On the Commutativity of Jumps

Timothy H. McNicholl

June 12, 2014

To appear in The Journal of Symbolic Logic

Abstract

We study the following classes:

• Q∗(r1A1, . . . , rkAk) which is defined to be the collection of allsets that can be computed by a Turing machine that on anyinput makes a total of ri queries to Ai for all i ∈ {1, . . . , k}.

• Q(r1A1, . . . , rkAk) which is defined like Q∗(r1A1, . . . , rkAk) ex-cept that queries to Ai must be made before queries to Ai+1 forall i ∈ {1, . . . , k − 1}.

• QC(r1A1, . . . , rkAk) which is defined like Q(r1A1, . . . , rkAk) ex-cept that the Turing machine must halt even if given incorrectanswers to some of its queries.

We show that if A1, . . . , Ak are jumps that are not too close together,then all three of these classes are identical and are not changed if wepermute (r1A1, . . . , rkAk). This extends a result of Beigel’s [?]. Sincethe second class is not affected by permutations, we say that these setscommute with each other. We also show that jumps that are too closetogether may not commute. We also characterize the commutativesequences of sets obtained by iterating the jump operation through anordinal notation.

1 Introduction

We address the following kind of question: Is there any advantage to query-ing ‘hard’ oracles before ‘easy’ oracles? We can make this more precise as

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follows. Let Q(A,B) be the class of all sets of natural numbers that canbe computed by a Turing machine that on any input first makes a query toA and then makes a query to B. If A is ‘easy’ with respect to B in somesense (e.g. A ≤T B), and if Q(A,B) and Q(B,A) are comparable, then isQ(A,B) a larger or smaller class than Q(B,A)? We define Q(A1, . . . , An) ina similar manner and ask similar questions about how this class is affectedby permuting (A1, . . . , An). Questions of this type were first considered inthe context of complexity theory by L. Hemaspaandra, H. Hempel, and G.Wechsung [?]. Here, we focus on a recursion-theoretic setting.

We show that if A1, . . . , An are jumps that are not too close together (ina sense to be spelled out later), then Q(A1, . . . , An) = Q(Aσ(1), . . . , Aσ(n))for all σ ∈ Sn. For this reason, we say that such a sequence of sets com-mutes. This extends an (unpublished) result due to R. Beigel that if forsome a, b ≥ 1 Ai ∈ {∅(a), ∅(b)} for all i ∈ {1, . . . , n}, then (A1, . . . , An) com-mutes. The methods used here however, are very different. Beigel’s prooffirst arranges (A1, . . . , An) into hardest first order without loss of computa-tional power. That is, if (B1, . . . , Bn) is a re-arrangement of (A1, . . . , An)such that B1 ≥T . . . ≥T Bn, then Q(A1, . . . , An) ⊆ Q(B1, . . . , Bn). Thehardest first ordering is then arranged into the least first order without lossof computational power, and then the least first ordering is arranged backinto the initial arbitrary ordering without loss of computational power. Theproof here of the commutativity of jumps that are not too close together firstarranges the arbitrary ordering into the least first ordering without loss ofcomputational power, and then arranges the least first ordering back into thearbitrary ordering without loss of computational power.

The chief advantage here is not that we can accomplish things with onefewer re-ordering. Rather, the main advantage appears when we move intothe realm of hyperarithmetical sets. Namely, we directly show that an arbi-trary ordering of sets of the form Ha for some a ∈ O can be put into leastfirst order without a loss of computational power. However, we also showthat these sets can not always be put into hardest first order without a lossof computational power. Since Beigel’s proof that a sequence composed of∅(a), ∅(b) can be put into least first order requires the sequence to first be putinto hardest first order, it does not extend to the hyperarithmetical hierarchy.

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2 Notations, definitions, and preliminary Propo-

sitions

Unless otherwise mentioned, all notations are as in Soare [?].Let≤lex be the lexicographic ordering of {0, 1}<ω. We say that σ ∈ {0, 1}n

is even (odd) if an even (odd) number of elements of {0, 1}n lexicographicallyprecede σ. If ∅ 6= S ⊆ {0, 1}<ω, then lexmin(S) is the lexicographicalminimum element of S.

If A is a set, then #A is the cardinality of A. We identify subsets of ωwith their characteristic functions. Thus, for all A ⊆ ω and x ∈ ω,

A(x) =

{1 if x ∈ A0 if x /∈ A.

We now wish to make a few remarks on terminology regarding Turingmachines. We let {φe}e∈ω be an effective enumeration of all Turing machines.Then, φe(x) is the output, if any, when Turing machine φe is run on input x.

φ(A)e (x) is the output, if any, when Turing machine φe is run on input x and

oracle A is used. That is, when the characteristic function of A is written onthe tape before the computation begins. Since we wish to consider situationswhere serveral oracles are queried, it might seem that we need to considernew classes of Turing machines where, say, more than one tape is used andthe characteristic function of each oracle to be queried is written down onone of the tapes. However, this is not necessary. For, if we wish k oraclesA1, . . . , Ak to be queried, we may write down the characteristic function oftheir join on the tape before the computation begins. That is, the singleoracle A1 ⊕ · · · ⊕ Ak may be used. Here, we define the join of k sets by

A1 ⊕ · · · ⊕ Ak = ∪j<k{ka+ j | a ∈ Aj}.

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

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

Definition 2.1 Let A1, . . . , Ak ⊆ ω, and let r1, . . . , rk ≥ 1.

1. We define FQ∗(r1A1, . . . , rkAk) to be the set of all functions Turingreducible to A1 ⊕ · · · ⊕ Ak via a Turing machine that on any inputmakes a total of ri queries to Ai.

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2. We define FQ(r1A1, . . . , rkAk) to be the set of all functions that arein FQ∗(r1A1, . . . , rkAk) via a Turing machine that on any input makesits queries to Ai before its queries to Ai+1 for all i < k.

3. We define FQC(r1A1, . . . , rkAk) to be the set of all functions that arein FQ(r1A1, . . . , rkAk) via a Turing machine that on all inputs haltson all possible computation paths.

We defineQ∗(r1A1, . . . , rkAk) to be the collection of all sets in FQ∗(r1A1, . . . , rkAk).We define Q(r1A1, . . . , rkAk) and QC(r1A1, . . . , rkAk) similarly. For all Φ ∈{Q,Q∗, QC, FQ, FQ∗, FQC} we let Φ(A1, . . . , Ak) be Φ(1A1, . . . , 1Ak).

The proofs of the following two propositions are left to the reader.

Proposition 2.2 Let r1, . . . , rk ≥ 1, and let A1, . . . , Ak, B1, . . . , Bk ⊆ ω.

1. FQC(r1A1, . . . , rkAk) ⊆ FQ(r1A1, . . . , rkAk) ⊆ FQ∗(r1A1, . . . , rkAk).

Let Φ ∈ {Q,QC,Q∗, FQ, FQC, FQ∗}.

2. If Ai ≤m Bi for all i ∈ {1, . . . , k}, then Φ(r1A1, . . . , rkAk) ⊆ Φ(r1B1, . . . , rkBk).

3. If Ai ≡m Bi for all i ∈ {1, . . . , k}, then Φ(r1A1, . . . , rkAk) = Φ(r1B1, . . . , rkBk).

Proposition 2.3 Let A1, . . . , Ak, B1, . . . , Bn ⊆ ω. Suppose that there existsan order-preserving injection f : {1, . . . , k} → {1, . . . , n} such that Bf(i) ≥mAi for all i ∈ {1, . . . , k}. Then, Φ(A1, . . . , Ak) ⊆ Φ(B1, . . . , Bn) for all Φ ∈{Q,FQ,QC, FQC}.

Here is some aditional terminology regarding Turing machines. If σ ∈{0, 1}<ω, then φ

(σ)e (x) is the output, if any, when Turing machine φe is run

on input x and σ is written on the tape before the start of the computation.Such a computation may diverge by virtue of the fact that σ may not containanswers to all of the queries made. Here is a variation on this idea. Let φ

[σ]e (x)

be the output, if any, when Turing machine φe is run on input x and for eachi < lh(σ), σ(i) is used to answer the i-th query made by this computation (ifthere is one). If lh(σ) = n, then such a computation may diverge by virtueof the fact that it may attempt to make an n + 1-st query. In this case, wedefine qe(σ, x) to be this (n + 1)-st query. This defines a partial functionqe from {0, 1}<ω × ω into ω. Now, when k is fixed and j < k, we can say

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that qe(σ, x) is a query to the j-th oracle (without knowing which oraclesare being used) when it is defined and congruent to j modulo k. We define

φ[σ]e,s, qe,s, etc. by limiting the number of computation steps to s. We define

the computation path taken by φ(A)e (z) to be the σ ∈ {0, 1}<ω such that for

all ρ ⊂ σ, qe(ρ, z) ↓ and

qe(ρ, z) ∈ A ⇔ σ(lh(ρ)) = 1.

Clearly, a function f is in FQ∗(r1A1, . . . , rkAk) if and only if there is

a Turing machine φe such that f = φ(A1⊕···⊕Ak)e and for all x ∈ ω and all

σ ∈ {0, 1}<ω for which φ[σ]e (x) ↓, the number of prefixes of σ at which the

j-th oracle is queried is rj for each j ∈ {1, . . . , k}. We call a Turing machinehaving the latter property an (r1, . . . , rk)-operator.

