Introduction Probabilistic solvability Arbitrary domains Enumerative induction Proofs Home Page Title Page Page 1 of 98 Go Back Full Screen Close Quit Advanced Topics in Inductive Logic 1 Eric Martin University of New South Wales Daniel Osherson Princeton University May 1, 2003 1 This work was supported by Australian Research Council Grant #A49803051 to Martin and by NSF Grant #IIS-9978135 to Osherson. We thank Scott Weinstein for generously checking proofs and offering improvements to several arguments. Thanks also to Sebastian Rahtz for developing the L A T E X package hyperref.sty, used to produce the current document. Authors’ addresses: [email protected], [email protected].
Transcript
Introduction
Probabilistic solvability
Arbitrary domains
Enumerative induction
Proofs
Home Page
Title Page
JJ II
J I
Page 1 of 98
Go Back
Full Screen
Close
Quit
Advanced Topics in Inductive Logic1
Eric MartinUniversity of New South Wales
Daniel OshersonPrinceton University
May 1, 2003
1This work was supported by Australian Research Council Grant #A49803051 to Martin and byNSF Grant #IIS-9978135 to Osherson. We thank Scott Weinstein for generously checking proofsand offering improvements to several arguments. Thanks also to Sebastian Rahtz for developingthe LATEX package hyperref.sty, used to produce the current document. Authors’ addresses:[email protected], [email protected].
The inductive logic developed in the second and third essays is limited in important ways.For example:
(a) the logic makes no provision for missing or misleading data;
(b) it gives the scientist no control over the evidence reaching him;
(c) revision-based scientist must work with theories written in the cramped idiom of first-order logic;
(d) the idea of efficient induction is only weakly expressed (in terms of “dominance”);
(e) solving a problem is an all-or-nothing affair (since all environments must lead tosuccess);
(f) only structures with countable domains are envisioned as potential realities;
(g) whereas revision-based scientists represent sophisticated schemes of scientific inquiry,no attention is given to the inductive powers of radically simpler methods of induction.
In the present essay we attempt to address the last three lacunae on the list. To addressshortcoming (e), Section 2 explores a quantified concept of success, based of probability.The idea is to impose a probability distribution on the environments for a given structure,and then to measure the class of environments on which the scientist succeeds. In Section3 we consider shortcoming (f). Structures of arbitrary cardinality are admitted into ourparadigm through an ordinal measure of their similarity to countable structures. Section4 responds to shortcoming (g). We investigate one of the simplest schemes for inquiry,namely, “induction by enumeration.”
The theorems of Sections 2 and 3 are natural generalizations of the results in essay#2. As such, they help to justify the definition of inquiry elaborated there. We can onlyhope that the interest of our findings will be perceptible across the veil of technicality thatsurrounds them. The results of Section 4 are somewhat less technical (at least, at first).Their underlying message is that simple-minded science only goes so far.
2. Probabilistic solvability
2.1. Success in enough environments
In the present section we address concerns that our requirements for success are excessive.Recall from Definition (23) of essay #2 that solving a proposition P requires the scientist toissue a noncontradictory proposition that implies P cofinitely often in every environmentfor a structure in P . Contrary behavior in even a single environment for P denies thescientist the laurel of success in any problem that includes P as a proposition. The refusalto allow failure in even a single environment has the ring of hysteria, however, especiallysince the odds of running into any one environment seem low. So why not relax a little,and settle for success in “almost” every environment? We now use the tools of probabilityto make this idea precise and explore it.
As a preliminary to the technicalities that follow, let us recall the motive for the present,“absolute” conception of success. By requiring scientists to succeed in every environmentfor a given problem, we prevent Nature from passing along illicit hints. To appreciate theissue, consider the problem P = {P`, Pn}, where P` is the class of total orders with a leastpoint, and Pn is the class of total orders with no least point. Proposition (57) of essay #2shows P to be unsolvable. It is nonetheless easy to specify scientist Ψ and uncountablecollections E`, En of environments such that:
• Ψ solves P` in every e ∈ E`, and solves Pn in every e ∈ En.
One way to achieve this is to distinguish which of the two variables v0, v1 appears first inan environment, and to define:
E` = {e | e is for P`, and v0 appears before v1 in e} ;
En = {e | e is for Pn, and v1 appears before v0 in e}.
The right answer is thus coded into the environments of E` ∪ En, and may be used inunderhanded fashion by a scientist designed just for them. All such coding schemes arefoiled in a single stroke by asking scientists to succeed in every environment for a givenproblem.
Still, such a categorical policy might be an over-reaction to the danger of illicit hints. Inparticular, it seems sufficient to construct environments according to some random processthat respects Nature’s choice of underlying structure. The randomization can then becounted on to block collusion while nonetheless allowing scientists to fail in the rare case.Let us now give substance to this idea.
2.2. Framework and conventions
To guide our conception of a randomly constructed environment, we shall picture environ-ments as created by a stochastic process. In particular, the elements of a model’s domainare assumed to be delivered for inspection via independent, identically distributed sam-pling according to a probability law which may be unknown to the scientist. Just as before,
the scientist’s mission is to discover the proposition from which her environment is drawn.However, this need only be achieved in “enough” environments. The latter expression isinterpreted probabilistically, in terms of the measure of the class of environments in whichthe scientist succeeds. The resulting paradigm classifies performance in probabilistic ratherthan all-or-none fashion, and thus differs from the absolute conception of scientific successconsidered up to now. The facts to be discussed below indicate the robustness of our origi-nal paradigm of inquiry. This is because the weaker, probabilistic conditions for solvabilityturn out to be equivalent to absolute conditions.
Just for this section, we rely on some conventions that will ease our notation. Here isthe first one.
(1) Convention: In the present section, the domain of every structure is taken to beN .
It is clear that Convention (1) has no impact on the generality of our results except insofaras finite models are excluded from the discussion. In light of this convention, in the presentsection we use the expression “full assignment” in place of “full assignment to S.” The classof full assignments is denoted H. Probabilities will be defined over subsets of H, and tomake this possible, we abide by the following convention.
(2) Convention: In the present section, members of H (that is, full assignmentsh : Var → N), will be conceived as drawn from the Baire space Nω, by identifyingeach vi ∈ Var with the integer i.
For example, 〈1, 0, 3, 2, 5, 4, 7, 6, 9, 8 · · ·〉 ∈ Nω represents the full assignment that sends v0to 1, v1 to 0, v2 to 3, v3 to 2, etc. Convention (2) is essential to what follows. Nevertheless,for h ∈ H and i ∈ N , we continue to write h(vi), not h(i).
(3) Convention: In the present section, it is assumed that Obs contains all identities(that is, formulas of the form t1 = t2, where t1, t2 are terms of L). It is also assumedthat Obs is closed under negation. Also, for the remainder of the section we fixan enumeration {αi | i ∈ N} of a subset of Obs such that for all ϕ ∈ Obs, eitherϕ = αi or ϕ = ¬αi for some i ∈ N .
Note that Convention (3) ensures that Obs is infinite. For some of the results that follow wemust also assume that Obs contains the atomic formulas of L. (By closure under negation,placing the atomic formulas in Obs places into Obs formulas that are equivalent to thebasic formulas).
It is clear that Convention (3) is consistent with taking Obs = Lbasic (our favorite choicefor Obs). You won’t miss anything essential if you focus just on Obs = Lbasic . In this case,{αi | i ∈ N} is given by an enumeration of Latomic .
2.3. Standard environments
Our goal is to measure the class of environments in which a given scientist succeeds. Tosimplify the process, we first impose a canonical ordering on the formulas that appear inan environment.
(4) Definition:
(a) We say that an environment e is standard just in case for all i ∈ N , eithere(i) = αi or e(i) = ¬αi. [The αi are the formulas fixed in Convention (3),above. Recall that the convention ensures that Obs is closed under negation.]
(b) For any structure S and h ∈ H there is a unique standard environment for Sand h. This environment is called the standard environment for S and h. It isdenoted dS, he.
(c) We say that scientist Ψ solves problem P in standard environments just in casefor every P ∈ P, S ∈ P and h ∈ H, ∅ 6= Ψ(dS, he[k]) ⊆ P for cofinitely manyk.
The following two facts motivate our decision to limit attention to standard environ-ments. The second follows directly from the first , whose proof is easy.
(5) Lemma: There is a (computable) function f : SEQ → SEQ such that:
(a) for all σ, τ ∈ SEQ, if σ ⊆ τ then f(σ) ⊆ f(τ);
(b) for all structures S, all h ∈ H and all environments e for S and h,⋃
k∈N f(e[k])is a standard environment for S and h.
Let scientist Ψ, proposition P , structure S in P , and h ∈ H be given. If Ψ solvesP in dS, he then Ψ ◦ f solves P in every environment for S and h.
(6) Corollary: A problem is solvable if and only if some scientist solves it in standardenvironments.
In view of the corollary, we see that the inductive challenge posed by a given problem ispreserved by considering only its standard environments. Variation among full assignmentscaptures the essential uncertainty about Nature’s choice of data presentation. For thisreason, we shall focus attention on just the standard environments when measuring theprobability that a scientist succeeds on a problem.
(7) Definition: Given finite sequence ζ ∈ N<ω we denote by Oζ the basic open set ofNω consisting of all h ∈ H that begin with ζ.
The class of all positive probability distributions over N is denoted M . (A distribu-tion µ over N is positive if for all n ∈ N , µ(n) > 0.) Given µ ∈ M , we extend µ tothe product measure over Nω in the usual way (as reviewed, for example, in [14, SectionVII.3.17]). Specifically, for ζ ∈ N<ω, the probability assigned to Oζ by µ is taken to beΠi<length(ζ)µ(ζ(i)); then, µ is extended to the collection of Borel subsets of Nω in the uniqueway possible. Finally, µ is extended (also in the only way possible) to sets of the form A∪B,where µ(A) is defined and B is a subset of set C for which µ(C) is defined to be 0. Forsuch sets, µ(A ∪B) is given the value µ(A).
The collection of measures that arise in the foregoing way from positive probabilitydistributions over N is also denoted by M . Context will make it clear whether µ ∈ Mrefers to a positive distribution over N , or to its completed product extension to Nω.
By our identification of assignments with members of Nω, each µ ∈ M induces proba-bilities over the measurable subsets of H. Observe that H does not exhaust Nω since thelatter includes sequences whose range is not all of N whereas H is limited to full assign-ments. However, in what follows we will be able to ignore Nω −H because of the followingfacts. (The following lemma is proved in Section 5.1, and its corollary in Section 5.2.)
(8) Lemma: Let µ ∈ M be given. Then µ(H) = 1.
(9) Definition: We call h ∈ Nω repetitive just in case for every i ∈ N,h−1[i] is infinite.
(10) Corollary: Let R = {h ∈ Nω |h is repetitive}. Then for every µ ∈ M , µ(R) = 1.
2.5. Probabilistic solvability defined
For a given scientist and structure we now wish to measure the class of full assignmentsthat lead the scientist to success. For this purpose we rely on the following definition.
(11) Definition: Let scientist Ψ, proposition P , and S ∈ P be given. The class ofh ∈ H such that Ψ solves P in dS, he is denoted: Hsolve(Ψ,S, P ).
For the definition to be useful, of course, Hsolve(Ψ,S, P ) must be measurable. The followinglemma (proved in Section 5.3) guarantees this.
(12) Lemma: Let scientist Ψ, proposition P , and S ∈ P be given. Then Hsolve(Ψ,S, P )is a Borel set.
From the lemma it follows that for all µ ∈ M , µ(Hsolve(Ψ,S, P )) is defined. Now wecan quantify success.
(13) Definition: Let scientist Ψ, problem P, M0 ⊆ M , and p ∈ [0, 1] be given. We saythat Ψ solves P on M0 with probability p just in case for all µ ∈ M0, all P ∈ P, andall S ∈ P , µ(Hsolve(Ψ,S, P )) ≥ p. In this case P is solvable on M0 with probabilityp.
The foregoing definition is specialized to individual propositions P and individual distribu-tions µ in the obvious way. Thus, Ψ solves P on µ with probability p just in case Ψ solves{P} on {µ} with probability p.
2.6. Probabilistic solvability under isomorphic closure
We now show that under a mild condition on problems, the probabilistic conception ofsolvability is equivalent to the absolute one.
(14) Theorem: Suppose that problem P consists of propositions that are closed underisomorphism.1 Let µ ∈ M be given. Then P is solvable if and only if P is solvableon µ with probability greater than .5.
The proof is given in Section 5.4.
We will see in Proposition (19) below that the condition of isomorphic closure cannotbe lifted in Theorem (14). The theorem nonetheless demonstrates the close proximity ofthe absolute and probabilistic criteria of solvability. When propositions are closed underisomorphism, the two criteria coincide. In this sense, our concept of (absolute) solvabilityis robust.
2.7. Solvability with low probability
Theorem (14) shows that solvability with probability greater than .5 is no less demandingthan absolute solvability, for problems closed under isomorphism. What about solvabilitywith probability of .5 or less? In contrast to Theorem (14), the following proposition andcorollary reveal that lowering the bar to .5 extends the scope of solvability.
(15) Proposition: Let n ≥ 1 be given, and suppose that problem P consists of just npropositions. Then P is solvable on M with probability equal to 1
n .1Proposition P is closed under isomorphism if and only if for all structures S, T , if S and T are isomorphic
The proof is given in Section 5.5. Specializing to the case n = 2, we have the followingcorollary, easily derived from the foregoing proposition and Proposition (41) of essay #2.
(16) Corollary: Suppose that Sym contains just a binary relation symbol. Let Obs =Lbasic . Then there is an isomorphically closed problem P such that:
(a) P is not solvable.
(b) P is solvable on M with probability equal to .5.
2.8. Probabilistic solvability under permutation
There is an interesting dual to Theorem (14). To state it, we rely on the following definition.
(17) Definition: Let µ ∈ M be given. The class of µ′ ∈ M such that for somepermutation π : N → N , µ′ = {( i, µ(π(i)) ) | i ∈ N} is denoted PERM(µ).
So, PERM(µ) is the class of distributions that apply the same probabilities as µ to N , butnot necessarily in the same order (in other words, not necessarily to the same numbers). Thenext theorem shows that no matter what distribution µ we start with, the class PERM(µ)is diverse enough to represent all of M . Indeed, we will see that a problem is solvable on Mwith probability greater than .5 if and only if it is solvable on PERM(µ) with probabilitygreater than .5. And it will be seen that solvability in the latter sense implies solvabilitytout court, that is, with respect to all environments. Observe that we do not suppose thatthe problem is isomorphically closed.
(18) Theorem: Let problem P and µ ∈ M be given. Then P is solvable if and only ifP is solvable on PERM(µ) with probability greater than .5.
2.9. Separating probabilistic and absolute solvability
Our results so far raise the possibility that solvability with high probability collapses intoabsolute solvability. So the question arises whether there are problems P and classes M0
of probabilities such that P is solvable on M0 but unsolvable absolutely. From Theorems(14) and (18), for this to be possible at least one member of P must not be closed underisomorphism, and M0 must not include PERM(µ) for any µ ∈ M . Under these constraintsthe following fact provides an affirmative answer to our question. Indeed, it suffices toconsider a “small” problem and a “large” class of distributions.
(19) Proposition: Suppose that Sym consists of a binary predicate, and that Obs =Lbasic . Then there exist disjoint propositions P1, P2 and uncountable M0 ⊆ M suchthat {P1, P2} is unsolvable, but {P1, P2} is solvable on M0 with probability 1.
The proof is given in Section 5.7.2
2.10. Reliable solvability with specified probability
To this point our results have concerned probability of success greater than .5. We wish nowto consider the competence of scientists whose probability of success is small. To formulateinteresting results requires reflection about the case in which success is not achieved. Ascientist who solves a given structure with small probability is worse than useless if sheexhibits high probability of misleading an external observer. In particular, it is misleading
2The idea behind the proof originates in [17] and has been extended in [1, 16, 15].
to converge to a false theory; for in this case the mistaken theory appears to be held withconfidence, and risks being accredited. If the probability that the scientist misleads us thisway is high, and the probability of genuine success low, it might be better to show him nodata at all. So we are led to the following definition.
(20) Definition: Let scientist Ψ, proposition P , structure S 6∈ P , and full assignmenth be given. We say that Ψ is unreliable on S, h and P if ∅ 6= Ψ(dS, he[k]) ⊆ P forcofinitely many k.
We denote by Hunrel (Ψ,S, P ) the class of h ∈ H for which Ψ is unreliable on S, hand P .
The ensuing definition of reliable, probable success includes the requirement that the chanceof being permanently misled is zero.
(21) Definition: Let scientist Ψ, problem P, M0 ⊆ M , and p ∈ [0, 1] be given. Wesay that Ψ reliably solves P on M0 with probability p just in case for every µ ∈ M0
and for every P ∈ P, the following holds:
(a) for all S ∈ P , µ(Hsolve(Ψ,S, P )) ≥ p ;
(b) for all S ∈⋃
P− P , µ(Hunrel (Ψ,S, P )) = 0.
In this case P is reliably solvable on M0 with probability p.
In particular, Ψ reliably solves P on M0 with positive probability just in case there is nochance that Ψ will mislead us in an environment for P and some chance that Ψ will succeed.We now show that reliable solvability in the foregoing sense embodies a surprising degreeof scientific competence.
(22) Theorem: Let problem P and M0 ⊆ M be given. Then P is solvable on M0 withprobability 1 if and only if P is reliably solvable on M0 with probability greaterthan 0.
For the proof, see Section 5.8.
2.11. Partial summary
The collection of problems of form (T , {θ0 . . . θn}), where T has no finite models, is a richand natural class for the theory of empirical inquiry. The results in the present chapterprovide alternative characterizations of its solvable subclass. To recapitulate, we state thecharacterizations in the following corollary, which is an immediate consequence of the resultsabove along with essay #2.
(23) Corollary: Let µ ∈ M and problem P of form (T , {θ0 . . . θn}) be given. Supposethat T has no finite models. Then the following conditions are equivalent.
(a) P is solvable.(b) For all i ≤ n, θi is equivalent in T to an ∃∀ sentence.(c) For all i ≤ n, θi is equivalent in T to a boolean combination of ∃ sentences.(d) P is solvable on µ with probability greater than .5.(e) P is reliably solvable on µ with probability greater than 0.
