Inductive Inquiry via Theory Revision1 · Introduction Theory revision Inquiry via revision Theories as parameters Eﬃcient revision Closure Iterated revision Proofs Home Page Title

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Inductive Inquiry via Theory Revision1

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].

http://www.princeton.edu/~osherson/IL/ILpage.htm

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1. Introduction

The achievements of modern science are stupendous but what does this imply about thescientists responsible for them? Here are two opposing reactions to the accomplishments ofthese fine women and men.

(a) How clever!(b) What luck!

Response (a) tacitly invokes a criterion of good inductive thinking and suggests thatscientists have so far scored well. Response (b) denies this, and perhaps even denies thatthere is a distinction between good and bad scientific reasoning (see [9]). The issue is notthe veracity and scope of current scientific doctrine. Even if modern doctrine is both trueand momentous it could be the result of fortune and happenstance. In particular, humangenetic endowment might be tuned to physical reality, predisposing even “wild guesses”towards accuracy. In this case, scientific success is due more to the quirks of evolutionaryhistory than to scientists’ problem-solving ability (see [6, Ch. 1]).

We might try to frame the opposition between (a) and (b) in terms of counterfactualclaims of the form: Had physical laws been so-and-so instead of what they actually are,then scientists would have figured them out anyway. That would make them seem clever.One difficulty with such a formulation is ambiguity surrounding reference to “scientists.”In counterfactual circumstances, are they just like actual scientists, or do they themselvesoperate according to the different laws so-and-so? To skirt the latter question, we can focuson the input-output function that a scientist embodies. Then it can be asked whether thisvery input-output function, applied to the data issuing from a different set of physical laws,would have converged to an accurate conjecture about those laws. (The function, beingabstract, does not interact with physical law.)


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Put this way, the choice between (a) and (b) hinges on the extent to which scientists’inductive strategies are general purpose. Can the strategies be used to solve a wide class ofinductive problems, or just the one problem that scientists were fortunate to find in theirpath? The answer, of course, depends on facts about human psychology that are presentlyunknown. There are nonetheless issues that can be addressed at the present juncture. Inparticular, it would be helpful to possess an example of a “general purpose” scheme forinductive inquiry. What would such a thing look like?

It is at this point that we wish to stick our (tiny) oar into the discussion. Supposethat the concept of “scientist” is interpreted in the extensional way envisioned in Section2.4 of the second essay; and suppose that “inductive problems” are sets of propositions,as discussed there in Section 2.2. Then it might be possible to provide at least one clearexample of a general purpose scientific strategy.

By “general purpose,” is not meant “able to solve all problems.” There are no generalpurpose scientists in such a sense because unsolvable problems [like (41) of the essay #2]lie outside the competence of every scientist. Nor can we hope (more modestly) for a singlescientist that solves all solvable problems. To see why, suppose that Obs = Lbasic , Sym = ∅,and let P1 = N, P2 = ∅ 6= D ⊆ N |D finite. In Exercise (47) of the second essay, youwere invited to demonstrate that the problem P1, P2 is not solvable. But each of theproblems P1, P2 is trivially solvable. (Since each contains but a single proposition, itsuffices to conjecture that one proposition on every datum.) This example can be adaptedto prove:

(1) Proposition: There is a pair of solvable problems whose union is not solvable.1

1Since the proposition invokes no hypotheses about Obs or Sym, it is intended to hold for any suchchoices. Small adjustments to the foregoing example suffice to show this. Proposition (1) is known as a“non-union” result. Theorems of this kind were first made prominent in the numerical paradigm of inductiveinference. See [4].


http://www.princeton.edu/~osherson/IL/essay2.pdf



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If P1, P2 is such a solvable pair then no scientist solves both since he would then solveP1 ∪P2, which is impossible.

It follows that a general purpose scientist must in reality be a parameterized family ofscientists. Each solvable problem would be solved by setting the parameter in the right way.But what kind of parameter are we talking about? The parameter could be interpreted insuch a way as to deprive the family of its interest. For example, it would not be instructive ifthe parameter amounted to specifying a whole scientist. We already know that the family ofall scientists is general purpose in the sense of solving every solvable problem! (For solvableproblem P, it suffices to set the parameter to some scientist that solves P.)

Our goal is therefore to exhibit a family of scientists whose free parameter is just afragment of a larger scheme for inductive inquiry. For this purpose, we shall considerscientists whose successive conjectures can be interpreted as the fruit of hypothesis revisionstarting from an initial scientific theory. Among other results, it will be shown that initialtheories can be used to parameterize a single scheme of inquiry that solves a broad class ofproblems. Specifically, for every solvable problem of form (T , θ0 . . . θn), the scientist thatresults from passing the parameter T to the scheme will be seen to solve (T , θ0 . . . θn).2Moreover, the scheme in question will be built around a form of hypothesis revision thatappears to meet some basic conditions of rational theory change.

If we keep these promises, the results might clarify the kind of facts needed to motivateresponse (a) above. At least, it will be seen how one kind of scientist could succeed in awide class of potential realities by relying on a sensible strategy of hypothesis selection.Whether any human scientists in fact approximate such a strategy then becomes a welldefined empirical issue.

So much for the big picture. Are you ready for the details? We proceed as follows.The next section introduces a class of revision functions, in other words, operations that

2For (T , θ0 . . . θn), see Definition (55) in essay #2.



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convert one theory into another as a function of the available evidence. Revision functionsare harnessed for the purpose of inquiry in Section 3, and basic facts are proved. In Section4 we explore the idea of a parameterized family of scientists based on theory revision. Ourprincipal results are stated there. Efficient inquiry via revision is taken up in Section 5.Revision of a theory in light of data is well defined whether or not the theory is deductivelyclosed. The impact of closure on successful inquiry via revision is the topic of Section 6.Section 7 considers iterated revision of the starting theory under the impact of the latestdatum (in place of repeated revision of the starting theory under accumulating data). Proofscan be found in Section 8.

2. Theory revision

Our overall goal is to exhibit a parameterized class of scientists that operate via rationaltheory revision. As a first step, we introduce the kind of theory revision that will animateour scientists. Our ideas are drawn in part from the voluminous literature on belief revision,a large fragment of which has been masterfully summarized and integrated in [17].3

There is a terminological disagreement with the larger literature that can be set asidehere. In Section 1 of the first essay we distinguished between belief and acceptance, andreserved the latter term for the kind of theoretical commitments that figure in scientificinquiry. What others call “belief revision functions” would therefore be more aptly called“functions for the revision of accepted theories” in the present context. As a compromise(and for brevity), we simply drop the qualifier “belief,” referring instead to “revision func-tions.”

3Our approach has been particularly influenced by [16], which generalizes [3]. We set things up somewhatdifferently, however. In particular, it will be convenient to define revision directly without the usual excursionthrough “contraction.”



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2.1. Revision functions

We shall attempt to exhibit inquiry within our paradigm as a process of rational revision ofa starting theory in the light of data. In order to give substance to this idea, we conceive ascientist as initiating inquiry with a set X of formulas that represent his provisional theoryprior to examining the environment. The arrival of data σ will then serve to modify Xaccording to some fixed scheme of revision. For now, deductive closure is not imposed ontheories; they are arbitrary subsets of Lform . The effect on inquiry of deductive closure isthe topic of Section 6, below.

The potential data that confront theories will be associated with the set SEQ, introducedin Section 2.3.6 of essay #2. Recall that the latter set represents all possible informationthat can be presented to scientists in our paradigm. A consequence of limiting attention toSEQ is that data are always assumed to be logically consistent. Let us recall from the earlierdiscussion that SEQ depends on the choice Obs. (Spefically, SEQ is the set of consistentfinite sequences over Obs.)

As discussed in the first essay (Section 4.1), there is typically more than one way torevise a theory in the face of data that contradict it. Any particular revision strategymay be understood as determining a function that maps pairs of the form (X,σ), whereX ⊆ Lform and σ ∈ SEQ, into new theories. In other words, revision is a function withdomain pow(Lform) × SEQ and co-domain pow(Lform). We typically denote a revisionfunction by the symbol + . The result of applying + to a pair in its domain is denotedby X +σ, and signifies the impact of the data σ on the theory X according to the revisionscheme + .

Not just any function from pow(Lform) × SEQ to pow(Lform) counts as a legitimatescheme of revision, however. For one thing, we require that the data appear in the successortheory. This is because in our paradigm, Nature can never misrepresent what’s true in herchoice of structure. The data don’t lie. So it makes sense to incorporate the data into the




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revised theory. In addition, the successor theory must include the “inoffensive” parts of thestarting theory. To make the latter idea clear, we rely on the following definition.

(2) Definition: Let B ⊆ Lform and σ ∈ SEQ be given. A formula ϕ ∈ B is calledσ-innocent (with respect to B) just in case for all A ⊆ B, if A ∪ content(σ) isconsistent then A ∪ ϕ ∪ content(σ) is consistent. We denote by Inn(B, σ) the setof all σ-innocent formulas in B.

In other words, ϕ is σ-innocent with respect to B just in case ϕ figures in no non-redundantproof of ¬

∧σ that includes just formulas from B. Using propositional logic to illustrate,

let B = p, p→ q, p ∨ r and σ = ¬q. Then Inn(B, σ) = p ∨ r.

When revising B to make content(σ) part of the new theory, the suppression of membersof Inn(B, σ) seems capricious; so the revision should leave behind at least Inn(B, σ). Onthe other hand, enough formulas must be removed from B to ensure the consistency of thesuccessor theory when the data are added. Finally, the new theory should include all ofσ but nothing else that goes beyond B. These boundary conditions are the only ones weimpose on legitimate revision.

(3) Definition: A mapping + from pow(Lform) × SEQ to pow(Lform) is a revisionfunction just in case for all B ⊆ Lform ,

(a) Inn(B, σ) ∪ content(σ) ⊆ B +σ ⊆ B ∪ content(σ);(b) B +σ is consistent.

The following lemma is immediate.

(4) Lemma: Let B ⊆ Lform and σ ∈ SEQ be such that B 6|= ¬∧σ. Then for all revision

functions + , B +σ = B ∪ content(σ).


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In other words, if new data are consistent with the current theory B, they are just dumpedin without touching anything else. Once again, this policy is justified by the credibility ofthe data, which has been a central feature of our paradigm. If the data are inconsistentwith B, they are dumped in nonetheless, at the price of displacing some formulas of B

The following Proposition (easily verified, but not essential to the sequel) gives someidea of the liberality of Definition (3) compared to other accounts of revision, e.g., [1]. Tostate the result, let us call a mapping ⊕ from pow(Lform) × SEQ to pow(Lform) a partialmeet revision just in case for all B ⊆ Lform and σ ∈ SEQ, B ⊕ σ is the union of content(σ)with the intersection of some class of ⊆-maximal subsets of B that are consistent withcontent(σ).4

(5) Proposition: The class of revision functions [defined in (3)] properly includes theclass of partial meet revisions.

2.2. Special kinds of revision

We’ve seen that Definition (3) embraces a broad class of revision functions. Indeed, theclass is broad enough to embrace some questionable policies. For example, draining thesuccessor theory of every formula except the innocent ones (plus the data), has a scorchedearth feel to it. So we are led to define more respectable subsets. A familiar idea (discussedin [2]) is to effect minimal change in the original theory, in the sense of abandoning no morethan is necessary to avoid contradiction with new data. One way of capturing this idea isthe following.

(6) Definition: A revision function + is maxichoice just in case for every B ⊆ Lform

and σ ∈ SEQ, no subset of B ∪ content(σ) that strictly includes B +σ is consistent.4See [11] for discussion of the partial meet idea.


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Thus, maxichoice revision functions keep a ⊆-maximal subset of the original theory that isconsistent with the data.

Although maxichoice revision is pleasingly conservative, it is still compatible with dis-orderly selection of theories. For example, suppose that σ, τ ∈ SEQ are distinct, and thatcontent(σ) is logically equivalent to content(τ). It is possible for + to be maxichoice yet for(B +σ)∪ (B + τ) to be inconsistent. Another kind of bad behavior was described in Section4.2 of essay #2. Let us repeat the earlier discussion. As before, we use propositional logicfor simplicity. Suppose that a revision function behaves as shown here:

(7)

Theory T1 : q, ¬p, (q ∧ r) → p Datum : r

Chosen revision R1 : q, ¬p, r .Rejected revision R2 : q, (q ∧ r) → p, r .

Thus, faced with datum r, the revision function retrenches Theory T1 by throwing out theconditional (q ∧ r) → p before adding in r. This yields the successor theory R1. A differentoption would have been to throw out ¬p from T1 before adding r, yielding the successorR2. The revision function has thereby revealed a preference for R1 compared to R2. (Ifthis way of phrasing the matter strikes you as too anthropomorphic, think of the revisionfunction as representing the preferences of a flesh-and-blood scientist.)

Now suppose that the same revision function behaves as summarized here:

(8)

Theory T2 : r, ¬p, (q ∧ r) → p Datum : q

Chosen revision R2 : q, (q ∧ r) → p , r .Rejected revision R1 : q, ¬p, r .



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We see that faced with datum q and Theory T2, the revision function retrenches T2 bythrowing out ¬p before adding q. This yields the revision R2. A different option wouldhave been to throw out (q ∧ r) → p from T2 before adding q, yielding R1. So now we mustconclude that there is a preference for R2 over R1.

Nothing prevents a maxichoice revision function from behaving as indicated in (7) and(8). Yet the two preferences are inconsistent since the revision function prefers the successortheory R1 to R2 in the first situation but R2 to R1 in the second situation. This seemsmuddled inasmuch as the different starting theories give no reason to invert one’s opinionabout the relative acceptability of the potential successors. To preclude such choices, wedefine a subclass of revision functions with a more orderly approach to theory selection.

(9) Definition: A revision function + is definite just in case there is a strict totalordering ≺ of pow(Lform) such that for all B ⊆ Lform and σ ∈ SEQ, B +σ is the≺-least consistent subset of B ∪ content(σ) that contains Inn(B, σ) ∪ content(σ).

Thus, definite revision is built upon a prior ordering of all potential theories. The orderingcan be interpreted in terms of preference, with earlier theories being prefered to later ones.Revision proceeds by choosing the most preferred theory that can legitimately serve assuccessor [in other words, the earliest theory that does not run afoul of Definition (3)]. Youcan see that the inconsistent preferences illustrated in (7) and (8) cannot be exhibited bydefinite revision.

It remains to show that definite revision functions exist [it is not evident that the neededordering can be constructed in such a way as to satisfy Definition (3)]. In fact, the followingproposition shows that there are revision functions that are simultaneously maxichoice anddefinite.

(10) Proposition: There are revision functions that are both maxichoice and definite.


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Such revision functions are not formal curiosities. The proof of the proposition (given inSection 8.1) shows that the class of revision functions that are simultaneously maxichoiceand definite is well populated.

Scientists that embody maxichoice, definite revision functions satisfy many formal cri-teria of rationality. They have transitive and connected preferences among theories, andrevise their theories so as to minimize change and maximize theory-acceptability.

Such is our model of theory-revision, and of rational theory-revision in particular. Wenow begin to explore its use in inductive inquiry. First, let it be emphasized that thereare many alternative accounts of revision, and that the inductive powers of any of themcould be investigated in place of ours. Although we have examined variants of the foregoingdefinitions, we do not know what fraction of the results reported below would survivetransplant to the various models reviewed in [17]. The model developed above is motivatedby its simplicity and generality. Also, it yields striking results about inductive inquiry. (Wehope you will soon agree.)

3. Inquiry via revision

3.1. From revision to science

In the present section we consider how theory revision can be harnassed for the purpose ofinquiry. The matter is simple in essence. Given a set B of formulas and a revision operator+ , the function λσ .B +σ maps SEQ into the power set of Lform .5 Intuitively, λσ .B +σconverts data into theory; data are represented by SEQ and theories by sets of formulas.With such a function we associate the scientist that maps σ ∈ SEQ into the class of models

5Following standard usage, λ in the expression “λσ .B +σ” indicates that σ serves as argument to thefunction, whereas B and + are fixed.


