sequent calculi for skeptical reasoning in predicate

Sequent Calculi for Skeptical Reasoning in Predicate

Default Logic and Other Nonmonotonic Logics ∗

Robert Saxon Milnikel ([email protected])Department of Mathematics, Kenyon College

Abstract. Sequent calculi for skeptical consequence in predicate default logic,predicate stable model logic programming, and infinite autoepistemic theories arepresented and proved sound and complete. While skeptical consequence is decidablein the finite propositional case of all three formalisms, the move to predicate orinfinite theories increases the complexity of skeptical reasoning to being Π1

1-complete.This implies the need for sequent rules with countably many premises, and such rulesare employed.

Keywords: Default logic, Stable models, Autoepistemic logic, Sequent calculus

AMS Subject Classifications: 03B42, 68N17, 68T27

1. Introduction

Skeptical consequence is a notion common to all forms of nonmono-tonic reasoning. Every nonmonotonic formalism permits different worldviews to be justified using the same set of facts and principles; theskeptical consequences of a framework are the notions common to allworld views associated with that framework. Our purpose in this paperis to present a Gentzen-style sequent calculus (incorporating some in-finitary rules) which will allow us to deduce the skeptical consequencesof a given framework. Such sequent calculi (with purely finite rules)were defined for several types of nonmonotonic systems by Bonatti andOlivetti in [6], but they restricted their attention to finite propositionalsystems for which skeptical consequence is decidable. We will adaptand extend their systems to accommodate infinite predicate systems.

We will focus on three types of nonmonotonic reasoning: stablemodel logic programming (due to Gelfond and Lifschitz, [8]), defaultlogic (due to Reiter, [22]), and autoepistemic logic (due to Moore, [19]).In all three cases, when one steps from the finite and propositionalto the predicate and potentially infinite, finding the set of skepticalconsequences of a framework goes from being decidable to being Π1

1-complete, at the same level of the computability hierarchy as truearithmetic. This result was proved for stable model logic program-

∗ This paper grew directly out of the author’s dissertation, written under thedirection of Anil Nerode.

c© 2004 Kluwer Academic Publishers. Printed in the Netherlands.

fulldefaultsequent.tex; 18/06/2004; 16:12; p.1

2 R.S. Milnikel

ming by Marek, Nerode, and Remmel in [13], but it translates to theother systems quite easily. (The reader should be aware that therewill be a few computability theoretic ideas and motivations discussedin this introductory section, but that they may be considered “deepbackground” and will not be a part of the exposition of the main ideas.)

Π11 sets correspond to finite-path computable (or Π0

1) subtrees ofω<ω in a very natural way. See [7] for an excellent exposition. Thismakes skeptical consequence a natural fit for sequent calculi with in-finitary rules, since a sequent proof is, at its core, a finite-path tree.Bonatti and Olivetti also addressed credulous consequence (“Can thisnotion be a part of some world view?”) in their paper, but in the casesthey were interested in, this question was also decidable. In our moregeneral context, credulous reasoning is Σ1

1-complete, not a natural typeof question to address with trees-as-proofs. (One could write a sequentcalculus for credulous reasoning in Π1

2 logic, but this would take us toofar afield.)

Why do we plan to address the nonmonotonic formalisms listedearlier and not others? Default logic, autoepistemic logic, and circum-scription are historically the most established frameworks, with stablemodel logic programming now as widely studied as those three. We willbegin with stable model logic programming because it has the simplestclassical monotonic base over which the nonmonotonicity is layered.This will permit us to focus on the distinctly nonmonotonic parts ofthe skeptical sequent calculus in our first encounter with such systems.The reason for excluding circumscription (due to McCarthy, [15]) iseasily stated: Skeptical consequence for predicate circumscription wasshown by Schlipf in [23] to be Π1

2-complete, not Π11-complete like the

other systems mentioned. As I said earlier, a Π12 sequent calculus would

take us far beyond the intended scope of this paper. We will also notbe discussing Marek, Nerode, and Remmel’s nonmonotone rule systems(see [12]) or McDermott and Doyle’s nonmonotonic modal logics (see[16]). The reasons are twofold in each case. Nonmonotone rule systemshave not entered the mainstream of the nonmonotonic reasoning liter-ature, and are also essentially isomorphic to propositional stable modellogic programming. The systems of McDermott and Doyle are also notas central to the field as some other formalisms, and would benefit froma more specific study in the context of Artemov’s logic of proofs (see[3]).

Why does the move from the propositional to the predicate entailsuch an enormous increase in complexity? Because in general, it alsoimplies a move from a finite set of rules to an infinite one. In predicatelogic programming, an open variable is viewed as an abbreviation forall values which that variable might take on. The standard approach


Predicate Nonmonotone Sequent Calculi 3

is to make each of those possible instantiations explicit, turning a fi-nite predicate logic program over an infinite domain into an infiniteprogram.

In default logic, there has been debate since Reiter first defined theframework in 1980 about how to treat unbounded variables in the rules.Reiter ([22]) advocated treating open variables (at least in the negativepremises of a default rule) in the same way that they are treated inlogic programs: as abbreviations for the same rule with each groundterm of the language substituted. Again, this can turn a finite defaulttheory into an infinite grounded one. This is the definition we will usein this paper. However, there is some quite justified criticism of this ap-

proach. Under these definitions, the default theory (: MP (x)P (x)

,¬P (a))

does not imply (∀x)[P (x) ↔ x 6= a]. Lifschitz, in [10], defines exten-sions (the possible world views associated with default logic) relativeto fixed domains. For finite theories and finite domains, everythingis decidable, but over infinite domains this is no longer the case. In[18], it is shown that skeptical reasoning over countable domains usingLifschitz’ definition of extension is Π1

2-complete, the same level as forcircumscription.

Because nonmonotonic logics deal not only with proof but with lackof proof, we will need not only standard monotone sequent calculi,but also rule systems for showing a lack of proof. (We will call theseantisequent calculi, using the terminology of Bonatti from [5].) Whilepropositional provability and lack of provability are decidable, predicateprovability is only recursively enumerable, and hence predicate non-provability is co-r.e. Just as Σ1

1 sets do not lend themselves naturallyto tree-based proofs, neither do co-r.e. sets. However, while Σ1

1 setsrequired a jump to Π1

2 logic, co-r.e. sets are easily accommodated inthe Π1

1 framework within which we will already be working. Bringingsuch enormously powerful logical machinery to bear on such a relativelysimple problem may seem like overkill, but it works out quite naturally.The reader is thus warned that infinitary proofs will appear throughoutthe paper, even when talking about something as simple as lack of astandard predicate logic proof.

When discussing autoepistemic logic, we limit ourselves to the propo-sitional case, but with a potentially infinite theory. This puts us at thesame Π1

1 level of complexity as for stable model logic programmingand default logic. Predicate autoepistemic logic is discussed in theliterature (see, for example, [9]), and the interested reader should notfind it difficult to combine the rules specific to predicate logic (fromthe discussion of default logic) with the results about propositionalautoepistemic logic.


4 R.S. Milnikel

We will begin in Section 2 with some preliminary definitions andresults about the types of nonmonotonic systems we will consider inthis paper. Section 3 presents rudimentary sequent and antisequentcalculi for Horn programs, continues with a sequent calculus for skep-tical reasoning in stable model logic programming, and concludes witha proof of the soundness and completeness of this calculus. Section 4concentrates on finding an antisequent calculus for classical predicatelogic, and concludes with a very brief treatment of a sequent calculusfor skeptical reasoning in default logic. The nonmonotonic aspects ofstable model logic programming and default logic are so similar thatthe sequent calculi are nearly identical and no proof is necessary for thesoundness and completeness theorems in this case. Section 5 presentsa sequent calculus for skeptical reasoning in infinite propositional au-toepistemic theories, and proves soundness and completeness theoremsfor this calculus. Section 6 looks toward further directions for research.

2. Preliminaries

We will be working with systems based on a variety of languages, bothpredicate and propositional. We will assume that the reader has somefamiliarity with classical propositional and predicate logic, includingthe notion of a Herbrand base for a given predicate language, thestandard sequent calculus LK for propositional logic, and the modaloperator L.

2.1. Stable Model Logic Programming

Logic programming is a broad and widely studied subfield of computerscience. We will concentrate here on only one aspect – finding modelsof a program given a particular interpretation of negation. For a moregeneral introduction, see [2], [11], or [21].

Definition 2.1. A logic program clause is an expression of the form

p← q1, . . . , qm,not r1, . . . ,not rn,

where p, the qi’s, and the rj ’s are atomic formulas of some predicate orpropositional language L. If p, all qi’s, and all rj ’s are ground atomicformulas, then the clause is called ground. A logic program is a set oflogic program clauses.

In general, logic programs may defined over a predicate language,but in practice, the first step in dealing with a predicate logic programP is to look at ground(P ), the grounding of P . To obtain the grounding



of a program, one replaces each clause with the set of propositionalclauses obtained by all ground substitutions whose domains include allvariables appearing in P . (Details can be found in any of the generalreferences cited above.) The principal result here is that a subset M ofthe Herbrand base of the general logic program is a Herbrand model ofthe program if and only if M is a model of its propositional grounding.One noteworthy effect of grounding a program is that a finite predicateprogram in a language with an infinite number of ground terms canhave an infinite grounding.

Example 2.2. Let L be a language with one constant 0, one unaryfunction S, three unary relations A, A, and B, and three binary re-lations G, L, and N . Following convention, we will let n stand for nS’s followed by a 0. Let finite program P consist of the following nineclauses:

− L(x, Sx)←

− L(x, Sy)← L(x, y)

− G(Sx, x)←

− G(Sx, y)← G(x, y)

− N(x, y)← L(x, y)

− N(x, y)← G(x, y)

− A(x)← not A(x)

− A(y)← A(x), N(x, y)

− B(0)← A(x)

The first six clauses make sure that N(x, y) will be true exactly ifx 6= y. The next finds some element x so that A(x) is not proven andestablishes A(x) for that x. The penultimate clause establishes A(y) forall y 6= x once A(x) has been established for some x. The final clauseestablishes B(0) if A(x) has been established for any x. Each of thesenine clauses has an infinite grounding. The program ground(P ) consistsof the union of the groundings of all of the clauses. We won’t addresseach, but let us look at A(x)← not A(x) and A(y)← A(x), N(x, y).

The grounding of A(x)← not A(x) consists of:

− A(0)← not A(0)

− A(1)← not A(1)


6 R.S. Milnikel

− A(2)← not A(2)

− A(3)← not A(3)...

The grounding of A(y)← A(x), N(x, y) consists of:

− A(0)← A(0), N(0, 0)

− A(1)← A(0), N(0, 1)

− A(0)← A(1), N(1, 0)

− A(2)← A(0), N(0, 2)

− A(1)← A(1), N(1, 1)

− A(0)← A(2), N(2, 0)

− A(3)← A(0), N(0, 3)

− A(2)← A(1), N(1, 2)

− A(1)← A(2), N(2, 1)

− A(0)← A(3), N(3, 0)...

Because we will come back to this example throughout this sectionand the next, let us also set up some shorthand, using HP to refer tothe Horn portion of ground(P ), the ground instantiations of all clausesof P except A(x)← not A(x).

Although there are debates about the proper way to deal with nega-tion in logic programs, the interpretation of Horn programs—programsconsisting entirely of Horn clauses—is straightforward. A logic programclause

p← q1, . . . , qm,not r1, . . . ,not rn,

is called Horn if m ≥ 0 and n = 0.

Definition 2.3. Let P be a Horn program in language L. We will saythat a set M ⊆ BH (where BH is the Herbrand base of L) is a Herbrandmodel of P if for each clause

p← q1, . . . , qm

of ground(P ), p ∈M whenever {q1, . . . , qm} ⊆M .



It is a well-known fact that Horn programs have least Herbrandmodels. The reader man see any of the general references on logicprograms for a proof. Also, because we will need the fact later, let usmention the analog of the Compactness Theorem for Horn programs.If p is in the least model of a Horn program P , then there is some finiteP ′ ⊆ ground(P ) such that p is in the least model of P ′.

