On the Commutativity of Jumps

On the Commutativity of JumpsAuthor(s): Timothy H. McNichollSource: The Journal of Symbolic Logic, Vol. 65, No. 4 (Dec., 2000), pp. 1725-1748Published by: Association for Symbolic LogicStable URL: http://www.jstor.org/stable/2695072 .

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THE JOURNAL OF SYMBOLIC LOGIC

Volume 65. Number 4. Dec. 2000

ON THE COMMUTATIVITY OF JUMPS

TIMOTHY H. MCNICHOLL

Abstract. We study the following classes:

* Q* (r, A, ..., ikAk) which is defined to be the collection of all sets that can be computed by a Turing

machine that on any input makes a total of ri queries to Ai for all i {1, k}.

* Q(r, A,,...,ikAk) which is defined like Q* (r, A,,...,'kAk) except that quer-ies to Ai must be made

before queries to Ai+I for all I {1,... k- 1}.

* QC(rIA1, ,...,r IkAk) which is defined like Q(r1A1.. 'kAk) except that the Turing machine must

halt even if given incorrect answers to some of its queries.

We show that if Al, . . Ak are jumps that are not too close together, then all three of these classes are

identical and are not changed if we permute (r1 A1 ,...rkAk). This extends a result of Beigel's [1]. Since

the second class is not affected by permutations, we say that these sets comnnnute with each other. We also

show that jumps that are too close together may not commute. We also characterize the commutative

sequences of sets obtained by iterating the jump operation through an ordinal notation.

?1. Introduction. We address the following kind of question: Is there any advantage to querying 'hard' oracles before 'easy' oracles? We can make this more precise as follows. Let Q(A, B) be the class of all sets of natural numbers that can be computed by a Turing machine that on any input first makes a query to A and then makes a query to B. If A is 'easy' with respect to B in some sense (e.g., A <T B), and if Q(A, B) and Q(B, A) are comparable, then is Q(A, B) a larger or smaller class than Q(B, A)? We define Q(A1 . An) in a similar manner and ask similar questions about how this class is affected by permuting (A 1 . An). Questions of this type were first considered in the context of complexity theory by L. Hemas- paandra, H. Hempel, and G. Wechsung [6]. Here, we focus on a recursion-theoretic setting.

We show that if A1,. , An are jumps that are not too close together (in a sense to be spelled out later), then Q(A1, . . ., A") = Q(A(1), . . , A,(n)) for all a E Sn. For this reason, we say that such a sequence of sets commutes. This extends an (unpublished) result due to R. Beigel that if for some a, b > 1 Ai E {0(a), 0(h)} for all i E {1, . . . }, n, then (A1, . . ., An) commutes. The methods used here however, are very different. Beigel's proof first arranges (A 1 . An) into hardest first order without loss of computational power. That is, if (B1,..., Bn) is a re-arrangement of (A,,..., An) such that B1 ?T ? * T B,7, then Q(A l, . . A,7 ) C Q(B , , B,1 ) The hardest first ordering is then arranged into the least first order without loss of computational power, and then the least first ordering is arranged back into

Received May 27, 1998; revised april 22, 1999.

? 2000. Association for Symbolic Logic

0022-4812/00/6504-001 4/$3.40

1725



1726 TIMOTHY H. MCNICHOLL

the initial arbitrary ordering without loss of computational power. The proof here of the commutativity of jumps that are not too close together first arranges the arbitrary ordering into the least first ordering without loss of computational power, and then arranges the least first ordering back into the arbitrary ordering without loss of computational power.

The chief advantage here is not that we can accomplish things with one fewer re-ordering. Rather, the main advantage appears when we move into the realm of hyperarithmetical sets. Namely, we directly show that an arbitrary ordering of sets of the form H,, for some a E a can be put into least first order without a loss of computational power. However, we also show that these sets can not always be put into hardest first order without a loss of computational power. Since Beigel's proof that a sequence composed of 0(a), 0(b) can be put into least first order requires the sequence to first be put into hardest first order, it does not extend to the hyperarithmetical hierarchy.

?2. Notations, definitions, and preliminary propositions. Unless otherwise men- tioned, all notations are as in Soare [8].

Let ?lex be the lexicographic ordering of {O, 1 }I<o . We say that a E {O, 1 }' is even (odd) if an even (odd) number of elements of {O, 1}'7 lexicographically precede a. If 0 74 S C {O, 1}<, then lexmin(S) is the lexicographical minimum element of S.

If A is a set, then #A is the cardinality of A. We identify subsets of co with their characteristic functions. Thus, for all A C co and x E co,

A 1x if x E A

l if x VA.

We now wish to make a few remarks on terminology regarding Turing machines. We let {/,e}e.Cw be an effective enumeration of all Turing machines. Then, 0e (x) is the output, if any, when Turing machine q5e is run on input x. +(A) (x) is the output. if any, when Turing machine 0e is run on input x and oracle A is used. That is, when the characteristic function of A is written on the tape before the computation begins. Since we wish to consider situations where several oracles are queried, it might seem that we need to consider new classes of Turing machines where, say, more than one tape is used and the characteristic function of each oracle to be queried is written down on one of the tapes. However, this is not necessary For, if we wish k oracles A1, . . ., Ak to be queried, we may write down the characteristic function of their join on the tape before the computation begins. That is, the single oracle AI e ... D Ak may be used. Here, we define the join of k sets by

Al * ...ED Ak =U{ka +j I a E AI}. .i<k

Thus, for fixed k and j < k, when we speak of a Turing machine making a query to the j-th oracle what we mean is that the number whose membership is being queried is congruent to J modulo k.

We can now define our order-of-query classes as follows.

DEFINITION 2.1. Let A1,...,A C co, and let rl,. . , rk > 1.



ON THE COMMUTATIVITY OF JUMPS 1727

(1) We define FQ* (r A1,...,rkAk) to be the set of all functions Turing reducible to Al e ... e Ak via a Turing machine that on any input makes a total of ri queries to Ai.

(2) We define FQ(riAi,...,rkAk) to be the set of all functions that are in FQ* (r A1,..., rkAk) via a Turing machine that on any input makes its queries to Ai before its queries to Ai+, for all i < k.

(3) We define FQC(r1A, A..., rkAk) to be the set of all functions that are in FQ(riAl,..., rkAk) via a Turing machine that on all inputs halts on all possible computation paths.

We define Q*(riAi,..., rkAk) to be the collection of all sets in

FQ*(riAl,..., rkAk)-

We define Q(riAi,..., rkAk) and QC(r1Ai,..., rkAk) similarly. For all

D E {Q. Q*, QC, FQ, FQ*, FQC}

we let D(A1,..., Ak) be F(lAI,..., lAk). The proofs of the following two propositions are left to the reader.

PROPOSITION2.2. Let r1,..., rk > 1, and let A1.-, Ak, B1, , Bk C co.

(1) FQC(r1A1,...,rkAk) C FQ(riAl,...,rkAk) C FQ*(riAi,...,rkAk).

Let F E {Q, QC, Q*, FQ, FQC, FQ*}.

(2) If Ai <m Bi for all i E {1 .., k}, then

D(rAi1,..., rkAk) C (D(riBi,...,rkBk).

(3) If Ai-m Bi for all i E {1 .I , k}, then

D(rAi1,..., rkAk) = (D(rlBi... rkBk).

PROPOSITION 2.3. Let A1.-, Ak, B1, . . ., B, C co. Suppose that there exists an order-preserving injection f: { 1, . . . -, k} { 1? , . . ., n} such that Bf (i) >1,7 Ai for all i E {1,. , k}. Then, D(A1,..., Ak) C D(B1,..., Bn) for all

(e {Q, FQ, QC, FQC}.

Here is some additional terminology regarding Turing machines. If a E {O, 1 }<W,

then q$() (x) is the output, if any, when Turing machine q5 is run on input x and a is written on the tape before the start of the computation. Such a computation may diverge by virtue of the fact that a may not contain answers to all of the queries made. Here is a variation on this idea. Let qP1 (x) be the output, if any, when Turing machine qe is run on input x and for each i < lh(a), a(i) is used to answer the i-th query made by this computation (if there is one). If lh (a) = n, then such a computation may diverge by virtue of the fact that it may attempt to make an n + 1-st query. In this case, we define qe(a, x) to be this (n + 1)-st query This defines a partial function qe from {O, 1}<(") x co into co. Now, when k is fixed and j < k, we can say that qe (a, x) is a query to the j-th oracle (without knowing which oracles are being used) when it is defined and congruent to j modulo k. We define

0g, qes, et cetera, by limiting the number of computation steps to s. We define the




computation path taken by q5(A)(Z) to be the a E {0, 1}W such that for all p c a, qe(p,z) t and

qe (P~Z) EEA X~ a (lh (p)) =1.