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of B +σ, and call such scientists revision-based (the official definition is given in Section3.3 below). Scientists of this kind are special in two ways. First, their outputs are notarbitrary propositions, but must instead be definable by a subset of Lform . In other words,each conjecture must have the form MOD(B) for some B ⊆ Lform instead of being justany collection of structures. Second, revision-based scientists are not arbitrary mappingsfrom SEQ to the definable propositions. Instead, they are constrained by Definition (3) ofrevision. Additional constraint arises if we require the + in λσ .B +σ to denote a restrictedkind of revision function, such as maxichoice and definite. We will see later that revision-based scientists employing maxichoice definite revision functions are canonical for a broadclass of problems; any solvable problem in the class is solved by a scientist of this kind.

In the preceding discussion we distinguished two special properties of revision-basedscientists. The first requires output propositions to be elementary classes of structures.Isolating just this property yields the set of linguistic scientists. They are “linguistic” inthe sense that the propositions they announce are expressible in our first-order language L.Thus, the revision-based scientists are a subset of the linguistic scientists, and informationabout the competence of the latter also provides insight about the former. So we start ourdiscussion with linguistic scientists. Revision-based scientists are introduced subsequently,in Section 3.3.

3.2. Linguistic scientists

Recall from Section 2.4 of the second essay that scientists issue classes of structures inresponse to members of SEQ. When the classes are definable from a subset of Lform , thescientists are called “linguistic.” Officially:

(11) Definition: Scientist Ψ is linguistic just in case there is ψ : SEQ → pow(Lform)such that for all σ ∈ SEQ, Ψ(σ) is defined iff ψ(σ) is defined, and when both are



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defined Ψ(σ) = MOD(ψ(σ)). In this case, we say that ψ underlies Ψ.

To lighten the notation we sometimes identify linguistic scientists with the functions ψ :SEQ → pow(Lform) that underlie them. In other words, if Ψ is linguistic, we allow ourselvesto write Ψ(σ) = B, for B ⊆ Lform , in place of the circumlocution: “ψ(σ) = B, where ψunderlies Ψ.”

Are linguistic scientists a canonical form of inquiry? Of course not. Suppose thatproposition P includes no isomorphically closed class of structures (for example, P mightbe a singleton, consisting of a single structure). Then no problem that includes P is solvableby linguistic scientist since no nonempty subset of P can be picked out by a set of formulas.[This is because MOD(B) is isomorphically closed, for any B ⊆ Lform .] There are plentyof solvable problems that include such emaciated propositions. Indeed, we can removefrom each proposition of any solvable problem all structures but one, and the resultingproblem will still be solvable.6 To illustrate, since Pf , Pb of Example (24)b of (essay #2)is solvable, so is (ω,<), (ω∗, <) (assuming Obs = Lbasic).7 No linguistic scientist cansolve the latter problem because no elementary class of structures is included in (ω,<)[or in (ω∗, <)].

On the other hand, linguistic scientists turn out to be well adapted to problems of form(T , P0, P1, . . .). Recall from Definition (65) of essay #2 that problem P has this form justin case P = P0, P1, . . . and

⋃P = MOD(T ). It is noteworthy that linguistic scientists

can solve any solvable problem of form (T , P0, P1, . . .). We record this fact in the nextproposition, which follows immediately from Proposition (67) of the second essay.

6There is a more general fact worth stating. If P = Pi | i ∈ N and Q = Qi | i ∈ N are problems suchthat Pi ⊆ Qi for all i, then P is solvable if Q is.

7Recall that ω denotes the set of natural numbers ordered naturally, and ω∗ denotes the natural numbersordered backwards, as · · · < 3 < 2 < 1 < 0.





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(12) Proposition: Every solvable problem of the form (T , P0, P1, . . .) is solved by alinguistic scientist. Indeed, the linguistic scientist may be chosen so that range(Ψ) ⊆pow(Lsen).

The coda to the proposition means that for problems of the form (T , P0, P1, . . .), nogenerality is lost by depriving linguistic scientists of the right to announce open formulas.8

3.3. Scientists based on revision

We now pass from linguistic scientists to their revision-based subset.

(13) Definition: Scientist Ψ is revision-based just in case there is B ⊆ Lform and revi-sion function + with the following property. For all σ ∈ SEQ, Ψ(σ) = MOD(B +σ).

Thus, faced with data σ, a revision-based scientist conjectures the proposition whose struc-tures satisfy the successor theory B +σ. As before, we allow ourselves to identify revision-based scientists with the functions λσ .B +σ that underlie them.

Since revision functions are total functions, so are revision-based scientists. Note alsothat the revision-based scientists are a proper subset of the linguistic scientists. It is easyto see that the inclusion is proper since linguistic scientists may be undefined on somedata. Even among the total linguistic scientists, some are not revision-based. Definition (3)implies, for example, that no revision-based scientist issues a conjecture that contradictshis data whereas this is possible for a linguistic scientist.

8This holds only for problems of form (T , P0, P1, . . .). In the general case, it is not difficult to exhibitan elementary class P such that P = MOD(B) implies B 6⊆ Lsen . For such a proposition P , the degenerateproblem P is solvable by a linguistic scientist but not by any linguistic scientist with range in pow(Lsen).


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The model of revision developed in Section 2 above invites us to view empirical inquiryas the process of revising one’s background theory by evidence that accumulates in theenvironment. We take the revision-based scientists to give formal substance to this idea.Moreover, if we choose the revision function + to be definite maxichoice, the revision-based scientist λσ .B +σ has some claim to the title “rational,” as we observed just afterProposition (10). As a consequence, we shall be particularly interested in the competence ofscientists of this form. We note that our enterprise would be vitiated if the starting theoryB could be taken to be inconsistent, for there is nothing commendable in starting from atheory that can be shown false prior to examining data. So our positive results will onlyconcern starting theories B that are consistent. When we use the term “maxichoice definitescientist” we mean a revision-based scientist whose starting theory is consistent and whoserevision function is maxichoice and definite.

The behavior of maxichoice definite scientists is distinctive. They don’t issue conjec-tures that contradict the data, for example, and they don’t revisit a conjecture that wasabandoned at an earlier stage of inquiry. Also, their preferences for theories are consistentin the sense discussed in Section 2.2. It would be nice to characterize the class of maxi-choice definite scientists in such terms (namely, as the class of scientists who never issueconjectures that contradict the data, etc.), but we have not pursued this question. When weexamine parameterized families of maxichoice definite scientists, moreover, the constraintson behavior become yet stricter.

Section 4 below is devoted to characterizing the competence of revision-based scientistswith respect to problems of the form (T , θ0 . . . θn). As a preliminary, the remainder ofthe present section brings out various facts that emerge at a more general level of analysis.Thus, we step back from problems of the special form (T , θ0 . . . θn) or (T , P0, P1, . . .),and consider the entire class of problems. In this more general setting, we hope to accomplishthree things. First, we show that the maxichoice revision functions are canonical for the classof problems that can be solved by revision-based scientists. That is, every problem that can


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be solved in the revision-based way can be solved by a revision-based scientist that relies ona maxichoice function + (Section 3.4).9 Next, we consider the interaction of computabilityand revision (Section 3.5). Finally, we define a partial order on the competence of revision-based scientists (Section 3.6).

These three topics have both mathematical and conceptual interest, but they are moretechnical and “theory internal” than the sequel. So you will be forgiven if you skip theremainder of the present section and proceed directly to Section 4.

3.4. The inductive power of revision

In the present subsection we establish that a single, maxichoice revision function is sufficientto define successful, revision-based scientists. The matter may be stated as follows.

(14) Theorem: Suppose that Obs = Lbasic . There is a maxichoice revision function +

with the following property. Let problem P be such that for some Y ⊆ Lform andrevision function ⊕ , λσ . Y ⊕σ solves P. Then there is a consistent X ⊆ Lform suchthat λσ .X +σ solves P.

Thus, one size fits all. A single maxichoice revision function allows the solution of anyproblem that is solved through the use of revision functions.10 The theorem is provedin Section 8.2. We believe that the hypothesis Obs = Lbasic is stronger than needed, inparticular, that it can be weakened to the supposition that Obs is closed under negation.The only demonstration of this claim known to us, however, is arduous and unrevealing.

9We do not know whether this maxichoice revision-based scientist can also be chosen to be definite.10We do not know whether this maxichoice revision function can be taken to be definite.


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In contrast, a simpler argument (given in Section 8.3) suffices to show that the maxi-choice scientist λσ .X +σ evoked in Theorem (14) solves its problem efficiently.11 Officially:

(15) Theorem: The revision-based scientist λσ .X +σ of Theorem (14) solves P effi-ciently.

Theorems (14) and (15) do not imply that every revision function is adapted to feasibleproblems, only that some of them are. In fact, not even the definite maxichoice revisionfunctions can all be made to work successfully. This is revealed by the following result.

(16) Theorem: Suppose that Sym is limited to a binary predicate and countably manyconstants and Obs = Lbasic . Then there exists a definite maxichoice revision func-tion + , and propositions P1, P2 that meet the following conditions.

(a) P1, P2 is solvable and each of P1, P2 is closed under elementary equivalence;(b) for all B ⊆ Lform , λσ .B +σ does not solve P1, P2.

Clause (a) ensures that both P1 and P2 can be conjectured by a revision-based scientist. Toconjecture Pi (i = 1, 2), it suffices to issue a set of sentences that defines a nonempty subsetof Pi; by clause (a), this is possible. Despite the nameability of P1, P2, there are definitemaxichoice revision functions that can’t be used to solve P1, P2. The theorem is provedin Section 8.4.

3.5. Scientists based on computable revision

From Definition (81) of the second essay, it follows that a revision-based scientist λσ .B +σis computable if and only if there is computable ψ : SEQ → N such that for all σ ∈ SEQ,

11Efficiency, you recall, involves non-dominated use of data, as defined in Section 6.1 of essay #2.




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B +σ = Wformψ(σ) . In this subsection we consider the consequences of requiring revision-based

scientists to be computer simulable. First, consider some good news. Recall from Definition(86) of essay #2 that a problem of form (T , θ0 . . . θn) is called “r.e.” just in case T is arecursively enumerable set of sentences. We have:

(17) Proposition: Suppose that Obs is closed under negation. For every solvable r.e.problem (T , θ0 . . . θn), there is consistent X ⊆ Lform with X |= T , and defi-nite maxichoice revision function + such that λσ .X +σ is computable and solves(T , θ0 . . . θn).

In other words, the solvable r.e. problems (T , θ0 . . . θn) fall within the inductive compe-tence of computable revision. The proof is given in Section 8.5.

In contrast to Proposition (17), it will now be seen that theory revision and computabil-ity do not always mix well. The reason is the consistency requirement expressed in Definition(3)b, which makes revision hard to calculate. The next proposition shows that even someproblems that are computably solvable are outside the reach of computable revision. Indeed,each proposition in such a problem can be taken to be strongly elementary, thus ensuringthat its propositions can be denoted by a single sentence.12 In sum:

(18) Proposition: Suppose that Sym is limited to the vocabulary of arithmetic (in-cluding 0 and a unary function symbol s) plus an additional constant. Also supposethat Obs = Lbasic . Then there is a problem P with the following properties.

(a) Every member of P is strongly elementary.

(b) P is computably solvable.12Recall from Definition (82) of essay #2 that a proposition is strongly elementary just in case it has the

form MOD(θ) for some θ ∈ Lsen .




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(c) some revision-based scientist solves P.

(d) For every B ⊆ Lform and revision function + , if λσ .B +σ is computable, thenit fails to solve P.

For the proof, see Section 8.6.

The foregoing result prompts the search for interesting subclasses of problems that canbe solved by computable, revision-based scientists. It also suggests the need to defineweaker senses of “computer simulable” so as to attenuate the impact of the consistencyrequirement expressed in Definition (3)b. We shall not pursue these projects here, however,preferring instead to work out the inductive logic of theory revision in the simpler contextof potentially ineffective functions.

3.6. Comparing the competence of revision-based scientists

A useful partial order on the competence of revision-based scientists may be defined asfollows. We say that revision function ⊕ “subsumes” revision function + just in case forall B ⊆ Lform and σ ∈ SEQ, B ⊕σ ⊇ B +σ. Definition (6) implies immediately that themaxichoice revision functions are maximal elements of the subsumes relation. That is, everyrevision function is subsumed by some maxichoice revision function, and no maxichoicefunction is subsumed “properly” by any other function.

Subsumption between revision functions is related to scientific competence in the fol-lowing way.

(19) Lemma: Suppose that revision functions + and ⊕ are such that ⊕ subsumes + .Let problem P and B ⊆ Lform be given. If λσ .B +σ solves P then λσ .B ⊕σ doesalso.


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The lemma is easy to verify, and defines a clear sense in which maxichoice revision functionsrepresent the most powerful scientific method based on hypothesis revision. If a problem issolvable by revision, then it is solvable by a revision-based scientist whose “ + ” is maxichoice.Conversely, for a given starting point B, the weakest, revision-based scientist is the revisionfunction that maps (B, σ) to Inn(B, σ)∪content(σ). The latter function may be called “fullmeet.” The revision function it underlies is subsumed by every other revision function. Sothe lemma implies that if the full meet scientist solves P then so does every scientist ofform λσ .B +σ. (The “full meet” terminology is adapted from [11].)

4. Theories as parameters

Now we are ready to make good on our promise about parameterized scientists (see page4). It will be seen how to parameterize revision-based scientists with the theory T figuringin problems of the form (T , θ0 . . . θn). If the problem is solvable then our parameterizedscientist will solve it.

You are probably thinking that our goal can be reached by constructing the right revisionfunction + , and then defining the family of scientists λσ . T +σ |T ⊆ Lsen, where T isthe free parameter. The revision function + would be chosen so that for any solvableproblem (T , θ0 . . . θn), the scientist λσ . T +σ would solve it. Matters won’t be quite sosimple. This is because we will shortly see problems (T , θ0 . . . θn) that are solved by norevision based scientist with starting theory T . This fact might seem discouraging, but it issurmounted by adding additional formulas to T . Specifically, it will be shown that scientistsof form λσ . T ∪ X +σ have wide inductive powers for problems of form (T , θ0 . . . θn)provided that the right supplement X is chosen to extend T . Indeed, provided that Obs isclosed under negation, a single supplement works for all solvable (T , θ0 . . . θn). In otherwords, there is X ⊆ Lform such that our desired family of scientists can be written as


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λσ . (T ∪X) +σ |T ⊆ Lsen, where T is still the only free parameter.13

For the details, let us first consider a large class of theory extensions that lead revision-based scientists to failure. This will help motivate the choice of successful theory-supplementintroduced subsequently.

4.1. Extensions of the background theory that are too modest

As noted above, to exploit theory revision for a problem (T , θ0 . . . θn), it is natural toconsider scientists of form λσ . T +σ. Such scientists attempt to revise the backgroundtheory T in the face of data coming from the environment. Unfortunately, there are simpleproblems of this kind that lead all such scientists to failure. Indeed, the following propositionshows that failure also results if T is extended to any consistent subset of Lsen . Recall thatthe latter set is limited to formulas without free variables.

(20) Proposition: Suppose that Sym is limited to the binary predicate R. Supposethat Obs = Lbasic . Let T = ∃x∀yRxy ↔ ¬∃y∀xRxy, and θ = ∃x∀yRxy. Then(T , θ,¬θ) is solvable, but for all revision functions + and all B ⊆ Lsen consistentwith T , λσ .B +σ does not solve (T , θ,¬θ).


Proposition (20) reveals that revision-based scientists cannot be made to work unlesstheir starting points extend the background theory T to include open formulas. The formu-las may be considered an “inductive leap” ready to be abandoned in whole or part accordingto the data encountered. Their free variables embody hypotheses about which of the objectsshown to the scientist have special properties involved in the problem under investigation.

13Recall from Definition (4) of essay #2 that Obs is closed under negation iff for every ϕ ∈ Obs, there isψ ∈ Obs that is logically equivalent to ¬ϕ.



