An Embedding into a Substructure of the r . e . Turing Degrees

Let (Im,≤) be the partial ordering of the m-introimmune r.e. Turing degrees. We wonder if such structure is an upper semi-lattice. We give a partial answer, by embedding some Boolean algebras into (Im,≤).


Introduction to the Problem
At the end of the sixties some recursion theorists wondered if there were sets of natural numbers without subsets of higher Turing degrees, and if so what would be their degree of unsolvability.This question was solved by Soare, who proved their existence in (Soare, 1969), and by Jockusch (Jockusch, 1973) and Simpson (Simpson 1978), who proved respectively that they cannot be neither arithmetic nor hyperarithmetic.A possible continuation of this study is to consider strong reducibilities ≤ r and to see whether or not there are arithmetical sets without subsets of higher r-degrees.We recall that a reducibility ≤ r is called strong reducibility if it implies the Turing reducibility ≤ T , that is ≤ r is different from ≤ T and A ≤ r B ⇒ A ≤ T B for every sets A, B of natural numbers.The unfamiliar reader on these reducibilities can see e.g.(Odifreddi, 1981), as well as a monograph on Computability Theory as, e.g., either (Soare, 1987), or (Odifreddi, 1989).A first positive answer to the above question for the conjunctive reducibiity ≤ c is contained in (Cintioli & Silvestri, 2003).In partuicular, in (Cintioli & Silvestri, 2003) has been defined the concept of r-introimmune set for a given reducibility ≤ r : an infinite set A is r-introimmune if for every B ⊆ A with |A − B| = ∞ it holds that A ̸ ≤ r B. So, r-introimmune sets have no subsets of higher r-degrees for all the reducibilities ≤ r studied in computability theory.Until today we know that for all the strong reducibilities ≤ r up to the truth-table reducibility ≤ tt there are r-introimmune arithmetical sets.In particular, there are tt-introimmune sets in ∆ 0 2 (Ambos-Spies, 2003) and there are m-introimmune sets in Π 0 1 (Cintioli, 2005).Whenever someone discover the existence of sets with a certain property, one possible natural continuation is to understand how they are distributed in some area.In this paper we consider the m-introimmune property, and the existence of mintroimmune sets in the class Π 0 1 suggests to study how they are distributed in the partial ordering (R, ≤) of the r.e.Turing degrees, which is one of the most important structures studied in computability theory.The natural way to proceed is to collect all the r.e.Turing degrees containing m-introimmune sets in the substructure I m and to study which properties I m owns.Let us define formally I m .
Definition 1.1 Let I m = (I m , ≤ I m ), where I m = {a ∈ R : a contains a m-introimmune set} and ≤ I m is the restriction of ≤ to I m .From now on we will write simply ≤ instead of ≤ I m .
Since every m-introimmune set is immune (Cintioli & Silvestri, 2003), the Turing degree 0 of recursive sets is not in I m .We know that I m has the maximum element 0 ′ (Cintioli, 2005), and that I m has no minimum element (Cintioli, 2014).In this paper we wonder whether or no I m is an upper semi-lattice.

Method
We know that cohesiveness implies m-introimmunity (Cintioli, 2005), and that each m-introimmune set is immune (Cintioli & Silvestri, 2003).Nevertheless, m-introimmunity is not hereditary under inclusion.Therefore, we cannot apply the Upward Closure Theorem (Jockusch, 1973) to deduce that I m is an upper semi-lattice.
However, the Upward Closure Theorem provides a sufficient condition.Let C be the class of all the cohesive r.e.Turing degrees, which is a proper subclass of I m .Then, by the Upward Closure Theorem of Jockusch (1973) A sufficient condition to establish that (I m , ≤) is an upper semi-lattice is to prove that given any two elements in I m − C their sup is in I m .We give here a partial result, by showing that a portion of (I m − C, ≤) is an upper-semi-lattice.We get this by embedding some uniformly recursive Boolean algebras into ((I m − C) ∪ {0}, ≤) preserving sups and infs.We preserve infs only for technical reasons, namely, to get one-one maps.We obtain the further results that every countable distributive lattice is embeddable into ((I m −C) ∪ {0}, ≤) preserving sups and infs, and that every countable partial ordering is embeddable in (I m − C, ≤).We give in the next section 2 some notations and terminology.In section 3 we prove that the m-introimmunity property is not hereditary under inclusion.In section 4 we give our main theorem on the embeddability of some uniformly recursive Boolean algebras in ((I m − C) ∪ {0}, ≤).We conclude the paper by observing that it is not possible to prove that (I m , ≤) is an upper semi-lattice by the direct sum of the representatives, that is if A and B are m-introimmune and Turing incomparable, then in general it is not true that A ⊕ B is m-introimmune.