3 A mind change lemma

The study of commutative oracles lies in the field of bounded queries whichis the study of classes of the form Q(n,A). See, for example [?] or [?].Many arguments in bounded queries are built around the observation thatthe limit of a computable binary sequence with at most 2n− 1 mind changescan be computed with n queries to the halting set. More formally, if f isa function which is defined as the pointwise limit of a uniformly recursivesequence of binary functions {fs}s∈ω, then f ∈ Q(n, ∅′) if for all s ∈ ω,#{x ∈ ω|fs(x) 6= fs(x + 1)} ≤ 2n − 1. This observation does not generalizeto sequences uniformly recursive in some oracle A. In later sections, we willneed 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 whatwe mean by a mind change in a partial binary sequence.

Definition 3.1 Let f be a partial function from ω into {0, 1}. The numberof mind changes in f is defined to be the cardinality of

{t ∈ dom(f)|∃t′ ∈ dom(f) [t′ > t ∧ f(t) 6= f(t′)∧

∀s ∈ dom(f)(t < s < t′ ⇒ f(s) = f(t))]}.

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Rather than look at functions defined as pointwise limits of sequencesof partial functions, it will simplify our formalism somewhat to look atthe equivalent situation of partial functions defined by limits of the formlimy→∞ φ(x, y) where φ is a partial function of two variables.

Definition 3.2

1. A partial (total) array is a partial (total) function from ω×ω into {0, 1}.

Let φ be a partial array.

2. For each z ∈ ω, the z-th row of φ is the partial function f given by

f(s) = φ(z, s).

3. Let z ∈ ω. If φ(z, t) ↓ for at least one t ∈ ω, then the initial number inrow z of φ is

φ(z,min{t ∈ ω|φ(z, t) ↓}).

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

Proposition 3.3 Let f : ω → ω be recursive, and let A ⊆ ω. Let k, k′ ∈ ω,and let θ = dlog2(k

′ − k + 1)e.

1. If k ≤ WAf(z) ≤ k′ for all z ∈ ω, then there exists h ∈ FQC(θA′) such

that h(z) = #WAf(z) for all z ∈ ω.

2. If k ≤ Wf(z) ≤ k′ for all z ∈ ω, then there exists g ∈ FQ(θ∅′) such thatDg(z) = Wf(z) for all z ∈ ω.

Proof sketch: The proof of (??) is by binary search. To prove (??), letz ∈ ω be given. We first compute h(z). We then wait for t ∈ ω such that#Wf(z),t = h(z). We then compute a cannonical index of Wf(z),t.

Lemma 3.4 Let f be a total array. Let (i, a) ∈ {0, 1} × ω be such that theinitial value of each row of f is i and the number of mind changes in anyrow of f is at most a. Then, the set f̃ defined by the equation

f̃(s) = limt→∞

f(s, t)

is in QC(dlog2(a+ 1)e, graph(f)′).

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Proof: Let A = graph(f). The key fact is that there exists recursiveg : ω → ω such that #WA

g(z) is the number of mind changes in row z of f forall z ∈ ω. One way to show this is to define a recursive operator Ψ such thatfor all z, y ∈ ω,

Ψ(X)(z, y) ↓

⇔ ∃y′ ∈ ω[y′ > y ∧ f(z, y) 6= f(z, y′) ∧ ∀x ∈ ω (y ≤ x < y′ ⇒ f(z, y) = f(z, x))].

The function g can then be obtained from the snm Theorem [?]. It followsfrom Proposition ?? that there exists h ∈ FQC(dlog2(a+ 1)e, A′) such thath(s) is the number of mind changes in row s of f for all s ∈ ω. It followsthat for all s ∈ ω

f̃(s) =

{i if h(s) is even

1− i otherwise.

It then follows that f̃ ∈ QC(dlog2(a+ 1)e, graph(f)′).

Definition 3.5 Let φ be a partial array.

1. A relative position function for φ is a map R : ω → {0, 1}×ω such thatfor all s ∈ ω the number of mind changes in row s of φ is no larger thanR2(s) and if equal to R2(s) then R1(s) is the initial number in row sof φ.

2. Let R be a relative position function for φ, and let M ∈ ω. We saythat R is bounded by M if R2(s) ≤M for all s ∈ ω.

Thus, for each z ∈ ω, a relative position function for φ specifies a guess atthe initial number in the z-th row of φ and an upper bound on the numberof mind changes in the z-th row of φ and gets this guess incorrect only if thisupper bound fails to be tight.

Lemma 3.6 (The Mind Change Lemma) Let n ∈ ω. Let φ be a partialarray with relative position function R that is bounded by 2n−1. Then, thereexists ψ ∈ FQC(n, graph(φ)′) such that for all s ∈ ω,

limt→∞

φ(s, t) exists⇒ limt→∞

φ(s, t) = ψ(s, R(s)).

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Proof: For each i ∈ {0, 1}, let fi be the total array defined as follows.Let fi(s, 0) = i for all s ∈ ω. Let t ∈ ω. If the number of mind changes in(fi(s, 0), . . . , fi(s, t)) is less than 2n − 1, then

fi(s, t+ 1) =df

{1− fi(s, t) if fi(s, t) 6= φ(s, t) ↓fi(s, t) otherwise.

If the number of mind changes in (fi(s, 0), . . . , fi(s, t)) is not less than 2n−1,then fi(s, t+ 1) =df fi(s, t).

We note that fi ≤T graph(φ). Hence, graph(fi)′ ≤m graph(φ)′. Further-

more, the number of mind changes in any row of fi is at most 2n − 1.For each s ∈ ω and (i, a) ∈ {0, 1} × ω, let ψ(s, (i, a)) =df limt→∞ fi(s, t).

It follows from Lemma ?? that ψ ∈ FQC(n, graph(φ)′). Let s ∈ ω. Supposethat limt→∞ φ(s, t) exists. It follows that if the number of mind changes inrow s of φ is less than 2n−1, then limt→∞ φ(s, t) = ψ(s, R(s)). Suppose thatthe number of mind changes in row s of φ is exactly 2n − 1. Then, R1(s) isthe initial number in row s of φ. It follows that

limt→∞

φ(s, t) = limt→∞

fR1(s)(s, t) = ψ(s, R(s)).

4 Commutativity of jumps

Definition 4.1 A sequence of jumps (A′1, . . . , A′k) is said to be well separated

if A′i ≤T Ai+1 for all 1 ≤ i < k.

The main result of this section is that any well-separated sequence ofjumps commutes. To begin the work of this section, we give a theorem thatallows us to move ‘harder’ oracles ahead of ‘easier’ oracles. We then show thatwhen our oracles are a well-separated sequence of jumps, we can rearrangethem into least-first order without a loss of computational power. We notethat the first result is considerably easier to obtain than the second.

Theorem 4.2 Let A,B ⊆ ω be such that B ≤T A. Then, for all Φ ∈{FQ,FQC}

Φ(A1, . . . , Ai, B,A′, Ai+1, . . . , Ak) ⊆ Φ(A1, . . . , Ai, A

′, B,Ai+1, . . . , Ak)

for all A1, . . . , Ak ⊆ ω and i ∈ {1, . . . , k − 1}.

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Proof Sketch: Let f ∈ Φ(A1, . . . , Ai, B,A′, Ai+1, . . . , Ak) via φe. Fix z ∈ ω.

Let σ ∈ {0, 1}k+2 be the computation path taken by φ(A1⊕···⊕Ai⊕B⊕A′⊕Ai+1...⊕Ak)e (z).

The first k bits of σ can be computed using one query to each of A1, . . . , Akin succession. Once these bits are known, we can determine the answer tothe A′ query without knowing the answer to the B query. For, the answerto the A′ query is 1 if and only if

∃s ∈ ω [(qe(σ, z) ∈ A1 ⊕ · · · ⊕ Ai ⊕B ⊕ A′ ⊕ Ai+1 . . .⊕ Ak

∧ qe,s(σ∧(1), z) ∈ A1 ⊕ · · · ⊕ Ai ⊕B ⊕ A′ ⊕ Ai+1 . . .⊕ Ak)

∨ (qe(σ, z) 6∈ A1 ⊕ · · · ⊕ Ai ⊕B ⊕ A′ ⊕ Ai+1 . . .⊕ Ak∧ qe,s(σ∧(0), z) ∈ A1 ⊕ · · · ⊕ Ai ⊕B ⊕ A′ ⊕ Ai+1 . . .⊕ Ak)].

Since B ≤T A, it follows that the matrix of the sentence above is A-recursiveand so the truth of the sentence can be obtained with a single query to A′.The remaining bits of σ can then be computed using one query to each ofB,Ai+1, . . . , Ak in succession.

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

First, fix an (r1, . . . , rk) operator e. Let n = r1 + · · ·+ rk. For all z ∈ ω,let

path(e, z) = lexmin{σ ∈ {0, 1}n | ∃t ∈ ω φ[σ]e,t(z) ↓ ∧ ∀s < t ∀τ ∈ {0, 1}n φ[τ ]

e,s(z) ↑}.

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

where φ[σ]e (z) ↑ for all σ. We therefore define an (r1, . . . , rk) operator to be

semitotal if for every z ∈ ω there is at least one σ for which φ[σ]e (z) ↓.

Let U ⊆ {1, . . . , k}. We define a Turing machine φ[U ]e . For input z and

oracles A1, . . . , Ak the computation of φ[U ](A1⊕···⊕Ak)e (z) proceeds as follows.

Run the computation of φ(A1⊕···⊕Ak)e (z). For i ∈ U , answer queries to the i-th

oracle with Ai, and for i 6∈ U and j < ri answer the (j + 1)st query to thei-th oracle with path(e, z)’s answer to the (j + 1)st query to the i-th oracle.

