higher (e + i)-state at subsequent stages.

Now we consider the requirements 3e. For each e < s with Wes n A= 0 and for each x such that

X C (We,s n C8+1) -As -{do+', djs+1 d. +}

we say that 3e wants to put x into AO. We set p(x, s + 1)T for all x V C-As. Otherwise x d'+' for some k, and

p(x, s + 1) is the least e < k (if any) such that either p(x, s)J, e or /e or Fe

wants to put x into Ao. Thus, the function p(x, s + 1) gives the priority currently assigned to putting x into AO. If there is no such e, let p(x, s + i)T.

We now follow the following steps for each x < s:




1. If p(x, s + 1)T, then q(x, s + 1)T also. 2. If p(x, s + 1) , but q(x, s)T, we ask if every oa < p(x, s + 1) satisfies either

x E S'+j U SS+j or x V T,,+1'. If so, set q(x, s + 1) = p(x, s + 1) + 1. If not, then

q(x, s + 1)T. 3. If p(x, s + 1)J and q(x, s)J, > p(x, s + 1), then set q(x, s + 1) to be the greatest

oa < p(x, s + 1) satisfying all four of the following conditions:

(a) Ss+1 n gs+l 0. (b) x V 5S+1

(c) X E Ts+C

(d) VB < ae, either ,B fails one of the three conditions (a)-(c), or ,= (i', j') and ae = (i, j) with i :+ i'.

Also, enumerate x in D s+'l. (For future reference, notice that if oa satisfies q(x,s?1)'

(a)-(c), then some ,B < o< with the same first coordinate as oa must satisfy (a)-(d).) If there is no such a, set q(x, s + 1) =-1. 4. If p(x, s + 1)1, and q(x, s) J with 0 < q(x, s) < p(x, s + 1), we ask whether

x E BS+l). If so, or if q(x,s) no longer satisfies the conditions (a)-(d), set

q(x, s + 1) to be the greatest ae < q(x, s) satisfying the conditions (a)-(d) above, and let x E Ds+' (If there is no such ae, let q(x, s + 1) -1.) Otherwise, let

q(x,s?1)'

q(x, s + 1) = q(x, s). 5. If p (x, s + 1) J and q (x,s)J. -1, enumerate x E As+', and let q (x, s + 1)1.

This completes our enumeration of As+l. Next we determine which elements to add to As+':

1. Find the least e < s (if any) such that de is not yet satisfied and there is an element x E Wet - We,t-i for some t < s such that w (t) > s, and there exists y < x such that y Cs+' - As+' and y V Atj U tdo, . .. des+ } and no 3i with i < e wants to put y into Ao. Put the greatest such y into As+'. This forces As+' [x :& At [x, satisfying de permanently. (If there is no such e, do nothing.)

2. Find the least e < s (if any) such that ge is not yet satisfied and there is an element x E Cs+' n (Wet - We,t-1) for some t < s with v(t) > s, such that x tds+', . . . ds+1 } and no 3i with i < e wants to put x into Ao. If no such x lies in As U As+' then put the least such x into As+'. This forces x E As+', satisfying ge permanently.

3. Find the least e < s (if any) such that Ve is not yet satisfied and there is an element x E (We,s+l n Cs+') - As+' with x . tds+l des+'}, such that no 3i with i < e wants to put x into Ao. Put this x into As+'. This satisfies Ve forever.

Let As+' = As+' U As+'. This completes the construction.

LEMMA 3.4. C - A is infinite.

PROOF. We prove by induction on e that de = lims ds exists. Assume that this holds for all markers di with i < e, and let so > e be a stage such that d7s0 = di for all i < e. Now each Sv, Vj, 9j, and &j with j > e cannot put any of the elements ds, ..., ds into A1 at stage s + 1, so none of these requirements ever moves the marker ds. Also, each i, i, and di with i < e puts at most one element into A,



692 RUSSELL MILLER

hence moves the markers at most once. Let si > so be a stage so large that no Vi, , or di with i < e moves any markers at any stage s > sj. By the construction, de can only be moved at stage s > si by a requirement JXi or

As with i < e. Furthermore, when Si (i < e) moves a marker, it puts an element into AO, so it is satisfied at that point. Before then it may have tried to put finitely many other elements into AO as well, and any of them may go into AO or Al at a later stage, moving markers in the process. However, since there are only finitely many such elements, de is moved only finitely many times on behalf of 57i.