2.12. Bayesian scientists
The definitions introduced in the present section may be exploited to contrast our absolute(nonprobabilistic) paradigm with the Bayesian approach to inquiry. For this purpose we
shall isolate the class of scientists that may fairly be said to exhibit “Bayesian behavior.”Then it will be asked whether every solvable problem is solved by one of them. The answerwill be negative.3 Before turning to the details, we indicate the importance of the questionto our enterprise.
The idea of “Bayesian behavior” will be defined in probabilistic terms, yet the successcriterion at issue is the absolute one proper to the first-order paradigm developed in thelast chapter: all environments must lead the scientist to success. Under these conditionsit will be possible to exhibit a simple, solvable problem that is not solved by any scientist(computable or otherwise) that behaves in Bayesian fashion. However, the reader mightobject that such a finding is irrelevant to the Bayesian since she is willing to settle forsuccess with probability 1 instead of success with certainty (as required in the absolute,first-order paradigm). In response, we note that our goal is only to show that within theabsolute paradigm, Bayesian scientists do not constitute a canonical form for inquiry, evenwith respect to some simple classes of problems. In this way we intend to further motivatethe developments of the third essay, which did provide canonical strategies for some broadclasses of problems.
For the remainder of the section we take Sym to be limited to the binary predicateR, and we assume that Obs = Lbasic . We let S` be the structure (N,≤), and Sg bethe structure (N,≥). (The subscript ` stands for “least point for R” whereas g refers to“greatest point for R.”) We let P∗ = { {S`}, {Sg} }. It follows immediately from Example(24)b of essay #2, that P∗ is solvable (indeed, it is computably solvable).
(24) Definition: Given p ∈ (0, 1) and µ ∈ M , we denote by Prµ,p the product measureon H×{S`,Sg} induced by the condition that for all U ∈ {S`,Sg} and µ-measurable
3For an alternative perspective on Bayesianism within a paradigm somewhat different from ours, see [10].
For U ∈ {S`,Sg}, we let U denote the event H × {U}.
The interpretation of p in the product measure is the prior probability of Nature choosingS` as the actual world; her probability of choosing Sg is 1−p. As before, for n ∈ N , µ(n) isthe (positive) probability that Nature chooses n when assigning a value to a variable. Thefollowing lemma is proved in Section 5.9.
(25) Lemma: Let p ∈ (0, 1), µ ∈ M , U ∈ {S`,Sg}, and σ ∈ SEQ be given. ThenPrµ,p({〈h,U〉 | U |= σ[h]}) is defined.
Conceiving of {〈h,U〉 | U |= σ[h]} as the conjunctive event of observing σ and U beingNature’s choice, leads to the following definition.
(26) Definition: Let p ∈ (0, 1), µ ∈ M , U ∈ {S`,Sg} and σ ∈ SEQ be given. Then wedefine:
Prµ,p(σ | U) =Prµ,p({〈h,U〉 | U |= σ[h]})
Prµ,p(U)
and
Prµ,p(σ) = Σ{Prµ,p(σ | U)× Prµ,p(U) | U ∈ {S`,Sg}}
(b) Scientist Ψ is Bayesian just in case Ψ is Bayesian with respect to some pair(p, µ), where p ∈ (0, 1), and µ ∈ M .
Note that if Prµ,p({〈h,S`〉 | S` |= σ[h]}) = Prµ,p({〈h,Sg〉 | Sg |= σ[h]}), we place noconstraint on the behavior of a scientist that is Bayesian with respect to (p, µ). We canfinally make the point indicated at the beginning of this subsection.
(28) Proposition: No Bayesian scientist solves P∗.
For the proof, see Section 5.10.
To exploit again our favorite example, let θ = ∃x∀yRxy, and let T 1 ⊆ Lsen be the theoryof total orders with either a greatest point or a least point, but not both. Proposition (57)of essay #2. shows that (T 1, {θ,¬θ}) is solvable. In contrast, since no Bayesian scientistsolves P∗, none solves (T 1, {θ,¬θ}) either. Therefore Bayesian scientists do not offer acanonical means of solving problems of the form (T , {θ0 . . . θn}), at least, not within thepresent paradigm.
3. Arbitrary domains
3.1. Inquiry in an uncountable setting
The inductive paradigm introduced in Section 1 of essay #2 makes essential use of thelimitation to countable structures, particularly in its simple conception of “environment”[see Definition (12) of the earlier essay]. It was for this reason that we introduced Convention(9) in essay #2, which binds the concept “structure” to its countable case. It must beadmitted, however, that legitimate complaints may arise about the exclusion of structureswith uncountable domains. If the universe has uncountably many “things” in it, then realscientists must deal with a domain that fits into none of the structures for which the first-order paradigm is currently adapted.4 Even without speculating about the numerical sizeof the universe, it is interesting to see how the first order paradigm can be extended tostructures with arbitrary domains. Such is the purpose of the present section.
4Things in the universe might include real numbers, e.g., every position between the Earth and the Sun ata particular moment. In this case the numerical size of the universe has at least the power of the continuum.
The present section is organized as follows. Section 3.2 introduces the technical machin-ery to be used in generalizing the original paradigm. The new paradigm is then presentedand motivated in Section 3.3. The relation between the original and generalized paradigmsis explored in Sections 3.4 - 3.6.
The following convention is central to the discussion. Whenever we wish to denote anobject in its general sense (without limitation to the countable case), we attach the subscriptg (for “general”). Thus, in particular:
(29) Convention: By a structureg is meant a model for Sym with an arbitrary domain(either countable or uncountable). We continue to use the term “structure” (withoutthe subscript g) in the sense of Convention (9) of essay #2, namely, as involving acountable domain.
Similarly, the class of modelsg (arbitrary domain!) for T ⊆ Lsen is denoted MOD(T )g, etc.The convention extends even to our name for the new paradigm; it will be qualified as:inductiveg.
3.2. Partial isomorphisms
To define scientific success within our inductiveg paradigm, we rely on the follow idea. Everyuncountable structureg S has many countable substructures. For the scientist to succeedon S we require that she succeed on a collection of its countable substructures that are“sufficiently similar” to S. To cash in the latter idea, we exploit a standard concept frommodel theory, namely “partial isomorphism.” Except for a few results to be cited fromother sources, our presentation of the relevant concepts is self-contained. For background,see [5, Ch. XII].
(30) Definition: Let structuresg S and T be given. Let p be an injective map from | S |into | T |. (Note: p need be neither total nor onto.) Then p is a partial isomorphismfrom S to T just in case the following conditions are satisfied.
(a) For every n-place predicate R in Sym, and every n elements x1 . . . xn indomain(p), (x1 . . . xn) ∈ RS iff (p(x1) . . . p(xn)) ∈ RT .
(b) For every n-ary function symbol f in Sym, and every n+1 elements x1 . . . xn, xin domain(p), fS(x1 . . . xn) = x iff fT (p(x1) . . . p(xn)) = p(x).
(c) For every individual constant c in Sym, and every x in domain(p), cS = x iffcT = p(x).
The class of partial isomorphisms between two structuresg can be used to assess theirsimilarity, where ordinal numbers serve as the similarity measure. The next definitionshows how such an ordinal conception of similarity arises.
(31) Definition: Let structuresg S and T be given. Denote by PI(S, T ) the class ofpartial isomorphisms from S to T . For each ordinal α we define a class Iα(S, T ) ofpartial isomorphisms from S to T . This is accomplished inductively, as follows.
I0(S, T ) = PI(S, T ).
Let ordinal α be given. Suppose that Iβ(S, T ) has been defined for all β < α. Wedistinguish two cases.
Case 1: α = β + 1. Then Iα(S, T ) is the class of all f ∈ PI(S, T ) suchthat:
(a) for all a ∈ | S | there is g ∈ Iβ(S, T ) with g ⊇ f and domain(g) =domain(f) ∪ {a};
(b) for all a ∈ | T | there is g ∈ Iβ(S, T ) such that g ⊇ f and content(g) =content(f) ∪ {a}.
Case 2: α is a limit ordinal. Then Iα(S, T ) =⋂
β<α Iβ(S, T ).
We say that S and T are finitely isomorphic just in case Iω(S, T ) 6= ∅.We say that S and T are partially isomorphic just in case Iα(S, T ) 6= ∅ for allordinals α.
Our definitions of “finitely isomorphic” and “partially isomorphic” are not standard, butthey are equivalent to those usually given, for example, in [5, Sec. XII.1]. For proof, seeSection 5.11.
The following examples should help in understanding the hierarchy introduced in Defi-nition (31). (See Section 5.12 for the proof.)
(32) Example:
Suppose that Sym is limited to a binary predicate.
(a) Let S = (N, succ) be the natural numbers with the successor relation andT = (N ⊕ Z, succ), be the disjoint union of the natural numbers and theintegers with the successor relation. Then Iω(S, T ) 6= ∅, but Iω+1(S, T ) = ∅.
(b) Let S = (Z,<) and T = (Z + Z,<), where Z is the set of integers, Z + Z istwo copies of the integers (one above the other), and < is the natural ordering.Then Iω+1(S, T ) 6= ∅, but Iω+2(S, T ) = ∅.
The following lemma draws together some facts to be used later. Section 5.13 gives theproof.
(a) The empty partial isomorphism belongs to I0(S, T ).(b) For all ordinals α, β, if α ≤ β then Iβ(S, T ) ⊆ Iα(S, T ).(c) Suppose that Sym is finite. Then S and T are finitely isomorphic iff S and T
are elementarily equivalent.(d) Suppose that | S | and | T | are countable. Then S and T are partially isomor-
phic iff S and T are isomorphic.
Thus, Lemma (33)c and Example (32) imply that (Z,<) and (Z+Z,<) are elementarilyequivalent (although not isomorphic).
3.3. The paradigm
Now we are ready to specify the inductiveg paradigm. This will be achieved by steppingthrough its key concepts (just as for the original paradigm; see Section 2 of essay #2). Thesubscript g continues to signal usage in the general sense.
3.3.1. Worlds
The potential realities of our paradigm are all the structuresg that interpret the symbol setSym.
3.3.2. Problems
The concepts of propositiong and problemg are straightforward generalizations of the count-able case. Officially:
(34) Definition: A nonempty class of structuresg is a propositiong. A problemg is acollection of disjoint propositionsg.
3.3.3. Environments
For the inductiveg paradigm there is no need to extend the concept of “environment” tostructuresg. Instead, we continue to rely on Definition (12) of the second essay, whichintroduced full assignments and environments just for (countable) structures. As before,we let SEQ denote the collection of proper initial segments of any environment.
We make no assumptions about Obs beyond what is stipulated in Convention (3) ofessay #2, of the original paradigm.5
3.3.4. Scientists
Again, the generalization from the countable case is straightforward. Officially:
(35) Definition: A scientistg is any partial or total mapping of SEQ into subclasses ofstructuresg.
3.3.5. Success
The key definition is now given. Subsequent to its presentation we provide some motivatingremarks.
5Namely, it is assumed that Obs is nonempty, and that ϕ[y/x] ∈ Obs for every ϕ ∈ Obs and x, y ∈ Var.
(36) Definition: Let scientistg Ψ and propositiong P be given. We say that Ψ solvesg
P just in case for all T ∈ P , there is ordinal α such that:
(a) there is countable substructure S of T with Iα(S, T ) 6= ∅;(b) for all countable substructures S of T with Iα(S, T ) 6= ∅ and for all environ-
ments e for S, ∅ 6= Ψ(e[k]) ⊆ P for cofinitely many k.
Let problemg P be given. We say that Ψ solvesg P just in case Ψ solvesg everyP ∈ P. In this case, P is solvableg.
Lemma (33)a implies that condition (36)a can always be met by choosing ordinal α smallenough.
Intuitively, for Ψ to solveg P in structureg T , Ψ must recognize that T ∈ P through theexamination of any countable substructure of T that is sufficiently similar to T . Moreover,the degree of similarity to T can be chosen arbitrarily, provided that there are countablesubstructures of T with the similarity in question. From Lemma (33)b we see that Ψis better off choosing a high degree of similarity to T if possible, since Ψ will then beresponsible for a smaller class of substructures. It is also immediate that solvabilityg is anonvacuous concept, admitting both the solvableg and unsolvableg case. This is becauseall singleton problemsg are solvableg, whereas any problemg that separates an uncountablestructureg from all of its countable substructures is unsolvableg.
In the remainder of this subsection, we attempt to motivate Definition (36) by consider-ing some of its special cases. For this purpose, let uncountable structureg T be given, andsuppose that some countable substructure of T is partially isomorphic to T . That this caseis possible is shown by the following example.
(37) Example: Suppose that Sym consists of a sole, binary predicate. Let R be thereals, and let Q be the rationals, both under their natural order. Then it is easy to
see that Q is a countable substructure of R that is partially isomorphic to R.
Among its countable substructures, a partially isomorphic substructure of T might beconsidered to be “as similar as possible” to T .6 So it is reasonable to require the scientistto succeed just on such substructures, when they exist. Fortunately, the following lemma[in conjunction with Lemma (33)b] shows this to be exactly the requirement imposed byDefinition (36). Recall that ω1 denotes the least, uncountable ordinal.
(38) Lemma: For all structuresg T and all countable substructures S of T , Iω1(S, T ) 6= ∅if and only if S is partially isomorphic to T .
The Lemma is proved in Section 5.14. To explain its point suppose that structureg Tin propositiong P has a countable, partially isomorphic substructure, and that scientistg
Ψ solvesg P . Then to behave successfully with respect to T it is necessary and sufficientthat Ψ issue a nonempty subset of P cofinitely often on any environment for a countable,partially isomorphic substructure of T . Note that any pair of such countable substructuresare isomorphic by Lemma (33)d.
If there were guaranteed to be partially isomorphic, countable substructures of T , wewould gladly restrict our definition of solvabilityg to this case. There is, however, no suchguarantee, as shown by the next lemma (proved in Section 5.15).
(39) Lemma: Suppose that Sym holds countably many unary predicates. Then thereis a structureg T with cardinality of the continuum such that I1(S, T ) = ∅, for allcountable substructures S of T .
6In discussing partially isomorphic structures M and N , Barwise [2, p. 292] remarks: “Some authorsprefer the more picturesque terminology potentially isomorphic, to suggest that M and N would becomeisomorphic if only they were to become countable, say in some larger universe of set theory.”
In the absence of partially isomorphic, countable substructures of T there may be agreatest ordinal α that satisfies (36)a. This situation is illustrated in extreme form by thepreceding lemma: for all countable substructures S of T , the maximal ordinal α for whichIα(S, T ) 6= ∅ is α = 0. When such a maximal α exists (be it 0 or a greater ordinal), the classof countable substructures S of T with Iα(S, T ) 6= ∅ can be considered to be maximallysimilar to T . It is thus appropriate that Definition (36) designates precisely this class asrelevant to success [as easily seen from Lemma (33)b]. It would thus be a satisfying stateof affairs if the two alternatives discussed so far exhausted the possibilities. In this case,either T would possess a partially isomorphic, countable substructure, or there would be amaximal ordinal α satisfying (36)a. Definition (36) could then be interpreted as isolatingjust the countable substructures of T that are most similar to T . However, the next lemmashows that matters are not so simple; neither alternative need obtain.
(40) Lemma: Suppose that Sym is limited to just a binary predicate. Then there isstructureg T such that:
(a) for all countable substructures S of T , Iω1(S, T ) = ∅;(b) for all ordinals α < ω1, there is a countable substructure S of T with Iα(S, T ) 6=
∅.
See Section 5.16 for the Proof.
In summary, when T possesses “most similar,” countable substructures, Definition (36)requires the scientist to behave appropriately on just this class. Otherwise, it is left to thescientist to choose the degree of similarity against which her performance will be measured.
3.4. The new paradigm as a generalization of the original one
The inductiveg paradigm is intended as a natural generalization of the original one. At theleast, such intention requires the two paradigms to coincide when domains are limited tothe countable case. The following proposition shows this desideratum to be fulfilled. Recallthat a problem P is the special kind of problemg in which every S ∈
⋃P is countable.
(41) Proposition: Let problem P be given. Then P is solvableg if and only if P issolvable.
The proof of the proposition (given in Section 5.17) reveals another way the inductiveg
paradigm generalizes the original one. From the proof we see that if scientistg Ψ solvesg
problem P, then the scientist that results from restricting Ψ’s outputs to (countable) struc-tures solves P.
3.5. Problemsg of form (T , {θ0 . . . θn})
Many of the results obtained in earlier sections carry over to the inductiveg paradigm, atleast in attenuated form. The theorem now to be discussed provides an illustration. Tostate it, we introduce a familiar notation, generalizing in a straightforward way Definition(55) of essay #2. Following our convention about the subscript g, note that for T ⊆ Lsen ,the class of structuresg that satisfy T is denoted by MOD(T )g.
(42) Definition: Let problemg P, T ⊆ Lsen , and θ0 . . . θn ∈ Lsen be given. We say thatP has the form (T , {θ0 . . . θn}) just in case:
(a) for every modelg S of T there is exactly one i ∈ {0 . . . n} such that S |= θi;
The following theorem provides a finitary condition for the solvabilityg of problemsg of theform (T , {θ0 . . . θn}), provided that Sym is finite. It is the same condition seen earlier forthe countable case (Section 5.1 of essay #2).
(43) Theorem: Suppose that Sym is finite. Let problemg of form (T , {θ0 . . . θn}) begiven. Then (T , {θ0 . . . θn}) is solvableg if and only if for every i ≤ n, θi is equivalentin T to an ∃∀ sentence.
The theorem follows directly from Corollary (58) of essay #2 and the following lemma, ofinterest in its own right (proved in Section 5.18).
(44) Lemma: Suppose that Sym is finite. Let enumeration {Xi | i ∈ N} of subsets ofpow(Lsen) be given. Then the problem {
⋃E∈Xi
MOD(E) | i ∈ N} is solvable if andonly if the problemg {
⋃E∈Xi
MOD(E)g | i ∈ N} is solvableg.
As an application of the lemma, we have the following.
(45) Example: Suppose that Sym contains just two binary function symbols. For k ∈ Neither 0 or prime, let propositiong Pk be the collection of all fieldsg of characteristick in this vocabulary. Then Lemma (44) and Exercise (71) of essay #2 imply that{Pk | k is either 0 or prime} is solvableg. Exercise (72) of the earlier essay can betransposed to the present setting in the same way.