Thus, φ[U ](A1⊕···⊕Ak)e is partial recursive in ⊕i∈UAi. In particular, if U = ∅,

then φ[U ](A1⊕···⊕Ak)e is partial recursive. If U = ∅ and φe is semitotal, then

φ[U ](A1⊕···⊕Ak)e is recursive.

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We now state and prove a Lemma whose main application will be tocommutativity in the hyperarithmetical hierarchy but which will make shortwork of the commutativity of well-separated sequences of jumps. In thestatement of the following Lemma, and throughout most of this paper, weregard sets as total 0-1 valued functions.

Lemma 4.3 (The Oracle Replacement Lemma) Let s1, . . . , sc, r1, . . . , rk ≥1, and let e be a semitotal (s1, . . . , sc, r1, . . . , rk) operator. Suppose that(A′1, . . . , A

′k) is well-separated and thatB1, . . . , Bc ≤T A1. Let U = {1, . . . , c}.

1. For all C1, . . . , Cd ⊆ ω, if φ[U ](B1⊕···⊕Bc⊕A′1⊕···⊕A′k)e can be extended to an

element of Q(C1, . . . , Cd), then φ(B1⊕···⊕Bc⊕A′1⊕···⊕A′k)e can be extended

to an element of Q(C1, . . . , Cd, r1A′1, . . . , rkA

′k).

2. For all C1, . . . , Cd ⊆ ω, if φ[U ](B1⊕···⊕Bc⊕A′1⊕···⊕A′k)e can be extended to an

element of QC(C1, . . . Cd), then φ(B1⊕···⊕Bc⊕A′1⊕···⊕A′k)e can be extended

to an element of QC(C1, . . . , Cd, r1A′1, . . . , rkA

′k).

Proof: We prove part 1. The proof of part 2 is almost identical. Let n =r1 + · · ·+ rk. For all i ∈ {1, . . . , k}, let {A′i,s}s∈ω be an Ai-recursive enumera-

tion of A′i. Let h ∈ Q(C1, . . . , Cd) be an extension of φ[U ](B1⊕···⊕Bc⊕A′1⊕···⊕A′k)e .

For all y ∈ {0, . . . , k} and z ∈ ω, let σy(z) ∈ {0, 1}n be the compu-

tation path taken by φ[{1,...,c+y}](B1⊕···⊕Bc⊕A′1⊕···⊕A′k)e (z) if this computation

converges, and let σy(z) be undefined if this computation does not con-verge. Let b0(z) = h(z). For all y ∈ {1, . . . , k} we define an element byof Q(C1, . . . , Cd, r1A

′1, . . . , ryA

′y) so that for all z ∈ ω

φ[σy(z)]e (z) ↓⇒ by(z) = φ[σy(z)]

e (z).

(We will arrange things so that if h ∈ QC(C1, . . . , Cd), then by ∈ QC(C1, . . . , Cd, r1A′1, . . . , ryA

′y).)

It follows that bk is the required extension of φ(B1⊕···⊕Bc⊕A′1⊕···⊕A′k)e .

Let y ∈ {0, . . . , k − 1}, and suppose that by has been defined and has allrequired properties. To define by+1 we proceed as follows. We will define apartial array φ and later a relative position function for it. This will enableus to use Lemma ??. For all z, s ∈ ω let ρ(z, s) ∈ {0, 1}n be the compuationpath taken by

φ[{1,...,c+y+1}](B1⊕···⊕Bc⊕A′1⊕···⊕A′y⊕A′y+1,s⊕A′y+2⊕···⊕A′k)e,s (z)

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

φ[ρ(z,s)]e,s (z) for all z, s ∈ ω. We note that for all z ∈ ω, σy+1(z) ↓ if and

only if lims→∞ ρ(z, s) exists in which case the two are equal. Hence, for

all z ∈ ω, φ[σy+1(z)]e (z) ↓ if and only if lims→∞ φ(z, s) exists in which case

the two are equal. Since B1, . . . , Bc, A′1, . . . , A

′y ≤ Ay+1, it follows that

φ, ρ are partial recursive in Ay+1. However, since the number of steps inthe computation of φ(z, s), ρ(z, s) is recursively bounded, it follows thatgraph(ρ), graph(φ) ≤T Ay+1. We define a relative position function R for φsuch thatR ∈ FQ(C1, . . . Cd, r1A

′1, . . . , ryA

′y). (We will arrange things so that

if h ∈ FQC(C1, . . . , Cd), then R ∈ FQC(C1, . . . , Cd, r1A′1, . . . , rkA

′k).) It will

turn out that R is bounded by 2ry+1−1. Let ψ be as given by the mind changeLemma. Then, ψ ∈ FQC(ry+1, A

′y+1). We can then let by+1(z) = ψ(z,R(z))

for all z ∈ ω. It follows that by+1 ∈ Q(C1, . . . , Cd, A′1, . . . , A

′y+1).

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

φ(z, s) ↓⇔ ρ(z, s) ↓ .

Let z, s ∈ ω be such that φ(z, s) ↓, φ(z, s′) ↓. Then, since {A′y+1,t}t∈ω is anAy+1-recursive enumeration of A′y+1, it follows that

s < s′ ⇒ ρ(z, s) ≤lex ρ(z, s′).

In fact, if we let θ(z, t) ∈ {0, 1}ry+1 be the sequence of answers given by ρ(z, t)to queries to the (y + 1)st oracle whenever ρ(z, t) ↓, then

s < s′ ⇒ θ(z, s) ≤lex θ(z, s′),

andρ(z, s′) 6= ρ(z, s)⇔ θ(z, s) 6= θ(z, s′).

It follows that the number of mind changes in any row of φ is at most 2ry+1−1.Therefore, we let R2(z) = 2ry+1 − 1 for all z ∈ ω. Let γ ∈ {0, 1}ry+1 be thesequence of path(e, z)’s answers to the (y + 1)st oracle. We then let

R1(z) =

{by(z) if γ is an even element of {0, 1}ry+1

1− by(z) otherwise.

It follows that R ∈ FQ(C1, . . . , Cd, r1A′1, . . . , ryA

′y). Let z ∈ ω and suppose

that there exist 2ry+1 − 1 mind changes in row z of φ. We show that R1(z)

11

is the initial number in the z-th row of φ. Let u = 2ry+1 − 1. It follows thatthere exist s0 < · · · < su such that for all i ∈ {0, . . . , u}, φ(z, si) ↓ and

i < u⇒ φ(z, si) 6= φ(z, si+1).

It follows that φ(z, s0) is the initial number in row z of φ and for all i ≤ u,

φ(z, s0) =

{φ(z, si) if i is even

1− φ(z, si) if i is odd.(1)

Furthermore,{θ(z, si) | i ∈ {0, . . . , u}} = {0, 1}ry+1 ,

and for all i ≤ u θ(z, si) is an even element of {0, 1}ry+1 if and only if i is even.There exists j ≤ u such that θ(z, sj) = γ. It follows that ρ(z, sj) = σy(z).Hence, φ(z, sj) = by(z). By making the appropriate substitutions in (??), wesee that

φ(z, so) =

{by(z) if γ is an even element of {0, 1}ry+1

1− by(z) otherwise.

Thus, R1(z) is the initial number in row z of φ.

Theorem 4.4 Let (A′1, . . . , A′k) be well-separated. Then, Q∗(r1A

′1, . . . , rkA

′k) =

QC(r1A′1, . . . , rkA

′k).

Proof: Let C ∈ Q∗(r1A′1 . . . rkA′k) via φe. Let U = ∅, d = 1, and C1 = ∅.Then, QC(1, C1) is the set of all recursive sets, and φ

[U ](A′1⊕···⊕A′k)e is easily

seen to be recursive. It follows that C ∈ QC(C1, r1A′1, ..., rkA

′k). But clearly

QC(C1, r1A′1, ..., rkA

′k) = QC(r1A

′1, ..., rkA

′k). Thus, Q∗(r1A1, . . . , rkAk) ⊆

QC(r1A′1, . . . , rkA

′k). But, by Proposition ??, QC(r1A

′1, . . . , rkA

′k)⊆Q∗(r1A′1, . . . , rkA′k).

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

Theorem 4.5 Let (A′1, . . . , A′k) be a well-separated sequence of jumps. Let

r1, . . . , rk ≥ 1. Then, for all σ ∈ Sk,

Q∗(rσ(1)A′σ(1), . . . , rσ(k)A

′σ(k))

= Q(rσ(1)A′σ(1), . . . , r

′σ(k)A

′σ(k))

= QC(rσ(1)A′σ(1), . . . , r

′σ(k)A

′σ(k)).

12

Proof: Let σ ∈ Sk. Clearly,

Q∗(rσ(1)A′σ(1), . . . , rσ(k)A

′σ(k)) = Q∗(r1A

′1, . . . , rkA

′k).

By Theorem ??,

Q∗(r1A′1, . . . , rkA

′k) ⊆ QC(r1A

′1, . . . , rkA

′k).

It follows from repeated applications of Theorem ?? that

QC(r1A′1, . . . , rkA

′k) ⊆ QC(rσ(1)A

′σ(1), . . . , rσ(k)A

′σ(k)).

But clearly

QC(rσ(1)A′σ(1), . . . , rσ(k)A

′σ(k)) ⊆ Q∗(r1A

′1, . . . , rkA

′k).

Thus,QC(r1A

′1, . . . , rkA

′k)

= QC(rσ(1)A′σ(1), . . . , rσ(k)A

′σ(k))

= Q∗(r1A′1, . . . , rkA

′k).