The basic idea for the interpretation of negation we will be interestedin, proposed in part by Apt ([1]) and more fully by Gelfond and Lifschitz([8]), is to interpret a clause

p← q1, . . . , qm,not r1, . . . ,not rn

as “if all qi are true (derived), and no ri is true (derivable), then p istrue (can be derived).” To formalize this interpretation, we will guessat a model of a program P . We will boldly assume that our guess is amodel of the interpretation we just outlined. Based on that assumption,we will be concerned only with clauses in which none of the rj ’s appearin our guessed-at model. If from the remaining clauses, we can deriveall and only the elements of our guessed-at model, our guess was a goodone.

Definition 2.4 (Reduct of a Logic Program). Let M be a subsetof BH , the Herbrand base of some predicate language.

1. A clausep← q1, . . . , qm,not r1, . . . ,not rn

is irrelevant with respect to M if at least one rj is in M .

2. Let P be a logic program. The reduct of P with respect to M ,denoted by PM , is obtained from ground(P ) by:

a) Removing all clauses that are irrelevant with respect to M .

b) Removing each premise not rj from the remaining clauses.

What remains after taking the reduct PM of a program P is a groundHorn program.

Definition 2.5 (Stable Model). Let P be a logic program and letM be a subset of BH , the Herbrand base of the language of P . The setM is a stable model of P if M coincides with the least model of thereduct PM .

Stable models of a program P are the “good guesses” for modelsalluded to earlier.


8 R.S. Milnikel

Example 2.6. Let us reexamine Example 2.2 in light of these twodefinitions to get a sense of what the stable models of P might be.

− Let M1 ⊇ {A(3), A(3)}. We can see that M1 is not a stablemodel of P , since PM1 will not include A(3) ←, the Horn reductof A(3) ← not A(3). There is no other rule in all of ground(P )with conclusion A(3), so PM1 is not sufficient to prove A(3).

− Let M2 ⊇ {A(1), A(3)}. We can see that M2 is not a stablemodel of P , since we require M2 to be closed under all rules inPM2 ⊇ HP . Because A(1) ∈M2, A(3)← A(1), N(1, 3) ∈ HP , andthere are also in HP clauses sufficient to prove N(1, 3), we see thatA(3) ∈M2, putting us back in the case of M1.

− Let M3 = {A(0), A(1), A(2), A(3), . . .}. In this case, PM3 = HP .The least model of HP includes no elements of the form A(n) orA(n) for any n. So M3 is not a stable model of P .

− Let M4 be the least model of HP . Because A(n) /∈ M4 for alln, PM4 = HP ∪ {A(0)← , A(1)← , A(2)← , A(3)← , . . .}.Thus both A(n) and A(n) are in the least model of PM4 for all n,and this is considerably more than was in M4 to begin with. Thus,M4 is not a stable model of P .

− Let M ′5 = {A(3), B(0)} ∪ {A(n)|n 6= 3}, and let M5 be M ′

5

together with the least model of HP . This makes PM5 = HP ∪{A(3) ← }. It turns out that the least model of PM5 is exactlyM5, and M5 is a stable model.

We will assume without any more formal proof that all stable modelsof P are similar to M5, with different numbers substituted for 3.

2.2. Default Logic

Default logic is one of the most intuitive and widely-studied nonmono-tonic formalisms. Default logics are built upon classical propositionalor predicate logic, and we will assume the reader is familiar with thelanguages and the basics of axiomatic treatments of these logics.

Definition 2.7 (Default). Let L be a predicate language. A de-fault is a triple 〈ϕ,Ψ, θ〉 where ϕ and θ are formulas from L andΨ = {ψ1, . . . , ψn} is a finite set of formulas from L. A default is usuallywritten

ϕ : Mψ1, . . . ,Mψm

θ,



with the intended interpretation “if ϕ is true and each ψi is possible,conclude θ.”

The formula ϕ will be called the prerequisite of the default, Ψ thejustifications, and θ the conclusion. If d contains formulas with un-bounded variables, we will refer to the default as open. A default thatis not open is closed.

A default theory is a pair (D,W ), where D is a set of defaults andW is a set of formulas of L. Note that we do not restrict D or W tobeing finite. If both D and W are finite, we will refer to (D,W ) as afinite default theory. If D contains open defaults, we will call (D,W )an open default theory. If D consists entirely of closed defaults, we willcall (D,W ) closed.

As was the case with logic programs, we will not want to workdirectly with default rules with open formulas, but will look at a de-fault with an open formula among its prerequisite, justifications, andconclusion as an abbreviation for the set of groundings of that formula.

Let d =ϕ;Mψ1, . . . ,Mψn

θbe an open default. The grounding of d,

denoted ground(d), will be the set of closed defaults obtained by re-placing each unbounded variable x occurring in d by some ground termt of L, and doing so uniformly throughout the default. (If d is not open,let ground(d) = {d}.) Let us define the grounding of a set D of defaultsto be the set of all defaults occurring in the grounding of some defaultin D. (So ground(D) =

⋃{ground(d)|d ∈ D}.) We will occasionally

abuse notation by writing ground((D,W )) for (ground(D),W ). Notethat D = ground(D) if D is closed.

One effect of grounding is that if L contains an infinite number ofground terms, then even when D is finite, ground(D) might be infinite.So just as for logic programs, a finite open default theory can lead toan infinite grounding.

It will simplify work we will do later to look at sets which containboth defaults and formulas. We can define the language Ldef as theunion of all formulas in L with the collection of all defaults constructedfrom formulas of L. This will allow us to express the default theory(D,W ) as the single set D ∪W in Ldef .

The intended reading of a default should put the reader in mind ofour interpretation of negation in logic programs. Indeed, default logic isbasically stable model logic programming using not only the languageof classical logic but the rules of deduction in that logic as well. (Manywould counter that the situation is in fact the reverse, that stable modellogic programming is default logic with the classical deduction removed.There are good arguments on each side and, happily, we do not needto resolve the issue in this forum.)


10 R.S. Milnikel

Default rules without justifications (ϕ :θ

) are monotonic and classi-cal in nature. It should not cause confusion if we conflate these withclassical rules of inference of the form

ϕ

θ. When such rules arise from

considering a defaultϕ : Mψ1, . . . ,Mψn

θin a context in which the

ψi’s are guaranteed to be possible, we will call them residues. We willdefine Lres analogously to Ldef , as the formulas of L taken togetherwith residues built from L.

Residues behave like Horn program clauses, and just as Horn pro-grams have least Herbrand models and classical sets of formulas havedeductive closures, so too we can define the closure of a subset of Lres.We will say that the closure of a set Γ ⊆ Lres, denoted Cl(Γ), is theleast set T of formulas of L which is deductively closed, Γ∩L ⊆ T , andhas the additional property that if

ϕ

θ∈ Γ and ϕ ∈ T then θ ∈ T .

We are now in a position to define reducts of default theories andextensions for default theories precisely on analogy with reducts andstable models of logic programs.

Definition 2.8 (Reduct of a Default Theory). Let S be a set offormulas of L.

1. A defaultϕ : Mψ1, . . . ,Mψn

θ

is irrelevant with respect to S if ¬ψj ∈ S for at least one ψj .

2. Let Γ be a closed default theory (in Ldef ). The reduct of Γ withrespect to S, denoted by ΓS , is obtained from Γ by:

a) Removing all defaults that are irrelevant with respect to S.b) Replacing each remaining default

ϕ : Mψ1, . . . ,Mψn

θ

with its residueϕ

θ.

What remains after taking the reduct ΓS of a default theory Γ is aresidue theory.

Definition 2.9 (Default Extension). Let Γ = D ∪W be a closeddefault theory. We say that a set of formulas S of L is a default extensionfor Γ if S = Cl(ΓS). We will say that S is an extension of open defaulttheory Γ if S is an extension of ground(Γ).



2.3. Stable Theories and Autoepistemic Logic

Moore’s autoepistemic logic ([19]) is an approach alternative to thoseabove which tries to capture the intuitions behind positive and negativeintrospection more directly. The difference between this approach andthe others we have presented so far is that we use our context bothpositively and negatively to reflect positive and negative introspection.In many ways, this is the most straightforward and intuitive approachone can take.

The language for autoepistemic logic will be LL, a propositionallanguage L extended by the modal operator L. However, we will not beinterested in traditional modal interpretations of LL, but in a strictlypropositional interpretation, in which every formula of the form Lϕis treated as an independently valued proposition. (Traditional modallogic will come up once, very briefly. The premise of a proposition willbe that a theory T ∈ LL is consistent with S5. If the reader is notfamiliar with modal logics in general, this proposition and its use inthe midst of the final proof of the paper can safely be skimmed.)

The analogue in autoepistemic logic of a stable model or extension isa stable expansion. A stable expansion is a special sort of stable theory.Before we define stable expansions, let us define stable theories and listsome of their elementary properties. Most of these results are due toMoore ([20]). Proofs can be found in [14].

Definition 2.10. A propositionally deductively closed theory T ⊆ LL

is called stable if it meets the following two criteria:

− For every ϕ ∈ LL, if ϕ ∈ T then Lϕ ∈ T .

− For every ϕ ∈ LL, if ϕ /∈ T then ¬Lϕ ∈ T .

To state the results we want, we will need several further definitions.

Definition 2.11. − The L-depth of a formula ϕ ∈ LL, denoteddL(ϕ), is defined recursively.

• If ϕ ∈ L, then dL(ϕ) = 0.

• If ϕ = ¬ψ, then dL(ϕ) = dL(ψ).

• If ϕ = ψ1 ∨ ψ2, ϕ = ψ1 ∧ ψ2, or ϕ = ψ1 → ψ2, then dL(ϕ) =max{dL(ψ1), dL(ψ2)}.

• If ϕ = Lψ, then dL(ϕ) = dL(ψ) + 1.

− LL,n = {ϕ ∈ LL|dL(ϕ) ≤ n}.

− Given T ⊆ LL, [T ]n = T ∩ LL,n.


12 R.S. Milnikel

Proposition 2.12. Let T ⊆ LL be stable. For every integer n ≥ 0,

[T ]n+1 = Th([T ]n ∪ {Lϕ|ϕ ∈ [T ]n} ∪ {¬Lϕ|ϕ ∈ LL,n \ [T ]n}) ∩ LL,n+1.

Proposition 2.13. Let U ⊆ LL be consistent with S5. Then there isa unique stable and consistent theory T such that U ⊆ T .

Proposition 2.14. If T is a stable consistent theory, then

T = Th([T ]0 ∪ {Lϕ|ϕ ∈ T} ∪ {¬Lϕ|ϕ /∈ T}).

Proposition 2.15. If T is a stable consistent theory, then for everyϕ ∈ LL, either Lϕ ∈ T or ¬Lϕ ∈ T , and not both.

We will now define stable expansions, the main object of study inautoepistemic logic.

Definition 2.16 (Stable Expansion). A set of formulas T ⊆ LL isa stable expansion of A ⊆ LL if and only if T is a consistent set offormulas for which T = Th(A ∪ {Lϕ|ϕ ∈ T} ∪ {¬Lϕ|ϕ /∈ T}).

2.4. Skeptical Consequence

If we think of stable models or extensions or stable expansions as coher-ent, justified points of view in a framework represented by the rules ofthe logic program or default theory or autoepistemic theory, there aretwo important sets of formulas we want to take note of: those which canbe part of some coherent, justified point of view, and those which mustbe part of any coherent, justified point of view. If you take as your setof conclusions things which are present in at least one stable model (orextension or stable expansion), you are reasoning credulously or bravely(both terms are widely used). If, on the other hand, you believe onlythose facts true in all stable models (or extensions or stable expansions),you are reasoning skeptically or cautiously. (The standard contrasts arecredulous vs. skeptical reasoning and brave vs. cautious reasoning.)