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4.2. A universal method of inquiry based on revision

If we are willing to extend our background theories with open formulas, then revision can beused successfully for inquiry. Indeed, in case Obs is closed under negation, it is possible tochoose a single extension and a single revision function that works for any solvable problemof the form (T , θ0 . . . θn). In this sense the next theorem exhibits a canonical method ofinquiry. Its proof exhibits a theory-extender X and a definite maxichoice revision function+ with the following property. Every solvable problem (T , θ0 . . . θn) is solved by thescientist λσ . (T ∪ X) +σ. Moreover, this scientist has some claim to the title “rational.”For, its starting theory is a consistent extension of the background theory T , and revisionoccurs on the basis of a definite maxichoice function.14

(21) Theorem: Suppose that Obs is closed under negation. There exists X ⊆ Lform

and definite maxichoice revision function + such that for all consistent T ⊆ Lsen ,the following holds.

(a) T ∪X is consistent.

(b) λσ . (T ∪X) +σ solves every solvable problem of the form (T , θ0 . . . θn).

The theorem exhibits the parameterized family of scientists hoped for in Section 1 above.The family is λσ . (T ∪ X) +σ, where T is the sole parameter. When confronted withproblem (T , θ0 . . . θn), putting T in for the parameter guides the resulting scientist tosuccess. Our “family” of scientists therefore looks more like a single scientist with thegood sense to establish his starting theory in light of the inductive problem he faces. Theinductive acumen of this single scientist is embodied in the fixed choice of revision function+ and theory-supplement X.

14That is, the revision occurs on the basis of transitive, connected preferences for theories that inflictminimal change on current opinion (as discussed in Section 2.2).


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So, what are the identities of the fixed + and X? They are revealed in the constructiveproof of (21) given in Section 8.8. We cannot claim that the choices of + and X in theproof are the simplest possible, only that nothing simpler has yet come to mind.

4.3. Supplementing as a function of the problem

Theorem (21) shows that some revision functions can be used for successful inquiry, but itdoes not show that all of them can. It thus leaves open the possibility that our conceptionof revision function is so lax as to embrace revision that is useless for inquiry. This would betrue if some revision functions were led to failure on a solvable problem (T , P0, P1, . . .) nomatter how T is extended. But such is not the case. It will now be shown that every revisionfunction can be made to succeed on every solvable problem of the form (T , P0, P1, . . .).

(22) Theorem: Suppose that solvable problem P is of form (T , P0, P1, . . .), and thatObs is closed under negation. Then there is a consistent extension X ⊆ Lform of Tsuch that for every revision function + , λσ .X +σ solves P.

In the present theorem we choose the theory-supplement X as a function of the problem tobe solved. This is what allows all revision functions to solve the problem (if it is solvableat all). In Theorem (21), X was chosen once and for all. Only a single revision functionis therefore invoked in the latter theorem. We also note the contrast with Theorem (16) ofSection 3.4. There we exhibited a revision function + and a solvable problem P such that+ could not be used to solve P. But the problem P in the earlier theorem was not of form(T , P0, P1, . . .). The present result affirms that all solvable problems of the latter formcan be solved using any revision function starting with a suitably chosen theory. For theproof of Theorem (22), see Section 8.9.


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Theorem (22) suggests that our definition of revision function is sufficiently strict, since itdoes not tolerate revision strategies that are inapt for science. The definition also appears tobe sufficiently liberal since it embraces revision functions that are rational and multipurposein the sense of Theorem (21). Let us therefore rejoice in the conviction that a sound andsufficient conception of revision has been achieved. Indeed, the moment for rejoicing iswell chosen since our satisfaction will prove shortlived! In Section 5 we will encounterrevision functions that cannot be used for the efficient solution of certain problems of form(T , θ,¬θ). For now, however, all is well. Indeed, things get even better when we specializeour problems to those whose background theories are finitely axiomatizable.

4.4. Supplementing as a function of finitely axiomatizable theories

In this subsection and the next we look more closely at the way background theories can besuccessfully extended. The extension X of T forseen in Theorem (22) depends not just onT but more generally on the problem (T , P0, P1, . . .). If we limit attention to problems ofthe form (T , θ0 . . . θn), assume that Obs is closed under negation, and add the hypothesisthat T is finitely axiomatizable, then T can be successfully extended without concern for thepartition imposed by θ0 . . . θn. This is the burden of the next proposition. In the succeedingsubsection we show that the hypothesis of finite axiomatizability cannot be lifted.

Let us tell you frankly that questions involving finite axiomatizability are not central tothe message of this essay. So the remainder of the present section (viz., 4.4 and 4.5) can beomitted on a first reading, and the discussion picked up in Section 5.

The proof of the following proposition is given in Section 8.10.

(23) Proposition: Suppose that Obs is closed under negation. Let T ⊆ Lsen beconsistent and finitely axiomatizable. Then there is consistent X ⊆ Lform with


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X |= T such that for every revision function + , λσ .X +σ solves every solvableproblem of form (T , θ0 . . . θn).

4.5. The inability to supplement independently of T

It would be nice to find a method of extending a theory T that allows all revision functionsto succeed on solvable problems of form (T , θ0 . . . θn). In this case the required extensionwould not depend on the sentences θ0 . . . θn, but only on the background theory T . Fulfillingthis desire would amount to lifting the hypothesis of finite axiomatizability from Proposition(23). Unfortunately, the next proposition shows this hope to be unrealizable.

(24) Proposition: Suppose that Sym is limited to a binary predicate, a constant, anda unary function symbol. Suppose that Obs = Lbasic . Then there exists T ⊆ Lsen

and definite maxichoice revision function + such that for all B ⊆ Lform , λσ .B +σfails to solve some solvable problem of form (T , θ,¬θ).

The proposition shows that in the general case, theory extensions must be chosen as afunction of the problem to be solved. Otherwise, some revision functions will be led toneedless failure. The proof is given in Section 8.11.

5. Efficient revision

One goal of the present essay is to represent inquiry as a process of rational theory revision.The theorems of the preceding section show that for a wide class of problems, revision-basedscientists can be made to succeed whenever success is possible in principle. We now consider


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efficient inquiry in the same terms. Our question is whether and to what extent efficientinquiry can be carried out using revision functions.

First it will be shown that revision functions are compatible with efficient solvabilityfor problems of the form (T , P0, P1, . . .). That is, for every solvable problem of thisform, some revision function can be made to work efficiently. In contrast, it will then bedemonstrated that not all revision functions are suitable for efficient discovery. Indeed, evensome maxichoice definite ones cannot be used to efficiently solve certain “easy” problems.The consequences of this finding for our theory of inquiry are then discussed.

Throughout the section we rely on the conception of efficient inquiry introduced inDefinition (76) of essay #2. To recapitulate, scientist Ψ is dominated on problem P if arival scientist succeeds at least as quickly as Ψ on all environments for P, and more quicklyon some. Conversely, Ψ is efficient on P if no rival scientist dominates Ψ.

5.1. Inductive efficiency by revision is possible

Theorem (15) of Section 3.4 provides a sense in which theory revision is compatible withefficient solvability. Indeed, it was there shown that a sole maxichoice revision functionsuffices to efficiently solve every problem, provided only that the problem be solvable inthe revision-based way. We now show that efficient inquiry can be carried out on a broadclass of problems by revision-based scientists whose starting points extend the backgroundtheory.

(25) Theorem: Suppose that Obs is closed under negation. For every solvable problemP of the form (T , P0, P1, . . .) there is a consistent extension X ⊆ Lform of T anda definite maxichoice revision function + such that λσ .X +σ solves P efficiently.

See Section 8.12 for the proof.



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If attention is limited to problems of the form (T , θ0 . . . θn), Theorem (25) can beimproved. We rely on the following definition, which isolates the revision functions thatstrive to hold onto a given theory T .

(26) Definition: Let revision function + and T ⊆ Lsen be given. We say that + isT -preserving just in case for all σ ∈ SEQ and X ⊆ Lform , if T ∪ content(σ) isconsistent then T ⊆ (T ∪X) +σ.

Thus, a T -preserving revision function maintains T in the successor theory if T is a subsetof the starting theory and is consistent with the data σ. The following lemma records thefact that T -preserving revision functions exist. It is proved in Section 8.13.

(27) Lemma: For all T ⊆ Lsen , there exists a T -preserving maxichoice definite revisionfunction.

The following theorem shows that the T -preserving revision functions can be used toefficiently solve any solvable problem of form (T , θ0 . . . θn). See Section 8.14 for the proof.

(28) Theorem: Suppose that Obs is closed under negation. For every solvable problemP of the form (T , θ0 . . . θn) there is a consistent extension X ⊆ Lform of T withthe following properties.

(a) For all revision functions + , λσ .X +σ solves P.

(b) For all T -preserving revision functions + , λσ .X +σ solves P efficiently.

Further improvement is possible in the case of finitely axiomatizable theories. Thefollowing corollary is proved in Section 8.15.


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(29) Corollary: Suppose that Obs is closed under negation. Let T ⊆ Lsen be con-sistent and finitely axiomatizable. Let solvable problem of form (T , θ0 . . . θn) begiven. Then there is consistent extensionX ⊆ Lform of T such that for every revisionfunction + , λσ .X +σ efficiently solves (T , θ0 . . . θn).

5.2. Inductive efficiency by revision is not inevitable

Theorem (25) shows that there are enough revision functions to carry out efficient inquiryon problems of form (T , P0, P1, . . .). Moreover, just the maxichoice definite subclass issufficient. It would be pleasing to report the converse fact as well, namely, that everyrevision function can be used efficiently. Theorem (22) would be reinforced thereby, sincethe latter states that every revision function can be used to solve any solvable problem ofform (T , P0, P1, . . .). Unfortunately, there is no such counterpart to Theorem (22). Somerevision functions cannot be brought to efficiently solve certain solvable problems. Worse,the guilty revision functions can be taken to be maxichoice definite and the recalcitrantproblem has a particularly simple form. Here is a precise statement of the situation.

(30) Theorem: Suppose that Sym is limited to countably many constants and Obs =Lbasic . Then there is a problem of form (T , θ,¬θ) and a definite maxichoicerevision function + with the following properties.

(a) T is recursive.

(b) (T , θ,¬θ) is solvable.

(c) For all B ⊆ Lform , if λσ .B +σ solves (T , θ,¬θ) then λσ .B +σ is dominatedon (T , θ,¬θ).



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Although the maxichoice definite revision functions possess strong credentials of ratio-nality, their properties do not guarantee that they can be used for efficient inquiry. This iswhat the foregoing theorem shows. To state the matter more generally, let us call a revisionfunction + “efficient” just in case for every solvable problem P of the form (T , θ,¬θ)there is consistent X ⊆ Lform such that λσ .X +σ solves P efficiently. Theorems (15) and(30) show that some but not all definite maxichoice revision functions are efficient.

The inefficient revision functions are in some way defective, since it is not rational toconduct inquiry in needlessly dilatory fashion. In particular, Theorem (25) shows thatinefficient revision can always be replaced by an efficient method of theory change, hencethere would seem to be no justification for using the inefficient function. The questionthus arises as to additional conditions that can be imposed on revision functions in viewof guaranteeing efficiency. Such conditions should be intuitively justifiable in terms ofmodifying theories in a rational way, and also pick out the efficient subset of functions.

One promising class of revision functions may be defined as follows.

(31) Definition:

(a) Given X ⊆ Lform , let sentence(X) denote X ∩Lsen , that is, the formulas of Xwithout free variables.

(b) We say that revision function + is sentence preserving just in case for allσ ∈ SEQ and X ⊆ Lform , if sentence(X) ∪ content(σ) is consistent thensentence(X) ⊆ X +σ.

A corollary to the proof of Lemma (27) reveals the existence of maxichoice definite,sentence preserving revision functions. From the proof of Theorem (28) we easily obtain:

(32) Corollary: Suppose that Obs is closed under negation. For every solvable prob-lem P of the form (T , θ0 . . . θn) there is a consistent extension X ⊆ Lform of T


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with the following properties.


(b) For all sentence preserving revision functions + , λσ .X +σ solves P efficiently.

Does Corollary (32) justify equating the maxichoice definite, sentence preserving revi-sion functions as “rational?” The difficulty with this suggestion is the status of sentencepreservation. Whereas the maxichoice definite revision functions appeal to intuitions aboutconservative revision and consistent preferences over theories (as discussed in Section 2.2),it is not clear what can be said on behalf of Definition (31), other than that it yields efficientinquiry.

6. Closure

Recall that the scientist’s theories are interpreted as arbitrary subsets of formulas, not nec-essarily closed under deduction. The present section explores the impact on revision-basedinquiry of requiring the starting theory B of scientist λσ .B +σ to be closed under somefragment of logic. We shall see that closed starting points are consistent with revision-basedsuccess. However, even a weak form of closure perturbs Theorem (22). That is, not everyrevision function can be used to solve every solvable problem of the form (T , P0, P1, . . .)if the starting point is required to include its own logical consequences.

We begin our discussion with some remarks about the epistemological issues surroundingdeductive closure of theories. The following notation will be used. Given B ⊆ Lform , we useCn(B) to denote the set of logical consequences of B. Thus, B ⊆ Cn(B), and B is closedunder Cn if and only if B = Cn(B).


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6.1. Closure operators

Within the literature on belief revision, two conceptions of belief states have emerged.According to the “coherence” view, belief states are deductively closed, and the agent iscommitted in the same way to statements whose conviction is rooted directly in experienceas she is to their logical consequences. The coherence approach is elaborated and defendedin [11]. In contrast, belief states within the “foundationalist” approach contain only thoseformulas which the agent has extra-logical reason to believe, and thus are not in generaldeductively closed. The foundationalist approach is developed in [10, 13, 15, 14, 20] and inwork cited there. For a particulary illuminating discussion of the issues, see [17, Ch. 1].

Both coherence and foundationalism advocate logical integrity in the following sense.A rational agent confronted with information that her belief state implies a falsehood isrequired to abandon some beliefs in order to remove the implication. However, founda-tionalism provides more guidance than coherence regarding what to abandon and whatto conserve. To see this, let φ, ψ be logically independent formulas, and consider A1 =φ, ψ +¬φ versus A2 = Cn(φ, ψ) +¬φ. For every revision function + , A1 = ¬φ, ψ. Incontrast, A2 can vary widely. Thus, for one choice of + , A2 includes ψ but not ψ → φ,whereas the situation is reversed for another choice. Giving such liberty to revision functionsleads some of them to mischief, as will be seen shortly.

Closure under Cn versus no closure at all can be viewed as extremes along a gradient.In order to consider intermediate points we rely on the following definition.

(33) Definition: By a closure operator is meant any function cl to and from subsets ofLform such that for all B ⊆ Lform : B ⊆ cl(B) ⊆ Cn(B).

At one end, we have the identity closure operator cl defined by cl(B) = B for every B ⊆Lform . At the other end is deductive consequence. When theories are closed under a closure


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operator cl, then revision-based scientists have the form λσ . cl(B) +σ, for some B ⊆ Lform

and revision function + . If B and + are chosen astutely, revision-based scientists can stillbe made to work under these conditions. Indeed, examination of the proof of Theorem (21)shows that it can be strengthened to the following.

(34) Theorem: Suppose that Obs is closed under negation. There exists X ⊆ Lform

and definite maxichoice revision function + such that for all closure operators cland all consistent T ⊆ Lsen , the following holds.

(a) T ∪X is consistent.(b) λσ . cl(T ∪X) +σ solves every solvable problem of the form (T , θ0 . . . θn).

The foregoing theorem provides a sense in which closure does no harm to revision-basedinquiry. Closure nonetheless complicates matters inasmuch as closed starting points requirethe revision function to be chosen with care if inquiry is to succeed. This is explained next.

6.2. Discovery under coherence-like starting points

Theorem (22) shows that every revision function can be used to solve any solvable problemof the form (T , P0, P1, . . .). Requiring belief states to be deductively closed, however,perturbs this result. In fact, the following theorem shows that it is enough to close beliefstates under the single rule: ψ, φ / ψ → φ. Belief state B ⊆ Lform is closed under the latterrule just in case ψ, φ ∈ B implies ψ → φ ∈ B. Notice how weak and innocent this conditionappears to be!