Now, since QC(r1A′1, . . . , rkA

′k) ⊆ Q(r1A

′1, . . . , rkA

′k), it follows that

Q∗(r1A′1, . . . , rkA

′k) ⊆ Q(r1A

′1, . . . , rkA

′k).

It follows from repeated applications of Theorem ?? that

Q(r1A′1, . . . , rkA

′k) ⊆ Q(rσ(1)A

′σ(1), . . . , rσ(k)A

′σ(k)).

But clearly

Q(rσ(1)A′σ(1), . . . , rσ(k)A

′σ(k)) ⊆ Q∗(r1A

′1, . . . , rkA

′k).

Thus,Q(r1A

′1, . . . , rkA

′k)

= Q(rσ(1)A′σ(1), . . . , rσ(k)A

′σ(k))

= Q∗(r1A′1, . . . , rkA

′k).

13

Corollary 4.6 Let A1, . . . , An ⊆ ω be such that for all i ∈ {1, . . . , n} thereexists j ∈ ω such that Ai is Σj-complete. Let σ, τ ∈ Sn. Then, the followingclasses are identical:

Q(A1, . . . , An),

Q(Aσ(1), . . . , Aσ(n)),

QC(A1, . . . , An), and

QC(Aτ(1), . . . , Aτ(n)).

Theorem ?? also allows us to find examples of commutative oracles out-side the arithmetical hierarchy. For example, we could take Ai to be Hai

where ai is a notation for a successor ordinal and |a1| < · · · < |ak|. Thisgives us a family of examples that are ‘above’ the arithmetical hierarchy Anexample that is incomparable with most of the hyperarithmetical hierarchycan be obtained as follows. First, note that by exercise 2.2 of Soare [?] thereexists A ⊆ ω such that A|T∅(n) for all n ≥ 1. It follows that A(m)|T∅(n) for allm ≥ 1 and n ≥ 2. We can then take A′1, . . . , A

′k to be any strictly increasing

subsequence of {A(n)}n∈ω. We also have the following proposition.

Proposition 4.7 There exists Y ⊆ ω such that Q(Y, ∅′′) = Q(∅′′, Y ) and forall n ∈ ω, Y is neither Σn nor Πn complete.

Proof: Let A be a low non-recursive r.e. set. Then, A′ ≤T ∅′. Thus, byTheorem ??, Q(A′, ∅′′) = Q(∅′′, A′). It is easily shown that for all n ∈ ω, A′

is neither Σn nor Πn 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 r.e.set whose jump does not commute with ∅′, and that there exists an r.e. setwhose jump does not commute with ∅′′.

We begin with two preliminary definitions. The first prescribes a methodfor generating from a set A an element of ΣA

1 . The second defines a ‘generic’element of Q(A,B).

Definition 5.1 For all A ⊆ ω, let D(A) =df {x ∈ ω| ∃y ∈ ω 〈x, 2y〉 ∈A ∧ 〈x, 2y + 1〉 6∈ A}.

14

Thus, D(A) ∈ ΣA1 . If A is r.e. and if {At}t∈ω is a recursive enumeration

of A, then D(A) = lim inft∈ωD(At). The usefulness of D(A) is that one canchange the value of χD(A)(x) an arbitrary number of times by enumeratingelements into A.

If A is r.e. and {At}t∈ω is a recursive enumeration of A, then A′t(e) =df

W(At)e,t (e) for all e ∈ ω. Thus, A′ = lim inft∈ω A

′t.

Definition 5.2 Let A,B ⊆ ω. For all x, y, y′ ∈ ω, we let 〈x, y, y′〉 ∈ V A,B

just in case(x 6∈ A ∧ y ∈ B) ∨ (x ∈ A ∧ y′ ∈ B)].

Clearly, V A,B ∈ QC(A,B) for all A,B ⊆ ω.We can form an effective enumeration {ψe}e∈ω of the Turing machines

that on any convgerent computation path make exactly two queries the firstof which is to the first oracle and the second of which is to the second oracle.Thus, for all C,A,B ⊆ ω, C ∈ Q(A,B) if and only if there exists e such thatC = ψA⊕Be .

Theorem 5.3 There exist low r.e. A,B ⊆ ω such that QC(D(A), B) 6⊆Q(∅′, A′) and QC(B,D(A)) 6⊆ Q(A′, ∅′).

Corollary 5.4 There exist low A ⊆ ω such that Q(A′, ∅′) 6⊆ Q(∅′, A′) andQ(∅′, A′) 6⊆ Q(A′, ∅′).

Proof of Theorem ?? : The proof is a finite injury argument. We constructlow A,B ⊆ ω such that V D(A),B 6∈ Q(∅′, A′) and V B,D(A) 6∈ Q(A′, ∅′). Foreach e ∈ ω we have the following requirements:

R4e : (∃<∞t ∈ ω φ(At)e,t (e) ↓)⇒ φ(A)

e (e) ↓

R4e+1 : (∃<∞t ∈ ω φ(Bt)e,t (e) ↓)⇒ φ(B)

e (e) ↓

R4e+2 : V D(A),B 6= ψ(∅′⊕A′)e

R4e+3 : V B,D(A) 6= ψ(A′⊕∅′)e

The R4e module (that is the procedure associated with satisfying require-ment R4e) uses the usual procedure for satisfying lowness requirements. At

15

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+ 1) =

{−1 if φ

(At)e,t (e) ↑

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

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

and x2e,0 = 〈u, v, w〉 where u, v, w are the least elements of ω[2e] in increasingorder. At stage t + 1, we proceed as follows. If t + 1 < 4e + 2, then we donothing at stage t+ 1. So, suppose that t+ 1 ≥ 4e+ 2. If the R4e+2 modulewas cancelled at stage t, then we let r(4e+ 2, 0) = −1 and x2e,t+1 = 〈u, v, w〉where u < v < w are the least three elements of ω[2e] that have not yet beenused by this module.

Suppose that the R4e+2 module was not cancelled at stage t. Let u, v, w besuch that x2e,t = 〈u, v, w〉. Let b = V D(At),Bt(x2e,t). The R4e+2 module does

nothing at stage t+ 1 if b 6= ψ(∅′t⊕A′t)e,t (x2e,t). Suppose that b = ψ

(∅′t⊕A′t)e,t (x2e,t).

We call t + 1 a stage of proper agreement for R4e+2. Let m be the query to

A′t made by the computation of ψ(∅′t⊕A′t)e,t (x2e,t). Let

r(4e+ 2, t+ 1) =

{−1 if φ

(At)m,t (m) ↑

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

Let τ ∈ {0, 1}2 be such that

(τ(0) = 1⇔ u ∈ D(At)) ∧ (τ(1) = 1⇔ α ∈ Bt)

where

α =

{v if u 6∈ D(At)w otherwise.

Case 1: τ = (0, 0).Enumerate v into Bt+1 unless it is less than or equal to max{r(e′, t)|e′ <

4e + 2} in which case the R4e+2 module is cancelled. If the R4e+2 module isnot cancelled, then it follows that case 2 holds at the next stage of properagreement.

Case 2: τ = (0, 1).Choose the least z ∈ ω such that 〈u, 2z〉 > max({r(e′, t)|e′ < 4e + 2} ∪

{r(4e + 2, t + 1)}) and enumerate 〈u, 2z〉 into At+1. It follows that case 3

16

holds at the next stage of proper agreement.

Case 3: τ = (1, 0).

Let σ be the path taken by the computation of φ(∅′t⊕A′t)e (x2e,t). It follows

that σ(0) = 1. If σ(1) = 1, then enumerate w into Bt+1 unless it is lessthan or equal to max{r(e′, t)|e′ < 4e+ 2} in which case the R4e+2 module iscancelled at stage t+ 1. Suppose that σ(1) = 0. Thus, r(4e+ 2, t+ 1) = −1.Since τ = (1, 0), there exists a unique z ∈ ω such that 〈u, 2z〉 ∈ At and〈u, 2z + 1〉 6∈ At. Enumerate 〈u, 2z + 1〉 into At+1 unless it is less than orequal to r(e′, t) for some e′ < 4e+ 2 in which case the R4e+2 module is can-celled at stage t+1. Suppose that the R4e+2 module is not cancelled at staget+ 1. If σ(1) = 1, then it follows that R4e+2 is satisfied at stage t+ 1 unlessa module for a higher priority requirement enumerates into A at some futurestage in which case case 4 will hold at the next stage of proper agreement.If σ(1) = 0, then it follows that case 2 holds at the next stage of properagreement and, unless this module is cancelled at some future stage, eithercase 3 or case 4 will hold at all future stages of proper agreement.

Case 4: τ = (1, 1).Cancel the R4e+2 module.

It follows by a standard finite injury argument that there are at mostfinitely many stages of proper agreement. Therefore, R4e+2 is eventuallysatisfied.

The R4e+3 module operates as follows. At stage 0, we let r(4e+3, 0) = −1and x2e+1,0 = 〈u, v, w〉 where u, v, w are the least three elements of ω[2e+1] inincreasing 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. Ifthe R4e+3 module was cancelled at stage t, then r(4e + 3, t + 1) =df −1 andx2e+1,t =df 〈u, v, w〉 where u < v < w are the least elements of ω[2e+1] thathave not yet been used by this module.

Suppose that the R4e+3 module was not cancelled at stage t. Let u, v, wbe such that x2e,t = 〈u, v, w〉, and let b = V Bt,D(At)(x2e,t). Do nothing if

b 6= ψ(A′t⊕∅′t)e,t (x2e,t). Suppose that b = ψ

(A′t⊕∅′t)e,t (x2e,t). We call t + 1 a stage of

proper agreement for R4e+2. Let τ ∈ {0, 1}2 be such that

(τ(0) = 1⇔ u ∈ Bt) ∧ (τ(1) = 1⇔ α ∈ D(At))

17

where

α =

{v if u ∈ Bt

w otherwise.