Clearly, a function f is in FQ*(rAl....rkAk) if and only if there is a Turing machine )e such that f = O(AeD3 .eAk) and for all x E co and all a E {0, }I<c for

which q$f7] (x) I, the number of prefixes of a at which the j-th oracle is queried is r1 for each j E {1 , k}. We call a Turing machine having the latter property an (rl,..., rk)-operator.

?3. A mind change lemma. The study of commutative oracles lies in the field of bounded queries which is the study of classes of the form Q(n, A). See, for example [3] or [4]. Many arguments in bounded queries are built around the observation that the limit of a computable binary sequence with at most 2'7 - 1 mind changes can be computed with n queries to the halting set. More formally, if f is a function which is defined as the pointwise limit of a uniformly recursive sequence of binary functions {ff}sIsco then f E Q(n, 0') if for all s E co, #{x E coIfs(x) 74 f (x + 1)} < 2n _ 1. This observation does not generalize to sequences uniformly recursive in some oracle A. In later sections, we will need to deal with this issue as well as with sequences of partial functions. The purpose of this section is to provide a facile way of doing so.

We begin with the pertinent definitions. We first formalize precisely what we mean by a mind change in a partial binary sequence.

DEFINITION 3.1. Let f be a partial function from co into {O, 1}. The number of mind changes inf is defined to be the cardinality of

{t E dom(f )I-t' E dom(f) [t' > t A f (t) 74 f (t')A

Vs E dom(f )(t < s < t' X f (s)=f (t))]}.

Rather than look at functions defined as pointwise limits of sequences of partial functions, it will simplify our formalism somewhat to look at the equivalent situation of partial functions defined by limits of the form limb+s q$(x, y) where q is a partial function of two variables.

DEFINITION 3.2.

(1) A partial (total) array is a partial (total) function from co x co into {O, 1}. Let q be a partial array.

(2) For each z E co, the z-th row of q is the partial function f given by

f (s) = q(z, s).

(3) Let z E co. If q(z, t) X for at least one t E co, then the initial number in row z of is

4(z,min{t E co(z, t) 4).

A key tool in using oracles to count mind changes is the following.

PROPOSITION 3.3. Let f: - -* be recursive, and let A C ow. Let k, k' E co, and




(1) If k < #WA) < k' for all z E co, then there exists h E FQC(OA') such that h(z) #WA ) for all z E co.

(2) If k < #Wf() < k' for all z E co, then there exists g E FQ(00') such that Dg(;) Jf( for all Zco.

PROOF SKETCH: The proof of (1) is by binary search. To prove (2), let z E co be given. We first compute h (z). We then wait for t E co such that #W()t h (z). We then compute a canonical index of Wf()t. A

LEMMA 3.4. Let f be a total array. Let (i, a) E {O, 1} x co be such that the initial value of each row of f is i and the number of mind changes in any row of f is at most a. Then, the set f defined by the equation

f (s) = lim f (s, t) t-oo

is in QC(0log2(a + 1)], graph(f)').

PROOF. Let A = graph(f). The key fact is that there exists recursive g: co - co such that # WA) is the number of mind changes in row z of f for all z E co. One way to show this is to define a recursive operator P such that for all z, y E co,

T( )(Zy) Ex ly e co[y > y A f(z,y) 4 f (Z,y')

A Vx wco(y <x<y' X f(z,y)= f(zx))].

The function g can then be obtained from the snm Theorem [8]. It follows from Proposition 3.3 that there exists h E FQC([log2(a + 1)1, A') such that h(s) is the number of mind changes in row s of f for all s E co. It follows that for all s E co

f(s) fi, if h(s) is even 1 - i, otherwise.

It then follows that f E QC([log2(a + 1)], graph(f)'). A

DEFINITION 3.5. Let 0 be a partial array.

(1) A relative position function for 0 is a map R: co {O, 1} x co such that for all s E co the number of mind changes in row s of q is no larger than R2(s) and if equal to R2 (s) then R1 (s) is the initial number in row s of q.

(2) Let R be a relative position function for q, and let M E co. We say that R is bounded by M if R2(s) < M for all s E co.

Thus, for each z E co, a relative position function for q specifies a guess at the initial number in the z-th row of q and an upper bound on the number of mind changes in the z-th row of q and gets this guess incorrect only if this upper bound fails to be tight.

LEMMA 3.6 (The Mind Change Lemma). Let n E co. Let X be a partial array with relative position function R that is bounded by 2n - 1. Then, there exists V E FQC(n, graph(Q)') such that for all s E co,

lim q$(s, t) exists X lim q$(s, t) =i(s, R(s)). t +oo t-+oo




PROOF. For each i E {0, 1}, let f i be the total array defined as follows. Let fi(s,O) = i for all s E co. Let t E co. If the number of mind changes in (f i(s, 0) I ... I f (s, t)) is less than 2' - 1, then

fP(s, t + 1) =df I -f (s, t) if fi (s, t) (

44s, t) X

f i (s, t) otherwise.

If the number of mind changes in (f i (s, 0), ... f X (s, t)) is not less than 21? - 1, then f(s, t + 1) =df f(s, t).

We note that f i <T graph(Q). Hence, graph(fi)' <m graph(Q)'. Furthermore, the number of mind changes in any row of f i is at most 2' - 1.

For each s E co and (i, a) E {0, 1} x co, let V (s, (i, a)) =df limto f i(s, t). It

follows from Lemma 3.4 that Vg E FQC(n, graph(q)'). Let s E co. Suppose that limit, 4(s, t) exists. It follows that if the number of mind changes in row s of q

is less than 2'" - 1, then limtOO q$(s, t) = V (s, R(s)). Suppose that the number of mind changes in row s of q is exactly 2n - 1. Then, R1 (s) is the initial number in row s of 0. It follows that

tliM +(S t) = tliM f R, (s)(S, t) =y(s, R (s)).

?4. Commutativity of jumps.

DEFINITION 4.1. A sequence of jumps (A . 4.)., Al is said to be well separated if Al <T Ai+, for all 1 < i < k.

The main result of this section is that any well-separated sequence of jumps commutes. To begin the work of this section, we give a theorem that allows us to move 'harder' oracles ahead of 'easier' oracles. We then show that when our oracles are a well-separated sequence of jumps, we can rearrange them into least-first order without a loss of computational power. We note that the first result is considerably easier to obtain than the second.

THEOREM 4.2. Let A, B C co be such that B <T A. Then, for all F E {FQ, FQC}

D(AI... I Ai, B, A', Ai+, ... I Ak) C F(A1,..., Ai, A', B, Ai+,.., Ak)

for all A1.-, Ak C wo and i E {1 .I. , k-1}.

PROOF SKETCH: Let f E F(A1, . . ., Ai, B, A', Ai+1, . . . Ak) via q,. Fix z E w0.

Leta E {0, 1 }k+2 be the computation path taken by q (A Ie ... eAeBeA'eA?...eAk(Z).

The first ki bits of a can be computed using one query to each of A1. A in succession. Once these bits are known, we can determine the answer to the A' query without knowing the answer to the B query. For, the answer to the A' query is 1 if and only if

Es E c [(qe(C [ ir z) E AlI..e Ai el B e A'. ( Ai+ ... Ak




Since B <T A, it follows that the matrix of the sentence above is A-recursive and so the truth of the sentence can be obtained with a single query to A'. The remaining bits of a can then be computed using one query to each of B, A?i+, Ak in succession. -1

We now give a few preliminary definitions and a Lemma which will be used heavily in other parts of the paper.

First, fix an (rl,..., rk) operator e. Let n = rl + + rk. For all z E co, let

path(e, Z) = lexmin{a E {o, I}In I1t Eco w q$,(z) Jt

A Vs < t V_ E {o, 1} In ] (Z) T}.

It may be that path(e, z) I, but this would only happen in the unusual case where

e$`7 (z) I for all c. We therefore define an (r1, . . ., rk) operator to be semitotal if for every z E co there is at least one a for which q$ ̀](z) 4.