Now Xo moves de at most 2e+1 times after stage s1: once to put do into Vo, possibly twice to put d1 into Vo, and so on. Once Io has finished moving de, Ai

moves it at most 2e more times, to put markers into V1. Similarly, once each J/ti has moved de for the last time, i?+1 may move it at most 2e-i more times. Hence we eventually reach a stage S2 after which de never is moved again. Possibly de2T, but since C is infinite and every di with i < e has already converged to its limit, we know that de will be defined at some stage t > S2. Since it never moves again, this yields dt - lims de.

LEMMA 3.5. For each e, the requirements e, ge, de, Se,, and fe are all satisfied.

PROOF. We proceed by induction on e. Assume the lemma holds for all i < e. We write oa for the pair coded by e, and prove first that A,, is satisfied. Suppose (B., n (S, -AO)) U Al = (D, n (S. -AO)) U Al and B., C C and S., LS., = C. Then F(av) holds and Z., is infinite. The construction of T., then guarantees that C C T.,. Let G., be the intersection of all those Vi with i < oa such that Vi is infinite, and let T, = T, n G.,. Thus C C Tc, since C C Vi whenever Vi is infinite.

SUBLEMMA 3.6. For each a and each n < a, there are only finitely many x E Tog such that Id, ever wants to put x into AO.

PROOF. First, if Vn is finite, then Id, will only want to put finitely many elements into AO. So we may assume that Vn is infinite, and hence that T., C Vn .

If 1/n wants to put x into AO at stage s, then x E Cs - As, so x = dis for some k. Moreover, there must be an i with n < i < k and a j > k such that u(n, di7, s) = u(n, djs, s) and di7 V Vns and djs E Vns. Furthermore, di is the leftmost marker which any X-requirement wants to put into AO at stage s, and n and j satisfy the minimality requirements of the construction.

Now if das V Vns, then das V Vn, since C \ Vn 0, and hence das , T.,. Therefore we may assume dks E Vns. (This guarantees k 7 i). Then minimality of n forces u(n, dic, s) > u(n, dfs, s), and minimality of j forces u(n, dic, s) > u(n, ds, s) (since dks E Vns). Hence there is some m < n such that c(m, dif, s) = u(m, ds, s) and df E V4m and dks V Vm. This forces di/ E Vm and da s Vn (since d Sff c C5 - VS).

If Vm is infinite, then dis f T.,. But if V,, is finite, then ds lies in the finite set

V UJ Vm m < n & Vn finite}.

Hence we need only find a stage t so large that for every d E V, either d E A' or 1/n wants to put d into AO at stage t or 1/n never wants to put d into AO. Then 1/n will never want to put into Ao any x > max(Ct) with x E To. H

We will show that the conclusion of Ar, holds for T.,:

(Ao n S, n ot,) U Al =* (B, n S, n o)U A1.




Once we have established this for all ar, clearly R (Ao, A1) itself must hold, since for each oa we can choose another Tc, which excludes the (finite) difference set of the two sides and still contains C.

Suppose first that x E AO n S, n To and x V AI, and assume that x is sufficiently large that:

* x > IZp I for every /3 < oa such that Zp is finite, and * No vii with i < oa ever tries to put x into AO, and * No JXi with i < oa ever tries to put x into AO.

The last condition is possible by Sublemma 3.6. Notice also that the first condition forces x V Tp for all / < oa with IZp finite.

Then for all s, either p(x, s) > oa or p(x, s)T. But since x E AO, we know that some p(x, s)J,. For the least such s we have x E CS, and hence x E T,8, since C n T. C T. \\ C.

