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A Framework for 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].
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The first essay on our web site provides an informal introduction to the acceptance-basedinductive logic elaborated in the sequel. We get down to business in the present essay. Aninductive paradigm is defined formally and some of its basic properties are investigated.The discussion presupposes only elementary mathematical logic (as in [8, 9]).
A roadmap might be useful. Section 1 establishes conventions and notation necessaryfor the rest of the discussion. Section 2 formalizes the inductive game that underlies ourtheory. The first two sections are thus indispensable for further reading. They suffice forunderstanding much of the discussion in the third essay, which bears on belief revision.Hence, you may wish to skip the remaining sections of the present essay. Section 3 makessome observations about the model of inquiry advanced in Section 2. The main technicalwork is carried out in Section 4, which offers a characterization of the winnable inductivegames. As noted below, these results serve as a tool for subsequent analyses but need notbe assimilated by readers who plan to skirt the proofs of theorems. Many of our results bearon inductive games with special properties that simplify their form. Section 5 discusses twokinds of special games. Efficient inquiry is the topic of Section 6. Scientists that can besimulated by a computable process occupy Section 7. Proofs are relegated to Section 8.
Be advised that the game terminology of the first essay will give way to more soberexpressions in the formal definitions of the present essay. For example, what were “games”will be called “problems,” and they will be “solved” rather than “won.” Games will still beinvoked, however, for informal explanations.
1. Logical setting
We start, inevitably, with definitions and conventions. We have endeavored to find the rightcompromise between generality and naturalness. Some generalizations of our basic setupare discussed in the fourth essay. Others will be indicated along the way in this essay (not
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all have been explored).
Section 2 defines a “paradigm” of inductive inference, by which is meant a formal modelof scientific inquiry. The paradigm is built from concepts and terminology adapted frommathematical logic. The latter material is presented in the present section.
1.1. A first-order framework
1.1.1. Language
Central to all our concepts is the choice of a first-order language. The language serves todelimit the kind of reality the scientist might confront and the kind of data he might receiveabout it. We use L to denote this language, and make the following assumptions about it.
(a) L is countable.
(b) L includes the identity symbol =,
(c) L includes a countably infinite set of (individual) variables (distinct fromall the other symbols).
The set of variables will be denoted by Var. The variables themselves are v0, v1, v2 · · ·,sometimes abbreviated to x, y, z when convenient.
To finish specifying L, it remains to choose its nonlogical vocabulary, denoted by Sym.But how do we choose the members of Sym? Our approach is to leave Sym as a freeparameter of inductive logic. Thus, you may choose whatever members of Sym seemappropriate, and then interpret our theory as relating to that choice. Some of our resultsimpose special hypotheses on Sym (typically, that it include predicates of various arities).Theorems stated without such hypotheses are true for any choice of (countable) Sym.
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We summarize as follows.
(1) Notation:
(a) Our first-order language is called L. It is countable and decidable, and includesidentity.
(b) The variables of L are v0, v1, v2 · · ·, collectively denoted by Var.
(c) The nonlogical vocabulary of L is brought together in Sym. It consists ofpredicates and function symbols of various arities, along with constants. Anyof these sets of symbols may be empty, finite, or countably infinite. We alsoassume that Sym is computably decidable.
By the latter clause we mean that it must be possible to determine via an algorithm whethera given symbol belongs to the class of unary predicates, etc. With this notation in hand,we can think of Sym and Var as fixed throughout the discussion.
Some familiar notation concerning our language L will be used in what follows.
(2) Notation:
(a) Lform denotes the set of formulas of L.
(b) Lsen denotes the set of sentences of L (no free variables).
(c) Latomic denotes the set of atomic formulas of Lform .
(d) Lbasic denotes the set of basic formulas of Lform .1
(e) The set of variables occurring free in ϕ ∈ Lform is denoted Var(ϕ).
1Reminder: a formula is basic iff it is an atomic formula or the negation of an atomic formula.
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(f) An ∃ formula is a formula equivalent to a formula in prenex form whose quan-tifier prefix is limited to existentials. The same terminology applies to ∃∀formulas and sentences, etc.
(g) Given ϕ ∈ Lform and variables x, y, the expression ϕ[y/x] denotes the result ofsubstituting y for every free occurrence of x in ϕ.
1.1.2. Potential data
There is another set to be fixed at the outset of our discussion. Recall our picture of scientificinquiry. Nature provides the scientist with an enumeration of facts about the reality chosento be “actual.” (See the first essay, Section 3.) But which facts get enumerated? In thefirst essay we took them to be basic formulas. This is a natural choice since it supports thepicture of data being the affirmation or denial of “atomic facts.” Such a picture may beexcessively positivistic, however, and our results typically support other choices. Often itsuffices that the data merely include the basic formulas, or even just the atomic formulas.Other times it is only necessary that the data be closed under negation. We thereforeproceed as for Sym, namely, by fixing a set of formulas to be used as data. The reader canchoose whatever set she likes for this purpose provided that it meets a few conditions. Weproceed officially as follows.
(3) Notation: We fix Obs ⊆ Lform to serve as potential observations available toscientists. It is assumed that Obs is nonempty, and that ϕ[y/x] ∈ Obs for everyϕ ∈ Obs and x, y ∈ Var.
The last clause means that Obs is not biased about variables. If Rv0v1 is a potentialobservation then so is Rv18v5. Which (if either) of the formulas ends up on Nature’s listdepends on what the variables name (see below). In general terms, Nature lists every
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ϕ ∈ Obs that is made true by the reality chosen at the outset of the game. If Obs consistsof the existential formulas, for example, then Nature reveals to the scientist an enumerationof the true existential formulas.2
Similarly to Sym, some of our results depend on assumptions about Obs that go beyond(3). Stronger assumptions will appear as needed in the formulation of theorems. Onecondition occasionally needed is that Obs be “closed under negation.” Let us be clearabout what this is supposed to mean.
(4) Definition: Let Obs ⊆ Lform be given. We say that Obs is closed under negationjust in case for every ϕ ∈ Obs, there is ψ ∈ Obs that is logically equivalent to ¬ϕ.
The point is that Obs can be closed under negation without being “closed under ¬.” Inthe latter case, every ϕ ∈ Obs would be accompanied by ¬ϕ, ¬¬ϕ, etc. To be closed undernegation it suffices that there be just one copy of each polarity of ϕ.
Certain choices of Obs can be used to model the distinction between “theoretical” and“observational” vocabulary. For example, it is allowed that certain predicates of Sym (e.g.,“is an electron”) appear in no formula of Obs. Such missing predicates might be consideredtheoretical in the sense that the “raw” data provide no direct information about theirapplication. Such a distinction, of course, yields only a crude approximation to scientificinquiry. In real inquiry, the observational/theoretical distinction is not fixed and sharp(if intelligible at all), but shifts with the character and stage of investigation. It maynonetheless be interesting to observe that many of our theorems make no assumptionsabout Obs [beyond (3)]. They are therefore true for any proposed distinction betweenobservational and theoretical vocabulary.
2Inductive paradigms with data varying in complexity were first explored in [13].
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Similar remarks apply to the issue of missing data. For example, to model a situationin which facts of the form f(x) = g(x) are not shown to scientists, it suffices to exclude allsuch formulas from Obs. This is not the same as excluding formulas with any occurrence off and g. The latter exclusion is the same as declaring f and g to be theoretical vocabularywhereas all that is intended in the present case is that certain facts involving these symbolsbe unavailable. Again, this kind of case is embraced by theorems formulated withoutassumptions on Obs.
We admit to fondness for the choice Obs = Lbasic . Our inductive logic was originallydeveloped under this hypothesis, then generalized subsequently. Many of the examples andresults below assume that Obs = Lbasic .
1.1.3. Structures
We turn now to the semantic side of our first-order framework. With one exception, ourconcepts are standard. Thus, structure S is a model of Γ ⊆ Lform just in case there is anassignment h : Var → | S | with S |= Γ[h]; in this case S is said to “satisfy” Γ.3 As usual,a set Γ of formulas implies a set ∆ just in case for every structure S and every assignmenth : Var → | S |, if S |= Γ[h] then S |= ∆[h]. We often rely on the following notation.
(5) Notation:
(a) The class of models of Γ ⊆ Lform is denoted MOD(Γ). In particular MOD(∅)is the class of all structures.
(b) For ϕ ∈ Lform , we write MOD(ϕ) in place of MOD({ϕ}).(c) A class of structures of the form MOD(Γ) for some Γ ⊆ Lform is called elemen-
tary.3The symbol | S | denotes the domain of the structure S.
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An example of an elementary class may be helpful. First, recall that a total order is reflexive,connected, transitive, and antisymmetric (like ≤ on the integers). To assert within L thata binary relation symbol R stands for a total order, it suffices to write the formulas:
(6) ∀xRxx (reflexivity)
∀x∀y(x 6= y → (Rxy ∨Ryx)) (connectedness)
∀x∀y∀z((Rxy ∧Ryz) → Rxz) (transitivity)
∀x∀y((Rxy ∧Ryx) → x = y) (antisymmetry)
Let T be the set of these four formulas. Then MOD(T ) is the class of structures thatinterpret R as a total order, and this class is elementary.
We often rely on two familiar facts about first-order logic, namely, the compactness andLowenheim-Skolem theorems. As a reminder, we state them here.
(7) Theorem: (Compactness) Let a set ∆ of formulas be given.
(a) ∆ is satisfiable iff every finite subset of ∆ is satisfiable.
(b) ∆ implies a given formula ϕ iff some finite subset of ∆ implies ϕ.
(8) Theorem: (Lowenheim-Skolem theorem, downward) A set ∆ of formulas is satis-fiable iff there is a structure with countable domain that satisfies ∆.4
There is one liberty we take with standard terminology. It simplifies much of our dis-cussion to limit attention to countable structures that interpret Sym, that is, to structures
4We depend here on Convention (1), requiring L to be countable.
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with finite or denumerable domains. In the fourth essay we generalize our paradigm tostructures of arbitrary cardinality, but until then the countability assumption is pivotal. Sowe record the following convention.
(9) Convention: By “structure” will always be meant a countable structure thatinterprets the symbol set Sym. Similarly, when we write MOD(Γ) for some Γ ⊆Lform , we refer to the class of countable structures that satisfy Γ. Such a class isstill called “elementary.”
Let T be the set of formulas in (6). Then according to our new convention, MOD(T ) is theclass of total orders over a countable domain with respect to the relation symbol R, and thisclass is elementary. In other words, Convention (9) means: just forget about uncountabledomains (until the fourth essay).
Does Convention (9) perturb the definition of implication between formulas (since thereare fewer potential counterexamples to an implication)? No. Theorem (8) ensures thatimplication is the same relation between sets of formulas whether are not we limit attentionto countable domains.
In the way of background logic, this is all we need to get started.
2. The paradigm
The present section formalizes the inductive game described in Section 3 of the first essay.To proceed, we step through five components of the game, defining each within the first-order setting established above. The components are as follows.
(a) Worlds, that is, the potential realities from which Nature chooses one to be “actual.”
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(b) Problem, that is, a partition of potential realities that defines the idea of a “correctconjecture.”
(c) Environment, that is, a stream of clues made available about reality.
(d) Scientist, that is, a method for converting clues into conjectures.
(e) Success, that is, a criterion for determining who wins the game.5
2.1. Worlds
This one is easy. The potential realities of our paradigm are all the (countable!) structuresthat interpret the symbol set Sym. Any such structure might turn out to be the scientist’sworld, so the scientist’s question is always roughly: “What’s true in my structure?”
2.2. Problems
It is standard fare in philosophy to construe a proposition as a collection of possible worlds[6]. The proposition corresponding to the assertion that lions are carnivores, for example, isidentified with the collection of worlds in which lions are carnivores. To exploit this snappyterminology (without taking a position on the associated philosophy), recall that worlds inthe present setting are just structures. We therefore take a proposition to be a collection ofstructures.
When does it make sense to call a proposition “true?” Well, a proposition is true ifit is entailed by the “facts.” The facts are embodied by the structure that Nature choseto be actual. Moreover, entailment between propositions comes down to inclusion: one
5The idea of analyzing inductive paradigms into these components is due to [17].
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proposition entails another if the first is a subclass of the second.6 Putting the piecestogether, proposition P is true iff {S} ⊆ P , where S is Nature’s choice of structure (the“real world”). That is, proposition P is true iff S ∈ P for this S.
By a “problem” is meant a choice among mutually exclusive propositions. Intuitively,the problem is to figure out which proposition is true. Let us record these concepts officially.
(10) Definition: A nonempty class of structures is a proposition. A problem is anonempty collection of disjoint propositions.
An example will make the definition clearer.7
(11) Example: Suppose that Sym consists of a binary predicate R. Let T be the theoryof total orders (with respect to R) with either a least point or a greatest point. butnot both.8 Let θ = ∃x∀yRxy and P = {MOD(T ∪ {θ}),MOD(T ∪ {¬θ})}. Then Pis a problem consisting of the propositions MOD(T ∪ {θ}) and MOD(T ∪ {¬θ}).
The two propositions in P are the class of total orders with a least but no greatest point andthe class of total orders with a greatest but no least point. In this example P is composedof propositions that are elementary classes, i.e., specified by sets of sentences. This wasthe kind of case envisioned in the informal discussion of the first essay (Section 3). In the
6This definition of entailment generalizes the usual idea of logical implication between formulas. In themore general setting we don’t need formulas to name classes of structures, which are then compared viainclusion. Instead of relying on formulas, the classes themselves serve as the relata of implication.
