Zeiuchr f mark Logik und Grundlagen d. Marh. Ed 37. S. 331-342 119911

@ 1991 Duch. Vlg. d. Wrrs.

MINIMAL DEGREES AND RECURSIVELY INSEPARABLE PAIRS OF RECURSIVELY ENUMERABLE SETS

by MANUEL LERMAN in Storrs. Connecticut (.U.S.A)')

1. Introduction

FRIEDMAN and SIMPSON [l] have asked if every degree d of a model of Peano Arithmetic (PA) bounds a minimal degree. KUEERA [2] answered this question positively for d 5 0'. SOL- OVAY, in a letter to SOARE, characterized the degrees of models of PA as the degrees which separate a pair of disjoint r.e. effectively inseparable sets. After obtaining his result, KUEERA asked if "effectively inseparable" could be replaced by "recursively inseparable" in his result, i.e., if d 5 0' has a set which separates a pair of disjoint r.e. recursively inseparable sets, must d bound a minimal degree. We give a negative answer to this question by constructing a set of degree d which separates a disjoint pair of r.e. recursively inseparable sets, but fails to bound a minimal degree. Our proof is similar to that in [3], where we construct a degree in H Z = (d S 0': d" = O"? which does not bound a minimal degree.

We will be using the following notation and definitions. (CP,: e E N) is a standard enumer- ation of all partial recursive functionals, approximated to by the recursive array of finite functionals {a:: e, S E N). We write @Ju; x) for @bh'"'(u; x), where u is a finite sequence from {O, I} and the function lh is defined below. We note that {(a, x) : ae(u, x) i } is recur- sive.

A string is a finite sequence of symbols from a set X . If X = {a, I}, then we let S2 denote the strings from X. If X = N, the set of natural numbers, then we let S denote the set of strings from X. We let N+= N u {a, m+} with n < < Q)+ for all n E N . We let S' denote the set of strings from N'. Strings are ordered by inclusion ( S ) in the usual way. We write uJr if, for some x, a(x)l 9 t ( x ) l . lh(cr) is the cardinality of (x: u ( x ) l } . The restriction of a string u to n, u 1 n, is the string r s u such that lh(7) = n. We let O k be the string of length k such that O'(x) = 0 for all x < k . If Ih(u) > 0, then (I- = u 1 Ih(u) - 1. Given strings cr and 7, we con- catenate them as cr * 7, the string of length Ih(u) + Ih(7) which starts with the sequence u and then follows with the sequence 7. If cr + 0 and u = u- * k . we let S ( U ) = U - * ( k + l), and p(o) = u- * (k - 1) if k * 0. This notation is extended to S' in the obvious way, producing functions s+(cr) and p'(a). We say that u is men if u* 0 and if u = U - * k, then either k is even or k = m ; u is odd if u + 0 and cr is not even.

2. A separating degree which does not bound a minimal degree

We construct a set A of degree a, and a disjont pair (Do , D, ) of r.e., recursively insepar- able sets such that A separates (Do , 0,) but a, the degree of A , does not bound a minimal degree. The construction will be a recursive approximation construction.

') Research supported by National Science Foundation grant DMS 85-21843 and DMS 89-00349.

332 M. LERMAN

We say that S separates (Uo, U,) if S(x) = i whenever x E Ui, for i E (0, I}. The string d separates ( p o , p l ) iff whenever i 5 1, u(x)J and p , ( x ) J = 1, then d x ) = i.

The construction will be a priority construction using a tree of trees. Each UE S3 will re- present a strategy predicated on the guess that u lies on the ?me path for the construction. Thus at stage s, we will specify a string y, as the current guess to the true path. We order strings lexicographically. r= lim inf, yJ will be the true path for the construction. We will make independent attempts to satisfy the requirement R, for each y such that lh(y) = n , and will show that the attempt made for y c r succeeds. Thus we decompose each requirement R, into a finite number of requirements {&: lh(y) = n}.

Fix a recursive enumeration (Z,: e E N) of all the recursive sets and finite strings. (Such an enumeration can easily be obtained by cutting down the set provided by the standard enu- meration of the r.e. sets.) In order to make (Do, D, ) recursively inseparable, we will satisfy the requirement P, for each e E N :

(2.1)

It will then be the case that either Do $€ z, or D1 0 Z,. Thus Z, will not separate (Do. 0,). The x used in (2.1) will be assigned during the construction as a location for P,. As men- tioned above, we will make a series of independent attempts to satisfy P, = R,, each set of at- tempts associated with some y E S3 such that lh(y) = n. Thus we decompose each P, into {Pe. ,,: lh(y) = n}, where P, = R,.

The paths through the tree of trees correspond to certain bushes on which the construction takes place. A bush is a subset of S2 which is closed under inclusion. A maximal element of the bush T is a string U E T which has no proper extension in T. We say that X c T if c-r c T for all uc X.

We will have certain algorithms for defining bushes. We let Id = S,, the identi@ bush, ap- proximated to by Id" = {UE S,: lh(u) < s}. If T is a bush and UE T, we let Ext(T, a) = { r e T : u s 7 ) and call Ext(T, u) the extension of T above u. In addition, we will be using e-splitting bushes and recursive computation bushes defined below. If T is the identity bush or an extension bush, we say that T releases (I if U E T.

