Splitting a Default Theory. Hudson Turner. University of Texas at Austin.

Transcription

1 Splitting a Default Theory Hudson Turner Department of Computer Sciences University of Texas at Austin Austin, TX , USA hudson@cs.utexas.edu Abstract This paper presents mathematical results that can sometimes be used to simplify the task of reasoning about a default theory, by \splitting it into parts." These so-called Splitting Theorems for default logic are related in spirit to \partial evaluation" in logic programming, in which results obtained from one part of a program are used to simplify the remainder of the program. In this paper we focus primarily on the statement and proof of the Splitting Theorems for default logic. We illustrate the usefulness of the results by applying them to an example default theory for commonsense reasoning about action. Introduction This paper introduces so-called Splitting Theorems for default logic, which can sometimes be used to simplify the task of reasoning about a default theory, by \splitting it into parts." These Splitting Theorems are related somewhat, in spirit, to \partial evaluation" in logic programming, in which results obtained from one part of a program are used to simplify the remainder of the program. In fact, the Splitting Theorems for default logic closely resemble the Splitting Theorems for logic programming introduced in (Lifschitz & Turner 994), despite complications due to the presence of arbitrary formulas in default theories. 2 Similar results for autoepistemic logic can be found in (Gelfond & Przymusinska 992). Related, independently obtained, results for default logic, restricted to the - nite case, appear in (Cholewinski 995). In this paper we focus on the statement and proof of the Splitting Theorems for default logic. 3 We illustrate See, for example, (Komorowski 99). 2 In (Lifschitz & Turner 994) we presented without proof Splitting Theorems for logic programs with classical negation and disjunction, under the answer set semantics (Gelfond & Lifschitz 99). The results for nondisjunctive logic programs follow from the Splitting Theorems for default logic. The denitions and proofs in this paper can be adapted to the more general case of disjunctive default logic (Gelfond et al. 99), from which the Splitting Theorems for disjunctive logic programs would follow as well. 3 We do not address the possible application of these the usefulness of the results by applying them to an example default theory for reasoning about action. An extended application of the Splitting Theorems to default theories for representing commonsense knowledge about actions will appear in (Turner 997). The paper is organized as follows. We present preliminary denitions. We then introduce the Splitting Set Theorem, followed by Splitting Sequence Theorem, which generalizes it. We exercise the Splitting Sequence Theorem on an example default theory for reasoning about action. We present an abridged proof of the Splitting Set Theorem, followed by a detailed proof of the Splitting Sequence Theorem. Preliminary Denitions Given a set U of atomic symbols (not including the special constants > and?), we denote by L(U) the language of propositional logic with exactly the atoms U f>;?g. 4 We say a set of formulas from L(U) is logically closed if it is closed under propositional logic. We write inference rules over L(U) as expressions of the form where and are formulas from L(U). When convenient, we identify a formula with the inference rule >. We say a set of formulas is closed under a set R of inference rules if for all 2 R, if 2 then 2. By Cn U (R) we denote the least logically closed set of formulas from L(U) that is closed under R. A default rule over L(U) is an expression of the form ; ; n () where ; ; ; n ; are formulas from L(U) (n ). Let r be a default rule of form (). We call the prerequisite of r, and denote it by pre(r). The formulas ; ; n are the justications of r; we write just(r) to denote the set f ; ; n g. We call the consequent of r, and denote it by cons(r). If pre(r) is > we often ideas to the problem of automated default reasoning. 4 Thus, L(;) consists of all formulas in which the only atoms are the constants > and?.