7We love this example. It already figured in the first essay as the game defined by Partition (9). Itappears again as we progress.
8More explicitly, T can be understood as the set of formulas in (6) along with the formula (∃x∀yRxy ∨∃x∀yRyx) ∧ ¬(∃x∀yRxy ∧ ∃x∀yRyx).
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present, more general setting, problems need not be elementary classes since they are justarbitrary collections of structures. For example, let P<ω be the collection of structureswhose domains are finite. It is well known that P<ω is not elementary.9 Yet P<ω is aperfectly legal proposition and may appear in a problem (e.g., the problem consisting ofP<ω and the collection of structures with countably infinite domain).
You see that a problem is a partition of some class of structures. The partitioned class isthe union of the propositions that make it up. The cells of the partition are the propositionsthemselves. In this respect, our set-up is similar to that of the Bayesians. A problem is apartition over an event-space, where the events are structures and each cell of the partitionis a proposition. A problem thus embodies the question: which of the propositions is truein my world? The partition embodied by a problem can be finite or infinite. It can evenhave just one member, although this makes the problem trivial. Notice, by the way, howExample (11) makes explicit the assumption we needed about Sym, namely, that it consistsof just the binary relation symbol R.
Finally, it is worth pointing out that Definition (10) rules the the empty set out of therealm of propositions. The empty set might have been named “the contradictory proposi-tion” but such a status leads to complications elsewhere. So we leave it as a non-proposition.
2.3. Environments
Let us now consider the information made available to a scientist about a given structure(namely, the structure chosen as the “actual” world). The information is organized asa stream of formulas called an “environment.”10 To explain environments we must firstexplain how Nature refers to the elements in her chosen structure.
9That is, there is no Γ ⊆ Lform with MOD(Γ) = P<ω. This fact follows from Theorem (7); see [8, IV.3].10We might have used the term “data-environment” except that it is cumbersome. Another possibility
would be “data stream.” Forget these possibilities, however; just remember “environment.”
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2.3.1. Variables as temporary names
Similarly to the informal games introduced in the first essay, each element of the structure’sdomain is given a temporary name, and the variables of L are used for this purpose. Taggingevery member of the domain with one or more variables amounts to an “assignment” inthe sense of model theory. Since every member must get tagged, the assignment is onto thedomain. Officially:
(12) Definition: Given structure S, a full assignment to S is any mapping of Var onto| S |.
Thus, a full assignment h to S provides temporary names for all the elements of | S |,namely, s ∈ | S | is assigned nonempty h−1(s) ⊆ Var.
Suppose that Nature chooses a full assignment h to name the domain elements of herchosen structure S. She is then required to provide information about the interpretation(S, h). For this purpose, she lists the members of Obs that are true in (S, h). Such is theessential idea behind the definition of an “environment.”
2.3.2. Pauses and data content
There is a slight complication, however. It will be technically useful to allow the flow of datato be interrupted by pauses from time to time. Perhaps the pauses reflect the scientist’ssleep or budgetary cycle, but whatever their provenance, pauses are moments at which nonew data are presented. They will be represented by the special symbol ], assumed to bedifferent from every other symbol appearing in L. Officially:
(13) Convention: We fix a symbol ], assumed to be disjoint from every symbol (logicaland nonlogical) appearing in L.
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Since pauses have no content, we’ll often need to ignore them when considering data.The following definition provides what’s needed.
(14) Definition: Let s be a sequence (finite or infinite) over Lform ∪ {]}. We letcontent(s) denote the range of the sequence without ]. That is, content(s) is theset of all formulas that occur in s.
For example, if s = 〈P v2, ], Rv1v3〉 then content(s) = {P v2, Rv1v3}.
2.3.3. Definition of environments
At last we can define the stream of data made available by Nature. Such streams are called“environments.”
(15) Definition: Let structure S and full assignment h to S be given.
(a) An environment for S and h is an infinite sequence e such that
content(e) = {β ∈ Obs | S |= β[h]}.
(b) An environment for S is an environment for S and h, for some full assignmenth to S.
(c) An environment is an environment for some structure.
(d) An environment for proposition P is an environment for some S ∈ P .
(e) An environment for problem P is an environment for some P ∈ P.
Here are some examples. Recall that N denotes the set {0, 1, 2, 3 · · ·} of natural numbers.
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(16) Example: Suppose that binary predicate R is the only member of Sym, and thatstructure S with | S | = N interprets R as <.
(a) Suppose that Obs = Lbasic , the set of basic formulas. Suppose also that fullassignment h to S is {(vi, i) | i ∈ N}. Then one environment for S and h beginsthis way:
v3 6= v4, ¬Rv0v0, Rv1v9, v9 = v9, Rv7v8, v0 6= v3, v5 = v5, ¬Rv23v8, · · ·If full assignment g to S is {(v2i, i), (v2i+1, i) | i ∈ N} then one environment forS and g begins this way:v2 = v3, ¬Rv4v5, Rv1v9, v9 = v9, Rv7v19, v0 6= v3, ¬Rv33v2, ¬Rv23v8, · · ·
If P is the proposition containing every strict total order, then this same en-vironment is for P .11 If P is a problem that includes P as a componentproposition, then the environment is also for P.
(b) Now suppose that Obs = Latomic , and that h is as above. Then one environ-ment for S and h begins this way:
Rv1v9, v9 = v9, Rv7v8, v5 = v5, Rv2v3, Rv7v8 · · ·
Let us note confusion that can be nipped in the bud. The assumption that Obs = Lbasic
doesn’t mean that a given environment includes every basic formula. Only the basic for-mulas true in the underlying interpretation (structure plus assignment) show up in theenvironment.
2.3.4. Remarks about environments
As a final aid to intuition, suppose that Sym contains the unary predicate H, the binarypredicate R, and the individual constant c. Suppose also that Obs = Lbasic . Then, abstract-
11A strict total order is irreflexive, connected, transitive, and asymmetric (like < on the integers).
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ing away from the arbitrary choice of temporary names for domain elements (embodied inthe underlying full assignment), an environment provides information like this:12
“The first object encountered falls under H. The second object encountereddoes not fall under H, and is related to the first object by R. The third objectencountered is identical to the first object. The fourth object encountered is notidentical to the second object, and is not related to the third object by R. Thething denoted by c is related to the second object. . . . ”
Here we picture temporary names as arising from ordinal position in the data stream.Alternatively, we can imagine the scientist affixing numbered PostIt notes to each objectfalling into his hands (each note has a unique number). Sometimes he discovers that thesame object has come back to him with multiple notes, leading to the affirmation of anidentity (“the object with PostIt #47 is the same object as the one with PostIt #82”).In the other cases, he affirms diversity (“the object with PostIt #19 is different from theobject with PostIt #22”). The point is that there is nothing essential about our use ofvariables as temporary names. Any scheme that allows the scientist to keep track of objectswould do as well. In the same connection, let us observe that our theory would not changeif we allowed Nature to use only a subset of the variables as names, or to use a new set ofconstants for this purpose. The use of full assignments in Definition (15) simply cuts downclutter later on. We could also require Nature to use v0 for the first name to appear in thedata, v1 for the second name, etc. Our theory would not be affected thereby.
Before returning to technical matters, let us address another conceptual issue. In ourinductive games, environments are controlled by Nature. The scientist plays a passive roleinasmuch as he receives but does not generate data. An alternative picture would empowerthe scientist at each stage to choose a formula. Nature would then have to declare it either
12The present remarks are similar to those on page 19 of the first essay.
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true or false. One way to formalize the latter picture is explored in [16, §3.4.3], but thematter is not pursued here. To keep things simple, we focus on the passive model of datacollection represented by the environments of Definition (15). Do not feel badly for thescientist. He will have enough to do!
2.3.5. The information available in environments
An environment for a structure offers considerable information about that structure. Indeed,if the environment includes every true atomic formula (relative to some full assignment),then it essentially determines the underlying structure. This is the content of the followinglemma. Its proof is easy (and also an immediate consequence of [12, Prop 3.2(i)]).
(17) Lemma: Let structures S and T be given.
(a) If S and T are isomorphic then the set of environments for S is identical tothe set of environments for T .
(b) Suppose that Latomic ⊆ Obs. If some environment is for both S and T then Sand T are isomorphic.
Lemma (17)b is also true if it is assumed that (Lbasic − Latomic) ⊆ Obs.
The lemma does not say that a scientist can infer the underlying structure (up toisomorphism) by progressively examining an environment for it (assuming Latomic ⊆ Obs).This is because no proper initial segment of an environment need provide enough informationto pin down the structure. For example, no finite piece of the environment indicated inExample (16)a reveals whether the underlying structure has a least point or a greatestpoint. (See also the discussion on page 20 of the first essay.)
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2.3.6. Data
Environments present their data neatly, offering one formula (or the pause symbol) perposition. Nothing hinges on such tidiness. We could just as well allow arbitrary finite setsof formulas in the positions currently reserved for just single formulas (in this case, pauseswould correspond to the empty set). Sticking with our current conception, we will oftenneed to denote the datum available in an environment at a given position, and also thesequence of data available up to that point. This is achieved as follows.
(18) Definition: Let environment e and k ∈ N be given.
(a) e(k) denotes the member of e that falls in its kth position.
(b) e[k] is the initial finite segment of e of length k.
Notice that e(k) comes right after e[k] in e.
(19) Example: If e is the environment of Example (16)a, then e(2) = Rv1v9, e[2] =(v3 6= v4,¬Rv0v0), e(0) = (v3 6= v4), and e[0] = ∅.
We can think of e[n] as the scientist’s data at stage n of his examination of environmente. His conjectures depend on nothing more than the available data, so it is convenientto have a notation for the collection of all such e[n]. The following definition serves thispurpose.
(20) Definition:
(a) We let SEQ denote the collection of proper initial segments of any environment.
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(b) Given σ ∈ SEQ, we denote by∧σ the conjunction (in order of appearance in
σ) of content(σ). To cover degenerate cases, we define∧σ to be ∀v0(v0 = v0)
in case σ contains no formulas (either because length(σ) = 0 or because only ]appears in σ).
(c) We denote by Var(σ) the set of free variables appearing in σ.
(d) Let problem P, proposition P and σ ∈ SEQ be given. We say that σ is for Pjust in case σ is an initial segment of an environment for P . We say that σ isfor P just in case σ is for some P ∈ P.
Thus, σ is for P if and only if∧σ is satisfiable in some member of P . Likewise, σ is for P
if and only if there is S ∈⋃
P that satisfies∧σ. If everything is clear to you, the following
observation should be evident.
(21) Observation: Let Obs be given. Then SEQ is the countable set of all finitesequences σ over Obs ∪ {]} such that content(σ) is consistent.
2.4. Scientists
We adopt the extensional approach to scientists, discussed in Section 2.3 of the first essay.Scientists are thus conceived as mappings from evidence to conjectures. How the mapping isimplemented does not concern us at the present stage (later on it will). Possible evidence isembodied by SEQ. Possible conjectures are embodied by the collection of all propositions.Scientists map one to the other. This is how we would like our definition to go, but thereis an annoyance. The empty set is not a proposition; see Definition (10). Yet it will beconvenient for us to give scientists the option of conjecturing the empty set. The officialdefinition of “scientist” is thus the following.
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(22) Definition: A scientist is a (possibly partial) mapping of SEQ into classes ofstructures.
The possibility of conjecturing the empty set will be ignored when there is no risk of confu-sion. So we can say informally that a scientist maps SEQ into the collection of propositions.Faced with evidence σ, Ψ asserts the proposition expressed by Ψ(σ). Should Ψ be undefinedon σ, then Ψ(σ) does not denote anything. (Scientists are allowed to be partial functions.)
2.5. Success
To solve a problem P, we require the scientist’s conjectures to stabilize on the one trueproposition of P, namely, the proposition holding the structure Nature chose at the outsetof the game. To express such belief it is sufficient to announce a consistent propositionthat implies the true one from P; there is no penalty for saying something stronger thanthe target proposition, provided it is consistent. This added flexibility will be exploited inour discussion of computable inquiry, and of belief revision. Before further comment, let uspresent the official definition of success.
(23) Definition: Let scientist Ψ be given.
(a) Let environment e for proposition P be given. We say that Ψ solves P in ejust in case for cofinitely many k, ∅ 6= Ψ(e[k]) ⊆ P . We say that Ψ solves Pjust in case Ψ solves P in every environment for P .
(b) Let problem P be given. We say that Ψ solves P just in case Ψ solves everymember of P. In this case we say that P is solvable, and otherwise unsolvable.
Hence, solving P requires solving every P ∈ P in every environment for P . Equivalently:Ψ solves P just in case for every P ∈ P, every S ∈ P , and every environment e for S, there
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are cofinitely many k such that ∅ 6= Ψ(e[k]) ⊆ P . Requiring success on all environmentsinstead of just some eliminates the possibility of communicating the correct answer to ascientist via a coding scheme involving the order in which formulas are presented. (Wediscuss such collusion at greater length in the fourth essay.) Let’s consider an example.
(24) Example: Suppose that Obs = Lbasic .
(a) Suppose that unary predicate H is the only member of Sym. Given n ∈ N ,let Pn be the class of all structures S such that card(HS) = n.13 Let P ={Pn |n ∈ N}. Then P is solvable. To see this, given σ ∈ SEQ, let t(σ) denotethe smallest n ∈ N such that σ implies there are at least n things falling in H.Define scientist Ψ such that for all σ ∈ SEQ, Ψ(σ) = Pt(σ). Then it is easy tosee that for all n ∈ N and all environments e for Pn, Ψ(e[k]) = Pn for cofinitelymany k. Thus, Ψ solves P.