In order to show that @,(A) does not have minimal degree, we build a reduction procedure 0, uniformly in @,(X), which, if @,(X) is total and nonrecursive, produces a partial charac- teristic function B , ( X ) . For X = A , Be = B,(A) will be total under the above assumptions, and if B i = Be n {2x + i: x E N} for i 5 1, then we will satisfy, for each k E N and i 5 1, the following requirement:

P,: dom(Z,)infinite- (3i 5 1) (3x E D,) (Z,(x) = 1 - i ) .

(2.2) Q c , k , r : @ k ( B : ) * B : - ' .

Since B:, B: 5 T Be s @&), if Q,. k, , is satisfied for all k E N and i 5 1, then @,(A) will not have minimal degree. We fix a recursive ordering {Rn: n E N} of requirements { P , : ~ E N ) U { Q ~ , ~ , , : ~ , ~ E N &isl}.

Because of our use of a tree of trees, we actually make a separate attempt to define 0, for each y such that lh(y) = e + 1. Similarly, if Q,. k, = R,, then we require that n 2 e + 2 and we make a series of independent attempts to satisfy R,, with each set of attempts predicated on the assumption that y c f for lh (y ) = n.

Each attempt at defining the reduction procedure 0, is made in terms of a bush Tc- 1. We assume that we have a recursive approximation {T:-l: s z so} to T , - l so that T,-, = U,T:-, and for each S E N , TS-, is finite. The reduction procedure 0, is defined by induction. At stage s, the reduction procedure will add finitely many triples (t, x, i ) , indicating that if

MINIMAL DEGREES AND RECURSIVELY INSEPARABLE PAIRS OF R.E. SETS 333

f S @,(A'), then B,(X; x ) = i. In addition, a bush T: S TZI ; will be specified at stage s, allow- ing the construction of A to control @,. Thus if ([, x , i ) E O:, then [E @,(a) for some u~ TS,. We define the length of S:, L,(s) = max((x: 35, i(([, x, i ) E @:I}).

The determination of the triples which are placed in 0,, - Oe.s- depends on certain par- ameters provided by the main construction, in addition to T:::; t is the initialstage, i.e., the stage at which we begin to define 0,; a is a root, and we require that if (5, x , i ) is placed in 0, at stage s, then U E T: and @,(a) is compatible with [; (do. 6 , ) will be an approximation to (Do. D1); we require that if ([, x , i ) is placed in 8, at stage s, then u separates (&, 6 , ) . And S is a set of focations on which 0, must "act independently". Thus 0, receives ( t , u, do,

If 0: = and the tuple received by 0, at stage s is consistent with @ : - I , then 0, may transmit a pair ( T , c ) at stage s. T 2 u and T is to be a revbed root for the construction at stage s, c is a code for the construction. If c = 1 , then the construction is instructed to try to com- pute @ , ( A ) recursively. Such a recursive computation is pursued as long as ( T , c ) continues to be transmitted, and this computation will impose certain restraints on enumerating loca- tions in Do and D1.

2.1. Cons t ruc t ion of 0,. We proceed by induction on s. Let 0: receive ( t . u. a0, a,, S, T:: :) . If s = 0 or s < t , set Ti = Q: = 0 and specify that 8: transmits noth- ing. Otherwise, we assume that (2.3)-(2.8) below hold, else we set T: = T:-', 0: = W-' , and specify that Q: transmits nothing.

(2.3)

(2.4) IS1 5 1 .

(2.5)

(2.6)

(2.7) U E Ti::

(2.8)

S, T;:;) at stage s.

a separates ( b0. 6,).

V x E S ( x 2 lh(u)).

V X E S V i s 1 ( ~ , ( X ) J . * ~ ~ ( X ) = 0).

I";-' E T:I: and Ti:: has released (1.

In addition. we assume the following induction hypotheses.

(2.9) v f , X , i ( (5 , X , i ) E @ : - ' * 3 T ( T E j":-'& t s @ e ( T ) ) ) .

(2.10) v ~ , x , i ( ( [ , x , i ) E @ : - ' * ~ y < ~ 3 t l ~ t 3 j ( ( t l , y , j ) ~ O S , - ' ) )

(2.11) t r [ , ~ , x , i , j ( t l s ~ & ( [ , x , i > , ( n , x , j ) ~ 8 S , - ' ~ i = j ) .

(2.12) V T , [, x , i(TE Ti-' - TS-' ~ & 5 compatible with @,(T) & ( f , x , i ) E 8:- ' * [& @,(T)).

Conditions (2.9) and (2.12) indicate that the definition of 0, is controlled by T,. (2.10) re- quires that for all X, QP,(X) is defined on an initial segment of N. And (2.11) requires that 0, be consistent, a condition necessary in order that 8, be a reduction procedure.

The construction proceeds through a sequence of 19 steps, defining at step u, a: ~ for i 5 5 . If @:-' is not extended at step s, @: will transmit a string a: to 7':- indicating a new root for the construction of remaining bushes, with u: 2 6. Until 0, is extended and assuming that the tuple received remains unchanged and (2.3)-(2.8) hold, the construction at stage s will duplicate all steps completed at stage s - 1 . If 8, is never extended after stage s under

334 M.LERMAN

the above assumption, then a; = a: for all r 2 s and either T,- is finite, or there are no e-splitting on T P d l which separate (Do, Dl), or there is no set X such that (7: t c X & yf 5 t} n Te-' is infinite and De(X) is total.