2 omit it, writing ; ; n instead. If just(r) = ;, we identify r with the corresponding inference rule.5 A default theory over L(U) is a set of default rules over L(U). Let D be a default theory over L(U), and E a set of formulas from L(U). The reduct of D by E, denoted by D E, is the following set of inference rules. pre(r) cons(r) r 2 D and for all 2 just(r); =2 E We say E is an extension of D if E = Cn U (D E ) Default logic is due to Reiter (Reiter 98). The (essentially equivalent) denitions given above follow (Gelfond et al. 99). Although the denitions in this section are stated for the propositional case, they are taken, in the standard way, to apply in the rst-order, quantier-free case as well, by taking each non-ground expression to stand for all of its ground instances. Splitting Sets Let D be a default theory over L(U) such that, for every rule r 2 D, pre(r) is in conjunctive normal form. (Of course any default theory can be easily transformed into an equivalent default theory, over the same language, satisfying this condition.) For any rule r 2 D, a formula is a constituent of r if at least one of the following conditions holds (i) is a conjunct of pre(r); (ii) 2 just(r); (iii) = cons(r). A splitting set for D is a subset A of U such that for every rule r 2 D the following two conditions hold. Every constituent of r belongs to L(A) L(U n A). If cons(r) does not belong to L(U n A), then r is a default rule over L(A). If A is a splitting set for D, we say that A splits D. The base of D relative to A, denoted by b A (D), is the default theory over L(A) that consists of all members of D that are default rules over L(A). Let U 3 = fa; b; c; dg. Consider the following default theory D 3 over L(U 3 ). b a a b a _ b a; b c _ d a ^ (c _ d) d d b ^ (c _ d) c Take A 3 = fa; bg. It's easy to verify that A 3 splits D 3, with b b A3 (D 3 ) = a ; a b Notice that default theory b A3 (D 3 ) over L(A 3 ) has two consistent extensions Cn A3 (fag) and Cn A3 (fbg). Given a splitting set A for D, and a set X of formulas from L(A), the partial evaluation of D by X with respect to A, denoted by e A (D; X), is the default theory over L(U n A) obtained from D in the following manner. For each rule r 2 D n b A (D) such that 5 Allowing the empty set of justications is mathematically convenient (Brewka 99; Marek & Truszczynski 993), although it seems Reiter (Reiter 98) may have meant to prohibit it. c every conjunct of pre(r) that belongs to L(A) also belongs to Cn A (X), and no member of just(r) has its complement in Cn A (X) there is a rule r 2 e A (D; X) such that pre(r ) is obtained from pre(r) by replacing each conjunct of pre(r) that belongs to L(A) by >, and just(r ) = just(r) \ L(U n A), and cons(r ) = cons(r). For example, it is easy to verify that > > ^ (c _ d) d e A3 (D 3 ; Cn A3 (fag)) = ; c _ d d and that > > ^ (c _ d) c e A3 (D 3 ; Cn A3 (fbg)) = ; c _ d c Let A be a splitting set for D. A solution to D with respect to A is a pair hx; i of sets of formulas satisfying the following two properties. X is a consistent extension of the default theory b A (D) over L(A). is a consistent extension of the default theory e A (D; X) over L(U n A). For example, given our previous observations, it is easy to verify that D 3 has two solutions with respect to A 3 h Cn A3 (fag) ; Cn U3nA 3 (fc; dg) i and h Cn A3 (fbg) ; Cn U3nA 3 (fc; dg) i Splitting Set Theorem. Let A be a splitting set for a default theory D over L(U). A set E of formulas is a consistent extension of D i E = Cn U (X ) for some solution hx; i to D with respect to A. Thus, for example, it follows from the Splitting Set Theorem that default theory D 3 has exactly two consistent extensions Cn U3 (fa; c; dg) and Cn U3 (fb; c; dg). Splitting Set Corollary. Let A be a splitting set for a default theory D over L(U). If E is a consistent extension of D, then the pair h E \ L(A) ; E \ L(U n A) i is a solution to D with respect to A. Splitting Sequences A (transnite) sequence is a family whose index set is an initial segment of ordinals f g. We say that a sequence ha i of sets is monotone if A A whenever, and continuous if, for each limit ordinal, A = S A. A splitting sequence for a default theory D over L(U) is a nonempty, monotone, continuous sequence ha i of splitting sets for D s.t. S A = U.

