2 Review: Stable models of propositional theories

We assume that formulas are built from propositional atoms and the symbol $\bot$ using the binary connectives $\land$ , $\lor$ , $\rightarrow$ ; $\neg F$ stands for $F\rightarrow\bot$ , and $F\leftrightarrow G$ stands for $(F\rightarrow G)\land (G\rightarrow F)$ . A propositional theory is a set of formulas. An interpretation is a set of atoms; we identify an interpretation I with the truth assignment that maps the elements of I to true and all other atoms to false.

The reduct $F^I$ of a formula F with respect to an interpretation I is the formula obtained from F by replacing every maximal subformula of F that is not satisfied by I with $\bot$ Ferraris (Reference Ferraris2005, Section 2.1). The reduct $T^I$ of a propositional theory T is the set of the reducts $F^I$ of all formulas F in T. An interpretation I is a stable model of a propositional theory T if it is minimal (with respect to set inclusion) among the models of $T^I$ .

Consider, for instance, the formulas:

(3) \begin{equation}\begin{array} lp\rightarrow q,\\q\land\neg r \rightarrow p,\end{array}\end{equation}

corresponding to rules (2). The reduct of each of them with respect to the interpretation $\emptyset$ is the tautology $\bot\rightarrow\bot$ ; since $\emptyset$ is a minimal model of this tautology, it is a stable model of theory (3). The reduct of (3) with respect to $\{p,q\}$ is

\begin{equation*}\begin{array} lp\rightarrow q,\\q\land\neg \bot \rightarrow p.\end{array}\end{equation*}

The interpretation $\{p,q\}$ is a model of this reduct, but it is not minimal: its subset $\emptyset$ is a model of the reduct as well. Consequently, $\{p,q\}$ is not a stable model of (3).

It is easy to check by induction that an interpretation I satisfies the reduct $F^I$ if and only if it satisfies F. It follows that every stable model of a propositional theory T is a model of T.

It is clear also that every atom occurring in $F^I$ belongs to I.

3 A tale of two graphs

A nondisjunctive rule is an implication whose consequent is an atom. Take a set T of nondisjunctive rules. What graph will we designate as the positive dependency graph of T? As far as the set of vertices is concerned, the decision is straightforward – we will include all atoms that occur in the members:

(4) \begin{equation}{Body}\rightarrow H\end{equation}

of T. How will we choose the edges of the graph? For every formula (4) in T, the graph will include edges going from H to some of the atoms occurring in Body. But how will we decide which of the atoms occurring in Body to choose as the heads of edges?

A subformula of a formula F is called strictly positive if it does not belong to the antecedent of any implication. For instance, in a conjunction of literals:

\begin{equation*}B_1\land\cdots\land B_m\land\neg B_{m+1}\land\cdots\neg B_n\end{equation*}

the atoms $B_1,\dots,B_m$ are strictly positive, and the atoms $B_{m+1},\dots,B_n$ are not (recall that $\neg B_i$ is shorthand for the implication $B_i\rightarrow\bot$ ). In our more general definition of the positive dependency graph, it would be natural to include, for every member (4) of T, the edges from H to all atoms that have

We will denote the graph formed from T according to this rule by $G^{sp}(T)$ . (The superscript sp stands for strictly positive.)

However, the publications mentioned in the introduction (Ferraris et al. Reference Ferraris, Lee and Lifschitz2006; Ferraris et al. Reference Ferraris, Lee and Lifschitz2011; Lifschitz and Yang Reference Lifschitz and Yang2013; Ferraris et al. Reference Ferraris, Lee, Lifschitz and Palla2009; Harrison and Lifschitz Reference Harrison and Lifschitz2016) use a different, and more complicated, definition of the positive dependency graph. A subformula of a formula F is called

• positive if the number of implications containing it in the antecedent is even, and

• nonnegated if it does not belong to the antecedent of any implication with the consequent $\bot$ .

The graph designated as the positive dependency graph of T in the publications mentioned above has the same vertices as $G^{sp}(T)$ , but its edges go from H to all atoms that have

for all members (4) of T. We will denote this graph by $G^{pnn}(T)$ . (The superscript pnn stands for positive nonnegated.) It is clear that $G^{sp}(T)$ is a subgraph of $G^{pnn}(T)$ . For example, if T is

\begin{equation*}((p\rightarrow q)\rightarrow r)\rightarrow s\end{equation*}

then the only edge of $G^{sp}(T)$ is (s,r); $G^{pnn}(T)$ has two edges, (s,r) and (s,p).

4 Which graph is right for your problem?

The definitions of $G^{sp}$ and $G^{pnn}$ in Section 3 are limited to sets of nondisjunctive rules. We will now extend them to arbitrary propositional theories; this generalization will be used in Sections 4.2–4.4.

A strictly positive occurrence of an implication ${Body}\rightarrow{Head}$ in a formula F is called a rule of F. For any propositional theory T, by $G^{sp}(T)$ we denote the directed graph such that

1. (a) its vertices are the atoms occurring in the members of T, and

2. (b) for every rule ${Body}\rightarrow{Head}$ of any member of T, it includes the edge (H,B) for every atom B that has at least one strictly positive occurrence in Body and every atom H that has at least one strictly positive occurrence in Head.

By $G^{pnn}(T)$ we denote the directed graph satisfying conditions (a) and

1. (b’) for every rule ${Body}\rightarrow{Head}$ of any member of T, it includes the edge (H,B) for every atom B that has at least one positive nonnegated occurrence in Body and every atom H that has at least one strictly positive occurrence in Head.

