Definability in the Enumeration Degrees

Definability in the Enumeration Degrees Theodore A. Slaman W. Hugh Woodin Abstract We prove that every countable relation on the enumeration degrees, E, is uniformly definable from parameters in E. Consequently, the first order theory of E is recursively isomorphic to the second order theory of arithmetic. By an effective version of coding lemma, we show that the first order theory of the enumeration degrees of the Σ 0 2 sets is not decidable. 1 Introduction Definition 1.1 The following terms specify the variations on relative enumerability. An recursive enumeration procedure U is a recursively enumerable collection of pairs a, b in which a and b are finite subsets of N. U(A) is equal to B if B = {n : ( a)( b)[a A and a, b U and n b]}. Say that B is enumeration reducible to A (A e B) if there is a recursive enumeration procedure U such that U(A) = B. Similarly, A is enumeration equivalent to B (A e B) if A e B and B e A. Definition 1.2 The enumeration degrees are defined as follows. The enumeration degree of A is its e equivalence class. E = 2 N / e, e is the partial ordering of the enumeration degrees. The reader might consult (Cooper 1990) for an introduction to the enumeration degrees. Notation. We will use upper case letters such as A and B to denote sets of natural numbers. We will use lower case boldface letters such as a and b to denote their enumeration degrees. We let A B denote the recursive join of A and B and let a b denote its enumeration degree. Note that a b is the least upper bound of the pair {a, b} in E and hence the binary function is definable in E. Slaman was partially supported by NSF Grant DMS 91-06714 and SERC Visiting Fellowship Research Grant ( Leeds Recursion Theory Year 1993/94 ) No. GR/H 91213. Woodin was supported by NSF Grant DMS. 1

2 The Coding Theorem This section is devoted to proving Theorem 2.11, the Coding Theorem for E. In a precise way, this theorem states that every countable relation on E is uniformly definable from parameters in E. It s proof follows the general line of the proof of the Coding Theorem for the Turing degrees; see (Slaman and Woodin 1986). We approach the final result by a sequence of lemmas. Our first step is to code countable antichains. Definition 2.1 Let (a) denote the principal ideal in E determined by a (a) = {x : a e x}. Let (A) denote the set of representatives of elements of (a). Lemma 2.2 Suppose that R is a countable antichain in E. There are reals B, F and G such that the elements of R are exactly the e-degrees of the e -minimal elements of the set {Y : Y e B and (Y F ) (Y G) (Y )}. Proof: Let R = R i : i N be a sequence of representatives for the elements of R. Let B be any upper bound on the elements of R. We find F and G by forcing. Definition 2.3 The forcing partial order P = P, P is defined as follows. A condition p in P consists of a finite initial segment R p = R 1,..., R lp of R; two sequences of finite enumeration procedures F p = F i,p : i l p and G p = G i,p : i l p such that for each i, the finite sets F i,p (R i ) and G i,p (R i ) enumerated by F i,p and G i,p relative to R i are equal; an integer k p, used as a parameter to limit the extensions of p. If p and q are conditions, q extends p (p P q) if R p is an initial segment of R q ; for each i less than or equal to the length of R p, F i,p F i,q and G i,p G i,q ; for each i, if a, b F i,q F i,p or a, b G i,q G i,p then the minimal element of b is greater than k p ; k q is greater than or equal to k p. Given a P-generic set G, let F i,g and G i,g be the unions of F i,p and G i,p such that p G. Then F i,g (R i ) and G i,g (R i ) are the sets enumerated by F i,g and G i,g relative to R i. For each i, F i,g (R i ) and G i,g (R i ) are equal. Further, this set is Cohen generic relative to B. Definition 2.4 If F is a sequence of enumeration procedures then F denotes its uniform recursive join of F. 2