Theorem (43) generalizes Theorem (59) of essay #2, but only in attenuated form becauseof the limitation to finite vocabulary. That this limitation is essential is shown by thefollowing proposition and its corollary.
(46) Proposition: Suppose that Sym consists of a denumerable set of unary predicates.Then there exists two structuresg S and T and a universal sentence θ such that:
(a) S |= θ and T |= ¬θ;(b) {{S}, {T }} is not solvableg.
See Section 5.19 for the proof. As an easy consequence of Theorem (59) of essay #2along with Proposition (46), we obtain:
(47) Corollary: Suppose that Sym consists of a denumerable set of unary predicates.Then there there exists θ ∈ Lsen such that:
(a) the problem (∅, {θ,¬θ}) is solvable;
(b) the problemg (∅, {θ,¬θ}) is not solvableg.
3.6. Solvabilityg under bounded similarity
As discussed in Section 3.2, to solveg a problemg P requires proper behavior on environmentsfor all countable substructures of T ∈
⋃P that are sufficiently similar to T . Choice of a
high criterion of similarity can often lighten the scientist’s load, by restraining the class ofsubstructures for which she is responsible. It turns out, however, that there is a fixed levelof similarity that is always high enough to solveg P, provided that all T ∈
⋃P possess
countable substructures with such similarity to T . To state the matter precisely, we relyon the following definition.
(48) Definition: Let ordinal α and problemg P be given.
(a) We say that P is α-saturated just in case for every T ∈⋃
P there is a countablesubstructure S of T such that Iα(S, T ) 6= ∅.
(b) We say that scientistg Ψ α-solvesg P just in case:
i. P is α-saturated, andii. for all P ∈ P, all T ∈ P , all countable substructures S of T with
Iα(S, T ) 6= ∅, and all environments e for S, ∅ 6= Ψ(e[k]) ⊆ P for cofinitelymany k.
In this case, we say that P is α-solvableg.
So, α-solvabilityg is a special case of solvabilityg. The next theorem shows, however, that2ω-solvabilityg is equivalent to solvabilityg tout court for the class of 2ω-saturated problemsg.
(49) Theorem: Let problemg P be 2ω-saturated. Then P is solvableg if and only if Pis 2ω-solvableg.
Let us apply the theorem to the countable context. In the following corollary, the term“problem” is used in the sense of our original paradigm; it thus designates a partition of aclass of countable structures.
(50) Corollary: Let problem P be given. Then P is solvable if and only if P is2ω-solvableg.
Theorem (49) and its corollary are proved in Section 5.20.
Enumerative induction may be illustrated in the context of the numerical paradigm intro-duced in Section 2 of essay #1. In our usual game, we imagine that Nature chooses a setS from the collection A = {N − {x} |x ∈ N}. She then presents you with an arbitraryenumeration of S. Upon examining each newly presented number, you issue a member of Awith the goal of stabilizing to S. One way to proceed is to make your own enumeration ofA. Then at each stage you conjecture the first member of your enumeration that includesthe finite set of numbers encountered so far. (Thus, if your enumeration puts N−{i} in theith position, your conjecture after seeing 5, 1, 0 would be N − {2}.) Such is “induction byenumeration” or “enumerative induction,” and it is easy to see that in this case it works:no matter which member S of A is chosen by Nature, and no matter the order in which itsnumbers are presented to you, your conjectures will be correct cofinitely often, that is, youwill stabilize to S. Moreover, you will succeed in this way no matter how you enumerate A.
The foregoing strategy appears to be the lowest form of inductive life with a semblanceof intelligence. So one is curious to determine its scope and limits. There are problems likethose above — in which the hypotheses form a countable collection of subsets of N — thatcan be solved but not by enumerative induction. One such collection is {N −{0 · · ·n} |n ∈N}, as you can easily prove. On the other hand, if the hypotheses form a countablecollection of total functions from N to N then enumerative induction always works (asreadily shown). The theory of enumerative induction is more challenging when projectedinto a recursion theoretic setting, in which inductive strategies must be implementable viacomputer. Aspects of the resulting theory are presented in [8].
Enumerative induction also becomes more interesting when considered in our more gen-eral paradigm, as developed in essay #2. The inductive competence of this strategy will
occupy us in the present section. We shall leave computational issues to one side, in order tofocus on the pure logic of enumerative induction. To keep the focus on main ideas, we alsolimit the discussion to the case Obs = Lbasic . To make sure that this limitation remains inmind, we enter the following convention.
(51) Convention: Throughout the present section we take Obs to be Lbasic , the set ofbasic formulas of L.
To proceed, our first task is to specify two kinds of inductive strategies, each of enu-merative character. The powers of the two strategies will then be analyzed and compared.Among other things we show that one of them is a canonical form for inductive inference:any solvable problem can be solved via the method.
4.2. Discrete enumerative induction
One idea for implementing the idea of enumerative induction is as follows. Given a countableproblem P, fix an enumeration E of the propositions in P. Then for any σ ∈ SEQ for P,conjecture the E-first proposition that is consistent with σ. It is clear, however, that thisstrategy fails to solve the problem described in Example (11) of essay #2. Recall that theproblem is MOD(T ∪ {θ}) and MOD(T ∪ {¬θ}) where (a) T is the theory of total orders(with respect to a binary relation symbol R) with either a least point or a greatest point.but not both, and (b) θ = ∃x∀yRxy. Our enumerative scheme fails for the simple reasonthat both propositions in P are consistent with every σ ∈ SEQ for P.
Given a problem P, a more promising method works as follows. First, a set of formulas isenumerated. Then for any σ ∈ SEQ, the first formula ψ is sought such that MOD(
∧σ ∧ ψ)∩⋃
P is a nonempty subset of some P ∈ P. This P is conjectured. Intuitively, the scientistproceeds down the ordering of formulas, looking for the first one that (in conjunction with
the data σ) picks out an admissible conjecture i.e., a member of P. Generalizing this ideaslightly leads to the following definition.
(52) Definition: Let problem P be given, and let O be a well ordering of a set offormulas. We define the scientist ∆[P, O] as follows. Let σ ∈ SEQ be given.
Case 1: There exists a first ψ ∈ O such that:
(*) for some P ∈ P, ∅ 6= {S ∈⋃
P |∧σ ∧ ψ is satisfiable in S} ⊆ P .
Then this P is unique (since the members of P are disjoint), and ∆[P, O] = P .
Case 2: There is no ψ ∈ O that satisfies (*). Then ∆[P, O] is undefined.
Let X ⊆ Lform be given. We define the discrete X-type of P as follows. If for everywell-ordering O of a subset of X, the scientist ∆[P, O] does not solve P, then thediscrete X-type of P is undefined. Otherwise, the discrete X-type of P is the firstordinal α such that for some well-ordering O of a subset of X,
(a) the well order type of O is α, and
(b) ∆[P, O] solves P.
The definition holds the promise of a bidimensional classification of solvable problems.Given a problem’s discrete X-type, one dimension records the quantifier complexity figuringin the formulas of X. The other records the lowest ordinal (if there is one) needed to arrangeX successfully. It will turn out, however, that both dimensions collapse considerably, andthat not every solvable problem can be sucessfully approached by discrete enumerativeinduction [see Proposition (70), below].
To define a more successful kind of enumerative induction we must allow the scientist topick out propositions via infinite sets of formulas (the discrete version offers just one formulaat a time). For this purpose we rely on the following notation.
(53) Definition: Let O be a well ordering of a set of formulas. Let σ ∈ SEQ and initialsegment s of O be given. We denote by satform(s, σ) the set of satisfiable formulasof the form
∧σ ∧ ψ, where ψ ∈ s.
Thus, satform(s, σ) gathers together all the formulas in the segment s of O that are consis-tent with σ. Such sets are useful because of the following (easy) fact.
(54) Lemma: Let problem P be given, and let O be a well ordering of a set of formulas.Let σ ∈ SEQ also be given. Then there is at most one P ∈ P such that for someinitial segment s of O, ∅ 6= {S ∈
⋃P | satform(s, σ) is satisfiable in S} ⊆ P .
So, given a problem P and an ordering O of formulas, the scientist can search in O for thefirst initial segment s such that satform(s, σ) picks out a proposition in P. This is the ideabehind the following definition.
(55) Definition: Let problem P be given, and let O be a well ordering of a set offormulas. We define the scientist Ψ[P, O] as follows. Let σ ∈ SEQ be given.
Case 1: There exists an initial segment s of O such that:(**) for some P ∈ P, ∅ 6= {S ∈
⋃P | satform(s, σ) is satisfiable in S} ⊆ P .
Then by Lemma (54) this P is unique and we set Ψ[P, O] = P .
Case 2: There is no initial segment s of O that satisfies (**). Then Ψ[P, O] isundefined.
Let X ⊆ Lform be given. We define the segmental X-type of P as follows. If forevery well-ordering O of a subset of X, the scientist Ψ[P, O] does not solve P, thenthe segmental X-type of P is undefined. Otherwise, the segmental X-type of P isthe first ordinal α such that for some well-ordering O of a subset of X,
(a) the well order type of O is α, and
(b) Ψ[P, O] solves P.
It is not immediately evident how to compare the inductive powers of discrete versussegmental methods. Given an ordering O, initial segment s of O, and σ ∈ SEQ, satform(s, σ)may be inconsistent, and thus be useless for picking out propositions. Perhaps there aresolvable problems for which this difficulty arises inevitably. But in fact such is not the case;we will see that every solvable problem can be solved via segmental enumerative induction.
For notational simplicity, we denote the class of universal formulas by ∀, and similarly forother quantifier prefixes. Observe that both the discrete and segmental types of unsolvableproblems are undefined (and if either is defined, the problem is solvable). We will see laterthat the discrete type of some solvable problems is also undefined.
(56) Example: Let P be as in Example (11) of essay #2. Given n ∈ N , set ψ2n =∀yRvny and ψ2n+1 = ∀xRxvn. Let E = {ψn |n ∈ N}. Then both ∆[P, E] andΨ[P, E] solve P. Indeed, let S ∈ MOD(T ∪ {θ}), full assignment h to S, andenvironment e for S and h be given (the proof is parallel if S ∈ MOD(T ∪ {¬θ})).So there is least n0 ∈ N such that S |= ψ2n0 [h] and there is k0 ∈ N such that for alln < 2n0,
∧e[k0] ∧ ψn is inconsistent. Hence for all k ≥ k0, 2n0 is the least n ∈ N
such that∧e[k]∧ψn is consistent. From this it is easy to verify that for all k ≥ k0,
∆[P, E](e[k]) = Ψ[P, E](e[k]) = MOD(T ∪ {θ}), which proves that both ∆[P, E]and Ψ[P, E] solve MOD(T ∪ {θ}) in e. Hence, both the discrete and segmental∀-types of P are bounded by ω.
We can exploit the example to make a useful point about types. The formulas that fix atype might require free variables; there may be no comparable ordering of sentences (closedformulas) that allows enumerative induction to proceed. The pitfall for sentences is that freevariables may be needed to refer to specific objects denoted by variables in an environment.This is illustrated in extreme form for the discrete case by the next proposition. It showsthat restriction to sentences can foreclose enumerative induction for a problem where itmight have been successfully applied.
(57) Proposition: Suppose that Sym consists of a binary predicate. Then there is aproblem P such that:
(a) the discrete ∀-type of P is defined (hence P is solvable);(b) the discrete Lsen -type of P is undefined.
The proof is given in Section 5.21.
4.4. Tip-off bases
To launch our investigation of discrete and segmental enumerative induction, a technicaltool is needed. It can be deployed when a problem enjoys tip-offs of a particularly simplecharacter.
(58) Definition: Let set X of ∀ formulas and problem P be given. We say that X is atip-off base for P just in case:
P and full assignment h to S, there is ϕ ∈ X such thatS |= ϕ[h];
(b) for all ϕ ∈ X, there is at most one P ∈ P such that ϕ is satisfiable in somemember of P .
The next two lemmas present some relevant facts about two familiar kinds of problems.They are proved in Section 5.22.
(59) Lemma: Every solvable problem of form (T , {θ0 . . . θn}) has a tip-off base.
(60) Lemma: No infinite problem of form (T , {P0, P1, . . .}) has a tip-off base.
Now we can begin to harness the concept of tip-off base. The next proposition (proved inSection 5.23) shows that every problem with a tip-off base lends itself easily to enumerativeinduction of the discrete kind.
(61) Proposition: Let set X of ∀ formulas be a tip-off base for problem P. Thenfor every enumeration E of ∀ formulas with X ⊆ content(E), the scientist ∆[P, E]solves P.
From Lemma (59) and Proposition (61) it is easy to derive the following.
(62) Corollary: Let solvable problem P of form (T , {θ0 . . . θn}) be given. Then forevery enumeration E of all ∀ formulas, the scientist ∆[P, E] solves P.
The corollary reveals that discrete enumerative induction is strikingly easy for problems ofform (T , {θ0 . . . θn}). There is no need for an astute choice of formulas nor for a clever wayto order them. Any ω-ordering of the universal formulas does the trick.
Proposition (61) informs us that the existence of a tip-off base is a sufficient conditionfor the solvability of a problem via discrete enumerative induction. But it is not necessary.As shown by the next proposition (whose proof is given at the end of Section 5.29), there areproblems without a tip-off base that can nevertheless be solved via enumerative inductionof the discrete kind.
(63) Proposition: Suppose that Sym consists of a unary function symbol and a con-stant. Then there is a solvable problem P with the following properties.
(a) P has no tip-off base.
(b) The discrete ∀-type of P is defined.
The problem to be exhibited in the proof of Proposition (63) has infinite discrete ∀-type.The next proposition shows this to be no accident.
(64) Proposition: Every problem with finite discrete ∀-type has a tip-off base.
See Section 5.24 for the proof.
4.5. Finite types
Discrete enumerative induction has an even more elementary character if it involves orderingno more than a finite number of universal formulas. We expect problems solvable by such a
method to be simple in some combinatorial sense. Proposition (64) satisfies this expectation;problems with finite discrete ∀-type enjoy tip-off bases in the sense of Definition (58).Does the same kind of simplicity characterize problems solvable by segmental enumerativeinduction over a finite ordering of universal formulas? The answer is affirmative becausethe same class of problems is at issue. Indeed, the next proposition shows that for everyn ∈ N , the discrete ∀-type of a problem is n if and only if the segmental ∀-type of theproblem is also n. (We subsequently show that all of these types are populated.)
(65) Proposition: For all problems P, the discrete ∀-type of P is finite iff the segmental∀-type of P is finite. Moreover, if finite, they are equal.
The proof is given in Section 5.25.
For every n ∈ N , the problems with discrete ∀-type equal to n are the same as thosewith segmental ∀-type equal to n. This is the content of the preceding proposition. Theproposition would not be informative if there were no such problems. But the finite typesare in fact rich, as the next proposition reveals. (It is proved in Section 5.26.)
(66) Proposition: Suppose that Sym = ∅. For all n ∈ N there is a problem whosediscrete and segmental ∀-types are n.
The proofs of Propositions (65) and (66) depend on the intimate connection betweenuniversal formulas and environments for a structure; any such formula false in the structureis inconsistent with each of its environments. [This is because the assignments underlyingenvironments are required to be onto.] One might hope that enumerative induction basedonly on finitely many formulas reduces to the case where all the formulas involved are ∀.But matters are not so simple. Indeed, it will be seen below that there are problems whosediscrete and segmental ∀∃ and ∃∀-types are 2 although their ∀-types are not finite. Can
anything general be said about finite types in this broader context? We can report onlythe following fact about the coincidence of finite types across the discrete/segmental divide.Here the quantifiers of enumerated formulas are not constrained at all. (See Section 5.27for the proof.)
(67) Proposition: For all problems P, if the segmental Lform -type of P is finite thenthe discrete Lform -type of P is finite.
4.6. Infinite types
The results of the previous subsection show that the discrete and segmental ∀-types coincidein the finite case. Does this situation extend to the infinite, and if so, for which ordinals?We’ll derive the following answer: when both are defined, the two kinds of ∀-types coincide,and they are never greater than ω. The next proposition covers much of the distance tothis result, and provides other useful information. The proof is given in Section 5.28.
(68) Proposition: For every solvable problem P, there exists a set X of ∀ formulassuch that for every enumeration E of X, the scientist Ψ[P, E] solves P.
We see from the proposition that segmental enumerative induction using ∀ formulas isa universal strategy of inference: every solvable problem can be solved this way. Moreover,designing a successful agent of this kind requires no more than specifying the right set Xof ∀ formulas. No further insight is required for their ordering since any enumeration willdo the job.
A similar order-independence characterizes discrete enumerative induction using ∀ for-mulas; and the ordinal is again bounded by ω. But it is necessary to qualify this fact by theproviso that the discrete ∀-type be defined for the problem in question; for, we will soon
see solvable problems with undefined discrete types at every level of quantifier complexity.The discrete parallel to the preceding proposition can thus be stated as follows.
(69) Proposition: Let solvable problem P have defined discrete ∀-type. Then thereexists a set X of ∀ formulas such that for every enumeration E of X, ∆[P, E] solvesP.
The proposition is proved in Section 5.29.
Let us now show that the “if defined” qualification in Proposition (69) cannot be elimi-nated. Indeed, the following result shows that discrete enumerative induction cannot alwaysbe made to work no matter what formulas are involved. So, unlike its segmental counterpart,discrete enumerative induction is not a universal method of inquiry within the first-orderparadigm. (For the proof, see Section 5.30.)
(70) Proposition: Suppose that Sym consists of a denumerable set of constants. Thenthere exists a solvable problem whose discrete Lform -type is undefined.
We summarize the relation between discrete and segmental enumerative induction usinguniversal formulas by the following consequence of Propositions (65), (68) and (69):
(71) Corollary: Let solvable problem P be given. The segmental ∀-type of P is ω atmost. If defined, the discrete ∀-type of P is ω at most, and equal to its segmental∀-type.
Proposition (70) provides an example of a solvable problem whose discrete Lform -typeis undefined, hence not finite. The following proposition shows that there is, in fact, a richcollection of solvable problems with nonfinite Lform -types.
(72) Proposition: Let solvable problem P be such that every σ ∈ SEQ for P is for atleast two members of P. Then the discrete Lform -type of P is not finite.
The proof is provided in Section 5.31.