3 Holds(Up(Left); S ) ^ Holds(Up(Right); S ) ^ Holds(Spilled; S ) Precedes(S ; S ) ^ Occurs(Raise(Left); S ) ^ Occurs(Raise(Right); S ) Precedes(S ; S 2 ) ^ Occurs(Lower(Left); S ) Occurs(Raise(x); s) ^ Precedes(s; s ) Holds(Up(x); s ) Holds(f; s) ^ Precedes(s; s ) Holds(f; s ) Holds(f; s ) Occurs(Lower(x); s) ^ Precedes(s; s ) Holds(Up(x); s ) Holds(Up(Left); s) 6 Holds(Up(Right); s) Holds(Spilled ; s) Figure Default theory D Holds(f; s) ^ Precedes(s; s ) Holds(f; s ) Holds(f; s ) The notion of a solution with respect to a splitting set is extended to splitting sequences as follows. Let A = ha i be a splitting sequence for D. A solution to D with respect to A is a sequence he i of sets of formulas that satises the following three conditions. is a consistent extension of the default theory b A (D) over L(A ). E For any such that +, E + is a consistent extension of the default theory over L(A + n A ). e ba+ (D); E A For any limit ordinal, E = Cn ; (;). We generalize the Splitting Set Theorem as follows. Splitting Sequence Theorem. Let A = ha i be a splitting sequence for a default theory D over L(U). A set E of formulas is a consistent extension of D if and only if E = Cn U E for some solution he i to D with respect to A. The proof of the Splitting Sequence Theorem, which appears in later in the paper, relies on the Splitting Set Theorem. We also have the following counterpart to the Splitting Set Corollary. Splitting Sequence Corollary. Let A = ha i be a splitting sequence for a default theory D over L(U). Let hu i be the sequence of pairwise disjoint subsets of U such that for all U = A n A If E is a consistent extension of D, then the sequence h E \ L(U ) i is a solution to D with respect to A. Example Splitting a Default Theory for Reasoning About Action We will illustrate the use of the Splitting Sequence Theorem by applying it to a default theory that formalizes commonsense knowledge about actions. The action domain we formalize is based on the \Soup Bowl" example from (Baral & Gelfond 993), which involves reasoning about the eects of concurrent actions. 6 The default logic formalization we consider relies, implicitly, on the notion of \static causal laws," recently investigated in (McCain & Turner 995; Turner 996). Default Theory D We will formalize the following action domain in default logic. There is a bowl of soup. There are actions that raise and lower its left and right sides. If one side of the bowl is up while the other is not up, the soup is spilled. In the initial situation S, both sides of the bowl are not up, and the soup is not spilled. We let S be the situation that would result from raising both sides of the bowl simultaneously. Notice that, intuitively speaking, it follows that in situation S both sides of the bowl are up and the soup is not spilled. We also consider an additional situation S 2 that would result from lowering the left side of the bowl, when in situation S. Intuitively speaking, we can conclude that in situation S 2 the left side of the bowl is not up, the right side is up, and the soup is spilled. We will use the Splitting Sequence Theorem to demonstrate that the default theory D, shown in Figure, is a correct formalization this action domain. More precisely, what we will show is that the literals Holds(Up(Left); S ), Holds(Up(Right); S ), and Holds(Spilled; S ) are among the consequences of default theory D, which establishes that D correctly describes the values of the uents in situation S. 6 We will not attempt in this paper a general analysis of the problem of representing commonsense knowledge about concurrent actions. Nor will we attempt a general justication of the style of representation employed in the example default theory itself. Such analysis and justication is an important problem that is beyond the scope of this paper.