Now oa satisfies conditions (a)-(c) in the construction at stage s, since F (e) holds and x E S,. So there must exist /3 (i, j') < o = (i, j) which satisfies (a)-(d) at stage s.

We claim that this /3 satisfies conditions (a)-(d) at every stage after s as well. Since x E Tps, we know that Zp is infinite and F (/3) holds, by choice of x. Hence (a) and (c) hold at all subsequent stages. Let t be the first stage at which q (x, t) converged. Then x E Ct, and x E T' since C \ Tp = 0. By the definition of q,

we must have had x E V U 4 But x V SV since (b) holds at stage s, and because s > t, this forces x E S', so (b) always holds of/.

To show that (d) always holds of /3, we choose an arbitrary y < / with the same first coordinate as fl. Since /3 satisfies (d) at stage s, y must fail one of (a)-(c) at stage s. If y fails (a) or (b) at stage s, then clearly it fails that same condition at every subsequent stage. Moreover, if y fails (c) at stage s, then x V T,8, and since x E C5, this forces x V T,. Thus /3 will always satisfy condition (d).

But since x E AO, there must also be a stage s' with q(x, s') = -1. Since (a)-(d) continue to hold of /3, the only way for q(x, s') < /3 to occur is for x to enter Bfl. (Recall that for all s, either p(x, s) > oa or p(x, s)T.) But Bq = Wi = Bo since /3 (i, j') and oa (i, j), so this forces x E B, Hence

(AO n Sc, n ) UAl C* (Bc, n s, nT,,) UAl.

Now suppose that x E Bc, n Sc, n i., and x V AI, and assume x is greater than max(do, ... d,), and also greater than the greatest finite IZp I with /3 < oa. (Thus x V Tp for all such /3.) Now x c C since S., C C, so at some stage so, x will enter C and be given a marker: say x = d s. So x E Cs', and since x E T.,, this forces x c Tas0.

If x V AO, then we must have x E Do,, since (Bc, n (s, - AO)) U Al = (D, n (S, - AO)) U Al and x V A1. (Notice that then x, being in C - A, eventually receives some permanent marker dk,, with k' > oa by choice of x.) For x to have entered Dc,, there must have been a stage sl > so with q(x, si) = y = (i, j'), where a = (i, j). (Also, then p(x, sl)J,, and since x V A1, p(x, s) for all s > sl.) But oa satisfies conditions (a)-(c) at all stages s > so, so by condition (d) on y, we must have y < a. The assumption x V AO UA1 then means that there is some S2 > sl such



694 RUSSELL MILLER

that q(x, s)J q(x, S2) for all s > S2. Let 3 = q(x, S2) < y. Then x E D -Bp, and furthermore /3 satisfies the conditions (a)-(d) at all stages s > S2.

Now x E Tf, to satisfy condition (c), so x < IZJ and /3 < y < oa. If a, then Zp is infinite since F (a) holds, and if / < ar, then Zp must be infinite, by our choice of x. Therefore F(f/) holds, and in particular Sp L Sp = C. Now x V SP by condition (b), so x E Sp. However, with x E Df - Bf, this contradicts F (/3). Hence x E AO, and

(Ao n sor n To) U Al C* (B S n So n iT) U A1.

This completes our proof that A, is satisfied. Now we continue with the other requirements. Let so be a stage such that no Hi,

di, 9i, or Vi with i < e tries to put any element into AO or AI at any stage after so. (7i is different from the other requirements in that it may try to put more than one element into AO. It only stops trying when one of those elements succeeds in entering AO. We choose so so that every element which Yi wants to put into AO either is in As, or never enters A.) Assume also that so is sufficiently large that

ds' = di for every i < e. Now if We \ A is infinite, then there must be an x in some We, - AS with s > so

and x V {do,... de}. No requirement of higher priority will need to put this x anywhere, except possibly some Xi, and according to our construction, Se does not respect the priority of the requirements XKi, so x E Asl+, and Se is satisfied.