To appreciate the import of Proposition (72), consider again the problem P defined inour running Example (11) of essay #2. It is easy to verify that P satisfies the conditionsof the proposition. Hence, P does not have finite discrete Lform -type. From Proposition(67) it follows that P does not have finite segmental Lform -type either. On the other hand,Example (56) shows P to have discrete and segmental ∀-types bounded by ω. The upshotis that P can be solved via enumerative induction of both the discrete and segmentalkind, but infinitely many formulas are required for this purpose. The following propositionsummarizes the situation.
(73) Proposition: Let P be as in Example (11) of essay #2. Then the discrete andsegmental ∀-types of P are both ω. The same is true of their Lform -types.
4.7. Existential, ∀∃-, and ∃∀-types
Once again, let P be as in Example (11) of the second essay. Although P can be solved byenumerative induction, infinitely many formulas are required for this purpose. In particular,Proposition (73) shows that no reduction in number is achieved by increasing the quantifiercomplexity of the enumerated formulas. This raises the general question: is the discrete,universal type of a problem the lowest possible, and similarly for segmental types? In thepresent section we provide a negative answer by exhibiting problems whose universal typesare either undefined or greater than their ∀∃- and ∃∀-types.
As a preliminary, let us ask whether simply switching from universal to existential
formulas can lower the type of a problem. In the discrete case the answer is No, as revealedby the following proposition (proved in Section 5.32).
(74) Proposition: For all solvable problems P, if the discrete ∃-type of P is definedthen the discrete ∀-type of P is 1 at most.
In contrast to the debility of existential formulas, we now give a sense in which the ascentto ∀∃ formulas allows maximal improvement in type. By Corollary (71), the worst universaltypes are ω in the segmental case, and undefined in the discrete case. Except in trivial cases,the best discrete and segmental ∀∃-types are 2. The following proposition shows there tobe problems that simultaneously have the worst ∀-types and the best ∀∃-types.
(75) Proposition: Suppose that Sym consists of a denumerable set of constants and abinary predicate. Then there is a solvable problem such that:
(a) the discrete ∀-type of P is undefined;
(b) the segmental ∀-type of P is infinite;
(c) the discrete and segmental ∀∃-types of P are 2.
For the proof, see Section 5.33.
Compared to ∀∃, the ascent to ∃∀ formulas does not yield quite the same improvementover universal formulas. The reason is that a problem’s discrete ∀-type is defined wheneverits ∃∀-type is defined. There can thus be no strict analogy to Proposition (75) with ∃∀replacing ∀∃.
(76) Proposition: For all solvable problems P, the discrete ∃∀-type of P is defined iffthe discrete ∀-type of P is defined.
In view of the last result and Corollary (71), what is the most drastic improvement tobe hoped for by using ∃∀ formulas instead of universal ones? There might turn out to bea problem whose discrete and segmental ∀-types are infinite but whose respective ∃∀-typesare 2. The following proposition (proved in Section 5.35) reveals the existence of just sucha problem.
(77) Proposition: Suppose that Sym consists of a binary predicate, a unary functionsymbol, and a constant. Then there exists a solvable problem P such that:
(a) the discrete and segmental ∀-types of P are infinite;
(b) the discrete and segmental ∀∃-types of P are 2;
(c) the discrete and segmental ∃∀-types of P are 2.
4.8. Concluding remarks
In Section 4.1 we imposed Convention (51), which fixed Obs as Lbasic . Most of our resultscan be generalized to any choice of Obs which closes it under negation. Indeed, when Obsis closed under negation, it is easy to adapt most results of this section with the refutableformulas playing the role of the ∀ formulas. On the other hand, when Obs is arbitrary, nolearning strategy based on enumeration is successful if it does not (a) dismiss some refutableformulas (inconsistent with the data), and (b) select some nonrefutable ones (implied bythe data).
An important challenge in the theory of scientific discovery is to provide a motivatedcharacterization of problem difficulty. Within the numerical paradigm, one approach tosuch characterization counts the number of times an hypothesis needs to be changed in
the course of solving a problem; see, for example, [11, 12, 13]. Other approaches include[4, 7, 9]. These perspectives are ingenious and well-developed but they are best suited toparadigms built around the recursivity of scientists and problems. Such is not the case forthe inductive paradigm studied in the present papers.
For our paradigm, it seems natural to measure problem difficulty in terms of consump-tion of an information resource. Segmental ∀-types illustrate the idea if we take the com-plexity of problem P to be the smallest ordinal κ such that P can be solved via segmentalenumerative induction using a κ-ordering of universal formulas. [Such is indeed the defini-tion of P’s segmental ∀-type; see Definition (55).] The difficulty for this approach is thatCorollary (71) largely trivializes it. There turns out to be just one level of infinite difficulty,concealing great differences in the intelligence needed to solve its different members. Incontrast, problems at finite levels all seem monotonously easy to solve. The situation iseven worse for the discrete case, since Lform -types in this sense may not even be defined[see Proposition (70)].
There nonetheless remain potentially interesting questions about enumerative induction.For one thing, we would like to have a revealing characterization of the class of problemswith discrete Lform -types. By Corollary (62), this class includes all solvable problems of form(T , {θ0 . . . θn}). By Lemma (59) and Proposition (63), there are yet other members, butwe have no independent description of them. A second question concerns the complexityof problems that do have discrete Lform -types. By Proposition (69), the complexity ofproblems with discrete ∀-types is bounded by ω. Is it similarly the case that every problemwith discrete Lform -type has Lform -type no greater than ω?
On the more speculative side, our results are consistent with the existence of a subsetX of Lform yielding a rich and interesting class of segmental X-types that embraces allsolvable problems. If the set X were natural or had other interesting properties, it wouldprovide a potentially useful measure of problem complexity. In view of Corollary (71), Xcould not include all ∀ formulas. But aside from this constraint it is presently unclear what
Perhaps something akin to induction by enumeration may yet prove useful for classifyingproblem complexity. At present, however, a key idea appears to be missing.
5. Proofs
5.1. Proof of Lemma (8)
(8) Lemma: Let µ ∈ M be given. Then µ(H) = 1.
Proof: Let m ∈ N be given. Set Em = (N − {m})ω. We show that µ(Em) = 0. For allk ∈ N , let Ak be the set of all e ∈ Nω such that m 6∈ content(e[k]). As a (countable) unionof basic open sets, every Ak is µ-measurable. Because µ is a product measure, for all k ∈ N ,µ(Ak) = (1− µ(m))k. Because µ(m) > 0 (since all our measures are positive), this impliesthat µ(Ak) converges to 0. Since Em ⊆ Ak for all k ∈ N , it follows that µ(Em) = 0. ButNω −H =
⋃m∈N Em, and a countable union of sets of null measure has null measure.
5.2. Proof of Corollary (10)
(10) Corollary: Let R = {h ∈ Nω |h is repetitive}. Then for every µ ∈ M ,µ(R) = 1.
(12) Lemma: Let scientist Ψ, proposition P , and S ∈ P be given. ThenHsolve(Ψ,S, P ) is a Borel set.
Proof: For k ∈ N , let Ak be the set of h ∈ H such that ∅ 6= Ψ(e[k]) ⊆ P , where e is thestandard environment for S and h. Ak is Borel since it is a (countable) union of basic opensets. It is easy to see that Hsolve(Ψ,S, P ) = lim infnAn =
⋃n∈N
⋂∞k=nAk, hence a Borel
set.
5.4. Proof of Theorem (14)
(14) Theorem: Suppose that problem P consists of propositions that are closedunder isomorphism.7 Let µ ∈ M be given. Then P is solvable if and only if Pis solvable on µ with probability greater than .5.
Proof of the theorem relies on the following lemma.
(78) Lemma: Suppose that scientist Ψ solves problem P on µ ∈ M with probabilitygreater than .5. Then for every P ∈ P and every S ∈ P , there are cofinitely manyk such that:
µ({h ∈ H | ∅ 6= Ψ(dS, he[k]) ⊆ P}) > .5.7Proposition P is closed under isomorphism if and only if for all structures S, T , if S and T are isomorphic
Proof: Let Ψ, P, and µ be as specified in the hypothesis of the lemma. Suppose that P ∈ Pand S ∈ P falsify the lemma’s conclusion. We derive a contradiction.
For k ∈ N , let Zk be the complement of {h ∈ H | ∅ 6= Ψ(dS, he[k]) ⊆ P}. By thefalsity of the lemma’s conclusion, for infinitely many k, µ(Zk) ≥ .5. It follows (via theBolzano-Weierstrass theorem) that:
(79) lim supk µ(Zk) ≥ .5.
In [3, Theorem 4.1(i)], it is shown that for any probability measure P and any sequence An
of measurable sets, lim supn P (An) ≤ P (lim supnAn). So from (79) we infer:
(80) µ(lim supk Zk) ≥ .5.
For every h ∈ lim supk Zk there are infinitely many k such that either Ψ(dS, he[k]) isundefined, equal to ∅, or not a subset of P . Hence, for every h ∈ lim supk Zk, Ψ fails tosolve P in dS, he So, from (80) if follows that Hsolve(Ψ,S, P ) < .5. By Definition (13), thiscontradicts our choice of Ψ.
Another lemma and corollary will facilitate proof of Theorem (14). First a definition.
(81) Definition: An environment e is bijective just in case there is structure S andbijective full assignment h such that e is for S and h.
Recall Convention (1), which stipulates that in the present section the domain of allstructures is taken to be N . Hence, bijective environments exist for every structure at issuein the present section. Keeping the same convention in mind, we record:
(82) Lemma: There is a (computable) function g : SEQ → SEQ such that:
(a) for all σ, τ ∈ SEQ, if σ ⊆ τ then g(σ) ⊆ g(τ);
(b) for all structures S, all h ∈ H and all environments e for S and h,⋃
k∈N g(e[k])is a bijective environment for S.
(83) Corollary: A problem P is solvable just in case there is a scientist Ψ such thatfor all P ∈ P, S ∈ P , and bijective h ∈ H, Ψ solves P in dS, he.
Proof: The left-to-right direction is immediate. For the other direction, let functions f, g :SEQ → SEQ be as specified in Lemmas (5) and (82), respectively. Suppose that scientist Ψis such that for all P ∈ P, S ∈ P , and bijective h ∈ H, Ψ solves P in dS, he. Then Ψ ◦ f ◦ gsolves P.
Finally, we record:
(84) Corollary: Suppose that problem P is isomorphically closed. Let h0 ∈ H be{(vi, i) | i ∈ N}. Then P is solvable if and only if there is a scientist Ψ such that forall P ∈ P and S ∈ P , Ψ solves P in dS, h0e.
Proof: Let Ψ be such that for all P ∈ P and S ∈ P , Ψ solves P in dS, h0e. By Corollary(83), it suffices to show that for all P ∈ P, T ∈ P , and bijective g ∈ H, Ψ solves P indT , ge.
Let T ∈ P and bijective g ∈ H be given. By the isomorphic closure of P , there is S ∈ Pisomorphic to T via h0 ◦ g−1. By assumption, Ψ solves P in dS, h0e. Hence, for cofinitelymany k ∈ N , ∅ 6= Ψ(dS, h0e) ⊆ P . But it is easy to see that dS, h0e = dT , ge. Therefore,for cofinitely many k ∈ N , ∅ 6= Ψ(dT , ge) ⊆ P . So Ψ solves P in dT , ge.
Proof of Theorem (14): The left-to-right direction of the theorem is immediate. For theother direction, suppose that scientist Ψ solves isomorphically closed problem P on µ ∈ M
with probability greater than .5. Let h0 ∈ H denote {(vi, i) | i ∈ N}. We will define scientistΨ′ such that for all P ∈ P and S ∈ P , Ψ solves P in dS, h0e. By Corollary (84), this sufficesto show that P is solvable.
Recall from Convention (3) our enumeration {αi | i ∈ N} of a subset of Obs, used todefine standard environments in Definition (4). For all n ∈ N denote by f(n) the leastinteger greater than all the indexes of the members of
⋃{Var(αi) | i < n}. For all n ∈ N
denote by SEQn the set of all τ ∈ SEQ such that length(τ) = n, and some standardenvironment extends τ . Note that for all n ∈ N , the indexes of variables appearing σ ∈SEQn are bounded by f(n).
Now we define Ψ′. Let σ ∈ SEQ be given. Suppose there is n ≤ length(σ) and (unique)P ∈ P such that the following holds.
(85) There is finite A ⊂ Nf(n) such that:
(a) µ(⋃{Oζ | ζ ∈ A}) > .5;
(b) for all ζ ∈ A there is τ ∈ SEQn such that ∅ 6= Ψ(τ) ⊆ P and for all structuresU , U |=
∧σ[h0] implies U |=
∧τ [ζ].
Then choose greatest n ≤ length(σ) and (unique) P ∈ P such that (85) holds, and setΨ′(σ) = P . Otherwise, Ψ′(σ) is undefined. Intuitively, Ψ′ conjectures P on σ if it appearsthat Ψ is on its way to conjecturing P on a collection of standard environments whoseunderlying full assignments have probability greater than .5.
Let P ∈ P, and S ∈ P be given. We must show that Ψ′ solves P in dS, h0e. By thechoice of Ψ and Lemma (78), there is k0 ∈ N such that:
(86) for all k ≥ k0, µ({h ∈ H | ∅ 6= Ψ(dS, he[k]) ⊆ P}) > .5.
Given k ∈ N and ζ ∈ Nf(k), let dS, ζe[k] ∈ SEQ be the (unique) initial segment of lengthk of any standard environment for S and h such that h extends ζ. Observe that for fixedk, {h ∈ H | ∅ 6= Ψ(dS, he[k]) ⊆ P} is a disjoint union of (countably many) basic open sets.Hence, (86) implies the existence of finite A ⊂ Nf(k0) with:
(87) (a) µ(⋃{Oζ | ζ ∈ A}) > .5, and
(b) for all ζ ∈ A, ∅ 6= Ψ(dS, ζe[k0]) ⊆ P .
Since A is finite, there is k1 ∈ N with content(h0[k1]) ⊇⋃{content(ζ) | ζ ∈ A}. Hence,
there is k1 ≥ k0 such that:
(88) for all ζ ∈ A, and all structures U , if U and h0 satisfy∧dS, h0e[k1] then U and ζ
satisfy∧dS, ζe[k0].
Now let k2 ≥ k1 be given. It suffices to show that Ψ′(dS, h0e[k2]) = P . From (87) and (88)we infer the existence of k ≤ k2 (for example: k0), finite A′ ⊆ Nf(k) (for example: A), andP ′ ∈ P (for example: P ) such that:
(a) µ(⋃{Oζ | ζ ∈ A′}) > .5;
(b) for all ζ ∈ A′ there is τ ∈ SEQk (for example: dS, ζe[k0]) such that:
i. ∅ 6= Ψ(τ) ⊆ P ′, and
ii. for all structures U , if U and h0 satisfy∧dS, h0e[k2], then U |=
∧τ [ζ].
So, there is greatest k3 ≤ k2, finite A′ ⊆ Nf(k3), and P ′′ ∈ P such that:
i. ∅ 6= Ψ(τ) ⊆ P ′′, andii. for all structures U , if U and h0 satisfy
∧dS, h0e[k3], then U |=
∧τ [ζ].
From the definition of Ψ′ we infer immediately that Ψ′(dS, h0e[k2]) = P ′′. It thus remainsto show that P ′′ = P . Suppose for a contradiction that P ′′ 6= P . Then (89) implies thatµ({h ∈ H | Ψ(dS, he[k3]) 6⊆ P }) > .5. In view of the fact that k3 ≥ k0, the latter claimcontradicts (86).
5.5. Proof of Proposition (15)
(15) Proposition: Let n ≥ 1 be given, and suppose that problem P consistsof just n propositions. Then P is solvable on M with probability equal to 1
n .
Proof: Let n ≥ 1 be given, and let infinite X0 . . . Xn−1 partition the variables of indexgreater than 0. For i < n, let xi
j , j ∈ N enumerate Xi. For i < n and j ∈ N , let E(i, j) bethe formula
v0 = xij
∧{v0 6= xk
j | 0 ≤ k < n and k 6= i}.
It is easy to see that
(90) for all µ ∈ M and structures S there is p > 0 such that for all i < n and j ∈ N ,
Given structure S and h ∈ H, let j(S, h) be the least j ∈ N (if any exists) such that(∃i < n)content(dS, he) |= E(i, j). Since i 6= k implies that {E(i, j), E(k, j)} is inconsistent,(90) and (91) show that:
Now suppose that problem P consists of propositions P1 . . . Pn. Let scientist Ψ be suchthat for all σ ∈ SEQ, Ψ(σ) = Pi for the unique i < n for which there is j ∈ N such that (a)content(σ) |= E(i, j), and (b) for all ` ≤ j, and k < n with k 6= i, content(σ) |= ¬E(k, `);Ψ(σ) is undefined if there is no such unique i. Then by (92) it is clear that for all µ ∈ M ,Ψ solves P on µ with probability 1
n .
5.6. Proof of Theorem (18)
(18) Theorem: Let problem P and µ ∈ M be given. Then P is solvable if andonly if P is solvable on PERM(µ) with probability greater than .5.
Proof: Let problem P and µ ∈ M be given. The left-to-right direction of the theoremis immediate. For the other direction, recall Definition (26) of essay #2 [introducing thenormal problems and the notation I(P )], and suppose that scientist Ψ solves P on PERM(µ)with probability greater than .5. Then, by Lemma (17)a of essay #2, it is easy to see that
P is normal. Hence, I(P) = {I(P ) |P ∈ P} is an isomorphically closed problem (inparticular, normality implies that its propositions are disjoint). We will show that Ψ solvesI(P) on µ with probability greater than .5. Since the latter problem is isomorphicallyclosed, Theorem (14) will then imply that I(P) is solvable, from which the solvability of Pfollows immediately [since the propositions of P are subsets of the propositions of I(P)].
The following notation will be used. Let permutation π : N → N be given. Wedenote by µπ the element {( i, µ(π(i)) ) | i ∈ N} of PERM(µ). Given h ∈ H, let hπ ={(vi, π(h(vi))) | i ∈ N}, and given H ⊆ H, let Hπ = {hπ |h ∈ H}. It follows easily that:
(93) For all µ-measurable H ⊆ H, µπ(H) = µ(Hπ).
Now let P ∈ P and S ∈ I(P ) be given. It suffices to show that:
(94) µ( Hsolve(Ψ,S, I(P )) ) > .5.