In particular, F G and G G are the uniform recursive joins of F i,g : i N and G i,g : i N. For each i, (R i F G ) (R i G G ) (R i ). Now we check that the parameters B, F G and G G satisfy the remaining conditions of Lemma 2.2. Suppose that Y is enumeration reducible to B and that U and V are recursive enumeration procedures. We must show that either U(Y F G ) is not equal to V (Y G G ), their common value is enumeration reducible to Y or there is an i such that R i is enumeration reducible to Y. Definition 2.5 Let p be a condition. Say that p is equal to R p, F p, G p, k p. Let l p be the length of R p. A finite sequence of finite enumeration procedures F is compatible with F p, k p if for each i less than or equal to l p, F i extends F i,p and for each i less than or equal to the length of F, if a, b F i F i,p then b is contained in the numbers greater than k p. Definition 2.6 Suppose that p P q and F is compatible with F p, k p. The amalgamation of q with F, q F, is the condition defined as follows. R q F is the sequence R 1,..., R l where l is the maximum of the the lengths of R q and F. For each i less than or equal to l, F i,q F is equal to F i,q F i, if i is less than or equal to the length of R q, and equal to F i, otherwise. For each i less than or equal to l, G i,q F is equal to G i,q (F i F i,q ), if i is less than or equal to the length of R q, and equal to F i, otherwise. k q F is equal to k q. Given p and q as above and a finite sequence G which is compatible with G p, k p, we define the amalgamation q G similarly, with the roles of F and G reversed. If F is compatible with F p, k p and q is stronger than p then q F is stronger than p. However, q F may not be stronger than q since F may include new ways to enumerate numbers less than k q. Conversely, if r extends p then F r is compatible with F p, k p. Note that the set of finite sequences of finite enumeration procedures which are compatible with p is recursive. The three cases of Lemma 2.2 occur as follows. We have an inequality between U(Y F G ) and V (Y G G ) if and only if there is an n and a condition in G that forces U(n, Y F G ) V (n, Y G G ). Since this case satisfies the claim of Lemma 2.2, assume that there is no such condition in G. We will conclude that the common value of U(Y F G ) and V (Y G G ) is reducible to Y in the case when there is a condition in G which decides U(Y F G ) at every argument. Suppose that p is such a condition. Then, for any n, n is an element of U(Y F G ) if and only if p n U(Y F G ). Fixing n, if there is an F such that F is compatible with F p, k p and n U(Y F ) then p cannot force n out of U(Y F G ). (Since n is in U(Y F p F ).) In this case, n must be in U(Y F G ). Similarly, if there is no such F then p cannot force n U(Y F G ) and so n is not in U(Y F G ). Thus, Y can enumerate n U(Y F G ) by finding a F that is compatible with F p, k p such that n U(Y F ). In this case, U(Y F G ) e Y. We complete the argument by further assuming that there is no condition in G which decides all of the values of U(Y F G ). We must show that there is an i such that R i e Y. Claim 2.7 Suppose that n is a number and q 0 and r 0 extensions of p such that the following conditions hold. R r0 and R q0 have the same length l 0. 3

For every i less than or equal to l 0, F i,r0 F i,q0 and G i,r0 G i,q0. k r0 is equal to k q0. n U(Y F q0 ) and r n U(Y F G ). Then there are conditions q and r extending p such that the following conditions hold. R q and R r have the same length l. There is an i less than or equal to l such that for all j less then or equal to l, if j i then F j,r = F j,q and G j,r = G j,q, there is a pair a, b such that F i,q = F i,r { a, b } and G i,q = G i,r { a, b }. k r is equal to k q. n U(Y F q ) and r n U(Y F G ). Further, for all i less than or equal to l 0, F i,r0 F i,r and G i,r0 G i,r ; consequently, F i,r0 F i,q and G i,r0 G i,q. Proof: For each j less than or equal to l 0, let a, b j,1,..., a, b j,mj be an enumeration of F j,q0 F j,r0. We may assume that k r0 is greater then the maximum number which appears in any b j,m. We work our way by recursion from r 0 toward q 0 as follows. Suppose that r i = R r0, F ri, G ri, k r is given and that r i n U(Y F G ). Proceeding lexicographically on subscript, let a, b j,m be the least pair which is not in F j,ri. Define F i = F 1,ri,..., F j,ri { a, b j,m },..., F n,ri. Consider the condition r i F i. If r i