Let m be the query to A′t made by the computation of ψ(A′t⊕∅′t)e,t (x2e,t). Let

r(4e+ 3, t+ 1) =

{−1 if φ

(At)m,t (m) ↑

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

Case 1: τ = (0, 0).We pick the least z ∈ ω such that 〈v, 2z〉 > max({r(e′, t)|e′ < 4e+ 3} ∪

{r(4e + 3, t + 1)}) and enumerate 〈v, 2z〉 into At+1. It follows that case 2holds at the next stage of proper agreement.

Case 2: τ = (0, 1).It follows that there exists unique z ∈ ω such that 〈v, 2z〉 ∈ At and

〈v, 2z+1〉 6∈ At. If 〈v, 2z+1〉 > max({r(e′, t)|e′ < 4e+3}∪{r(4e+3, t+1)}),then we enumerate 〈v, 2z + 1〉 into At+1 in which case case 1 holds at thenext stage of proper agreement. Otherwise, we enumerate u into Bt+1 unlessit is less than or equal to max{r(e′, t)|e′ < 4e + 3} in which case the R4e+3

module is cancelled at stage t+ 1. If the R4e+3 module is not cancelled, thencase 3 holds at the next stage of proper agreement.

Case 3: τ = (1, 0).Choose the least z ∈ ω such that 〈w, 2z〉 > max({r(e′, t)|e′ < 4e + 3} ∪

{r(4e + 3, t + 1)}) and enumerate 〈w, 2z〉 into At+1. It follows that case 4holds at the next stage of proper agreement. Note that since cases 1 and 2have held previously m ∈ A′t and therefore r(4e+ 3, t+ 1) 6= −1.

Case 4: τ = (1, 1).There exists unique z ∈ ω such that 〈w, 2z〉 ∈ At and 〈w, 2z + 1〉 6∈

At. Enumerate 〈w, 2z + 1〉 into At+1 unless it is less than or equal tomax({r(e′, t)|e′ < 4e+ 3} ∪ {r(4e+ 3, t+ 1)}) in which case the R4e+3 mod-ule is cancelled 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 higherpriority requirement enumerates into A at some future stage in which casecase 3 holds at the next stage of proper agreement. It follows by a standardfinite injury argument that R4e+3 is eventually satisfied.

18

Theorem 5.5 There exists r.e. A,B such that Q(D(B), D(A)) 6⊆ Q(A′, ∅′′).

Corollary 5.6 There exists r.e. A ⊆ ω such that Q(∅′′, A′) 6⊆ Q(A′, ∅′′).

Proof of Theorem ?? : We construct A,B so that V D(B),D(A) 6∈ Q(A′, ∅′′).For each e ∈ ω we satisfy the requirement

Re : V D(B),D(A) 6= ψ(A′⊕∅′′)e .

Doing so may require infinitely many numbers to be enumerated into A.Thus, we use an infinite injury priority argument. Our notations and generalset up are in the spirit of chapter 14 of Soare [?]. A minor departure is thatwe let ω[e] be {〈e, x〉 | x ∈ ω} rather than the e-th column of ω. For allη ∈ {0, 1, 2}<ω, let Rη = Rlh(η).

At each stage, the Rη module will either do nothing, or enumerate anumber into exactly one of ∪x∈ω[η]ω[2x], ∪x∈ω[η]ω[2x+1] in order to change thestatus of some number’s membership in D(A). An Rη module may have toenumerate an infinite number of elements into one of these sets. At stage t,we form a binary sequence δt ∈ {0, 1}t which reflects our best guess as towhat the Rη modules for η ⊂ δt are going to do.

Let δ0 = ∅. Let t ∈ ω. For all η ∈ {0, 1, 2}<ω let

vt(η) = max({0} ∪ {s ≤ t|δs ⊇ η}).

We define δt+1 to be the element of {0, 1}t+1 such that for all η ∈ {0, 1, 2}<ω,

η∧(2) ⊆ δt+1

⇔ η ⊆ δt+1 ∧ (At+1 − At) ∩ ∪{ω[2x+i] | x ∈ ω[η], i ∈ {0, 1}} = ∅.

and for all i ∈ {0, 1}

η∧(i) ⊆ δt+1 ⇔ η ⊆ δt+1 ∧ (At+1 − At) ∩ (∪x∈ω[η]ω[2x+(1−i)]) 6= ∅.

We say that t ∈ ω is an η stage if η ⊆ δt.Let z ∈ ω, and let η ∈ {0, 1, 2}<ω. Let i ∈ {0, 1}, and let x, y ∈ ω be

such that z = 〈2x + i, y〉. We say that z is not believed by η at stage t + 1if there exists η′ such that η′∧(i) ⊆ η, x = xη′,t, and 〈2x + (1 − i), y′〉 6∈ Atfor all y′ ≥ y. Let v ∈ ω, and consider the computation φ

(At)v,t (v). We say

that this computation is η-correct at stage t + 1 if it diverges or if every

19

z < u(v,At, v, t) is believed by η at stage t + 1. Let z ∈ ω, and consider

a computation of the form ψ(A′t⊕∅′′t )e,t (z). We say that this computation is η

correct at stage t+ 1 if it converges and φ(At)v,t (v) is η correct at stage t where

v is the query to A′t made by the computation of ψ(A′t⊕∅′′t )e (y).

Fix η ∈ {0, 1, 2}<ω. The Rη module operates as follows. At stage 0, we letxη,0 = min(ω[η]) and r(η, 0) = −1. Let t ∈ ω. If the η-module was cancelledat stage t, then we let xη,t+1 = min{z ∈ ω[η]|z > xη,t}. Suppose that the Rη

module was not cancelled at stage t. If t+ 1 is not an η stage and δt+1 <L η,then the Rη module is not cancelled at stage t+ 1. If t+ 1 is not an η-stageand δt+1 >L η, then the Rη module does nothing at stage t+1. Suppose that

t + 1 is an η stage. Let y = 〈xη,t, 2xη,t, 2xη,t + 1〉. Let b = ψ(A′t⊕∅′′t )e,t (y). We

call b the value of the η module at stage t . If ψ(A′t⊕∅′′t )e,t (y) does not converge

η-correctly or converges to 1− b, then the Rη module does nothing at staget+ 1. So, suppose that this computation convereges η-correctly to b. We callt + 1 a stage of proper agreement . Let v be the query to A′t made by the

computation of ψ(A′t⊕∅′′t )e (y). We let

r(η, t+ 1) =

{−1 if φ

(At)v,t (v) ↑

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

Let σ be the path taken by the computation of ψA′t⊕∅′′te (y). Let τ ∈ {0, 1}2

be such that

(τ(0) = 1⇔ xη,t ∈ D(Bt)) ∧ (τ(1) = 1⇔ α ∈ D(At))

where

α =

{2xη,t + 1 if xη,t 6∈ D(Bt)2xη,t + 2 if xη,t ∈ D(Bt).

Case 1: τ = (0, 0).Thus, b = 0 and xη,t 6∈ D(Bt). We want to change the value of the

Rη module to 1. Let z be the least element of ω such that 〈2xη,t, 2z〉 >max({r(η′, t)|η′ <L η} ∪ {r(η, t+ 1)}) and 2z > max{z′ ∈ ω|〈2xη,t, z′〉 ∈ At}.We then enumerate 〈2xη,t, 2z〉 into At+1.

Case 2: τ = (0, 1).Thus b = 1 and xη,t 6∈ D(Bt). It follows that there exists unique z ∈ ω

such that 〈2xη,t, 2z〉 ∈ At and 〈2xη,t, 2z + 1〉 6∈ At. We want to change the

20

value of the Rη module to 0. One way to do this is to enumerate 〈2xη,t, 2z+1〉into At+1. This may be either unwise or unecessary according to the following

subcases. Let ρ = (σ(0), 0). If ψ[ρ]e,t(y) = 0 or if 〈2xη,t, 2z + 1〉 ≤ r(η, t + 1),

then we enumerate 〈xη,t, 2z′〉 into Bt+1 where z′ ∈ ω is the least number suchthat 〈xη,t, 2z′〉, 〈xη,t, 2z′ + 1〉 6∈ Bt. It follows that case 3 holds at the next

stage of proper agreement. If ψ[ρ]e,t(y) 6= 0 and 〈2xη,t, 2z + 1〉 > r(η, t + 1),

then we enumerate 〈2xη,t, 2z + 1〉 into At+1 unless it is less than or equal tomax{r(η′, t) | η′ <L η} in which case we cancel the Rη module at stage t+ 1.

Case 3: τ = (1, 0)Thus, b = 0 and xη,t ∈ D(Bt). It follows that there exists unique z′ ∈ ω

such that 〈xη,t, 2z′〉 ∈ Bt and 〈xη,t, 2z′ + 1〉 6∈ Bt. Let ρ = (σ(0), 0). If

ψ[ρ]e,t(xη,t) = 0, then we enumerate 〈xη,t, 2z′+ 1〉 into Bt+1 and do nothing else

at this stage. Otherwise, we choose the least z ∈ ω such that

〈2xη,t + 1, 2z〉 > max({r(η′, t)|η′ <L η} ∪ {r(η, t+ 1)})

and 2z > max{z′′ ∈ ω|〈2xη,t+1, z′′〉 ∈ At}. We then enumerate 〈2xη,t+1, 2z〉into At+1. It follows that case 4 holds at the next stage of proper agreement.