Notations
For the terminology we refer to the Soare's book (Soare, 1987).Letter N denotes the set of natural numbers, and all the sets considered in this paper will be subsets of N. For any set A, A = N − A, and deg(A) denotes the Turing degree of the set A. With ⟨•, •⟩ : N 2 → N and ⟨•, •, •⟩ : N 3 → N we denote two recursive bijections.Given two sets A and B, their direct sum is A ⊕ B := {⟨x, 0⟩ : x ∈ A} ∪ {⟨x, 1⟩ : x ∈ B}.For every nonempty set I ⊆ N and for every collection We fix an acceptable numbering φ 0 , φ 1 , . . . of all the Turing computable unary functions.W 0 , W 1 , . . . is the corresponding enumeration of all the recursively enumerable (r.e.) sets.For every e ∈ N and every set X ⊆ N, let φ X e be the unary function computed by the oracle Turing machine whose gödelian is e, with the aid of the oracle X.For every numbers e, s, x and for every oracle X we define φ X e,s (x) := φ X e (x) if there exists t ≤ s such that the e-th oracle Turing machine on input x with oracle X halts in exactly t steps; in this case we say that φ X e,s (x) is defined and we write φ X e,s (x) ↓; we say that φ X e,s (x) is undefined otherwise.We recall that the computation of φ X e,s (x) asks the oracle X only numbers less than s, that is s is an upper bound of the use function u(X; , e, x, s) = the maximum element asked to X in the computation of φ X e,s (x).Finally, given two sets A, B ⊆ N, A is many-one reducible to B, in short A ≤ m B, if there exists a recursive function f : N → N such that for every x ∈ N, x ∈ A if and only if f (x) ∈ B.

No hereditary under Inclusion
We recall that a property P of infinite sets is hereditary under inclusion if for every infinite set A, if A has property P then each infinite subset of A has property P. In (Jockusch, 1973), Jockusch proved that: "If P is any property of infinite sets which is hereditary under inclusion and enjoyed by some arithmetical set, then the class of P-degrees is closed upward".So, if the m-introimmunity property was hereditary under inclusion, then (I m , ≤) would be an upper semi-lattice.However, the m-introimmunity property is not hereditary under inclusion.
Theorem 3.1 The m-introimmunity property is not hereditary under inclusion.
Proof We construct an r.e.set A with A m-introimmune and containing an infinite not m-introimmune set.We describe the strategy of the construction.

Strategy
It is enough to ensure the existence of an infinite set B such that B ⊕ B ⊆ A. B ⊕ B is not m-introimmune because the recursive function f (⟨x, 0⟩) = ⟨x, 1⟩ and f (⟨x, i⟩) = ⟨x, i⟩ for every i 0 is an m-reduction of B ⊕ B to its co-infinite subset {⟨x, 1⟩ : x ∈ B}.The set A will be constructed by the finite-extension method with finite injury.At every stage s ≥ 0 we will define the finite set A s .The construction ensures that A s ⊆ A s+1 for every stage s ≥ 0, and A = ∪ s≥0 A s .It is enough to meet the following requirements, for every e ≥ 0.
The fulfillment of all the requirements {N 3e+1 } e≥0 guarantees the existence of an infinite set B with B ⊕ B ⊆ A, namely, the set B = {x : ⟨x, 0⟩, ⟨x, 1⟩ ∈ A}.
The fulfillment of all the requirements {P 3e } e≥0 and |A| = ∞ guarantees the immunity of A. Immunity of A together with the fulfillment of all the requirements {N 3e+2 } e≥0 imply A m-introimmune, as stated in the following lemma.
Lemma 3.2 If all the requirements {N 3e+2 } e≥0 are satisfied and A is immune, then A is m-introimmune.
Proof For the sake of contradiction, let us suppose that A is not m-introimmune, and let φ e be a recursive function that m-reduces A to some its subset X with |A − X| = ∞.Then, the set {φ e (x) : φ e (x) x ∧ x ∈ A} is infinite, as stated in the following lemma.
Lemma 3.3 (Cintioli, 2011) Let C be immune, and By hypothesis N 3e+2 is satisfied; since {φ e (x) : φ e (x) x ∧ x ∈ A} is infinite, it follows that there is an element u ∈ A such that φ e (u) A, hence φ e does not m-reduce A to X, which is a contradiction.This concludes the proof of Lemma 3.2.