4 Holds(Up(Left); S ) ^ Holds(Up(Right); S ) ^ Holds(Spilled; S ) Precedes(S ; S ) ^ Occurs(Raise(Left); S ) ^ Occurs(Raise(Right); S ) Precedes(S ; S 2 ) ^ Occurs(Lower(Left); S ) Holds(Up(Left); S ) 6 Holds(Up(Right); S ) Holds(Spilled; S ) Figure 2 Default theory b A (D ) Holds(Up(Left); S ) Holds(Up(Right); S ) Holds(Up(Left); S ) 6 Holds(Up(Right); S ) Holds(Spilled ; S ) Holds(Up(Left); S ) Holds(Up(Right); S ) Holds(Up(Left); S ) Holds(Up(Right); S ) Holds(Spilled; S ) Holds(Spilled ; S ) Figure 3 Default theory e A (b A (D ); E ) Holds(Up(Left); S 2 ) Holds(Up(Left); S 2 ) 6 Holds(Up(Right); S 2 ) Holds(Spilled; S 2 ) Holds(Up(Left); S 2 ) Holds(Up(Right); S 2 ) Holds(Up(Left); S 2 ) Holds(Up(Right); S 2 ) Holds(Spilled; S 2 ) Holds(Spilled; S 2 ) Figure 4 Default theory e A (b A2 (D ); E E ) Similarly, we will also show that D entails the literals Holds(Up(Left); S 2 ), Holds(Up(Right); S 2 ), and Holds(Spilled; S 2 ), which establishes that D also correctly describes situation S 2. Splitting Default Theory D Before applying the Splitting Sequence Theorem, we must be more precise about the language of D Let U consist of all ground atoms of the following many-sorted, rst-order language L. The sorts of L are situation, action, uent and side. There are object constants S, S and S 2 of sort situation, Left and Right of sort side, and Spilled of sort uent. There is a unary function symbol Up of sort side uent, and unary function symbols Raise and Lower of sort side action. There is a binary predicate symbol Holds of sort uent situation, a binary predicate symbol Occurs of sort action situation, and a binary predicate symbol Precedes of sort situation situation. We take D to be the (propositional) default theory over L(U ) consisting of all ground instances in L of the rules shown in Figure. Now we can specify a splitting sequence for D, which will allow us to break D into simpler parts. To begin, let A consist of all ground instances in L of the following atoms Holds(f; S ) ; Occurs(a; s) ; Precedes(s; s ) Notice that A splits D, with b A (D ) the default theory over L(A ) shown in Figure 2. We obtain A by adding to A all ground instances in L of Holds(f; S ). Finally, let A 2 = U. One easily checks that ha ; A ; A 2 i is a splitting sequence for D. Let X consist of the following literals Holds(Up(Left); S ) ; Holds(Up(Right); S ) ; Holds(Spilled; S ) ; Precedes(S ; S ) ; Occurs(Raise(Left); S ) ; Occurs(Raise(Right); S ) ; Precedes(S ; S 2 ) ; Occurs(Lower(Left); S ) Take E = Cn A (X ) Notice that E is the unique extension of b A (D ). The default theory e A (b A (D ); E ) over L(A na ) is (essentially) as shown in Figure 3. Thus, if we take X to consist of the literals Holds(Up(Left); S ) ; Holds(Up(Right); S ) ; Holds(Spilled; S ) ; and let E = Cn AnA (X ) we nd that E is the unique extension of default theory e A (b A (D ); E ). Finally, observe that the default theory e A (b A2 (D ); E E ) over L(A 2 n A ) is essentially as shown in Figure 4. Let X 2 consist of the literals Holds(Up(Left); S 2 ) ; Holds(Up(Right); S 2 ) ; Holds(Spilled ; S 2 ) Take E 2 = Cn A2nA (X 2 ) It is not dicult to verify that E 2 is the unique extension of e A (b A2 (D ); E E ). We have shown that he ; E ; E 2 i is a solution to default theory D with respect to ha ; A ; A 2 i. In fact it is the unique solution. It follows by the Splitting Sequence Theorem that Cn U (E E E 2 ) is the unique consistent extension of D. This shows that the literals Holds(Up(Left); S ), Holds(Up(Right); S ), and Holds(Spilled; S ) are among the consequences of D, which, as we discussed previously, is intuitively correct. We have similarly established that the literals Holds(Up(Left); S 2 ), Holds(Up(Right); S 2 ), and Holds(Spilled; S 2 ) are among the consequences of D, which is again the intuitively correct result.