F i n U(Y F G ) then define r i+1 to be r i F i and proceed with the recursion. Otherwise, extend r i F i to a condition q such that n U(Y F q ). And then, let F r equal F 1,qi,..., F j,q { a, b j,m },..., F n,q, G r equal G 1,qi,..., G j,q { a, b j,m },..., G n,q and r equal R r0, F r, G r, k r0. Now, r is a stronger condition than r 0 and so r n U(Y F G ). By construction, n U(Y F q ). These two conditions satisfy the claim. Now, we claim that there must be an i such that the second case holds for r i. If not, then at the last step of the recursion we would produce a condition r such that F q0 = F r and r n U(Y F G ). This is impossible since k U(Y F q0 ). Since p does not decide every element of U(Y F G ) there are n, q 0 and r 0 extending p such that q 0 n U(Y F G ) and r 0 n U(Y F G ). Since it true of some extension of q 0, we can assume that n is an element of U(Y F q0 ). We may assume that R r0 is equal to R q0. Since enumeration procedures are positive, we may also assume that for each j less than or equal to l r0, the length of R r0, F j,r0 F j,q0. Finally, we may assume that k r0 is equal to k q0. In other words, there are a number n and a pair of conditions stronger than p which satisfy the hypotheses of Claim 2.7. Let i, q and r be fixed as in the conclusion of Claim 2.7. Since both q and r extend p and p U(Y F G ) = V (Y G G ), q n V (Y G G ) and r n V (Y G G ). Let a, b be the pair such that F i,q = F i,r { a, b }. Note, that if a were not a subset of R i then substituting F i,r { a, b } for F i,r in r would produce a condition s extending p. The additional axiom in F i,s would not contradict the fact that F i,s (R i ) = G i,r (R i ) since it would not apply to R i. But then s would be a condition extending p such that n U(Y F s ) but there is no extension of G s which is compatible with G s, k s and enumerates n into V (Y G G ). In other words, we would have s U(Y F G ) V (Y G G ), a contradiction. 4

Let k be the maximum of k r and k q. For m N, let G r i m, b be the sequence obtained from G r by replacing G i,r with G i,r { m, b }. If m is in R i then s m = R, F q, G r i m, b, k is a condition stronger than p. Since n U(Y F q ), s m forces n U(Y F G ) and so, s m forces n V (Y G G ). Thus, there is a G that is compatible with G r i m, b, k such that n V (Y G). This is a positive Σ 0 1(Y ) property which is satisfied by every element of R i. On the other hand, suppose that m is not in R i. Then R, F r, G r i m, b, k is also an extension of p and this condition forces n U(Y F G ). Thus, there cannot be a G that is compatible with G r i m, b, k such that n V (Y G). Thus, the positive Σ 0 1(Y ) property of all the elements of R i is not satisfied by any of the elements of the complement of R i. Hence, R i e Y, as desired. Corollary 2.8 There is a formula ϕ(x, b, f, g) in the language of E such that for any countable antichain A in E there are parameters b, f and g such that for all x, x is in A if and only if E = ϕ(x, b, f, g). Proof: Define ϕ(x, b, f, g) as follows ϕ(x, b, f, g) [ x b and (x) (x f) (x g) and ( w < x)[(w) = (w f) (w g)] ]. If A is a countable antichain then Lemma 2.2 states that there are B, F and G whose degrees define A in E using ϕ. In the next lemma, we document a well known property of mutually Cohen generic reals. Lemma 2.9 Suppose that B is a real and {C i : i N} is a countable set of reals that are mutually Cohen generic with respect to meeting all dense sets that are arithmetic in B. For all X and Y enumeration reducible to B and all i and j in N, if X C i e Y C j then i = j and X e Y. Proof: First suppose that U is a recursive enumeration procedure and p is a Cohen condition such that p U(Y C j ) = C i. Note that, we can decide the values of U(Y C j ) by extending p only on its jth coordinate. Since p is finite, let x be an integer such that p does not decide the value of C i (x). If i were different from j we could extend p on the jth coordinate to decide U(x, Y C j ) and then extend p on the ith coordinate to give C i (x) the opposite value. This condition would force U(Y C j ) C i, contradicting the assumption on p. Secondly, suppose that p V (Y C j ) = X. If p were to