A question arises: What is the set of skeptical consequences of alogic program P if P has no stable models? One might be temptedto consider the set of skeptical consequences empty, but it is generallyaccepted that this is not the approach to use. On the contrary, if P hasno stable models, every element p of the Herbrand base is considered askeptical consequence of P . The most intuitive argument for this is tolook at the sentence “p is a member of every stable model of P”. It isvacuously true if there are no stable models of P . Of course, this sameargument applies equally well to default and autoepistemic logic.

We will be concerned in this paper only with skeptical consequence,so we will limit the formal definitions to that term.



Definition 2.17 (Skeptical Consequence).

− A ground atomic formula p in predicate language L is in the setof skeptical consequences of a logic program P if for every stablemodel M of P , p ∈M .

− A formula ϕ predicate language L is in the set of skeptical con-sequences of a default theory (D,W ) if for every extension S of(D,W ), ϕ ∈ S.

− A formula ϕ in propositional modal language LL is in the set ofskeptical consequences of autoepistemic theory A if for every stableexpansion T of A, ϕ ∈ T .

Example 2.18. We will continue to exploit Example 2.2. We sawin Example 2.6 that stable models of P all look like {A(n0), B(0)} ∪{A(n)|n 6= n0}, along with some statements about G, L, and N . Theonly items common to all of these are B(0) plus the consequences ofHP , making {B(0)} ∪ Cl(HP ) the set of skeptical consequences of P .There is no compactness theorem for skeptical consequence. By thatwe mean that although in each particular stable model M of P , somefinite portion of PM was used to prove B(0), there is no finite portionof ground(P ) which is responsible for B(0) being present in all stablemodels.

3. An Infinitary Sequent Calculus for Skeptical StableModel Logic Programming

Throughout this section, we will assume that we are working in acountable language L with Herbrand base BH .

The definition of a stable model insists that certain members ofBH be derivable (that PM be sufficient to derive all members of M)and that others not be derivable (that PM not be able to derive anymember of BH \M). Thus, as we accumulate information about poten-tial stable models by backtracking through a sequent proof, we willfind that we need to establish, at various points, both derivabilityand non-derivability. Because of the compactness of classical monotonederivability, establishing derivability will be straightforward. Estab-lishing non-derivability, on the other hand, will be quite complicated.However, when we limit ourselves to finite sets of premises, derivabil-ity and non-derivability are decidable and relatively straightforwardto establish. Following Bonatti and Olivetti’s model ([6]), as we will


14 R.S. Milnikel

do for most of the rest of the paper, let us begin our explorationof nonmonotonic inference with rules for monotone derivability andnon-derivability.

3.1. Finite Monotone Sequents and Antisequents for LogicPrograms

We will be interested in when a finite set of ground program clausesΓ can and can not prove some element (or one of several elements) ofBH . We will define a monotone logic program sequent as a pair 〈Γ,∆〉,usually written Γ ` ∆, where Γ is a finite set of ground Horn programclauses and ∆ is a finite subset of BH . We will say that a sequent Γ ` ∆is true if some element of ∆ is in the unique least model of Γ.

Just as it was safe when talking about residues to conflate justification-free defaults

ϕ :θ

with propositional rules of inferenceϕ

θ, so it will be

safe here to conflate elements p ∈ BH with program clauses p←.Provability in Horn Logic Programming is so straightforward that

we will need only axiom and one sequent rule. A sequent Γ ` ∆ is anaxiom if and only if Γ ∩∆ 6= ∅. Our only rule of inference will be

Γ ` q1,∆ . . . Γ ` qm,∆ Γ, p ` ∆Γ, (p← q1, . . . , qm) ` ∆

.

Proposition 3.1. Monotone logic program sequent Γ ` ∆ is derivableif and only if it is true.

Proof. To show that derivable sequents are true, we must establishthe soundness of our axiom and of our rule. The soundness of theaxiom is apparent. If p ∈ Γ ∩∆, then p ← is a rule of program Γ andp will be an element of any model of Γ. The soundness of our rule isalmost as apparent. Assume that Γ ` qi,∆ is true for 1 ≤ i ≤ m, andthat Γ, p ` ∆ is true, but that Γ, (p ← q1, . . . , qm) ` ∆ is not true.There is a Herbrand model M of Γ, (p ← q1, . . . , qm) which excludesall elements of ∆. By the definition of Herbrand model, any model ofΓ, (p← q1, . . . , qm) is a model of Γ, so M is a model of Γ excluding allelements of ∆. By the truth of each of the sequents Γ ` qi,∆, any modelof Γ which excludes all of ∆ must therefore include {q1, . . . , qm}. ThusM is a model of Γ, (p← q1, . . . , qm) which contains all of {q1, . . . , qm}.That means p ∈M and M is a model of Γ, p. By the truth of Γ, p ` ∆,M must include some element of ∆. This is our desired contradiction.

We will show by induction on the number n of rules p← q1, . . . , qmin Γ with m > 0 that if Γ ` ∆ is not provable, it is not true. If n = 0and Γ ` ∆ is not provable, then both Γ and ∆ are, in effect, subsets of



BH and Γ∩∆ = ∅. In that case, Γ itself is a model of Γ which excludesall members of ∆.

Now let us assume that n > 0. If Γ, (p ← q1, . . . , qm) ` ∆ is notprovable, then either Γ, p ` ∆ is not provable or Γ ` qi,∆ is notprovable for some 1 ≤ i ≤ m. By our induction hypothesis, any of thesequents just listed which is unprovable is also not true. Thus, eitherthere is a model M of Γ, p which excludes all of ∆ or there is a modelM of Γ which excludes not only all of ∆ but also excludes some qi. Ineither case, M is a model of Γ, (p ← q1, . . . , qm) which excludes all of∆ and hence Γ, (p← q1, . . . , qm) ` ∆ is not true.

Example 3.2. Continuing to use the language of Example 2.2, let usgive a sequent proof that A(0), B(0)← A(0) ` B(0).

A(0) ` B(0), A(0) A(0), B(0) ` B(0)A(0), B(0)← A(0) ` B(0)

Both premises, in this case, are axioms.

Antisequents will be defined in a way quite similar to sequents. Amonotone logic program antisequent is a pair 〈Γ,∆〉 usually writtenΓ 0 ∆ with Γ a finite ground Horn program and ∆ a finite subset ofBH . An antisequent Γ 0 ∆ will be considered true if there is a modelof Γ which excludes all of ∆.

Although we will again need only one axiom scheme, Γ ` ∆ withΓ ⊆ BH and Γ ∩∆ = ∅, we will need two antisequent rules:

Γ 0 ∆ Γ 0 qiΓ, (p← q1, . . . , qm) 0 ∆

(for any 1 ≤ i ≤ m) andΓ, p 0 ∆

Γ, (p← q1, . . . qm) 0 ∆.

Proposition 3.3. Monotone logic program antisequent Γ 0 ∆ is prov-able if and only if it is true.

Proof. Again, the soundness of the axiom is apparent. If Γ ⊆ BH andΓ∩∆ = ∅, then Γ itself is a model of Γ excluding ∆. The soundness ofeach of the rules is also easy to see. If there are models M ′ and M ′′ ofΓ excluding all of ∆ and some qi respectively, then M , the least modelof Γ, will exclude all of ∆ and at least one of the qi’s. That meansthat M is a model of Γ, (p← q1, . . . , qm) which excludes all of ∆. Evenmore simple is the other rule. Any model of Γ, p is also a model ofΓ, (p← q1, . . . , qm), so if there is a model of Γ, p which excludes ∆, thesame model will also be a model of Γ, (p← q1, . . . , qm) which excludes∆.

To show that any antisequent Γ 0 ∆ which is not derivable is falsewe will again use induction on the number n of rules in Γ which are of


16 R.S. Milnikel

the form p ← q1, . . . , qm with m > 0. If n = 0, Γ ⊆ BH but Γ 0 ∆ isnot provable, so Γ ∩∆ 6= ∅. Of course any p ∈ Γ (technically p← is inΓ) will be a member of any model of Γ, so if Γ∩∆ 6= ∅, no model of Γwill exclude all of ∆.

If n > 0, assume that Γ, (p← q1, . . . , qm) 0 ∆ is not derivable. Thismeans that either Γ 0 ∆ is not provable or Γ 0 qi is not provable forall 1 ≤ i ≤ m (and by inductive hypothesis not true). Thus, eitherit is the case that every model of Γ contains some element of ∆ orit is the case that every model of Γ contains all of {q1, . . . , qm}. Bythe non-derivability of Γ, (p ← q1, . . . , qm) 0 ∆ we also know thatΓ, p 0 ∆ is not derivable and by induction not true. This means thatevery model of Γ, p contains some element of ∆. Let M be any modelof Γ, (p ← q1, . . . , qm). Because M is a model of Γ, either it containssome element of ∆, or it contains all of {q1, . . . , qm}. If M is a modelof p← q1, . . . , qm and {q1, . . . , qm} ⊆M , then p ∈M . If M is a modelof Γ and p ∈ M , then M is a model of Γ, p and thus contains someelement of ∆. We have just shown that any model of Γ, (p← q1, . . . , qm)contains some element of ∆ and thus that Γ, (p ← q1, . . . , qm) 0 ∆ isnot true.

Example 3.4. Continuing to use the language of Example 2.2, let usshow that A(3), (A(0)← A(6), N(6, 0)), (G(7, 4)← G(6, 4)) 0 A(0).

A(3), G(7, 4) 0 A(0) A(3), G(7, 4) 0 A(6)A(3), G(7, 4), (A(0)← A(6), N(6, 0)) 0 A(0)

A(3), (G(7, 4)← G(6, 4)), (A(0)← A(6), N(6, 0)) 0 A(0)

Both top sequents are axioms.

3.2. Skeptical Sequent Calculus

One can think of Gentzen proof systems as failed exhaustive searchesfor countermodels. Thus, what we will want to do as we search for acountermodel to the claim “All stable models of program P contain p”is keep track of which elements of BH are in and out of our potentialcountermodel to the claim. We will want to make sure that all elementsof BH we would like to see in our potential countermodel do, in fact,have proofs; and we want also to make sure that all elements of BH

we plan to exclude do not have proofs. Finally, we will use our increas-ing information about the potential countermodel to determine whichclauses of P will be dismissed as irrelevant and which will be retainedas Horn clauses in the reduct.



The sequents for skeptical reasoning for logic programs will be triples〈Σ,Γ,∆〉, usually notated Σ; Γ|∼∆. This notation is drawn directly from[6]. The sets Γ and ∆ are fairly straightforward. Γ is a ground logicprogram and ∆ is a subset of BH , neither necessarily finite. The setΣ is more complicated. In Bonatti and Olivetti’s formulation, it was afinite collection provability constraints of the form Lp or ¬Lp where p ∈BH . The intention was to suggest the modal operator L, and indicatewhether p was in or out of our potential countermodel, as it existsso far. We will need to extend the notion of a provability constraintto be more explicit than “p can be proved.” We will need to be ableto say “p can be proved from these specific rules.” Thus, in additionto provability constraints of the form Lp and ¬Lp (which we will callimplicit provability constraints), we will also include explicit provabilityconstraints LΓp where Γ is a finite ground Horn program. The intendedmeaning of LΓp is “Γ ` p.” Together, explicit and implicit provabilityconstraints will be known as general provability constraints. We cannow finally describe Σ as a finite set of general provability constraints.

We will say that a M ⊆ BH satisfies Lp if p ∈M and satisfies ¬Lpif p /∈ M . We will say that M satisfies LΓp if p ∈ M and in additionΓ ` p.

The reader may wonder why we do not need explicit provabilityconstraints of the form ¬LΓp. We will be making claims both of theform “p has a proof” and of the form “p has no proof”. To contradicta claim of the form “p has no proof,” it is necessary simply to exhibit asingle proof of p. On the other hand, to contradict a claim of the form“p has a proof,” we must look at all possible proofs of p, that is, all(relevant) explicit proof constraints.

Before we proceed to the axioms and rules for the skeptical sequentcalculus, let us establish the meaning of the sequents. We will say thatΣ; Γ|∼∆ is true if every stable modelM of the program Γ which satisfiesall constraints in Σ includes at least one member of ∆. Thus, p is askeptical consequence of the logic program P if ∅; ground(P )|∼p is atrue sequent. (As usual, we will abuse notation by writing p for {p}and ∆, p for ∆ ∪ {p} in sequents.)