5 Proof of the Splitting Set Theorem Due to space constraints, we present an abridged proof of the Splitting Set Theorem, omitting several intermediate lemmas, and proofs of the remaining lemmas. Lemma Let U; U be disjoint sets of atoms. Let R be a set of inference rules over L(U), and R a set of inference rules over L(U ). Let X = Cn UU (R R ). If X is consistent, then X \ L(U) = Cn U (R). X = Cn UU (Cn U (R) R ). In proving the Splitting Set Theorem it is convenient to introduce a set of alternative denitions, diering very slightly from those used in stating the theorem. (We nonetheless prefer the original denitions, because they are more convenient in applications of the Splitting Theorems.) Let D be a default theory over L(U) split by A. We dene the following. t A (D) = f r 2 D cons(r) 2 L(U n A) g b (D) = A D n t (D) A The advantage of these alternative denitions is captured in the following key lemma, which fails to hold for their counterparts b A (D) and D n b A (D). Lemma 2 Let D be a default theory over L(U) with splitting set A. For any set X of formulas from L(U), b A (D)X = b A (DX ) and t A (D)X = t A (DX ). The proof will also make use of the following additional alternative denitions. Given a set X of formulas from L(A), let e A (D; X) be the default theory over L(U n A) obtained from D in the following manner. For each rule r 2 t A (D) s.t. every conjunct of pre(r) that belongs to L(A) also belongs to Cn A (X), and no member of just(r) has its complement in Cn A (X) there is a rule r 2 e A (D; X) such that pre(r ) is obtained from pre(r) by replacing each conjunct of pre(r) that belongs to L(A) by >, and just(r ) = just(r) \ L(U n A), and cons(r ) = cons(r). Notice that e diers from A e A only in starting with the rules in t A (D) instead of the rules in D n b A(D). Finally, let s A (D) be the set of all pairs hx; i s.t. X is a consistent extension of b A (D), and is a consistent extension of e A (D; X). The following lemma shows that these alternative denitions are indeed suitable for our purpose. Lemma 3 If a default theory D over L(U) is split by A, then s A (D) is precisely the set of solutions to D with respect to A. Now we present the two main lemmas, and then use them to prove the Splitting Set Theorem. Lemma 4 Let D be a default theory over L(U) with splitting set A. Let E be a consistent set of formulas from L(U). E = Cn U (D E ) if and only if E \ L(A) = Cn A b A (D)E\L(A) ] and E = Cn U (E \ L(A)) t A (D)E ]. Lemma 5 Let D be a default theory over L(U) split by A. Let E be a logically closed set of formulas from L(U), with X = E \ L(A) and = E \ L(U n A). We have Cn U X e A (D; X) ] = Cn U X t A (D)E ] Proof of Splitting Set Theorem. Given a default theory D over L(U) with splitting set A, we know by Lemma 3 that s A (D) is precisely the set of solutions to D with respect to A. We will show that E is a consistent extension of D if and only if E = Cn U (X ) for some hx; i 2 s A (D). ()) Assume E is a consistent extension of D. Let X = E \ L(A) and = E \ L(U n A). By Lemma 4, X = Cn A (b A (D)X ) and E = Cn U (X t A (D)E ). By Lemma 5, E = Cn U (X e A (D; X) ). By Lemma, we can conclude that = Cn UnA (e A (D; X) ). So we have established that hx; i 2 s A (D). We can also conclude by Lemma that E = Cn U X Cn UnA (e A (D; X) )]. And since = Cn UnA (e A (D; X) ), we have E = Cn U (X ). (() Assume E = Cn U (X ) for some hx; i 2 s A (D). Since hx; i 2 s A (D), we have = Cn UnA (e A (D; X) ). Hence E = Cn U X Cn UnA (e A (D; X) )]. By Lemma, we can conclude that E = Cn U (X e A (D; X) ). Thus, by Lemma 5, E = Cn U (X t A (D)E ). By Lemma, E \ L(A) = Cn A (X), and since X is logically closed, Cn A (X) = X. So E \ L(A) = X. Since hx; i 2 s A (D), we have X = Cn A (b A (D)X ). We can conclude by Lemma 4 that E = Cn U (D E ). 2 Proof of Splitting Set Corollary. Assume that E is a consistent extension of D. By the Splitting Set Theorem, there is a solution hx; i to D with respect to A such that E = Cn U (X ). Since X L(A) and L(U n A), we can conclude by Lemma that E \ L(A) = Cn A (X). And since X is logically closed, Cn A (X) = X. So E \ L(A) = X. A symmetric argument shows that E \ L(U n A) =. 2 Proof of Splitting Sequence Theorem Lemma 6 Let D be a default theory over L(U) with splitting sequence A = ha i. Let E be a set of formulas from L(U). Let X = hx i be a sequence of sets of formulas from L(U) such that X = E \ L(A ), for all s.t. +, X + = E \ L(A + n A ), for any limit ordinal, X = Cn ; (;). If E is a consistent extension of D, then X is a solution to D with respect to A.