have two extensions that force different values for V (Y C j ) then one of these values would disagree with X. Then some condition extending p would force V (Y C j ) X, an impossibility. Thus, for any argument n, every extension of of p must force the same value for V (n, Y C j ), namely X(n). Consequently, we may enumerate n X relative to Y by finding a finite subset of Y and a finite set compatible with the jth coordinate of p which would be sufficient to enumerate x into V (Y C j ). Thus, X e Y. In the following lemma, we let R denote the cardinality of R. Lemma 2.10 Suppose that R is a countable subset of E, B is a e -upper bound on the representatives of elements of R, and {C i : i R } is a set of reals which are mutually Cohen generic relative to all dense sets which are arithmetic in B. Let C be the set of enumeration degrees represented in {C i : i R } and let ψ be a bijection, mapping R to C. Then, ψ, C and R are uniformly definable from parameters in E. Proof: Let b denote the enumeration degree of B. By Lemma 2.9, C is an antichain. By Corollary 2.8, it is definable from parameters in E. Let R + C be the set {x ψ(x) : x R}. Applying the same lemmas, R + C is an antichain and so is definable from parameters in E. 5

Apply Lemma 2.9 once more. For any x below b, x is in R if and only if there is a c in C such that x c is in R + C. This gives a definition of R from parameters in E. Finally, ψ is definable from parameters in E since ψ(x) = c if and only if x R, c C and x c R + C. Theorem 2.11 For each n there is a formula ϕ(x 1,..., x n, y 1,..., y m ) such that for each countable n-place relation R on E there is a sequence of parameters a 1,..., a m such that R is defined by ϕ(x 1,..., x n, a 1,..., a m ) in E. Proof: Suppose that R is a countable subset of E n. For each m smaller than n, let R(m) be defined by R(m) = {x : ( x 1,..., x n R)(x m = x)}. Let b be an upper bound on the union of the R(m). Let B be a representative of b. Let C be the enumeration degrees of a set of reals that are mutually Cohen generic with regard to meeting all of the dense sets arithmetic in B so that C as a disjoint union of sets C(m), each of which has the same cardinality as the set of enumeration degrees R(m). Fix bijections ψ m between R(m) and C(m). By Lemma 2.10 each ψ m, R(m), C(m) is definable from parameters in E. Define A by A = { c 1 c 2... c n : ψ 1 1 (c 1), ψ 1 2 (c 2),..., ψ 1 n (c n ) R }. By Lemma 2.10 A is definable from parameters in E. Further, by Lemma 2.9, for each element a of A there is a unique sequence c 1,..., c n of elements from C such that a = c 1... c n. Now, { ( ) } ( m n) x R = x 1, x 2,..., x n : m R(m) and. ψ 1 (x 1 ) ψ 2 (x 2 )... ψ n (x n ) A gives the desired definition of R in E. The required uniformity in the form of the definition follows from the uniformity of Corollary 2.8 and Lemma 2.10. Corollary 2.12 The first order theory of E is recursively isomorphic to the second order theory of arithmetic. Proof: The usual second order characterization of a standard model N of arithmetic involves specifying a countable set N, a distinguished element 0 N, and a unary function s N, such that N = N, 0 N, s N satisfies second order induction. The terms of this characterization and all of the objects which appear in the second order theory of N can be represented by countable relations on E; the second order variables over N are interpreted by first order variables over the degrees using the translation provided by Theorem 2.11. Question 2.13 Is the -theory of E decidable? 3 The Enumeration Degrees of the Σ 0 2 Sets In this section, we will implement enough of the machinery in the proof of Theorem 2.11 to conclude the undecidability of the first order theory of the enumeration degrees of the Σ 0 2 sets of integers. Let E(Σ 0 2) denote this structure. The desired undecidability result is a consequence of the following technical lemma, which shows that each of a sufficient collection of effectively presented antichains can be defined from parameters within E(Σ 0 2). 6