We are now finally in a position to define the sequent calculus forskeptical reasoning in stable model logic programming. This calculusincorporates three sorts of sequents, in fact: monotone logic programsequents, monotone logic program antisequents, and the skeptical rea-soning sequents just defined. The sequent calculus for skeptical rea-soning will include all axioms and rules of the monotone sequent andantisequent calculi, plus five new rules. (No additional axioms will benecessary. The leaves of every proof tree will be monotone sequent andantisequent axioms.)


18 R.S. Milnikel

Definition 3.5 (Skeptical Sequent Calculus—Logic Program-ming). The axioms of the skeptical sequent calculus are monotonelogic program sequents Γ ` ∆ with Γ ⊆ BH and Γ∩∆ 6= ∅; and mono-tone logic program antisequents Γ 0 ∆ with Γ ⊆ BH and Γ ∩∆ = ∅.The rules are:

0. The three rules of the monotone sequent and antisequent calculi.

1.Σ′,Γ′ ` ∆′

Σ; Γ|∼∆where Γ′ ⊆ Γ is a finite ground Horn program, Σ′ ⊆

({p|Lp ∈ Σ} ∪ {p|LΓ′′p ∈ Σ for some Γ′′}), and ∆′ ⊆ ∆ is finite.

2.Σ′,Γ′ ` p¬Lp,Σ; Γ|∼∆

where Γ′ ⊆ Γ is a finite ground Horn program, Σ′ ⊆

({p|Lp ∈ Σ} ∪ {p|LΓ′′p ∈ Σ for some Γ′′}).

3.Γ0 0 p

LΓ0p,Σ; Γ|∼∆where Γ0 is a finite ground Horn program.

4.{LΓ′

0p,Σ,Σ′

0; Γ′0, (Γ \ Γ′)|∼∆|Γ′ ⊆ Γ is finite}Lp,Σ; Γ|∼∆

where Γ′0 = {p ←

q1, . . . , qm| p← q1, . . . , qm, not r1, . . . ,not rn ∈ Γ′}, Σ′0 = {¬Lr|r =

rj for some p ← q1, . . . , qm, not r1, . . . ,not rn ∈ Γ′ and some1 ≤ j ≤ n}.

5.¬Lr1, . . . ,¬Lrn,Σ; Γ, (p← q1, . . . , qm)|∼∆ Lr1,Σ; Γ|∼∆ · · · Lrn,Σ; Γ|∼∆

Σ; Γ, (p← q1, . . . , qm,not r1, . . . ,not rn)|∼∆

The reader may recall from the introduction that we noted thatinfinitary rules of inference would be necessary in some cases. Whenapplied to a sequent with an infinite Γ, rule 4 has infinitely manypremises.

Let us examine what each of these rules accomplishes, thinking ofourselves as traversing a completed proof backwards, from conclusionsto premises, examining each branch as a failed attempt to find a coun-terexample to the assertion made at the root of the tree. Just as weare moving backwards through the proof, let us also move backwardsthrough the rules.

Rule 5 says: “Either clause p ← q1, . . . , qm,not r1, . . . ,not rn isrelevant or it is not. If it is relevant, make sure that the context reflectsthat, and put the reduct p ← q1, . . . , qm into our list of usable Hornclauses. If it is not relevant, it must be because one of the rj ’s is in thecontext.”

Rule 4 says: “If we are asserting that p is in the stable model weare trying to build, it must have a proof from the available clauses.



Because in different stable models, it might have different proofs, wewill need to examine each possible proof independently. If non-Hornclauses from Γ are used, update the context to reflect their usabilityand replace them in Γ with their reducts.”

Rule 3 says: “If we have said that Γ0 ` p and yet we can show thatΓ0 0 p, show this and stop.”

Rule 2 says: “If we have said that p has no proof, and yet from whatwe already know about the potential stable model we are building wecan show that that model must contain p, show this and stop.”

Rule 1 says: “If from what we already know about the potentialstable model we are building, we can show that a member of ∆ mustbe in that stable model, show this and stop.”

Before we state and prove the soundness and completeness theoremsfor the skeptical sequent calculus, let us finish the journey started inExample 2.2 and continued in Examples 2.6, 2.18, 3.2, and 3.4. Becauseof the complexity of the example, the long lists of clauses involved, andthe infinite numbers of premises, we will not write out a complete proof,or even assemble the parts we will look at in detail. We will considerthe sequents we will need on an individual basis and leave it to thereader to assemble the proof in something like its entirety.

Example 3.6. We stated in Example 2.18 that B(0) is in the set ofskeptical consequences of program P from Example 2.2. Let us nowshow how our skeptical sequent calculus would prove this. We willagain use the abbreviation HP to refer to the entire Horn portion ofground(P ). The sequent we want to prove, then is:

;HP , (A(0)← not A(0)), (A(1)← not A(1)), (A(2)← not A(2)), . . . |∼B(0).

We will prove this sequent by rule 5. The two premises we will needare:

¬LA(0);HP , A(0), (A(1)← not A(1)), (A(2)← not A(2)), . . . |∼B(0)

and

LA(0);HP , (A(1)← not A(1)), (A(2)← not A(2)), . . . |∼B(0).

These sequents will have very different proofs. We can prove theformer by means of rule 2, since A(0) and B(0) ← A(0) are amongthe Horn clauses in the Γ of the sequent, and we saw in Example 3.2a proof of A(0), (B(0) ← A(0)) ` B(0). The latter we will prove bymeans of rule 4, which will have infinitely many premises, each of theform

LΓ′0,Σ′

0; Γ′0, (Γ \ Γ′)|∼B(0)


20 R.S. Milnikel

where

Γ = HP∪{(A(1)← not A(1)), (A(2)← not A(2)), (A(3)← not A(3)), . . .}.

We will look at a few representative premises. In each case, writing outthe sequent with Γ0 and the other terms expanded to fit the particularcase would be unwieldy, so we will simply use the template above andrefer to Γ′

0 and so forth by name.

− In the case that Γ′ consists of A(3) ← not A(3) and A(0) ←A(3), N(3, 0) plus enough of HP to prove N(3, 0), Γ′

0 will consistof A(3) and A(0) ← A(3), N(3, 0) together with the rest of theHorn portions of Γ′. Σ′

0 will consist of ¬LA(3).

In this case, the premise can be proved using rule 1, since A(3) ∈ Γ′0

and B(0)← A(3) ∈ (Γ \ Γ′), and A(3), (B(0)← A(3)) ` B(0).

− In the case that Γ′ consists ofA(3)← not A(3),A(0)← A(3), N(3, 0),and A(4) ← not A(4), plus enough of HP to prove N(3, 0), wecould proceed just as in the above case. We do have another optionopen to us in this case, though, which will illustrate the use of rule2. Γ′

0 will include both A(3) and A(4), while Σ′0 will consist of both

¬LA(3) and ¬LA(4).

By taking A(3) from Γ′0 and the relevant portions ofHP from Γ\Γ′,

we can show that

A(3), (A(4)← A(3), N(3, 4)), (N(3, 4)← L(3, 4)), L(3, 4) ` A(4)

and use rule 2 to prove the desired sequent.

− In the case that Γ′ is drawn entirely from HP , say Γ′ consists ofG(7, 4) ← G(6, 4) and A(0) ← A(6), N(6, 0), we see that Γ′

0 = Γ′,and we can show Γ′

0 0 A(0). That is to say we can prove

(G(7, 4)← G(6, 4)), (A(0)← A(6), N(6, 0)) 0 A(0)

and use rule 3 to prove our sequent.

We have seen that we can derive at least some of the sequents whichare rule 4 premises using rules 1, 2, and 3. In fact, all of the infiniteset of premises of our particular application of rule 4 can be provedusing these three rules. One application of rule 4 gets us one of our twopremises of our desired case of rule 5, and the other was proved directlyby rule 2. With one application of rule 5, our deduction is complete.

Theorem 3.7. If a stable model logic programming skeptical reasoningsequent Σ; Γ|∼∆ is provable then it is true.



Proof.We will establish the soundness of each of our rules:

0. The soundness of the monotone sequent and antisequent rules hasalready been established.

1. Any stable model M of Γ satisfying Σ will contain Σ′ and will beclosed under all clauses of Γ′, so if any subset of BH containing Σ′

and closed under Γ′ must contain an element of ∆, then of courseM must contain an element of ∆.

2. Any stable model M of Γ satisfying Σ will contain Σ′ and will beclosed under all clauses of Γ′ and so must contain p. Thus, there willbe no stable model of Γ satisfying Σ and ¬Lp, and the conclusionis vacuously true.

3. If Γ0 0 p, then it is impossible for any subset M of BH to satisfyLΓ0p, and thus the conclusion is vacuously true.

4. Suppose that the conclusion of the rule, Lp,Σ; Γ|∼∆, were false.That is, assume that there is a stable model M of Γ satisfying Σwith M ∩ ∆ = ∅ and p ∈ M . Because p is in the least Herbrandmodel of ΓM , there must be some finite Γ′

0 ⊆ ΓM such that p is inthe least Herbrand model of Γ′

0. Define Γ′ as the set of clauses ofΓ whose M -reducts form Γ′

0. Because Γ′ consists entirely of ruleswhich survive the M -reduction process, and because of the way Σ′

0

was defined, we get essentially for free that M is a stable modelof Γ′

0 ∪ (Γ \ Γ′) satisfying Σ′0 ∪ Σ. M is therefore a witness that

LΓ′0p,Σ′

0,Σ; Γ′0, (Γ \ Γ′)|∼∆ is false. Thus, if all premises of rule 4

are true, so is the conclusion.

5. Suppose that the conclusion of the rule,

Σ; Γ, (p← q1, . . . , qm,not r1, . . . ,not rn)|∼∆,

were false. That is, suppose that there is a stable model M ofprogram Γ, (p← q1, . . . , qm,not r1, . . . ,not rn) satisfying the con-straints Σ and M ∩∆ = ∅. If {r1, . . . , rn}∩M = ∅, then M is also astable model of Γ, (p← q1, . . . , qm). (Because {r1, . . . , rn}∩M = ∅,the reduct p ← q1, . . . , qm of p ← q1, . . . , qm,not r1, . . . ,not rnwould appear in ΓM , so M must satisfy p ← q1, . . . , qm.) Thus, if{r1, . . . , rm}∩M = ∅, M witnesses that ¬Lr1, . . . ,¬Lrm,Σ; Γ, (p←q1, . . . , qm)|∼∆ is false. If, on the other hand rj ∈M , then M is alsoa stable model of Γ (because p ← q1, . . . , qm,not r1, . . . ,not rnwould be excluded in any reduction using M as context). Thus, if


22 R.S. Milnikel

rj ∈M , M witnesses that Lrj ,Σ; Γ|∼∆ is false. What we have justshown is that if the conclusion of rule 5 is false, so is some premise.This completes the argument for the soundness of the rules.

Theorem 3.8. A stable model logic programming skeptical reasoningsequent Σ; Γ|∼∆ is true then it is provable.

Proof. We will show the contrapositive of our completeness theorem.Let us assume that sequent Σ0; Γ0|∼∆0 has no proof. We will build afailed attempt at a proof which will be guaranteed to have at least onebranch not terminating in an axiom. The context developed on thisbranch will be a witness to the falsehood of Σ0; Γ0|∼∆0.

Choose an arbitrary well-ordering of the tree ω<ω, and let us workon extending the attempt at a proof we have so far by working on thesequent whose position is least in that ordering. We will also want towell-order BH and the rules of Γ0 in some way.