(b) Suppose that binary predicate R is the only member of Sym. Set:
Pf = {〈N,�〉 | � is isomorphic to less-than-or-equals on ω} ,
Pb = {〈N,�〉 | � is isomorphic to less-than-or-equals on ω∗} .
(Think of the subscripts as “forward” and “backward,” respectively.) Then itis easy show that {Pf , Pb} is solvable.14 Observe that each of Pf , Pb containsuncountably many structures. For example, Pf contains the ordering 1 � 0 �3 � 2 � 5 � 4 · · ·.
(c) Again suppose that binary predicate R is the only member of Sym. Let Pbe as specified in Example (11). Then P is solvable. (If the solvability of Pisn’t clear to you, review Section page 21 of the first essay.) This fact implies
13HS is the set interpreting H in the structure S. The cardinality of that set is card(HS).14ω∗ is the set of natural numbers ordered backwards, that is, as · · · 4, 3, 2, 1, 0.
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that {Pf , Pb} is solvable since Pf ⊂ MOD(T ∪ {θ}) and Pb ⊂ MOD(T ∪ {¬θ}),where T and θ are given in (11).
The solvability of these problems depends on the assumption that Obs = Lbasic . Otherassumptions lead to unsolvability. For example, suppose that Obs contains nothing butexistential and universal sentences (no free variables). Then {Pf , Pb} in (24)b is unsolvable.(You should be able to figure out why.)
One aspect of Definition (23) needs further discussion. Let P be the proposition con-sisting of every total order (over a binary predicate R). Let e be an environment for 〈N,≤〉,and suppose that scientist Ψ issues the class of dense total orders on cofinitely many initialsegments of e.15 Our success criterion credits Ψ with solving P in e, yet Ψ is systematicallymistaken in one respect about the structure underlying e, namely, it is not a dense order.Is the criterion too liberal? We think not. If Ψ’s behavior is deemed unsatisfactory inthis example, the excessive liberality is inherent in P , not in our success criterion. When,for particular purposes, density is an important property of orders, the problem should bedefined in such a way that dense and non-dense orders belong to different cells. In contrast,if the class of all total orders is one of the propositions of a problem, this can only signifythat density is irrelevant to the inquiry in question. The criterion introduced in Definition(23) allows the problem itself to determine what counts as an accurate conjecture.
3. Three observations
That’s all there is to it! Our basic inductive paradigm is henceforth in place. (We realizethat the journey through the preceding definitions was arduous, but now you’re done.) We
15A total order is dense iff for every pair of objects in the relation there is a third object between them.The rational numbers ordered by ≤ are dense whereas the integers are not.
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are therefore free to consider what kinds of problems are solvable and by what kinds ofscientists. This task will occupy much of the remainder of the essay. First, we make threeobservations about our paradigm.
3.1. Countability
Let Ψ be a given scientist. Then {Ψ(σ) |σ ∈ SEQ} is a countable set because SEQ is count-able and Ψ is a function. Suppose that problem P contains uncountably many propositions(pairwise disjoint, of course). Then because of the excess cardinality there is P ∈ P suchthat for no σ ∈ SEQ:
∅ 6= Ψ(σ) ⊆ P.
It follows immediately that Ψ does not solve P because it can’t even conjecture one of itspropositions (no matter what evidence it receives). We thus have:
(25) Lemma: Every solvable problem is countable.
Of course, by a problem being “countable” is meant that it contains countably many propo-sitions. Each of the propositions may have arbitrary cardinality (or even be a “proper class,”as will often be the case in our examples).
3.2. Normal problems
If two structures are isomorphic then every environment for one is also for the other. Insuch circumstances it is impossible for the scientist to detect which of the two structureswas chosen by Nature. It is therefore impossible to solve a problem that places the twostructures in different propositions. The point is worth stating precisely.
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(26) Definition: For any proposition P , let I(P ) denote the closure of P under isomor-phism.16 Problem P is normal just in case for all P1, P2 ∈ P, I(P1) ∩ I(P2) = ∅.
Thus, in a normal problem there are no isomorphic structures appearing in different propo-sitions. Directly from Lemma (17)a:
(27) Lemma: Only normal problems are solvable.
Normal problems will be the focus of the our discussion. It remains that case, however thata problem is any collection of disjoint propositions.
3.3. Separation
Let Pf , Pb be as specified in Example (24)b. Suppose Obs = Lbasic . Let e be an environmentfor Pf ∪ Pb. Then no formula
∧e[k], with k ∈ N , “separates” Pf and Pb, in the sense of
being satisfiable in just one of them. Indeed, every initial segment of e is satisfiable in everymember of Pf ∪ Pb. The solvability of {Pf , Pb} thus illustrates that within our paradigm,separability is not required for successful inquiry.17 The paradigm differs in this respectfrom the framework established in [10], in which separation plays a major role. Roughlyspeaking, one motivation for the developments to follow is to understand inquiry withoutrequiring the scientist’s data to rule out any theoretically possible world.
16In other words, S ∈ I(P ) implies that every structure isomorphic to S is also in I(P ).17We made a similar point in the first essay on page 20.
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4. Characterization
Which problems are solvable? To respond nontrivially to this question, we seek a necessaryand sufficient condition that does not simply repeat Definition (23) by invoking the behaviorof scientists. The present section offers such a condition. It will be used immediately todemonstrate the unsolvability of some simple problems. A wider range of examples isconsidered in Section 4.4.
To be truthful with you, our condition has a slightly technical character. It is importantmainly as a tool for demonstrating more memorable results, e.g., about belief revision (seethe third essay). Consequently, we have organized the material so that you can jump overthe present section, and pick up the discussion in Section 5.
4.1. Locking pairs
As a first step towards characterizing the class of solvable problems, we need the idea ofdata that “lock” a scientist onto the correct conjecture.
(28) Definition: Let scientist Ψ, proposition P , S ∈ P , σ ∈ SEQ, and finite assignmenta : Var → | S | be given. We say that (σ, a) is a locking pair for Ψ, S, and P just incase the following conditions hold.
(a) domain(a) ⊇ Var(σ).
(b) S |=∧σ[a].
(c) For every τ ∈ SEQ, if S |= ∃x∧
(σ∗τ)[a], where x contains the variables inVar(τ)− domain(a), then ∅ 6= Ψ(σ∗τ) ⊆ P .
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You might think that locking pairs would be reserved for scientists suffering from anidee fixe. To the contrary, every successful scientist is prey to them.
(29) Lemma: Let scientist Ψ, proposition P , and S ∈ P be given. Suppose that Ψ solvesP in every environment for S. Then there is a locking pair for Ψ, S, and P .
The proof is given in Section 8.1, below.18
4.2. Tip-offs
With Lemma (29) in hand, we are ready to characterize solvability. The solvability of aproblem requires that each of its propositions be associated with a faithful signal.19 Thesignals are called “tip-offs,” and defined in the present subsection. After tip-offs are defined,we show that their existence is both sufficient and necessary for solvability. A preliminarydefinition is needed.
(30) Definition: We say that ϕ ∈ Lform is refutable just in case for all structures S andfull assignments h to S, the following holds. Suppose that S 6|= ϕ[h]. Then thereexists finite D ⊆ Obs such that S |= D[h] and D |= ¬ϕ.
Thus, a refutable formula is contradicted by any environment for a structure in whichthe formula is false. To get a grip on the idea, let us see what refutability means whenenvironments are composed of basic formulas or atomic formulas. (These are perhaps themost natural choices of Obs.) The following lemma is demonstrated in Section 8.2.
18It is inspired by an argument appearing in [2] for a similar lemma in a numerical setting.19The idea of such signals comes from [1], which investigates a numerical paradigm.
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(31) Lemma:
(a) Suppose that Obs = Latomic . Then an invalid formula is refutable iff it islogically equivalent to a formula of form ∀x1 . . .∀xn¬ψ, where ψ is a positiveformula.20
(b) Suppose that Obs = Lbasic . Then a formula is refutable iff it is logicallyequivalent to a ∀ formula.
Now let us define the key combinatorial concept, embodying the “signal” that mustaccompany each proposition of a solvable problem.
(32) Definition: Let problem P and P ∈ P be given. A tip-off for P in P is a countablecollection t of subsets of Lform such that:
(a) Each member of t has the form Y ∪ Z, where (i) Y is a finite subset of Obs,(ii) every member of Z is refutable, and (iii) the set of free variables occurringin Y ∪ Z is finite.
(b) For every S ∈ P and full assignment h to S, there is π ∈ t with S |= π[h].(c) For all U ∈ P ′ ∈ P with P ′ 6= P , all full assignments g to U , and all π ∈ t,
U 6|= π[g].
If every member of P has a tip-off in P, then we say that P has tip-offs.
When Obs is closed under negation (as when Obs = Lbasic), tip-offs have a simplerform.21 This is shown by the following lemma and example. They follow directly from
20Aformula is “positive” if it is equivalent to a formula with no conditionals or biconditionals and in whichall atomic subformulas appear under the scope of an even number of negations.
21Recall from Definition (4) that Obs is closed under negation just in case for every ϕ ∈ Obs, there isψ ∈ Obs that is logically equivalent to ¬ϕ.
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Definition (32).
(33) Lemma: Suppose that Obs is closed under negation. Let problem P and P ∈ Pbe given. Then if there is a tip-off for P in P, there is one with Y = ∅ in Definition(32)a.
(34) Example: Suppose that Obs = Lbasic , and that P in P has a tip-off. Then byLemma (33), there is a tip-off t for P in P of the following form.
(a) Each member of t is a set of universal formulas with only finitely many freevariables occuring in the set.
(b) For every S ∈ P and full assignment h to S, there is π ∈ t with S |= π[h].
(c) For all U ∈ P ′ ∈ P with P ′ 6= P , all full assignments g to U , and all π ∈ t,U 6|= π[g].
The next example shows what tip-offs look like in case Obs = Latomic .
(35) Example: Suppose that Obs = Latomic . Let problem P and P ∈ P be given. Thena tip-off for P in P is a countable collection t of subsets of Lform such that:
(a) Each member of t has the form Y ∪Z, where (i) Y is a finite subset of Latomic ,(ii) every member of Z is equivalent to a formula of form ∀x1 . . .∀xn¬ϕ, whereϕ is positive quantifier-free formula, and (iii) the set of free variables occurringin Y ∪ Z is finite.
(b) For every S ∈ P and full assignment h to S, there is π ∈ t with S |= π[h].
(c) For all U ∈ P ′ ∈ P with P ′ 6= P , all full assignments g to U , and all π ∈ t,U 6|= π[g].
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Here is an example of a problem with tip-offs, and one without. Their proofs are givenin Section 8.3.
(36) Example: Suppose that Lbasic ⊆ Obs. Then the problem defined in Example (11)has tip-offs.
(37) Example: Suppose that Obs = Lbasic , and that binary predicate R is the onlysymbol of Sym. Let S = 〈N,�〉 be isomorphic to ω, let T = 〈Z,�∗〉, with Z theset of integers, be isomorphic to ω∗ω, and let disjoint propositions P1, P2 be suchthat S ∈ P1 and T ∈ P2. Then {P1, P2} does not have tip-offs.
4.3. Tip-offs and solvability
Countable problems are solvable if and only if they have tip-offs. We state the matter intwo propositions, first concerning sufficiency then necessity.
(38) Proposition: If problem P is countable and has tip-offs, then P is solvable.
The proof is given in Section 8.4. As an application, from Example (36) we obtain thefollowing.
(39) Proposition: Suppose that Lbasic ⊆ Obs. Then the problem defined in Example(11) is solvable.
Now necessity:
(40) Proposition: Every solvable problem has tip-offs.
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See Section 8.5 for the proof. As an application, we obtain the following from Example(37).
(41) Proposition: Suppose that binary predicate R is the only symbol of Sym, andthat Obs = Lbasic , Let S = 〈N,�〉 be isomorphic to ω, and let T = 〈Z,�∗〉, withZ the set of integers, be isomorphic to ω∗ω. Suppose that propositions P1, P2 aresuch that S ∈ P1 and T ∈ P2. Then {P1, P2} is not solvable.
Along with Lemma (25), Propositions (38) and (40) yield the following characterizationof solvability.
(42) Theorem: A problem is solvable if and only if it is countable and has tip-offs.
4.4. Examples
Let’s put Theorem (42) to work by considering a range of examples. Actually, we preferto put you to work by treating the examples as exercises. If you can’t stand this kind ofthing, just skip to Section 5.
For the first two exercises, suppose that Sym consists of a binary predicate R and thatObs = Latomic .
(43) Exercise: Let T be the theory of strict total orders (with respect to R) with eithera least point or a greatest point, but not both. Let θ = ∃x∀y(x = y ∨ Rxy) andP = {MOD(T ∪ {θ}),MOD(T ∪ {¬θ})}. Show that P is solvable.
(44) Exercise: Let T be the theory of total orders (with respect to R) with eithera least point or a greatest point, but not both. Let θ = ∃x∀yRxy and P ={MOD(T ∪ {θ}),MOD(T ∪ {¬θ})}.
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(a) Show that P is not solvable(b) Suppose that environment e is for structure S and full assignment h. We say
that e is bijective just in case h is a bijection between Var and | S |. Specify ascientist Ψ such that for all P ∈ P, Ψ solves P in every bijective environmentfor P .
For the remaining exercises, suppose that Obs = Lbasic .