Let U C {1,. k }, . We define a Turing machine q1U]. For input z and oracles

Al,., Ak the computation of 4[U](Ae ..

eAk) (z) proceeds as follows. Run the

computation of q(AD .

eAk) (z). For i E U, answer queries to the i -th oracle with Ai, and for i , U and j < ri answer the (j + I)st query to the i-th oracle with path(e, z)'s answer to the (j + I)st query to the i-th oracle. Thus, e[U](Ae .Ak) is partial recursive in BiEuAi. In particular, if U 0, then [U](Ae E..Ak) is partial

recursive. If U 0 and Oe is semitotal, then eU](A Fat) is recursive. We now state and prove a Lemma whose main application will be to commu-

tativity in the hyperarithmetical hierarchy but which will make short work of the commutativity of well-separated sequences ofjumps. In the statement of the following Lemma, and throughout most of this paper, we regard sets as total 0-1 valued functions.

LEMMA 4.3 (The Oracle Replacement Lemma). Let sI, . . ., sc, r , rk > 1, and let e be a semitotal (si, . . ., sc, ri, . rk) operator. Suppose that (A ..., A') is well-separated and that B1, . . . ?, B <T A1. Let U {1, . . ., c

(1) For all C1. , Cd C co, if +[U](B3 ..BeAleak) cans be extended to an

element of Q(C1 . .., Cdl), then I (BeeBdeA ak) can be extended to an element of Q(C1,...,Cd, r1Al,...,rk~k).

(2) For all C1, . Cd C cf U](BQfe. BcAie...eAk) can be extended to an

element of QC(C1, . .. Cd), then 0(B 1

kflBeAe...fAk) can be extended to an element of QC(C1, . Cd, rAl,.. , rk~A).

PROOF. We prove part 1. The proof of part 2 is almost identical. Let n= ri + .. + rk. For all i E {1 . . ,k}, let {A$s}sec, be an AB-recursive enumeration

of A'. Let h E Q(C1,...,Cd) be an extension of 0[U](Beff *elB@eA' eAt). For all y E {0. k} and z C c, let if}(z) E {0, 1} k be the computation path taken

by {f [{ l !c+Y}](B' @...eBcefflA'e lAB)(z) if this computation converges, and let (/(z)

be undefined if this computation does not converge. Let bo(z) h(z). For all y E {1. k} wedefineanelementby of Q(CQ, . . . rA .r )sothatfor




all z E co

~e (7 ) At by (z) = W (

(We will arrange things so that if h E QC(C1, C ), then

by E QC(C1I ... . Cd, rjA ..., r A').) (Bi D ..eB~aeA'a .. DA')

It follows that bk is the required extension of 4e

Let y E {O. . . , k - 1}, and suppose that by has been defined and has all required properties. To define by+, we proceed as follows. We will define a partial array q and later a relative position function for it. This will enable us to use Lemma 3.6. For all z, s E co let p(z, s) E {O, I}" be the computation path taken by

[.c,...,c+y+I}](BjDe .DBeDA'D ... eA3XeA 3eA'?2e... eA3)()

if this computation converges, and let it be undefined otherwise. Let (z, s)

0rP(zs)](z) for all z,s E co. We note that for all z E co, U)+,(z) 1 if and only if limo p(z, s) exists in which case the two are equal. Hence, for all z E co,

O[e1 I(I)](Z) t if and only if limse (z, s) exists in which case the two are equal. Since B1, ... , B, A.... ,A < Ay+l, it follows that X, p are partial recursive in

Ay+l. However, since the number of steps in the computation of O(z, s), p(z, s) is recursively bounded, it follows that graph(p), graph(X) <T A3,+1. We define a relative position function R for q such that R E FQ(C1, .. . Cd, riAl ... , ryAy). (We will arrange things so that if h E FQC(CI, C(d), then

R E FQC(Ci,..., Cd, rIA, ., rkAk).)

It will turn out that R is bounded by 2r-?+' - 1. Let qi be as given by the mind change Lemma. Then, qi E FQC(ry+i, A+,). We can then let by+, (z) =i(zR(z)) for all z E co. It follows that by+I E Q(C1, . . ., Cd, A..., A

To define R we first note a few facts. First, for all z, s E co,

q$(z, s) J j#p(z, s) 1.

Let z, s E co be such that (z, s) l, +(z, s') l. Then, since {A'y+I }tE is an Ay+1- recursive enumeration of A +,+1' it follows that

s < s ?s p(z, s) <lex p(z, s')

In fact, if we let O(z, t) E {O, 1}'+'? be the sequence of answers given by p(z, t) to queries to the (y + 1)st oracle whenever p(z, t) l, then

S < S' => 0(z, S) <lex 0(z, S'),

and

p (z, S') f p (z, S) X- 0 (z, s) 4 0 (z, s').

It follows that the number of mind changes in any row of X is at most 2r-,+' - 1. Therefore, we let R2(z) >-'+2 - 1 for all z E co. Let y E {O, 1}'IY+' be the sequence of path(e, z)'s answers to the (y + 1)st oracle. We then let

R { (Z) if y is an even element of {O, 1 }Ir?,+

l1z I -by((z) otherwise.




It follows that R E FQ(C1,.. ., Cd, rIA', ...., r A,). Let z E co and suppose that there exist 2'Y+ - 1 mind changes in row z of A. We show that RI(z) is the initial number in the z-th row of A. Let u =2"Y+ - 1. It follows that there exist so <... < sl, such that for all i E {O. . . , U (z, si) J and

i < U X~ 0( , Si) 7 (z, si+I)-

It follows that /(z, so) is the initial number in row z of X and for all i < u,

(1) 0( ) { 1~~~~ - O(z, si) if i is odd.

Furthermore,

{O(z, si) I i E {O...,}} - {O, 1 }'+1,

and for all i < u O(z, si) is an even element of {O, I}r+i" if and only if i is even. There exists j < u such that O(z, sj) = y. It follows that p(z, sj) = v) (z). Hence,

(z, sj) = b (z). By making the appropriate substitutions in (1), we see that

0(z, so) by (Z) if y is an even element of {O, I}"?+'

1 - by (z) otherwise.

Thus, RI (z) is the initial number in row z of q. -

THEOREM 4.4. Let (A ....4. ) be well-separated. Then, Q*(riAj....rk)= QC(rIA .... rkAk).

PROOF. Let C E Q*(rIA' ... rkkA') via 0e. Let U 0, d = 1, and C1 = 0.

Then, QC(1, Cl) is the set of all recursive sets, and [ 1 k is easily seen to be recursive. It follows that C E QC(C1, riAl,..., rkAk). But clearly

QC(C1, rIA', * *., rkAk) - QC(rAl, ... , rkAi6.

Thus, Q*(riAi,..., rkAk) C QC(rIA .... rkA'). But, by Proposition 2.2,

QC(rAl,....ArkA) C Q* (rAll, ,rk~k). A

We can now finish proving that a well-separated sequence of jumps commutes.

THEOREM 4.5. Let (A .....Al) be a well-separated sequence of jumps. Let rl...rk>1.Thenforall aESk,

Q*(r()A . rA') = . , rAlA' )

o() (k) o(k))

QC(rA' ...A , r' A'l) = r a(l), **, (k) a(k))-

PROOF. Let a E Sk. Clearly,

Q*(r~l)A/ .) , rU(k)A(k)) Q*(riA A'.ra).

By Theorem 4.4,

Q(rIA1, ...,rk k) C QC(rIA',...,rkAk).

It follows from repeated applications of Theorem 4.2 that

QC(rAl, ... , rk k) C QC(r,(l)A'(l). r(k)Al(k)).




But clearly

QC(r,(l)A'l). r(k)A (k)) C Q*(rjA.., rkAk).

Thus,

QC(rlA', * , rkAl) - QC(rq(l)A<l). =Q*(rA Arklk).

Now, since QC(ri A l.... rkAl) C Q(riA l. ... rkA ), it follows that

Q*(rAl, ... , rkAk) C Q(riA,. , rkAk).

It follows from repeated applications of Theorem 4.2 that

Q(rlA * rkA') C Q(rg(l)A' (l).rU(k)A(k))r

But clearly

Q(rU(l)A5(l9)., rU(k)A (k)) C Q*(rA A.rklk).