6 Proof. There are three things to check. First, by the Splitting Set Corollary, we can conclude that E \ L(A ) is a consistent extension of b A (D). Second, choose such that +. We must show that X + is a consistent extension of e ba+ (D); X A (2) Let = +. By the Splitting Set Corollary, E \ L(A ) is a consistent extension of b A (D). Let D = b A (D) and let E = E \ L(A ). By the Splitting Set Corollary, since A splits D, E \ L(A n A ) is a consistent extension of e A (D ; E \ L(A )). It is easy to verify that X + = E \ L(A n A ). It is not dicult to verify also that e A (D ; E \ L(A )) is the same as (2). Third, for any limit ordinal, X = Cn ; (;). 2 Lemma 7 Let D be a default theory over L(U) with splitting sequence A = ha i. Let he i be a solution to D with respect to A. For all Cn E A is a consistent extension of b A (D). Proof. For all, let X = Cn E A Proof is by induction on. Assume that for all, X is a consistent extension of b A (D). We'll show that X is a consistent extension of b A (D). There are two cases to consider. Case is not a limit ordinal. Choose such that + =. By the inductive hypothesis, X is a consistent extension of b A (D). We also know that E is a consistent extension of S e A (b A (D); E ). Let D = b A (D). It is clear that b A (D) = b A (D ). It is not dicult to verify that S e A (b A (D); E ) is the same as e A (D ; X ). So we've shown that X is a consistent extension of b A (D ) and that E is a consistent extension of e A (D ; X ). By the Splitting Set Theorem, it follows that Cn A (X E ) is a consistent extension of D. And since it's easy to check that Cn A (X E ) = X, we're done with the rst case. Case 2 is a limit ordinal. First we show that X is closed under b A (D) X. So suppose the contrary. Thus there is an r 2 b A (D) X such that pre(r) 2 X and cons(r) =2 X. Since A is continuous and is a limit ordinal, we know there must be a such that r 2 b A (D) X. Since b A (D) is a default theory over L(A ), we have b A (D) X = b A (D) X. So r 2 b A (D) X. Furthermore, it follows that pre(r) 2 X and cons(r) =2 X. This shows that X is not closed under b A (D) X, which contradicts the fact that, by the inductive hypothesis, X is a consistent extension of b A (D). So we have shown that X is closed under b A (D) X. Now, let E = Cn A (b A (D) X ). We will show that E = X, from which it follows that X is a consistent extension of b A (D). Since X is logically closed and closed under b A (D) X, we know that E X. Suppose E 6= X, and consider any formula 2 X n E. Since A is continuous and is a limit ordinal, there must be a such that is from L(A ) and therefore 2 X. Thus, X is a proper superset of E \ L(A ). By the inductive hypothesis, we know that X is a consistent extension of b A (D). Thus, X = Cn A (b A (D) X ). And since b A (D) X = b A (D) X, we have X = Cn A (b A (D) X ). Since E = Cn A (b A (D) X ) and b A (D) X b A (D) X, we know that E is closed under b A (D) X. Moreover, since b A (D) X is a default theory over L(A ), E \ L(A ) is closed under b A (D) X. But X is the least logically closed set closed under b A (D) X, so X E \L(A ), which contradicts the fact that X is a proper superset of E \ L(A ). We can conclude that E = X, which completes the second case. 2 Let D be a default theory over L(U) with splitting sequence A = ha i. The standard extension of A is the sequence B = hb i + such that for all, B = A, and B = U. Notice that the standard extension of A is itself a splitting sequence for D. Lemma 8 Let D be a default theory over L(U) with splitting sequence A = ha i. Let B = hb i + be the standard extension of A. Let X = hx i be a sequence of sets of formulas from L(U). Let = h i + be dened as follows. For all, = X. = Cn ; (;). If X is a solution to D with respect to A, then is a solution to D with respect to B. Proof. First, it's clear that is a consistent extension of b B (D), since = X, b B (D) = b A (D), and X is a consistent extension of b A (D). Similarly, it's clear that for any such that +, + is a consistent extension of e bb+ (D); A We also know that for any limit ordinal, = Cn ; (;). It remains to show that we handle correctly. There are two cases to consider. Case is a limit ordinal. In this case we must show that = Cn ; (;), which it does.