(45) Exercise: Suppose that Sym is the vocabulary of Boolean algebra. Let T ⊆ Lsen
be such that MOD(T ) is the class of Boolean algebras, and let θ ∈ Lsen be true ofa Boolean algebra iff it is atomic. Show that {MOD(T ∪ {θ}),MOD(T ∪ {¬θ})} isnot solvable. (For Boolean algebra, see [14, Ch. VIII].)
(46) Exercise: Suppose that Sym is limited to a unary constant 0, a unary functionsymbol s, and two binary function symbols ⊕,⊗. Let Q be the conjunction of theseven axioms of “Robinson’s arithmetic” (see [3, Ch. 14]). Show that the problem{MOD(Q),MOD(¬Q)} is not solvable.
(47) Exercise: Suppose that Sym = ∅. Let P1 = {N}, P2 = {∅ 6= D ⊆ N |D finite}.Show that {P1, P2} is not solvable.
(48) Exercise: Suppose that Sym is limited to the binary predicateR. Let T be the the-ory of strict total orders (with respect to R). Denote by θ the sentence ∀xz[Rxz →∃y(Rxy ∧Ryz)] (“R is dense”). Show that P = {MOD(T ∪ {θ}),MOD(T ∪ {¬θ})}is not solvable.
(49) Exercise: Suppose that Sym is limited to a binary predicate R. Say that structureS is standard if | S | = N . We specify a countable collection {Sj | j ∈ N} of standard
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structures by specifying the extension RSj of R for all j ∈ N . RS0 is the relation{(i, i + 1) | i ∈ N}. For j > 0, RSj is the finite relation {(i, i + 1) | i < j}. LetP1 = {S0} and P2 = {Sj | j > 0}. Show that {P1, P2} is not solvable.
(50) Exercise: Given a well-ordering ≺ over collection K of structures, we define anassociated scientist Ψ≺ as follows. For all σ ∈ SEQ, Ψ≺(σ) is the ≺-least memberS of K such that
∧σ is satisfiable in S. Ψ≺(σ) is undefined if there is no such.
Problem P is solvable by enumeration just in case there is a well-ordering ≺ of⋃
Psuch that Ψ≺ solves P. Exhibit structures S and T such that {{S}, {T }} is solvable,but not by enumeration.
(51) Exercise: Let scientist Ψ and problem P be given. We say that Ψ is P-conservativejust in case for all P ∈ P, σ for P and β ∈ Lbasic , if ∅ 6= Ψ(σ) ⊆ P and every structurein Ψ(σ) satisfies {
∧σ, β}, then ∅ 6= Ψ(σ∗β) ⊆ P . Exhibit structures S and T such
that P = {{S}, {T }} is solvable, but no P-conservative scientist solves {{S}, {T }}.
(52) Exercise: Call problem P “separated” (respectively “elementary separated”) justin case for all P1 6= P2 ∈ P, S ∈ P1, and T ∈ P2, there is no isomorphic (respectivelyelementary) embedding from S into T . [For background on embeddings, see [11,Secs. 1.2, 2.5].]
(a) Exhibit a separated, unsolvable problem.
(b) Show that every solvable problem is elementary separated.
The present exercise thus strengthens Lemma (27).
(53) Exercise: Call problem P elementary just in case for all S,U ∈⋃
P, S ≡ U .Exhibit an elementary, solvable problem consisting of more than one proposition.
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(54) Exercise: Scientist Ψ weakly solves problem P just in case for every environmente for P ∈ P, the following conditions are met:
(a) There are infinitely many k such that ∅ 6= Ψ(e[k]) ⊆ P .
(b) For every P ′ ∈ P with P ′ 6= P , there are only finitely many k such that∅ 6= Ψ(e[k]) ⊆ P ′.
In this case P is said to be weakly solvable.
(a) Exhibit a problem that is weakly solvable but not solvable.
(b) Show that no unsolvable problem consisting of finitely many propositions isweakly solvable.
5. Special Problems
The preceding section was devoted to characterizing solvability in terms of a combinatorialconcept called “tip-offs.” In the present section we exploit the characterization in the serviceof analyzing the solvability of a restricted class of problems.
5.1. Problems of form (T , {θ0 . . . θn})
The next definition introduces notation for problems of an appealing form. They becomeincreasingly important as our theory develops.
(55) Definition: Let problem P, T ⊆ Lsen , and θ0 . . . θn ∈ Lsen be given. We say thatP has the form (T , {θ0 . . . θn}) just in case:
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(a) for every model S of T there is exactly one i ∈ {0 . . . n} such that S |= θi;(b) P = {MOD(T ∪ {θi}) | 0 ≤ i ≤ n}.
By Definition (10) of “problem,” P can have the form (T , {θ0 . . . θn}) only if MOD(T ∪ {θi})is nonempty for all i ≤ n. Hence T ∪ {θi} must be consistent for all i ≤ n, which impliesthat T is consistent. Intuitively, given problem (T , {θ0 . . . θn}), we may think of T as anaccepted background theory that leaves open which among the θi’s is true. The θi’s thuspartition the models of T . The informal overview provided by the first essay concentratedon problems of form (T , {θ0 . . . θn}).
A yet more special case arises when the partition has just two members. The problemthen has the form (T , {θ,¬θ}). In contrast, the more general form (T , {θ0, θ1, . . .}) is of nouse since an easy compactness argument shows that there is no infinite partition of MOD(T )of the form {MOD(T ∪ {θi}) | i ∈ N}.
To make sure that our notation is clear, let us consider an example.
(56) Example: Suppose that Sym is limited to a binary function symbol ◦, and let Tbe the theory of groups. Let θ be the sentence ∀xy(x ◦ y = y ◦x). Then (T , {θ,¬θ})is the problem whose propositions are:
• the class of abelian groups, namely, MOD(T ∪ {θ});• the class of non-abelian groups, namely, MOD(T ∪ {¬θ}).
If Obs includes Lbasic , then (T , {θ,¬θ}) is trivially solvable.
For a more substantial illustration, Propositions (39) and (41) yield the following.
(57) Proposition: Suppose that Sym is limited to the binary predicate R and thatObs = Lbasic . Let θ = ∃x∀yRxy, let T 0 ⊆ Lsen be the theory of total orders over R,
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and let T 1 ⊆ Lsen be the theory of total orders with either a greatest point or a leastpoint, but not both. Then the problem (T 1, {θ,¬θ}) is solvable but (T 0, {θ,¬θ}) isnot.
We can exploit Proposition (57) to reformulate a point about inquiry that we made inSection 2.3.5 above. Let a problem of form (T , {θ0 . . . θn}) be given, and suppose that e isone of its environments. If there is k ∈ N such that T ∪ {
∧e[k]} implies the right choice
among θ0 . . . θn, then the problem can be solved in e by simply issuing MOD(T ∪ {∧e[k]})
at each stage k of inquiry. If the same is true of all the environments, then it may be saidthat (T , {θ0 . . . θn}) is solved by “waiting for deduction to work.” For, it suffices to wait forthe background theory and current data to imply the right answer. Our inductive paradigmwould have a trivial character if the solvability of (T , {θ0 . . . θn}) implies solvability via sucha strategy. But in fact, awaiting deduction is not sufficient. This can be seen from the factthat (T 1, {θ,¬θ}) of Proposition (57) is solvable, with Obs = Lbasic , even though for allσ ∈ SEQ,
∧σ implies neither θ nor ¬θ in the models of T 1. So, some kind of genuinely
inductive strategy is needed to solve (T 1, {θ,¬θ}).
5.2. Solvability of problems of form (T , {θ0 . . . θn})
Theorem (42) offers an infinitary condition on solvability since the tip-offs it evokes areinfinite sets of formulas. The restricted form of problems (T , {θ0 . . . θn}), however, leads usto hope for deeper characterization of their solvability. In fact, for such problems we canprove various finitary conditions depending on the choice of Obs. For example, we willdemonstrate the following.
(58) Proposition: Suppose that Obs = Lbasic . Then a problem of form (T , {θ0 . . . θn})is solvable if and only if for every i ≤ n, θi is equivalent in T to an ∃∀ sentence.
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The proposition follows from a more general theorem.
(59) Theorem: A problem of form (T , {θ0 . . . θn}) is solvable if and only if for everyi ≤ n, θi is equivalent in T to the existential closure of any of the following kinds offormulas.
(a) a formula built from Obs using only conjunctions and disjunctions,
(b) a refutable formula, or
(c) a formula of form χ ∧ ϕ, where χ is built from Obs using only conjunctionsand disjunctions and ϕ is refutable.
For the proof, see Section 8.6. We have the following corollaries.22
(60) Corollary: A problem of form (T , {θ0 . . . θn}) is solvable if and only if for everyi ≤ n, θi is equivalent in T to the existential closure of a formula of form χ or ∀y¬ϕor χ∧∀y¬ϕ, where χ and ϕ are built from Obs using conjunctions and disjunctionsonly.
(61) Corollary: Suppose that Obs is closed under negation. A problem of form(T , {θ0 . . . θn}) is solvable if and only if for every i ≤ n, θi is equivalent in T to theexistential closure of a refutable formula.
Proposition (58), given above, follows from Theorem (59) in conjunction with Corollary(31)b. From (59) and (31)a, we get:
22The first follows directly from the theorem and Lemma (92). The latter is proved in Section 8.2. Thesecond comes from from Theorem (59), Lemma (91)a, and Corollary (93). The latter two results are provedin Section 8.2.
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(62) Corollary: Suppose that Obs = Latomic . Then a problem of form (T , {θ0 . . . θn})is solvable if and only if for every i ≤ n, θi is equivalent in T to the existentialclosure of a positive quantifier-free formula or of a formula of form χ∧∀y¬ϕ, whereχ and ϕ are positive quantifier-free formulas.
Here is another useful way to formulate Proposition (58).
(63) Corollary: Suppose that Obs = Lbasic . Then a problem of form (T , {θ0 . . . θn})is solvable if and only if for all i ≤ n, θi is equivalent in T to a boolean combinationof ∃ sentences.
To get (63) from (58), we rely on the following fact, which is an easy adaptation of [5,Theorem 3.1.16].
(64) Fact: Let T ⊆ Lsen and θ ∈ Lsen be given. Suppose that θ is equivalent in T bothto an ∃∀ sentence and to a ∀∃ sentence. Then θ is equivalent in T to a booleancombination of ∃ sentences.
5.3. Problems of form (T , {P0, P1, . . .})
We now define another special class of problems.
(65) Definition: Let T ⊆ Lsen be given. We say that problem P has the form(T , {P0, P1, . . .}) just in case P = {P0, P1, . . .} and
⋃P = MOD(T ).
A problem can have the form (T , {P0, P1, . . .}) only if T is a consistent theory (otherwisethe propositions Pi are empty). For consistent T , a problem of form (T , {P0, P1, . . .}) is a
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partition of the models of T . Plainly, any problem of form (T , {θ0 . . . θn}) also has the moregeneral form (T , {P0, P1, . . .}). For examples, see Section 5.5.
5.4. Solvability of problems of form (T , {P0, P1, . . .})
The solvable problems of form (T , {P0, P1, . . .}) share an interesting logical feature. Let usfirst state the matter under the assumption that Obs = Lbasic . Then we will formulate amore general result.
(66) Proposition: Suppose that Obs = Lbasic . Let there be given a solvable problem Pof form (T , {P0, P1, . . .}). Then for all P ∈ P, there is an enumeration {Xi | i ∈ N}of sets of ∃∀ sentences such that P =
⋃i∈N MOD(T ∪Xi).
The matter might be put this way: first-order sentences are sufficient to distinguish amongany collection P of propositions that partition an elementary class, provided only that P issolvable. It is striking that no further constraint need be placed upon the propositions.
Proposition (66) follows directly from the next fact, along with Corollary (31).
(67) Proposition: Let solvable problem P of form (T , {P0, P1, . . .}) be given. Then forall P ∈ P, there is an enumeration {Xi | i ∈ N} of sets of existential closures offormulas of form
∧D ∧ ϕ, where D is a (possibly empty) finite subset of Obs and
ϕ a refutable formula, such that P =⋃i∈N MOD(T ∪Xi).
The proposition is proved in Section 8.7.
Two other easy corollaries of Proposition (67) include the following.23
23Corollary (68) follows from Proposition (67) and Lemma (92). The latter is proved in Section 8.1.Corollary (69) follows from Proposition (67) and Lemma (31).
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(68) Corollary: Let solvable problem P of form (T , {P0, P1, . . .}) be given. Then forall P ∈ P, there is an enumeration {Xi | i ∈ N} of sets of existential closures offormulas of form χ or ∀y¬ϕ or χ ∧ ∀y¬ϕ, where χ and ϕ are built from Obs usingconjunctions and disjunctions only, such that P =
⋃i∈N MOD(T ∪Xi).
(69) Corollary: Suppose that Obs = Latomic . Let solvable problem P of form(T , {P0, P1, . . .}) be given. Then for all P ∈ P, there is an enumeration {Xi | i ∈ N}of sets of existential closures of positive quantifier-free formulas or of formulas ofform χ ∧ ∀y¬ϕ, where χ and ϕ are positive quantifier-free formulas, such thatP =
⋃i∈N MOD(T ∪Xi).
For technical interest, let it be noted that the converse of Proposition (67) is false. Tosee this, suppose that Sym = ∅, Obs = Lbasic , and for i > 0 let χi be an ∃∀ sentencetrue in just the structures of cardinality i. Let X = {∃x0 . . . xn
∧0≤i<j≤n xi 6= xj |n > 0}.
Then P= {MOD(X),MOD(χ1),MOD(χ2) . . .} partitions the class of all structures, hencehas the form (T , {P0, P1, . . .}). But an easy consequence of Exercise (47) shows P to beunsolvable.