S t e p 1 . We search for a string 5 1 a such that T E TS,I:, 7 separates (do, a,), and @,(T; x ) l for all x 5 L,(s - 1). If no such t is found, let TZ = Tf-' , 0, = and let Of transmit (a, 2 ) . (In this case, if a c A E T,- and 0, lies along the true path, then @,(A) will not be total.) Otherwise, fix the first such 5, let a: = t for i 4 5 , and proceed to the next step.

S t ep 2 . We search for a pair of strings to, t, 2 ah such that to/ rl, to, 71 E TS,: i and if X E S , isl, then t , ( x ) J = i . If no such pair (to, tl) is found, let T i = T : - ' , O f = 0"-' , , k t 0: transmit (ah, 2 ) , and go to the next stage. (In this case either x < lh(ai) in which case the requirement which provided the location x is asked to provide a new location and returns with x 2 lh(ak) at a later stage, or Te- will be finite and so 0, will not lie along the true path.) Otherwise fix the first such pair (to, tl) which is found and let a: = a: = to,

a: = U: = tl and a', = a: = 0, and proceed to the next step. S t ep r = 3 + 2 j , O s j 1 5 . Set u = O i f j s 2 , u = l i f j ~ I 3 . 4 ) and u = 2 i f j = 5 . Search

for an e-splitting (to, tl) such that uL-'E to, 71 E T",; and to, t1 separate (do, 6,). If (to, 71) is not found, set T; = T;-' , 0: = @;-I, let 0; transmit (a:-', 1) and proceed to the next stage. (In this case, we will begin trying to compute @ , ( A ) recursively if @,(A) is total. This requires a separate procedure as there may be e-splittings on T,- which do not separate (do , 6'). If this step is never completed and T e - l lies along the true path, then a:-' c A . ) Otherwise, we fix the first such pair (to, tl) which we find letting (to, tl) e-split on xJ, and set a: = a:-' for i 5 3 , a; = to and a; = tl, and proceed to the next step.

S t ep r = 4 + 2j, 0 sj s 4. Set u = 1 i f j = 0 , u = 2 i f j E (1,3} and u = 3 i f j E { 2 , 4 , 5 } . We search for v such that a:-' S v E TS,::, v separates (do , 6,) and @,(v; x,)L. If no such v is found, let T i - Tf-', 0: = @ ; - I , let Of transmit (a:-', 2) and proceed to the next stage. (In this case, if A E T,- and v is never found and a:- ' S A , then @,(A) will not be total.) 0 th- erwise, fix the first such v. Let u be chosen as in Step r - 1 . Set a; = a; = 0 , a: = v , a: = a;-' if @,(a;-'; x,) * Qe(v; xJ) and a:= a;-' otherwise, and a; = u;- for k s 3, k * u , u. We now proceed to the next step. (It is now the case that (a:, a:) e-split. At the end of Step 14, it will be the case that for all u, u 6 3 if u * u, then ( at4, a:) e-split.)

S t e p r = 1 5 + j , 0 s j s 3 . L e t y = m a x ( { x k : k ~ 5 } u { L , ( s - 1 ) } ) . Search f o r 7 such that a;-* E q E T::: , 71 separates (do, 8 , ) and @,(q; x ) J for all x s y . If no such 7 is found, set T: = TZ-', 0: = @;-I , let 0: transmit (a:-', 2 ) and go to the next stage. (In this case, if T,- lies along the true path and a:- E A E T,- then @,(A) is not total.) Otherwise, fix the first such q found. Let a: = a:- if i 3 and i * j , a: = 7, a; = a; = 0 and proceed to the next step.

0: transmits nothing. Fix y as in Step 18. Let 5, = @,(a:*) f y + 1 for j s 3 . It follows from Steps 15-18 that lh(5,) = y + 1 . Forj s 3 , fix the greatest zJ such that for some ( S S, and i, (F, z,, i ) E @:-'. By Step 1 of the construction, the E found will satisfy f S @,(a;) so that z ,=zJ for i + j ; let z = z o . Place ( & , z + l , O ) , ( & , z + l , O ) , ( F l , z + l , l ) and ( & , z + 1 , 1 ) in @ : Z @;-'. Note that by Step 2, if ~ E S , then

(2.13)

We leave it to the reader to verify induction hypotheses (2.9)-(2.12).

S t e p 1 9 . T: releases a. We now extend T, and 0,. Set T: = T:-' u {t: 3 i 5 3 ( t S

V k s 1 3 , j S 3(a;'(x) = O & CT:~(X) = 1 & (&, z + 1 , k ) , ( 8 , z + 1 , k ) E 0:").

MINIMAL DEGREES AND RECURSIVELY INSEPARABLE PAIRS OF R.E. SETS 335

This completes the construction. The comments made during the construction should en- able the reader to verify the following lemma.

2.2. Lemma. LetTe=UsT:beconstructedusin2.1. LetO:receiue ( t s , u s , S ~ , S ; , S s , T " , _ : ) and assume that (2.3)-(2.8) hold for all t 2 ts and that T,- = u { T t : u 2 rJ. Then:

(i) (ii) u z s & r E T : " - T:*rsepurates (S~,S; ) . (iii) Vu(T: transmits ( 7 , c ) * 7 2 uu) .