(35) Theorem: Let Sym consist of a binary predicate, and suppose that Obs = Lbasic .Then there exists a problem of form (T , θ,¬θ) (with T finite), and maxichoicerevision function + such that:


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(a) (T , θ,¬θ) is solvable;

(b) for all B ⊆ Lform , if B is closed under ψ, φ / ψ → φ, then λσ .B +σ does notsolve (T , θ,¬θ).


6.3. Remarks on closure and successful revision

Let us consider what conclusions may be drawn from Theorem (35). The theorem does notshow that deductive closure of theories is incompatible with scientific success via theoryrevision. After all, Theorem (34) assures us that some revision function succeeds no matterwhat closure operator is in force. Rather, the lesson of Theorem (35) seems to bear on theissue of commitment to individual assertions. That the theory of an ideally rational agentshould be deductively closed seems difficult to challenge. However, such an ideal agent maynonetheless be more attached to her background theory T and to its extension X than totheir nontrivial, logical consequences. Indeed, this kind of situation arises whenever wediscover an implausible consequence of our strong beliefs, and suffer doubt about which toabandon. Often our desire is to conserve as much as possible of the original beliefs, withno attempt to measure how large a fragment of their consequences will be lost. With thisin mind, Theorem (35) suggests that when an agent faces contradictory data, she shouldseek to salvage a subset of her starting theory and then let the logical consequences fallwhere they may. For, as the theorem shows, a revision function that does not adopt thispolicy may be doomed to fail on some solvable problems, starting from any theory that isdeductively closed.

Alternatively, if closed starting theories have high epistemological priority, then furtherconditions should be placed on revision functions. The conditions should ensure that allconforming revision functions can be used successfully on solvable problems, starting from


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closed starting points. Otherwise, one is left in the uncomfortable position of insistingthat starting theories be deductively closed, while admitting that some acceptable meansof revising those theories lead to needless scientific failure.

Another observation may serve to put a finer point on the distinction between coherenceand foundationalist approaches to representing the epistemological state of an idealizedscientist Ψ. Suppose Ψ solves problem P, and consider the P ∈ P that is implied cofinitelyoften by Ψ’s conjectures on some environment e for P . It is tempting to say that Ψ ends upknowing that P is true in this situation because (a) Ψ believes P (at least, he consistentlyannounces P at the “end” of the game), (b) P is true in the world giving rise to e, and (c)it is not just lucky that Ψ ends up believing P in E since Ψ embodies a reliable strategythat succeeds in any environment for any proposition in P. Thus, the inductive logic builtaround our paradigm of discovery is well suited to a reliabilist account of knowledge (fordiscussions of reliabilism, see [18, 12]). One well-known difficulty with reliabilism, however,is deductive closure. For, a procedure that can reliably discover the truth of both thesentences ϕ and θ may not be reliable for discovering the truth of their conjunction.15

In response, it is natural to define the known sentences at a given stage of inquiry to bethe deductive closure of whatever sentences are true, and believed to be so by dint of theapplication of a reliable process. So, once again we see a distinction between a “core” set ofknown sentences versus the result of closing them via deduction. Theorem (35) underlinesthe importance of making this kind of distinction.

15This is clearest in a probabilistic setting since the discovery-rates for the two sentences might be stochas-tically independent, thereby lowering the discovery rate for both together. But even if a scientific strategyhas perfect reliability for each of ϕ and θ it does not follow that the same is true of their conjunction.


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7. Iterated revision

7.1. Another conception of inquiry

The scientists discussed so far in this essay revise their starting theory by confrontingit anew with each initial segment of the environment. An alternative conception allowsstarting theories to evolve with time, confronting only the latest datum at any given stageof inquiry. To explain, let B ⊆ Lform be a starting theory. Faced with data 〈δ1, δ2〉 ∈ SEQ,the scientist λσ .B +σ conjectures B + δ1 at the first step of inquiry and B + 〈δ1, δ2〉 at thesecond. In contrast, an iterated model of inquiry conceives the scientist as first conjecturingB + δ1, as before, but then conjecturing (B + δ1) + δ2. In general, at step n + 1 of iteratedinquiry, the theory facing the (n + 1)th datum has already been modified n times. Toformalize this idea we define the “iterated extension” of a revision function.

(36) Definition: Let revision function + be given. The iterated extension of + isdefined to be the function +ie : pow(Lform)× SEQ → pow(Lform) such that for allB ⊆ Lform , σ ∈ SEQ and k ∈ N ,

B +ie σ =B if σ = ∅,[B +ie (β0 . . . βk−1)] + βk if σ = (β0 . . . βk).

Iterative revision resembles memory-limitation within the numerical paradigm, discussedin Section 2.5 of the first essay. In both cases the latest datum plays a special role inhypothesis selection, whereas earlier data are preserved only through their impact on thehypothesis of the preceding stage.

Iterated revision represents in a natural manner the influence of earlier hypotheses onthe choice of later ones. This is because the fragment of the starting theory that survives



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long enough to confront a non-initial datum depends on decisions about what to preservein the face of earlier data. In contrast, non-iterative revision allows the following kind ofcut-and-paste operation. Let Ψ1 and Ψ2 be revision-based scientists with the same startingpoint, B. Then the following scientist Ψ3 is also revision-based. For all σ ∈ SEQ:

Ψ3(σ) =

Ψ1(σ) if length(σ) < 1000;Ψ2(σ) otherwise.

Scientist Ψ3 is revision-based because revision functions + impose no constraint on therelation between B +σ and B + (σ∗τ). [See Definition (3).]

We now consider what can and cannot be achieved using iterative revision. Our firsttwo propositions show that iterative revision is compatible with successful inquiry, and evenwith efficient inquiry. However, it will then be demonstrated that some revision functionscannot be used in iterative fashion to solve certain simple problems, no matter what startingpoint is chosen.

7.2. Achievements of iterative revision

The following two propositions show that iterative revision can be used successfully forinquiry. In particular (and similarly to before), a sole revision function and theory-extendersuffice to solve every solvable problem of form (T , θ0 . . . θn).

(37) Proposition: Suppose that Obs is closed under negation. There exists X ⊆ Lform

and definite maxichoice revision function + such that for all consistent T ⊆ Lsen ,the following holds.

(a) T ∪X is consistent.


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(b) λσ . (T ∪X) +ie σ solves every solvable problem of the form (T , θ0 . . . θn).

The proof is given in Section 8.18. We see that Proposition (37) is the iterative counter-part to Theorem (21). The analogy extends to efficient inquiry insofar as the next proposi-tion shows that iterative revision can be used to efficiently solve every solvable problem ofform (T , P0, P1, . . .). [Compare Theorem (25).] The proof is given in Section 8.19.

(38) Proposition: Suppose that Obs is closed under negation. For every solvableproblem P of the form (T , P0, P1, . . .) there is a consistent extension X ⊆ Lform

of T and a definite maxichoice revision function + such that λσ .X +ie σ solves Pefficiently.

Despite the achievements recorded in the foregoing propositions we shall now provide asense in which hypothesis revision is less successful in the iterative context than before.

7.3. Iterative revision is not robust

Given B ⊆ Lform and σ ∈ SEQ, Lemma (4) shows that B +σ |=∧σ. In contrast, it is

easy to construct examples to show that B +ie σ may contradict∧σ, although it must

imply σ’s last member. Such license to ignore past data gives iterative revision so muchflexibility that some revision functions can no longer be used to solve simple problems.The infirmity afflicts even the class of maxichoice definite functions. This is shown by thefollowing theorem, which may be contrasted with Theorem (22). To state our result, lettwo be the sentence ∃xy[x 6= y ∧ ∀z(z = x ∨ z = y)], asserting that there are exactly twoindividuals. It is easy to verify that (∅, two,¬two) is solvable. However:

(39) Theorem: Suppose that Sym = ∅ and Obs = Lbasic . Then there exists a definitemaxichoice revision function + such that for all B ⊆ Lform , λσ .B +ie σ does not


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solve (∅, two,¬two).

See Section 8.20 for the proof. Theorem (22) and Proposition (39) imply the existence of amaxichoice, definite revision function + such that +ie 6= ⊕ for every iterated extension ⊕of a revision function. In this sense, iterated and non-iterated theory revision yield differentapproaches to scientific discovery.

7.4. Iterative scientists under deductive closure

Having examined deductive closure of belief states in Section 6, and iterated belief revisionin the present section, it is natural to wonder about the consequences of combining the twofeatures of inquiry.16 To give substance to this idea requires closing each belief state thatemerges at successive steps of iterated revision. It is not enough to close only the startingbelief state since its successors may be left open by the revision process. We thereforeformalize the matter as follows.

(40) Definition: Let revision function + be given. The closed iterated extension of +

is defined to be the function +Cnie : pow(Lform)× SEQ → pow(Lform) such that for

all B ⊆ Lform , σ ∈ SEQ and k ∈ N ,

B +Cnie σ =

B if σ = ∅,Cn(B +

Cnie (β0 . . . βk−1)) + βk if σ = (β0 . . . βk).

It will now be seen that iterative revision under closure, like its unclosed counterpart,suffers from the existence of maxichoice definite revision functions unable to solve simpleproblems. As before, let two be the sentence ∃xy[x 6= y ∧ ∀z(z = x ∨ z = y)]. Then wehave the following parallel to Theorem (39). It is proved in Section 8.21.

16The present subsection is not central to our discussion, and may be omitted on a first reading.


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(41) Proposition: Suppose that Sym = ∅ and Obs = Lbasic . Then there exists adefinite maxichoice revision function + such that for all B ⊆ Lform , λσ .B +

Cnie σ

does not solve (∅, two,¬two).

8. Proofs

8.1. Proof of Proposition (10)

(10) Proposition: There are revision functions that are both maxichoice anddefinite.

In the present subsection we establish the existence of definite maxichoice revision func-tions, thereby proving Proposition (10). For this purpose we rely on a construction thatwill be pivotal in later proofs. Here and elsewhere we rely on elementary facts about ordinalnumbers (see [8, 19] for background).

(42) Definition: Let nonnull ordinal κ be given. Let enumeration S = Sα | 0 < α < κof members of pow(Lform) also be given. We specify a function +S : pow(Lform)×SEQ → pow(Lform). Let B ⊆ Lform and σ ∈ SEQ be given. Then B +S σ is definedby induction over α < κ. Specifically, for each α < κ we define the set Y α, and putB +S σ =

⋃α<κ Y

α. The inductive definition of Y α is as follows.

Set Y 0 = Inn(B, σ) ∪ content(σ). Let nonnull ordinal α < κ be given,and suppose that Y β is defined for all β < α. Then Y α = Sα if Sα ⊆B ∪ content(σ) and (

⋃β<α Y

β) ∪ Sα is consistent; otherwise Y α = ∅.

Intuitively, B +S σ is constructed as follows. At stage 0 you dump Inn(B, σ) ∪ content(σ)into B +S σ. At stage 1 you examine S1. If S1 ⊆ B ∪ content(σ) and S0 ∪ S1 is consistent,


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then you dump S1 into B +S σ; otherwise, you leave B +S σ as it was after the precedingstage. Keep going in this way. Denote by S the portion of B +S σ constructed at stages0, 1, 2 . . .. At stage ω, you examine Sω. Suppose that Sω ⊆ B ∪ content(σ) and S ∪ Sω isconsistent. Then you dump Sω into B +S σ; otherwise, leave B +S σ as S. Keep going . . .

The important fact about this construction is that it yields a definite maxichoice revisionfunction, provided only that the enumeration Sα | 0 < α < κ contains all singleton setsϕ, ϕ ∈ Lform . Precisely:

(43) Proposition: Let ordinal κ be given. Let enumeration S = Sα | 0 < α < κ ofmembers of pow(Lform) contain all singleton sets. Then +S is a definite maxichoicerevision function.

Proof: Let B ⊆ Lform and (consistent) σ ∈ SEQ be given. Clearly, Inn(B, σ)∪content(σ) ⊆B +S σ. An easy application of compactness shows that B +S σ is consistent. Moreover,because χ ∈ S for all χ ∈ Lform , the construction in Definition (42) ensures that for allχ ∈ (B∪ content(σ))− (B +S σ), (B +S σ)∪χ is inconsistent. This implies easily that +S

is a maxichoice revision function.

Define the following strict total order ≺# on pow(Lform). Given X,Y ⊆ Lform , X ≺# Yiff the following condition holds.

There is nonnull α < κ such that Sα ⊆ X, Sα 6⊆ Y , and for all nonnull β < α,Sβ ⊆ X iff Sβ ⊆ Y .

Indeed, ≺# is a strict total ordering of pow(Lform) because S contains all singleton sets.We need to show that B +S σ is the ≺# -least consistent subset of B ∪ content(σ) thatcontains Inn(B, σ) ∪ content(σ). Let Y α, α < κ, be the subsets of B ∪ content(σ) definedfrom B and σ as in Definition (42). For a contradiction suppose that C ⊆ Lform is suchthat:


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(44) (a) Inn(B, σ) ∪ content(σ) ⊆ C,

(b) C ≺#⋃α<κ Y

α,

(c) C ⊆ B ∪ content(σ),

(d) C is consistent.

By (44)b, let nonnull α < κ be such that Sα ⊆ C, Sα 6⊆⋃γ<κ Y

γ , and for all nonnullβ < α, Sβ ⊆ C iff Sβ ⊆

⋃γ<κ Y

γ . By (44)c, Sα ⊆ B ∪ content(σ). By (44)a,d, Inn(B, σ) ∪content(σ) ∪

⋃Sβ |Sβ ⊆ C and 0 < β < α is consistent. These facts, along with the

construction in Definition (42) imply that Sα ⊆ Y α, contradiction.

So, there are plenty of definite maxichoice revision functions.

8.2. Proof of Theorem (14)

(14) Theorem: There is a maxichoice revision function + with the followingproperty. Let problem P be such that for some Y ⊆ Lform and revision function⊕ , λσ . Y ⊕σ solves P. Then there is a consistentX ⊆ Lform such that λσ .X +σsolves P.

To prove the theorem we rely on an evident fact, namely: for a revision-based scientistto solve a problem P, it must be possible to produce sets of formulas that imply the propo-sitions of P while remaining consistent with the incoming data. The following definitionstates this condition precisely.

(45) Definition: Problem P is feasible just in case for all P ∈ P, and all σ ∈ SEQ forP , there is Y ⊆ Lform such that ∅ 6= MOD(Y ∪ content(σ)) ⊆ P .


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We will now demonstrate two things. First, if a problem is not feasible then it is notsolvable by a revision-based scientist. Second, there is a maxichoice revision function +

such that if a problem is both solvable and feasible then + suffices to solve the problem ina revision-based way. First, revision-based solvability implies feasibility:

(46) Proposition: Suppose that problem P is not feasible. Then for all B ⊆ Lform andrevision functions + , λσ .B +σ does not solve P.

Proof: Let problem P, P ∈ P, and σ ∈ SEQ for P be such that for all Y ⊆ Lform eitherMOD(Y ∪ content(σ)) = ∅ or MOD(Y ∪ content(σ)) 6⊆ P . Then for all τ ∈ SEQ:

(47) for all Y ′ ⊆ Lform , either

MOD(Y ′ ∪ content(σ∗τ)) = ∅, or

MOD(Y ′ ∪ content(σ∗τ)) 6⊆ P.

Let X ⊆ Lform , revision function + , and environment e for P with σ ⊆ e be given. Then(47) implies immediately that λσ .X +σ does not solve P in e.

The second part of the proof of Theorem (14) is formulated as follows.

(48) Proposition: There is a maxichoice revision function + with the following prop-erty. For every solvable, feasible problem P there is consistent X ⊆ Lform such thatλσ .X +σ solves P.


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Thus, the class of scientists of form λσ .X +σ with X consistent and + maxichoice is acanonical strategy for the feasible problems.

The usual kind of completion argument shows the following.

(49) Lemma: For every revision function ⊕ there is a maxichoice revision function +

with the following property. For all B ⊆ Lform and σ ∈ SEQ, B ⊕σ ⊆ B +σ.