3.1 Coding Effectively Presented Antichains Proposition 3.1 Suppose that B is low (i.e., B = 0 ) and R = R i : i N is a sequence of e - incomparable reals which is uniformly recursive in B. Let R be the enumeration degrees represented in R. Then there are Σ 0 2 enumeration degrees b, f, and g such that R is the collection minimal elements of {r : r e b and (r f) (r g) (r)}. Proof: Let P be the forcing introduced in Definition 2.3 relative to R and B. In the proof of Lemma 2.2, we used a P generic set G to define sets F G and G G whose enumeration degrees satisfy the conditions of Proposition 3.1. By a priority argument relative to 0, we will enumerate F G and G G. Our construction is organized by a recursion on stages. At the end of each stage s, we will specify p[s], a condition from P. While these conditions need not be compatible, we will ensure that if s > t then F i,p[s] F i,p[t] and G i,p[s] G i,p[t]. (We adopt this notation as the conditions produced by our strategies in their limit states are compatible and determine a filter G.) Thus, F G and G G, the unions of F p[s] and G p[s] over all stages s, will be Σ 0 2 sets. By the definition of P, F G and G G will be uniform recursive joins of enumeration procedures F i,g and G i,g. For each i, the definition of P ensures that F i,g (R i ) = G i,g (R i ). Requirements. We divide conditions to be satisfied by F G and G G into the following list of requirements. For each i, n, enumeration procedure U and Y e B, there is a condition p in P which applies to F G and G G such that R i R p, either n U(Y F p ) or p n U(Y F p ). We call this requirement G(i, n, U, Y ). For each enumeration procedure W and each i, W (R i ) F i,g (R i ). Call this requirement D(i, W ). For each Y e B and each pair of enumeration procedures U and V, one of the following conditions holds U(Y F G ) e Y, ( i)[r i e Y ], U(Y F G ) V (Y G G ). Call this requirement M(Y, U, V ). In the previous section we showed that each of the requirements is forced by the trivial condition in P. Our problem is that the dense sets which would have to be met by a generic set in order to conclude that the requirements are satisfied are not recursive in 0. In particular, p U(Y F G ) = V (Y G G ) is a Π 0 2 property of p. Since we wish to produce Σ 0 2 sets, we cannot merely point to the most natural F and G, those belonging to a sufficiently generic G. The first requirement is that the G we construct is sufficiently generic so that forcing atomic statements about sets enumeration reducible to Y F G or to Y G G is equivalent to their being true. The second requirement ensures that for each i, (R i F G ) (R i G G ) is not equal to (R i ). The final requirement ensures that for every Y enumeration reducible to B, if (Y F G ) (Y G G ) is not equal to (Y ) then there is an i such that Y e R i. Thus, if the requirements are satisfied then the elements of R will be the e -minimal solutions below B to (Y F G ) (Y G G ) (Y ). So it is sufficient to satisfy the requirements in order to prove Proposition 3.1. We now describe the strategies for our priority construction. 7

Genericity: g(i, n, U, Y ). Suppose that p is given. It is trivial to extend p to a condition q(p) so that R i is an element of R q. Say that l p is less than i. Set R q(p) equal to R 1,..., R i. For j less than or equal to l p, set F j,q(p) = F j,p and G j,q(p) = G j,p. For each j greater than m and less than or equal to i, set F j,q(p) and G j,q(p) equal to the empty set. Define k q(p) to be k p. Now we extend q(p) to r(p) to decide whether n is an element of U(Y F G ). Recall that for each Y and n, q(p) n U(Y F G ) if and only if there is no F which is compatible with F q(p), k q(p) such that n U(Y F ). If q(p) n U(Y F G ) then let r(p) equal q(p). If not, then find F compatible with F q(p), k q(p) such that n U(Y F ). Define r(p) to be q(p) F, the amalgamation of q(p) with F. By construction, n U(Y F r(p) ). In either case, p P q(p) P r(p). The condition under which r(p) is equal to q(p) is Π 0 1(B). If this condition fails then the search for a r(p) as in the second case is recursive in B, as Y e B. Since B is low, r(p) can be found effectively in 0. The strategy g(i, n, U, Y ) operates as follows during stage s + 1. 1. Let p[s + 1] be r(p[s]). Go to 2. 