If the sequent Σ; Γ|∼∆ under consideration has a proof under rule1, 2, or 3 from true premises, make Σ; Γ|∼∆ the conclusion of that ruleand finish out the proof above that point. (If the premises of the rulewere true, they are guaranteed to have finite proofs by the completenessof the sequent and antisequent calculi, so completing the proof abovethat point can be done in a finite number of steps.) 1

If the sequent Σ; Γ|∼∆ whose branch we are trying to extend hasat least one Lp ∈ Σ that is not a LΓ′p, or has at least one non-Hornrule in Γ, we will be guaranteed that there are rules of type 4 or 5 withconclusion Σ; Γ|∼∆. If no rules of type 1, 2, or 3 with the appropriateconclusion (and true premises) can be found, extend the proof attemptwith either rule 4 or rule 5, if possible. If there are an even number ofsequents below the present one (and if it may be applied) use rule 4;if there are an odd number of sequents below the present one (and itmay be applied) use rule 5. When using rule 4, work with the Lp whichappeared earliest in the development of the proof attempt. (If morethan one appeared at the same time, choose the one whose p is least inthe chosen ordering of BH .) When using rule 5, work with the clausefrom Γ which is least in your ordering of Γ0. (Since no non-Horn ruleever gets added as we extend our proof attempt, all non-Horn rules inΓ will come from Γ0.)

1 This procedure is not effective, but a careful inspection will show it to be Π01,

if there is some priority assigned to the possible deductions using rules 1, 2, and 3.Thus, a proof can be considered as a finite-path Π0

1 tree. We knew that such a proofsearch existed from the fact that skeptical reasoning is in Π1

1 and from facts in [7].One can also find in [7] a procedure for turning Π0

1 tree searches into computabletree searches.



Of course, if no rule at all has conclusion Σ; Γ|∼∆, leave the sequentalone.

Since we assumed that the sequent Σ0; Γ0|∼∆0 has no proof, therewill be at least some branch of our attempted proof tree which was ex-panded using only rules 4 and 5 and which therefore does not terminatein an axiom. Select one of these branches.

Let us define Γ∗ to be the set of all Horn clauses which appear inthe Γ of a sequent Σ; Γ|∼∆ anywhere along the selected branch. Wewill show that M = Cl(Γ∗) is a stable model of Γ satisfying Σ andexcluding ∆ for every sequent Σ; Γ|∼∆ in our selected branch. Oncethis is accomplished, the proof will be complete, since M will be awitness that the sequent Σ0; Γ0|∼∆0 which we could not prove is false.

To establish thatM is a stable model of Γ satisfying Σ and excludingall of ∆, we need to establish the following:

(A) that M satisfies Lp (and LΓ′p) for each such in Σ,

(B) that M satisfies ¬Lp for each such in Σ,

(C) that Cl(ΓM ) = M , and

(D) that M ∩∆ = ∅.

Let us address each of these points.

(A) For each p for which Lp appears in Σ, LΓ′p appears somewherein the selected path, with Γ′ a subset of the set of Horn clausesappearing in the sequent in which it first appears. (This is due torule 4.) If Γ′ 0 p, we could terminate the branch of the tree with avalid proof of Γ′ 0 p based on rule 3. We chose this branch becausesuch a termination could not occur, so Γ′ ` p. By the way Γ∗ wasdefined, Γ′ ⊆ Γ∗, so we know that p ∈ Cl(Γ∗) = M . Thus, Msatisfies the requirements Lp. (The additional condition that weneed to satisfy the requirement LΓ′p for requirements of that typein Σ is that Γ′ ` p, for which we just presented a justification.)

(B) Assume that ¬Lp is in Σ and that p ∈ Cl(Γ∗). By the compactnessof proofs from Horn clauses, there is some finite subset Γ′ of Γ∗ suchthat Γ′ ` p. Because the length of our selected branch is at most ω,any finite subset of Γ∗ must be entirely present in the Γ′′ of a sequentΣ′′; Γ′′|∼∆′′ appearing at a finite stage of the construction. If bothof our assumptions were true, we could terminate this branch with(a proof of) the monotone sequent Γ′ ` p, but we chose this branchbecause that was impossible. Thus, if ¬Lp ∈ Σ, it must be thatp /∈ Cl(Γ∗) and so M satisfies the requirement ¬Lp.


24 R.S. Milnikel

(C) We will show that ΓM = Γ∗. This is sufficient to establish point (C):Cl(ΓM ) = Cl(Γ∗) = M . As a preliminary observation, let us notethat for no p are both Lp and ¬Lp in Σ′ for any sequent Σ′; Γ′|∼∆′

in our selected branch, for if they were, we could terminate ourbranch immediately with the monotone sequent p ` p by rule 2.

Let us assume that the Horn reduct of rule

p← q1, . . . , qm,not r1, . . . ,not rn ∈ Γ

is in Γ∗ but not in ΓM . That means that at some point in thedevelopment of our selected branch, either by rule 4 or by rule 5,p← q1, . . . , qm was placed into a sequent Σ′; Γ′|∼∆′. When either ofthose rules is used to expand a potential proof, {¬Lr1, . . . ,¬Lrn}is added to Σ′. On the other hand, if p ← q1, . . . qm is not in ΓM ,it must be because rj ∈ M for some 1 ≤ j ≤ n. This would meanthatM does not satisfy Σ′ for the sequent Σ′; Γ′|∼∆′ on our selectedbranch. In demonstrating point (B), we showed that this can nothappen.

Now let us assume that the Horn reduct of rule

p← q1, . . . , qm,not r1, . . . ,not rn ∈ Γ

is in ΓM but not in Γ∗. That it is not in Γ∗ means that at somepoint in the development of our selected branch, the clause wasdeleted when we moved to sequent Σ′; Γ′|∼∆′ by rule 5. That couldnot happen without Lrj being added to Σ′ at the same time. Onthe other hand, because p ← q1, . . . , qm is in ΓM , it must meanthat no rj is in M . That means that M does not satisfy Σ′ for thesequent Σ′; Γ′|∼∆′ on our selected branch. In demonstrating point(A), we showed that this can not happen.

(D) Assume that p ∈ ∆ ∩ Cl(Γ∗). Repeating the argument from part(B), we arrive at the same contradiction, this time based on rule 1instead of rule 2. This tells us that ∆ ∩M = ∅.

All conditions for M = Cl(Γ∗) to be a stable model of Γ satisfyingΣ and disjoint from ∆ are satisfied, and we see that any unprovablesequent is not true.



4. Skeptical Sequent Calculus for Default Logic

In moving from logic programs to default logic, the nonmonotonic as-pects are unchanged. It is only the monotone proofs in the underlyinglogic which are now more complicated. This will be reflected in thefact that the monotone sequent calculi will be more complicated, butthe rules added for skeptical reasoning will be essentially identical tothose used for stable model logic programming. The similarity is sogreat that we will state the soundness and completeness theorem forthe skeptical sequent calculus for default logic without proof, since theproof would be a nearly word-for-word repetition of the proof for theskeptical sequent calculus for logic programming.

Throughout this section, we will be working in a predicate languageL without equality.

4.1. An Antisequent Calculus for Predicate Logic

Because of the way extensions of open default theories are defined,the general case of skeptical default reasoning for even finite predicatedefault theories will necessitate infinitary sequent rules. For this reason,we will not hesitate to bring the enormous Π1

1 power of infinitary se-quent rules to bear on our comparatively simple Π0

1 problem of showingthat there is no proof in the sequent calculus LK of the sequent Γ ` ∆.

Bonatti in [5] presented an antisequent calculus for propositionallogic, and our antisequent calculus will be the one of that paper ex-tended by four rules, the counterparts of the four rules for quanti-fiers from Gentzen’s LK. We assume that the reader is familiar withGentzen’s sequent calculus LK. However, because there are many mi-nor variations of LK, and because we will be referring to specific rulesoften in this section, we will explicitly state the version we are using,drawn from [4]. In this formulation, a sequent Γ ` ∆ is an axiom if andonly if Γ∩∆ 6= ∅. The rules are to be found in Table I. (In the table, tis a term of L and we must restrict v to variables which do not occurfree in Γ ∪∆.)

An antisequent is a pair 〈Γ,∆〉 of finite sets of formulas, denotedΓ 0 ∆. We will call Γ 0 ∆ true if there is a model of Γ in which all ofthe formulas of ∆ are false. We have the benefit of the Soundness andCompleteness Theorems for LK, which tell us that Γ 0 ∆ is true if andonly if Γ ` ∆ is false if and only if Γ ` ∆ is not derivable in LK.

An antisequent Γ 0 ∆ will be considered an axiom of our antisequentcalculus if Γ ∪ ∆ consists entirely of atomic formulas and Γ ∩ ∆ = ∅.The rules for the antisequent calculus can be found in Table II.


26 R.S. Milnikel

Table I. Rules of the Sequent Calculus LK

(¬ `)Γ ` ∆, ϕ

Γ,¬ϕ ` ∆(` ¬)

Γ, ϕ ` ∆

Γ ` ∆,¬ϕ

(∧ `)Γ, ϕ, ψ ` ∆

Γ, ϕ ∧ ψ ` ∆(` ∧)

Γ ` ∆, ϕ Γ ` ∆, ψ

Γ ` ∆, ϕ ∧ ψ

(∨ `)Γ, ϕ ` ∆ Γ, ψ ` ∆

Γ, ϕ ∨ ψ ` ∆(` ∨)

Γ ` ∆, ϕ, ψ

Γ ` ∆, ϕ ∨ ψ

(→`)Γ ` ∆, ϕ Γ, ψ ` ∆

Γ, ϕ→ ψ ` ∆(`→)

Γ, ϕ ` ∆, ψ

Γ ` ∆, ϕ→ ψ

(∀ `)Γ, ϕ(t) ` ∆

Γ, (∀x)ϕ(x) ` ∆(` ∀) Γ ` ∆, ϕ(v)

Γ ` ∆, (∀x)ϕ(x)

(∃ `)Γ, ϕ(v) ` ∆

Γ, (∃x)ϕ(x) ` ∆(` ∃) Γ ` ∆, ϕ(t)

Γ ` ∆, (∃x)ϕ(x)

The usual proviso that v may not be free in Γ ∪ ∆ applies to 0 ∀and ∃ 0.

Showing that the antisequent calculus is sound is the counterpartto showing the classical sequent calculus LK complete, and vice versa.Fortunately, we have the Soundness and Completeness Theorems ofLK to rely on in our proof.

Theorem 4.1. If antisequent Γ 0 ∆ is provable, then it is true.

Proof. To show that if an antisequent is derivavable, it is true, wewill limit ourselves to an in-depth examination of the soundness of onlythree rules, (∧ 0), (0 •∧), and (∀ 0). Proving the soundness of the lastthese rules will rely heavily on the soundness and completeness of LK.

− To show (∧ 0) sound, let us assume that the premise is true andshow the conclusion is true. For Γ, ϕ, ψ 0 ∆ to be true, there mustbe a model of Γ ∪ {ϕ,ψ} in which every formula of ∆ is false.This same structure is also a model of Γ ∪ {ϕ ∧ ψ} in which everyformula of ∆ is false, so the antisequent Γ, ϕ ∧ ψ 0 ∆ is true.

− To show (0 •∧) sound, let us assume that the premise is true andshow that the conclusion is true. For Γ 0 ∆, ϕ to be true, there



Table II. Rules of the Antisequent Calculus

(¬ 0)Γ 0 ∆, ϕ

Γ,¬ϕ 0 ∆(0 ¬)

Γ, ϕ 0 ∆

Γ 0 ∆,¬ϕ

(∧ 0)Γ, ϕ, ψ 0 ∆

Γ, ϕ ∧ ψ 0 ∆(0 •∧)

Γ 0 ∆, ϕ

Γ 0 ∆, ϕ ∧ ψ

(0 ∧•) Γ 0 ∆, ψ

Γ 0 ∆, ϕ ∧ ψ

(•∨ 0)Γ, ϕ 0 ∆

Γ, ϕ ∨ ψ 0 ∆(0 ∨)

Γ 0 ∆, ϕ, ψ

Γ 0 ∆, ϕ ∨ ψ

(∨• 0)Γ, ψ 0 ∆

Γ, ϕ ∨ ψ 0 ∆

(• →0)Γ 0 ∆, ϕ

Γ, ϕ→ ψ 0 ∆(0→)

Γ, ϕ 0 ∆, ψ

Γ 0 ∆, ϕ→ ψ

(• →0)Γ, ψ 0 ∆

Γ, ϕ→ ψ 0 ∆

(∀ 0){Γ, ϕ(t) 0 ∆|t is a term in L}

Γ, (∀x)ϕ(x) 0 ∆(0 ∀) Γ 0 ∆, ϕ(v)

Γ 0 ∆, (∀x)ϕ(x)

(∃ 0)Γ, ϕ(v) 0 ∆

Γ, (∃x)ϕ(x) 0 ∆(0 ∃) {Γ 0 ∆, ϕ(t)|t is a term in L}

Γ 0 ∆, (∃x)ϕ(x)

must be a model of Γ in which every formula of ∆ ∪ {ϕ} is false.Of course, if ϕ is false in the model, so is ϕ∧ψ, so the same modeltells us that the sequent Γ 0 ∆, ϕ ∧ ψ is also true.