Thus,

Q(riA', * *., rkAk) = Q(rg(l)Ar(l).rC(k)A5(k)) Q*(rAl, ... rkAk).

COROLLARY4.6. Let A1,. A,7 C co be such that for all i E {1,...,n} there exists j E co such that Ai is Yj-complete. Let a, z E Sn. Then, the following classes are identical:

Q(A1,. , An),

Q(A,(1), . , A,(n))

QC(A1,..., An),

and QC(A,(1), ., AT(1))

Theorem 4.5 also allows us to find examples of commutative oracles outside the arithmetical hierarchy. For example, we could take Ai to be Ha, where ai is a notation for a successor ordinal and Ial I < < jak 1. This gives us a family of examples that are 'above' the arithmetical hierarchy An example that is incomparable with most of the hyperarithmetical hierarchy can be obtained as follows. First, note that by exercise 2.2 of Soare [8] there exists A C o such that A T0O(1) for all n > 1. It follows that A(n) IT0(n) for all m > 1 and n > 2. We can then take A ..., A' to be any strictly increasing subsequence of {A(n) }Jn~ We also have the following proposition.

PROPOSITION 4.7. There exists Y C co such that Q( Y 0") = Q(0", Y) and for all n E c, Y is neither In nor I-n complete.

PROOF. Let A be a low non-recursive recursively enumerable set. Then, A' <T 0'. Thus, by Theorem 4.4, Q(A, 0//) = Q(0"1, A'). It is easily shown that for all n E O, A' is neither In nor I1,, complete. -




?5. Non-commutative oracles. The question now arises as to whether every sequence of jumps commutes. We answer this question negatively by showing that there exists a low recursively enumerable set whose jump does not commute with 0', and that there exists an recursively enumerable set whose jump does not commute with 0".

We begin with two preliminary definitions. The first prescribes a method for generating from a set A an element of Sj'. The second defines a 'generic' element of Q(A, B).

DEFINITION 5.1. For all A C Co, let

D(A)=df {X E C01 3y E C(x,2y) E A A (x,2y + 1) f A}.

Thus, D(A) E S1. If A is recursively enumerable and if {At~tco) is a recursive enumeration of A, then D (A) = lim inf,co D (A,). The usefulness of D (A) is that one can change the value of XD(A) (x) an arbitrary number of times by enumerating elements into A.

If A is recursively enumerable and {At }tCt(o is a recursive enumeration of A, then Al (e) -df Wj(A,) (e) for all e E co. Thus, A' = lim inft c,, A.

DEFINITION 5.2. Let A, B C co. For all x, y, y' E co, we let (x, y, y') E VAB just in case

(x X A A y E B) V (x C A A y' E B)].

Clearly, VAB E QC(A, B) for all A, B C co. We can form an effective enumeration { qil of the Turing machines that on

any convgerent computation path make exactly two queries the first of which is to the first oracle and the second of which is to the second oracle. Thus, for all C, A, B C co, C E Q(A, B) if and only if there exists e such that C= V<A B.

THEOREM 5.3. There exist low recursively enumerable A, B C co such that

QC (D(A), B) 5X Q(0', A/)

and QC(B,D(A)) g Q(A',0').

COROLLARY 5.4. There exist low A C co such that Q(A', 0') Q(0', A') and Q(W, A') g Q (A', 0').

PROOF OF THEOREM 5.3: The proof is a finite injury argument. We construct low A, B C co such that VD(A) B X Q(0', A') and VBD(A) f Q(A', 0'). For each e E co we have the following requirements:

R4e :(3<,t (E co o(A,)'(e) B = (A) (e)

R4e+i (3<t E C (Bt)(j,) (e) = /(B)(e)

R4e+2 :VD(A)eB 7 q/01EDA')

R4e+3 :VBD(A) V /1(A'EDO)

The R4e module (that is the procedure associated with satisfying requirement R4e) uses the usual procedure for satisfying lowness requirements. At stage 0, we let




r(4e, 0) -1. At stage t + 1, we do nothing if t + 1 < 4e. Otherwise, we let

r(4e, t~l) 1 if OIA) (e)1

u(e, At, e, t) otherwise.

The R4e+l module operates similarly. The R4e+2 module operates as follows. At stage 0, we let r(4e + 2, 0) = -1 and

X2eO = (U, v, w) where u, v, w are the least elements of o0[2e] in increasing order. At stage t + 1, we proceed as follows. If t + 1 < 4e + 2, then we do nothing at stage t + 1. So, suppose that t + 1 > 4e + 2. If the R4e+2 module was canceled at stage t, then we let r(4e + 2,0) = -1 and X2et+I (u, v, w) where u < v < w are the least three elements of (W[2e] that have not yet been used by this module.

Suppose that the R4e+2 module was not canceled at stage t. Let u, v, w be such that X2et =(u, v, w). Let b = VD(A,)B, (X2et). The R4e+2 module does nothing

at stage t + 1 if b 74 q10t tAt)(X2et)* Suppose that b = qu(Dt )(X2e1)* We call t + 1 a stage of proper agreementfor R4e+2. Let m be the query to A' made by the

computation of ( t )(X2eAt). Let

r(4e +2, t +1) if(t Lu(m, At, m, t) otherwise.

Let z E {0, 1}2 be such that

(z(0) =1 X u E D(At)) A (T(1) =1 X a E Bt)

where

fv if u f D(At) w otherwise.

Case]. -=(0,0). Enumerate v into B1+I unless it is less than or equal to max{r(e', t) e' < 4e + 2}

in which case the R4e+2 module is canceled. If the R4,+2 module is not canceled, then it follows that Case 2 holds at the next stage of proper agreement. Case2. r=(0,1).

Choose the least z E co such that

(u, 2z) > max({r(e', t) le' < 4e + 2} U {r(4e + 2, t + 1)})

and enumerate (u, 2z) into A,+,. It follows that Case 3 holds at the next stage of proper agreement. Case 3. r=(1,0).

Let a be the path taken by the computation of ( t3A )(X2et). It follows that v(0) = 1. If v(1) 1, then enumerate w into B,+i unless it is less than or equal to max{r (e', t) Ie' < 4e + 2} in which case the R4e+2 module is canceled at stage t + 1. Suppose that a (1) = 0. Thus, r (4e + 2, t + 1) = -1. Since z = (1, 0), there exists a unique z E co such that (u, 2z) E A, and (u, 2z + 1) f A. Enumerate (u, 2z + 1) into A,+I unless it is less than or equal to r (e', t) for some e' < 4e + 2 in which case the R4e+2 module is canceled at stage t + 1. Suppose that the R4e+2 module is not canceled at stage t + 1. If a (1) = 1, then it follows that R4e+2 is satisfied at stage t + 1 unless a module for a higher priority requirement enumerates into A at some




future stage in which case Case 4 will hold at the next stage of proper agreement. If a (1) = 0, then it follows that Case 2 holds at the next stage of proper agreement and, unless this module is canceled at some future stage, either Case 3 or Case 4 will hold at all future stages of proper agreement.

Case 4.- z=(1, 1). Cancel the R4e+2 module. It follows by a standard finite injury argument that there are at most finitely many

stages of proper agreement. Therefore, R4e,+2 is eventually satisfied. The R4,+3 module operates as follows. At stage 0, we let r(4e + 3, 0) = -1 and

X2e+1O0 = (u, v, w) where u, v, w are the least three elements of (0[2e+l] in increasing order. At stage t + 1 we proceed as follows. If t + 1 < 4e + 3, then we do nothing at stage t + 1. So, suppose that t + 1 > 4e + 3. If the R4e+3 module was canceled at stage t, then r(4e + 3, t + 1) =df -1 andXe+l1t =df (u, v, w) where u < v < w are the least elements of (W[2e+l] that have not yet been used by this module.

Suppose that the R4e?3 module was not canceled at stage t. Let u, v, w be such

thatx9et = (u, v, w), and let b = VB D(At) (x?,..t). Do nothing if b =4 Vet t (X2e, t)

Suppose that b = ,(t t) (X2?e t). We call t + 1 a stage of proper agreementfJr R4,+2. Let z E {0, 1}2 be such that

(-r(0) =1 X u E Bt) A (-r(1) =1 X a E D(At))

where

{v ifucBt w otherwise.

Let m be the query to A' made by the computation of (A0t) (X26,t). Let

r(4e?+3,t?+ 1) ={u(mAm, t) otherwise.