Case 4: τ = (1, 1)Thus, b = 1, and xη,t ∈ D(Bt). We note that case 3 held at the previ-

ous stage of proper agreement. It follows that there exists a unique z ∈ ωsuch that 〈2xη,t + 1, 2z〉 ∈ At and 〈2xη,t + 1, 2z + 1〉 6∈ At. It follows by ashort argument that hinges on the notion of η correctness and case 3 that〈2xη,t + 1, 2z + 1〉 > r(η, t + 1). We then enumerate 〈2xη,t + 1, 2z + 1〉 intoAt+1 unless it fails to be greater than max{r(η′, t) | η′ <L η} in which casethe Rη module is cancelled at stage t+ 1. If the Rη module is not cancelledat stage t + 1, then it follows that case 3 holds at the next stage of properagreement.

Let f = lim inft∈ω δt. That is, for all η ∈ {0, 1, 2}<ω

η ⊂ f ⇔ (∃∞t δt ⊇ η) ∧ (∃<∞t δt <L η).

Let η ⊂ f . Let e = lh(η). By induction, suppose that rη′ =df limt→∞ r(η′, t)

exists for all η′ ⊂ η. It follows that xη =df limt→∞xη,t exists. Let t0 bethe stage at which this limit is reached. If there are only finitely many

21

stages of proper agreement, then Rη is satisfied and rη exists. We thereforeassume that there are infinitely many stages of proper agreement. Let y =〈xη, 2xη, 2xη + 1〉. Let

v =q(=2)e (∅, y)

2.

It follows via a short argument that hinges on cases 3 and 4 and the notionof η correctness that rη exists. Let t1 > t0 be a stage of proper agreementsuch that r(η, t1) = rη. Therefore, A′t1(v) = A′(v).

We claim that one of the following holds.

1. The Rη module eventually oscillates between cases 1 and 2 . That is,at every sufficiently large stage of proper agreement t either case 1 orcase 2 holds at t and if case i holds at t then case 3 − i holds at thenext stage of proper agreement.

2. The Rη module eventually oscillates between cases 2 and 3 . That is,at every sufficently large stage of proper agreement t, either case 2 orcase 3 holds at t and if case i holds at t then case 5 − i holds at thenext stage of proper agreement.

3. The Rη module eventually oscillates between cases 3 and 4 .

To see how to prove this, take a proper stage of agreement t2 > t1. Let σ

be the path taken by the computation of ψ(A′t2−1⊕∅′′t2−1)e . Let ρ = (σ(0), 0). If

case 1 or case 2 holds at stage t2, then (1) holds if ψ[ρ]e (xη) 6= 0; otherwise (2)

holds. If case 3 or 4 holds at stage t2, then (2) holds if ψ[ρ]e (xη) = 0; otherwise

(3) holds.Now, if (1) holds, 2xη 6∈ D(A), xη 6∈ D(B), and therefore V D(B),D(A)(y) =

0. If (2) holds, then xη 6∈ D(B), 2xη ∈ D(A), and therefore V D(B),D(A)(y) =1. If (3) holds, then xη ∈ D(B), 2xη+1 6∈ D(A), and therefore V D(B),D(A)(y) =0. Whether (1),(2), or (3) holds, ρ(0) = σ(0), and so

v′ =dfq(=2)e (ρ � 1, y)− 1

2

is defined and is not in ∅′′. If either (1) or (3) holds, ψ[ρ]e (y) = 1 and

ψ[(ρ(0),1)]e (y) = 0. Hence, since v′ 6∈ ∅′′, ψ(A′⊕∅′′)

e (y) = 1. If (2) holds,

ψ[ρ]e (y) = 0 and ψ

[(ρ(0),1)]e (y) = 1. Therefore, since v′ 6∈ ∅′′, ψ(A′⊕∅′′)

e (y) = 0.

22

6 Commutative oracles in the hyperarithmetic

hierarchy

We use Kleene’s system of notations, O. For all a ∈ O we define Ha in theusual way (see [?]). For a ∈ O, let |a| be the ordinal represented by a. Leta1, . . . , an ∈ O. The fundamental result of this section is that (Ha1 , . . . , Han)commutes if and only if a1, . . . , an are all notations for sucessor ordinals orexactly one of |a1|, . . . , |an| is a limit ordinal and this ordinal is the minimumof {|a1|, . . . , |an|}. We prove this by showing that these sets can always beput into least-first order and then classifying when they can be put intohardest first order. This classification will then be used to classify whichsequences can be put into an arbitrary order. We will make use of logicalclosure properties which are defined as follows.

Definition 6.1 Let A,B ⊆ ω.

1. We say that B is closed under conjunctions with A if A×B ≤m B.

2. We say that B is closed under disjunctions with A if B is closed underconjunctions with A.

3. We say that B is closed under conjunctions (disjunctions) if B is closedunder 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 leftto the reader.

Proposition 6.2 1. The jump of any set is closed under conjunctionsand disjunctions.

2. For all a ∈ O, if |a| is a limit ordinal, then Ha is closed under negations.

3. If B ≤T X, then X ′ is closed under conjunctions and disjunctions withB and B.

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

23

Definition 6.3 Let r1, . . . , rk ∈ ω, and let e be an (r1, .., rk) operator. Fixi ∈ {1, . . . , k} and j ∈ {1, . . . , ri}. For all z ∈ ω and σ ∈ {0, 1}<ω we letnodeij(σ, e, z) be the j-th prefix of σ at which the i-th oracle is queried ifthere is one and otherwise nodeij(σ, e, z) ↑. We define nodeij,s to be like nodeijexcept that all compuations involved are limited to s steps.

We regard nodeij as the j-th component function of nodei. We note thatnodeij and nodeij,s are partial recursive functions. However, the domain ofthe latter function is recursive.

Definition 6.4 Let r1, . . . , rk, e, i, j be as in definition ??. For all z ∈ ω andσ ∈ {0, 1}<ω we let answerij(σ, e, z) be the answer given by σ to the j-thquery to the i-th oracle if there is one and otherwise answerij(σ, e, z) ↑. Wedefine answerij,s(σ, e, z) by limiting all computations involved to at most ssteps.

We regard answerij as the j-th component function of answeri. We notethat answerij and answerij,s are partial recursive functions and that the do-main of the latter is recursive.

Definition 6.5 Let r1, . . . , rk, e be as in definition ??. Let A1, . . . , Ak ⊆ ω.For all X ⊆ {1, . . . , k} and z ∈ ω we say that σ ∈ {0, 1}<ω is consistent with

(A1, . . . , Ak) on (X, e, z) if φ[σ]e (z) ↓ and for all i ∈ X and j ∈ {1, . . . , ri},

answerij(σ, e, z) = 1⇔ qe(nodeij(σ, e, z), z) ∈ A1 ⊕ · · · ⊕ Ak.

The term consistent at stage s is defined similarly by limiting all computa-tions to s steps.

Definition 6.6 Let r1, . . . rk, e be as in definition ??. Let ρ ∈ {0, 1}<ω. Forall X ⊆ {1, . . . , k} and z ∈ ω we say that σ ∈ {0, 1}<ω is consistent with ρ on

(X, e, z) if φ[σ]e (z) ↓, φ[ρ]

e (z) ↓, and for all i ∈ X and j ∈ {1, . . . , ri} σ and ρgive the same answer to the j-th query to the i-th oracle. The term consistentat stage s with ρ on (X, e, z) is defined similarly by replacing answerij withanswerij,s and requiring the corresponding function values to be defined.

24

Thus, for all U ⊆ {1, . . . , k}, φ[U ](A1⊕... Ak)e (z) ↓ if and only if

∃τ ∈ {0, 1}nφ[τ ]e (z) ↓

∧τ is consistent with (A1, . . . , Ak) on (U, e, z)

∧ τ is consistent with path(e, n, z) on ({1, . . . , k} − U, e, z).

We will use the following Lemma.

Lemma 6.7 Let e be an (r1, . . . , rk) operator. Let a1, . . . , ak ∈ O be suchthat 1 ≤ |a1| ≤ · · · ≤ |ak|, and let j ∈ {1, . . . , k} be such that |aj| is a limit

ordinal. Then φ[{1,...,j}](Ha1⊕···⊕Hak )e extends to an element of Q((r1 + · · · +

rj)∅′, Haj). If e is semitotal, then this function extends to an element ofQ((r1 + · · ·+ rj − 1)∅′, Haj). If e is globally total, then this function extendsto an element of QC(1, Haj).

Proof: Let n = r1 + · · ·+ rk. Let h : ω → {0, 1} be such that

h(z) = 0⇔ φ[{1,...,j}](Ha1⊕···⊕Hak )e (z) = 0.

Clearly, h is a total function extending φ[{1,...,j}](Ha1 ,...,Hak )e . The tricky part is

to show that h ∈ Q((r1 + · · · + rj)∅′, Haj). The pieces of this puzzle are asfollows. There exists recursive f : ω → ω such that for all z ∈ ω

Wf(z) = {nodejrj(τ, e, z)|τ ∈ {0, 1}n ∧ φ[τ ]

e (z) ↓

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

Let s = r1 + · · ·+ rj. It is easy to show that #Wf(z) ≤ 2s−1 for all z ∈ ω. Ife is semitotal, then 1 ≤ #Wf(z) for all z ∈ ω. It follows from Proposition ??that there exists g ∈ FQ(s∅′) such that Dg(z) = Wf(z) for all z ∈ ω. If eis semitotal, then we may take g to be an element of FQ((s − 1)∅′). If e isglobally total, then we may take g to be recursive.