5.5. Examples
Here are some examples and facts related to the material in this section. We leave them asexercises.
(70) Exercise: Suppose that Sym consists of a unary function symbol s and a constant0. The term that results from n applications of s to 0 is denoted n. Let P0 bethe class of models of {n 6= 0 |n > 0}, and for i > 0, let Pi be the class of modelsof {n 6= 0 | 0 < n < i} ∪ {i = 0}. It is easy to verify that if Latomic ⊆ Obs then(∅, {P0, P1, . . .}) is solvable.
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(71) Exercise: Suppose that Sym contains just two binary function symbols. For i ∈ Neither 0 or prime, let proposition Pi be the collection of all fields of characteristic i inthis vocabulary. Show that if Obs = Lbasic , {Pi | i is either 0 or prime} is solvable.
(72) Exercise: Suppose that Sym is limited to a binary function symbol. Let Gbe all groups, T all torsion groups, and F all torsion-free groups. Show that ifObs = Lbasic then {F,G− F} is solvable but that {T,G−T} is not.
(73) Exercise: Suppose Obs = Lbasic . Show that any problem of form (T , {θ0 . . . θn})is solvable if T is model-complete. (For model-completeness, see [11, §8.3].)
(74) Example: It is interesting to see how Theorem (59) — in particular, its corollary(58) — can be generalized to the case of problems of arbitrary form. We just considerthe case Obs = Lbasic . Show that a problem P is solvable iff it consists of countablymany propositions, each of which is equivalent in
⋃P to an Lω1ω-sentence of form:∨
i<ω
∃xi∧j<ω
∀yjϕi,j ,
where χi and ϕi,j are quantifier-free members of Lform , for all i, j < ω.24
6. Efficiency
The solvability criterion advanced in Section 2.5 imposes no requirement of speedy conver-gence, so is open to the objection that tardy success is often no better than failure. We
24For Lω1ω, see [8, §IX.2].
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are therefore led to reinforce our conception of success by including a standard of efficiency.Specifically, efficient investigation will be defined as inquiry whose use of data cannot beuniformly reduced. Efficiency in this sense will be relevant to our study of belief revision inthe third essay.
The needed definitions are introduced in the next subsection. Then we prove that theclass of solvable problems coincides with the class of problems that can be solved efficiently.
6.1. Success points, dominance, and efficiency
Given proposition P and environment e for P , scientist Ψ solves P in e just in case Ψ’ssuccessive conjectures ultimately become non-void subsets of P [see Definition (23)]. Theparticular propositions that Ψ announces, however, may vary indefinitely as Ψ progressesthrough e. So, success does not imply convergence to a single proposition. There is nonethe-less a sense in which success requires convergence to the right proposition. The matter canbe formulated as follows.
(75) Definition: Let proposition P , environment e for P , and scientist Ψ be given. Wesay that Ψ is P -correct on e at k ∈ N just in case ∅ 6= Ψ(e[k]) ⊆ P . If Ψ solves Pin e, we write SP(Ψ, e, P ) to denote the least k0 ∈ N such that Ψ is P -correct on eat k for all k ≥ k0.
Such a k0 may be called the “success point” of Ψ on e, relative to P . The earlier the successpoint, the fewer data required for the scientist to figure out Nature’s choice of proposition(even if the scientist remains uncertain whether convergence to the correct conjecture hasin fact been achieved).
It would be nice if we could associate a single success point with a given scientist.But this is not possible because of the multiplicity of environments for a given structure.
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An environment that begins with many pauses or repetitions, for example, will typicallyyield a greater success point than an environment which presents denser information. Forthis reason, we proceed indirectly, defining efficient induction in terms of a strong form ofinefficiency.
(76) Definition: Suppose that scientist Ψ solves problem P. We say that Ψ is domi-nated on P just in case there is scientist Ψ′ such that:
(a) Ψ′ solves P;(b) for all P ∈ P and environments e for P , SP(Ψ′, e, P ) ≤ SP(Ψ, e, P );(c) for some P ∈ P and environment e for P , SP(Ψ′, e, P ) < SP(Ψ, e, P ).
We say that Ψ solves P efficiently just in case no scientist dominates Ψ on P. Inthis case, P is solvable efficiently.
An efficient scientist, in other words, cannot be improved uniformly. No other scientistsometimes consumes less data and never consumes more.
6.2. Solvability implies efficient solvability
Requiring scientists to be efficient as well as successful does not reduce the class of solvableproblems. This is the content of the following proposition
(77) Proposition: Every solvable problem is solvable efficiently.
Let us head off a misunderstanding. The proposition does not assert that every scientistwho solves a problem does so efficiently. It says only that if some scientist solves a givenproblem then some scientist (not necessarily the same one) solves it efficiently.
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To prove Proposition (77), it will be convenient to demonstrate a stronger result. Thisrequires a stronger concept of efficiency that is interesting in its own right.
(78) Definition: Suppose that scientist Ψ solves problem P. We say that Ψ solves Pstrongly efficiently just in case for every scientist Ψ′ that solves P, if there is anenvironment e0 for P0 ∈ P with k0 = SP(Ψ′, e0, P0) < SP(Ψ, e0, P0) then there isP1 ∈ P, S ∈ P1, and full assignment h to S such that:
(a) S |=∧e0[k0][h] and
(b) SP(Ψ, e, P1) ≤ k0 < SP(Ψ′, e, P1) for every environment e for S and h thatextends e0[k0].
In this case, P is solvable strongly efficiently.
Intuitively, Ψ is strongly efficient if it succeeds faster than any rival on many environments— where a “rival” is a scientist that succeeds faster than Ψ on some environment. Plainly,strong efficiency implies efficiency. Proposition (77) thus follows from:
(79) Theorem: Every solvable problem is solvable strongly efficiently.
The theorem is demonstrated in Section 8.8.
7. Computability
It is an engaging (but far from proven) hypothesis that human scientific behavior is Turingsimulable. If the hypothesis is true, then our paradigm would benefit from greater realism
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by isolating the computable scientists for separate study. Our first goal is thus to formalizethe idea that a scientist [in the sense of Definition (22)] is Turing simulable. The questionwill then arise: Are there solvable problems that cannot be solved by computable scientists?
7.1. Finitization of inputs and outputs
In order to conceive of a scientist as Turing simulable, we must find a way to representhis inputs and outputs as finite objects. It might be thought that inputs pose no problemsince SEQ contains just finite sequences of formulas, hence is a countable set. But thereis a difficulty. If Obs is sufficiently rich (e.g., including all formulas with two quantifiers)then SEQ will not be effectively decidable. This is because SEQ contains just the consistentfinite sequences over Obs [see Definition (20)]. To give our computable processes a moresuitable domain we therefore introduce an extension of SEQ.
(80) Definition: The set of all finite sequences over Lform ∪{]} is denoted SEQ∗. As inDefinition (14), for σ ∈ SEQ∗ we let content(σ) denote the set of formulas appearingin σ (] is excluded).
The inputs to the computable processes used to simulate scientists will thus be drawn fromSEQ∗.
Now consider the scientist’s outputs, namely propositions. Propositions, it will be re-called, are vast objects that comprehend an arbitrary class of structures. [see Definition(10)]. To represent them finitely we limit attention to elementary classes of structures, thatis, to classes composed of every structure that satisfies some given set of formulas. However,if the set of formulas is not recursively enumerable (r.e.), then such a proposition remainsineffable (at least, in one important sense) by computable scientists. So we must furtherrestrict propositions to classes of structures that are definable by an r.e. set S of formulas.
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The r.e. index of S can then be used to express the proposition defined by S. Finitizingoutputs in this way makes it possible for some scientists to be simulated by computablefunctions from SEQ∗ to the set of indices. These will be called the “computable scientists.”
To keep things tidy, it is necessary to distinguish the indices used for r.e. sets of formulasfrom those used for r.e. sets of numbers. So we fix for the remainder of our discussion anacceptable indexing {Wform
i | i ∈ N} of the r.e. subsets of Lform , along with an acceptableindexing {Wnum
i | i ∈ N} of the r.e. subsets of N .25 Observe that for i ∈ N with Wformi
consistent, MOD(Wformi ) is a proposition; namely, it is the class of structures that satisfy
all the formulas enumerated by the ith Turing Machine. [Also note that MOD(Wformi )
is the universal proposition (all structures) if Wformi = ∅.] Hence, given mapping ψ :
SEQ∗ → N and σ ∈ domain(ψ), MOD(Wformψ(σ) ) is also a proposition (when nonempty),
namely, MOD(Wformi ) for i = ψ(σ). In this way, ψ uses i to express its view that a certain
proposition is true.
7.2. Computable scientists
Pursuant to the preceding discussion, the class of computable scientists is defined as follows.
(81) Definition: Scientist Ψ is computable just in case there is computable ψ : SEQ∗ →N such that for all σ ∈ SEQ, ψ(σ) is defined if Ψ(σ) is defined, and when both aredefined Ψ(σ) = MOD(Wform
ψ(σ) ). In this case, we say that ψ underlies Ψ. A problemthat is solved by a computable scientist is computably solvable.
25For “acceptable indexing,” see [4, 15].
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So, a computable scientist Ψ is associated with a computable function ψ : SEQ∗ → N thatsimulates Ψ’s behavior. If Ψ issues proposition P on input σ ∈ SEQ, then ψ issues an indexi on σ with P = MOD(Wform
i ). It makes no difference how ψ behaves on SEQ∗ − SEQ.(The latter set holds the inconsistent sequences of formulas.)
Computability imposes a double constraint on scientists. On the one hand, conjecturesmust be expressed via (indices for) recursively axiomatizable theories.26 On the otherhand, the conversion of data into theories must be Turing simulable. The two conditionsare necessary and jointly sufficient for computer implementation of a scientist, at least inprinciple.
It is evident that not every scientist is computable according to Definition (81). Indeed,there are only countably many computable scientists (one for each Turing Machine) butuncountably many scientists in general.
7.3. Competence of computable scientists
Now we may consider the inductive competence of computable scientists. Let us start withthe most basic question, namely: Is there a solvable problem that is solved by no computablescientist? Yes, of course. Let T be a complete, nonrecursive theory. Then the degenerateproblem {MOD(T )} is trivially solvable but not computably. This is because there is noconsistent, r.e. extension of T , hence no means whereby a computable scientist can expressa nonempty subset of MOD(T ). It is a more trenchant fact that computable scientists havelimited competence even when there is no computability obstacle to their announcing anadequate theory. To prove the point, we rely on the following extension of our notation (5).
26Let us recall that a set S of formulas is r.e. if and only if S is recursively axiomatizable, i.e., if and onlyif S is a equivalent to a recursive subset of Lform [7]. Therefore, S has a recursive set of axioms if and onlyif S has an r.e. set of axioms.
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(82) Definition: A proposition is strongly elementary just in case it has the formMOD(θ) for some θ ∈ Lsen .
Thus, strongly elementary propositions can be expressed by just a single sentence. Afortiori, for every proposition P that is strongly elementary, there is an index i withP = MOD(Wform
i ). Indeed, it suffices to take i such that Wformi = {θ}, where θ de-
fines P . The upshot is that problems consisting of strongly elementary propositions poseno difficulty specifically linked to expressing the propositions. This leaves open the pos-sibility that a collection of strongly elementary propositions is difficult for a computablescientist to manipulate, and indeed we shall see that such collections can be solvable butnot computably. The following result is proved in Section 8.9.
(83) Theorem: Suppose that Sym is limited to a binary predicate, two constants, anda unary function symbol. Let Obs = Lbasic . Then for every countable collection Σof scientists there is a problem P with the following properties.
(a) Every member of P is strongly elementary.
(b) P is solvable.
(c) No member of Σ solves P.
As noted above, the computable scientists form a countable set, since their number isbounded by the number of Turing Machines. So we obtain the immediate corollary:
(84) Corollary: Let Sym and Obs be as in Theorem (83). Then there is a solvableproblem whose members are strongly elementary but which is not solvable com-putably.
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7.4. Efficiency and computability
We have so far conceived efficiency in terms of the number of data examined prior toconvergence (see Section 6). In the context of computable scientists, however, a varietyof additional efficiency concepts may be defined. The new concepts are based on the timethat a given scientist spends examining individual σ ∈ SEQ. For example, given a functionf : SEQ → N , it may be said that scientist Ψ is “f -fast” just in case the running timeof Ψ on σ ∈ SEQ is bounded by f(σ). The running-time conception of efficiency can bestudied in conjunction with the data-use conception to provide an overall picture of resourceconsumption during inquiry. For our part, however, we shall continue to focus just on data-use, leaving running-time to one side. The reason is the shape of the ensuing theory ofrevision-based inquiry, to be developed in the next essay. Only data-use will be relevant.
Recall that Proposition (77) shows every solvable problem to be solvable efficiently.This reassuring fact does not survive the transition to computable solvability. Instead, thefollowing proposition shows that there are computable, solvable problems that cannot besolved efficiently by computable scientist.
(85) Proposition: Suppose that Sym is limited to the vocabulary of arithmetic (in-cluding 0 and a unary function symbol s) plus the additional constant a. Supposealso that Obs includes all identities (that is, formulas of form t1 = t2, for termst1, t2). Then there is a problem P with the following properties.
(a) Every member of P is strongly elementary.
(b) P is solvable computably.
(c) Every computable scientist that solves P is dominated on P.
The proof is given in Section 8.10.