Requirements Requiring Attention and Active Requirements
The construction of our set A will be by infinitely many stages.At each stage s we try to satisfy one requirement, if possible, among those not yet satisfied at stage s.From now on letter R will denote any requirement.At every stage s there will be a set of requirements requiring attention; we call a requirement R m active at stage s if m is the minimum index such that R m requires attention at s. Throughout the construction we will use a restraint function r : N 2 → N.For every negative requirement N q , r(q, s) = −1 means that the requirement is not yet satisfied at stage s.We will not use the restraint function for any positive requirement P 3e , therefore we will pose r(3e, s) = −1 for every e, s ≥ 0. We now define formally when a requirement requires attention.For every e, s ≥ 0, we say that • requirement P 3e requires attention at stage s + 1 ≥ 3e if W e,s ∩ A s = ∅ and there is an element u ∈ W e,s with u > max{r(n, s) : n < 3e}; • requirement N 3e+1 requires attention at stage s

Actions to Fulfill Requirements
We describe here the actions to fulfill all the requirements.
• To fulfill P 3e we wait for a stage s + 1 ≥ 3e at which P 3e is active w.r.t.some x, and we enumerate the minimum such x in A s+1 .
• To fulfill N 3e+2 , we wait for a stage s + 1 ≥ 3e + 2 at which N 3e+2 is active w.r.t.some x.Then we consider the minimum such x, enumerate φ e,s (x) in A s+1 and restrain x in A s+1 by setting r(3e For all the requirements R m that are not active at stage s + 1 we set r(m, s + 1) = r(m, s).

Injured Requirements
We say that a negative requirement N m is injured at stage s + 1 if an element u ≤ r(m, s) is enumerated in A s+1 ; in this case r(m, s + 1) is settled to −1.For those requirements N m which are not injured at stage s + 1, we set r(m, s + 1) = r(m, s).
• Stage s + 1 > 0. Let A s be the set constructed up to the end of stage s.If there are no requirements requiring attention, then set A s+1 := A s and go to the next stage s + 2. Otherwise let R n 0 be the active requirement, and distinguish the following three cases on n 0 : 1. n 0 = 3e.Pick the minimum x for which P 3e requires attention.Set A s+1 = A s ∪ {x}.

Verifications
The proof that all the requirements require attention at most finitely often does not involve special arguments outside of the finite-extension method with finite injury, so we omit it.We prove only that each N 3e+2 is met; the proofs of the other cases are similar.
Proof Let us suppose that each requirement requires attention at most finitely often, and let us suppose that requirements P 3m and N 3m+1 are met for every m ≥ 0. For the sake of contradiction, let us suppose that some N 3e+2 is not met, and let s 0 be the minimum stage after which no requirement of higher priority than N 3e+2 requires attention.Let X be a subset of A with |A − X| = ∞ and A ≤ m X via the recursive function φ e .It must necessarily be r(3e + 2, s 0 ) = −1, because if it was r(3e + 2, s 0 ) = u > 0 then at some stage s ′ + 1 < s 0 the number φ e,s ′ (u) has been enumerated in A s ′ +1 and N 3e+2 was not injured up to s 0 .Since N 3e+2 will not be injured anymore after s 0 , it would be u ∈ A and φ e (u) A, contrary to the assumption that φ e m-reduces A to X.The set A is immune because all the requirements P 3e and N 3e+1 are met.Thus, by Lemma 3.3 the set {φ e (x) : φ e (x) x ∧ x ∈ A} is infinite.Let s ′′ + 1 ≥ s 0 be the minimum stage such that there exists x ≤ s ′′ + 1 with • φ e,s ′′ (x) ↓ x and • φ e,s ′′ (x) > max{r(n, s ′′ ) : n < 3e + 2}.
At stage s ′′ + 1 N 3e+2 requires attention via x, and by hypothesis is active.If x ′ is the minimum w.r.t.N 3e+2 requires attention, then the number φ e,s ′′ (x ′ ) is enumerated in A s ′′ +1 and x ′ is restrained in A s ′′ +1 by setting r(3e + 2, s ′′ + 1) = x ′ .This action will not be injured anymore, therefore x ′ ∈ A and φ e (x ′ ) A, contrary to the assumption that φ e is a mreduction of A to X.This concludes the proof of Lemma 3.4.