(iv) Suppose that for all u 2 u 2 ts, I , = t , , 0: transmie the same pair (T, c ) from the same step of the construction, T c X E T, - and X separates (Do. D1 ) . Then either (a) ~ P c X V Y ( P S Y E T ) , - l * v c X ) , (b) c = 2 & 3 u c X V Y ( u c Y E Tc-I*@p,(Y) not total), (c) c = 1 & 3 a c X V u z u ( a E T i - l & there is no e-split extension ( T o , sl) on T:-l

r f T: releases u at stage s, then there is an a E T: such that a 2 u and @,(a; y ) l for all Y < L,(s).

Let ( T i : s E N) and (0:: s E N) be as above. We define B),.S(a) for a E T i as follows: We set B,.,(a; x ) = m if there is a (t, x, m ) E 0; such that Fs @,(a). We note that Be,s(a) is a string. We decompose Be, s(a), setting B;, s(a; x ) = Be, Ja; 2x + i ) .

The construction of A will eliminate 2.2 (iva) as an alternative. If 2.2 (ivc) holds and @,(A) is total, then we will want to show that @),(A) is recursive. We introduce recursive computa- tion bushes to achieve this goal. The purpose of these bushes is to impose retraint on the enu- meration of locations in Do and DI which could cause a previously computed value for @,(a; x ) to change. Approximations a to A are allowed to change, and we wait for @,(a; x) to converge. If this never happens, then @,(A) is not total. If we find convergence to the pre- vious value, then the restraint on Do and D1 is removed. Otherwise, we will have found an e-splitting and so no longer need this bush.

{T,": u 2 tr) is a recursive sequence & V u 2 ts(T: E T::;).

such that both T~ and T, separate (S;, 8 : ) ) .

(v)

2.3. Const ruc t ion of recursive e - computa t ion bushes. We proceed by induction on s. Let T: receive ( t , 6, do, S1, T;::). If s = 0 or s < t or T:: has not released u, then T:= 0 = A : and T: transmits nothing. Otherwise, we let L:(s) = = max {x: V y < x(@,(A:- ' ;y)J)) . We search for 7 such that U S T E Ti:; , T separates (So, S,), @e(~;y f~ fo ra11y5L: (~ )and a C ( 7 ; y ) = @,(1:-';y)foraIly<L:(s).Ifnosuch t is found, set T: = T:- ', A: = A;-' and let T: transmit (u, 2 ) . If such a f exists, fix the first such T. Let A: = T and T: = Tt-' u (e: e

This completes the construction. We note that if T: * Tz- I , then L?(s + 1) > L:(s) and by induction, if T is the unique longest string on T: - T i - ', then for all 1 E T;-' and x < L:(s) if @),(A; x ) J , then @),(A; x) = @),(T; x ) . Hence if T, is infinite and A E Te, then@,(A; x) can be computed recursively by finding the least stage s such that L ; ( s ) > x and noting that Ge(A; x ) = @&:-I; x ) . With this in mind, the reader should be able to verify the following lemma.

T}. I": transmits nothing, and releases u.

2.4. Lemma. Let T,=U,T: be constructed as in 2.3. Suppose that @J, receives ( t s , u , , S & 6 f , T::;) andthat T , - l = U { T ~ - , : u 2 t s } . Then:

(i) IT:: u 2 ts} is u recursive sequence & V u 2 t J T ; S T::;).

336 M.LERMAN

(ii) uz t , & T E T : + ' - T:=zseparates ( d i , S;). (iii) V u (T: transmits ( 7, c ) 3 7 2 a,). (iv) X E Te* ae(X) is recursive. (v) Suppose that for all u 2 u 2 t,, t , = t , , T i transmits (7, 2 ) , 7 c X E T,- , , X separates

(Do, D1 ) and there are no e-splittings extending 7 on T, - which separate (Do, D1 ) . Then @,(X) is not total.

(vi) If T: releases u at stage s, then there is an a E T: such that a 2 u and @,(a; y)J for all

The priority ordering for our construction is a combination of priority orderings for our tree of trees and for the requirements. For u, T E S, we say that u has higher priority than T if u precedes 7 in the lexicographical ordering. We now fix a recursive ordering ( R , : n E N) of all requirements in {P,: e E N} u (Q,, k, i : e, k E N & i I}, with the provise that if Q,, k, = R , , then n 2 e + 2 . We say that R i has higher priority than R j if i < j .

The types of requirements we deal with in our construction are of the form (R,,, a) for n E N and U E S3 such that lh(u) = n. We denote such a requirement as R,. We say that R, has higher priority than R, if (I has higher priority than t.

Requirements will be declared to be satisfied at some stages of the construction, but this satisfaction may be removed at later stages. The determination of whether or not a require- ment is satisfied depends on our approximations to A , Do and D1 as well as to the bushes we define. These bushes are indexed by elements of S3. To find the main tree on which a re- quirement is satisfied, we define @(a, e ) = u 1 e + 1.

If R, = P , k.,, , is declared to be satisfied at stage s, one of the following situations will exist. The first is u(e) * 0 and a, E T;;a!e). The other is u ( e ) = 0 and a, E T;;u,le, and there will be an x such that @,{&,,,; x ) J * &.',,(x).

If R, = P, is declared to be satisfied at stage s, then there will be a location x assigned to P , , at stage s such that x < lh(a,), lh(d$), lh(6;) and d : ( x ) = 1 for some i s 1, and for this i,

As for t > s, it may not be the case that a,2 a,, the situations causing satisfaction may no longer exist, so the satisfaction of requirements will be declared to be cancelled.