(50) Definition: A linguistic scientist Λ is consistent just in case for all σ ∈ SEQ,Λ(σ) ∪ content(σ) is consistent.

We also have:

(51) Lemma: Every solvable, feasible problem is solved by a consistent linguistic scien-tist.

Proof: Let scientist Ψ solve feasible problem P. By feasibility, for every P ∈ P and for everyσ ∈ SEQ for P , choose Y (σ, P ) ⊆ Lform such that (a) ∅ 6= MOD(Y (σ, P )) ⊆ P and (b)Y (σ, P ) ∪ content(σ) is consistent. Define linguistic scientist Λ such that for all σ ∈ SEQ,Λ(σ) = MOD(Y (σ, P )) if σ is for P and ∅ 6= Ψ(σ) ⊆ P ; otherwise, Λ(σ) = MOD(σ). It isclear that Λ solves P, and that Λ is consistent.

Proof of Proposition (48): By (49), let maxichoice revision function + have the followingproperty. For all B ⊆ Lform and σ ∈ SEQ, if σ 6= ∅ and B ∩

∧σ → ϕ |ϕ ∈ Lform is

consistent with content(σ), then B ∩ ∧σ → ϕ |ϕ ∈ Lform ⊆ B +σ. Let solvable and

feasible problem P be given. There is nothing to prove if P = ∅, so choose P0 ∈ P. Byfeasibility, there is Y1 ⊆ Lform such that ∅ 6= MOD(Y1) ⊆ P0. Let Y0 ⊆ Lform be the resultof doubling the index of every variable appearing in Y1. Then also ∅ 6= MOD(Y0) ⊆ P0. So


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we may choose S0 ∈ P0 and full assignment h with S0 |= Y0[h]. Let environment e0 be forS0 and h. Then we have:

(52) S0 |= Y0 ∪ content(e0)[h].

Since P is solvable and feasible, by Lemma (51) there is linguistic scientist Λ that solves Pand satisfies:

(53) for all σ ∈ SEQ, Λ(σ) is consistent with content(σ).

Define X0 = ∧σ → ϕ | ∅ 6= content(σ) ⊆ content(e0) and ϕ ∈ Y0. Define X1 =

∧σ →

ϕ | content(σ) 6⊆ content(e0) and ϕ ∈ Λ(σ). We take X = X0 ∪ X1. Observe that S0 |=X0[h] because by (52) S0 |= ϕ[h] for all ϕ ∈ Y0. Also S0 |= X1[h] because by (52) againS0 6|=

∧σ[h] for all σ ∈ SEQ with content(σ) 6⊆ content(e0). So X is consistent. It remains

to show that λσ .X +σ solves P. Let environment e for P1 ∈ P be given. There are twocases.

Case 1: content(e) = content(e0). Let k > 0 be given. By (52), Y0 ∪ content(e[k])is consistent. So the definition of + and X implies that X + e[k] =

∧e[k] → ϕ |ϕ ∈

Y0 ∪ content(e[k]) |= Y0. Since e is for S0, e is for both P0 and P1, and we infer fromthe assumption that P is solvable that P0 = P1. Since ∅ 6= MOD(Y0) ⊆ P0, it follows thatλσ .X +σ solves P1 in e.

Case 2: content(e) 6= content(e0). Let k0 > 0 be least such that content(e[k0]) 6⊆content(e0). Then by (53) and the definition of + and X, for all k ≥ k0, X + e[k] =∧e[k] → ϕ |ϕ ∈ Λ(e[k]) ∪ content(e[k]), hence X + e[k] |= Λ(e[k]). Since Λ solves P1 on

e, so does λσ .X +σ.


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(15) Theorem: Suppose that Obs = Lbasic . There is a maxichoice revisionfunction + with the following property. Let problem P be such that for someY ⊆ Lform and revision function ⊕ , λσ . Y ⊕σ solves P. Then there is a consis-tent X ⊆ Lform such that λσ .X +σ solves P efficiently.

Proof: Let problem P satisfy the hypothesis of the theorem. Then P is solvable andfeasible. By Theorem (79) of the second essay, we can assume that the linguistic scientist Λconsidered in the proof of Theorem (14) solves P strongly efficiently. Let revision function +

and X ⊆ Lform be the revision function and formula-set defined in the proof of Proposition(48). Using Lemma (108) of essay #2, and the fact that Λ solves P strongly efficiently, weinfer easily from the proof of Proposition (48) that λσ .X +σ solves P strongly efficiently.Hence, λσ .X +σ solves P efficiently.


(16) Theorem: Suppose that Sym is limited to a binary predicate and count-ably many constants and Obs = Lbasic . Then there exists a definite maxichoicerevision function + , and propositions P1, P2 that meet the following conditions.

(a) P1, P2 is solvable and each of P1, P2 is closed under elementary equiva-lence (hence, P1, P2 is feasible).

(b) for all B ⊆ Lform , λσ .B +σ does not solve P1, P2.

Proof: Let R be the binary predicate of Sym. Let n |n ∈ N enumerate the constantsof Sym. Let S be the structure with domain N that interprets n as n, for all n ∈ N , and




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such that RS is the usual total order on N . Denote by θ a constant-free sentence such thatS |= θ, and U |= ¬θ for all finite structures U . Define P1 to be the set of all structureselementary equivalent to S, P2 the set of all finite structures. It is immediate that eachof P1, P2 are closed under elementary equivalence. Moreover, it is straightforward to showthat P1, P2 has tip-offs, so Proposition (38) of the second essay implies that P1, P2 issolvable.

Now we define + . Fix an environment e for S. For all k ∈ N , let S(ω×k)+i | i ∈N, k + i 6= 0 be an enumeration of all finite D ⊆ Lform such that D ∪ content(e[k]) isconsistent and D ∪ content(e[k]) |= θ. Fix an enumeration ϕi | i ∈ N of Lform , and setSω2+i = ϕi for all i ∈ N . Let S = Sα | 0 < α < ω2 + ω, and let + be constructed fromS as in Definition (42). So + is definite maxichoice by Proposition (43). Let B ⊆ Lform begiven, and suppose that λσ .B +σ solves P1 in e. To verify (16)b, it suffices to show thatfor some environment e′ for P2, λσ .B +σ does not solve P2 in e′. For this purpose, we needthe following fact.

(54) Fact: Let consistent and finite D ⊆ Lform be given. Then there is U ∈ P2 and fullassignment h to U such that U |= β ∈ Lbasic |D |= β[h].

Proof of Fact (54): Let consistent and finite D ⊆ Lform be given. Let X1 = D ∪ 0 =n |n does not appear in D. Then X1 is consistent. Let w0, w1 . . . be an enumeration with-out repetition of the cofinitely many variables that do not appear in D. Then X2 =X1 ∪ wn = n |n ∈ N is consistent. Let X3 be a maximally consistent extension of X2 inX2 ∪Lbasic . By a familiar construction due to Leon Henkin (used for proving the existenceof models [7]) it is clear that there is finite structure U and full assignment h to U such thatU |= (X3 ∩ Lbasic)[h]. So, U and h satisfy Fact (54).

Since λσ .B +σ solves P1 in e, MOD(B + e[k]) ⊆ P1 for some k ∈ N . Hence thereis smallest k0 ∈ N such that for some Y ⊆ B, Y ∪ content(e[k0]) is consistent and Y ∪



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content(e[k0]) |= θ. By compactness, let i0 ∈ N (with i0 6= 0 if k0 = 0) be smallest suchthat:

(55) (a) S(ω×k0)+i0 ⊆ B and S(ω×k0)+i0 ∪ content(e[k0]) is consistent;

(b) S(ω×k0)+i0 ∪ content(e[k0]) |= θ.

By (55)a and Fact (54), let structure U ∈ P2 and full assignment h to U be such thatU |= β ∈ Lbasic |S(ω×k0)+i0 ∪ content(e[k0]) |= β[h]. Let e′ be an environment for U andh that extends e[k0]. It follows from (55)a and the choice of e′ that for all k ≥ k0, thereis Y ⊆ B such that Y ∪ content(e′[k]) is consistent and S(ω×k0)+i0 ⊆ Y . This with thechoice of k0, i0 implies for all k ≥ k0, S(ω×k0)+i0 ⊆ B + e′[k]. With (55)b we infer that forall k ≥ k0, B + e′[k] |= θ. However, since | U | is finite, U 6|= θ by hypothesis. Hence for allk ≥ k0, B + e′[k] 6⊆ P2. Hence λσ .B +σ does not solve P2 in e′, as required.


(56) Proposition: (17) Proposition: Suppose that Obs is closed under negation. Forevery solvable r.e. problem (T , θ0 . . . θn), there is consistent X ⊆ Lform with X |=T , and definite maxichoice revision function + such that λσ .X +σ is computableand solves (T , θ0 . . . θn).

We need the following lemma.

(57) Lemma: Let B ⊆ Lform , σ ∈ SEQ, and revision function + be given. Suppose thereis C ⊆ B such that: (a) for every χ ∈ C, χ |= ¬

∧σ, and (b) B −C ∪ content(σ) is

consistent. Prove that B +σ = (B − C) ∪ content(σ).


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Proof: Condition (a) implies that Inn(B, σ) ∩ C = ∅. With Condition (b) it follows thatInn(B, σ) = B − C. The exercise follows immediately.

Proof of Proposition (17): Let T 0 be an r.e. axiomatisation of T . By Corollary (61) essay#2, let refutable formula ϕ0(x0) (with free variables x0) be such that T |= ∃x0ϕ0(x0) → θ0

and T ∪ ϕ0(x0) is consistent. Let ϕi(xi) | i ∈ N be a recursive enumeration of allrefutable formulas ϕ(x) such that for some i ≤ n, T |= ∃xϕ(x) → θi. Note that theenumeration starts off with the formula ϕ0(x0) chosen to be consistent with T . Set X =T 0 ∪ ϕ0(x0) ∨ . . . ∨ ϕi(xi) | i ∈ N. Since T 0 ∪ ϕ0(x0) is consistent, so is X.

Now we define + . Let partial function f : SEQ → N be defined as follows. Let σ ∈ SEQbe given. If there is no i ∈ N such that T ∪ϕ0(x0)∨ . . .∨ϕi(xi)∪content(σ) is consistent,then f(σ) is undefined; otherwise, f(σ) is the least i ∈ N such that T ∪ ϕ0(x0) ∨ . . . ∨ϕi(xi) ∪ content(σ) is consistent. Since T is decidable and since ϕi(xi) | i ∈ N is arecursive enumeration, f is a partial recursive function. Let S0 = T 0 and for all i ∈ N , letSi+1 = ϕ0(x0) ∨ . . . ∨ ϕi(xi). Let S = Si | i < ω, and let + be constructed from S asin Definition (42). So + is maxichoice and definite by Proposition (43). It is easy to verifythat for all σ ∈ SEQ, the following holds. If f(σ) is undefined then X −¬

∧σ = T 0.

If f(σ) is defined then X −¬∧σ = T 0 ∪ ϕ0(x0) ∨ . . . ∨ ϕi(xi) | i ≥ f(σ). Since f is

partial recursive and since T 0 is r.e., it follows immediately that λσ .X +σ is computable.

Now we show that for every revision function + (whether or not it is computable),λσ .X +σ solves (T , θ0 . . . θn). This suffices to complete the proof. Let revision function+ , j ≤ n, S ∈ MOD(T ∪ θj), full assignment h to S, and environment e for S and h begiven. We must show that:

(58) for cofinitely k, X + e[k] |= T ∪ θj.




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By Theorem (59) of the second essay, let i0 be least with S |= ϕi0(xi0)[h]. So, because theθi’s are mutually exclusive in T , we have:

(59) T |= ϕi0(xi0) → θj .

By the choice of i0, S |= T ∪ ϕ0(x0)∨ . . .∨ϕi0(xi0)[h], and for all i < i0, S |= ¬ϕi(xi)[h].Since the ϕi’s are ∀ formulas, and since h is onto | S |, it follows that there is k0 ∈ N suchthat for all k ≥ k0:

(60) (a) for all i < i0, ϕ0(x0) ∨ . . . ∨ ϕi(xi) |= ¬∧e[k], and

(b) T 0 ∪ ϕ0(x0) ∨ . . . ∨ ϕi(xi) | i ≥ i0 6|= ¬∧e[k].

With Lemma (57), this implies that for all k ≥ k0, X + e[k] = T 0∪ϕ0(x0)∨ . . .∨ϕi(xi) | i ≥i0 ∪ content(e[k]). Hence for all k ≥ k0, T 0 ∪ ϕ0(x0) ∨ . . . ∨ ϕi0(xi0) ∪ content(e[k]) ⊆X + e[k]. With (60)a, we infer that for all k ≥ k0, X + e[k] |= T ∪ϕi0(xi0). This with (59)yields (58).


(18) Proposition: Suppose that Sym is limited to the vocabulary of arithmetic(including 0 and a unary function symbol s) plus an additional constant. Alsosuppose that Obs = Lbasic . Then there is a problem P with the followingproperties.

(a) Every member of P is strongly elementary.

(b) P is computably solvable.



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(c) some revision-based scientist solves P.

(d) For every B ⊆ Lform and revision function + , if λσ .B +σ is computable,then it fails to solve P.

Proof: In view of Proposition (48), clauses (a) and (b) of the present proposition entail and(c). So we prove parts (a), (b), and (d).

Let a be the additional constant of Sym. For n ∈ N , let n be the result of n applicationsof s to 0. Let Q be the seven axioms of Robinson’s arithmetic (see [5, Ch. 14]) and let Nbe the standard model of arithmetic. It is well known that every recursively enumerablesubset of N is weakly representable in Q by a formula which defines it in N . So, let Y ⊆ Nbe a non-recursive, recursively enumerable set and let ϕ(x) ∈ Lform have one free variablex, exclude a, and be such that for all n ∈ N , both

(61) Q |= ϕ(n) iff n ∈ Y,

and

(62) N |= ϕ(n) iff n ∈ Y.

For n ∈ N , if n ∈ Y then set proposition Pn equal to MOD(∧Q ∧ a = n); otherwise, if

n 6∈ Y , set proposition Pn equal to MOD(∧Q ∧ a = n ∧ ¬ϕ(a)). It follows immediately

from (62) that for all n ∈ N , Pn 6= ∅.

Let P = Pn |n ∈ N. The verification of (18)a is immediate. For (18)b, we describe acomputable scientist Ψ that solves P. Let σ ∈ SEQ be given. If for all n ∈ N ,

∧σ 6|= a = n,

then Ψ(σ) is undefined. Otherwise, let n ∈ N be least such that∧σ |= a = n. Then Ψ(σ)

is an index for ∧Q ∧ a = n ∧ ¬ϕ(a) if n does not appear in some standard enumeration


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of Y within length(σ) steps; otherwise, Ψ(σ) is an index for ∧Q ∧ a = n. It is easy to

see that Ψ solves every member of P.

For (18)d, we proceed as follows. Let T be the set of σ ∈ SEQ such that N satisfiesthe set of a-free formulas contained in content(σ). It is easy to verify that T is recursivelyenumerable. For a contradiction, suppose that B ⊆ Lform and revision function + are suchthat λσ .B +σ is computable and solves P. Let n ∈ N be given with n ∈ Y, and supposethat the first member of nonvoid σ ∈ T is a = n. Then, by (61),

∧σ |=

∧Q → ϕ(a), so

Definition (3)b implies that B +σ 6|=∧Q ∧ ¬ϕ(a). This shows:

(63) If n ∈ Y , then for all σ ∈ T that begin with a = n, B +σ 6|=∧Q ∧ ¬ϕ(a).

Now suppose that n 6∈ Y . Let S ∈ Pn be given such that N is the a-free reduct of S [suchan S exists by (62)]. Let e be an environment for S that begins with a = n. Then, sinceλσ .B +σ solves Pn, there is k ∈ N such that B + e[k] |=

∧Q∧¬ϕ(a). Moreover, N satisfies

every a-free formula of content(e[k]). This shows:

(64) If n 6∈ Y , then there is σ ∈ T that begins with a = n and is such that B +σ |=∧Q ∧ ¬ϕ(a).