Case 1. z= (0, 0). We pick the least z E co such that

(v, 2z) > max({r(e', t)le' < 4e + 3} U {r(4e + 3, t + 1)})

and enumerate (v, 2z) into At+,. It follows that Case 2 holds at the next stage of proper agreement.

Case 2: r =(0,1). It follows that there exists unique z E co such that (v, 2z) E A, and (v, 2z + 1) f

At. If (v, 2z + 1) > max({r(e', t)Ie' < 4e + 3} U {r(4e + 3, t + 1)}), then we enumerate (v, 2z + 1) into At+, in which case Case 1 holds at the next stage of proper agreement. Otherwise, we enumerate u into Bt+1 unless it is less than or equal to max{r(e', t)Ie' < 4e + 3} in which case the R4e+3 module is canceled at stage t + 1. If the R4e+3 module is not canceled, then Case 3 holds at the next stage of proper agreement.

Case 3: - = (1, 0). Choose the least z E co such that

(w, 2z) > max({r(e', t) le' < 4e + 3} U {r(4e + 3, t + 1)})




and enumerate (w, 2z) into A,+,. It follows that Case 4 holds at the next stage of proper agreement. Note that since Cases 1 and 2 have held previously m E A' and therefore r(4e + 3, t + 1) 74 -1.

Case 4. z = (1, 1). There exists unique z E co such that (w, 2z) E At and (w, 2z?+ 1) f A1t. Enumerate

(w, 2z - 1) into At+, unless it is less than or equal to max({r(e', t)Ie' < 4e + 3} U {r(4e + 3, t + 1)}) in which case the R4e+3 module is canceled at stage t + 1. We note that (w, 2z + 1) > r (4e + 3, t + 1). It follows that R4e+3 is satisfied at this point unless a module for a higher priority requirement enumerates into A at some future stage in which case Case 3 holds at the next stage of proper agreement. It follows by a standard finite injury argument that R4e+3 is eventually satisfied. -

THEOREM 5.5. There exist recursively enumerable A,B such that

Q (D (B), D (A)) g Q (A/, 0 ").

COROLLARY 5.6. There exists recursively enumerable A C co such that Q (0", A') 9 Q(A', 0").

PROOF OF THEOREM 5.5: We construct A, B so that VD(B)D(A) i Q(A', 0"). For each e E co we satisfy the requirement

R, : VD(B),D(X) 74 V/(A'ff(0"/)

Doing so may require infinitely many numbers to be enumerated into A. Thus, we use an infinite injury priority argument. Our notations and general set up are in the spirit of Chapter 14 of Soare [8]. A minor departure is that we let co[e] be {(e, x) I x e co} rather than the e-th column of wo. For all q E {0, 1, 2}1', let Re1 = Rlh(q).

At each stage, the R., module will either do nothing, or enumerate a number into exactly one of U c.[l] (O21 U x ['] 9(O[2+1] in order to change the status of some number's membership in D (A). An R., module may have to enumerate an infinite number of elements into one of these sets. At stage t, we form a binary sequence 6t E {O, I}t which reflects our best guess as to what the R,, modules for q C ct are going to do.

Let o0 0. Lett e co. For all q E {0, 1,2}1 ̀let

vt(q) = max({0} U {s < t 16, D q1})

We define 6,+1 to be the element of {0, 1 }t+1 such that for all q E {0, 1, 2}1w,

il (2) C 6t+I

- q C 6t+1 A (At+, - At) n Uwr[2x+i] I X E CON], i E {o, I}} 0.

andforalli e {0,1}

sa (i) C 6t+1 C 6t+1 A (At+, - A,) n (U cow[2x+(-i)] 74 0.

vY G (O [E1 ]

We say that t E co is an q stage if q C 6, Let z e co, and let q e {0, 1, 2}<0. Let i e {0, 1}, and let x,y e co be such that

z = (2x + i, y). We say that z is not believed by q at stage t + 1 if there exists q' such




that q1'A(i) C A, x = x*,t, and (2x 4 (1 - i), y') X A, for all y' > y. Let v E co, and consider the computation (A,)' (v). We say that this computation is i-correct at stage t + 1 if it diverges or if every z < u(v, A, v, t) is believed by q at stage t + 1.

Let z E co, and consider a computation of the form (A 0') (z). We say that this computation is q correct at stage t + 1 if it converges and 0(')(v) is q correct at

stage t where v is the query to A' made by the computation of qjje ') (y) Fix tE {O ,1, 2}1<'. The R,, module operates as follows. At stage 0, we let

X,1 o = min(co['"]) and r(t7, 0) -1. Let t E co. If the i-module was canceled at stage t, then we let x*,1t+l = min{z E oi'l] 1z > x~,}t }. Suppose that the R,, module was not canceled at stage t. If t + 1 is not an t7 stage and 6,+1 <AL }i, then the R, module is not canceled at stage t + 1. If t + 1 is not an hi-stage and 6,+1 >L q, then the R,, module does nothing at stage t + 1. Suppose that t + 1 is an q stage. Let

y (x,1 , 2x,,,,, 2x*1 , + 1). Let b = ,uE ' )(y). We call b the value of the q module

at stage t . If V4 fr' e)(y) does not converge I-correctly or converges to 1 - b, then the R,, module does nothing at stage t + 1. So, suppose that this computation convereges n-correctly to b. We call t + 1 a stage of proper agreement. Let v be the query to A' made by the computation of (AV 0' )(y). We let t

e~~~~~~~~~~~~tl

r(q~, t ? ) -iif$t()I ) u(v A, v, t) otherwise.

Let a be the path taken by the computation of v'l t (y). Let z C {O, 1}2 be such that

(z(0) = 1 X X,It E D(Bt)) A (z(1) =1 X a E D(At))

where

f2x,,t + I if x,,,t D(Bt)

2x,,t + 2 if x,,t E D(Bt).

Case 1 - - (0O 0). Thus, b = 0 and x,1jt X D(B,). We want to change the value of the Ru module

to 1. Let Z be the least element of co such that (2x,1 t, 2z) > max({r(q', t) 1 ' <

ui} U {r(, t + 1)}) and 2z > max{z' E coI(2x, t, z') E A,}. We then enumerate

(2x,,,t 27) into At+,. Case2. r=(0,1).

Thus b = 1 and x,,,t X D (B,). It follows that there exists unique z E co such that (2x,1 t, 2z) E A, and (2x,1 t, 2z + 1) X At. We want to change the value of the R,, module to O. One way to do this is to enumerate (2x,1 t, 2z+ 1) into A,+1. This may be either unwise or unnecessary according to the following subcases. Let p = (a (0), 0) . If que)t](y) 0 or if (2x,,, 2z + 1) < r(q, t + 1), then we enumerate (x* 1, 2z') into Bt+j where z' E co is the least number such that Nx,,W, 2z'), (x,,.t, 2z' + 1) X Bt. It follows that Case 3 holds at the next stage of proper agreement. If VIP] (y) 74 0 and (2x,,1t, 2Z + 1) > r(r, t + 1), then we enumerate (2x,,1t, 2z + 1) into At+, unless it is less than or equal to max{r(q', t) I a' < a} in which case we cancel the R,, module at stage t + 1.




Case 3. cz-(1, 0) Thus, b = 0 and xv,,t E D (Bt). It follows that there exists unique z' E co such that

(x,11 , 2z') E Bt and (x,,,t, 2z' + 1) , Bt. Let p (a(0), 0). If qfc4' (x,,,t) = 0, then we enumerate Kx,,,t, 2z' + 1) into B,+i and do nothing else at this stage. Otherwise, we choose the least z E co such that

(2x,1,, + 1, 2z) > max({r(j', t) jq' < q} U {r(j, t + 1)})

and 2z > max{z" E col(2x, t + 1, z") E A,}. We then enumerate (2x,,, + 1, 2z) into A,+,. It follows that Case 4 holds at the next stage of proper agreement.

Case 4: r=(1, 1) Thus, b = 1, and x,.t E D(B,). We note that Case 3 held at the previous

stage of proper agreement. It follows that there exists a unique z E co such that

(2x*,,, + 1, 2Z) E A, and (2x,1,t + 1, 2z + 1) , A,. It follows by a short argument that hinges on the notion of q correctness and Case 3 that (2x,1,, + 1, 2z + 1) > r (u, t + 1). We then enumerate (2x,1,,t + 1, 2z + 1) into A,+, unless it fails to be greater than max{r(j', t) I ' < q} in which case the R1 module is canceled at stage t + 1. If the R,, module is not canceled at stage t + 1, then it follows that Case 3 holds at the next stage of proper agreement.