Each R, may have associated with it at stage s, a location x , and a guide p;. We will define A = lim{pS,: uc r}. The location x , may become realized at stage s depending on the action of a certain computation. If R, = P,, and x , is associated with R,, then x, is realized at stage s + 1 if 2:' ( x , ) J . If x , is associated with R, = Q,, k. i. then the guide p', will also be asso- ciated with a coguide v:. x, is realized at stage s + 1 if there is a @ 2 v: such that x, < lh(B) s and @,(@; x , ) J . Such a @ will then be established as a target for P,,,,.

Action at stage s + 1 for R, cannot take place unless the guide p i is released by an appro- priate tree T:. We say that p: has been released by T, by stages if T: receives the initial stage t and there is a stage r such that t 5 r 5 s and p i is released by T, at stage r .

We write p:+' +p: if either p:J but was cancelled at stage s + 1 (even if p:+' is then newly defined to take the value p i ) or exactly one of { p i + ' , p:] is defined, or both are de- fined but are unequal.

Action at stage s + 1 of the construction will be governed by the highest priority require- ment which requires attention. We say that R , = P,. requires attention at stage s + 1 if R, is not satisfied at the end of stage s and one of the following conditions holds: (2.14) R, has no location assigned to it at the end of stage s and the guide p i is defined

and is released by T, by stage s.

y < L : W .

Z,(X) * i .

MINIMAL DEGREES AND RECURSIVELY INSEPARABLE PAIRS OF R.E. SETS 337

(2.15)

(2.16)

R, has location x assigned to it at the end of stage s, lh(p ' , ) > x and p: is released by T, at stage s.

Neither (2.14) nor (2.15) holds but R, has a guide p', for some TE u such that s(lh(7) - 1) = 0, T: and T:-' have the same initial stage and both receive the same tuples with p', as root, and T: proceeds to a step beyond that at which the construc- tion of T:-' ended.

We say that R, = Q,, k. ,, , requires attention at stage s + 1 if R, is not satisfied at the end of

R, has no location assigned to it at the end of stage s and the guide p', is defined and is released by Te(,,c, at stage s.

R, has realized location x assigned to it at the end of stage s and some a 3 p', is re- leased by Tot,. e) at stage s.

R, is potentially satisfied or is presatisfied and the guide p', is defined and has been released by T, at stage s.

None of (2.17)-(2.19) hold, but R, has a guide p: and for some 7s u such that r(lh(t) - 1) s 1, T: and T:- ' have the same initial stage and both receive the same tuples with p: as root, and T: proceeds to a step beyond the step at which the con- struction of ?':-' ended.

We now present the construction. All quantities are undefined at stage s unless we specif- ically define them at stage s. Requirements satisfied at the end of stage s remain satisfied at the end of stage s + 1 unless this satisfaction is cancelled during stage s + 1 .

Stage 0. Set a. = yo = 6: = 6; = p i = 0 and let TB = Id . No requirements are satisfied. Stage s + 1. There are two steps; the satisfaction of requirements and the extension of

bushes.

S tep 1. Let R, be the requirement of highest priority which requires attention at stage s + 1 . If no such requirement exists, set ys+ = ys and Sf" = S; for j 6 1. All guides, co- guides, targets and locations associated with a given requirement at the end of stage s remain associated with that requirement.

Suppose that R, exists. For all R, of lower priority than R,, cancel all guides, coguides, targets and locations associated with R, and cancel the satisfaction of R,. For all R, of higher priority than R,, all guides, coguides, targets and locations associated with R, at the end of stage s remain associated with R,. Set ys + , = u. We now proceed by cases, depending on the nature of the requirement R,.

Case 1. R,= Pe,,. We adopt the appropriate subcase below, depending on which of

Case 1 a. (2.14) holds. Fix the least x > s + 1 such that x is not currently assigned as a lo- cation to a requirement of higher priority than R,. Assign x as a location to R,. Set SJ + ' = 6s for j s 1. Fix a 3 p: extending a maximal string on T: and separating (a;, Sf). Let p r ' =a.

Case l b . (2.15) holds. If x is unrealized, set p::: = p:" = p ' ,and6" '=6j f o r j l l . Suppose that x is realized. If p:(x) = Z,(x), we must modify p;. Fix the lowest priority 7 c ~7

such that either 7 = 0 or t(lh(r) - 1) = 0. Then there will be an a E T: such that c( separates (a:, Sf), a ( x ) * Z,(x) and pL;E a for all [2 6 such that p i is defined. (For the sake of

stage s and one of the following conditions holds.

(2.17)

(2.18)

(2.19)

(2.20)

(2.14)-(2.16) holds.

22 Ztschr. f. math. Logik

338 M.LERMAN

completeness, if no such a exists, then we terminate the construction, defining all uncan- celled quantities to take the value they currently have at all stages t > s.) Fix such an a. Set p",' = a and Sf' = 8; for j 5 1. Finally, suppose that p",x) * Z,(x) = i. R, becomes satis- fied. Set p::: = p:" = p: and S f " = 8: . For z E N, define

(S;-i(z) if S";_,(z)J and z * x,

I? otherwise.

Case l c . (2.16) holds. Set p:+' = ,u; and a;+' = Sf f o r j s 1. Fix the highest priority T sa- tisfying (2.16). If 0: transmits ( q , j ) , set p:?; = q and establish s + 1 as the initial stage for Ti:$. If 0: has no transmission, then T: releases p',. Since none of the previous cases ob- tain, there is a highest priority e such that T E e c a and p: is not released by Ti. 0: will then transmit some pair ( q , j ) as above. We proceed as above with e in place of t.