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7.5. R.e. problems of form (T , {θ0 . . . θn})
Problems of form (T , {θ0 . . . θn}) are central to our theory so it is interesting to consider theircomputable solvability. Of course, if T is not recursively axiomatizable then (T , {θ0 . . . θn})may not be computably solvable for want of the ability to name its propositions. To setaside this kind of inexpressible case, we rely on the following definition.
(86) Definition: A problem of form (T , {θ0 . . . θn}) is called r.e. just in case T is arecursively enumerable set of sentences.
All the propositions of an r.e. problem of form (T , {θ0 . . . θn}) are recursively axiomatizable,so computable scientists can express any of them. Do there remain further obstacles tosolving such problems, obstacles that are specific to computable scientists? Surprisingly,this question receives a negative answer. In other words, the computable solvability of anr.e. problem of form (T , {θ0 . . . θn}) follows from its solvability tout court. The matter maybe summarized as follows.
(87) Theorem: Suppose that Obs is recursive. Then every solvable, r.e. problem ofform (T , {θ0 . . . θn}) is computably solvable.
For the proof, see Section 8.11. A little adjustment to the proof allows the hypothesis ofTheorem (87) to be weakened to: Obs is r.e.
Here is a nice exercise to test your grasp of the concepts in the present subsection.
(88) Exercise: Assume that Obs is a recursive set. Suppose that scientist Ψ solvesproblem P. We say that Ψ respects P just in case for all σ ∈ SEQ for P, Ψ(σ) is anonempty subset of some P ∈ P. Show that there is a computably solvable problemP such that no computable scientist that solves P respects P.
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8. Proofs
8.1. Proof of Lemma (29)
(29) Lemma: Let scientist Ψ, proposition P , and S ∈ P be given. Suppose thatscientist Ψ solves P in every environment for S. Then there is a locking pair forΨ, S, and P .
Proof: Suppose there is no locking pair for Ψ, S, and P . Then:
(89) For every σ ∈ SEQ and finite assignment a : Var → | S |, if domain(a) ⊇ Var(σ)and S |=
∧σ[a], then there is τ ∈ SEQ and finite extension a′ : Var → | S | of a
such that:
(a) Var(τ) ⊆ domain(a′),
(b) S |=∧τ [a′], and
(c) either Ψ(σ∗τ) is not defined or Ψ(σ∗τ) = ∅, or Ψ(σ∗τ) 6⊆ P .
We shall use (89) to construct an environment e for S such that Ψ does not solve P in e.This suffices to finish the proof.
Recall our enumeration {vi | i ∈ N} of Var. Let {si | i ∈ N} enumerate | S |. Let{αi | i ∈ N} enumerate Obs. Set a−1 = ∅ and σ−1 = ∅. The construction of e proceeds bydefining σk ∈ SEQ and finite assignment ak : Var → | S | for each k ∈ N such that:
(90) (a) ak extends ak−1;
(b) vk ∈ domain(ak) and sk ∈ content(ak);
(c) σk ⊃ σk−1;
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(d) Var(σk) ⊆ domain(ak) and S |=∧σk[ak];
(e) αk ∈ content(σk) iff S |= αk[ak].
(f) either Ψ(σk) is not defined or Ψ(σk) = ∅, or Ψ(σk) 6⊆ P .
By (90)a,b,⋃k∈N a
k is a full assignment to S. By (90)c-e,⋃k∈N σ
k is an environment e forS and
⋃k∈N a
k. By (90)f, Ψ does not solve P in e.
Let k ∈ N be given, and suppose that a−1 . . . ak−1, σ−1 . . . , σk−1 have been defined. Wedefine ak and σk. Let n ≥ k be least such that every variable appearing in αk has index lessthan or equal to n. Let a be any finite extension of ak−1 such that {vi | i ≤ n} ⊆ domain(a)and sk ∈ content(a) ⊆ | S |. Define σ ∈ SEQ to be σk−1∗αk if S |= αk[a]; otherwise, defineσ to be σk−1∗]. By (89), let τ ∈ SEQ and finite extension a′ : Var → | S | of a be suchthat Var(τ) ⊆ domain(a′), S |=
∧τ [a′], and either Ψ(σ∗τ) is not defined or Ψ(σ∗τ) = ∅, or
Ψ(σ∗τ) 6⊆ P . Define ak = a′ and σk = σ∗τ . It is easy to verify that ak and σk satisfy (90).
8.2. Proof of Lemma (31)
(31) Lemma:
(a) Suppose that Obs = Latomic . Then an invalid formula is refutable iff itis logically equivalent to a formula of form ∀x1 . . .∀xn¬ψ, where ψ is apositive formula.
(b) Suppose that Obs = Lbasic . Then a formula is refutable iff it is logicallyequivalent to a ∀ formula.
We obtain Lemma (31) from the following lemmas, which will also be useful later on.The first is evident.
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(91) Lemma:
(a) A conjunction of refutable formulas is refutable.
(b) A disjunction of refutable formulas is refutable.
(92) Lemma: An invalid formula is refutable iff it is logically equivalent to a formula ofform ∀x1 . . .∀xn¬ψ, where ψ is a formula built from Obs using conjunctions anddisjunctions only and x1 . . . xn is a (possibly empty) sequence of variables.
Proof of Lemma (92): Let invalid ϕ ∈ Lform be given. Suppose that ϕ is refutable.Let X be the set of pairs (S, h), where S is a structure, h is a full assignment to S, andS 6|= ϕ[h]. For all (S, h) ∈ X we can choose by hypothesis a finite subset DS,h of Obs suchthat S |= DS,h[h] and DS,h |= ¬ϕ. Given (S, h) ∈ X set ψS,h = ∀x1 . . .∀xn¬
∧DS,h, where
x1 . . . xn are all the variables that occur free in DS,h but not free in ϕ. It is easy to verifythat {ψS,h | (S, h) ∈ X} ∪ {¬ϕ} is not satisfiable. Since only finitely many variables occurfree in that set and X is nonempty, it follows by compactness that there exists a nonempty,finite subset Y of X such that {ψS,h | (S, h) ∈ Y } ∪ {¬ϕ} is inconsistent. We show that ϕis logically equivalent to
∧S,h∈Y ψS,h. Trivially,
∧S,h∈Y ψS,h |= ϕ. Let (S, h) ∈ Y be given.
Since ϕ |= ¬∧DS,h by the definition of DS,h, it follows from the definition of ψS,h that
ϕ |= ψS,h. Clearly∧S,h∈Y ψS,h is logically equivalent to a formula of form ∀x1 . . .∀xn¬ψ,
where ψ is a disjunction of conjunctions of members of Obs and x1 . . . xn is a (possiblyempty) sequence of variables.
Conversely, suppose that ϕ is logically equivalent to ∀x1 . . .∀xn¬ψ, where ψ is a formulabuilt from Obs using conjunctions and disjunctions only and x1 . . . xn is a (possibly empty)sequence of variables. Without loss of generality we can assume that ψ = ψ1 ∨ . . . ∨ ψp,where ψ1 . . . ψp are conjunctions of members of Obs. Let structure S and full assign-ment h to S be such that S 6|= ϕ[h]. Hence there exists χ ∈ {ψ1 . . . ψp} such that
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S 6|= ∀x1 . . .∀xn¬χ[h]. So S |= ∃x1 . . .∃xnχ[h], and we can choose variables y1 . . . yn suchthat S |= χ[y1/x1 . . . yn/xn][h]. By Convention (3), χ[y1/x1 . . . yn/xn] is a conjunction ofmembers of Obs. Moreover, since ϕ |= ∀x1 . . .∀xn¬ψ, we infer that ϕ |= ∀x1 . . .∀xn¬χ,hence ϕ |= ¬χ[y1/x1 . . . yn/xn], hence χ[y1/x1 . . . yn/xn] |= ¬ϕ, and we conclude that ϕ isrefutable.
The following corollary is immediate. So is Lemma (31), which we wanted to prove.
(93) Corollary: Suppose that Obs is closed under negation. An invalid formula isrefutable iff it is logically equivalent to a formula of form ∀x1 . . .∀xnψ, where ψis a boolean combination of members of Obs and x1 . . . xn is a (possibly empty)sequence of variables.
8.3. Proof of Examples (36) and (37)
(36) Example: Suppose that Lbasic ⊆ Obs. Then the problem defined inExample (11) has tip-offs.
Proof: Let P be the problem defined in Example (11). As a tip-off for MOD(T ∪ {θ}) in Pwe may take {{∀xRvix} | i ∈ N}. A tip-off for MOD(T ∪ {¬θ}) in P is {{∀xRxvi} | i ∈ N}.
(37) Example: Suppose that Obs = Lbasic , and that binary predicate R is theonly symbol of Sym. Let S = 〈N,�〉 be isomorphic to ω, let T = 〈Z,�∗〉, withZ the set of integers, be isomorphic to ω∗ω, and let disjoint propositions P1, P2
be such that S ∈ P1 and T ∈ P2. Then {P1, P2} does not have tip-offs.
Proof: We show that there is no tip-off for P2 in {P1, P2}. Let n ∈ N , variables x0 . . . xn,and set π consisting of ∀ formulas all of whose free variables are taken from {x0 . . . xn}
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be given. Suppose that π is satisfiable in T . By Example (34), and since S ∈ P1 andT ∈ P2, we can conclude that there is no tip-off for P2 in {P1, P2} if we show that π issatisfiable in S. Choose ao . . . an ∈ Z such that for all ϕ ∈ π, T |= ϕ[a0/x0 . . . an/xn].Choose k ≥ 0 such that {ai + k | i ≤ n} ⊆ N . It is easy to verify that for all ϕ ∈ π,T |= ϕ[a0 + k/x0 . . . an + k/xn]. Since ∀ formulas are preserved in substructures (see [11,Cor. 2.4.2]), it follows that for all ϕ ∈ π, S |= ϕ[a0 + k/x0 . . . an + k/xn]. Hence π issatisfiable in S.
8.4. Proof of Proposition (38)
(38) Proposition: If problem P is countable and has tip-offs, then P is solvable.
Proof: Let {Pj | j < κ} be a repetition-free enumeration of the countably many propositionsin P, where κ = card(P). Let {tj | j < κ} enumerate tip-offs corresponding to the Pj . SinceP is nonempty and countable, and since each tip-off is countable, we may enumerate
⋃j<κ tj
as {πi | i ∈ N}. For i ∈ N , we let πi = Yi∪Zi, where Yi and Zi are as specified in Definition(32)a. By the same definition we have:
(94) (a) For every S ∈⋃
P and full assignment h to S, there is i ∈ N such thatS |= πi[h].
(b) For every i such that πi is satisfiable in some S ∈⋃
P, there is exactly one jsuch that πi ∈ tj .
We specify scientist Ψ that solves P. Let σ ∈ SEQ be given. If for all i, {∧σ} ∪ πi is not
satisfiable in any S ∈⋃
P or∧σ does not imply Yi, then Ψ(σ) is undefined. Otherwise,
let i be least such that {∧σ} ∪ πi is satisfiable in some S ∈
⋃P and
∧σ implies Yi. Then
Ψ(σ) = Pj , where by (94)b, j is unique with πi ∈ tj . To see that Ψ solves P, let j < κ,
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S ∈ Pj , full assignment h to S, and environment e for S and h be given. By (94)a, let i0 beleast with S |= πi0 [h]. Then by Definition (32), j is unique with πi0 ∈ tj . We need to showthat Ψ(e[k]) = Pj for cofinitely many k. By the definition of Ψ, it suffices for this purposeto show:
(95) (a) there are cofinitely many initial segments σ of e such that∧σ implies Yi0 ;
(b) for all i < i0 there are cofinitely many initial segments σ of e such that {∧σ}∪πi
is inconsistent or∧σ does not imply Yi0 .
(95)a follows immediately from the facts that Yi0 ⊆ Obs, e is an environment for S and h,and S |= πi0 [h]. To prove (95)b, let i < i0 be given. If S 6|= Yi[h], then all initial segmentsσ of e are such that
∧σ 6|= Yi. So suppose that S |= Yi[h]. Then S 6|= (πi − Yi)[h]. So there
is ψ ∈ πi − Yi such that S 6|= ψ[h]. Since ψ ∈ πi − Yi, ψ is refutable by Definition (32)a. Soby Definition (30) there is finite D ⊆ Obs such that S |= D[h] and D |= ¬ψ. Hence, sincee is an environment, there is k ∈ N such that
∧e[k] |= D[h]. So {
∧e[k], ψ} is inconsistent.
So {∧e[k′], ψ} is inconsistent for all k′ ≥ k, implying (95)b.
8.5. Proof of Proposition (40)
(40) Proposition: Every solvable problem has tip-offs.
Proof: Let problem P and scientist Ψ be such that Ψ solves P. Let P ∈ P be given. Weexhibit a tip-off for P in P. Given σ ∈ SEQ, define Xσ to be the collection of pairs (S, a)such that S ∈ P , a is a finite assignment to S, and (σ, a) is a locking pair for Ψ, S and P .Define Πσ to be the set consisting of content(σ) together with all refutable formulas ψ suchthat:
(a) Var(ψ) ⊆ Var(σ);
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(b) for all (S, a) ∈ Xσ, S |= ψ[a].