2. Take no further action. Let D(i, n, U) be the dense set of conditions {r : ( p P )[r(p) P r]}. Diagonalization: d(i, W ). Similarly, we can effectively meet the dense subsets of P which ensure that for all R i and all recursive enumeration procedures W, W (R i ) F i (R i ). Let p[s] be given. The strategy d(i, W ) operates as follows during stage s. 1. Choose an n which is greater than k p so that n F i,p[s] (R i ). If n W (R i ) then form p[s+1] by letting F i,p[s+1] = F i,p[s] {, {n} }, G i,p[s+1] = G i,p[s] {, {n} } and otherwise keeping p[s + 1] identical with p[s]. Otherwise, form p[s + 1] by letting k p[s+1] = n + 1 and in all other respects keeping p[s + 1] identical with p[s]. Go to 2. 2. Take no further action. Here, d(i, W ) is a direct diagonalization that can be recursively implemented relative to 0. Minimality: M(Y, U, V ). follows during stage s. We satisfy M(Y, U, V ) by the finite injury strategy m(y, U, V ), which acts as 1. Let s 0 be the current stage s + 1. Define p[s + 1] to be p[s]. Go to 2. 2. (a) If there is an n less than or equal to s such that p[s] n U(Y F G ) and p[s 0 ] n U(Y F G ) then go to 3. If d(y, U, V ) stays in this state permanently then Y can correctly enumerate the condition n U(Y F G ) by looking for a F which is compatible with F p [s 0 ], k p [s 0 ] for which n U(Y F ). (b) Otherwise, end the stage s activity for d(y, U, V ). 3. Let s 1 be the current stage. Let F be compatible with F p[s0], k p[s0] with n U(Y F ) and p[s] n U(Y F G ). Let l be the length of F. Let r and q be the conditions obtained by applying Claim 2.7 with p given by p[s 0 ], q 0 given by p[s] F and r 0 given by p[s]. Then both q and r extend p[s 0 ] in P; F p[s] F r and G p[s] G r ; q is different from r only in that F i,q = F i,r { a, b } and G i,q = G i,r { a, b }; r n U(Y F G ); and n U(Y F q ). Let p[s 1 ] equal r. (If s 1 = s 0 = s + 1, this may involve changing the value of p[s + 1].) At the next stage, go to 4. Note, p[s 0 ] P p[s 1 ]. 4. (a) If there is an m less than or equal to s such that m is not in R i and there is a G which is compatible with G p[s1] i m, b, k p[s1] such that n V (Y G) then go to 5a. 8

(b) If there is an m less than or equal to s such that m is an element of R i and R p[s], F p[s] i a, b, G p[s] i m, b, k p[s] forces n V (Y G G ) then go to 5b (c) Otherwise, end the stage s activity of m(y, U, V ). 5. (a) Choose G as in 4. Define p[s + 1] so that R p[s+1] is equal to R 1,..., R l where l is the maximum of the lengths of R p[s] and G. For each j less than or equal to l, define F i,p[s+1] by if j i and F j,p[s] (G j G j,p[s], j l p[s] ; F j,p[s+1] = F i,p[s] [G i (G i,p[s] { {m}, b })], if j = i; G j, otherwise. For each j less than or equal to l, define G i,p[s+1] by G j,p[s] G j, if j i and j l p[s] ; G j,p[s+1] = G j, otherwise. Define k p[s+1] to be equal to k p[s]. Here, there is no F compatible with F p[s+1], k p[s+1] such that n U(Y F ) and n V (Y G). In this outcome, n U(Y F G ) and n V (Y G G ). (b) Define p[s + 1] to be R p[s], F p[s] i a, b, G p[s] i m, b, k p[s]. Here, there is no G compatible with G p[s+1], k p[s+1] such that n V (Y G) and n U(Y F ). In this outcome, n V (Y G G ) and n U(Y F G ). Take no further action for the remainder of the construction. We first note that the conditions under which m(y, U, V ) acts and the actions that it takes are uniformly recursive in 0. So it is appropriate to use m(y, U, V ) in a priority construction relative to 0. There are three cases in which m(y, U, V ) determines p[s + 1]. In Step 1, m(y, U, V ) sets the default value for p[s + 1] to be p[s]. In Step 3, there is an n and a stage s such that p[s] forces n U(Y F G ) but p[s 0 ] does not force n U(Y F G ). In this case, m(y, U, V ) applies Claim 2.7 to produce two conditions q and r which involve extending F p[s] and G p[s] and force incompatible values for U(n, Y F G ). In either Step 5a or 5b, m(y, U, V ) adds an axiom m, b to G i,p[s] and other axioms to F p[s] and G p[s] to obtain a condition which forces U(Y F G ) to be different from V (Y G G ). In each of these cases, F p[s+1] contains F p[s] and G p[s+1] contains G p[s]. Thus, it is appropriate to use m(y, U, V ) in a priority