2
This concludes the proof of Theorem 3.1.

Embedding
We know that every cohesive set is m-introimmune (Cintioli, 2005), and that the class of cohesive Turing degrees is upward closed (Jockusch, 1973).Therefore, by taking any two elements a, b ∈ I m of which at least one cohesive, the least upper bound a ⊕ b is in I m .But we do not know if the least upper bound is in I m in case both a and b are not cohesive.Let C be the class of all the cohesive r.e.Turing degrees, which is a proper subclass of I m .We ask if given any two elements We give here a partial answer by showing that every uniformly recursive Boolean algebra of subsets of N having ∅ as minimum is lattice embeddable in ((I m − C) ∪ {0}, ≤).We will obtain also that every countable distributive lattice is embeddable into ((I m −C) ∪ {0}, ≤) preserving sups and infs, and that every countable partial ordering is embeddable in (I m − C, ≤).Let us first fix some terminology.We recall that a lattice L = ({L i } i∈N , ⊆, ∪, ∩) of subsets of N is uniformly recursive if the set {⟨x, i⟩ : x ∈ L i } is recursive.Given a partial ordering (P, ≤ P ) and given a one-one map f : P → I m we say that: • f is an embedding of (P, ≤ P ) into (I m , ≤) if x ≤ P y ⇔ f (x) ≤ f (y) for every x, y ∈ P; • if f is an embedding of (P, ≤ P ) into (I m , ≤) and f preserves sups and infs, that is for every x, y ∈ P, f The strategy to prove that the uniformly recursive Boolean algebras above are lattice embeddable in ((I m − C) ∪ {0}, ≤) is to construct a collection {B i } i∈N of infinite subsets of N in such a way that ⊕ i∈N B i is m-introimmune, co-r.e. and low 1 .The co-r.e.property together with the lowness property guarantee that ⊕ i∈I B i is co-r.e. and not cohesive for every nonempty recursive set I.Then, the m-introimmunity of ⊕ i∈N B i implies the m-introimmunity of ⊕ i∈I B i for every nonempty recursive set I. We prove the latter in the following lemma.

Lemma 4.1 Let
⊕ i∈N A i be m-introimmune, and let I ⊆ N be any recursive set such that Proof For the sake of contradiction, let us suppose that ⊕ i∈I A i is not m-introimmune, and let X ⊆ We get a contradiction by proving that Let b 0 Y be fixed.Then, ( ⊕ ) ≤ m Y via f defined in the following way: a) for every i ∈ I, f (⟨x, i⟩) = g(⟨x, i⟩), if g(⟨x, i⟩) = ⟨y, j⟩ for some j ∈ I; f (⟨x, i⟩) = b 0 otherwise; b) for every i I, f (⟨x, i⟩) = ⟨x, i⟩.
Given any ⟨x, k⟩ ∈ N let us prove that ⟨x, k⟩ ∈ (3) • (⇒) Let us suppose that ⟨x, k⟩ ∈ ⊕ i∈N A i ; we have two cases: k ∈ I and k I. Proof We show that there is a one-one map f : The strategy of the proof is the following.

Strategy
We construct a collection {B i } i∈N of infinite sets such that the set ⊕ i∈N B i will have the following properties: is low 1 , co-r.e., m-introimmune, and for every α i , α j of B, for every set D, Then, we define the map f as for every α of B. We have to show that f is the desired function satisfying all the conditions (a), (b) and (c).
Proof From the fact that ⊕ i∈N B i is low 1 and co-r.e. it follows that ⊕ i∈N B i is not cohesive, because a cohesive co-r.e.set is necessarily high 1 .Then, for every recursive set α it holds that ⊕ i∈α B i is co-r.e. and Turing reducible to ⊕ i∈α B i is low 1 , hence not cohesive.Moreover, by Lemma 4.1 it holds that for every recursive nonempty set α, ⊕ i∈α B i is m-introimmune.In conclusion, for every nonempty recursive set α the set ⊕ i∈α B i is not cohesive, co-r.e. an m-introimmune, therefore its Turing degree deg ( ⊕ This concludes the proof of the claim.
, which is true by the following known argument.Without loss of generality, let us suppose that α m α n , and for the sake of contradiction let us assume that ) and by hypothesis ( ⊕ . Therefore, by ( 5) and ( 6) Proof For the condition (a), (⇒) is obvious, while the argument of the proof of Claim 4.4 proves (⇐).Condition (b) is obvious.Condition (c) follows form ( 5), ( 6) and by the obvious facts that This concludes the proof of the claim.