Since S ∈ I(P ) there is permutation π : N → N and Sπ ∈ P such that for all α ∈ Obs andfull assignments h, S |= α[h] if and only if Sπ |= α[hπ]. Hence:
(95) Every environment for S and h is an environment for Sπ and hπ, and vice versa.
By our assumption on Ψ, µπ( Hsolve(Ψ,Sπ, P ) ) > .5. Since P ⊆ I(P ), this immediatelyimplies:
(19) Proposition: Suppose that Sym consists of a binary predicate, and thatObs = Lbasic . Then there exist disjoint propositions P1, P2 and uncountableM0 ⊆ M such that {P1, P2} is unsolvable, but {P1, P2} is solvable on M0 withprobability 1.
Proof: Let the binary predicate of Sym be R. We specify a countable collection {Sj | j ∈ N}of structures by specifying the extension RSj of R for all j ∈ N . RS0 is the successor function{(i, i + 1) | i ∈ N}. For j > 0, RSj is the finite relation {(i, i + 1) | i < j}. Let P1 = {S0}and P2 = {Sj | j > 0}. To prove the proposition it suffices to show:
(98) (a) {P1, P2} is not solvable.
(b) Let M0 ⊂ M be any class of measures such that for all i ∈ N , glb{µ(i) |µ ∈M0} > 0. Then {P1, P2} is solvable on M0 with probability 1.
Exercise (49) of essay #2 yields (98)a. We prove (98)b. Towards specifying a scientist thatsolves P on M0 with probability 1, we introduce some notation and facts.
Given i, j ∈ N , let B(i, j) be the collection of h ∈ H such that not all of 0, 1, . . . , i occurin {h(v0) . . . h(vj)}. By the assumption on M0, let strictly increasing f : N → N be suchthat for all i ∈ N and µ ∈ M0, µ(B(i, f(i))) < 1
2i . So for each µ ∈ M0, Σiµ(B(i, f(i)))converges. Hence, by the first Borel-Cantelli lemma ([3, Theorem 4.3]):
(100) For every µ ∈ M0 the class of h ∈ H such that {0 . . . i} 6⊆ {h(v0) . . . h(vf(i))} forinfinitely many i ∈ N has µ-measure 0.
Call σ ∈ SEQ “complete through m” if m is the greatest number such for all i, j ≤ m,either
∧σ |= Rvivj or
∧σ |= ¬Rvivj .
We now define scientist Ψ that solves P on M0 with probability 1. Let σ ∈ SEQ begiven. If σ is not complete through f(0), then Ψ(σ) is undefined. Otherwise, suppose that nis greatest such that σ is complete through f(n). If
∧σ implies the existence of an R-chain
of length at least n, then Ψ(σ) = P1; otherwise, Ψ(σ) = P2. Because f is strictly increasing,Ψ is well defined.
To prove that Ψ solves P2 on M0 with probability 1, let standard environment e forP2 be given. Then for cofinitely many n ∈ N , content(e) does not imply the existence ofan R-chain of length at least n. So Ψ(e[k]) = P2 for cofinitely many k. Hence Ψ solvesP2. Hence Ψ solves P2 on M0 with probability 1. To prove that Ψ solves P1 on M0
with probability 1, call h ∈ H “bad” just in case for infinitely many k, dS0, he is completethrough f(k) but
∧dS0, he does not imply the existence of an R-chain of length at least
k. Let µ ∈ M0 be given. It follows directly from (100) that the class of bad h ∈ H hasµ-measure 0. Hence, there is a class X ⊆ H of µ-measure 1 such that for every h ∈ X,Ψ(dS0, he[k]) = P1 for cofinitely many k. That is, µ(Hsolve(Ψ,S0, P1)) = 1. Hence Ψ solvesP1 on M0 with probability 1.
(22) Theorem: Let problem P and M0 ⊆ M be given. Then P is solvableon M0 with probability 1 if and only if P is reliably solvable on M0 withprobability greater than 0.
To prove the theorem, we develop some preliminary material. Given n ∈ N and h ∈ H,we denote by hn the assignment {(vi, h(vi+n)) | i ∈ N}. Thus, hn is the assignment generatedby “starting over” at n. If h is repetitive, then hn is a full assignment. It is easy to verifythe following lemma.
(101) Lemma: For every n ∈ N there is a (computable) function Fn : SEQ → SEQ suchthat:
(a) for all σ, τ ∈ SEQ with σ ⊆ τ , Fn(σ) ⊆ Fn(τ);
(b) for all structures S and repetitive h ∈ H,⋃
k∈N Fn(dS, he[k]) = dS, hne.
Let scientist Ψ, proposition P , and S ∈ P be given. Because members of H areconstructed via independent draws from N , it is easy to see that for every distinct n,m ∈ Nand every µ ∈ M , the conditional µ-probability that Ψ solves P in dS, hme given that Ψsolves P in dS, hne is equal to the µ-probability that Ψ solves P in dS, hme. This is becausethe repetitive full assignments have µ-measure 1 by Corollary (9), and ignoring the first nversus m draws from N changes nothing in the remaining draws. In other words:
(102) Lemma: Let scientist Ψ, proposition P , S ∈ P , µ ∈ M , and distinct n,m ∈ N begiven. Then the events
{h ∈ H | ∅ 6= Ψ(dS, hne[k]) ⊆ P for cofinitely many k ∈ N},
{h ∈ H | ∅ 6= Ψ(dS, hme[k]) ⊆ P for cofinitely many k ∈ N}
are independent and have µ-probability equal to µ(Hsolve(Ψ,S, P )).
We can now prove the following, key lemma.
(103) Lemma: Let scientist Ψ reliably solve P on M0 with probability greater than 0.Then for all P ∈ P, all S ∈ P , and all µ ∈ M0 the following holds.
(a) Let X be the class of all repetitive h ∈ H such that for some n ∈ N , Ψ solvesP on dS, hne. Then µ(X) = 1.
(b) Let Z be the class of all repetitive h ∈ H such that for some n ∈ N , Ψ isunreliable on S, hn, and P . Then µ(Z) = 0.
Proof: Part (a) is an easy consequence of Lemma (102). By Definition (21), Z is a countableunion of sets of µ-measure 0. This proves Part (b).
Proof of Theorem (22): The left-to-right direction is immediate. To prove the otherdirection, let scientist Ψ reliably solve P on M0 with probability greater than 0. Withoutloss of generality, we can assume:
(104) Ψ is total and for all σ ∈ SEQ, Ψ(σ) ∈ P.
We describe a scientist Ψ′ that solves P on M0 with probability 1.
Given σ ∈ SEQ and k ≤ length(σ), let σ[k] be the initial sequence of length k in σ. ForP ∈ P, define the “P -score” of σ ∈ SEQ to be length(σ)−k0 for the smallest k0 ≤ length(σ)such that Ψ(σ[k]) = P for all k with k0 ≤ k ≤ length(σ). By (104), the P -score of anyσ ∈ SEQ is well defined.
Then Ψ′(σ) = P . Otherwise, Ψ′(σ) is undefined. Intuitively, faced with environmentdS, he, Ψ′ examines Ψ’s behavior on initial segments of environments generated from theinformation in the “tails” of dS, he. Ψ′ issues the proposition that Ψ issues the most timesin a row.
Now let P ∈ P, S ∈ P , and µ ∈ M0 be given. By Lemma (103)a:
µ({h ∈ H | (∃n0)Ψ(Fn0(dS, he[k]) = P for cofinitely many k ∈ N}) = 1.
In contrast, By Lemma (103)b, for all Q ∈ P with Q 6= P ,
µ({h ∈ H | (∃n0)Ψ(Fn0(dS, he[k]) = Q for cofinitely many k ∈ N}) = 0.
It follows easily from the definition of Ψ′ that:
µ(h ∈ H |Ψ′(dS, he[k]) = P for cofinitely many k ∈ N}) = 1,
as desired.
5.9. Proof of Lemma (25)
(25) Lemma: Let p ∈ (0, 1), µ ∈ M , U ∈ {S`,Sg}, and σ ∈ SEQ be given.Then Prµ,p({〈h,U〉 | U |= σ[h]}) is defined.
Proof: For i, j ∈ N , let β be the basic formula Rvivj . Then {〈h,S`〉 | S` |= β[h]} is thecountable union of sets of the form Oζ ×{S`}, where length(ζ) = max(i, j) and ζ(i) ≤ ζ(j).A similar fact holds for other basic formulas and Sg. The set {〈h,U〉 | U |= σ[h]} is a finiteintersection of such countable unions, hence Prµ,p-measurable.
5.10. Proof of Proposition (28)
(28) Proposition: No Bayesian scientist solves P∗.
Proof: Suppose that scientist Ψ is Bayesian with respect to p ∈ (0, 1), and µ ∈ M . Letσ ∈ SEQ, and finite assignment a : Var → N be given with domain(a) ⊇ Var(σ) andS` |=
∧σ[a]. We shall exhibit τ ∈ SEQ such that:
(105) (a) S` |= ∃x∧
(σ∗τ)[a], where x contains the variables in Var(τ)−domain(a), and
(b) Ψ(σ∗τ) = {Sg}.
This suffices to show that there is no locking pair for Ψ, S`, and {S`}, in the sense ofDefinition (28) of essay #2. It then follows from Lemma (29) of essay #2 that Ψ does notsolve {S`} in every environment for S`, hence does not solve P∗.
Choose i to be greater than any index for a variable in σ. Let G(n,m) ∈ SEQ be auniformly chosen finite sequence with range {Rvi+j vi+j+1 | j ≤ n}∪ {vi+n+1 = vi+n+j | 1 <j ≤ m}. Intuitively, S` interprets G(n,m) as: “there is a long ascending sequence whose lastmember must be far from 0, and is sampled many times,” whereas Sg interprets G(n,m) as:“there is a long descending sequence whose last member may be close to 0 and is sampledmany times.” Since µ must ultimately attach smaller probabilities to large numbers thanto 0, as n,m get large, G(n,m) becomes much less likely under S` than under Sg. Using
Since S` and Sg satisfy the same members of SEQ, and S` |=∧σ[a], both of the numbers
Prµ,p({〈h,S`〉 | S` |= σ[h]}) and Prµ,p({〈h,Sg〉 | Sg |= σ[h]}) are greater than 0. The sameis true of p, (1− p) by Definition (27)a. So from (106) we infer:
(107) There are n,m ∈ N such that p× Prµ,p({〈h,S`〉 | S` |= σ∗G(n,m)[h]}) < (1− p)×Prµ,p({〈h,Sg〉 | Sg |= σ∗G(n,m)[h]}).
Let τ = G(n,m) for n,m that satisfy (107). Since S` and Sg satisfy the same existentialsentences, the choice of i implies (105)a. Definition (27)a and (107) imply (105)b.
(108) Fact: Structuresg S and T are said to be finitely isomorphic in the standard sensejust in case there is a sequence {Jn |n ∈ N} of nonempty sets of partial isomorphismsfrom S to T such that for all n ∈ N the following holds:
(a) For every f ∈ Jn+1 and a ∈ | S | there is g ∈ Jn such that g extends f anda ∈ domain(g).
(b) For every f ∈ Jn+1 and a ∈ | T | there is g ∈ Jn such that g extends f anda ∈ domain(g).
Show that S and T are finitely isomorphic in the standard sense iff they are finitelyisomorphic in the sense of Definition (31).
Proof: As an immediate consequence of Definition (31), we have:
(109) For all ordinals α, Iα(S, T ) 6= ∅ iff the empty partial isomorphism belongs toIα(S, T ).
It is also immediate from the definition that S and T are finitely isomorphic iff for alln ∈ N , In(S, T ) 6= ∅, which by (33)a is equivalent to Iω(S, T ) 6= ∅.
(110) Fact: Structuresg S and T are said to be partially isomorphic in the standardsense just in case there is a nonempty set J of partial isomorphisms from S to Tsuch that the following holds.
(a) For every f ∈ J and a ∈ | S | there is g ∈ J such that g extends f anda ∈ domain(g).
(b) For every f ∈ J and a ∈ | T | there is g ∈ J such that g extends f anda ∈ domain(g).
Show that S and T are partially isomorphic in the standard sense iff they arepartially isomorphic in the sense of Definition (31).
Proof: Suppose that S and T are partially isomorphic in the standard sense. Let nonemptyI ⊆ PI(S, T ) be such that the following holds:
(a) for all a ∈ | S | and f ∈ I, there is g ∈ I such that g extends f and domain(g) =domain(f) ∪ {a};
(b) for all a ∈ | T | and f ∈ I, there is g ∈ I such that g extends f and content(g) =content(f) ∪ {a}.
Then it is easy to verify by induction on the class of ordinals that for all ordinals α,I ⊆ Iα(S, T ). For the converse, suppose that Iα(S, T ) 6= ∅ for all ordinals α. SincePI(S, T ) is a set and not a proper class, it follows from (33)b that there is ordinal α suchthat Iα(S, T ) = Iβ(S, T ) for all ordinals β ≥ α. Hence we may choose ordinal α such thatIα = Iα+1 6= ∅. Let f ∈ Iα be given. Since f ∈ Iα+1, it follows that:
(a) for all a ∈ | S | there is g ∈ Iα(S, T ) such that g extends f and domain(g) =domain(f) ∪ {a};
(b) for all a ∈ | T | there is g ∈ Iα(S, T ) such that g extends f and content(g) =content(f) ∪ {a}.
Suppose that Sym is limited to a binary predicate.
(a) Let S = (N, succ) be the natural numbers with the successor relationand T = (N ⊕ Z, succ), be the disjoint union of the natural numbersand the integers with the successor relation. Then Iω(S, T ) 6= ∅, butIω+1(S, T ) = ∅.
(b) Let S = (Z,<) and T = (Z + Z,<), where Z is the set of integers, Z + Zis two copies of the integers (one above the other), and < is the naturalordering. Then Iω+1(S, T ) 6= ∅, but Iω+2(S, T ) = ∅.
Proof: We prove Example (32)a. In order to show that Iω is nonempty, it suffices to showthat In is nonempty for each n ∈ ω. For each n ∈ ω, we describe a nonempty set of partialisomorphisms Jn which is easily seen to be contained in In. Let a, b ∈ | S | and c, d ∈ T .We say that < a, b > is n-separated as < c, d > just in case either the distance from a to bin S is equal to the distance from c to d in T or both distances are greater than n. We saythat a finite function f from | S | to | T | is n-satisfactory just in case 0 is in the domain off and for every a, b in the domain of f,< a, b > is n-separated as < f(a), f(b) > . We thenlet Jn be the set of finite functions from | S | to | T | which are n-satisfactory. On the otherhand, it is easy to see that Iω+1(S, T ) = ∅, for otherwise, there would be some f ∈ Iω withf(i) ∈ Z, for some i ∈ N. But this is obviously impossible, since no such f can be containedin fi+1.
(a) The empty partial isomorphism belongs to I0(S, T ).
(b) For all ordinals α, β, if α ≤ β then Iβ(S, T ) ⊆ Iα(S, T ).
(c) Suppose that Sym is finite. Then S and T are finitely isomorphic iff Sand T are elementarily equivalent.
(d) Suppose that | S | and | T | are countable. Then S and T are partiallyisomorphic iff S and T are isomorphic.
Proof: The proofs of (33)a,b are trivial. We obtain (33)c directly from Fraısse’s theorem,[5, Thm XII.2.1]. Similarly, (33)d is a generalization of a famous result of Cantor, provedin [5, Thm XII.1.5d].
5.14. Proof of Lemma (38)
.
(38) Lemma: For all structuresg T and for all countable substructures S of T ,Iω1(S, T ) 6= ∅ if and only if S is partially isomorphic to T .
Proof: It follows from the results described in [2, Sec. VII.6] (see especially the material fromCorollary 6.4 to Theorem 6.8) that for every countable structure S, there is a countableordinal α such that for every structureg T , Iα(S, T ) 6= ∅, if and only if, S is partiallyisomorphic to T . The lemma follows immediately from this fact and Lemma (33)b.
(39) Lemma: Suppose that Sym holds countably many unary predicates. Thenthere is a structureg T with cardinality of the continuum such that I1(S, T ) = ∅,for all countable substructures S of T .
Proof: Let the unary predicates be enumerated as Pi, i ∈ N . Let structureg T be suchthat | T | = pow(N), and for all i ∈ N , P T
i = {X ⊆ N | i ∈ X}. To prove the lemma, letcountable substructure S of T be given. Since there are uncountably many subsets of N ,we may choose X0 ⊆ N such that:
(111) for all X ∈ | S | there is i ∈ N such that X ∈ PSi iff X0 6∈ P T
i .
Directly from Definition (31), (111) is enough to show that I1(S, T ) = ∅.
5.16. Proof of Lemma (40)
(40) Lemma: Suppose that Sym is limited to a binary predicate. Then thereis structureg T such that:
(a) for all countable substructures S of T , Iω1(S, T ) = ∅;(b) for all ordinals α < ω1, there is a countable substructure S of T with
Iα(S, T ) 6= ∅.
Proof: It is proven in [6] that:
(112) (a) for all ordinals α, if 0 < α < ω1 then Iω1(α, ω1) = ∅;
Let countable substructure S of T be given. Since S is isomorphic to a countableordinal, Iω1(S, T ) = ∅ by (112)a. This proves (40)a. As for (40)b, it follows immediatelyfrom (112)b and the fact that for all ordinals α < ω1, ωα is a countable substructure of ω1.
5.17. Proof of Proposition (41)
(41) Proposition: Let problem P be given. Then P is solvableg if and only ifP is solvable.
Proof: Suppose that scientistg Ψ solvesg P. Let scientist Ψ′ be such that for all σ ∈ SEQ,Ψ′(σ) = {S ∈ Ψ(σ) | S is countable}. By Lemma (33)d, Iα(S,S) 6= ∅ for all ordinals α, soit follows directly from Definition (36) that Ψ′ solves P.
Conversely, suppose that P is solvable, and let (countable) T ∈⋃
P be given. ByLemma (38), for all countable substructures S of T , Iω1(S, T ) 6= ∅ if and only if S ispartially isomorphic to T . Since T is countable, Lemma (33)d implies that for all countablesubstructures S of T , Iω1(S, T ) 6= ∅ if and only if S is isomorphic to T . Obviously,Iω1(T , T ) 6= ∅, hence the first clause of Definition (36) is satisfied. The second clausefollows immediately from the hypothesis that P is solvable.