7 Case 2 is not a limit ordinal. In this case, choose such that + =. We must show that is a consistent extension of the default theory e bb (D); A (3) over L(B n B ). Since A is a splitting sequence for a default theory over L(U), we know that S A = U. Moreover, since A is monotone and is not a limit ordinal, it follows that A = U. And since B = A, we know that b B (D) = D. It follows that default theory (3) is empty. It also follows that B n B = ;, so the language of (3) is L(;). Since = Cn ; (;), we have shown that is a consistent extension of (3). 2 Proof of Splitting Sequence Theorem. ()) Assume that E is a consistent extension of D. By Lemma 6, there is a solution he i to D with respect to ha i for which it is not dicult to verify that E = Cn U E (() Assume that X = hx i is a solution to D with respect to ha i. Let E = Cn U X Let B = hb i + be the standard extension of ha i. By Lemma 8, we know there is a solution h i + to D with respect to B such that E = Cn U + Moreover, we know there is an + such that B = U. Thus b B (D) = D and E = Cn A It follows by Lemma 7 that E is a consistent extension of D. 2 Proof of Splitting Sequence Corollary. Assume that E is a consistent extension of D. By the Splitting Sequence Theorem, there is a solution hx i to D S with respect to A such that E = Cn U X. We will show that for all, E \ L(U ) = X. Let X = S X. Consider any. We have X L(U ), X n X L(U n U ), and E = Cn U (X X n X ). Thus, by Lemma we can conclude that E \ L(U ) = Cn U (X ). And since X is logically closed, we have Cn U (X ) = X. 2 Acknowledgments Many thanks to Vladimir Lifschitz and Norm McCain. This work was partially supported by National Science Foundation grant IRI References Baral, C., and Gelfond, M Representing concurrent actions in extended logic programming. In Proc. of IJCAI-93, 866{87. Besnard, P An Introduction to Default Logic. Springer-Verlag. Brewka, G. 99. Nonmonotonic Reasoning Logical Foundations of Commonsense. Cambridge University Press. Cholewinski, P Reasoning with stratied default theories. In Proc. of 3rd Int'l Conf. on Logic Programming and Nonmonotonic Reasoning, 273{286. Gelfond, M., and Lifschitz, V. 99. Classical negation in logic programs and disjunctive databases. New Generation Computing 9365{385. Gelfond, M., and Przymusinska, H On consistency and completeness of autoepistemic theories. Fundamenta Informaticae XVI59{92. Gelfond, M.; Lifschitz, V.; Przymusinska, H.; and Truszczynski, M. 99. Disjunctive defaults. In Allen, J.; Fikes, R.; and Sandewall, E., eds., Principles of Knowledge Representation and Reasoning Proc. of the 2nd Int'l Conference, 23{237. Komorowski, J. 99. Towards a programming methodology founded on partial deduction. In Proc. of ECAI-9. Lifschitz, V., and Turner, H Splitting a logic program. In Van Hentenryck, P., ed., Proc. Eleventh Int'l Conf. on Logic Programming, 23{37. Marek, W., and Truszczynski, M Nonmonotonic Logic Context-Dependent Reasoning. Springer- Verlag. McCain, N., and Turner, H A causal theory of ramications and qualications. In Proc. of IJCAI- 95, 978{984. Poole, D Default logic. In Gabbay, D.; Hogger, C.; and Robinson, J., eds., The Handbook of Logic in AI and Logic Programming, volume 3. Oxford University Press. 89{25. Reiter, R. 98. A logic for default reasoning. Arti- cial Intelligence 3(,2)8{32. Turner, H Representing actions in default logic A situation calculus approach. In Working Papers of the Third Symposium on Logical Formalizations of Commonsense Reasoning. Turner, H Representing actions in logic programs and default theories A situation calculus approach. Journal of Logic Programming. Forthcoming.