− To show (∀ 0) sound, let us assume that the conclusion is false.We will show that some premise must also be false. To say thatthe conclusion of the rule Γ, (∀x)ϕ(x) 0 ∆ is false is to say thatΓ, (∀x)ϕ(x) ` ∆ is true. By the completeness of LK, we know thatΓ, (∀x)ϕ(x) ` ∆ has a derivation. Let us show by induction on thedepth n of the derivation of Γ, (∀x)ϕ(x) ` ∆ that for some term tin L it must also be the case that Γ, ϕ(t) ` ∆ is derivable. Thatwill finish the argument for (∀ 0), because Γ, ϕ(t) ` ∆ derivablemeans Γ, ϕ(t) ` ∆ true, which in turn means Γ, ϕ(t) 0 ∆ false. So


28 R.S. Milnikel

by assuming the conclusion of (∀ 0) false, we can show that somepremise must also have been false.

If n = 0, then Γ, (∀x)ϕ(x) ` ∆ is an axiom of LK. Either (∀x)ϕ(x) ∈∆ or Γ ∩∆ 6= ∅. Of course, if Γ ∩∆ 6= ∅, then Γ, ϕ(t) ` ∆ for anyterm t. On the other hand, if (∀x)ϕ(x) ∈ ∆, let ∆ = ∆′, (∀x)ϕ(x).The following is a deduction of Γ, ϕ(x) ` ∆ as long as x is not freein Γ ∪∆′:

Γ, ϕ(x) ` ∆′, ϕ(x)Γ, ϕ(x) ` ∆′, (∀x)ϕ(x)

(` ∀).

(If x is free in Γ∪∆′, rename variables to get around the problem.)In this case, the term t is x.

If n = k + 1, we again face two cases, only one of which hasmuch content. If the last step of the derivation of Γ, (∀x)ϕ(x) ` ∆involved the use of the LK rule (∀ `) on formula ϕ, then thedefinition of that rule tells us immediately that for some term t,there must have been a derivation of Γ, ϕ(t) ` ∆. On the otherhand, if the last step of the derivation of Γ, (∀x)ϕ(x) ` ∆ involvedformulas entirely from Γ and/or ∆, then the derivation with thelast step excluded is a k-step derivation of Γ′, (∀x)ϕ(x) ` ∆′ forsome Γ′ and ∆′. By induction hypothesis, there is a derivation ofΓ′, ϕ(t) ` ∆′ for some term t. Apply the same rule that finishedthe k+ 1 step proof to extend the proof of Γ′, ϕ(t) ` ∆′ to a proofof Γ, ϕ(t) ` ∆.

The demonstration of soundness for each of the other propositionalrules is as straightforward as for the first two above, and one can makevery similar inductive arguments for the remaining three quantifierrules.

The reader may notice that the induction in the argument for thesoundness of (∀ 0) bears a resemblance to portions of the proof ofthe cut-elimination theorem, often a major ingredient in proofs of thecompleteness of LK. Just as the proof of the soundness of the antise-quent calculus evokes the proof of the completeness of LK, the proof ofthe completeness of the antisequent calculus will mirror the proof of thesoundness of LK. The completeness proof will involve a naıve inductionon the total number of connectives, unary and binary, in an unprovableantisequent Γ 0 ∆. The inductive portion of our argument will breakdown into cases which match up precisely with the rules for connectivesin LK. Again, we will look at only three cases of the inductive step.The reader interested in the details of each step is invited to considerany standard proof of the soundness of LK, see how our argumentparallels that proof in those cases, and extrapolate.



Theorem 4.2. If antisequent Γ 0 ∆ is true, then it can be proved.

Proof. Let us show that if there is no derivation of Γ 0 ∆, thenit is not true. We will show this by induction on the total number nof connectives and quantifiers in Γ 0 ∆. If n = 0, then all formulasof Γ 0 ∆ are atomic, but Γ 0 ∆ is not an axiom. That means thatΓ ∩∆ 6= ∅, and so it would be impossible to have a model of Γ whichdid not satisfy at least one formula of ∆, making Γ 0 ∆ false.

If n = k + 1, let us look at the various possibilities for the k + 1st

connective or quantifier. It could be any of ¬,∧,∨,→,∀,∃, and it couldbe on the left or right side of the 0. Let us look at a few of these casesin detail. The pattern of reasoning should become obvious.

− If the k + 1st connective is ∧ on the left side of the 0, thenΓ = Γ′, ϕ ∧ ψ, and Γ′, ϕ ∧ ψ 0 ∆ is not derivable. This meansthat Γ′, ϕ, ψ 0 ∆ is also not derivable, since if it were, we’d have aderivation of Γ′, ϕ ∧ ψ 0 ∆. Since Γ′, ϕ, ψ 0 ∆ has only k connec-tives, we can use our inductive hypothesis to say that Γ′, ϕ, ψ 0 ∆is false. If Γ′, ϕ, ψ 0 ∆ is false, then Γ′, ϕ, ψ ` ∆ is true (andderivable). By the soundness of LK, Γ′, ϕ ∧ ψ ` ∆ is also true,making Γ′, ϕ ∧ ψ 0 ∆ false.

− If the k + 1st connective is ∧ on the right side of the 0, then∆ = ∆′, ϕ∧ψ, and Γ 0 ∆′, ϕ∧ψ is not derivable. This means thatneither Γ 0 ∆′, ϕ nor Γ 0 ∆′, ψ are derivable. By our inductivehypothesis, neither antisequent is true, and so both Γ ` ∆, ϕ andΓ ` ∆, ψ are true (and hence derivable). By the soundness of LK,Γ ` ∆, ϕ ∧ ψ is also true, making Γ 0 ∆′, ϕ ∧ ψ false.

− If the k + 1st connective is ∀ on the right side of the 0, thenΓ = Γ′, (∀x)ϕ(x) and Γ′, (∀x)ϕ(x) 0 ∆ is not derivable. Thistells us that for some term t in language L, Γ′, ϕ(t) 0 ∆ is notderivable. By our inductive hypothesis, this antisequent is not true,and so Γ′, ϕ(t) ` ∆ is true and derivable. By the soundness of LK,Γ′, (∀x)ϕ(x) ` ∆ is true, and so Γ′, (∀x)ϕ(x) 0 ∆ is false.

Those three cases plus an inspection of how closely the rules of theantisequent calculus mirror those of LK should give the reader theflavor of the full argument.

Example 4.3. Let us show that (∀x)(P (x) ∨ Q(x)) 0 (∀y)(P (y)) ∨(∀z)(Q(z)) for unary relations P and Q. The following is a partialproof, with an infinite number of premises remaining at the top.


30 R.S. Milnikel

{P (t) ∨Q(t) 0 P (y), Q(z)|t is a term of L}(∀x)(P (x) ∨Q(x)) 0 P (y), Q(z)

(∀x)(P (x) ∨Q(x)) 0 P (y), (∀z)(Q(z))(∀x)(P (x) ∨Q(x)) 0 (∀y)(P (y)), (∀z)(Q(z))

(∀x)(P (x) ∨Q(x)) 0 (∀y)(P (y)) ∨ (∀z)(Q(z))

We are left with an infinite number of premises of the form P (t) ∨Q(t) 0 P (y), Q(z) to prove. By rules (•∨ 0) and (∨• 0), we need tobe able to show only either P (t) 0 P (y), Q(z) or Q(t) 0 P (y), Q(z). Aslong as t 6= y, P (t) 0 P (y), Q(z) is an axiom. If t = y, Q(t) 0 P (y), Q(z)is an axiom. This provides us with proofs of all of the infinitely manypremises of the form P (t)∨Q(t) 0 P (y), Q(z) and completes the proof.

4.2. A Sequent Calculus and an Antisequent Calculus forResidues

The sequent calculus for monotone proofs based on predicate logic willbe the standard propositional sequent calculus LK extended by tworules for dealing with residues, very closely related to our one-rulesequent calculus for Horn programs.

A residue sequent is a pair 〈Γ,∆〉 where both Γ ⊆ Lres and ∆ ⊆ Lare finite, and is usually written Γ ` ∆. We say that Γ ` ∆ is true if∨

∆ ∈ Cl(Γ).If we extend the classical predicate sequent calculus LK (restricted

to L) by the following two rules about residues, we obtain a sequentcalculus for residues.

Γ ` ∆Γ, ϕ

θ ` ∆Γ ` ϕ Γ, θ ` ∆

Γ, ϕθ ` ∆

This sequent calculus for residues was defined by Bonatti and Olivettiin [6], and they proved this theorem about it:

Theorem 4.4. Γ ` ∆ is derivable in the sequent calculus for residuesif and only if it is true.

Their proof was based on the soundness and completeness of propo-sitional rather than predicate logic, but the proof in either case isidentical, and quite similar to the proof of Proposition 3.1.

We can also extend antisequents to deductions from residues. Aresidue antisequent will be a pair of finite sets Γ ⊆ Lres and ∆ ⊆ Lwritten Γ 0 ∆. And just as the residue sequent Γ ` ∆ was consideredtrue if

∨∆ ∈ Cl(Γ), we will consider the residue antisequent Γ 0 ∆

to be true if∨

∆ /∈ Cl(Γ). Just as we extended the classical sequent



calculus (limited to L) by two rules to produce a sound and completeresidue sequent calculus, we will also extend the predicate antisequentcalculus (again limited to L) by the two rules below to produce a soundand complete residue antisequent calculus.

Γ 0 ∆ Γ 0 ϕ

Γ, ϕθ 0 ∆

Γ, θ 0 ∆Γ, ϕ

θ ` ∆

Bonatti and Olivetti proved the following in [6]:

Theorem 4.5. Antisequent Γ 0 ∆ is derivable in the antisequentcalculus of residues if and only if it is true.

Again, their proof was for a residue antisequent calculus built overpropositional rather than predicate logic, but the same proof workshere. It is very similar to the proof of Proposition 3.3.

4.3. Skeptical Sequent Calculus

As was alluded to in Section 2, stable model logic programming anddefault logic differ only in the complexity of the monotone logic un-derlying each. The nonmonotone parts of all definitions are identical,modulo changes in notation. For this reason, the definitions of skepticalsequents and of the skeptical sequent calculus for default logic willbe essentially identical to the analogous definitions for stable modellogic programming. The proof of the soundness and completeness ofthe skeptical sequent calculus would be identical to the one for logicprograms (nearly no changes in notation would even be necessary), soit will be omitted.

The sequents for skeptical reasoning for default logic will be triples〈Σ,Γ,∆〉, usually notated Σ; Γ|∼∆. Γ is a closed predicate default the-ory, and ∆ is a set of formulas of L, neither necessarily finite. Σ is onceagain a finite set of provability constraints. In the default logic context,provability constraints are of the form Lϕ, LΓϕ, or ¬Lϕ where ϕ ∈ Land Γ ∈ Lres is finite.

We will say that a theory T ⊆ L satisfies Lϕ if ϕ ∈ T and satisfies¬Lϕ if ϕ /∈ T . We will say that T satisfies LΓϕ if ϕ ∈ T and in additionΓ ` ϕ.