5.18. Proof of Lemma (44)
(44) Lemma: Suppose that Sym is finite. Let enumeration {Xi | i ∈ N} ofsubsets of pow(Lsen) be given. Then the problem {
Proof: For the right-to-left direction, suppose that {⋃
E∈XiMOD(E)g | i ∈ N} is solvableg.
Then trivially, {⋃
E∈XiMOD(E) | i ∈ N} is solvableg. So {
⋃E∈Xi
MOD(E) | i ∈ N} issolvable by Proposition (41).
For the other direction, suppose that scientist Ψ solves {⋃
E∈XiMOD(E) | i ∈ N}. Let
i ∈ N , E ∈ Xi and T ∈ MOD(E)g be given. Observe that Ψ is also a scientistg. So byDefinition (36) and the fact that
⋃E∈Xi
MOD(E) ⊆⋃
E∈XiMOD(E)g it suffices to show:
(113) (a) there is countable substructure S of T with Iω(S, T ) 6= ∅;(b) for all countable substructures S of T with Iω(S, T ) 6= ∅ and for all environ-
ments e for S, ∅ 6= Ψ(e[k]) ⊆⋃
E∈XiMOD(E), for cofinitely many k.
By Lemma (33)c and the assumption on Sym, (113) follows directly from:
(114) (a) there is countable substructure S of T that is elementarily equivalent to T ;
(b) for all countable substructures S of T that are elementarily equivalent to T ,and for all environments e for S, ∅ 6= Ψ(e[k]) ⊆
⋃E∈Xi
MOD(E), for cofinitelymany k.
The downward Lowenheim-Skolem theorem implies (114)a. We obtain (114)b. from the factthat Ψ solves
⋃E∈Xi
MOD(E), since the latter class is closed under elementarily equivalence[with respect to (countable) structures].
(46) Proposition: Suppose that Sym consists of a denumerable set of unarypredicates. Then there exists two structuresg S and T and a universal sentenceθ such that:
(a) S |= θ and T |= ¬θ;(b) {{S}, {T }} is not solvableg.
Proof: Let Q and Pi, i ∈ N enumerate the unary predicates of Sym. Let structuresg S andT be defined as follows.
(115) (a) | S | = | T | = pow(N);
(b) QS = pow(N);
(c) QT = {∅};(d) for all i ∈ N , PS
i = P Ti = {X ⊆ N | i ∈ X}.
From (115)a-c we have that S |= ∀xQx and T |= ¬∀xQx. From (115)d it follows thatuncountably many elements are distinguished by the Pi’s in both S and T . It follows easilythat for all (countable) structures U , I1(S,U) = ∅ and I1(T ,U) = ∅. This with Lemma(33)a,b implies that for all (countable) structures U and all ordinals α:
Trivally, U is a substructure of S and U is a substructure of T . Suppose for a contradictionthat scientistg Ψ solvesg {{S}, {T }}. Then Definition (36) togehter with (116) implies thatfor every environment e for U , Ψ(e[k]) = {S} for cofinitely many k, and Ψ(e[k]) = {T } forcofinitely many k, contradiction.
5.20. Proof of Theorem (49) and its corollary
(49) Theorem: Let problemg P be 2ω-saturated. Then P is solvableg if andonly if P is 2ω-solvableg.
Corollary: Let problem P be given. Then P is solvable if and only if P is2ω-solvableg.
Let us first prove the corollary from the theorem.
Proof of the corollary: Since P is a problem, every S ∈⋃
P is countable. It followsimmediately from Definition (48)a that P is α-saturated for all ordinals α. So the corollaryfollows directly from Proposition (41) and Theorem (49).
To prove the theorem we need a technical lemma bearing on the original, countableparadigm. Theorem (59) of essay #2 can be generalized to the case of problems of arbitraryform. Indeed, Theorem (42) of essay #2 can be interpreted as expressing an existential-universal relation between solvability and expressibility in the infinitary language Lω1ω.8
Indeed, as an immediate consequence of Theorem (42) and Lemma (92) of the earlier essay,we have:
(117) Corollary: A problem P is solvable iff it consists of countably many proposi-tions, each of which is equivalent in
⋃P to an Lω1ω-sentence which is a countable
disjunction of members of Lsen of form:
∃xχ or ∃x∧i<ω
∀yi¬ϕi or ∃x(χ ∧∧i<ω
∀yi¬ϕi),
where χ and ϕi are built from Obs using conjunctions and disjunctions only, for alli < ω.
Proof of Theorem (49) also hinges on the following lemma.
(118) Lemma: Let structuresg S and T be such that I2ω(S, T ) 6= ∅. Let X be anycollection of formulas all of whose free variables are drawn from the same finite set.Then X is satisfiable in S iff X is satisfiable in T .
Proof: Let n ∈ N and variables x0 . . . xn be such that for all ϕ ∈ X, Var(ϕ) ⊆ {x0 . . . xn}.Using standard techniques to eliminate constants and function symbols (see [5]), there existssymbol set Sym? which consists exclusively of predicates, Sym?-structures S? and T ?, andset X? of Sym?-formulas such that:
(119) (a) | S | =| S? |, | T | =| T ? |, and PI(S, T ) = PI(S?, T ?);
(b) all free variables occuring in any member of X? are drawn from {x0 . . . xn};(c) X is satisfiable in S iff X? is satisfiable in S?;
(d) X is satisfiable in T iff X? is satisfiable in T ?.
From (119)a and the hypothesis that I2ω(S, T ) 6= ∅, we infer that I2ω(S?, T ?) 6= ∅. Fromthis and (119)b-d, it follows that we can, without loss of generality, suppose that Symconsists exclusively of predicates.
Now suppose that X is satisfiable in S. Choose a0 . . . an ∈ | S | such that for all ϕ ∈ X,S |= ϕ[a0/x0 . . . an/xn]. Since I2ω(S, T ) 6= ∅, choose p ∈ Iω+n+1(S, T ). By Definition (31),p can be extended to q ∈ Iω(S, T ) with domain(q) ⊇ {a0 . . . an}. Since the quantifier ranksof the formulas in X are all finite, it follows immediately from [5, Lemma XII.3.2] thatT |= ϕ[q(a0)/x0 . . . q(an)/xn], for all ϕ ∈ X.
Proof of Theorem (49): We first note the following, immediate consequence of Lemma (118).concerning the language Lω1ω. (See [5, Section IX.2] for Lω1ω.)
(120) Let structuresg S and T be such that I2ω(S, T ) 6= ∅. Then S and T satisfy thesame extended-∃∀ sentences of Lω1ω.
Proof of Theorem (49): Trivially, if P is 2ω-solvableg then P is solvableg. For the otherdirection, suppose that scientistg Ψ solvesg problemg P. Since SEQ is countable, content(Ψ)is countable, so it follows immediately that P is countable. Because Ψ solvesg P, it followsfrom Definition (36) and Lemma (33)b that for each P ∈ P and T ∈ P we may choosea countable substructure T ? of T such that: for some α ≥ 2ω, Iα(T ?, T ) 6= ∅ and forall environments e for T ?, ∅ 6= Ψ(e[k]) ⊆ P for cofinitely many k. For all P ∈ P, setP ? = {T ? | T ∈ P}. Set P? = {P ? |P ∈ P}. It follows immediately that P? is a problem(that is, the P ? ∈ P? are disjoint sets of countable structures), and that P? is solvable. ByCorollary (117), for all P ∈ P there is an extended-∃∀ sentence of Lω1ω that is true in allT ? ∈ P ?, and false in all T ? ∈
⋃P? − P ?. So it follows from (120) and the choice of the
(121) for all P ∈ P there is an extended-∃∀ sentence of Lω1ω that is true in all T ∈ P ,and false in all T ∈
⋃P− P .
For all T ∈⋃
P, define P (T ) to be the set of countable substructures S of T withI2ω(S, T ) 6= ∅. For all P ∈ P, set P ◦ =
⋃{P (T ) | T ∈ P}, and set P◦ = {P ◦ |P ∈ P}. It
follows from (120) and (121) that for all P ∈ P there is an extended-∃∀ sentence of Lω1ω
that is true in all T ∈ P ◦, and false in all T ∈⋃
P◦ − P ◦. The P ◦ ∈ P◦ are thus disjoint,so P◦ is a problem. Moreover, since P is countable, so is P◦. It follows from anotherapplication of Corollary (117) that P◦ is solvable. With Definition (36), we conclude thatP is 2ω-solvableg.
5.21. Proof of Proposition (57)
(57) Proposition: Suppose that Sym consists of a binary predicate. Thenthere is a problem P such that:
(a) the discrete ∀-type of P is defined (hence P is solvable);
(b) the discrete Lsen -type of P is undefined.
Proof: Let R be the binary predicate of Sym. Let T be the theory of total orders (withrespect to R) with either a least point or a greatest point (but not both). Let θ = ∃x∀yRxy.We show that P = {MOD(T ∪ {θ}),MOD(T ∪ {¬θ})}, satisfies the claim of the proposition.Clause (a) has been proved in Example (56). We prove (b). Let O be any well orderingof some set X of sentences. If no sentence in X is true in any member of
⋃P, then
∆[P, O] = ∅ and we are done. Otherwise, let sentence ψ be the O-least member of Xwhich is true in some S ∈
⋃P. Suppose that S ∈ MOD(T ∪ {θ}) (the proof is parallel if
S ∈ MOD(T ∪ {¬θ})). Let σ ∈ SEQ be for P. Because all the models of T are infinite,∧σ
is satisfiable in S, hence∧σ∧ψ is satisfiable in S (since ψ is closed). This shows that for all
σ ∈ SEQ which are for P, either ∆[P, O](σ) = MOD(T ∪ {θ}) or ∆[P, O](σ) is undefined.Hence ∆[P, O] does not solve MOD(T ∪ {¬θ}), so ∆[P, O] does not solve P.
5.22. Proofs of Lemmas (59) and (60)
(59) Lemma: Every solvable problem of form (T , {θ0 . . . θn}) has a tip-off base.
(60) Lemma: No infinite problem of form (T , {P0, P1, . . .}) has a tip-off base.
Proof of Lemma (59): Let a solvable problem of form (T , {θ0 . . . θn}) be given. ByCorollary (58) of essay #2, for all m ≤ n, θm is equivalent in T to an ∃∀ sentence. It followseasily that the problem has a tip-off base.
Proof of Lemma (60): Let T ⊆ Lsen and disjoint propositions P0, P1 . . . be such thatMOD(T ) =
⋃{Pi | i ∈ N}. Suppose for a contradiction that set X = {ϕi | i ∈ N} of ∀
formulas is a tip-off base for P = {P0, P1 . . .}. For all i ∈ N , let ψi be the conjunction ofthe universal closure of ¬ϕ0 ∧ . . . ∧ ¬ϕi−1 with the existential closure of ϕi. Trivially:
(122) For all i, j ∈ N , if i 6= j then MOD(ψi) ∩MOD(ψj) = ∅.
Let P ∈ P and S ∈ P be given. By Definition (58)a, there is least i ∈ N such that ϕi
is satisfiable in S. Hence ψi is true in S ∈ P . By Definition (58)b, ϕi is satisfiable in noP ′ ∈ P with P ′ 6= P . Hence ψi is false in all members of P ′ ∈ P, P ′ 6= P . Since P isinfinite, this proves that:
(123) There are infinitely many i ∈ N with MOD(T ∪ {ψi}) 6= ∅.
From (122) and (123), we infer that T ∪ {¬ψ0 . . .¬ψi} is satisfiable for all i ∈ N . Withcompactness, this implies that T ∪ {¬ψi | i ∈ N} is satisfiable, which contradicts (124).
5.23. Proof of Proposition (61)
(61) Proposition: Let set X of ∀ formulas be a tip-off base for problem P.Then for every enumeration E of ∀ formulas with X ⊆ content(E), the scientist∆[P, E] solves P.
Proof: Let X be a tip-off base for P, and let E = {ϕi | i ∈ N} be an enumeration of ∀formulas with X ⊆ content(E). Let P ∈ P, S ∈ P , full assignment h to S, and environmente for S and h be given. We will show that ∆[P, E] solves P in e, thus proving that ∆[P, E]solves P.
By Definition (58), there is least i0 ∈ N such that:
(a) S |= ϕi0 [h], and
(b) ϕi0 is satisfiable in no P ′ ∈ P with P ′ 6= P .
Let k0 ∈ N be such that for all i ≤ i0, if S |= ¬ϕi[h] then∧e[k0] |= ¬ϕi. By the choice of
i0:
(a) for all i < i0 and all k ≥ k0, if∧e[k] 6|= ¬ϕi then ϕi is satisfiable in at least two
structures taken from two distinct propositions in P;
(b) for all k ≥ k0, content(e[k])∪ {ϕi0} is satisfiable in S ∈ P , and ϕi0 is satisfiable in nomember of P ′, for all P ′ ∈ P, P ′ 6= P .
This implies immediately that for all k ≥ k0, ∆[P, E](e[k]) = P . So ∆[P, E] solves P in e.
5.24. Proof of Proposition (64)
(64) Proposition: Every problem with finite discrete ∀-type has a tip-off base.
Proof: Let problem P with finite discrete ∀-type be given. Let finite enumeration E of ∀formulas be such that ∆[P, E] solves P. Define X to be the set of all formulas of form∧σ ∧ ϕ, σ ∈ SEQ, ϕ ∈ E, satisfying the following:
(125) there is at most one P ∈ P such that∧σ ∧ ϕ is satisfiable in some member of P .
It suffices to show that X is a tip-off base for P. By (125) this is proved if we show thatfor all S ∈
⋃P and full assignments h to S, there is ψ ∈ X with S |= ψ[h]. So let P ∈ P,
S ∈ P , and full assignment h to S be given. Since ∆[P, E] solves P, there is k0 ∈ N andϕ ∈ E such that:
(126) for all k ≥ k0, ∅ 6= {T ∈⋃
P |∧e[k] ∧ ϕ is satisfiable in T } ⊆ P .
We derive immediately from (126) that∧e[k0] ∧ ϕ belongs to X. Hence it suffices to show
that S |= (∧e[k0] ∧ ϕ)[h]. From (126) again we infer that
∧e[k] ∧ ϕ is satisfiable for all
k ∈ N . Hence by compactness, content(e)∪ {∧e[k0]∧ϕ} is satisfiable. Since
∧e[k0]∧ϕ is
a ∀ formula and e is an environment for S and h, this implies that S |= (∧e[k0] ∧ ϕ)[h], as
(65) Proposition: For all problems P, the discrete ∀-type of P is finite iff thesegmental ∀-type of P is finite. Moreover, if finite, they are equal.
Proof: Let problem P be given. The discrete ∀-type of P is equal to 0 iff the segmental∀-type of P is equal to 0 iff P = ∅. So suppose P 6= ∅. If P consists of a sole proposition,and if this proposition is the class of all structures, then it is easy to verify that the discreteand segmental ∀-types of P are both equal to 1. So suppose otherwise. It suffices to showthat if the discrete ∀-type of P is equal to p > 0, then the segmental ∀-type of P is at mostequal to p, and if the segmental ∀-type of P is equal to p > 0, then the discrete ∀-type of Pis at most equal to p. Let n ∈ N and enumeration E = {ϕi | i ≤ n} of ∀ formulas be suchthat ∆[P, E] solves P. We show that Ψ[P, E] solves P, thus proving that if the discrete∀-type of P is finite, then the segmental ∀-type of P is at most equal to the latter. LetP ∈ P, S ∈ P , full assignment h to S, and environment e for S and h be given. It sufficesto show that Ψ[P, E] solves P in e. Since ∆[P, E] solves P (and E is finite), there is k0 ∈ Nand n0 ≤ n such that:
(127) for all k ≥ k0, ∅ 6= {T ∈⋃
P |∧e[k] ∧ ϕn0 is satisfiable in T } ⊆ P .
In particular, (127) implies that∧e[k] ∧ ϕn0 is satisfiable for all k ∈ N . Hence by com-
pactness, content(e)∪{ϕn0} is satisfiable. Since ϕn0 is a ∀ formula and e is an environmentfor S and h, we infer that S |= ϕn0 [h]. Let X be the set of all ϕ ∈ {ϕ0 . . . ϕn0} such thatS |= ϕ[h]. We have thus shown that ϕn0 ∈ X. With (127), this implies that:
Let k1 ≥ k0 be such that for all ϕ ∈ {ϕ0 . . . ϕn0} −X,∧e[k1] |= ¬ϕ. Then for all k ≥ k1
and for all initial segments s of E, {∧e[k]∧ϕ |ϕ ∈ {ϕ0 . . . ϕn0}−X}∩ satform(s, e[k]) = ∅.
With (128), this implies that for all k ≥ k1, Ψ[P, E](e[k]) = P . Hence Ψ[P, E] solves P ine, as required.
Conversely, let n ∈ N and enumeration E = {ϕi | i ≤ n} of ∀ formulas be such thatΨ[P, E] solves P. For all i ≤ n, denote by ψi the disjunction of all the conjunctions ofsubsets of {ϕ0 . . . ϕn} whose cardinality is equal to n + 1 − i (conjunctions are written inascending order of indexes of the ϕi’s). For instance:
ψ0 =∧i≤n
ϕi, ψ1 =∨i≤n
∧j≤n
j 6=i
ϕj , and ψn =∨i≤n
ϕi.
Note that all of the ψi’s are ∀ formulas. Set F = {ψi | i ≤ n}. We show that ∆[P, F ] solvesP, thus proving that if the segmental ∀-type of P is finite, then the discrete ∀-type of P isat most equal to the latter. Let P ∈ P, S ∈ P , full assignment h to S, and environment efor S and h be given. It suffices to show that ∆[P, F ] solves P in e. Since Ψ[P, E] solvesP (and E is finite), there is k0 ∈ N and X ⊆ {ϕ0 . . . ϕn} such that:
(129) for all k ≥ k0, ∅ 6= {T ∈⋃
P | content(e[k]) ∪X is satisfiable in T } ⊆ P .
Since by hypothesis P does not consist of a sole proposition equal to the class of all struc-tures, it is clear from Definition (55) that:
(130) X 6= ∅.
Moreover, (129) implies that content(e[k]) ∪ X is consistent for all k ∈ N . Hence bycompactness, content(e) ∪ X is satisfiable. Since X is a set of ∀ formulas and e is an
environment for S and h, we infer that S |= X[h]. Let Y + (respectively Y −) be the set ofall ϕ ∈ {ϕ0 . . . ϕn} such that S |= ϕ[h] (respectively S 6|= ϕ[h]). We have thus shown that:
(131) X ⊆ Y +.