2
We now describe the construction of the collection {B i } i∈N .We will construct an infinite and co-infinite low 1 r.e.set A in such a way that A = ⊕ i∈N B i , with A m-introimmune.The construction is by infinitely many stages; at every stage s we will define the finite set A s , with A 0 = ∅ and A s ⊆ A s+1 for every s ≥ 0. The final set is A = ∪ s≥0 A s .For every i, s ∈ N, let Then, for every i, s ∈ N is , where B i = lim s→∞ B i,s .To ensure that the set A = ⊕ i∈N B i will have the required properties, it suffices to construct A in order to satisfy the following requirements for every e, m, n, i ∈ N.

Requirements
For every natural numbers e, i, m, n ≥ 0 we define the following requirements: -N 5⟨e,m,n⟩+4 : We show that all the above requirements imply all the requested properties for A = ⊕ i∈N B i .Lemma 4.6 Requirements P 5e , N 5⟨i,e⟩+3 and N 5⟨i,e⟩+3 guarantee that A is m-introimmune.
The proof of this lemma is essentially the same of that of Lemma 3.2, so we omit it.
Lemma 4.7 Requirements N 5e+2 guarantee that A is low 1 .
Proof The statement follows from the known proposition.
Proof The natural requirement to get (5) and ( 6) is N 5⟨a,b,m,n⟩+4 : However, under the hypothesis that all the requirements P 5e , N 5e+1 , N 5⟨i,e⟩+3 and N 5⟨e,m,n⟩+4 are satisfied, it follows that all the requirements (7) are satisfied.The proof is an argument due to Posner (see Soare, (1987) and let For the sake of contradiction, assume that there are i 0 and j 0 such that Let e = e(x 0 , i 0 , j 0 ) be such that for every oracle X and every Then clearly φ , contrary to the hypothesis that N 5⟨e,m,n⟩+4 is met.

Requirements Requiring Attention and Active Requirements
In this section we formulate the conditions under which a requirement requires attention.Similarly to the previous Theorem 3.1, we use a restraint function r : N 2 → N.For every negative requirement N q with q 5n + 4 for every n, r(q, s) = −1 means that the requirement is not yet satisfied at the stage s.We do not need the use of the restraint function for the positive requirements P 5e , therefore we will pose r(5e, s) = −1 for every e, s ≥ 0. For every e, i ≥ 0, • P 5e requires attention at stage s + 1 ≥ 5e if W e,s ∩ A s = ∅ and there exists x ∈ W e,s such that x > max{r(n, s) : n < 5e}; • N 5e+1 requires attention at stage s+1 ≥ 5e+1 if r(5e+1, s) = −1 and there exists x ∈ A s , x ≤ s+1, with φ e,s (x) ↓ x and φ e,s (x) > max{r(n, s) : n < 5e + 1}; • N 5e+2 requires attention at stage s + 1 ≥ 5e + 2 if φ A s e,s (e) ↓ and r(5e + 2, s) = −1; • N 5⟨i,e⟩+3 requires attention at stage s + 1 ≥ 5⟨i, e⟩ + 3 if r(5⟨i, e⟩ + 3, s) = −1.
We say that a requirement R m above is active at stage s + 1 if m is the minimum for which R m requires attention at stage s + 1.