We will say that Σ; Γ|∼∆ is true if every extension S of Γ ⊆ Ldef

which satisfies all constraints in Σ includes at least one member of∆. Thus, ϕ is a skeptical consequence of the default theory (D,W ) if∅; ground((D,W ))|∼ϕ is a true sequent.

As before, this calculus incorporates three sorts of sequents: residuesequents, residue antisequents, and the skeptical reasoning sequentsjust defined. The sequent calculus for skeptical reasoning will include


32 R.S. Milnikel

all axioms and rules of the residue sequent and antisequent calculi,plus five new rules. (No additional axioms will be necessary. The leavesof every proof tree will be classical predicate sequent and antisequentaxioms.)

Definition 4.6 (Skeptical Sequent Calculus—Default Logic).The axioms of the skeptical sequent calculus are classical predicatesequents Γ ` ∆ with Γ∩∆ 6= ∅; and predicate antisequents Γ 0 ∆ withΓ ∪∆ all atomic formulas and Γ ∩∆ = ∅. The rules are:

0. The rules of LK and the antisequent rules from Table II, all limitedto sequents and antisequents in L; plus the two pairs of additionalrules for residue sequents and antisequents.

1.Σ′,Γ′ ` ∆′

Σ; Γ|∼∆where Γ′ ⊆ Γ ∩ Lres is finite, Σ′ ⊆ ({ϕ|Lϕ ∈ Σ} ∪

{ϕ|LΓ′′ϕ ∈ Σ for some Γ′′}), and ∆′ ⊆ ∆ is finite.

2.Σ′,Γ′ ` ϕ¬Lϕ,Σ; Γ|∼∆

where Γ′ ⊆ Γ ∩ Lres is finite, Σ′ ⊆ ({ϕ|Lϕ ∈ Σ} ∪

{ϕ|LΓ′′ϕ ∈ Σ for some Γ′′}).

3.Γ0 0 ϕ

LΓ0ϕ,Σ; Γ|∼∆where Γ0 ⊆ Lres is finite.

4.{LΓ′

0ϕ,Σ,Σ′

0; Γ′0, (Γ \ Γ′)|∼∆|Γ′ ⊆ Γ is finite}Lϕ,Σ; Γ|∼∆

where

Γ′0 = {ϕ

θ|ϕ;Mψ1, . . . ,Mψn

θ∈ Γ′} and Σ′

0 = {¬L¬ψ|ψ = ψj for

someϕ : Mψ1, . . . ,Mψn

θ∈ Γ′ and some 1 ≤ j ≤ n}.

5.¬L¬ψ1, . . . ,¬L¬ψn,Σ; Γ,

ϕ

θ|∼∆ L¬ψ1,Σ; Γ|∼∆ · · · L¬ψn,Σ; Γ|∼∆

Σ; Γ,ϕ : Mψ1, . . . ,Mψn

θ|∼∆

Theorem 4.7. A default logic skeptical reasoning sequent Σ; Γ|∼∆ istrue if and only if it is provable.

5. Skeptical Sequent Calculus for Autoepistemic Logic

Since autoepistemic logic—as we will consider it—is built over a baseof propositional logic in the modal language LL, we do not need to



establish any more monotone sequent or antisequent calculi. We canuse the propositional portions of LK and its antisequent counterpartgiven in Tables I and II in Section 4.1. The difference in dealing withautoepistemic logic is that the provability constraints, expressed in ourprevious types of sequents as L-formulas, are the very objects of thetheory. Thus, we do not want to construct our sequents in three parts,but only two, the traditional Γ and ∆ of a sequent Γ|∼∆. However, indealing with lack of proof, we will need not only to assert that such-and-such proposition is proved, but to state explicitly how it was proved.Were we dealing explicitly with a full nonmonotonic modal logic, wemight want to incorporate Artemov’s logic of proofs ([3]), but for ourpurposes, it will be sufficient to build our skeptical sequents out of acombination of formulas and classical sequents.

A sequent for skeptical reasoning in autoepistemic logic will be apair 〈Γ,∆〉, usually written Γ|∼∆, where Γ ⊆ LL ∪ {[Γ0 ` ∆0]|Γ0 ∪∆0 ⊆ LL is finite} and ∆ ⊆ LL. (Because, as before, our sequentcalculus rules will incorporate monotone sequent and antisequent rules,we enclose the classical sequents which are elements of a larger sequentin [brackets]. This is not an ideal notational situation, but [Γ0 ` ∆0]will be true as part of a larger sequent if and only if Γ0 ` ∆0 is trueas its own independent sequent, so there should not be any confusion.)The meaning of a sequent Γ|∼∆ will be: If monotone sequents in Γ aretrue, then each stable expansion of Γ ∩ LL contains some member of∆. The other notion that we will need is that of an L-subformula. ψis an L-subformula of ϕ if ψ is a subformula of ϕ of the form Lϕ′. Wewill denote by LS(ϕ) the set of subformulas of ϕ. We will also defineLS(Φ) =

⋃{LS(ϕ)|ϕ ∈ Φ}.

A last standard bit of shorthand: LΦ = {Lϕ|ϕ ∈ Φ} and ¬LΦ ={¬Lϕ|ϕ ∈ Φ}.

Definition 5.1 (Skeptical Sequent Calculus—Autoepistemic Logic).The axioms of the skeptical sequent calculus for autoepistemic logic arethe axioms of the classical sequent calculus LK and of its counterpartantisequent calculus, all restricted to the modal propositional languageLL. The rules of the sequent calculus are:

0. The propositional rules of the classical sequent and antisequentcalculi.

1.Γ′ ` ∆′

Γ|∼∆where Γ′ ⊆ Γ ∩ LL is finite and ∆′ ⊆ ∆ is finite.

2.¬Lϕ,Γ′ ` ϕ¬Lϕ,Γ|∼∆

where Γ′ ⊆ Γ ∩ LL is finite.


34 R.S. Milnikel

3.Γ0 0 ϕ

[Γ0 ` ϕ],Γ|∼∆where Γ0 ⊆ LL is finite.

4.{[Γ0, LΦ,¬LΨ ` θ], LΦ,¬LΨ,Γ|∼∆|Γ0 ⊆ Γ ∩ LL, LΦ ∪ LΨ ⊆ LS(Γ ∪ {Lθ}) are finite}

Lθ,Γ|∼∆.

5.Lϕ,Γ|∼∆ ¬Lϕ,Γ|∼∆

Γ|∼∆where Lϕ ∈ LS(Γ ∪∆).

We will conclude this section with the expected soundness and com-pleteness theorems for the skeptical sequent calculus for autoepistemiclogic.

Theorem 5.2. If autoepistemic logic skeptical reasoning sequent Γ|∼∆is derivable, then it is true.

Proof.We will establish the soundness of each of our rules:

0. The soundness of the classical propositional sequent and antise-quent calculi have already been established.

1. Because any stable expansion T of Γ∩LL will contain Γ′ and will beclosed under propositional provability, T must contain an elementof ∆.

2. Any stable model T of {¬Lϕ} ∪ Γ ∩ LL will contain Γ′ and willbe closed under propositional provability, and hence will containϕ. Because T is a stable model it will also contain Lϕ. Thus, Twill contain Lϕ and ¬Lϕ, making T inconsistent. Because stablemodels are, by definition, consistent, there is no such T and theconclusion is vacuously true.

3. If Γ0 0 ϕ, then it is not the case that Γ0 ` ϕ, and the conclusion isvacuously true.

4. Suppose that the conclusion of the rule, Lθ,Γ|∼∆, were false. Thatis, assume that there is a stable expansion T of Γ∩LL with T∩∆ = ∅and θ ∈ T . (We need Lθ ∈ T . Were T stable and θ /∈ T , then¬Lθ ∈ T and T would be inconsistent.) Because θ is a propositionalconsequence of Γ ∩ LL ∪ {Lϕ|ϕ ∈ T} ∪ {¬Lϕ|ϕ /∈ T}, it must be apropositional consequence of some finite subset thereof. It should beclear to the reader familiar with propositional logic that formulasLϕ and ¬Lϕ which are not in LS(Γ∪{Lθ}) will play no part in thispropositional proof. Choose a specific proof of θ. Let Φ be the set ofϕ ∈ T such that Lϕ is used in this proof, let Ψ be the set of ψ /∈ T



such that ¬Lψ is used in the proof, and let Γ0 be the formulas fromΓ∩LL used in the proof. Obviously, [Γ0, LΦ,¬LΨ ` θ] is true by ourchoice of Φ, Ψ, and Γ0. T itself will be a witness to the falsehood of[Γ0, LΦ,¬LΨ ` θ], LΦ,¬LΨ,Γ|∼∆, because T is a stable model ofΓ∩LL ∪{Lϕ|ϕ ∈ Φ}∪{¬Lψ|ψ ∈ Ψ} which has empty intersectionwith ∆. This will be a premise of rule 4 which is false. Thus, if allpremises of rule 4 are true, so is the conclusion.

(In this rule and the next, we don’t have to worry about classicalsequent [Γ′

0 ` θ′] ∈ Γ being false, since we assumed the conclusionof rule 4 false, and had any classical sequent in Γ been false, theskeptical sequent would have been vacuously true.)

5. Suppose that the conclusion of the rule, Γ|∼∆, were false. That is,suppose that there is a stable expansion T of Γ∩LL with T ∩∆ = ∅.Either ϕ ∈ T or ϕ /∈ T . If ϕ ∈ T , then T is a stable expansionof Lϕ,Γ with empty intersection with ∆. If ϕ /∈ T , then T is astable expansion of ¬Lϕ,Γ with empty intersection with T . If theconclusion of rule 5 is false, so is one of its two premises. Thus ifboth premises are true, so is the conclusion.

Theorem 5.3. An autoepistemic logic skeptical reasoning sequent Γ|∼∆is true then it is provable.

Proof. The proof of this theorem will be somewhat similar to that ofTheorem 3.8. Again, we will show the contrapositive of our complete-ness theorem. Let us assume that sequent Γ0|∼∆ has no proof. We willbuild a failed attempt at a proof which will be guaranteed to have atleast one branch not terminating in an axiom. The Γ developed on thisbranch will be a witness to the falsehood of Γ0|∼∆.

We will start with Γ0|∼∆ and build a proof attempt, expandingeach sequent with the premises of some rule having that sequent as aconclusion, if possible. We will be constructing a countably branchingtree, so we will need to work in some suitable ordering of ω<ω.

If the sequent Γ|∼∆ under consideration has a proof under rule 1,2, or 3 from true premises, make Γ|∼∆ the conclusion of that rule andfinish out the proof above that point.

If the sequent Γ|∼∆ whose branch we are trying to extend has atleast one Lθ ∈ Γ or has at least one Lϕ ∈ LS(Γ∪∆) such that neitherLϕ nor ¬Lϕ is in Γ, we will be guaranteed that there are rules oftype 4 or 5 with conclusion Γ|∼∆. If no rules of type 1, 2, or 3 withthe appropriate conclusion (and true premises) can be found, extendthe proof attempt with either rule 4 or rule 5, if possible. If there arean even number of sequents below the present one (and if it may be


36 R.S. Milnikel

applied) use rule 4; if there are an odd number of sequents below thepresent one (and it may be applied) use rule 5. When using rule 4, workwith the Lθ which appeared earliest in the development of the proofattempt. (If more than one appeared at the same time, choose the onewhose θ is least in some ordering of LL.) When using rule 5, work withthe ϕ least in your ordering of LL.

Of course, if no rule at all has conclusion Γ|∼∆, leave the sequentalone.

Since we assumed that the sequent Γ0|∼∆ has no proof, there will beat least some branch of our attempted proof tree which was expandedusing only rules 4 and 5 and which therefore does not terminate in anaxiom. Select one of these branches. (Note that neither rule 4 nor rule5 affects ∆, so while Γ will expand as we traverse the branch, ∆ willremain constant.)

Let us define Γ∗ to be the set of all formulas of LL which appear inthe Γ of a sequent Γ|∼∆ anywhere along the selected branch. We willshow that Th(Γ∗) can be extended to a stable theory which is a stableexpansion of Γ for each sequent Γ|∼∆ in our selected branch and whichexcludes all of ∆. Let us select an arbitrary Γ|∼∆.