Set i0 = card(Y −). By (130) and (131), i0 ≤ n. By a finite pigeon hole argument it followseasily that:
(132) (a) for all i < i0, every subset of {ϕ0 . . . ϕn} whose cardinality is equal to n+1− ihas a nonempty intersection with Y −;
(b) Y + is the only subset of {ϕ0 . . . ϕn} whose cardinality is equal to n + 1 − i0which has an empty intersection with Y −.
Let k1 ≥ k0 be such that for all ϕ ∈ Y −,∧e[k1] |= ¬ϕ. It follows immediately from (132)
and the definition of the ψi’s, i ≤ n, that:
(133) (a) for all i < i0 and k ≥ k1,∧e[k] ∧ ψi is unsatisfiable;
(b) for all k ≥ k1,∧e[k] ∧ ψi0 is equivalent to
∧e[k] ∧
∧Y +.
From (129), (131) and (133), we infer that ∆[P, F ](e[k]) = P for all k ≥ k1. Hence ∆[P, F ]solves P in e, as required.
5.26. Proof of Proposition (66)
(66) Proposition: Suppose that Sym = ∅. For all n ∈ N there is a problemwhose discrete and segmental ∀-types are n.
Proof: The empty problem satisfies the claim of the proposition for n = 0. Given n > 0,denote by ϕn the sentence ∀x0 . . . xn(
∨0≤i<j≤n xi = xj) (“there are at most n elements”).
For all n > 0, let Pn be the class of structures whose cardinality is n. Let n > 0 be given.By Proposition (65) it suffices to show that the problem P = {P1 . . . Pn} has discrete ∀-typeequal to n. Set E = {ϕ1 . . . ϕn}. We first show that ∆[P, E] solves P, thus proving thatthe discrete ∀-type of P is at most equal to n. Let 1 ≤ m ≤ n and environment e forPm be given. It suffices to show that ∆[P, E] solves Pm in e. Let k0 ∈ N be such that∧e[k0] implies that there are at least m distinct elements in the domain of the underlying
structure. Trivially, for all k ≥ k0,∧e[k] ∧ ϕp is unsatisfiable for all 1 ≤ p < m, and ∅ 6=
{S ∈⋃
P |∧e[k]∧ϕm is satisfiable in S} ⊆ Pm. Hence for all k ≥ k0, ∆[P, E](e[k]) = Pm.
So ∆[P, E] solves Pm in e, as required.
We now show that the discrete ∀-type of P is at least equal to n, thus completing theproof. Let m > 0 and enumeration F = {ψ1 . . . ψm} of ∀ formulas be such that ∆[P, F ]solves P. It suffices to show that n ≤ m. Suppose that sequence {σp | 1 ≤ p ≤ n} ofmembers of SEQ and function f : {1 . . . n} → {1 . . .m} satisfy:
(134) (a) for all 1 ≤ p < n, σp ⊆ σp+1;
(b) for all 1 ≤ p ≤ n, ∅ 6= {S ∈⋃
P |∧σp ∧ ψf(p) is satisfiable in S} ⊆ Pp.
Since the Pp’s are pairwise disjoint, it is easy to see that f is one-to-one, which implies thatn ≤ m. So we only have to build a sequence {σp | 1 ≤ p ≤ n} of members of SEQ and afunction f : {1 . . . n} → {1 . . .m} that satisfy (134). We proceed by induction on p. Forp = 1 let e1 be an environment for P1. Since ∆[P, F ] solves P there is k ∈ N and 1 ≤ i ≤ msuch that ∅ 6= {S ∈
⋃P |
∧e1[k]∧ψi is satisfiable in S} ⊆ P1. Set σ1 = e1[k] and f(1) = i.
Observe that σ1 is an initial segment of an environment for P1. For the induction step p > 1suppose that σ1 . . . σp−1 and f(1) . . . f(p− 1) have been defined and that σp−1 is an initialsegment of an environment for Pp−1. Let environment e for Pp be given, extending σp−1;
there is such an environment since the cardinality of structures in Pp is greater than thatfor structures in Pp−1. Since ∆[P, F ] solves P, there is k > length(σp−1) and 1 ≤ i ≤ mwith ∅ 6= {S ∈
⋃P |
∧e[k] ∧ ψi is satisfiable in S} ⊆ Pp. Set σp = e[k] and f(p) = i. We
see that {σp | 1 ≤ p ≤ n} and f satisfy (134).
5.27. Proof of Proposition (67)
(67) Proposition: For all problems P, if the segmental Lform -type of P is finitethen the discrete Lform -type of P is finite.
Proof: Let problem P have finite segmental Lform -type. Let enumeration E = {ϕ0 . . . ϕn}of formulas be such that Ψ[P, E] solves P. Without loss of generality we can suppose thatϕ0 = (v0 = v0). Denote by F the enumeration of {ϕi0∧ . . .∧ϕip | p ∈ N, 0 ≤ i0 < . . . < ip ≤n} in lexicographical order. (For instance, if n = 5 then ϕ0 ∧ϕ2 comes before ϕ0 ∧ϕ2 ∧ϕ3,which comes before ϕ2.) It suffices to show that ∆[P, F ] solves P. Let P ∈ P, S ∈ P , fullassignment h to S, and environment e for S and h be given. It suffices to show that ∆[P, F ]solves P in e. Since Ψ[P, E] solves P [and E is finite and begins with (x = x)], there ism ≤ n and nonempty X ⊆ {ϕ0 . . . ϕm} such that the following holds for cofinitely many k:
(135) (a) ∅ 6= {T ∈⋃
P | content(e[k]) ∪X is satisfiable in T } ⊆ P ;(b) for all ∅ 6= X ′ ⊂ X, {T ∈
⋃P | content(e[k]) ∪X ′ is satisfiable in T } 6⊆ P ;
(c) for all p ≤ m, if ϕp 6∈ X then∧e[k] ∧ ϕp is not satisfiable.
Let ψ ∈ F be the conjunction of all members of X (written in ascending order of indexesof the ϕi’s). Let χ ∈ F come before ψ in F . By the definition of F there are two cases:
Case 1: There exists p ≤ m such that ϕp occurs in χ but not in ψ. Then by (135)c,∧e[k] ∧ χ is unsatisfiable for cofinitely many k.
Case 2: For all p ≤ n, if ϕp occurs in χ then ϕp occurs in ψ. Since all members ofP are pairwise disjoint, it then follows from (135)a,b that for cofinitely many k, {T ∈⋃
P |∧e[k] ∧ χ is satisfiable in T } 6⊆ P ′ for all P ′ ∈ P.
By (135)a and the definition of ψ, ∅ 6= {T ∈⋃
P |∧e[k] ∧ ψ is satisfiable in T } ⊆ P
for cofinitely many k. We conclude from the preceding facts that ∆[P, F ](e[k]) = P forcofinitely many k. Hence ∆[P, F ] solves P in e, as required.
5.28. Proof of Proposition (68)
(68) Proposition: For every solvable problem P, there exists a set X of ∀formulas such that for every enumeration E of X, the scientist Ψ[P, E] solvesP.
To facilitate the proof (and some others), we first consider a consequence of Convention(51), which sets Obs equal to Lbasic . The convention implies that tip-offs for solvableproblems have a characteristic form. We indicate the form using the following terminology.
(136) Definition: By a π-set is meant any collection of ∀ formulas all of whose freevariables are drawn from the same finite set.
From Example (34) of essay #2, the definition implies:
(137) Lemma: Suppose that P in P has a tip-off. Then there is a tip-off t for P in Pcomposed entirely of π-sets. Moreover:
(a) For every S ∈ P and full assignment h to S, there is π ∈ t with S |= π[h].
(b) For all U ∈ P ′ ∈ P with P ′ 6= P , all full assignments g to U , and all π ∈ t,U 6|= π[g].
Proof of Proposition (68): Let solvable problem P = {Pj | j ∈ N} be given. For allj ∈ N let tj be a tip-off for Pj in P. By Lemma (137) and the countability of tip-offs,let the π-sets in
⋃j∈N tj be enumerated as {πi | i ∈ N}. For all i ∈ N fix an enumeration
{ϕni |n ∈ N} of πi. Then set:
X = {(ϕ00 ∧ . . . ∧ ϕ
n00 ) ∨ . . . ∨ (ϕ0
i ∧ . . . ∧ ϕnii ) | i, n0 . . . ni ∈ N}.
Denote by E any enumeration of X. Since X consists of ∀ formulas, to prove the propositionit suffices to show that Ψ[P, E] solves P. Let j ∈ N , S ∈ Pj , full assignment h to S, andenvironment e for S and h be given. It suffices to show that Ψ[P, E] solves Pj in e. Leti0 ∈ N be least such that S |= πi0 [h]. For i < i0, S 6|= πi[h], so for each i < i0 we maychoose c(i) ∈ N such that:
(138) (a) S |= (ϕ0i ∧ . . . ∧ ϕ
c(i)−1i )[h];
(b) S 6|= ϕc(i)i [h].
Since for all i < i0, ϕc(i)i is a ∀ formula, and since h is onto | S |, it follows from (138)b that
there is k0 > 0 such that:
(139) for all k ≥ k0 and for all i < i0,∧e[k] ∧ ϕc(i)
i is unsatisfiable.
Let Y = {(ϕ00 ∧ . . .∧ϕ
n00 )∨ . . .∨ (ϕ0
i ∧ . . .∧ϕnii ) | i < i0, n0 ≥ c(0) . . . ni ≥ c(i)}. We deduce
from (139) that:
(140) for all k ≥ k0 and for all ϕ ∈ Y ,∧e[k] ∧ ϕ is unsatisfiable.
From (138)a and the fact that S |= πi0 [h] it follows easily that S |= X − Y [h]. So:
(141) For all k ≥ k0, content(e[k]) ∪ (X − Y ) is satisfiable.
Let Z = {(ϕ00 ∧ . . . ∧ ϕ
c(0)0 ) ∨ . . . ∨ (ϕ0
i0−1 ∧ . . . ∧ ϕc(i0−1)i0−1 ) ∨ (ϕ0
i0∧ . . . ∧ ϕn
i0) |n ∈ N}. From
(139) we obtain content(e[k]) ∪ Z |= {ϕni0|n ∈ N} (= πi0) for all k ≥ k0. Hence, since
Z ⊆ X − Y :
(142) for all k ≥ k0, content(e[k]) ∪ (X − Y ) |= πi0 .
For k ∈ N , let E(e[k]) be the set of satisfiable formulas of the form∧e[k]∧ψ, ψ ∈ E [as in
Definition (53)]. It follows immediately from (140), (141) and (142) that:
(143) for all k ≥ k0, E(e[k]) is consistent and E(e[k]) |= πi0 .
We infer from (143) that for all k ≥ k0, there is an initial segment s of E such that
∅ 6= {T ∈⋃
P | satform(s, e[k]) is satisfiable in T } ⊆ Pj .
It follows immediately from Definition (55) that for all k ≥ k0, Ψ[P, E](e[k]) = Pj . HenceΨ[P, E] solves Pj in e, as required.
5.29. Proof of Propositions (69) and (63)
(69) Proposition: Let solvable problem P have defined discrete ∀-type. Thenthere exists a set X of ∀ formulas such that for every enumeration E of X,∆[P, E] solves P.
Proposition (69) follows directly from the proof of the following lemma, of interest inits own right.
(144) Lemma: For all solvable problems P, the discrete ∀-type of P is defined if and onlyif the following condition holds:
(*) For all P ∈ P, and all σ ∈ SEQ for P , there is a ∀ formula ϕ suchthat ∅ 6= {S ∈
⋃P |
∧σ ∧ ϕ is satisfiable in S} ⊆ P .
Proof: Let solvable problem P be given. For the “only if” direction, suppose that (*)does not hold. Let P ∈ P and σ ∈ SEQ for P be such that for all ∀ formulas ϕ, either{S ∈
⋃P |
∧σ ∧ ϕ is satisfiable in S} = ∅ or {S ∈
⋃P |
∧σ ∧ ϕ is satisfiable in S} 6⊆ P .
Then for all τ ∈ SEQ extending σ:
(145) for all ∀ formulas ϕ, either
{S ∈⋃
P |∧τ ∧ ϕ is satisfiable in S} = ∅, or
{S ∈⋃
P |∧τ ∧ ϕ is satisfiable in S} 6⊆ P.
Let environment e for P with σ ⊆ e be given. Then (145) implies immediately that forevery set X of ∀ formulas, for every well ordering O of X, and for every k ≥ length(σ),∆[P, O](e[k]) is undefined. Hence ∆[P, O] does not solve P in e, so ∆[P, O] does not solveP. It follows that the discrete ∀-type of P is not defined.
For the “if” direction, suppose that (*) holds. Suppose that P 6= ∅ (otherwise thediscrete ∀-type of P is trivially equal to 0). We will define a set X of ∀ formulas and showthat for every enumeration E of X, ∆[P, E] solves P. First we define X. Let κ ≤ ω and
propositions Pj , j < κ, be such that P = {Pj | j < κ} and for all j < j′ < κ, Pj andPj′ are distinct. Since P is solvable, for all j < κ let tj be a tip-off for Pj in P. By thecountability of tip-offs, let the π-sets in
⋃j<κ tj be enumerated as {πi | i ∈ N}. Without
loss of generality we can assume that for all i ∈ N , πi is satisfiable in some member of⋃
P.
Let i ∈ N be given. Fix an enumeration {ϕni |n ∈ N} of πi and an enumeration
{αni |n ∈ N} of all α ∈ Lbasic such that πi |= α. By the definition of tip-offs, and since Pj and
Pj′ are distinct for all j < j′ < κ, there is a unique j < κ such that πi is satisfiable in somemember of Pj . Fix an enumeration {σn
i |n ∈ N} of all σ ∈ SEQ such that πi∪ content(σ) issatisfiable in some member of Pj . Let n ∈ N be given. We define a formula ψn
i as follows.It is easy to verify that there is σ ∈ SEQ such that content(σ) = {α0
i . . . αni } ∪ content(σn
i )and σ is satisfiable in some member of Pj . Hence, σ is for Pj . By (*) there exists a ∀formula ϕ such that ∅ 6= {S ∈
⋃P |α0
i ∧ . . . ∧ αni ∧
∧σn
i ∧ ϕ is satisfiable in S} ⊆ Pj . Wechoose one such ∀ formula ϕ and denote it by ψn
i .
Given i, j, n0 . . . ni ∈ N , define the formula of form (i, n0 . . . ni) to be:
(146) (ϕ00 ∧ . . . ∧ ϕ
n00 ) ∨ . . . ∨ (ϕ0
i−1 ∧ . . . ∧ ϕni−1
i−1 ) ∨ (α0i ∧ . . . ∧ α
nii ∧
∧σni
i ∧ ψnii ).
Let X be the set of formulas of form (i, n0 . . . ni), i, n0 . . . ni ∈ N . Note that X consists of∀ formulas. Let E be any enumeration of X. We show that ∆[P, E] solves P. Let j < κ,S ∈ Pj , full assignment h to S, and environment e for S and h be given. It suffices to showthat ∆[P, E] solves Pj in e. Let i0 ∈ N be least such that S |= πi0 [h] (there must be suchan i0 by the definition of the πi’s). For i < i0, S 6|= πi[h], πi is a set of ∀ formulas and h isonto | S |. So for each i < i0 we may choose c(i), d(i) ∈ N such that:
i and the ψni are ∀ formula, and since h is onto | S |, it follows
from (147)b,c and the definition of the ψni that there is k0 > 0 such that the following holds.
(148) (a) For all k ≥ k0 and for all i < i0,∧e[k] ∧ ϕc(i)
i is unsatisfiable.
(b) For all k ≥ k0 and for all i < i0,∧e[k] ∧ αd(i)
i is unsatisfiable.
(c) Let i < i0 and (unique) P ∈ P be such that πi is satisfiable in some member ofP . If P 6= Pj then for all k ≥ k0 and n < d(i),
∧e[k]∧α0
i ∧ . . .∧αni ∧
∧σn
i ∧ψni
is unsatisfiable.
Let k ≥ k0 be given. Since S |= πi0 [h], we deduce from (147)a, (148) and the definition ofthe ψn
i0that for all i, n0 . . . ni ∈ N , the following holds.
(149) (a) If 0 < i ≤ i0 and if n0 < c(0) or . . . or ni−1 < c(i − 1), then the conjunctionof
∧e[k] with the formula of form (i, n0 . . . ni) is satisfiable in S.
(b) If i < i0 and if n0 ≥ c(0) and . . . and ni−1 ≥ c(i − 1) and ni ≥ d(i), then theconjunction of
∧e[k] with the formula of form (i, n0 . . . ni) is not satisfiable.
(c) If i < i0 and if n0 ≥ c(0) and . . . and ni−1 ≥ c(i − 1) and ni < d(i), thenthe conjunction of
∧e[k] with the formula of form (i, n0 . . . ni) is either not
satisfiable or satisfiable in some member of Pj .
(d) If i = i0 and if n0 ≥ c(0) and . . . and ni−1 ≥ c(i− 1), then the conjunction of∧e[k] with the formula of form (i, n0 . . . ni) is either not satisfiable or satisfiable
in some member of Pj .
(e) If i > i0 then the conjunction of∧e[k] with the formula of form (i, n0 . . . ni) is
then P = Pj . So to finish the proof it suffices to exhibit ϕ ∈ X such that:
(150) ∅ 6= {T ∈⋃
P |∧e[k] ∧ ϕ is satisfiable in T } ⊆ Pj .
Since S |= πi0 ∪ content(e[k])[h], there is n ∈ N such that e[k] = σni0
. We infer immediately
from (148)a and the definition of ψni0
that ϕ = (ϕ00∧ . . .∧ϕ
c(0)0 )∨ . . .∨(ϕ0
i0−1∧ . . .∧ϕc(i0−1)i0−1 )∨
(α0i0∧ . . . ∧ αn
i0∧
∧σn
i0∧ ψn
i0) satisfies (150), as required.
The lemma also allows us to derive Proposition (63).
(63) Proposition: Suppose that Sym consists of a unary function symbol anda constant. Then there is a solvable problem P with the following properties.
(a) P has no tip-off base.
(b) The discrete ∀-type of P is defined.