construction in which F G and G G are recursively enumerated relative to 0. The construction. We let G(i, n, U, Y ) 1, G(i, n, U, Y ) 2,... ; D(i, W ) 1, D(i, W ) 2,... ; and M(Y, U, V ) 1, M(Y, U, V ) 2,... be recursive orderings of all of the above requirements. We assign priority to the requirements in decreasing order according to the list G(i, n, U, Y ) 1, D(i, W ) 1, M(Y, U, V ) 1, G(i, n, U, Y ) 2, D(i, W ) 2, M(Y, U, V ) 2,.... Our construction goes by recursion on stages. During stage s + 1, we define a condition p[s + 1] and monitor the states of the active strategies. We say that the strategy σ acts during stage s if σ changes from one step to another during stage s. If σ acts during stage s + 1 then we say that each strategy of lower priority is injured. We say that a strategy σ requires attention during stage s + 1 if the instructions of σ interpreted on input p[s] require it to change state during stage s + 1. 9

We start by setting R p[0], F p[0] and G p[0] to be empty and k p[0] to be 0. Then we define p[0] to be R p[0], F p[0], G p[0], k p[0]. No strategy acts during stage 0. We begin stage s + 1 by letting σ 1, σ 2,..., σ n be the sequence of strategies which have been active during an earlier stage and were not injured in the most recent stage during which they were active, presented in order of priority and in the states they occupied at the end of stage s. We let σ n+1 be the strategy associated with the n + 1st requirement in our priority list. We put σ n+1 in Step 1. We say that σ 1,..., σ n+1 are active during stage s + 1. We proceed in stage s + 1 as follows. We let k be the least index of an active strategy which requires attention. There will be such a strategy since any strategy in Step 1 requires attention and σ n+1 is in Step 1. We follow the instructions in σ k to define p[s + 1]. All of the strategies with index greater than k are said to be injured. We end stage s + 1 with all the strategies of index less than k in the same state as at the end of stage s and with σ k in the state it achieved during stage s + 1. Lemma 3.2 For each requirement S there is a stage s + 1 during which a strategy σ for S is active, the construction respects the instructions in σ during every subsequent stage, and σ is not injured during any stage greater than or equal to s + 1. Further, for every u greater than or equal to s + 1, if t is the greatest element of [s + 1, u] during which σ required attention then p[t] P p[u]. Proof: The proof goes by induction on the priority ordering. At Stage 1, g(i, n, U, Y ) 1 is active. Since there is no strategy of higher priority, g(i, n, U, Y ) 1 can never be injured. By induction, let s be the last stage during which a strategy of higher priority than S requires attention. Since every strategy requires attention when it is made active in Step 1, we cannot activate a strategy of higher priority than S in Step 1 during any stage after stage s. Thus, during stage s + 1, we activate a strategy σ for S. Since no strategy of higher priority requires attention during or after stage s + 1, whenever σ requires attention the instruction within σ is carried out. Finally, σ can only require attention finitely often: once if it is of the form g(i, n, U, Y ) or d(i, W ) and at most three times if it is of the form m(y, U, V ). Once σ has required attention for the last time, its successor will never again be injured. The second claim of the lemma follows from two observations. First, if either g(i, n, U, Y ) or d(i, W ) acts during stage s then p[s + 1] P p[s]. Second, if m(y, U, V ) is in Step 1 during stage s + 1 and acts without being injured during subsequent stages then the conditions that it produces are all below p[s + 1] in P. The sequence of conditions produced by m(y, U, V ) may not be compatible but they are all compatible with the initial condition p[s + 1]. Let F G and G G be the sets enumerated during the above construction. Lemma 3.3 The sets F G and G G satisfy the requirements. Proof: By Lemma 3.2, it is enough to show that once a strategy stops being injured it ensures that its associated requirement is satisfied. Given p[s], g(i, n, U, Y ) produces a