The first step in this procedure will be to show Th(Γ∗) consistentwith S5. To accomplish this, we’ll use several claims.

− Claim 1: Γ∗ is consistent.

Justification: If Γ∗ weren’t consistent, then it would have to havebeen inconsistent by some finite stage in the development of thebranch. We could have used rule 1 to terminate the branch at thepoint where the inconsistency entered.

− Claim 2: If Lϕ ∈ Th(Γ∗), then Lϕ ∈ Γ∗.

Justification: Γ∗ consists of Γ plus some formulas of the form Lψand ¬Lψ. If Lϕ ∈ Th(Γ∗), but Lϕ /∈ Γ∗, it must have been becausesome formula of Γ was used to prove Lϕ. (Th refers to propositionalprovability and Lϕ is treated as a propositional atom, so this is theonly explanation.) That means that Lϕ is the conclusion of someimplication in Γ, and is therefore in LS(Γ). Because of the way Γ∗

was defined, for every Lψ ∈ LS(Γ), either Lψ ∈ Γ∗ or ¬Lψ ∈ Γ∗.Because Γ∗ is consistent, if Lϕ ∈ LS(Γ) ∩ Th(Γ∗), then Lϕ ∈ Γ∗.

− Claim 3: If ¬Lϕ ∈ Th(Γ∗), then ¬Lϕ ∈ Γ∗.

Justification: Same as for claim 2.

− Claim 4: If Lϕ ∈ Γ∗, then ϕ ∈ Th(Γ∗).



Justification: By rule 4, if Lϕ ∈ Γ∗, then so is [Γ′ ` ϕ] for somefinite Γ′ ⊆ Γ∗. Were Γ′ ` ϕ false, then the branch could have beenterminated using rule 3. Since the branch could not be terminated,it must be that Γ′ ` ϕ, and hence ϕ ∈ Th(Γ∗).

− Claim 5: If ¬Lϕ ∈ Γ∗, then ϕ /∈ Th(Γ∗).

Justification: If ϕ were in Th(Γ∗), then for some Γ′|∼∆ on theselected branch, it would be the case that Γ′ ` ϕ and ¬Lϕ ∈ Γ′,and the branch could have been terminated using rule 2.

− Claim 6: If Lϕ ∈ Th(Γ∗), then ϕ ∈ Th(Γ∗).

Justification: Immediate from claims 2 and 4.

− Claim 7: If ¬Lϕ ∈ Th(Γ∗), then ϕ /∈ Th(Γ∗).

Justification: Immediate from claims 3 and 5.

We can now easily show Th(Γ∗) to be consistent with S5.

− To be inconsistent with axiom k, we would need L(ϕ → ψ), Lϕ,and ¬Lψ in Th(Γ∗). By claims 6 and 7 above, this would putϕ → ψ and ϕ in Th(Γ∗) and leave ψ out of Th(Γ∗). So Th(Γ∗) isconsistent with axiom k.

− To be inconsistent with axiom t, we would need Lϕ ∈ Th(Γ∗) andϕ /∈ Th(Γ∗), contradicting claim 6.

− To be inconsistent with axiom 4, we would need Lϕ ∈ Th(Γ∗) and¬LLϕ ∈ Th(Γ∗). Claim 7 tells us that Lϕ /∈ Th(Γ∗), which isimpossible, so Th(Γ∗) is consistent with axiom 4.

− To be inconsistent with axiom 5, we would need ¬L¬Lϕ and ¬Lϕboth in Th(Γ∗). The former, with claim 7, tells us that ¬Lϕ /∈Th(Γ∗), so it must be that Th(Γ∗) is consistent with axiom 5.

We can now say by Proposition 2.13 that Th(Γ∗) can be extendedto a unique consistent stable theory T . Finally, we need to show T tobe a stable expansion of Γ.

Let us define S = Th(Γ ∪ {Lϕ|ϕ ∈ T} ∪ {¬Lϕ|ϕ /∈ T}). If we canshow that S = T , then we will have shown T to be a stable expansionof Γ.

That S ⊆ T is fairly clear. We know that Γ ⊆ Th(Γ∗) ⊆ T . BecauseT is stable, both {Lϕ|ϕ ∈ T} ⊆ T and {¬Lϕ|ϕ /∈ T} ⊆ T and T ispropositionally closed.

To show T ⊆ S, let us first show that for each Lϕ ∈ Γ∗, ϕ ∈ T .Because T extends Γ∗, if Lϕ ∈ Γ∗, then Lϕ ∈ T . If it were the case


38 R.S. Milnikel

that ϕ /∈ T , then by the stability of T , we would also have ¬Lϕ ∈ T .Because we know T to be consistent, this can not happen. A nearlyidentical argument shows that for each ¬Lϕ ∈ Γ∗, ϕ /∈ T .

Now, to show T ⊆ S, we’ll make use of Proposition 2.14, which saysthat

T = Th([T ]0 ∪ {Lϕ|ϕ ∈ T} ∪ {¬Lϕ|ϕ /∈ T}).Thus, all we need to show is that [T ]0 ⊆ S. Because Th(Γ∗) was consis-tent with S5, so is [Th(Γ∗)]0. Since each U ⊆ LL consistent with S5 iscontained in a unique stable theory T , it must be that [T ]0 = [Th(Γ∗)]0.So what we need to show is that any propositional formula in Th(Γ∗)can be proved from Γ, {Lϕ|ϕ ∈ T}, and {¬Lϕ|ϕ /∈ T}. We just arguedthat any L and ¬L formulas in Γ∗ are to be found in the latter twosets, and of course any formula of Γ∗ which is not of one of these twotypes must have come from Γ itself. We have shown not only that[Th(Γ∗)]0 ⊆ S, but that Th(Γ∗) ⊆ S.

One last step will complete the proof. We need to show that Texcludes all formulas of ∆. By rule 5, we know that Γ∗ contained eitherLϕ or ¬Lϕ for all formulas Lϕ ∈ LS(∆). We also know that [T ]0 =[Th(Γ∗)]0. Were it possible to prove any formula of ∆ from T , it wouldhave been possible to prove that formula from Γ∗. Were it possible toprove some formula of ∆ from Γ∗, it would have been possible to provethat formula from Γ′ for some sequent Γ′|∼∆ on our chosen branch.Had this been possible, we could have terminated the branch usingrule 1. Since we could not, it must be that Th(Γ∗) ∩∆ = ∅, and henceT ∩∆ = ∅.

6. Further Directions For Research

We have now seen sequent calculi for skeptical reasoning in stablemodel logic programming, default logic, and autoepistemic logic. Oneobvious direction to go to extend this work would be to do the same forpredicate circumscription. As we noted in the introduction, this wouldrequire a Π1

2 sequent calculus. Just as, we hope, the calculi presentedhere will aid in the understanding of the subtlety of skeptical reasoningin the logics discussed, a Π1

2 sequent calculus for circumscription mightoffer insights into that framework.

One distinguishing feature of the calculi presented above is that ineach case, the assertion that p or ϕ simply has a proof is not enough.We must look at all possible proofs of p (or ϕ). This necessity to takean assertion of the existence of a proof and explicate it with an actualproof fairly calls out for a connection with Artemov’s logic of proofs



(see [3]). Because the logic of proofs has such a strong connection withmodal logic, McDermott and Doyle’s nonmonotonic modal logics arethe natural candidates for the first study of such a connection.

Acknowledgements

Although this paper is newly written and the notation is so modifiedas to make the connections obscure, the core ideas presented hereinwere to be found in my doctoral dissertation, [17], written under thedirection of Anil Nerode. I would like to take this opportunity to thankhim for the support, help, and encouragement he offered throughoutmy graduate career.

References

1. Apt, K. R.: 1987, ‘Introduction to Logic Programming’. Technical ReportTR-97-35, University of Texas.

2. Apt, K. R.: 1990, ‘Logic Programming’. In: J. van Leeuwen (ed.): Handbook ofTheoretical Computer Science. Cambridge, MA: MIT Press, pp. 493–574.

3. Artemov, S. N.: 2001, ‘Explicit Provability and Constructive Semantics’. Bull.Symbolic Logic 7, 1–36.

4. Barwise, J.: 1977, ‘An Introduction to First-Order Logic’. In: J. Barwise (ed.):Handbook of Mathematical Logic. Amsterdam: North-Holland, pp. 5–46.

5. Bonatti, P.: 1993, ‘A Gentzen System for Non-Theorems’. Technical ReportCD-TR 93/52, Christian Doppler Labor fur Expertensysteme.

6. Bonatti, P. and N. Olivetti: 2002, ‘Sequent Calculi for Propositional Nonmono-tonic Logics’. ACM Trans. Comput. Log. pp. 226–278.

7. Cenzer, D. and J. B. Remmel: 1998, ‘Π01 Classes in Mathematics’. In: Y. L.

Ershov, S. S. Goncharov, V. W. Marek, A. Nerode, and J. B. Remmel (eds.):Handbook of Recursive Mathematics, Vol. 2. Amsterdam: North-Holland, pp.623–821.

8. Gelfond, M. and V. Lifschitz: 1988, ‘The Stable Semantics for Logic Programs’.In: R. A. Kowalski and K. A. Bowen (eds.): Proceedings of the 5th AnnualSymposium on Logic Programming. pp. 1070–1080.

9. Konolige, K.: 1994, ‘Autoepistemic Logic’. In: D. M. Gabbay, C. J. Hogger, andJ. A. Robinson (eds.): Handbook of Logic in Artificial Intelligence and LogicProgramming, Vol. 3. Oxford: Clarendon Press, pp. 217–296.

10. Lifschitz, V.: 1990, ‘On Open Defaults’. In: J. W. Lloyd (ed.): ComputationalLogic. Symposium Proceedings. pp. 80–95.

11. Lloyd, J. W.: 1987, Foundations of Logic Programming. Berlin: Springer-Verlag,second edition.

12. Marek, V. W., A. Nerode, and J. B. Remmel: 1990, ‘Nonmonotonic RuleSystems I’. Ann. Math. Art. Int. 1, 241–273.

13. Marek, V. W., A. Nerode, and J. B. Remmel: 1994, ‘The Stable Models of aPredicate Logic Program’. J. Log. Prog. 21, 129–154.


40 R.S. Milnikel

14. Marek, V. W. and M. Truszczynski: 1993, Nonmonotonic Logic: Context-Dependent Reasoning. Berlin: Springer Verlag.

15. McCarthy, J.: 1980, ‘Circumscription — A form of Nonmonotonic Reasoning’.Art. Int. 13, 27–39.

16. McDermott, D. and J. Doyle: 1980, ‘Nonmonotonic Logic I’. Art. Int. 13,41–72.

17. Milnikel, R. S.: 1999, ‘Nonmonotonic Logic: A Monotonic Approach’. Ph.D.thesis, Cornell University.

18. Milnikel, R. S.: 2003, ‘The Complexity of Predicate Default Logic Over aCountable Domain’. Ann. Pure Appl. Logic 120, 151–163.

19. Moore, R. C.: 1984, ‘Possible-World Semantics for the Autoepistemic Logic’.In: R. Reiter (ed.): Proceedings of the Workshop on Non-Monotonic Reasoning.pp. 344–354.

20. Moore, R. C.: 1985, ‘Semantical Considerations on Non-Monotonic Logic’. Art.Int. 25, 75–94.

21. Nerode, A. and R. A. Shore: 1997, Logic for Applications. Berlin: Springer-Verlag, second edition.

22. Reiter, R.: 1980, ‘A Logic for Default Reasoning’. Art. Int. 13, 81–132.23. Schlipf, J. S.: 1987, ‘Decidability and Definability with Circumscription’. Ann.

Pure Appl. Logic 35, 173–191.

Address for Offprints: Robert MilnikelDepartment of MathematicsKenyon CollegeGambier, OH 43022 USA


sequent calculi for skeptical reasoning in predicate

Documents