Let f = lim inf,,CE, 6. That is, for all i E {0, 1, 2}1'

,/ C f X (3??t bt D /) A (3<'t t <LI').

Let q c f . Let e = lh (q). By induction, suppose that r,1i =df limto r(j', t) exists for all q' c q. It follows that x1 =df limtrno xlt exists. Let to be the stage at which this limit is reached. If there are only finitely many stages of proper agreement, then R,1 is satisfied and r. exists. We therefore assume that there are infinitely many stages of proper agreement. Let y =(X, 2x, , 2x, + 1). Let

v =2) (0' y)

2 It follows via a short argument that hinges on cases 3 and 4 and the notion of q correctness that r,1 exists. Let t1 > to be a stage of proper agreement such that r (, t1 ) = r,1 . Therefore, A', (v) = A'(v).

We claim that one of the following holds.

(1) The R, module eventually oscillates between cases 1 and 2. That is, at every sufficiently large stage of proper agreement t either Case 1 or Case 2 holds at t and if Case i holds at t then Case 3 - i holds at the next stage of proper agreement.

(2) The R,1 module eventually oscillates between Cases 2 and 3 . That is, at every sufficiently large stage of proper agreement t, either Case 2 or Case 3 holds at t and if Case i holds at t then Case 5 - i holds at the next stage of proper agreement.

(3) The R,1 module eventually oscillates between cases 3 and 4.

To see how to prove this, take a proper stage of agreement t2 > tj. Let a be the path

taken by the computation of 0e ). Let p = (a(0), 0). If Case 1 or Case 2 holds at stage t2, then (1) holds if [Ve](x,) 78 0; otherwise (2) holds. If Case 3 or 4 holds at stage t2, then (2) holds if if1 (x,1) = 0; otherwise (3) holds.




Now, if (1) holds, 2x,1 , D(A), x,1 f D(B), and therefore VD(B)D(A)(y) = 0. If

(2) holds, then x,1 , D(B), 2x,1 E D(A), and therefore VD(B)^D(A)(y) = 1. If (3) holds, then x,1 E D (B), 2x,, +1 D (A), and therefore VD (B) D (A) (y) = 0. Whether (1), (2), or (3) holds, p(0) = a(0), and so

/ q(v=2)(p I 1y)

- 1 V =df 2

is defined and is not in 0". If either (1) or (3) holds, V[PI (y) = 1 and /[(P)1)] (0 ) = O. Hence, since v' , 0", ( AE0/)(Y) = 1. If (2) holds, 1f[,](y) = 0 and [V(P(O).l)]()I 1. Therefore, since v' , 0", (A/E0D)(Y) = 0.

?6. Commutative oracles in the hyperarithmetic hierarchy. We use Kleene's system of notations, A. For all a E a we define Hr. in the usual way (see [7]). For a E let aI be the ordinal represented by a. Let a I....an E &. The fundamental result of this section is that (HAu ...., Ha,,r) commutes if and only if a ., a, are all notations for successor ordinals or exactly one of IaI 1, . . ., Ian I is a limit ordinal and this ordinal is the minimum of {IaI 1 . . . }, an 1}- We prove this by showing that these sets can always be put into least-first order and then classifying when they can be put into hardest first order. This classification will then be used to classify which sequences can be put into an arbitrary order. We will make use of logical closure properties which are defined as follows.

DEFINITION 6. 1. Let A, B C co. (1) We say that B is closed under conjunctions with A if A x B <m B. (2) We say that B is closed under disjunctions with A if B is closed under conjunc-

tions with A. (3) We say that B is closed under conjunctions (disjunctions) if B is closed under

conjunctions (disjunctions) with itself. (4) We say that B is closed under negations if B <?m B.

The proof of the following proposition is a technical exercise and is left to the reader.

PROPOSITION 6.2. (1) The jump of any set is closed under conjunctions and disjunctions.

(2) For all a E &, if la I is a limit ordinal, then Ha is closed under negations. (3) If B <T X, then X' is closed under conjunctions and disjunctions with B

and B.

We now define a few useful technical devices for later use.

DEFINITION6.3. Let r, . . .,rk E co, and let e be an (ri, .., rk) operator. Fix E {1,...,k} and j E {1,...,ri}. For all z E c and a E {0,1}I<" we let

node' (a, e, z) be the j-th prefix of a at which the i-th oracle is queried if there is one and otherwise node' (a, e, z) T. We define node'. to be like node' except that all computations involved are limited to s steps.

We regard node" as the j-th component function of node'. We note that node' J J and node'. are partial recursive functions. However, the domain of the latter function is recursive.



1742 TIMOTHY H. MCN1CHOLL

DEFINITION 6.4. Let rl,...,rk , e, i, j be as in definition 6.3. For all z E co' and a E {O, 1 }<'" we let answer' (a, e, z) be the answer given by a to the ]-th query to the i -th oracle if there is one and otherwise answer' (a, e, z) T. We define answer> 5 (a, e, z) by limiting all computations involved to at most s steps.

We regard answer' as the ]-th component function of answer'. We note that answer and answer'js are partial recursive functions and that the domain of the latter is recursive.

DEFINITION 6.5. Let rl,..., rk, e be as in definition 6.3. Let A1, . . ., Ak C co'. For all X C {1, . . ., k} and z E co we say that a E {O, 1}"' is consistent with

(A,,..., Ak) on (X, e, 7) if 4 [,](z) t and for all i E X and E E {1L. rij, answer' (a, e, z) = 1 X qe,(node'i (v, e, z), z) E Al .. A,.

The term consistent at stage s is defined similarly by limiting all computations to s steps.

DEFINITION 6.6. Let ri, .... rk, e be as in definition 6.3. Let p E {O, 1}"'. For all X C {1, . . ., k} and z E co) we say that a E {O, 1}<LO is consistent with p on (X, e, z) if qQ'](z) l, 4'P](z) A, and for all i E X and j e {1, ..., ri} a and p give the same answer to the j-th query to the i-th oracle. The term consistent at stage s with p on (X, e, z) is defined similarly by replacing answer with answer', and requiring the corresponding function values to be defined.

Thus, for all U C {1,., k}, q[U](AjE. Ad) (z) 1 if and only if

-r E {O. 1}n [e, (Z) t

A r is consistent with (A 1, . . ., Ak) on (U. e, z)

A r is consistent with path(e, n, z) on ({1, . . . U, e, ).

We will use the following Lemma.

LEMMA 6.7. Let e be an (ri, . . ., rk) operator. Let a ., ak E a be such that 1 j ail < ...< ?ak, and let j E {1,. k} be such that aj I is a limit ordinal.

Then q[V{ j I}.(H, ..

E@HAk) extends to an element of Q((rI + + r1)0', Ho; ). Ife is

semitotal, then this function extends to an element of Q((r1 + * + rj - 1)0', Hr,-,). If e is globally total, then this function extends to an element of QC(1, Ho;).

PRooF. Let n = rl + * * * + rk. Let h: co -* {O ,1 } be such that

h(z) = X <=~ 0

[ I .... I] (HE... j]Hkf)(z) 0.

[I j}]( ,I_ Hk Clearly, h is a total function extending . HK' ) The tricky part is to show that h E Q((ri + + rj)0', Haj). The pieces of this puzzle are as follows. There exists recursive f co --* co such that for all z E co

Wf( ) = {node' (c, e, -) r E {0, 1}" A q?j] (Z) t

A r is consistent with path(e, z) on ({ij + 1,., k}, e, z)}.

Lets rI + + rj. It is easy to show that #Wf() < 2"' for all z E cc'. If e is semitotal, then 1 < # W7 for all 7 E Co. It follows from Proposition 3.3 that there




exists g E FQ(s0') such that Dg(,) WIt () for all z E co. If e is semitotal, then we may take g to be an element of FQ((s - 1)0'). If e is globally total, then we may take g to be recursive.