Similarly, if We is infinite, then there must be an x and an s > so such that x E Wes - We,s1 and x E cv(s), by prompt simplicity of C. If this x is not already in Av(s)-1, then the construction puts it into Av(s), so Me holds. Also, there must be an x and an s > so with x E We,s - We,s-i such that Cs [x :& CW(s) [x, by promptness of C. Thus there is a y < x which entered C at some stage t with s < t < w(s). We must have y V At-i since At-i C Ct-i. But now y V {do, . . . d' }, since these markers had reached their limits by stage so and y only entered C at stage t. Hence the construction will put this y into A', and

Aw(s) [x :& As [x, satisfying de. Continuing with the induction, we need a sublemma to handle 7e.

SUBLEMMA 3.7. For this e andfor all sufficiently large x, if Ye wants to put x into AO at some stage, then x E AO.

PROOF. Choose x so large that it satisfies all of the following:

1. x > max{ jZpj : < e & Zp is finite }. 2. No 5i, Wi, Hi, or di with i < e ever wants to put x into AO or Al. 3. x {do,... .de}.

Suppose 3e wants x to enter AO at stage so. Then x = ds0 for some k and p(x, so) J < e. Now no Vj, Mj, or tj with > ? e ever manages to put x into Al, since 9e takes priority over these. (Since x :& de, the only way to have k < e is for x eventually to enter AO. Hence we may assume k > e.) Also, for every /3 < e, either x V Tp (if IZJ < x) or F(/3) holds (if Zp is infinite). Hence there is an si > so such that q(x, sj) I and q(x, si + 1) < e.

Now suppose q(x, s) = / for some s > si (so /3 < e). If F(f3) failed, then Zf would have to be finite, so x V Tp (since IZp I < x) and q(x, s) would never equal /3.




Therefore, F(/3) must hold. Suppose x V AO. If x V Sp, then x E Sp by F(/3) and so q(x, Sfl) < / for some sp > Si. Otherwise x E DA n (Sp - AO) C Bp by F(/3), so x E Bpf for some sP > SI, and hence q(x, Sfl) < /3. Thus, by induction on /3 < e,

eventually we must have q(x, s) =-1, and so x E A0?l, proving the sublemma. -1

Now if We \ A is infinite, then 7e has infinitely many elements at its disposal to try to put into AO. Hence once we find a sufficiently large x E We \ A, we know by the sublemma that this x will eventually enter AO, thus satisfying 3e. This completes the induction of Lemma 3.5. -

LEMMA 3.8. The requirements "'e are all satisfied by our construction.

PROOF. Suppose that C C We. To prove that Jfe holds, we must show A C* We. By induction we assume that Xi holds for all i < e. Let

a - { i < e : C C Wi }

Now if i E a, then also C C Vi, so by inductive hypothesis A C* Vi, whereas if i V a (and i < e), then Vi is finite. Hence for all but finitely many k we have

ar(e, dk) = a. Now let VJ = Ve n (fl vi: i E a }). Then C C VT. But C, being promptly

simple, is noncomputable, so VJ7 C must be infinite. Choose y so large that no element > y can be held out of AO forever by any requirement A,, with oa < e, and let so be a stage such that C? [y = C [y.

Suppose for a contradiction that Ve n (C - A) is infinite. Then there exists p such that dp V Ve with p so large that dp V CS and with a(e, dq) = a. (Hence dp > y.) Let sI be a stage with dpl = dp and a(e, dp, si) = a. Now since VJ7 C is infinite, there will be a stage s > sI at which some element x E V,'s-1 enters C and is assigned the marker dqs (with q > p since dpo = dp). Moreover, we may assume that q is sufficiently large that not only is d4s in VT, but that a(e, d4s, s) = a, since every Vi with i < e and i V a is finite. Since dE Vie C Ve and dp V Ve, -Xe will want to put dp into AO at stage s, and since dp > y, no negative requirement will keep dp out of AO. Possibly dp will be diverted into AI by some requirement 9y, Iy, or tj, since these do not respect the priority of Ide. If so, then dp will enter Al; if not, then dP will enter AO. Either way, dp enters A, contradicting our assumption that the marker dp had reached its limit at stage so.