However, in view of the recursive enumerability of T and the computability of λσ .B +σ,the conjunction of (63) and (64) yields a positive test for the complement of Y, whichcontradicts the hypothesis that Y is not recursive.


(20) Proposition: Suppose that Sym is limited to the binary predicate R.Suppose that Obs = Lbasic . Let T = ∃x∀yRxy ↔ ¬∃y∀xRxy, and θ =


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∃x∀yRxy. Then (T , θ,¬θ) is solvable, but for all revision functions + and allB ⊆ Lsen consistent with T , λσ .B +σ does not solve (T , θ,¬θ).

Proof: Corollary (58) of essay #2 shows that (T , θ,¬θ) is solvable. Choose revisionfunction + , and let B ⊆ Lsen be consistent with T . Because B ∪ T is a consistent set ofclosed formulas, we may choose structure S and full assignment h such that S |= B ∪ T [h].Suppose that S |= ¬θ (the argument is parallel for the case S |= θ). We show that λσ .B +σdoes not solve MOD(T ∪ θ). Let σ ∈ SEQ be given, and suppose that S |=

∧σ[h]. Then

there is an interpretation of B and∧σ that does not satisfy θ. Along with Definition (3)a

this proves:

(65) For all σ ∈ SEQ, if S |=∧σ[h], then B +σ 6|= θ.

By the choice of T and the fact that S |= T ∪ ¬θ, S |= ∃y∀xRxy. Let U be such that| U | = | S | and for all x, y ∈ | U |, (x, y) ∈ RU if and only if (y, x) ∈ RS . Then it can beseen that:

(66) (a) U |= T ∪ θ.(b) For all σ ∈ SEQ, U |=

∧σ[h] if and only if S |=

∧σ[h].

Let e be an environment for U and h. By (65) and (66)b, for all k ∈ N , B + e[k] 6|= θ. Soby (66)a, λσ .B +σ does not solve MOD(T ∪ θ) in e.


(21) Theorem: Suppose that Obs is closed under negation. There exists X ⊆Lform and definite maxichoice revision function + such that for all consistentT ⊆ Lsen , the following holds.



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(a) T ∪X is consistent.(b) λσ . (T ∪X) +σ solves every solvable problem of the form (T , θ0 . . . θn).

Proof: First we define X, then show it to satisfy (21)a. Next we define + , then show it tosatisfy (21)b.

8.8.1. Definition of X.

Let (ϕ1n(x), ϕ2

n(x)) |n ∈ N enumerate all pairs of refutable formulas having all free vari-ables present in the sequence of variables x. For every n ∈ N we fix an enumerationvin | i ∈ N of all sequences of variables such that the following properties hold.

(67) (a) For all i ∈ N , vin is of length equal to the length of x in ϕ1n(x) (and hence of

x in ϕ2n(x)).

(b) No variable is repeated in v0n.

(c) For all m ∈ N , if n 6= m then v0n and v0

m share no variables.

For all n, i ∈ N , we use the following abbreviations.

Hn is: ∃xϕ1n(x) ↔ ¬∃xϕ2

n(x).F in is:

∨r≤i[ ϕ

1n(v

rn) ∨ ϕ2

n(vrn) ].

Now we set X = Hn → F in |n, i ∈ N.

8.8.2. Proof of (21)a.

Let consistent T ⊆ Lsen be given. Let structure S be such that S |= T . We exhibit anassignment h to S such that S |= X[h]. (It is not assumed that h will end up being onto


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|S|.) For all n ∈ N such that S |= Hn, if p = length(v0n) and if variables x1 . . . xp are such

that v0n = (x1 . . . xp), then choose s1 . . . sp ∈ |S| with S |= ϕ1

n ∨ ϕ2n[s1/x1 . . . sp/xp], and set

h(xi) = si for all i, 0 < i ≤ p. This condition on h can be satisfied because of (67). For allvariables x such that h(x) is not defined by the foregoing, h(x) may be selected arbitrarily.Hence for all n ∈ N , if S |= Hn then S |= F 0

n [h]. Since for all n, i ∈ N , F 0n |= F in, this proves

that S |= X[h].

8.8.2.1. Definition of + . Fix an enumeration θi | i > 0 of Lsen , an enumerationϕi | i ∈ N of X, and an enumeration ψi | i ∈ N of Lform . For all i > 0, set Si = θi,and for all i ∈ N , set Sω+i = ϕi, and S(ω×2)+i = ψi. Set S = Sα | 0 < α < ω× 3, andlet +S be constructed from S as in Definition (42). So +S is definite and maxichoice byProposition (43). As an immediate consequence we have the following.

(68) Let T ⊆ Lsen and σ ∈ SEQ be such that T ∪ content(σ) is consistent. ThenT ⊆ (T ∪X) +S σ.

8.8.2.2. Proof of (21)b. Let solvable problem of form (T , θ0 . . . θn) be given. Letm ≤ n also be given. By Corollary (61) of essay #2, there is n ∈ N such that T |=θm ↔ ∃xϕ1

n(x) and T |= ¬θm ↔ ∃xϕ2n(x). Hence T |= Hn. Let S ∈ MOD(T ∪ θm)

and full assignment h to S be given. (The case S ∈ MOD(T ∪ ¬θm) is parallel.) Thenwe may choose the least i ∈ N such that S |= ϕ1

n(vin)[h]. So S |= Hn → F in[h]. Let e

be an environment for S and h. In view of Definition (3)b (which ensures consistency), tocomplete the proof it suffices to show:

(69) (a) for cofinitely many k, (T ∪X) +S e[k] |= Hn;

(b) for cofinitely many k, (T ∪X) +S e[k] |= Hn → F in;



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(c) for cofinitely many k, (T ∪X) +S e[k] |= ¬F i−1n ∧ ¬ϕ2

n(vin).

Claim (69)c follows immediately from the choice of i and the choice of e. Claim (69)afollows immediately from (68), the choice of e, and the fact that T |= Hn. So it remains todemonstrate (69)b.

Let j ∈ N be least such that ϕj = Hn → F in. Let X0 be the set of all ϕp, p < j suchthat S |= ϕp[h]. By our choice of e, and the fact that S |= T and S |= Hn → F in[h], wehave:

(70) For all k > 0, T ∪X0 ∪ Hn → F in 6|= ¬∧e[k].

Since the pairs (ϕ1n(x), ϕ2

n(x)) |n ∈ N are refutable [see Definition (30) of essay #2].Lemma (91)b of essay #2]. implies that for all i, n, F in is refutable. Since h is a fullassignment to S we may choose k0 > 0 such that:

(71) For all p < j, if ϕp = Hn′ → F i′n′ then the following holds: S |= F i

′n′ [h] iff F i

′n′ 6|=

¬∧e[k0].

Let k ≥ k0 be given. Consider Y ω+j in the construction of (T ∪ X) +S e[k] via Definition(42). By (68), (71), and the choice of k, Y β ⊆ T ∪ X0 for all nonnull β < ω + j. So by(70), Y β ∪ Hn → F in 6|= ¬

∧e[k] for all nonnull β < ω + j. Hence, Hn → F in ∈ Y ω+j ⊆

(T ∪X) +S ¬∧e[k], verifying (69)b.


(22) Theorem: Suppose that solvable problem P is of form (T , P0, P1, . . .),and that Obs is closed under negation. Then there is a consistent extensionX ⊆ Lform of T such that for every revision function + , λσ .X +σ solves P.




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Proof: By Proposition (40) of the second essay, and our hypothesis on Obs, for all j ∈ Nlet tj be a tip-off for Pj in P that satisfies the condition stated in Lemma (33) of essay#2, By the countability of tip-offs, let

⋃j∈N tj be enumerated as πi | i ∈ N. Without

loss of generality we may assume that each πi is consistent with T . For all i ∈ N fix anenumeration ϕni |n ∈ N of the refutable formulas in πi. Then set:

X = T ∪ (ϕ00 ∧ . . . ∧ ϕ

n00 ) ∨ . . . ∨ (ϕ0

i ∧ . . . ∧ ϕnii ) | i, n0 . . . ni ∈ N.

We show that X satisfies the claim of the theorem.

Note that T∪ϕn0 |n ∈ N is consistent by hypothesis. This implies that X is consistent.

Let revision function + , j ∈ N , S ∈ Pj , full assignment h to S, and environment e forS and h be given. To finish the proof we must show that:

(72) for cofinitely k, ∅ 6= MOD(X + e[k]) ⊆ Pj .

By Definition (32) of essay #2, let i0 be least with S |= πi0 [h]. For i < i0, S 6|= πi[h], so foreach i < i0 we may choose c(i) ∈ N such that:

(73) (a) S |= (ϕ0i ∧ . . . ∧ ϕ

c(i)−1i )[h];

(b) S 6|= ϕc(i)i [h].

Since for all i < i0, ϕc(i)i is refutable, and since h is onto | S |, it follows from (73)b that

there is k0 > 0 such that:

(74) for all k ≥ k0 and for all i < i0, content(e[k]) ∪ ϕc(i)i is inconsistent.

Let Y = (ϕ00 ∧ . . .∧ϕ

n00 )∨ . . .∨ (ϕ0

i ∧ . . .∧ϕnii ) | i < i0, n0 ≥ c(0) . . . ni ≥ c(i). We deduce

from (74) that:






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(75) for all k ≥ k0 and for all ϕ ∈ Y , content(e[k]) ∪ ϕ is inconsistent.

We deduce from (73)a and the fact that S |= πi0 [h] that:

(76) for all k ∈ N , content(e[k]) ∪ (X − Y ) is consistent.

Then (75), (76), and Lemma (57) imply that:

(77) for all k ≥ k0, X + e[k] = content(e[k]) ∪ (X − Y ).

Now note that Z = (ϕ00∧. . .∧ϕ

c(0)0 )∨. . .∨(ϕ0

i0−1∧. . .∧ϕc(i0−1)i0−1 )∨(ϕ0

i0∧. . .∧ϕni0) |n ∈ N ⊆

X − Y . Moreover, (74) implies that content(e[k]) ∪ Z |= ϕni0 |n ∈ N for all k ≥ k0. Thiswith (77) implies that X + e[k] |= T ∪ϕni0 |n ∈ N for all k ≥ k0. Hence X + e[k] |= T ∪πi0for all k ≥ k0. This with Definition (32) of essay #2, and the hypothesis that S |= πi0 [h]implies (72).


(23) Proposition: Suppose that Obs is closed under negation. Let T ⊆ Lsen

be consistent and finitely axiomatizable. Then there is consistent X ⊆ Lform

with X |= T such that for every revision function + , λσ .X +σ solves everysolvable problem of form (T , θ0 . . . θn).

We need:

(78) Lemma: Let B ⊆ Lform , σ ∈ SEQ and revision function + be given. Suppose thatfor all ψ, χ ∈ B, either ψ |= χ, or χ |= ψ. Show that B +σ = (B − Y ) ∪ content(σ),where Y = ψ ∈ B |ψ 6|= ¬

∧σ.



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Proof: Trivially, Inn(B, σ) ∩ Y = ∅. So it suffices to show that B − Y 6|= ¬∧σ. Suppose

otherwise. By compactess there is finite D ⊆ B − Y such that D |= ¬∧σ. By hypothesis

there is ψ ∈ D such that for all χ ∈ D, ψ |= χ. Hence ψ |= ¬∧σ, so ψ ∈ Y , contradiction.

Proof of Proposition (23): First we define X, using some notation. Fix an enumerationϕm |m ∈ N of all refutable formulas. Let (n1 . . . nk) ∈ Nk be a strictly increasingsequence of k numbers. By induction, we associate a formula χni with each ni appearingin (n1 . . . nk). It will turn out that χni is a disjunction over ϕm |m ∈ N, and that ϕni isthe disjunct of highest index in χni .

Basis step: χn1 is ϕ0 ∨ . . . ∨ ϕn1 . Observe that χn1 is a disjunction over ϕm |m ∈ N,and that ϕn1 is the disjunct of highest index in χn1 .

Induction step: Let 0 < i < k be given. Then ni+1 > ni. Suppose that χni has beendefined to be a disjunction over ϕm |m ∈ N such that ϕni is the disjunct of highest indexin χni . Let χ′ be the result of suppressing ϕni in χni . Then χni+1 is χ′ ∨ϕni+1 ∨ . . .∨ϕni+1 .Observe that χni+1 is a disjunction over ϕm |m ∈ N, and that ϕni+1 is the disjunct ofhighest index in χni+1 .

Finally, we associate the formula χn1 ∧ . . . ∧ χnkwith (n1 . . . nk). For example, the

formula associated with (2, 5, 7) is:

(ϕ0 ∨ ϕ1 ∨ ϕ2) ∧ (ϕ0 ∨ ϕ1 ∨ ϕ3 ∨ ϕ4 ∨ ϕ5) ∧ (ϕ0 ∨ ϕ1 ∨ ϕ3 ∨ ϕ4 ∨ ϕ6 ∨ ϕ7).

Let t ∈ Lsen axiomatize T . Then we define X to be the set of all consistent formulas ofform t ∧ χ, where χ is any formula associated with some nonempty, increasing sequence ofnatural numbers.

The key to our construction is the following fact.


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(79) Fact: For every δ, δ′ ∈ X, δ |= δ′ or δ′ |= δ.

Proof of Fact (79): Let γ and γ′ be distinct, nonempty, finite, increasing sequences ofnumbers. Then there is λ ∈ N<ω and n ∈ N such that exactly one of the following holds:

(a) λ∗n is an initial segment of γ and for all m ≤ n, λ∗m is not an initial segment of γ′;

(b) λ∗n is an initial segment of γ′ and for all m ≤ n, λ∗m is not an initial segment of γ.

Suppose λ and n satisfy (a); the other case is exactly parallel. Let formula χ be associatedwith γ, and formula χ′ be associated with γ′. Then it can be seen that the first length(λ)conjuncts in χ are the same as the first length(λ) conjuncts in χ′, and that all of the disjunctsappearing in the length(λ) + 1st conjunct of χ appear as disjuncts in each of the conjunctsof χ′ that come after the first length(λ) ones. So χ |= χ′. Hence t ∧ χ |= t ∧ χ′, and allδ ∈ X are of the form t ∧ χ.

We now show that X satisfies the claim of the proposition. Since every member of Xis consistent, it follows from Fact (79) that X is consistent. Let there be given solvableproblem of form (T , θ0 . . . θn), revision function + , and i ≤ n. To finish the proof weshow that λσ .X +σ solves MOD(T ∪ θi). Let S ∈ MOD(T ∪ θi), full assignment h toS, and environment e for S and h be given. By our hypothesis on Obs and Corollary (61)of essay #2, there is θ ∈ Lsen such that θ is the existential closure of a refutable formulaand:

(80) T |= θi ↔ θ.

Since h is onto | S |, this implies the existence of m0 ∈ N such that the existential closureof ϕm0 is θ, and S |= ϕm0 [h]. From Lemma (78) and Fact (79) we deduce that:



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(81) for all k ∈ N , X + e[k] includes the set of all formulas in X that are consistent with∧e[k].

Denote by γ the increasing sequence of integers that ends with m0 and such that for allm ≤ m0, m occurs in γ if and only if S |= ϕm[h]. Denote by χ the formula associated withγ. It is immediate that t ∧ χ ∈ X, and that for all k ∈ N , t ∧ χ is consistent with

∧e[k].

We can thus deduce from (81) that:

(82) for all k ∈ N , t ∧ χ belongs to X + e[k].

Since h is onto | S |, there is k0 ∈ N such that for all k ≥ k0 and for all m ≤ m0, ifm 6∈ content(γ) then

∧e[k] |= ¬ϕm. From the definition of χ this can be seen to imply that

for all k ≥ k0 and for all m ∈ content(γ), content(e[k]) ∪ χ |= ϕm. In particular:

(83) for all k ≥ k0, content(e[k]) ∪ χ |= ϕm0 .

From (80), (82), (83), and the definition of ϕm0 we deduce that for all k ≥ k0, X + e[k] |=T ∪ θi.