Case 2. R, = Q,, c. i, ,. We adopt the appropriate subcase below depending on which of

Case 2 a. (2.17) holds. If a(e + 1) = 0, then by Lemma 2.2 (v), there is an a E Ti(,. e) such that a 2 u and (Qe(a, y)J, for all y < L,(s). Fix the first such a found. Establish BJa) as a coguide for R,. Let y be the least number 2 lh(B, Ja)) such that for some x, y = 2x + 1 - i; establish this x as a Iocation for R,. In all cases, let p:+' = p:, and 8 j L 1 = 6; for j s 1. R, be- comes potentially satisfied.

Case 2b . (2.18) holds. Set p;+' = @; if the target B: is defined. Otherwise, we establish the target B:+' as follows. Since Case 2a has already been followed, it follows from the uni- formity of the cancellation procedure that R, has a coguide v; in existence at the end of stage s. Fix the first B 2 v', such that x < lh(B) 5 s and Ok(B; x)L = m. Define

(2.17)-(2.20) holds.

v;(z) if z<lh(v",), I - m if z = 2 x + l - i ,

D ( u ) B:+"(z) = 0 if z = 2u + 1 - i for some u > x, v < lh(B),

if z = 2v + i and u < lh(P), !? otherwise.

We assume by induction that v: = B,,,(pJo). Since the location x is assigned, a(e + 1) = 0, so T, = Tot,, c) is an e-splitting bush. By Lemma 2.2 (ii) and (v), there is an a E Ti such that a separates (S;, S;), v: c Bc,,(a) E and a is maximal on 7':. Fix such an a. Let

If v:' = D;' ' , then R, becomes presatisfied at stage s + 1.

at stage s + 1 un lys a(e + 1) = 0 and R, is not presatisfied at the end of stage s.

v s + l , = Bp,l+l(a) and maintain x as a location for R,. Set 6;'' = Sf f o r j s 1 , and pi+ ' = a.

Case 2c. (2.19) holds. Setp:+'=p",i=p:and Sf"=Sf f o r j s 1. R, becomessatisfied

Case 2d . (2.20) holds but none of the previous cases hold. Proceed as in Case lc.

S t e p 2. T:' ' is undefined unless p:' is defined. If T = a * 0 , then T",: receives the in- itial stage s + 1. We proceed by induction on Ih(r), higher priority requirements first.

Case 1. t =0 . Set Ids+'. Case 2. T = 7- * O . T:" has the same initial stage t as T: unless it has previously been as-

signed the initial stage t = s + 1. Find the highest priority e 1 T, if any, such that either p i + ' e T:, or pic1 J * p i , or Q = yJ+' and there is no 5 such that p $ + ' J = p;, and pi' ' sepa- rates (Si+ l , 8;' l ) . If no such e exists, set T:' l = T:. Otherwise, if R, = R, is assigned a lo-

MINIMAL DEGREES AND RECURSIVELY INSEPARABLE PAIRS OF R.E. SETS 339

cation x by stage s + 1, set S = {x} ; S = 0 otherwise. Let T:+ extend T: as in Construction 2.1, with @:+' receiving ( t , p : + l , d:+l, d i " ' , S, T : - ) , where 8, corresponds to 0, for e = lh(z) - 1.

C a s e 3 . z = 7- * 1. T:+' has the same initial stage t as T: unless it has previously been assigned the initial stage t = s + 1. If t < s + 1, let T:' extend T: as in Construction 2.3 with T:+' receiving the same tuple as T:. If t = s + 1, the T:-,o transmits a pair (?,I, 1) . We begin constructing T:+' as in Construction 2.3, with T:" receiving ( 1 , ?,I, 6 ; + ' , h i+ ' , T ; - ) .

C a s e 4 . t = 7 * 2 . Then T:-., transmits a pgir ( 7 , 2 ) for exactly one jsl. Let

This completes the construction. Let A = lim,a,, r= lirn inf,y,, and for j 5 1, we let 0, = {x: 3s(6:(x) = l)] . Since the construction can be carried out recursively, Do and Dl are r.e.

We extend priorities to include r. Thus 6 has higher priority than f if 6 has higher priority

T S C 1 = ~

Ext (T:- ; 7) .

than some y c f. Otherwise 6 has lower priority than f.

We now prove a sequence of lemmas which will enable us to show that A , Do and D , have the desired properties. We first note the following properties of the construction. ((2.21) be- low follows from Lemma 2.2 (iii) and Lemma 2.4(iii).)

(2.21) For all t , u, ?,I a n d j s 1, if 7 is the root of Tf-.j and p = P C - . ~ + I ) is defined, then p 3 7.

(2.22) For all 1 and u, if p f is defined, then pf- is defined and p!,- E pf.

Furthermore, it follows from the cancellation procedure that

(2.23) If p!, is defined and p f * p ! , - ' , then pl is undefined for all E of lower priority than u.

2.5. L e m m a . For all t~ N and u~ S,: (i) There is at most one T 2 u such that p : is defined and is not released by T i . Furthermore, for 7 as in (i), (ii) If E 2 CT and T has higher priority than t, then p [ is undefined.

Proof. We proceed by induction. The lemma clearly holds for t = 0. We note from the construction that for any stage r , there is a t most on q such that p i L * p i - ' . Fix t satisfying (i) and (ii) for u. We show that (i) and (ii) hold for t + 1 in place of t .