condition p[s + 1] which is in D(i, n, U). By Lemma 3.2, for every t greater than s + 1, p[t] extends p[s + 1]. Thus, F G and G G satisfy G(i, n, U, Y ). Given p[s], the strategy d(i, W ) produces a condition p[s+1] such that p[s+1] forces W (n, R i ) F i (n, R i ). The inequality is achieved in one of two ways. If n W (R i ) then n F i,p[s+1] (R i ). No axiom is ever removed from F i so we may conlude that if n W (R i ) then n F i,g (R i ). Similarly, if n W (R i ) then n F i,p[s+1] (R i ) and k p[s+1] is greater than n. By Lemma 3.2, the strategies of lower priority will only act to produce conditions which extend p[s + 1] in P. Thus, for every t greater than or equal to s + 1, p[t] P p[s + 1]. In particular, for every t greater than or equal to s + 1, there is no axiom in F i,p[t] enumerating n into F i,p[t] (R i ). Thus, if n W (R i ) then n is not in F i,g (R i ). 10

Now we consider m(y, U, V ). It has four possible outcomes in the construction. It could reach a limit in Step 2, in Step 4c, in Step 5a or in Step 5b. In the first case, for every s and every n less than or equal to s, if p[s + 1] n U(Y F G ) then p[s 0 ] n U(Y F G ). Since F G and G G satisfy G(i, n, U, Y ), for each n, either p[s 0 ] forces n U(Y F G ) or n U(Y F G ). But then Y e U(Y F G ). Now consider the second case. If m is not in R i and there cannot be a G which is compatible with G p[s1] i m, b, k p[s1] for which n V (Y G) or m(y, U, V ) would go to Step 5a. Conversely, if m is in R i then R p[s], F p[s] i a, b, G p[s] i m, b, k p[s] cannot force n V (Y G G ) or m(y, U, V ) would go to Step 5b. Consequently, there is a G which is compatible with G p[s] i m, b, k p[s] for which n V (Y G). By Lemma 3.2, p[s 1 ] P p[s] and so this G is also compatible with G p[s1] i m, b, k p[s1]. In this case, m is an element of R i if and only if there is a G which is compatible with G p[s1] i m, b, k p[s1] for which n V (Y G). Hence, Y e R i. Finally, m(y, U, V ) could reach a limit in Step 5. In 5a, n V (Y G p[s+1] ) and there is no F compatible with F p[s+1], k p[s+1] for which n U(Y F ). In 5b, n U(Y F p[s+1] ) and there is no G compatible with G p[s+1], k p[s+1] for which n V (Y G). In either case, U(Y F G ) and V (Y G G ) disagree at n. By the previous remarks, we may conclude Proposition 3.1 from Lemma 3.3. 3.2 Undecidability Theorem 3.4 The first order theory of E(Σ 0 2) is not decidable. Proof: By (Lavrov 1963), it is enough to show that every finite symmetric irreflexive binary relation can be uniformly interpreted within E(Σ 0 2). Such a relation is given by a model G = G, E in which G is a finite set and E is a binary relation on G with the appropriate properties. We can represent G within E(Σ 0 2) as follows. Recall that a set of degrees is independent if no element of the set is below the join of finitely many of the others. We fix a bijection ψ between G and an independent set I of elements from E(Σ 0 2) whose uniform join is low. We omit the proof but such independent sets can be obtained from sequences of mutually 1-generic reals below 0. Since I is independent, for each pair a and b of distinct elements of I, a b is in J, where J is equal to {ψ(x) ψ(y) : x G and y G and E(x, y)} if and only if there are a and b in G such that a = ψ(a) and b = ψ(b). By Proposition 3.1, both I and J are uniformly definable from parameters in E(Σ 0 2). Further, G is isomorphic to the struture I, E I where a, b belongs to E I if and only if a b is an element of J. Question 3.5 Is the first order theory of E(Σ 0 2) recursively isomorphic to the first order theory of arithmetic? References Cooper, S. B. (1990). Enumeration reducibility, nondeterministic computations and relative computability of partial functions. In K. Ambos-Spies, G. Müller, and G. E. Sacks (Eds.), Recursion Theory Week, Oberwolfach 1989, Volume 1432 of Lecture Notes in Mathematics, Heidelberg, pp. 57 110. Springer Verlag. Lavrov (1963). Effective inseparability of the sets of identically true formulae and finitely refutable formulae for certain theories. Algebra i Logika 2, 5 18. Slaman, T. A. and W. H. Woodin (1986). Definability in the Turing degrees. Illinois J. Math. 30 (2), 320 334. 11