It follows from Proposition 6.2 that there exists partial recursive Vi/ such that for all r, z

[(Vc E {1,...,j} nodeC(me,z) 1) # I q(-,z) t] A [(c is consistent with (Ha,..., Hjk) on ({1,.. * , e, z) X V/1(,Z) E Ha, )]

(To define V'i (r, z) wait for node' (r, e, z) to converge for all c E {1,.]..} and then use the reductions whose existence is guaranteed by Proposition 6.2 to form the appropriate query to Haj.) It also follows from Proposition 6.2 that there exists recursive V2 such that for all r, z

V/2 Qr, z) E 0' (qTf](z) 0

A r is consistent with path(e, z) on ({j + 1, . . ., k}, e, z)).

Note that nodec(c, e, n, z) t for all c E {1, . . ., j} if r D p for some p E Dg(). It follows that

h(z) = 0 = e {0, I}%np E Dg(z)(T D p A qI (r, z) E Haj A V/2(r, Z) E 0')

It follows from Proposition 6.2 that h is in the required class. -

We can now show that a sequence of the form (Ha, ..., Ha,, ) can always be put into least first order.

THEOREM 6.8. Let a1,..a E a be such that 1 < IaII < < ake. Let rl,..., rk > 1. Then, Q* (riHCa1 ..., rkHak) =Q(riH,..., rkHak). Furthermore if lay I is the maximum limit ordinal in { Ia1 l, . . , Iak|}, then

Q* (r Hal, , rkHa,) = Q(uO', HJa, rj+ aj+,,..., rkH(Ik)

where u = r, + + rj -1.

PROOF. If there are no limit ordinals in a 1. ak }, then this follows from Theorem 4.4. So, suppose that at least one of aI 1, . lak is a limit ordinal, and let j be the maximum element of { 1,...,k} such that IajI is a limit ordinal. Clearly, Q(u0', Haj, rj+lHa1j+,,..., rkHak) C Q*(rjHal,,..., rkH,,). Let C E Q*(riHa,,.., rkHa,) via 0,. Thus, /e is semitotal. Let U {1, . . ., j}. By

Lemma 6.7, 4U[](H H ...) extends to an element h of Q(u0', Haj). It then follows from the Oracle Replacement Lemma that C E Q(u0', Ha,, r?+l Hclj+,,..., rkHC,).

COROLLARY 6.9. Let a 1. an E a be such that 1 < a I <... < an 1. Then, for all o E Sn, Q(Ha(j),..., Hca(,,)) C Q(Hal,... Ha).

PROOF. By Theorem 6.8. 1

We now want to classify which sequences of H-sets can be put into hardest- first order without loss of computational power. We will need the following two Lemmas.




LEMMA6.10. LetA1,...,Ak Co. Let a I . , a,,,, g E & be such that

lalI....Ia,,. < jgl

and Ig is a limit ordinal. Then,

Q(A I,-, Ak . Ha, HHl) = Q(Aij... . Ak , Hg)

PROOF. Let n = k + m + 1. Let C E Q(Ai. . Ak, H,, H H(I..., H,,, ) via qc . Let (Xo,...,Xn.1) = (A1...AHHgHel...HH). For all z E w), let h(z) be the sequence of answers to the first k queries made by the computation of

e(,Xo. .E0x>)( Thus, h e FQ(A1. Ak). Furthermore, for all - E Co,

C(z) = 0 X 3a E {0, 1}y [qej](z) = A a D h(z)

A a is consistent with (X0, . . ., X,,) on ({k + 1 .... n}, e, z)}].

Let Z E co, and let a be a binary n-tuple that is prefixed by h (z). Note that

answer4l+(a,e, )7

Then, a is consistent with (X0, . . ., X,1) on ({k + 1, . .. , n}, e, I) if and only if

(answer k+I (a, e, z) =1 q,.( node k+l (a, e, z), z) E HE )

AEs E co [a is consistent at stage s

with (Xo... , X,11,) on ({k + 2 ..., n}, )].

We note that the relation

{(a, s, z) I a is consistent at stage s with (X0, . X,1_1) on ({k + 2, . n}. , )}

is recursive in Hf e3 ffl H1,. Let v E {a . a,,.} be such that

lvl = max{la, I. a,,,1

Then, H,,, Ha,,, -T H,. Since lgl is a limit ordinal and lgl > lvl, it follows that Hg is closed under conjunctions with (H,)'. It is also closed under negations. It follows that C E_ Q(A1, . . ., A, H). -

LEMMA6.11. Let dl,...,dk,g E & be such that lgl is a limit ordinal, and .dll,..., d/|Ijaresuccessorordinalsand gI < Idil < < ? dk l. Then, forallm > 1,

PROOF. It follows from Theorem 6.8 that

Q(mHg, Hd,, . ., Hdj ) C Q((m - 1)0', HI, H(,,,. .., H(/, ).

Let n = m + k. Let {fr )}eC,,, be an effective enumeration of all (m - 1, (I)k+1)

operators. There exists an (m - 1, (I)k+1) operator Oe such that qO(A)(x) 1 - F(A)(x) for all x E co and A C co. There exists a semitotal (m, (l)k+1) operator OCe, such that for all z E co

(0' Hg, D H 0 0

(0'Q E E H. HQ E... E H,,)() - . E E(0O Hg E HO ED ...O HdI)()




It follows from Lemma 6.7 and the Oracle Replacement Lemma that (0'efHgeHil f~I

Oe'

extends to an element C of Q(m0', Hs, H.1,..... H,1). It follows that

C 0 Q((M - I)O/ ]H~ I H(11 I .. Hal/ ,

We can now classify the sequences of H-sets that can be put into hardest-first order.

THEOREM 6.12. Let a,,..., an E &. Let b . b, be the elements of {a, I.a, } in an order such that lb, 1 > ... > Ibn 1. Then, Q(HUI, ..., H1,,) C Q(HbI,..., Hib,,) if and only if for all y, if y is the maximum limit ordinal in {aIa, I.... Ian j}, then ail > JajJ foralli,j E {1,...,n}suchthati<j and JajJ y.

PROOF. (<=): If there are no limit ordinals in aI 1, I I a I}, then this follows from Corollary 4.6. Suppose that there is a limit ordinal in {aI 1, . . . } Ian 1, and let y be the maximum such limit ordinal. It follows from Theorem 4.2 that there exists a rearrgangement, (Cl I ... c,1), of (al, . . . a,,) such that Q(HC,, ..., H1,,,) C Q(HC.I ..., H(.,,) and if k is the maximum element of {1, ... n} such that JCkJ Y, then for all i E co

k < i < n 4 ci < yn

andl < i < j < k IciI > Jc1.

It follows from Lemma 6.10 that

Q(HC.,..., Hc.,,) C Q(HCI, Hck)

Note that (Cl, C) prefixes (b1. b,). It follows from Proposition 2.3 that Q(Hc(I,.II IH6,,,,) CQ(Hl,v,..., HI,)j

(X): We prove the contrapositive. Thus, we assume that there exists a maximum limit ordinal y E aaI 1, . . . I1I} and that there exists i, j E {1, . . . n} such that i < ? and Iaj < y a 1X. Let m #{i' E {1, . . .4 n ai, y y. It follows from Theorem 4.2 that there exists a rearrangement of (al, a,,), (cl . c,) such that Q(HCj,l. .., Hc,,, D Q(Hcl, . .., H(.,,) and

i' < j'A lci,C|, 1cj,| > y ?> Jci,| < lcj,|.

for all i', j' E {1, . .. I n}. Let u be the maximum element of {1, . .n} such that

I ciI = y. Since JaiI < y and IajI = y, it follows that u > m + 1. Let k be the maximum element of { 1, . . . n} such that Ibk Y. Since u > m + 1, it follows from Proposition 2.3 and Proposition 2.2 that

Q(H(.,..., H(,,,) D Q(mI',Hbk I Hbk_, ,HOl.

On the other hand, by Theorem 6.10

Q(H,..., H, ) C Q(H, ..., Hbk).

By Theorem 6.8

Q(Hb,***, HbO) C Q(Hb, I Hbk- * bl)-

And by Lemma 6.11

Q(mO', Hbk. HO) Q(MHbk, Hbk. HO).




Thus, Q(Hal,..., Ha,,) g Q(Hh,.., Hh,,) A Theorem 6.12 allows us to classify the commutative sequences of H-sets.

THEOREM 6.13. Let a1. a,a E a be such that 1 < Ia1. a,1. Then, the following are equivalent.