If there is no (2 u such that p y 'J * p i , then (i) and (ii) are immediate for t + 1 in place of t . Otherwise, fix such a E. By (ii), there must then be an 7 such that either t f = 7 or ?,I has higher priority than z and R , requires attention at stage f + 1. If q has higher priority than 7

then p : is cancelled at stage t + 1. Suppose that t] = 7. Since t 2.u, R a ' must release p :+ I .

The lemma now follows from (2.23) in both cases.

2.6. Le m m a. Let y, 1 and t be given such that y has higher priority than 1 and p and p are both defined. Xken p'y E p l .

Proof. Immediate from (2.21), (2.22) and (2.23) and the fact that if p t l * @ ; - I and R6 is the highest priority requirement to require attention at stage t , then y E 6 and there is a

340 M.LERMAN

vS 6 such that lh(v).= l h ( y ) , v has higher priority than y, T i transmits (pb, c ) for some c , and p i is the root of TL. (note that the cancellation procedure implies that there is no 7 such that 6 has higher priority than 7, 7 has higher priority than y and p : is defined.)

2.7. Lemma. Let y have higherpriority than f. Then there is a stages such that for all t 2 s: (i) r f TI is defined, then Ti has initial stage r < s. (ii) R, does not require attention at stage t . (iii) p I = p;.

Proof. We proceed by induction on l h ( y ) and then by priority. Assume that the lemma holds for all 6 such that lh(6) < l h ( y ) and all 6 such that lh(6) = lh(y) and 6 has higher prior- ity than y.

We note that if R , is the highest priority requirement to require attention at stage s + 1 , then ys+ = y. Hence if y has higher priority than f, then

(2.24) { t : 6 a y & 6 has higher priority than p & Rd requires attention at stage r} is finite.

Hence (i)-(iii) follow by induction if ya f, as in this case R, can require attention only finitely often. Thus we may assume that y c f.

The initial stage of T, can be changed at stage t only if some requirement of higher priority than R, requires attention at stage t . Hence by (2.24) and induction, there is an r such that, for all sufficiently large t , if TL is defined, then TI has initial stage r .

Thus we may fix a stage so such that for all t 2 so no requirement of higher priority than R, requires attention at stage t and Ti has initial stage 5 so if it has an initial stage. We note by Constructions 2.1 and 2.3 that R, can require attention only finitely often through (2.16) or (2.20) as long as ,uiJ = p i - ' , since after finitely many successive steps, each tree will release p i . Hence if R, requires attention infinitely often, some other clause from (2.14)-(2.20) must cause R, to require attention infinitely often. R, can require attention through (2.14) at most once after stage so, after which it has a location for all subsequent stages. R, can then require attention at most three times through (2.15), at most once before x is realized and then twice when is released by R,, the second time causing the satisfaction of R,. (Note for the stages between those at which R, requires attention through (2.14) and (2.15), p i re- mains unchanged.) Thus (ii) will hold for all sufficiently large t if R, = Pe,,,.

Suppose that R,= Q , . k , i . v . Again p t - ' = g;-l if R, requires attention through (2.20). R, can require attention at most once through (2.17), at which point it becomes potentially satis- fied. If y ( e ) * 0, then R, will never require attention through (2.18), and if it ever requires at- tention through (2.19), R, becomes satisfied and will never again require attention. Suppose y ( e ) = 0. If R, requires attention through (2.18) at some least t > so, then R, is assigned a target S i and S: = S: for all u t . Furthermore, if t 5 v 5 u and R, requires attention through (2.18) at stages v and u, then there are coguides v ; and v ; for R, such that v ; c v ; E SL. Since this can occur only finitely often, there must be a last stage at which R, requires attention through (2.18). If R, is not presatisfied at this stage, then R, never again requires attention. R, will not require attention through (2.19) more than twice - once when it is potentially satisfied and released by T, and once after it is presatisfied and is released by T,. Hence again (ii) holds for all sufficiently large t .

If p Y 1 * p i for t 2 so, then R, requires attention at stage t . Hence by (ii), p Y 1 = p: for all sufficiently large t .

MINIMAL DEGREES AND RECURSIVELY INSEPARABLE PAIRS OF R.E. SETS 341

For y of higher priority that f, it follows from Lemma 2.17 (iii) that py = lim, p t exists. Suppose that yc f. We note that

(2.25) p i . , is undefined for all i 5 2 unless pb has been released by T:.

2.8. Lemma. Let aand t be given such that p i J . Then p: separates (ao, 6 , ) .

Proof. Fix a and t as in the hypothesis of the lemma. By Lemma 2.6, we may assume that u is the lowest priority string such that p fJ . The lemma follows easily by induction from the construction, Lemma 2.2(ii) and Lemma 2.4(ii), except when R, becomes satisfied in Case lb of the construction.

Suppose that R, becomes satisfied at stage t of the construction through Case lb. By in- duction, it suffices to show that pi t l (x ) = 0 if 6i(x) = 1 and p : + ' ( x ) = 1 if SS"(x) = 1. Fix i such that Z,(x) = i . Then

pL",+'(x) = ps,(x) * ZC(X),

so pS,+'(x) = 1 - i . Furthermore 6, -,(x) = 1. Since p i separates (a:, 6 ; ) by induction and 6 J + ' = , d:, p i + ' must separate (&,+I, a;+,).