It follows from Proposition ?? that there exists partial recursive ψ1 suchthat for all τ, z

[(∀c ∈ {1, . . . , j} nodec(τ, e, z) ↓) ⇒ ψ1(τ, z) ↓]

∧[(τ is consistent with (Ha1 , . . . , Hak) on ({1, . . . , j}, e, z)⇔ ψ1(τ, z) ∈ Haj)].

25

(To define ψ1(τ, z) wait for nodec(τ, e, z) to converge for all c ∈ {1, . . . , j}and then use the reductions whose existence is guaranteed by Proposition ??to form the appropriate query to Haj .) It also follows from Proposition ??that there exists recursive ψ2 such that for all τ, z

ψ2(τ, z) ∈ ∅′ ⇔ (φ[τ ]e (z) = 0

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

Note that nodec(τ, e, n, z) ↓ for all c ∈ {1, . . . , j} if τ ⊇ ρ for some ρ ∈ Dg(z).It follows that

h(z) = 0⇔ ∃τ ∈ {0, 1}n∃ρ ∈ Dg(z)(τ ⊇ ρ ∧ ψ1(τ, z) ∈ Haj ∧ ψ2(τ, z) ∈ ∅′).

It follows from Proposition ?? that h is in the requried class.

We can now show that a sequence of the form (Ha1 , . . . , Han) can alwaysbe put into least first order.

Theorem 6.8 Let a1, . . . , ak ∈ O be such that 1 ≤ |a1| < · · · < |ak|. Letr1, . . . , rk ≥ 1. Then, Q∗(r1Ha1 , . . . , rkHak) = Q(r1Ha1 , . . . , rkHak). Further-more, if |aj| is the maximum limit ordinal in {|a1|, . . . , |ak|}, then

Q∗(r1Ha1 , . . . , rkHak) = Q(u∅′, Haj , rj+1Haj+1, . . . , rkHak)

where u = r1 + · · ·+ rj − 1.

Proof: If there are no limit ordinals in {|a1|, . . . , |ak|}, then this followsfrom Theorem ??. So, suppose that at least one of |a1|, . . . , |ak| is a limit or-dinal, and let j be the maximum element of {1, . . . , k} such that |aj| is a limitordinal. Clearly, Q(u∅′, Haj , rj+1Haj+1

, . . . , rkHak) ⊆ Q∗(r1Ha1 , . . . , rkHak).Let C ∈ Q∗(r1Ha1 , . . . , rkHak) via φe. Thus, φe is semitotal. Let U =

{1, . . . , j}. By Lemma ??, φ[U ](Ha1 ,...,Hak )e extends to an element h ofQ(u∅′, Haj).

It then follows from the Oracle Replacement Lemma that C ∈ Q(u∅′, Haj , rj+1Haj+1, . . . , rkHak).

Corollary 6.9 Let a1, . . . , an ∈ O be such that 1 ≤ |a1| ≤ . . . ≤ |an|. Then,for all σ ∈ Sn, Q(Haσ(1) , . . . , Haσ(n)) ⊆ Q(Ha1 , . . . , Han).

26

Proof: By Theorem ??.

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

Lemma 6.10 Let A1, . . . , Ak ⊆ ω. Let a1, . . . , am, g ∈ O be such that|a1|, . . . , |am| < |g| and |g| is a limit ordinal. Then,

Q(A1, . . . , Ak, Hg, Ha1 , . . . , Ham) = Q(A1, . . . , Ak, Hg).

Proof: Let n = k+m+1. Let C ∈ Q(A1, . . . , Ak, Hg, Ha1 , . . . , Ham) via φe.Let (X0, . . . , Xn−1) = (A1, . . . , Ak, Hg, Ha1 , . . . , Ham). For all z ∈ ω, let h(z)be the sequence of answers to the first k queries made by the computationof φ

(X0⊕···⊕Xn−1)e (z). Thus, h ∈ FQ(A1, . . . , Ak). Furthermore, for all z ∈ ω,

C(z) = 0⇔ ∃σ ∈ {0, 1}n [φ[σ]e (z) = 0 ∧ σ ⊇ h(z) ∧

σ is consistent with (X0, . . . , Xn−1) on ({k + 1, . . . , n}, e, z)}].

Let z ∈ ω, and let σ be a binary n-tuple that is prefixed by h(z). Note thatanswerk+1

1 (σ, e, z) ↓. Then, σ is consistent with (X0, . . . , Xn−1) on ({k +1, . . . , n}, e, z) if and only if

(answerk+11 (σ, e, z) = 1⇔ qe(node

k+11 (σ, e, z), z) ∈ Hg)

∧∃s ∈ ω [σ is consistent at stage s with (X0, . . . , Xn−1) on ({k+2, . . . , n}, z)].

We note that the relation

{(σ, s, z) | σ is consistent at stage s with (X0, . . . , Xn−1) on ({k+2, . . . , n}, e, z)}

is recursive in Ha1 ⊕ · · · ⊕ Ham . Let v ∈ {a1, ..., am} be such that |v| =max{|a1|, . . . , |am|}. Then, Ha1⊕· · ·Ham ≡T Hv. Since |g| is a limit ordinaland |g| > |v|, it follows that Hg is closed under conjunctions with (Hv)

′. Itis also closed under negations. It follows that C ∈ Q(A1, . . . , Ak, Hg).

Lemma 6.11 Let d1, . . . , dk, g ∈ O be such that |g| is a limit ordinal, and|d1|, . . . , |dk| are successor ordinals and |g| < |d1| ≤ . . . ≤ |dk|. Then, for allm ≥ 1,

Q(m∅′, Hg, Hd1 , . . . , Hdk) 6⊆ Q(mHg, Hd1 , . . . , Hdk).

27

Proof: It follows from Theorem ?? that Q(mHg, Hd1 , . . . , Hdk) ⊆ Q((m−1)∅′, Hg, Hd1 , . . . , Hdk). Let n = m+k. Let {Γ( )

e }e∈ω be an effective enumer-ation of all (m−1, (1)k+1) operators. There exists an (m−1, (1)k+1) operator

φe such that φ(A)e (x) = 1 − Γ

(A)x (x) for all x ∈ ω and A ⊆ ω. There exists a

semitotal (m, (1)k+1) operator φe′ such that for all z ∈ ω

φ(∅′

⊕Hg

⊕Hd1

⊕...

⊕Hdk )

e (z) ↓

⇒ φ(∅′

⊕Hg

⊕Hd1

⊕...

⊕Hdk )

e′ (z) = φ(∅′

⊕Hg

⊕Hd1

⊕...

⊕Hdk )

e (z).

It follows from Lemma ?? and the Oracle Replacement Lemma that φ(∅′⊕Hg⊕Hd1 ···⊕Hdk )e′

extends to an element C of Q(m∅′, Hg, Hd1 , . . . , Hdk). It follows that C 6∈Q((m− 1)∅′, Hg, Hd1 , . . . , Hdk).

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

Theorem 6.12 Let a1, . . . , an ∈ O. Let b1, . . . , bn be the elements of {a1, . . . , an}in an order such that |b1| ≥ . . . ≥ |bn|. Then, Q(Ha1 , . . . , Han) ⊆ Q(Hb1 , . . . , Hbn)if and only if for all γ, if γ is the maximum limit ordinal in {|a1|, . . . , |an|},then |ai| ≥ |aj| for all i, j ∈ {1, . . . , n} such that i < j and |aj| = γ.

Proof: (⇐): If there are no limit ordinals in {|a1|, . . . , |an|}, then this fol-lows from Corollary ??. Suppose that there is a limit ordinal in {|a1|, . . . , |an|},and let γ be the maximum such limit ordinal. It follows from Theorem ??that there exists a rearrgangement, (c1, . . . , cn), of (a1, . . . , an) such thatQ(Ha1 , . . . , Han) ⊆ Q(Hc1 , . . . , Hcn) and if k is the maximum element of{1, . . . , n} such that |ck| = γ, then for all i ∈ ω

k < i ≤ n⇒ |ci| < γ, and

1 ≤ i < j ≤ k ⇒ |ci| ≥ |cj|.

It follows from Lemma ?? that

Q(Hc1 , . . . , Hcn) ⊆ Q(Hc1 , . . . , Hck)

Note that (c1, . . . , ck) prefixes (b1, . . . , bn). It follows from Proposition ??that Q(Ha1 , . . . , Han) ⊆ Q(Hb1 , . . . , Hbn)

28

(⇒): We prove the contrapositive. Thus, we assume that there ex-ists a maximum limit ordinal γ ∈ {|a1|, . . . , |an|} and that there existsi, j ∈ {1, . . . , n} such that i ≤ j and |ai| < γ = |aj|. Let m = #{i′ ∈{1, . . . , n} | |ai′ | = γ}. It follows from Theorem ?? that there exists a rear-rangement of (a1, . . . , an), (c1, . . . , cn) such thatQ(Ha1 , . . . , Han) ⊇ Q(Hc1 , . . . , Hcn)and

i′ < j′ ∧ |ci′|, |cj′ | ≥ γ ⇒ |ci′| ≤ |cj′ |.

for all i′, j′ ∈ {1, . . . , n}. Let u be the maximum element of {1, . . . , n} suchthat |cu| = γ. Since |ai| < γ and |aj| = γ, it follows that u ≥ m + 1. Let kbe the maximum element of {1, . . . , n} such that |bk| = γ. Since u ≥ m+ 1,it follows from Proposition ?? and Proposition ?? that

Q(Hc1 , . . . , Hcn) ⊇ Q(m∅′, Hbk , Hbk−1, . . . , Hb1).

On the other hand, by Theorem ??