Hence Fe n (C-A) is finite, and A C (C-A) U C C* Ve C We. Thus A is satisfied, and the lemma is proven. -1

Knowing that the requirements are all satisfied, we can easily complete the proof of the theorem. The construction ensured that AO n AI = 0, and the conjunction of all the 3i and Wi implies that AO U A1 is a Friedberg splitting of A. (See pp. 181-182 of [16].) The requirements Hi together make A a promptly simple set, by definition, and the di together allow A1 to satisfy the Promptly Simple Degree Theorem (Thm. XIII.1.6 of [16]), so that Al is of prompt degree. To prove that R(A0, A1) holds, we note that the requirements IX(i, along with Lemma 3.4, show that A = LO U AI is a major subset of C. Moreover, given a B = Wi and a pair (Sjl, S9j1) with Sj, U Sj, = C, we have the Di and T, (with a = (i, (j', j"))) constructed above. If

(Bi n (Sjl - AO)) U AI = (Di n (sjl - AO)) U AI,



696 RUSSELL MILLER

then F (as) holds. Since A', is satisfied, we know that there exists a T with C C T such that

(Ao n Sj, n T)UAl=* (Bi n Sj, n T)UAl. So we can pick a sufficiently large nO, and let

T' ={ x E T :x > na } U { x E C :x < n}

Then C C T' and also (AOnSjn T')UA1 = (Bin Sj, n T')UA1, since S' n C 0. Thus R(AO, A1) holds. Finally, since A is a major subset of the set C, A must be of high degree (see [10], page 214). H

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[3] R. DowNEY and M. STOB, Jumps of hemimaximal sets, Zeitschriftfir Math. Logik Grundlagen, vol. 37 (1991), pp. 113-120.

[4] , Automorphisms of the lattice of recursively enumerable sets: Orbits, Advances ini Mathe- matics, vol. 92 (1992), pp. 237-265.

[5] , Friedberg splittings of recursively enumerable sets, Annals of Pure and Applied Logic, vol. 59 (1993), pp. 175-199.

[6] R. M. FRIEDBERG, Two recursively enumerable sets of incomparable degrees of unsolvability. Pro- ceedings of the National Academy of Sciences, USA, vol. 43 (1957), pp. 236-238.

[7] L. HARRINGTON and R. I. SoARE, Post's program and incomplete recursively enumerable sets, Proceedings of the National Academy of Sciences, USA, vol. 88 (1991), pp. 10242-10246.

[8] , The A?-automorphism method and noninvariant classes of degrees, Journal of the American Mathematical Society, vol. 9 (1996), pp. 617-666.

[9] , Definable properties of the computably enumerable sets, Annals of Pure and Applied Logic, vol. 94 (1998), pp. 97-125.

[10] C. G. JOCKUSCH, JR., Review of Lerman [11], Mathematical Reviews, vol. 45 (1973), #3200. [11] M. LERMAN, Some theorems on r-maximal sets and major subsets of recursively enumerable sets,

this JOURNAL, vol. 36 (1971), pp. 193-215. [12] W MAAss and M. STOB, The intervals of the lattice of recursively enumerable sets determined by

major subsets, Annals of Pure and Applied Logic, vol. 24 (1983), pp. 189-212. [13] A. A. MUCHNIK, On the unsolvability of the problem of reducibility in the theory of algorithms,

DokL. Akad. Nauk SSSR, N. S., vol. 109 (1956), pp. 194-197, in Russian. [14] J. MYHILL, The lattice of recursively enumerable sets, this JOURNAL, vol. 21 (1956), pp. 215, 220. [15] H. ROGERS, JR., Theory of recursive functions and effective computability, The MIT Press, Cam-

bridge, Massachusetts, 1987. [16] R. I. SOARE, Recursively enumerable sets and degrees, Springer-Verlag, New York, 1987.

DEPARTMENT OF MATHEMATICS

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Current address: Department of Mathematics, Cornell University, Ithaca, New York 14853, USA E-mail: russell~math.cornell.edu



definable incompleteness and friedberg splittings

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