Proof of Proposition (63): Let s be the function symbol and 0 the constant of Sym. Forn ∈ N , let n be the result of n applications of s to 0. Set:
We claim that P = {P0, P1 . . .} witnesses the proposition.
Clearly, for all i ∈ N , Pi 6= ∅ and for all distinct i, j ∈ N , Pi ∩ Pj = ∅, hence P ={P0, P1 . . .} is an infinite problem. It is equally immediate that P is solvable, and that
⋃P
is the class of all structures. So (a) follows directly from Lemma (60).
It remains to show (b). Let P ∈ P and σ ∈ SEQ be for P . By Lemma (144) it sufficesto show that for some ∀ formula ϕ:
(151) ∅ 6= {S ∈⋃
P |∧σ ∧ ϕ is satisfiable in S} ⊆ P .
If P = Pn for some n > 0 then ϕ = (∧
0<m<n(m 6= 0))∧(n = 0) satisfies (151). Suppose thatP = P0. We can choose 0 < m < n such that content(σ)∪{p 6= q | 0 ≤ p < q < n}∪{n = m}is consistent. Moreover, by a simple induction on terms of the form p, {p 6= q | 0 ≤ p < q <n}∪{n = m} |= {p 6= 0 | p > 0}. Hence ϕ = (
∧0≤p<q<n(p 6= q))∧ (n = m) satisfies (151).
5.30. Proof of Proposition (70)
(70) Proposition: Suppose that Sym consists of a denumerable set of con-stants. Then there exists a solvable problem whose discrete Lform -type is unde-fined.
Proof: Let {n |n ∈ N} enumerate the constants of Sym. Set P0 = MOD({n = 0 |n ∈ N}).Let P1 be the class of all structures that do not belong to P0. We show that P = {P0, P1}satisfies the claim of the proposition. Obviously, P is solvable. It remains to prove that forno well ordering O of a subset of Lform does ∆[P, O] solve P. So let O well order a subsetof Lform . Choose T ∈ MOD({n = 0 |n ∈ N} ∪ {∃x(x 6= 0)}) ⊂ P0. Fix an environment efor T . To finish the proof we show that ∆[P, O] does not solve P0 in e.
There exists k0 ∈ N such that∧e[k0] |= ∃x(x 6= 0). Let k ≥ k0 be given, and let ψ be
any formula such that∧e[k] ∧ ψ is satisfiable. Let n ∈ N be such that n does not appear
in∧e[k] ∧ ψ. Since
∧e[k] |= ∃x(x 6= 0), {
∧e[k] ∧ ψ, n 6= 0} is consistent. Hence
∧e[k] ∧ ψ
is satisfiable in some member of P1. So we have shown that for every k ≥ k0 and for everyformula ψ, if
∧e[k] ∧ ψ is satisfiable then {S ∈
⋃P |
∧e[k] ∧ ψ is satisfiable in S} 6⊆ P0.
By Definition (52) this implies immediately that for cofinitely many k, either ∆[P, O](e[k])is undefined or ∆[P, O](e[k]) 6⊆ P0. Hence ∆[P, O] does not solve P0 in e.
5.31. Proof of Proposition (72)
(72) Proposition: Let solvable problem P be such that every σ ∈ SEQ for Pis for at least two members of P. Then the discrete Lform -type of P is not finite.
Proof: Suppose for a contradiction that there is n ∈ N and enumeration E = {ϕ0 . . . ϕn} offormulas such that ∆[P, E] solves P. We will build by induction on i a sequence {σi | i ∈ N}of members of SEQ that are for P, a sequence {Pi | i ∈ N} of members of P, and a totalfunction f : N → {0 . . . n} such that for all i ∈ N :
(152) (a) f(i) is the least m ≤ n with ∅ 6= {S ∈⋃
P |∧σi ∧ϕm is satisfiable in S} ⊆ P
for some P ∈ P;
(b) σi ⊆ σi+1;
(c) ∅ 6= {S ∈⋃
P |∧σi ∧ ϕf(i) is satisfiable in S} ⊆ Pi;
(d) Pi 6= Pi+1.
Before we build those items, we show how to derive a contradiction from (152), thus com-pleting the proof. Let i0 ∈ N be such that content(f) ⊆ {f(0) . . . f(i0)}. Then:
Proof of (153): Let i0 ≤ i1 ≤ i2 be given, and suppose for a contradiction that f(i2) <f(i1). By (152)a applied twice, since f(i2) ∈ {f(0) . . . f(i0)} there exists j ≤ i0 suchthat ∅ 6= {S ∈
⋃P |
∧σj ∧ ϕf(j) is satisfiable in S} ⊆ P for some P ∈ P, ∅ 6= {S ∈⋃
P |∧σi2 ∧ ϕf(i2) is satisfiable in S} ⊆ P for some P ∈ P, and f(j) = f(i2). Hence∧
σi2 ∧ ϕf(i2) is satisfiable and ∅ 6= {S ∈⋃
P |∧σj ∧ ϕf(i2) is satisfiable in S} ⊆ P for
some P ∈ P. Since j ≤ i1 ≤ i2, it follows from (152)b that σj ⊆ σi1 ⊆ σi2 . Thus:
(154) ∅ 6= {S ∈⋃
P |∧σi1 ∧ ϕf(i2) is satisfiable in S} ⊆ P for some P ∈ P.
But (152)a applied to i = i1 implies that f(i1) is the least m ≤ n such that ∅ 6= {S ∈⋃P |
∧σi1 ∧ ϕm is satisfiable in S} ⊆ P for some P ∈ P. This contradicts (154) together
with the hypothesis that f(i2) < f(i1).
Now we use (153) to obtain the desired contradiction. Since all members of P arepairwise disjoint, we infer from (152)b–d that for all i ∈ N , ϕf(i) 6= ϕf(i+1). Hence [sincecontent(f) is finite] there is i1 ≥ i0 with f(i1 + 1) < f(i1), which contradicts (153).
So to finish the proof we build {σi | i ∈ N}, {Pi | i ∈ N} and f : N → {0 . . . n} satisfying(152). Let P be any member of P. Fix an environment e for P . Since ∆[P, E] solves P, thereexists k > 0 such that m ≤ n is least with ∅ 6= {S ∈
⋃P |
∧e[k]∧ϕm is satisfiable in S} ⊆
P ′ for some P ′ ∈ P, and ∅ 6= {S ∈⋃
P |∧e[k]∧ϕm is satisfiable in S} ⊆ P . Set σ0 = e[k],
P0 = P , and f(0) = m. Note that σ0 is for P. Let i ∈ N be given, and suppose thatσ0 . . . σi, P0 . . . Pi and f(0) . . . f(i) have been defined, and that σi is for P. By hypothesisthere exists P ∈ P such that P 6= Pf(i) and σi is for P . Let environment e for P extendσi. Since ∆[P, E] solves P, there exists k ≥ length(σi) such that m ≤ n is least with∅ 6= {S ∈
⋃P |
∧e[k] ∧ ϕm is satisfiable in S} ⊆ P ′ for some P ′ ∈ P, and ∅ 6= {S ∈
∧e[k] ∧ ϕm is satisfiable in S} ⊆ P . Set σi+1 = e[k], Pi+1 = P , and f(i + 1) = m.
Plainly this construction satisfies (152).
5.32. Proof of Proposition (74)
(74) Proposition: For all solvable problems P, if the discrete ∃-type of P isdefined then the discrete ∀-type of P is 1 at most.
Proof: Let nonempty, solvable problem P be given (if P = ∅ then the ∀-type of P is0). Suppose that the discrete ∃-type of P is defined. Let well ordering O of set X of ∃formulas be such that ∆[P, O] solves P. Let P ∈ P and environment e for P be given.Let k0 ∈ N be such that ∆[P, O](e[k0]) = P . Then there is ϕ ∈ X such that ∅ 6= {S ∈⋃
P |∧e[k0]∧ϕ is satisfiable in S} ⊆ P . Since ϕ is an ∃ formula, there is k1 ≥ k0 such that∧
e[k1] |= ϕ. This implies immediately that for all k ≥ k1, ∅ 6= {S ∈⋃
P |∧e[k1] ∧ (x =
x) is satisfiable in S} ⊆ P . Hence ∆[P, {x = x}] solves P, and the discrete ∀-type of P isequal to 1.
5.33. Proof of Proposition (75)
(75) Proposition: Suppose that Sym consists of a denumerable set of con-stants and a binary predicate. Then there is a solvable problem such that:
(a) the discrete ∀-type of P is undefined;
(b) the segmental ∀-type of P is infinite;
(c) the discrete and segmental ∀∃-types of P are 2.
Proof: Let {n |n ∈ N} enumerate the constants of Sym, and let R be its binary predicate.Let P0 consist of all members of MOD({n = 0 |n ∈ N}) in which the interpretation ofR is a strict total order without greatest point. Let P1 consist of all finite members of⋃
n>0 MOD(n 6= 0) in which the interpretation of R is a strict total order. Then P ={P0, P1} is a solvable problem. We show that P satisfies the claim of the proposition,starting with (a). Let satisfiable ∀ formula ϕ be such that ϕ |= ∃x(x 6= 0). Let X be theset of all n > 0 such that n does not appear in ϕ. Denote by n0 the least member of X.By the choice of ϕ, {ϕ, n0 6= 0} is consistent. Hence W = {ϕ, n0 6= 0} ∪ {n = n0 |n ∈ X} isconsistent. Choose structure S and full assigment h to S with S |= W [h]. Let A ⊆ | S | bethe union of {h(x) |x ∈ Var(ϕ)} with the set of interpretations of the constants n, n ∈ N .Let T be the restriction of S to A. Since ∀ formulas are preserved in substructures, W issatisfiable in T . Note that | T | is finite, and T |= (n0 6= 0). Hence T ∈ P1. So we haveshown the following:
(155) Every satisfiable ∀ formula that implies ∃x(x 6= 0) is satisfiable in some member ofP1.
Now let e be an environment for a structure S in P0 that satisfies S |= ∃x(x 6= 0). Letk0 ∈ N be such that
∧e[k0] |= ∃x(x 6= 0). It follows from (155) that for all ∀ formulas ϕ and
for all k ≥ k0, either∧e[k] ∧ ϕ is unsatisfiable or
∧e[k] ∧ ϕ is satisfiable in some member
of P1. This implies immediately that for all well orderings O of some set of ∀ formulas andfor all k ≥ k0, ∆[P, O](e[k]) 6= P0. Hence for all well orderings O of some set of ∀ formulas,∆[P, O] does not solve P0 in e. Hence the discrete ∀-type of P is undefined.
Clause (b) follows immediately from clause (a) and Propositions (68) and (65). Forclause (c), set E = {∀x∃yRxy, x = x}. If e is an environment for P0, then trivially∆[P, E](e[k]) = Ψ[P, E](e[k]) = P0 for all k ∈ N . Let environment e for P1 be given. Letn > 0 and k0 ∈ N be such that
∧e[k0] |= (n 6= 0). Then trivially, ∆[P, E](e[k]) =
Ψ[P, E](e[k]) = P1 for all k ≥ k0. So both ∆[P, E] and Ψ[P, E] solve P, and the
discrete and segmental ∀∃-types of P are 2 at most. They cannot be equal to 1. In-deed, let ψ ∈ Lform be given. Suppose that ∆[P, {ψ}] (respectively Ψ[P, {ψ}]) solvesP. Let environment e0 for P0 be given. Then there exists k0 ∈ N such that ∅ 6= {S ∈⋃
P |∧e0[k0] ∧ ψ is satisfiable in S} ⊆ P0. Let environment e for P1 extend e0[k0]. Then
for all k ≥ k0, ∆[P, {ψ}](e[k]) (respectively Ψ[P, {ψ}](e[k])) is not equal to P1.
5.34. Proof of Proposition (76)
(76) Proposition For all solvable problems P, the discrete ∃∀-type of P isdefined iff the discrete ∀-type of P is defined.
Proof: Let solvable problem P be given. Trivially, if the discrete ∀-type of P is defined thenthe discrete ∃∀-type of P is defined. Suppose that the discrete ∃∀-type of P is defined. Letset X of ∃∀ formulas and well ordering O of X be such that ∆[P, O] solves P. Let P ∈ Pand σ ∈ SEQ for P be given. By Lemma (144) it suffices to exhibit a ∀ formula ϕ suchthat:
(156) ∅ 6= {S ∈⋃
P |∧σ ∧ ϕ is satisfiable in S} ⊆ P .
Fix an environment e for P extending σ. Since ∆[P, O] solves P, there is k ≥ length(σ)and ψ ∈ X such that ∅ 6= {S ∈
⋃P |
∧e[k] ∧ ψ is satisfiable in S} ⊆ P . Since ψ is
an ∃∀ formula, we can choose p ∈ N , variables x1 . . . xp, and ∀ formula χ such that{x1 . . . xp} ∩ Var(
∧e[k] ∧ ψ) = ∅ and |= ∃x1 . . . xpχ ↔ ψ. Then ∅ 6= {S ∈
⋃P |
∧e[k] ∧
χ is satisfiable in S} ⊆ P , and ϕ =∧e[k] ∧ χ satisfies (156), as required.
(77) Proposition: Suppose that Sym consists of a binary predicate, a unaryfunction symbol, and a constant. Then there exists a solvable problem P suchthat:
(a) the discrete and segmental ∀-types of P are infinite;(b) the discrete and segmental ∀∃-types of P are 2;(c) the discrete and segmental ∃∀-types of P are 2.
Proof: Let R be the binary predicate, s the function symbol, and 0 the constant of Sym.For n ∈ N , let n be the result of n applications of s to 0. Set:
T = {(n = 0) → (∃x∀yRxy ∧ ∀x∃yRxy) |n > 0}.
Set:
P0 = MOD({n 6= 0 |n > 0}).
For all n > 0 set:
Pn = MOD(T ∪ {m 6= 0 | 0 < m < n} ∪ {n = 0}).
Clearly, for all i ∈ N , Pi 6= ∅ and for all distinct i, j ∈ N , Pi∩Pj = ∅, hence P = {P0, P1 . . .}is an infinite problem. It is equally immediate that P is solvable. We prove that P satisfiesthe claim of the proposition, starting with (a). Since {n 6= 0 |n > 0} |= T , it is easy tosee that
⋃P = MOD(T ). Hence, by Lemma (60), Propositions (64) and Corollary (71), it
suffices to show that the discrete ∀-type of P is defined. Let P ∈ P and σ ∈ SEQ be for P .By Lemma (144) it suffices to show that for some ∀ formula ϕ:
0<m<n(m 6= 0)) ∧ (n = 0) satisfies trivially (157).Suppose that P = P0. We can choose 0 < m < n such that content(σ) ∪ {p 6= q | 0 ≤ p <q < n} ∪ {n = m} is consistent. It is easy to verify that ϕ = (
∧0≤p<q<n(p 6= q)) ∧ (n = m)
satisfies (157).
We prove (b). Set E = {∀x∃y¬Rxy, x = x}. Observe that every initial segment of anenvironment for P is consistent with ∀x∃y¬Rxy. Using this fact and the definition of T , itis easy to verify that if e is an environment for P0, then ∆[P, E](e[k]) = Ψ[P, E](e[k]) = P0
for all k ∈ N . Let n > 0 and environment e for Pn be given. Let k0 ∈ N be such that∧e[k0] |= (
for all k ≥ k0. So both ∆[P, E] and Ψ[P, E] solve P, and the discrete and segmental∀∃-types of P are 2 at most. They cannot be equal to 1. Indeed, let ψ ∈ Lform be given.Suppose that ∆[P, {ψ}] (respectively Ψ[P, {ψ}]) solves P. Let environment e0 for P0 besuch that content(e0) |= {m 6= n |m,n ∈ N, m 6= n}. Then there exists k0 ∈ N such that∅ 6= {S ∈
⋃P |
∧e0[k0]∧ψ is satisfiable in S} ⊆ P0. Let n > 0 be such that for all m ≥ n,
m does not occur in∧e0[k0]. Let environment e for Pn extend e0[k0]. Then for all k ≥ k0,
∆[P, {ψ}](e[k]) (respectively Ψ[P, {ψ}](e[k])) is not equal to Pn. This ends the proof of(b). The proof of (c) is similar, using {∃x∀y¬Rxy, x = x} instead of E.
References
[1] D. Angluin. Identifying languages from stochastic examples. Technical Report 614,Yale University, 1988.
[2] J. Barwise. Admissible Sets and Structures. Springer-Verlag, Berlin, 1975.
[3] P. Billingsley. Probability and Measure (Second Edition). John Wiley and Sons, NewYork, 1986.
[4] R. Daley and C. Smith. On the complexity of inductive inference. Information andComputation, 69:12–40, 1986.
[5] H. D. Ebbinghaus, J. Flum, and W. Thomas. Mathematical Logic, Second Edition.Springer-Verlag, Berlin, 1994.
[6] R. Fraisse. Cours de logique mathematique, Tome 2. Gauthier-Villars, Paris, 1972.
[7] R. Freivalds, E. Kinber, and C. H. Smith. On the Intrinsic Complexity of Learning.Information and Computation, 123(1):64–71, 1995.
[8] S. Jain, D. Osherson, J. Royer, and A. Sharma. Systems that Learn, 2nd Edition. MITPress, Cambridge MA, 1999.
[9] S. Jain and A. Sharma. The Structure of Intrinsic Complexity of Learning. Journal ofSymbolic Logic, 62:1187–1201, 1997.
[10] C. Juhl. Bayesianism and reliable scientific inquiry. Philosophy of Science, 60:302–319,1993.
[11] E. Kinber and F. Stephan. Language learning from texts: mind changes, limitedmemory and monotonicity. Information and Computation, 123:224–241, 1995.
[12] S. Lange and T. Zeugmann. Language learning with a bounded number of mindchanges. In Proc. 10th Annual Symposium on Theoretical Aspects of Computer Science,pages 682–691. Springer–Verlag, 1993. Lecture Notes in Computer Science 665.
[13] S. Lange and T. Zeugmann. Trading monotonicity demands versus efficiency. Bull.Inf. Cybern, 27(1):53–83, 1995.
![More Homological Approach to Composition of Subfactors · Composition of Subfactors 603 By[C,Remark6.8],Obs(α)istheclassofthe3-cocycleκwhichisdefinedfor0≤i,j,k](https://static.documents.pub/doc/80x56/5fdb12694da2f92aff6043af/more-homological-approach-to-composition-of-subfactors-composition-of-subfactors.jpg)