Given σ ∈ SEQ and f : Var(σ) → Var, set Πσ,f = {ξ[f(x)/x, x ∈ Var(σ)] | ξ ∈ Πσ}. Weclaim that t = {Πσ,f |σ ∈ SEQ, f : Var(σ) → Var} is a tip-off for P in P. By Convention(3), for all ψ ∈ content(σ), ψ[f(x)/x, x ∈ Var(σ)] belongs to Obs, and it is easy to verifythat for every refutable formula ψ, ψ[f(x)/x, x ∈ Var(σ)] is also refutable. It follows easilythat t is a countable collection of sets of form Y ∪ Z, where Y and Z are as stipulated inDefinition (32)a. To prove clause (b) of Definition (32), let S ∈ P and full assignment h toS be given. By Lemma (29), let σ ∈ SEQ and finite assignment a : Var → | S | be such that(σ, a) is a locking pair for Ψ, S, and P . Then (S, a) ∈ Xσ and S |=
∧σ[a], so S |= Πσ[a].
Since h is onto | S |, we can choose f : Var(σ) → Var such that S |= Πσ,f [h]. Hence, Πσ,f
is the set in t that witnesses (32)b.
To finish the proof, we show that (32)c is satisfied. Let σ ∈ SEQ, f : Var(σ) → Var,U ∈ P ′ ∈ P with P ′ 6= P , and full assignment g to U be given. Suppose for a contradictionthat U |= Πσ,f [g]. Then we can choose full assignment g′ to U such that:
(96) U |= Πσ[g′].
Since content(σ) ⊆ Πσ, there is an environment e for U and g′ such that σ ⊂ e. We willshow that for all k ≥ length(σ), ∅ 6= Ψ(e[k]) ⊆ P . Since e is for U ∈ P ′ ∈ P with P ′ 6= P ,this implies that Ψ does not solve P ′, contradicting our choice of Ψ. So let k ≥ length(σ)be given. Choose τ ∈ SEQ such that e[k] = σ∗τ . We must show:
(97) ∅ 6= Ψ(σ∗τ) ⊆ P .
For this purpose, we establish:
(98) Claim: There is (S, a) ∈ Xσ such that S |= ∃x∧
(σ∗τ)[a], where x contains thevariables in Var(τ)− domain(a).
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Proof of (98): For a contradiction, suppose that for all (S, a) ∈ Xσ, S |= ∀x¬∧
(σ∗τ)[a].Let (S, a) ∈ Xσ be given. Since Var(σ) ⊆ domain(a), S |= ((¬
∧σ) ∨ (∀x¬
∧τ))[a]. With
the fact that S |=∧σ[a], this implies that S |= ∀x¬
∧τ [a]. By Lemma (92), ∀x¬
∧τ is
refutable. Hence we infer that ∀x¬∧τ ∈ Πσ, and it follows from (96) that U |= ∀x¬
∧τ [g′].
However, this is impossible since e was chosen to be for U and g′, and σ∗τ ⊂ e.
We now use (98) to prove (97). For any (S, a) ∈ Xσ, (σ, a) is a locking pair for Ψ, S,and P . Hence, by Claim (98) and clause (c) of Definition (28), ∅ 6= Ψ(σ∗τ) ⊆ P .
8.6. Proof of Theorem (59)
(59) Theorem: A problem of form (T , {θ0 . . . θn}) is solvable if and only if forevery i ≤ n, θi is equivalent in T to the existential closure of any of the followingkinds of formulas.
(a) a formula built from Obs using only conjunctions and disjunctions,
(b) a refutable formula, or
(c) a formula of form χ∧ϕ, where χ is built from Obs using only conjunctionsand disjunctions and ϕ is refutable.
Proof: Let P be a problem of form (T , {θ0 . . . θn}). Call a sentence special if it is theexistential closure of a formula of form
∧D ∧ ϕ where D is a finite subset of Obs, or of
a refutable formula, or of a formula of form∧D ∧ ϕ, where D is a finite subset of Obs
and ϕ is a refutable formula. By Convention (3) and Lemma (91)b, it suffices to showthat P is solvable if and only if for every i ≤ n, θi is equivalent in T to a disjunction ofspecial sentences. For the right-to-left direction, let i ≤ n be given, and suppose that θi isequivalent in T to ψ0∨ . . .∨ψm, where ψ0 . . . ψm are special sentences. Then every memberof MOD(T ∪ θi) is a model of ψp for some p ≤ m. Moreover, for all j ≤ n with j 6= i and
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for all p ≤ m, no model of MOD(T ∪ θj) is a model of ψp. It follows easily from Proposition(38) that if for all i ≤ n, θi is equivalent in T to a disjunction of special sentences, then Pis solvable.
For the left-to-right direction, suppose that P is solvable. Let i ≤ n be given. ByProposition (40), let t be a tip-off for MOD(T ∪ {θi}) in (T , {θ0 . . . θn}). By Definition (32),for all j ≤ n with j 6= i, every member of t is unsatisfiable in every U ∈ MOD(T ∪ {θj}).Since the θj ’s partition the models of T , it follows that T ∪ π |= θi for all π ∈ t. Bycompactness, for all π ∈ t there is finite Y ⊆ π with T ∪ Y |= θi. From this, the fact that aconjunction of refutable formulas is refutable [Lemma (91)a], and Definition (32)a, we infer:
(99) For every S ∈ MOD(T ∪ {θi}) there is a special χS ∈ Lsen such that:
(a) S |= χS .(b) T ∪ {χS} |= θi.
Let Σ = {χS | S ∈ MOD(T ∪ {θi})}. We show that θi is equivalent over T to some finitedisjunction of members of Σ. By (99)b, for every disjunction ρ of members of Σ, T∪{ρ} |= θi.Hence, it suffices to show that for some such ρ, T ∪ {θi} |= ρ. For a reductio suppose thatfor every disjunction ρ of members of Σ, T ∪ {θi} 6|= ρ. Then for every finite ∆ ⊆ Σ,T ∪ {θi} ∪ {¬ϕ |ϕ ∈ ∆} is satisfiable. Hence by the compactness and Lowenheim-Skolemtheorems there is (countable) S ∈ MOD(T ∪ {θi}) such that S |= {¬ϕ |ϕ ∈ Σ}. But thisimplies that there is S ∈ MOD(T ∪ {θi}) such that S 6|= χS , contradicting (99)a.
8.7. Proof of Proposition (67)
(67) Proposition: Let solvable problem P of form (T , {P0, P1, . . .}) be given.Then for all P ∈ P, there is an enumeration {Xi | i ∈ N} of sets of existential
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closures of formulas of form∧D∧ϕ, where D is a (possibly empty) finite subset
of Obs and ϕ a refutable formula, such that P =⋃i∈N MOD(T ∪Xi).
Proof: Let P ∈ P be given. By Proposition (40), let {πi | i ∈ N} be an enumeration ofthe members of a tip-off for P in P. For all i ∈ N denote by Xi the set of existentialclosures of all formulas of form
∧D, with D a nonempty, finite subset of πi. By Definition
(32) and Lemma (91)a, and since valid formulas are refutable, it suffices to show thatP =
⋃i∈N MOD(T ∪Xi). To show the left-to-right inclusion, let S ∈ P be given. By
Definition (32)b let i ∈ N be such that πi is satisfiable in S. Hence, T ∪Xi is satisfiable inS. So P ⊆
⋃i∈N MOD(T ∪Xi). For the other direction, suppose for a contradiction that⋃
i∈N MOD(T ∪Xi)−P 6= ∅. Let structure S and i ∈ N be such that S ∈ MOD(T ∪Xi)−P .Since
⋃P = MOD(T ), S belongs to some P ′ ∈ P with P ′ 6= P . By Proposition (40) let t
be a tip-off for P ′ in P, and let π ∈ t be such that π is satisfiable in S. Without loss ofgenerality we may suppose that πi and π share no free variables. It is clear that T ∪ πi ∪ πis unsatisfiable, since otherwise some model of T satisfies members from distinct tip-offs,contradicting Definition (32). Summarizing:
(100) (a) all variables that occur free in πi do not occur free in π;
(b) π is satisfiable in S;
(c) T ∪ πi ∪ π is unsatisfiable.
By (100)c and compactness, let nonempty, finite D ⊆ πi and finite D′ ⊆ π be such that:
(101) T ∪D ∪D′ is unsatisfiable.
Denote by ϕ the existential closure of∧D and by ϕ′ the universal closure of ¬
∧D′. It
follows from (100)a and (101) that:
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(102) T |= ϕ→ ϕ′.
We deduce from (100)b that S 6|= ϕ′, which with (102) and S |= T implies that S |= ¬ϕ.Since ϕ ∈ Xi this contradicts the hypothesis that S ∈ MOD(Xi).
8.8. Proof of Proposition (79)
(79) Theorem: Every solvable problem is solvable strongly efficiently.
Proof of the theorem relies on two lemmas. Here is the first.
(103) Lemma: Let solvable problem P and scientist Ψ that solves P be given. Then Ψis not dominated on P if and only if for all σ ∈ SEQ, if σ is for P then there existsP ∈ P and environment e for P such that:
(a) e extends σ, and
(b) for every k ≥ length(σ), ∅ 6= Ψ(e[k]) ⊆ P .
Proof: Suppose that Ψ is not dominated on P. Let σ be for P, and suppose for a contra-diction that:
(104) for all P ∈ P and for all environments e for P that extend σ, there is k ≥ length(σ)such that either Ψ(e[k]) is not defined or Ψ(σ) = ∅ or Ψ(e[k]) 6⊆ P .
Since some environment for P extends σ and Ψ solves P, there is least k0 ∈ N with thefollowing property:
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(105) There is P ∈ P and environment e for P such that e extends σ and ∅ 6= Ψ(e[k]) ⊆ Pfor all k ≥ k0.
It follows from (104) and (105) that:
(106) k0 > length(σ).
Let scientist Ψ′ satisfy the following conditions, for all τ ∈ SEQ. If σ ⊆ τ ⊆ e[k0] thenΨ′(τ) = P . If Ψ(τ) is defined but either σ 6⊆ τ or τ 6⊆ e[k0], then Ψ′(τ) = Ψ(τ).
It is immediate that Ψ′ solves P. Let environment e′ for P ′ ∈ P be given. If e′ doesnot extend σ then SP(Ψ′, e′, P ′) = SP(Ψ, e′, P ′) by the definition of Ψ′. If e′ extends σ ande′ 6= e, then SP(Ψ, e′, P ′) ≥ k0 by the definition of k0, and SP(Ψ′, e′, P ′) ≤ SP(Ψ, e′, P ′) bythe definition of Ψ′. Finally from (105), (106), and the definition of Ψ′ it is easy to see thatSP(Ψ′, e, P ′) ≤ length(σ) < k0 = SP(Ψ, e, P ′). This contradicts the hypothesis. So we haveshown that if Ψ is not dominated on P, then conditions (a) and (b) are satisfied.
For the opposite suppose that conditions (a) and (b) are satisfied. Let scientist Ψ′ solveP, and let environment e for P ∈ P be such that SP(Ψ′, e, P ) < SP(Ψ, e, P ) (if there isno such Ψ′, P and e, then there is nothing left to prove). Since SP(Ψ′, e, P ) < SP(Ψ, e, P )there is k0 ∈ N such that:
(107) It is not true that ∅ 6= Ψ(e[k0]) ⊆ P and ∅ 6= Ψ′(e[k0]) ⊆ P .
By hypothesis there exists P ′ ∈ P and environment e′ for P ′ such that e′ extends e[k0]and ∅ 6= Ψ(e′[k]) ⊆ P ′ for all k ≥ k0. This, (107), and the fact that Ψ solves P implythat SP(Ψ, e′, P ′) ≤ k0 < SP(Ψ′, e′, P ′). With Definition (76) we conclude that Ψ is notdominated on P, as required.
Here is the second lemma.
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(108) Lemma: Suppose that scientist Ψ solves problem P. Then Ψ solves P stronglyefficiently if and only if for every σ ∈ SEQ, if σ is for P then there exists P ∈ P,S ∈ P , and full assignment h to S such that:
(a) S |=∧σ[h], and
(b) for every environment e for S and h that extends σ, for every k ≥ length(σ),∅ 6= Ψ(e[k]) ⊆ P .
Proof: Suppose that Ψ solves P strongly efficiently. Let σ ∈ SEQ be for P. We show thatconditions (a),(b) are satisfied for some P ∈ P, S ∈ P , and full assignment h to S. Plainly,Ψ is not dominated on P. So by Lemma (103) there is P ∈ P such that
∧σ is satisfiable
in some member of P and ∅ 6= Ψ(σ) ⊆ P . Suppose that for all P ′ ∈ P, if P ′ 6= P then∧σ is satisfiable in no member of P ′. Let scientist Ψ′ have the following properties, for all
τ ∈ SEQ. If σ ⊆ τ then Ψ′(τ) = P . If σ 6⊆ τ and Ψ(τ) is defined, then Ψ′(τ) = Ψ(τ). It iseasy to see that Ψ′ solves P, and that for every P ′ ∈ P and for every environment e for P ′:
(a) SP(Ψ′, e, P ′) ≤ SP(Ψ, e, P ′), and
(b) if e extends σ and there is k > length(σ) such that either Ψ(e[k]) is undefined orΨ(e[k]) = ∅ or Ψ(e[k]) 6⊆ P ′, then SP(Ψ′, e, P ′) < SP(Ψ, e, P ′).
It follows from Definition (76) that ∅ 6= Ψ(e[k]) ⊆ P for every environment e for P thatextends σ and for every k ≥ length(σ). So, conditions (a),(b) of the lemma are satisfied.Thus, to conclude this direction of the proof, suppose that there is P ′ ∈ P with P ′ 6= Psuch that
∧σ is satisfiable in some member of P ′. Let P ′ ∈ P and environment e′ for
P ′ be such that P ′ 6= P and σ ⊆ e′. Let scientist Ψ′ be defined as follows, for everyγ ∈ SEQ. If γ 6⊆ e′ then Ψ′(γ) = Ψ(γ). If γ ⊂ σ then Ψ′(γ) = ∅. If σ ⊆ γ ⊆ e′ thenΨ′(γ) = P ′. Trivially, Ψ′ solves P and length(σ) = SP(Ψ′, e′, P ′) < SP(Ψ, e′, P ′). From this
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and Definition (78), we deduce that there is S ∈ P and full assignment h to S such thatS |=
∧σ[h], and ∅ 6= Ψ(e[k]) ⊆ P for every environment e for S and h that extends σ and
for every k ≥ length(σ). Hence conditions (a),(b) of the lemma are satisfied.