2.9. Lemma. ( t : 3a ( R , requires attention at stage t ) } ir infinite.

Proof. We note that R B will require attention at stage 1 as Id' releases p i = pi = 0. Sup- pose that R, is the highest priority requirement to require attention at stage t . Then one of the following cases must hold:

(2.26)

(2.27) T2-I releases pi-' and * pf- ' .

(2.28)

(2.29)

If either (2.26) or (2.27) holds, then by Lemma 2.5 (i), T: has p i as root at stage t + 1 and has not released p : (let 7 = a for (2.27)). By Lemma 2.8 and the definition of S, (2.3)-(2.8) will hold for some [ G 7. Hence R, will require attention through (2.16) or (2.20) if it does not re- quire attention through another clause.

If x is a location for Pc, and Pc, is not satisfied at stage r, then if Sl(x)J for any j 5 1, then 6 f ( x ) = 0. Hence by Lemma 2.8, (2.28) cannot occur.

Suppose that R, becomes potentially satisfied, through Case 2a of the construction, at stage t . If Ti does not release R,, then R, will require attention through (2.16) or (2.20) at stage t + 1 if it does not require attention through another clause. Otherwise Ti releases R,, and R, requires attention through (2.19) at stage t + 1. Thus the lemma follows by induction.

Some p : is appointed and Ti is given initial stage t .

The construction terminates in Case lb.

R, becomes potentially satisfied (Case 2a) at stage t .

U

2.10. Lemma. (i) h(T) = 00. (ii) A = lim[py: yc f} hm infinite domain. (iii) A S T @ ' .

Proof. (i) is immediate from Lemma 2.9, Lemma 2.7(ii) and the definition 0f.T. (ii) fol- lows from (2.21) and (2.22) and the fact that there are infinitely many e for which no e-split- ting exist, and so T(e + 1) = 0. (iii) follows from (ii), Lemma 2.7(iii) and the fact that the construction is recursive. o

342 M.LERMAN

2.11. Lemma. (Do, 0,) is recursively inseparable.

Proof . It follows from Lemma 2.7 and Lemma 2.10 that R, releases p, for all a c f. Fix a recursive set Z,. Fix a c T such that R, = P,, ,. Since R, releases p,, R, must become satis- fied. But when R , becomes satisfied, there is a j such that 6f(x) = 1 and Z,(x) = 1 - j . Thus x E Dj and so Z, cannot separate (Do, D, ) .

2.12. Lemma. A does not bound a minimal degree.

Proof. It follows from Lemma 2.7 and Lemma 2.10 that T, releases p, for all a c f. Fix e such that @,(A) is total. There are three cases, depending on the value of T ( e ) .

Case 1. f ( e ) = 0 . By Lemma 2.2, we can define Be 5 T @ , ( A ) . It suffices to show that for all k and all i 1 , if Q k ( B f ) is total, then there is an x such that Ok(B;; x)J * Bl-'(x)J. Fix k and i such that Gr(Bj) is total. Let R, = Q,, k , ;, ,. Since T, releases p,, we must establish a final location x for R, by Lemma 2.7. Since Ok(B;) is total, x must eventually become real- ized. It now follows from Case 2 of the construction and the fact T, releases pg (and hence Tpt,, c) releases p,) that R, becomes satisfied and this satisfaction is never cancelled. When R, becomes satisfied, we define pu", such that p: = pu", for all t 2 s and so pu", = p, c A . Fur- thermore i f / ? { = & ( p , ) , then @k(/?f,s; x)J +/?:.;'(x)J. Hence Qk(B6; x)J * B : - ' ( x ) .

Case 2 . T(e) = 1. Fix a such that lh(a) = e + 1 and a c r. Let t= 6- * O . Then 0: trans- mits (p, , 1) at all sufficiently large stages s, and A E T,. By Lemma 2.4 (iv), @,(A) is recur- sive.

Case 3 . T ( e ) = 2. Fix uc T such that lh(a) = e . Then for all sufficiently large s, either Ou",-., transmits (p,, 2) or Tu",-.l, transmits (k , , 2 ) . Since T,- releases p,, it follows from Lemma 2.2(ivb) and Lemma 2.4(v) that for all XE T,, @,(A7 is not total. Hence @,(A) is not total, contrary to our assumption. Thus this case cannot hold. 0

Our theorem now follows from Lemmas 2.8, 2.10, 2.11 and 2.12.

2.13. T h e o r e m : There is a degree a < 0' containing a set which separates a pair of r.e., recur- sively inseparable sets, but does not bound a minimal degree.

References

[l] FRIEDMAN, H., and S. SIMPSON, Minimal degrees, hyperimmune degrees, and complete extension of arithmetic (preliminary report). Abstract # 781-E10, Abstracts Arner. Math. SOC. 1 (1980), 546.

[2] KUCERA, A., An alternative priority-free solution to Post's problem. In: Procedings MFCS '86 (J. GRUSKA, B. ROVAN and J. WIEDERMANN, eds.), Lecture Notes in Computer Science 233 (1986),

[3] LERMAN, M., Degrees which do not bound minimal degrees. Ann. Pure Applied Logic 30 (1986), 493-500.

249-276.

M. Lerman University of Connecticut Department of Mathematics U-9, Room 111, 196 Auditorium Road Storrs, Connecticut 06269-3009 U.S.A.

(Eingegangen am 18. Januar 1990)