(24) Proposition: Suppose that Sym is limited to a binary predicate, a con-stant, and a unary function symbol. Suppose that Obs = Lbasic . Then thereexists T ⊆ Lsen and definite maxichoice revision function + such that for allB ⊆ Lform , λσ .B +σ fails to solve some solvable problem of form (T , θ,¬θ).


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Proof: Let R be the binary predicate, 0 the constant, s the unary function symbol of Sym.For n ∈ N , let n be the result of n applications of s to 0.

Let χi | i > 0 enumerate Lform , and for all i > 0 set Si = χi. Let S = Si | 0 <i < ω, and let +S be constructed from S as in Definition (42). So +S is definite andmaxichoice by Proposition (43). Fix a nonrecursive subset E of N , and define T to be:

(∀yR2ny) → 2n = 2m+ 1 |n ∈ N and m ∈ E ∪ (∀yR2ny) → 2n 6= 2m+ 1 |n ∈ N and m 6∈ E

Let B ⊆ Lform be given. By our assumption on Obs and Corollary (58) of essay #2, forevery universal sentence θ, (T , θ,¬θ) is solvable. So there would be nothing left to provewithout assuming:

(84) For every universal sentence θ, λσ .B +S σ solves (T , θ,¬θ).

We use (84) to derive a contradiction. Specifically, an environment e for T will be exhibitedsuch that for infinitely many k, there exists n,m ∈ N with:

(85)B +S e[k] |= (∀yR2ny) ∧ 2n 6= 2m+ 1 if m ∈ E,B +S e[k] |= (∀yR2ny) ∧ 2n = 2m+ 1 if m 6∈ E.

This will imply that for infinitely many k, B +S e[k] 6|= T . So, trivially, λσ.B +S σ does notsolve (T , ∀x(x = x),¬∀x(x = x)), contradicting (84).

Fix an enumeration αi | i ∈ N of all atomic formulas. To exhibit the promised envi-ronment e we will build by induction on i ∈ N a sequence τi | i ∈ N of members of SEQand a sequence ni | i ∈ N such that for all i ∈ N :



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(86) (a) for all j < i, τj ⊆ τi;

(b) exactly one of αi or ¬αi occurs in τi;

(c)∧τi |= ¬∀yR2iy;

(d) B +S τi |= ∀yR2niy;

(e) for some m ∈ E, B +S τi |= 2ni 6= 2m+ 1, or for some m 6∈ E, B +S τi |= 2ni =2m+ 1.

Then we will set e =⋃i∈N τi. It follows from (86)a,b that e is an environment, which by

(86)c is for T . It follows from (86)d,e that (85) is satisfied. Let i ∈ N be given. Supposethat τj and nj have been defined for all j < i and satisfy (86) for i = j. We define τi andni that satisfy (86). Choose ni 6= i such that 2ni does not appear in τj for any j < i. Wemay choose an environment d for T ∪ ¬∀yR2iy, ∀yR2niy that extends τj for all j < i.By (84) we deduce the existence of k0 ∈ N with:

(87) (a) for all j < i, d[k0] extends τj ;

(b) exactly one of αi, ¬αi occurs in d[k0];

(c)∧d[k0] |= ¬∀yR2iy;

(d) B +S d[k0] |= ∀yR2niy.

By (87)d and compactness, let finite D ⊆ (B +S d[k0]) ∩ B be such that D ∪ ∧d[k0] |=

∀yR2niy. Denote by p the greatest integer such that χp ∈ D, and set D′ = χ0 . . . χp ∩(B +S d[k0]) ∩B. So:

(88) D ⊆ D′ ⊆ B +S d[k0].


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Since E is non recursive there is m ∈ E with D′ ∪ ∧d[k0] 6|= 2ni = 2m+ 1, or there is

m 6∈ E with D′ ∪ ∧d[k0] 6|= 2ni 6= 2m+ 1. Choose such an m, and set τi = d[k0]∗(2ni 6=

2m+ 1) if m ∈ E, τi = d[k0]∗(2ni = 2m+ 1) if m 6∈ E. From (87)a-c we infer (86)a-cimmediately. By (88), the definition of +S and the choice of m we obtain: D′ ⊆ (B +S τi)∩B, hence B +S τi |= ∀yR2niy, verifying (86)d. By Lemma (4), B +S τi |= (2ni 6= 2m+ 1) ifm ∈ E, and B +S τi |= (2ni = 2m+ 1) if m 6∈ E, verifying (86)e.


It will be convenient to prove something a little stronger, namely:

(25) Strengthened Theorem: Suppose that Obs is closed under negation.For every solvable problem P of the form (T , P0, P1, . . .) there is a consistentextension X ⊆ Lform of T and a definite maxichoice revision function + suchthat λσ .X +σ solves P strongly efficiently.17

Proof: Let solvable problem P = (T , P0, P1, . . .) be given. Recall from the proof ofTheorem (22) the enumeration πi | i ∈ N and the enumeration ϕni |n ∈ N of πi. Recallalso the definition of X ⊆ Lform . For all i ∈ N , set Si+1 = T ∪(ϕ0

0∧ . . .∧ϕn00 )∨ . . .∨ (ϕ0

i ∧. . .∧ϕni

i ) |n0 . . . ni ∈ N. Fix an enumeration ψi | i ∈ N of Lform , and set Sω+i = ψi forall i ∈ N . Let S = Sα | 0 < α < ω × 2, and let + be constructed from S as in Definition(42). So, + is definite maxichoice by Proposition (43). The proof of Theorem (22) showsthat λσ .X +σ solves P. To complete the proof of the theorem we show that λσ .X +σsolves P efficiently.

Let σ ∈ SEQ be for P. By Definition (32) of essay #2, there is least i0 ∈ N such thatT ∪ πi0 ∪ content(σ) is consistent. Let P ∈ P, S ∈ P , and full assignment h to S be such

17Strong efficiency was introduced in Definition (78) of essay #2.




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that S |= πi0 ∪ content(σ)[h]. Let τ ∈ SEQ extend σ and be such that S |=∧τ [h]. As an

immediate consequence of our choice of + , we have:

Si0+1 ⊆ X + τ .

Moreover by compactness, for all i < i0 we may choose c(i) ∈ N such that T ∪ content(σ)∪ϕ0

i ∧ . . .∧ϕc(i)i is inconsistent. Since (ϕ0

0∧ . . .∧ϕc(0)0 )∨ . . .∨(ϕ0

i0−1∧ . . .∧ϕc(i0−1)i0−1 )∨(ϕ0

i0∧

. . . ∧ ϕni0) |n ∈ N ⊆ Si0+1, this implies that Si0 ∪ content(σ) |= ϕ0i0∧ . . . ∧ ϕni0 |n ∈ N.

With Si0 ⊆ X + τ we deduce that X + τ |= T ∪πi0 which implies via Definition (32) of essay#2 (and for the case σ = 0, the fact that X is consistent) that ∅ 6= MOD(X + τ) ⊆ P .It follows immediately from Lemma (108) of essay #2 that λσ .X +σ solves P stronglyefficiently.

8.13. Proof of Lemma (27)

(27) Lemma: For all T ⊆ Lsen , there exists a T -preserving definite maxichoicerevision function.

Proof: Let T ⊆ Lsen be given. Fix an enumeration ϕi | i > 0 of T and an enumerationψi | i ∈ N of Lform − T . For all i > 0, set Si = ϕi. For all i ∈ N , set Sω+i = ψi. LetS = Sα | 0 < α < ω × 2, and let + be constructed from S as in Definition (42). So, + isdefinite maxichoice by Proposition (43). For all σ ∈ SEQ and X ⊆ Lform , if T ∪ content(σ)is consistent then Inn(T ∪X,σ) ∪ T ∪ content(σ) is consistent. This together with thedefinition of + implies immediately that for all σ ∈ SEQ, if T ∪ content(σ) is consistentthen Inn(T ∪X,σ) ∪ content(σ) ∪ T ⊆ T +σ. Hence + is T -preserving.





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Actually, we’ll prove something a little stronger, namely:

(28) Strengthened Theorem: Suppose that Obs is closed under negation.For every solvable problem P of the form (T , θ0 . . . θn) there is a consistentextension X ⊆ Lform of T with the following properties.


(b) For all T -preserving revision functions + , λσ .X +σ solves P strongly effi-ciently.18

Proof: Let solvable problem P of the form (T , θ0 . . . θn) be given. By Corollary (61) of es-say #2, let refutable formula ϕ0(x0) (with free variables x0) be such that T |= ∃x0ϕ0(x0) →θ0 and T ∪ ϕ0(x0) is consistent. Let ϕi(xi) | i ∈ N be an enumeration of all refutableformulas ϕ(x) such that for some j ≤ n, T |= ∃xϕ(x) → θj . Note that the enumerationstarts off with the formula ϕ0(x0) chosen to be consistent with T . For all i ∈ N , denoteby ψi the formula ϕ0(x0) ∨ . . . ∨ ϕi(xi). Set X = T ∪ ψi | i ∈ N. Since T ∪ ϕ0(x0) isconsistent, so is X. We show that X satisfies clauses (a) and (b) of the theorem.

We prove part (a). Let revision function + , j ≤ n, S ∈ MOD(T ∪ θj), full assignmenth to S, and environment e for S and h be given. We must show that:

(89) for cofinitely k, X + e[k] |= T ∪ θj.

By Corollary (61) of essay #2, let i0 be least with S |= ϕi0(xi0)[h]. So, because the θi’s aremutually exclusive in T , we have:







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(90) T |= ϕi0(xi0) → θj .

By the choice of i0, S |= T ∪ ψi0[h], and for all i < i0, S |= ¬ϕi(xi)[h]. Since the ϕi’s arerefutable formulas, and since h is onto | S |, it follows that there is k0 ∈ N such that for allk ≥ k0:

(91) (a) for all i < i0, ψi |= ¬∧e[k], and

(b) T ∪ ψi | i ≥ i0 6|= ¬∧e[k].

With Exercise (57), this implies that for all k ≥ k0, X + e[k] = T∪ψi | i ≥ i0∪content(e[k]).Hence for all k ≥ k0, T ∪ ψi0 ∪ content(e[k]) ⊆ X + e[k]. With (91)a, we infer that for allk ≥ k0, X + e[k] |= T ∪ ϕi0(xi0). This with (90) yields (89).

We prove part (b). Let T -preserving revision function + be given. Let σ ∈ SEQ be for(T , θ0 . . . θn). Let i0 ∈ N be least such that:

(92) T ∪ ψi0 6|= ¬∧σ.

By the definition of the ϕi’s, there is j ≤ n such that:

(93) T ∪ ϕi0 |= θj .

By (92), let structure S and full assignment h to S be such that S |=∧σ[h] and:

(94) S |= ψi0 [h].

Let environment e for S and h extend σ, and let k ≥ length(σ) be given. The definition ofi0, the fact that e is an environment for S and h, and (94) then imply that:


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(95) (a) for all i < i0, T ∪ ψi |= ¬∧e[k], and

(b) T ∪ ψi | i ≥ i0 6|= ¬∧e[k].

Since + is T -preserving, T ⊆ X + e[k]. Together with (95), this implies that X + e[k] =T ∪ ψi | i ≥ i0. Hence ψi0 ∈ X + e[k]. With the definition of i0, it follows that X + e[k] |=T ∪ ϕi0. With (93), we conclude that X + e[k] |= T ∪ θj. So we have shown that for allk ≥ length(σ), ∅ 6= MOD(X + e[k]) ⊆ MOD(T ∪ θj). We conclude with Lemma (108) ofessay #2.

8.15. Proof of Corollary (29)

Once again, we prove something slightly stronger.

(29) Strengthened Corollary: Suppose that Obs is closed under negation.Let T ⊆ Lsen be consistent and finitely axiomatizable. Let solvable problem ofform (T , θ0 . . . θn) be given. Then there is consistent extension X ⊆ Lform ofT such that for every revision function + , λσ .X +σ strongly efficiently solves(T , θ0 . . . θn).19

Proof: Recall the enumeration ψi | i ∈ N defined from (T , θ0 . . . θn) in Theorem (28).Let t ∈ Lsen axiomatize T . Set X = T ∪ t ∧ ψi | i ∈ N. Let revision function + be given.It is easy to verify that for all σ ∈ SEQ, if σ is consistent with T then T ⊆ Inn(X,σ).An immediate adaptation of the proof of Theorem (28) then shows that λσ .X +σ stronglyefficiently solves (T , θ0 . . . θn).





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(30) Theorem: Suppose that Sym is limited to countably many constantsand Obs = Lbasic . Then there is a problem of form (T , θ,¬θ) and a definitemaxichoice revision function + with the following properties.

(a) T is recursive.

(b) (T , θ,¬θ) is solvable.

(c) For all B ⊆ Lform , if λσ .B +σ solves (T , θ,¬θ) then λσ .B +σ is domi-nated on (T , θ,¬θ).

Proof: Let a, b, and n, n ∈ N , be distinct constants. (In view of (1) of essay #2 weassume that this can be done via a total computable isomorphism between N and the setof constants.) Choose a recursive subset E of N that is not primitive recursive. We take θto be a = b, and T to be the following set of sentences:

θ ↔ a = n, for all n ∈ E,θ ↔ a 6= n, for all n 6∈ E.

It is immediate that (30)a,b are satisfied. For (30)c, we say that B ⊆ Lform disagrees with Tif there is n ∈ N with either (n ∈ E and B |= a 6= n) or (n ∈ E and B |= a = n); otherwise,we say that B agrees with T . Fix an enumeration ϕi | i > 0 of Lform , and set Si = ϕifor all i > 0. Let S = Si | 0 < i < ω, and let + be constructed from S as in Definition(42). So, + is definite maxichoice by Proposition (43). To verify (30)c, let B ⊆ Lform begiven, and define Ψ = λσ .B +σ. We suppose that Ψ solves (T , θ,¬θ) since otherwisethere is nothing left to prove. To complete the proof it must be shown that Ψ is dominatedon (T , θ,¬θ). Let P1 = MOD(T ∪ θ), P2 = MOD(T ∪ ¬θ), and let scientist Θ bedefined as follows.



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(96) For all σ ∈ SEQ,

Θ(σ) =

P1 if T ∪ content(σ) |= θ.P2 if T ∪ content(σ) |= ¬θ.P1 if T ∪ content(σ) 6|= θ, T ∪ content(σ) 6|= ¬θ, and neither

∅ 6= Ψ(σ) ⊆ P1, nor ∅ 6= Ψ(σ) ⊆ P2.Ψ(σ) otherwise.

It follows that Θ solves (T , θ,¬θ). Hence, for all environments e for P ∈ P1, P2,SP(Θ, e, P ) and SP(Ψ, e, P ) are well defined, and SP(Θ, e, P ) ≤ SP(Ψ, e, P ), as easily veri-fied. Thus, by Definition (76) of essay #2, it suffices to show that there is an environmente for some P ∈ P1, P2 such that SP(Θ, e, P ) < SP(Ψ, e, P ). This is demonstrated via thefollowing, exhaustive cases.

(97) (a) B |= θ iff B |= ¬θ.(b) B is consistent, B |= θ, and B disagrees with T .

(c) B is consistent, B |= θ, and B agrees with T .

(d) Same as (b) except that θ is replaced by ¬θ.(e) Same as (c) except that θ is replaced by ¬θ.

If (97)a, let e be any environment for P1 such that for some n ∈ E, e(0) = (a = n).By Definition (3), B + e[0] = B + ∅ = B, so by (97)a, neither ∅ 6= Ψ(e[0]) ⊆ P1 nor ∅ 6=Ψ(e[0]) ⊆ P2. Since T = T ∪ content(e[0]) 6|= θ and T ∪ content(e[0]) 6|= ¬θ, the third clauseof (96) implies that Θ(e[0]) = Θ(∅) = P1. Moreover, for all k > 0, T ∪ content(e[k]) isconsistent and implies θ. Hence, the first clause of (96) implies SP(Θ, e, P1) = 0. On theother hand, since it is not the case that ∅ 6= Ψ(e[0]) ⊆ P1, SP(Ψ, e, P1) > 0.