Q(Hb1 , . . . , Hbn) ⊆ Q(Hb1 , . . . , Hbk).

By Theorem ??

Q(Hb1 , . . . , Hbk) ⊆ Q(Hbk , Hbk−1, . . . , Hb1).

And by Lemma ??

Q(m∅′, Hbk , . . . , Hb1) 6⊆ Q(mHbk , Hbk−1, . . . , Hb1).

Thus, Q(Ha1 , . . . , Han) 6⊆ Q(Hb1 , . . . , Hbn).

Theorem 6.12 allows us to classify the commutative sequences of H-sets.

Theorem 6.13 Let a1, . . . , an ∈ O be such that 1 ≤ |a1|, . . . , |an|. Then, thefollowing are equivalent.

1. Q(Ha1 , . . . , Han) = Q(Haσ(1) , . . . , Haσ(n)) for all σ ∈ Sn.

2. There is at most one limit ordinal in {a1, . . . , an}, and if γ is this limitordinal then γ = min{|a1|, . . . , |an|}.

29

Proof: If none of {|a1|, . . . , |an|} are limit ordinals, then this followsfrom Theorem ??. So, suppose that γ is the maximum limit ordinal in{|a1|, . . . , |an|}. If (??) fails, then it follows from Theorem ?? that there existsσ ∈ Sn such that Q(Haσ(1) , . . . , Haσ(n)) 6⊆ Q(Hb1 , . . . , Hbn) where (b1, . . . , bn)is a rearrangement of (a1, . . . , an) such that |b1| ≥ . . . ≥ |bn|. Thus, (??)implies (??). Suppose that (??) holds. Without loss of generality, supposethat |a1| ≤ . . . ≤ |an|. Let k be the maximum integer such that |ak| = γ.Thus, |ak+1|, . . . , |an| are successor ordinals, and |a1| = . . . = |ak| = γ.Let σ ∈ Sn. It follows from repeated application of Theorem ?? thatQ(Ha1 , . . . , Han) ⊆ Q(Haσ(1) , . . . , Haσ(n)).

We note that not only did well-separated sequences of jumps commute,but their Q and QC classes coincided. In classifying the commutative se-quence 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? Weanswer the second question first. We first deal with the case where at leastone limit ordinal is involved.

Theorem 6.14 Let a1, . . . , an ∈ O be such that 1 ≤ |a1| ≤ . . . ≤ |an| and atleast one of {|a1|, . . . , |an|} is a limit ordinal. Let j be the maximum elementof {1, . . . , n} such that |aj| is a limit ordinal. Then, for all σ ∈ Sn,

QC(Haσ(1) , . . . , Haσ(n)) = QC(Haj , Haj+1, . . . , Han).

Proof: Let C ∈ QC(Haσ(1) , . . . , Haσ(n)) via φe. Thus, φe is a globallytotal (r1, ..., rk)-operator. Let U = {i ∈ {1, . . . , k}|aσ(i) ∈ {a1, . . . , aj}}.By Lemma ??, φ

[U ](Haσ(1) ,...,Haσ(n) )

e extends to an element of QC(1Haj). Theconclusion then follows from the Oracle Replacement Lemma.

Corollary 6.15 Let a1, . . . , an ∈ O. Then, for all σ ∈ Sn, QC(Ha1 , . . . , Han) =QC(Haσ(1) , . . . , Haσ(n)).

Proof: If none of |a1|, . . . , |an| are limit ordinals, then this follows fromTheorem ??. Otherwise, this follows from Theorem ??.

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

30

Theorem 6.16 Let a, b1, . . . , bk ∈ O be such that 1 ≤ |a| ≤ |b1| < · · · < |bk|and |a| is a limit ordinal and |b1|, . . . , |bk| are successor ordinals. Then, forall r1, . . . , rk ≥ 1,

Q(Ha, r1Hb1 , . . . , rkHbk) = QC(Ha, r1Hb1 , . . . , rkHbk).

Proof: Let n = r1 + · · · + rk + 1. Let C ∈ Q(Ha, r1Hb1 , . . . , rkHbk) viaφe. Thus, φe is a globally total (1, r1, ..., rk)-operator. By Lemma ??, there

exists h ∈ QC(Ha) which extends φ[{1}](Ha,Hb1 ,...,Hbk )e . The conclusion then

follows from the Oracle Replacement Lemma.

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

Theorem 6.17 Let a1, . . . , an ∈ O be such that 1 ≤ |a1|, . . . , |an|. Then,

Q(Ha1 , . . . , Han) = QC(Ha1 , . . . , Han).

if and only if the following two conditions hold.

1. There is at most one j ∈ {1, . . . , n} such that |aj| is the maximum limitordinal in {|a1|, . . . , |an|}.

2. For all j ∈ {1, . . . , n}, if |aj| is the maximum limit ordinal in {|a1|, . . . , |an|},then |ai| > |aj| for all i ∈ {1, . . . , j − 1}.

Proof: (⇒): Suppose that either condition ?? or condition ?? fails. Letγ be the maximum limit ordinal in {|a1|, . . . , |an|}. Let (c1, . . . , cn) be arearrangement of (a1, . . . , an) such that |c1| ≤ . . . ≤ |cn|. Let k be themaximum element of {1, . . . , n} such that |ck| = γ. Then, |ck| < |ck+1| ≤. . . ≤ |cn| and |ck+1|, . . . , |cn| are successor ordinals. By Lemma ??,

Q(∅′, Hck , . . . , Hcn) 6⊆ Q(Hck , . . . , Hcn).

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

(a) Q(∅′, Hck , . . . , Hcn) ⊆ Q(Ha1 , . . . , Han).

(b) Q(Hck , . . . , Hcn) ⊇ QC(Ha1 , . . . , Han).

31

The proof of (a) is as follows. It follows from Theorem ?? that there isa rearrangement of (a1, . . . , an), (b1, . . . , bn), such that Q(Ha1 , . . . , Han) ⊇Q(Hb1 , . . . , Hbn) and

i < j ∧ |bi|, |bj| ≥ γ ⇒ |bi| ≤ |bj|

for all i, j ∈ {1, . . . , n}. Furthermore, since either (1) or (2) fails, if i isthe maximum element of {1, . . . , n} such that |bi| = γ, then i > 1. And, if|bi| > |γ|, then j > i. It follows from Proposition ?? that

Q(∅′, Hbi , . . . , Hbn) ⊇ Q(∅′, Hck , . . . , Hcn).

Since i > 1, it follows from Proposition ?? thatQ(∅′, Hbi , . . . , Hbn) ⊆ Q(Hb1 , . . . , Hbn).Thus, (a) holds.

By Theorem ??, QC(Ha1 , . . . , Han) = QC(Hck , . . . , Hcn) and (b) followsimmediately.

(⇐): Suppose that condition ?? and condition ?? hold. If there is no limitordinal in {|a1|, . . . , |an|}, then the conclusion follows from Corollary ??. So,suppose that there is a limit ordinal in {|a1|, . . . , |an|}. Let γ be the maxi-mum such limit ordinal. Hence, any ordinal in {|a1|, . . . , |an|} that is greaterthan γ is a successor ordinal. It follows from Theorem ?? that there ex-ists a rearrangement of (a1, .., an), (b1, . . . , bn), such that Q(Ha1 , . . . , Han) ⊆Q(Hb1 , . . . , Hbn), and

i < j ∧ |bi|, |bj| ≥ γ ⇒ |bi| ≥ |bj|

for all i, j ∈ {1, . . . , n}. Furthermore, if i ∈ {1, . . . , n} is such that γ =|bi|, then |bi+1|, . . . , |bn| < γ and |b1| ≥ . . . ≥ |bi−1| > |bi|. By Theorem??, Q(Hb1 , . . . , Hbn) ⊆ Q(Hb1 , . . . , Hbi). By Theorem ??, Q(Hb1 , . . . , Hbi) ⊆Q(Hbi , . . . , Hb1). By Theorem ??, Q(Hbi , . . . , Hb1) = QC(Hbi , . . . , Hb1). But,by Theorem ??, QC(Hbi , . . . , Hb1) = QC(Ha1 , . . . , Han).

7 Acknowledgements

I thank William Gasarch for proofreading this paper. I also thank RichardBeigel for suggesting logical closure properties to me.

32

References

[1] R. Beigel, personal communication.

[2] R. Beigel, R. Chang, Commutative Queries , to appear in Informationand Computation

[3] R. Beigel, W. Gasarch, M. Kummer, G. Martin, T. McNicholl, F.Stephan, On the query complexity of classes, Proceedings of the 21stInternational Symposium, Mathematical Foundations of Com-puter Science. Lecture Notes in Computer Science 1113, 206 - 217(1996).

[4] W. Gasarch, G. Martin, Bounded Queries in Recursion Theory,New York: Springer Verlag 1998.

[5] E. Hemaspaandra, L. Hemaspaandra, H. Hempel, Query Order in thePolynomial Hierarchy, Proceedings of the International Sympo-sium on Fundamentals of Computation Theory. Lecture Notes inComputer Science 1279, 222-232 (1997).

[6] L. Hemaspaandra, H. Hempel, G. Weschung, Query Order, to appear inSIAM Journal on Computing.

[7] G.E. Sacks, Higher Recursion Theory, 1st ed. (Springer-Verlag, Heidel-berg 1990).

[8] R.I. Soare, Recursively Enumerable Sets and Degrees, 1st ed. (Springer-Verlag, Berlin, Heidelberg, 1987).

Department of MathematicsUniversity of DallasIrving, Texas 75062 USAemail: tmcnicho@acad.udallas.edu

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