Conversely, suppose that for all σ ∈ SEQ, if σ is for P then conditions (a),(b) are satisfiedfor some P ∈ P, S ∈ P , and full assignment h to S. We show that Ψ solves P stronglyefficiently. Suppose that scientist Ψ′ solves P. Suppose there is P0 ∈ P and environmente0 for P0 such that SP(Ψ′, e0, P0) < SP(Ψ, e0, P0). (If there is no such Ψ′, P0 and e0, thenthere is nothing left to prove.) Let k0 ∈ N be such that k0 = SP(Ψ′, e0, P0). Choose S ∈ Pand full assignment h to S such that S |=
∧e0[k0][h] and for every environment e for S
and h that extends e0[k0], for every k ≥ k0, ∅ 6= Ψ(e[k]) ⊆ P . Suppose that P = P0.Then e0 is for P , and SP(Ψ, e0, P0) ≤ k0 = SP(Ψ′, e0, P0), contradiction. Hence P 6= P0.Since Ψ′(e0[k0]) = P0 6= P , we infer that for every environment e for S and h that extendse0[k0], SP(Ψ, e, P ) ≤ k0 < SP(Ψ′, e, P ). With Definition (78) this completes the proof ofthe lemma.
Proof of Theorem (79): Let problem P be solvable, and let scientist Ψ be as specifiedin the proof of Proposition (38). It follows easily that for every σ ∈ SEQ for P, we maychoose P ∈ P, S ∈ P , and full assignment h to S with the following properties:
(109) (a) S |=∧σ[h];
(b) Ψ(e[k]) = P , for every k ≥ length(σ), and every environment e for S and hthat extends σ.
Relying on Lemma (108) we conclude from (109) that Ψ solves P strongly efficiently.
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8.9. Proof of Theorem (83)
(83) Theorem: Suppose that Sym is limited to a binary predicate, two con-stants, and a unary function symbol. Let Obs = Lbasic . Then for every count-able collection Σ of scientists there is a problem P with the following properties.
(a) Every member of P is strongly elementary.
(b) P is solvable.
(c) No member of Σ solves P.
Proof: Let R be the predicate, 0, a be the constants, and s be the unary function symbolof Sym. We denote n applications of s to 0 by n. Let δ ∈ Lsen be true in a structure iff (a)R is interpreted as a discrete total ordering of the entire domain, and (b) s is interpreted asa one-to-one function [that is, the structure must satisfy ∀x∀y(sx = sy → x = y)]. Giveni ∈ N , let θ+
i be i = a ∧ δ ∧ ∃x∀yRxy, and let θ−i be i = a ∧ δ ∧ ¬∃x∀yRxy.
LetX ⊆ N be given. We denote by PX the set {MOD(θ+i ) | i ∈ X}∪{MOD(θ−i ) | i 6∈ X}.
It is obvious that PX is a solvable problem whose members are strongly elementary. Solet countable collection Σ of scientists be given. To finish the proof it suffices to show thatthere is X ⊆ N such that no member of Σ solves PX . Because there are uncountably manysubsets of N , it thus suffices to show:
(110) If X and Y are distinct subsets of N then PX ∪PY is not solvable.
To demonstrate (110), let X,Y ⊆ N and i ∈ N be such that i ∈ X iff i 6∈ Y . SetP1 = MOD(θ+
i ) and P2 = MOD(θ−i ). Observe that {P1, P2} ⊆ PX ∪PY . Hence it sufficesto show that {P1, P2} is not solvable. By Proposition (40), for this purpose it suffices toshow that:
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(111) P2 does not have a tip-off in {P1, P2}.
Let S = 〈N,≤, 0, i, successor〉, and let T = 〈Z,≤, 0, i, successor〉, where Z is the set ofintegers, 0 interprets 0, i interprets a, and the successor function interprets s. Then S ∈ P1
and T ∈ P2, and an easy adaptation of Example (37) yields (111).
8.10. Proof of Proposition (85)
(85) Proposition: Suppose that Sym is limited to the vocabulary of arithmetic(including 0 and a unary function symbol s) plus the additional constant a.Suppose also that Obs includes all identities (that is, formulas of form t1 = t2,for terms t1, t2). Then there is a problem P with the following properties.
(a) Every member of P is strongly elementary.
(b) P is solvable computably.
(c) Every computable scientist that solves P is dominated on P.
Proof: Let n applications of s to 0 be denoted n. Let Q be the finite set of axioms ofRobinson’s Arithmetic. By standard results [3, Ch. 14], let φ(x, y) ∈ Lform have two freevariables x, y, exclude a, and be such that for all i ∈ N , Q |= ∃xφ(x, i) iff i ∈ Wnum
i . Giveni ∈ N , set P+
i = MOD(Q ∪ {a = i,∃xφ(x, i)}), and P−i = MOD(Q ∪ {a = i,¬∃xφ(x, i)}).
We claim that P = {P+i | i ∈ Wnum
i } ∪ {P−i | i 6∈ Wnum
i } witnesses the proposition.
It is immediate that the propositions of P are strongly elementary. To show that Pis computably solvable, define computable ψ : SEQ → N as follows. Let σ ∈ SEQ begiven. Suppose that i ∈ N is unique with a = i ∈ content(σ). Then if i has not appearedin Wnum
i , within length(σ) steps of its standard enumeration, ψ(σ) equals an index forQ ∪ {a = i,¬∃xφ(x, i)}; and if i has appeared in Wnum
i within length(σ) steps of its
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standard enumeration, ψ(σ) is an index for Q ∪ {a = i,∃xφ(x, i)}. For all other σ ∈ SEQ,ψ(σ) is undefined. It is easy to verify that if ψ underlies Ψ, then Ψ solves P.
Finally, for a contradiction, suppose that computable ψ : SEQ → N underlies a scientistthat solves P and is not dominated on P. Then Definition (76) is easily seen to imply thefollowing.
(112) For all i ∈ N ,
(a) if i ∈ Wnumi then ψ(a = i) is an r.e. index for consistent X ⊆ Lform with
X |= ∃xφ(x, i),
(b) if i 6∈ Wnumi then ψ(a = i) is an r.e. index for consistent X ⊆ Lform with
X |= ¬∃xφ(x, i).
But (112) yields a decision procedure for the halting problem.
8.11. Proof of Theorem (87)
(87) Theorem: Suppose that Obs is recursive. Then every solvable, r.e. prob-lem of form (T , {θ0 . . . θn}) is computably solvable.
Proof: Let solvable, r.e. problem of form (T , {θ0 . . . θn}) be given. We denote the problemby (T , {θ0 . . . θn}) and exhibit a computable scientist Ψ that solves it. For this purpose, letA be a computable proof procedure with the following property.
Given n ∈ N , σ, τ ∈ SEQ with σ ⊆ τ , and formula ϕ with∧σ |= ϕ, if A
generates ϕ from∧σ in at most n steps of computation, then A generates ϕ
from∧τ in at most n steps of computation.
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To specify computable ψ : SEQ → N underlying Ψ, we rely on some terminology andnotation. Let t be an r.e. index for T . Call a formula “inferable” just in case it is built fromObs using only conjunctions and disjunctions. Let {ϕj | j ∈ N} be a recursive enumerationof all formulas ϕ such that:
(a) ϕ is either refutable or the conjunction of an inferable formula with a refutable formula;
(b) for some 0 ≤ i ≤ n, θi is equivalent in T to the existential closure of ϕ.
Since Obs is recursive and T is r.e., it is easy to verify that such a recursive enumerationexists (and that the refutable subformulas can be effectively distinguished from the inferableones).
Define computable ψ : SEQ → N as follows. Let σ ∈ SEQ be given. Suppose thereexists least j0 ∈ N such that either
(113) (a) ϕj0 is refutable and∧σ does not imply ¬ϕj0 in length(σ) steps of computation
(using A), or
(b) ϕj0 is the conjunction of an inferable formula χ with a refutable formula ξ,∧σ
implies χ in at most length(σ) steps of computation, and∧σ does not imply
¬ξ in length(σ) steps of computation (again, using A).
Then ψ(σ) is an index for T ∪ {ϕj0}. Otherwise ψ(σ) is undefined. (Using t and j0, it isclear that an index for T ∪ {ϕj0} can be constructed uniformly effectively.)
Suppose that ψ underlies Ψ. To prove that Ψ solves (T , {θ0 . . . θn}), let i0 ≤ n, S ∈MOD(T ∪ {θi0}), full assignment h to S, and environment e for S and h be given. ByTheorem (59), there is least j0 ∈ N such that S |= ϕj0 [h]. First we show:
(114) For cofinitely many k, ψ(e[k]) is an index for T ∪ {ϕj0}.
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Let j < j0 be given. Then S 6|= ϕj [h].
Case 1: ϕj is refutable. Then by Definition (30) and the choice A of proof procedure,there is n ∈ N such that A shows that
∧e[k] |= ¬ϕj for all k > n. Hence, for cofinitely
many k, ψ(e[k]) is not an index for T ∪ {ϕj}.
Case 2: ϕj is the conjunction χ ∧ ξ of an inferable formula χ with a refutable formulaξ. Then either (a) S 6|= χ[h] or (b) S 6|= ξ[h]. In case (a), by the definition of “inferable,”∧e[k] 6|= χ for all k ∈ N , and in case (b)
∧e[k] |= ¬ξ for cofinitely many k (as in Case 1).
Hence, for cofinitely many k, ψ(e[k]) is not an index for T ∪ {ϕj}.
In light of (113) and the foregoing cases, to prove (114) it remains only to show that:
(115) (a) If ϕj0 is refutable then for cofinitely many k,∧e[k] does not imply ¬ϕj0 in k
steps of computation (using A), and(b) if ϕj0 is the conjunction of an inferable formula χ with a refutable formula ξ,
then for cofinitely many k,∧e[k] implies χ in at most k steps of computation,
and∧e[k] does not imply ¬ξ in k steps of computation (again, using A).
Case 1: ϕj0 is refutable. Then because S |= ϕj0 [h], for all k ∈ N ,∧e[k] 6|= ¬ϕj0 .
Case 2: ϕj0 is the conjunction χ ∧ ξ of an inferable formula χ with a refutable formulaξ. Then because S |= ϕj0 [h], (a) for all k ∈ N ,
∧e[k] 6|= ¬ξ, and (b) for cofinitely many k,∧
e[k] implies χ in at most k steps of computation (by the definition of “inferable” and thechoice of A).
This proves (115), hence (114). So it remains only to show that T ∪ {ϕj0} is consistentand implies T ∪ {θi0}.
Since S |= ϕj0 [h], and S ∈ MOD(T ), it follows that T ∪ {ϕj0} is consistent. Finally,since the θi’s partition the models of T , and the existential closure of ϕj0 is equivalent in Tto one of the θi’s, it follows that T ∪ {ϕj0} |= T ∪ {θi0}.
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References
[1] D. Angluin. Inductive inference of formal languages from positive data. Informationand Control, 45:117–135, 1980.
[2] L. Blum and M. Blum. Toward a mathematical theory of inductive inference. Infor-mation and Control, 28:125–155, 1975.
[3] G. Boolos and R. Jeffrey. Computability and Logic (Third Edition). Cambridge Uni-versity Press, Cambridge, England, 1989.
[4] D. S. Bridges. Computability: A Mathematical Sketchbook. Springer-Verlag, Berlin,1994.
[5] C. C. Chang and H. J. Keisler. Model Theory (2nd Edition). North Holland, Amster-dam, 1977.
[6] G. Chierchia and S. McConnell-Ginet. Meaning and Grammar: An Introduction toSemantics (2nd edition). MIT Press, Cambridge MA, 2000.
[7] W. Craig. On axiomatizability within a system. Journal of Symbolic Logic, 18:30–32,1953.
[8] H. D. Ebbinghaus, J. Flum, and W. Thomas. Mathematical Logic, Second Edition.Springer-Verlag, Berlin, 1994.
[9] H. Enderton. A Mathematical Introduction to Logic. Academic Press, New York, 1972.
[10] H. Gaifman and M. Snir. Probabilities over rich languages. Journal of Symbolic Logic,47:495–548, 1982.
[11] W. Hodges. Model Theory. Cambridge University Press, Cambridge, England, 1993.
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[12] H. J. Keisler. Fundamentals of model theory. In Jon Barwise, editor, Handbook ofMathematical Logic, pages 47–104. North Holland, Amsterdam, 1977.
[13] K. Kelly and C. Glymour. Theory Discovery from Data with Mixed Quantifiers. Journalof Philosophical Logic, 19:1–33, 1990.
[14] A. Levy. Basic Set Theory. Springer-Verlag, Berlin, 1979.
[15] M. Machtey and P. Young. An Introduction to the General Theory of Algorithms.North-Holland, New York, 1978.
[16] Eric Martin and Daniel Osherson. Elements of scientific inquiry. M.I.T. Press, 1998.
[17] K. Wexler and P. Culicover. Formal Principles of Language Acquisition. M.I.T. Press,Cambridge MA, 1980.