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If (97)b, then either there is n ∈ E with B |= a 6= n or n ∈ E with B |= a = n. Weconsider these subcases in turn. Suppose n is such that n ∈ E and B |= a 6= n. Let e bean environment for P2 such that e(0) = (a 6= n). Then by Lemma (4) and the fact thatB is consistent, Ψ(e[1]) = B + e[1] |= B, so Ψ(e[1]) |= θ. It follows that SP(Ψ, e, P2) > 1.In contrast, the definition (96) of Θ implies that SP(Θ, e, P2) = 1. Now suppose n is suchthat n ∈ E and B |= a = n. Let e be an environment for P2 such that e(0) = (a = n).Then again by Lemma (4) and the fact that B is consistent, Ψ(e[1]) = B + e[1] |= B, soΨ(e[1]) |= θ. It follows that SP(Ψ, e, P2) > 1, whereas (96) implies SP(Θ, e, P2) = 1.

Suppose that (97)c holds. We rely on the following fact, an immediate corollary to theproof that the monadic predicate calculus is decidable (see [5, Ch. 25]).

(98) Fact: Suppose that Sym is limited to constants. Then the function that associatesto every φ ∈ Lform the value 1 if |= φ, and the value 0 if 6|= φ, is primitive recursive.

Let i0 > 0 be least such that D = B ∩ ϕi | 0 < i < i0 |= θ. Then there exists n ∈ E suchthat D 6|= a = n, or there exists n ∈ E such that D 6|= a 6= n. Indeed, if this were notthe case then (98) would imply that the characteristic function of E is primitive recursive,contradicting the choice of E. We consider the two subcases in turn. Assume n is such thatn ∈ E and D 6|= a = n. Let e be an environment for P2 such that e(0) = (a 6= n). Thenby the choice of i0 and the definition of + , D ⊆ B + e[1] = Ψ(e[1]), hence Ψ(e[1]) |= θ. SoSP(Ψ, e, P2) > 1. In contrast, SP(Θ, e, P2) = 1 since T ∪ content(e[1]) |= ¬θ. Now assume nis such that n ∈ E andD 6|= a 6= n. Let e be an environment for P2 such that e(0) = (a = n).Then, once again, by the choice of i0 and the definition of + , D ⊆ B + e[1] = Ψ(e[1]), henceΨ(e[1]) |= θ. So SP(Ψ, e, P2) > 1, whereas SP(Θ, e, P2) = 1.

The argument for (97)d is parallel to that for (97)b, and the argument for (97)e isparallel to that for (97)c.


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(35) Theorem: Let Sym consist of a binary predicate, and suppose that Obs =Lbasic . Then there exists a problem of form (T , θ,¬θ) (with T finite), andmaxichoice revision function + such that:

(a) (T , θ,¬θ) is solvable;

(b) for all B ⊆ Lform , if B is closed under ψ, φ / ψ → φ, then λσ .B +σ doesnot solve (T , θ,¬θ).

Proof of the theorem relies on the following lemma.

(99) Lemma: Suppose that Y ⊆ Lform is closed under ψ, φ / ψ → φ. Let σ ∈ SEQ andθ ∈ Lsen be given with Y ∪ content(σ) inconsistent and content(σ) 6|= θ. Then thereis maximally consistent Z ⊆ content(σ) ∪ Y such that content(σ) ⊆ Z and Z 6|= θ.

Proof: Suppose that Y ⊆ Lform is closed under ψ, φ / ψ → φ, and let σ, θ be such that:

(100) (a) Y ∪ content(σ) is inconsistent.

(b) content(σ) 6|= θ.

Let Ω = X ⊆ content(σ) ∪ Y |content(σ) ⊆ X ∧X 6|= θ. By (100)(b), Ω is nonempty.It follows immediately from the Compactness Theorem that Ω is closed under unions ofchains and hence, by Zorn’s Lemma, that Ω contains maximal elements (with respect to⊆). So let Z ∈ Ω be such a maximal element, that is, for every ϕ ∈ Y, if ϕ 6∈ Z, thenZ ∪ ϕ |= θ. We must show that Z is a maximal consistent subset of content(σ) ∪ Y suchthat content(σ) ⊆ Z and Z 6|= θ. By construction, Z ⊆ content(σ) ∪ Y , content(σ) ⊆ Z,


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and Z 6|= θ (thus Z is consistent). To show that Z is maximal with the foregoing properties,let ϕ ∈ Y − Z. We must show that Z ∪ ϕ is inconsistent.

By (100)(a) (and because content(σ) ⊆ Z), to show that Z ∪ ϕ is inconsistent, itsuffices to show for every ψ ∈ Y , ϕ → ψ ∈ Z. So fix ψ ∈ Y . By the closure condition onY , ϕ→ ψ ∈ Y . Since Z is a maximal element of Ω, to prove that ϕ→ ψ ∈ Z, it suffices toshow that Z ∪ ϕ→ ψ 6|= θ.

Since Z ∪ ϕ |= θ and Z 6|= θ, there is a structure S and assignment h such thatS |= Z∪¬θ[h] and S 6|= ϕ[h]. So also, S |= ϕ→ ψ[h]. Hence, S, h witness Z∪ϕ→ ψ 6|= θ.

We now exploit Lemma (99) to prove the theorem.

Proof of Theorem (35): Let R be the binary predicate of Sym. Let T = ∃x∀yRxy ↔∀xy(x = y) and θ = ∃x∀yRxy. We claim that (T , θ,¬θ) witnesses the proposition.Clause (35)a is evident. For Clause (35)b, observe:

(101) for every σ ∈ SEQ, content(σ) 6|= ¬θ.

To finish the proof we define a maxichoice revision function + such that for all B ⊆ Lform ,if B is closed under ψ, φ / ψ → φ, then λσ .B +σ does not solve (T , θ,¬θ). It followsdirectly from (101) and Lemma (99) that there exists a maxichoice revision function + withthe following property.

(102) Suppose that B ⊆ Lform is closed under ψ, φ / ψ → φ. Then for every σ ∈ SEQ, ifB ∪ content(σ) is inconsistent, B +σ 6|= ¬θ.

Let environment e be for MOD(T ∪ θ). By compactness:

(103) T ∪ content(e) 6|= θ.


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Let B ⊆ Lform be closed under ψ, φ / ψ → φ. If B + e[k] 6|= θ for all k ∈ N , then λσ .B +σdoes not solve MOD(T ∪ θ), and we are done. So let σ ⊂ e be such that

(104) B +σ |= θ.

By (103) there is a structure U ∈ MOD(T ∪ ¬θ) and a full assignment h to U such thatU |= content(σ)[h]. So we can choose environment e′ for U via h such that σ ⊂ e′. Letk > length(σ) be given. It suffices to show that B + e′[k] 6|= ¬θ. For this purpose wedistinguish two cases.

Case 1: B ∪ content(e′[k]) is consistent. Then by (104), Lemma (4), and the fact thatσ ⊂ e′, B + e′[k] |= θ. Hence, by Lemma (4), B + e′[k] 6|= ¬θ.

Case 2: B ∪ content(e′[k]) is inconsistent. Then (102) implies B + e′[k] 6|= ¬θ.


(37) Proposition: Suppose that Obs is closed under negation. There existsX ⊆ Lform and definite maxichoice revision function + such that for all consis-tent T ⊆ Lsen , the following holds.

(a) T ∪X is consistent.(b) λσ . (T ∪X) +ie σ solves every solvable problem of the form (T , θ0 . . . θn).

Proof: Let X be defined as in the proof of Theorem (21). Fix an enumeration χi | i > 0 ofLbasic ∪ Lsen , an enumeration ϕi | i ∈ N of X, and an enumeration ψi | i ∈ N of Lform .For all i > 0, set Si = χi. For all i ∈ N , set Sω+i = ϕi, and S(ω×2)+i = ψi. LetS = Sα | 0 < α < ω × 3, and let + be constructed from S as in Definition (42). So + isdefinite maxichoice by Proposition (43). Then it is straightforward to adapt the proof ofTheorem (21) for the present proposition.


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(38) Proposition Suppose that Obs is closed under negation. For everysolvable problem P of the form (T , P0, P1, . . .) there is a consistent exten-sion X ⊆ Lform of T and a definite maxichoice revision function + such thatλσ .X +ie σ solves P efficiently.

Proof: Let solvable problem P = (T , P0, P1, . . .) be given. Recall from the proof of The-orem (22) the enumeration πi | i ∈ N and the enumeration ϕni |n ∈ N of πi. Withoutloss of generality, we may suppose that ϕ0

0 6∈ Lbasic . Recall from the proof of Theorem (25)the enumeration Si | i > 0 of subsets of Lform . Fix an enumeration βi | i ∈ N of Lbasic

and an enumeration ψi | i ∈ N of Lform . For all i > 0 set S′i = βi, S′ω+i = Si. Forall i ∈ N set S′(ω×2)+i = ψi. Let S = S′α | 0 < α < ω × 3, and let + be constructedfrom S as in Definition (42). So + is definite maxichoice by Proposition (43). Then it isstraightforward to adapt the proof of Theorem (25) for the present proposition.


(39) Theorem: Suppose that Sym = ∅ and Obs = Lbasic . Then there exists adefinite maxichoice revision function + such that for all B ⊆ Lform , λσ .B +ie σdoes not solve (∅, two,¬two).

Proof of the theorem relies on two lemmas. The first is a general fact about iterativerevision, straightforward to verify.

(105) Lemma: Suppose that Obs = Lbasic . Let B ⊆ Lform and σ ∈ SEQ be given. ThenB +ie σ ⊆ B ∪ content(σ).


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(106) Lemma: Suppose that Obs = Lbasic . Let finite, consistent D ⊆ Lform be such thatD |= two. Then β ∈ Lbasic | D |= β and 6|= β is a finite set of equalities andinequalities among variables.

Proof of Lemma (106): Suppose for a contradiction that Lemma (106) is false. Then sinceD is finite there is β ∈ β ∈ Lbasic | D |= β and 6|= β that contains a variable x that doesnot appear in D. Up to logical equivalence, β has either the form x = y or x 6= y, withy distinct from x. In the first case D |= ∀x(x = y), which is a contradiction since D isconsistent and D |= two. In the second case D |= ∀x(x 6= y), which is also a contradictionsince D is consistent.

Proof of Theorem (39): Fix an enumeration Di | i ∈ N of all finite, consistent D ⊆Lform such that D |= two, an enumeration βi | i ∈ N of Lbasic , and an enumerationϕi | i ∈ N of Lform . For all j ∈ N , set Sj+1 = D0 − βj , βj+1 . . .. For all i > 0and j ∈ N , set S(ω×i)+j = Di − βj , βj+1 . . .. Set Sω2+i = ϕi for all i ∈ N . LetS = Sα | 0 < α < ω2 + ω, and let + be constructed from S as in Definition (42). So + isdefinite maxichoice by Proposition (43). Let B ⊆ Lform be given. To prove that λσ .B +ie σdoes not solve (∅, two,¬two), we distinguish two cases.

Case 1: For all i ∈ N , Di − Lbasic 6⊆ B. Then by Lemma (105) and compactness,λσ .B +ie σ solves no environment for MOD(two).

Case 2: For some i ∈ N , Di−Lbasic ⊆ B. Let i0 ∈ N be least such that Di0−Lbasic ⊆ B.Let n ∈ N and integers j0 < . . . < jn−1 be such that Di0 = (Di0 − Lbasic) ∪ βj0 . . . βjn−1.With the definition of + and the choice of i0, it is easy to verify by induction on m ≤ nthat (Di0 − Lbasic) ∪ βj0 . . . βjm−1 ⊆ B +ie (βj0 . . . βjm−1). In particular:

(107) Di0 ⊆ B +ie (βj0 . . . βjn−1).


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By Lemma (106) there exists structure S and full assignment h to S such that S 6|= twoand S |= β ∈ Lbasic | Di0 |= β[h]. We may choose environment e for S and h that beginswith (βj0 . . . βjn−1). Since Di0 ∪ content(e) is consistent, it follows from the definition of + ,the choice of i0, and (107) that for all k ≥ n, Di0 ⊆ B +ie e[k]. Hence B +ie e[k] |= two forall k ≥ n. We conclude that λσ .B +ie σ does not solve MOD(¬two) in e.


(41) Proposition: Suppose that Sym = ∅ and Obs = Lbasic . Then thereexists a definite maxichoice revision function + such that for all B ⊆ Lform ,λσ .B +

Cnie σ does not solve (∅, two,¬two).

Proof: Fix an enumeration ϕi | i > 0 of Lform with ϕ1 = two, and set Si = ϕi for alli > 0. Let S = Si | i < ω, and let + be constructed from S as in Definition (42). SoProposition (43) implies that + is definite maxichoice. Let B ⊆ Lform be given, and lete0 be an environment for MOD(two). Suppose that λσ .B +

Cnie σ solves MOD(two) in e0

(otherwise we are done). Hence we may choose k0 ∈ N such that B +Cnie e0[k0] |= two. This

implies:

(108) two ∈ Cn(B +Cnie e0[k0]).

It easy to verify the existence of an environment e such that:

(109) (a) e0[k0] ⊂ e.

(b) e is for the structure 0, 1, 2.


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Now for all β ∈ Lbasic , two, β is consistent (recall that Sym = ∅). So Definition (40),(108), (109)a, and the definition of + imply that for all k > k0, two ∈ B +

Cnie e[k]. Thus,

(109)b implies that λσ .B +Cnie σ does not solve MOD(¬two) in e.

References

[1] C. E. Alchourron, P. Gardenfors, and D. Makinson. On the logic of theory change:Partial meet contraction and revision functions. Journal of Symbolic Logic, 50:510–530,1985.

[2] C. E. Alchourron and D. Makinson. Hierarchies of regulation and their logic. InR. Hilpinen, editor, New Studies in Deontic Logic. D. Reidel Publishing Company,Dordrecht, 1981.

[3] C. E. Alchourron and D. Makinson. On the logic of theory change: Safe contraction.Studia Logica, 44:405–422, 1985.

[4] L. Blum and M. Blum. Toward a mathematical theory of inductive inference. Infor-mation and Control, 28:125–155, 1975.

[5] G. Boolos and R. Jeffrey. Computability and Logic (Third Edition). Cambridge Uni-versity Press, Cambridge, England, 1989.

[6] N. Chomsky. Reflections on Language. Pantheon Books, New York, 1975.

[7] H. Enderton. A Mathematical Introduction to Logic. Academic Press, New York, 1972.

[8] H. Enderton. Elements of Set Theory. Academic Press, New York, NY, 1977.

[9] P. Feyerabend. Against Method. Verso Press, London, 1975.


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[10] A. Fuhrmann. Theory contraction through base contraction. Journal of PhilosophicalLogic, 20:175–203, 1991.

[11] P. Gardenfors. Knowledge in Flux: Modeling the Dynamics of Epistemic States. MITPress, Cambridge MA, 1988.

[12] A. Goldman. Epistemology and Cognition. Harvard University Press, Cambridge MA,1986.

[13] S. O. Hansson. A dyadic representation of belief. In P. Gardenfors, editor, BeliefRevision, pages 89–121. Cambridge University Press, New York, 1992.

[14] S. O. Hansson. Changes of disjunctively closed bases. Journal of Logic, Language, andInformation, 2:225–284, 1993.

[15] S. O. Hansson. Reversing the Levi Identity. Journal of Philosophical Logic, 22(6):637–669, 1993.

[16] S. O. Hansson. Kernel contraction. Journal of Symbolic Logic, 59(3):845–859, 1994.

[17] S. O. Hansson. A Textbook of Belief Dynamics: Theory Change and Database Updating.Kluwer Academic Publishers, Dordrecht, 1999.

[18] H. Kornblith, editor. Naturalizing Epistemology. MIT Press, Cambridge MA, 1985.

[19] A. Levy. Basic Set Theory. Springer-Verlag, Berlin, 1979.

[20] A. C. Nayak. Foundational Belief Change. Journal of Philosophical Logic, 23(5):495–534, 1994.


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