(1) Q(HaiX . . . Xflu,,l) = Q(Ha,(1) . . ., Ha,,(,,)) for all a E S,1. (2) There is at most one limit ordinal in {a ., a,n}, and if y is this limit ordinal

then y = min{ lai, I., la, [ .

PROOF. If none of aI 1, . . .a, I } are limit ordinals, then this follows from The- orem 4.5. So, suppose that y is the maximum limit ordinal in { a, I, a, a|. If (2) fails, then it follows from Theorem 6.12 that there exists a CE S, such that Q (IIa,,, M I.. Ha(,, ) )7 Q (HbI, .. . , HI,,) where (b1,..., b,) is a rearrangement of

(a1,...,a,,) such that IbI > ...> b, 1. Thus, (1) implies (2). Suppose that (2) holds. Without loss of generality, suppose that aI < ... < Ian1. Let k be the maximum integer such that I ak I= y. Thus, I ak+ I 1, . . ., I a, I are successor ordinals, and la, I = a.= lak I= y. Let a E S,. It follows from repeated application of Theorem 4.2 that Q(Ha,, . . ., Ha,,,) C Q(Haa(l)I ... I Hc~aa(,,)). -

We note that not only did well-separated sequences of jumps commute, but their Q and QC classes coincided. In classifying the commutative sequence of H-sets, only Q classes have been used. This raises two questions. For which sequences of H-sets are the Q and QC classes identical? Also, which sequences of H-sets commute with respect to the QC classes? We answer the second question first. We first deal with the case where at least one limit ordinal is involved.

THEOREM 6.14. Let a1I....aI E a be such that 1 < a I < all and at least one of {aI,.. .a, Ia, I} is a limit ordinal. Let j be the maximum element of {1, . . .}, n such that laj I is a limit ordinal. Then, for all a E Sn,

QC(Hla(l), ..., H,,,(,,)) = QC(HJjI Hc,+, ... HC1,,).

PROOF. Let C E QC(H,(I) H{ ,Ha(,,)) via q, Thus, by is a globally total (ri,., rk)-operator. Let U = {i E {1,., k} a,,(i) E {al, a1}}. By Lemma

6.7, TIC awl) a'"' extends to an element of QC(IHaj). The conclusion then follows from the Oracle Replacement Lemma. A

COROLLARY 6.15. Let al,..., an E A. Then, for all a E S,, QC(Ht,..., Ht.,,) QC(HtUO (I) I . . Ht,(,l, )o

PROOF. If none of IaI 1, . an I are limit ordinals, then this follows from Theo- rem 4.5. Otherwise, this follows from Theorem 6.14. -A

We now classify the sequences of H-sets for which the Q and QC classes coincide. We will need the following preliminary result.

THEOREM6.16. Let a,bl,...,bk E abe such that 1 < IaI < IbII < . < bkI and aI is a limit ordinal and bI . bkI are successor ordinals. Then, for all r ., .... rk > 1,

Q(H,,, rlHb,,... rkHbk) = QC(HC,. r1Hl,,... rkHk).




PROOF. Let n = ri + + rk + 1. Let C E Q(Ha,,rlHb,,..., rkHbk) via ,e* Thus, 4e is a globally total (1, r1, . . ., rk)-operator. By Lemma 6.7, there exists

h E QC(H) which extends o[{1}](H Hb.Hb) The conclusion then follows from the Oracle Replacement Lemma. -1

We can now classify the sequence of H-sets for which the corresponding Q and QC classes coincide.

THEOREM 6.17. Let al,..., all E a be such that 1 < la,1,., Ian 1. Then,

if and only if the following two conditions hold.

(1) There is at most one j E {f 1,. n } such that IajI is the maximum limit ordinalin la, I ., IanI}.

(2) For all j E {1, . . ., n}, if Iaj I is the maximum limit ordinal in { al . a 1, then I a; I > I aj I for all i Ez{ 1, . . ., j - 1}.

PROOF. (z$): Suppose that either condition 1 or condition 2 fails. Let y be the maximum limit ordinal in { a I, . . . }, Ian 1. Let (cl, . . . , c]) be a rearrangement of (a,,...,an)suchthat I c < < Icn . Letkbethemaximumelementof{1,...,n} such that ICk I= y. Then, ICk I < ICk+1 l <? I Cn I and ICk+1 1, cIn are successor ordinals. By Lemma 6.11,

Q(0W, HCk I ... Hc, ) 9 Q(Hck. * *HC,,).

We complete the proof of this direction by showing that the following hold.

(a) Q(0', Hc,. II, HC,,) C Q(Hc,,., ,H . (b) Q (Hck I ... I HIn) D QC(Hcj,,..., rHcl,)

The proof of (a) is as follows. It follows from Theorem 4.2 that there is a rearrangement of (ai,...,an), (bl . bn), such that Q(Hal, 'Ia,) D Q(Hb1,...,Hh,) and

i < j A lbj , lbjI > y X lbiI < lbjI

for all i, j E { 1, n }. Furthermore, since either (1) or (2) fails, if i is the maximum element of {1, . . . ,n} such that IbiI = y, then i > 1. And, if bi > y 1, then j > i. It follows from Proposition 2.3 that

(0',Hb,,) D Q(o',HskH,..

Since i > 1, it follows from Proposition 2.3 that

Q (, Hb, ---Hb,,) C Q(Hb,,..., HI,,).

Thus, (a) holds. By Theorem 6.14, QC(Ht,,...., Ha,,,) =QC(H(.,..., H,, ) and (b) follows imme-

diately. (#): Suppose that condition 1 and condition 2 hold. If there is no limit ordinal

in { ail,..., }anI, then the conclusion follows from Corollary 4.6. So, suppose that there is a limit ordinal in aI1,..., Ian }. Let y be the maximum such limit ordinal. Hence, any ordinal in { I a1. . Iand } that is greater than y is a successor




ordinal. It follows from Theorem 4.2 that there exists a rearrangement of (a1, .., an)I

(b . b,,), such that Q(Hal, . . ., Ha,,,,) C Q(Hb,, . . ., Hh,), and

i< j A Jb, , 1bI > y z Jbj > ?b1I

for all i, e { 1, n}. Furthermore, if i E { 1, . . ., n } is such that y bi , then b . I,., Ibn I< y and Jbil > .> Jbi-q1 > Jb l. ByTheorem6.10,

Q(Hbl, . . ., Hh,,) C Q(Hh,. . ., Hb,)-

By Theorem 6.8, Q(Hbl,..., Hb.) C Q(Hb .... Hb, ). By Theorem 6.16,

Q(Hb. ....Hb) -- QC(Hb. , Hb,).

But, by Theorem 6.14, QC(Hj .... Hb,) = QC(HaI,..., Ha,). A

?7. Acknowledgements. I thank William Gasarch for proofreading this paper. I also thank Richard Beigel for suggesting logical closure properties to me.

REFERENCES

[1] R. BEIGEL, personal communication. [2] R. BEIGEL and R. CHANG, Commutative queries, Information and Computation, (to appear). [3] R. BEIGEL, W GASARCH, M. KUMMER, G. MARTIN, T. MCNICHOLL, and F STEPHAN, On the query

complexity of classes, Proceedings of the 21st International Symposium, Mathematical Foundations of Computer Science, Lecture Notes in Computer Science, vol. 11 13, 1996, pp. 206-217.

[4] W GASARCH and G. MARTIN, Bounded queries in recursion theory, Springer Verlag, New York, 1998.

[5] E. HEMASPAANDRA, L. HEMASPAANDRA, and H. HEMPEL, Query order in the polynomial hierarchy, Proceedings of the International Symposium on Fundamentals of Computation Theory, Lecture Notes in Computer Science, vol. 1279, Springer-Verlag, 1997, pp. 222-232.

[6] L. HEMASPAANDRA, H. HEMPEL, and G. WESCHUNG, Query order, SIAM Journal on Computing, (to appear).

[7] G.E. SACKS, Higher recursion theory, 1st ed., Springer-Verlag, Heidelberg, 1990. [8] R.I. SOARE, Recursively enumerable sets and degrees, 1st ed., Springer-Verlag, Berlin, Heidelberg,

1987.

DEPARTMENT OF MATHEMATICS

UNIVERSITY OF DALLAS

IRVING, TEXAS 75062, USA

E-mail: tmcnichogacad.udallas.edu



On the Commutativity of Jumps

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