Actions to Fulfill Requirements
We describe here the actions to fulfill requirements.Let e, i ≥ 0.
• For P 5e : we wait for a stage s + 1 ≥ 5e at which it is active w.r.t. to some x; then we enumerate the minimum such x in A s+1 .
• For N 5e+1 : we wait for a stage s + 1 ≥ 5e + 1 at which it is active w.r.t.some x.Then we consider the minimum such x, enumerate φ e,s (x) in A s+1 and restrain x in A s+1 by setting r(5e + 1, s + 1) = x.
• For N 5e+2 : we employ the classical method of to obtain a low 1 set, namely, we wait for a stage s + 1 ≥ 5e + 2 such that it is is active.Then we preserve the computation φ A s e,s (e) by restraining all the queries made to the oracle by setting r(5e + 2, s + 1) = s.
• For N 5⟨i,e⟩+3 : we wait for a stage s + 1 ≥ 5⟨i, e⟩ + 3 at which it is active and we restrain ⟨x, i⟩ in A s+1 by setting r(5⟨i, e⟩ + 3, s + 1) = ⟨x, i⟩, where x is the minimum number such that ⟨x, i⟩ ∈ A s and x ≥ e.
• For N 5⟨e,m,n⟩+4 : we use the generalization of the Yates' minimal-pair method due to Thomason (Thomason, 1971), Lerman and Lachlan (see Soare, (1987), page 157 et seq.).At every stage s of the construction of A we define the length function l (⟨e, m, n⟩, s) for every e, m, n ∈ N as: Then, at every stage s ≥ 0 we define the restraint function r(5⟨e, m, n⟩ + 4, s) by induction on ⟨e, m, n⟩ according to the minimal-pair method, where we recall that the number s is an upper bound of the use function.
is a 0-expansionary stage max{t : t < s and t is a 0-expansionary stage} otherwise, where a stage s is 0-expansionary if either s = 0 or l(0, s) > max{l(0, t) : t < s}.
Given r(5⟨e, m, n⟩ + 4, s), define r(5(⟨e, m, n⟩ + 1) + 4, s) as the maximum of: (1) r(5⟨e, m, n⟩ + 4, s), • For N 5e+1 , the proof is very similar to that of the previous Lemma 3.4.Let s 0 be the minimum stage such that for every s ≥ s 0 no requirement of higher priority than N 5e+1 requires attention at stage s.For the sake of contradiction, let us suppose that N 5e+1 is not met.So, there is a subset X of A with |A − X| = ∞ and A ≤ m X via φ e .Then necessarily it has to be r(5e + 1, s 0 ) = −1.In fact, if it were r(5e + 1, s 0 ) = u > −1 for some u, then there is a stage s ′ + 1 < s 0 at which the number φ e,s ′ (u) has been enumerated into A s ′ +1 and N 5e+1 was not injured up to s 0 .Since by hypothesis N 5e+1 will not be injured anymore after s 0 , it would be u ∈ A and φ e (u) X, contrary to the assumption that φ e m-reduces A to X. Let k 5e+1 = lim inf s→∞ (max{r(i, s) which is finite by Corollary 4.12.Requirements P 5e and N 5⟨i,e⟩+2 are met for every e, i ≥ 0, therefore the set A is immune.By Lemma 3.3 the set {φ e (x) : φ e (x) x ∧ x ∈ A} is infinite.By (11) there are infinitely many stages s such that max{r(t, s) : t < 5e + 1} = k 5e+1 .
(13) Let x ′ be the minimum x satisfying (13).This means that N 5e+1 requires attention at stage s ′ + 1 via x ′ , and N 5e+1 has the highest priority by hypothesis, that is N 5e+1 is active.At stage s ′ + 1 the number φ e,s ′ (x ′ ) is enumerated into A s ′ +1 and r(5e + 1, s ′ + 1) = x ′ .After stage s ′ + 1 N 5e+1 will not be injured anymore, therefore x ′ ∈ A and φ e (x ′ ) A, contrary to the assumption that A ≤ m X via φ e .
• Now, let us prove that each N 5⟨e,m,n⟩+4 is met.Let e, m, n be fixed, and let us suppose that both φ We have to prove that D can be decided with the aid of the oracle ⊕ i∈α m ∩α n B i .First, we fix some terminology and parameters.For every set X and every natural number n, X n denotes the set X ∩ {0, 1, . .Let s 0 be the minimum stage after which no requirement of higher priority than N 5⟨e,m,n⟩+4 is active after s 0 and such that, by (15), for every s ≥ s 0 r(5(⟨e, m, n⟩ − 1) + 4, s) ≥ k. (17) Finally, for every x ∈ N, we say that the computation φ where u α m = u( ⊕ i∈α m B i,s ; e, x, s) < s.
Similarly, we define the computation φ , given any two elements a, b ∈ I m of which at least one in C, the least upper bound a ⊕ b exists in C ⊂ I m .But we do not know if the least upper bound exists in I m in case both a and b are not in C.
which implies B h recursive.But by Lemma 4.1 for I = {h} it follows that ⊕ i∈{h} B i is m-introimmune, which implies B h m-introimmune, hence nonrecursive, a contradiction.This concludes the proof of the claim. 2 Claim 4.5 Conditions (a), (b) and (c) hold.