For any formula F, we will write $G^{sp}(\{F\})$ as $G^{sp}(F)$ , and similarly for $G^{pnn}$ .

4.1 Supported models

A model I of a set T of nondisjunctive rules is supported if every atom A in I is the consequent of some member ${Body}\rightarrow A$ of T such that I satisfies ${Body}$ . Supported models are important because of their relation to program completion (Clark Reference Clark1978; Lloyd and Topor Reference Lloyd and Topor1984): for any finite set T of nondisjunctive rules, an interpretation I is a model of the completion of T if and only if I is a supported model of T (Apt et al. Reference Apt, Blair and Walker1988).

Every stable model of a set of nondisjunctive rules is supported, but the converse is, generally, not true. For instance, $\{p,q\}$ is a supported model of (3), but it is not stable. From published work on generalizations of Fages’ theorem, we know that the stability of all supported models can be asserted for the sets T of nondisjunctive rules such that the graph $G^{pnn}(T)$ has no infinite paths Lifschitz and Yang (Reference Lifschitz and Yang2013), Electronic Appendix B. (For finite T, this is the same as assuming that the graph is acyclic.) We will show that the graph $G^{sp}(T)$ has the same property:

Theorem 1 For any set T of nondisjunctive rules, if the graph $G^{sp}(T)$ has no infinite paths then every supported model of T is stable.

Thus cycles and other infinite paths in $G^{pnn}(T)$ containing edges that are not included in $G^{sp}(T)$ are harmless – they do not destroy the match between stable models and supported models. For instance, let T be the pair of formulas:

(5) \begin{equation}\begin{array} lp\rightarrow q,\\((q\rightarrow r)\rightarrow r)\rightarrow p.\end{array}\end{equation}

The graph $G^{sp}(T)$ has two edges, (q,p) and (p,r), and it is acyclic. Consequently, the stable models of T are identical to its supported models $\emptyset$ , $\{p,q\}$ . The graph $G^{pnn}(T)$ is not acyclic in this case because of the additional edge (p,q).

4.2 Loops

For any formula F and any set Y of atoms occurring in F, the “negated external support” formula $\hbox{NES}_F(Y)$ is defined recursively, as follows:

• for an atom A, $\hbox{NES}_A(Y)$ is $\bot$ if $A\in Y$ , and A otherwise;

• $\hbox{NES}_\bot(Y) = \bot$ ;

• $\hbox{NES}_{F\land G}(Y) = \hbox{NES}_F(Y) \land \hbox{NES}_G(Y)$ ;

• $\hbox{NES}_{F\lor G}(Y) = \hbox{NES}_F(Y) \lor \hbox{NES}_G(Y)$ ;

• $\hbox{NES}_{F\rightarrow G}(Y) = (\hbox{NES}_F(Y) \rightarrow \hbox{NES}_G(Y)) \land (F\rightarrow G)$

Ferraris et al. (Reference Ferraris, Lee and Lifschitz2006), Section 3. A set I of atoms occurring in F is a stable model of F iff it satisfies both F and the loop formulas:

(6) \begin{equation} \bigwedge_{A\in Y}(A \rightarrow \neg\hbox{NES}_F(Y)) \end{equation}

for all sets Y of atoms occurring in F Ferraris et al. (Reference Ferraris, Lee and Lifschitz2006), Theorem 2. Furthermore, according to the same theorem, there is no need to check all loop formulas (6). A set Y of atoms occurring in F is called a loop for F if the subgraph of $G^{pnn}(F)$ induced by Y is strongly connected. If I satisfies both F and the loop formulas (6) for all loops Y of F, then I is a stable model of F.

The discussion in Section 4.1 suggests the question: will the last result remain true if we replace the graph $G^{pnn}(F)$ in the definition of a loop by the smaller graph $G^{sp}(F)$ ? The answer to this question is no. A counterexample is given by the formula:

(7) \begin{equation}(p\rightarrow q)\land(((q\rightarrow p)\rightarrow p)\rightarrow p)\end{equation}

as F, and $\{p,q\}$ as I. Indeed, the edges of the graph $G^{sp}(F)$ in this case are (q,p) and (p,p), and the sets Y for which the subgraph of $G^{sp}(F)$ induced by Y is strongly connected are $\{p\}$ and $\{q\}$ . Calculations show that each of the formulas

\begin{equation*}\hbox{NES}_F(\{p\}),\ \hbox{NES}_F(\{q\})\end{equation*}

is equivalent to $\neg p\land\neg q$ , so that each of the loop formulas

\begin{equation*}p\rightarrow\neg \hbox{NES}_F(\{p\}),\ q\rightarrow\neg \hbox{NES}_F(\{q\})\end{equation*}

is a tautology. Thus, I is a model of F that satisfies these loop formulas, although it is not stable.

The graph $G^{pnn}(F)$ , on the other hand, has one more edge, (p,q). The subgraph of this graph induced by $\{p,q\}$ is strongly connected, and the corresponding loop formula eliminates the model I.

4.4 Splitting

Splitting a logic program (Lifschitz and Turner Reference Lifschitz and Turner1994) allows us to relate its stable models to stable models of its parts. The form of splitting described below is a special case of published results on splitting first-order formulas (Ferraris et al. Reference Ferraris, Lee, Lifschitz and Palla2009) and infinitary propositional theories (Harrison and Lifschitz Reference Harrison and Lifschitz2016), expressed in a form convenient for our present purposes.

Let $\{P,Q\}$ be a partition of the set of atoms occurring in a formula $F\land G$ . If

1. (i) every atom that has a strictly positive occurrence in F belongs to P, and

2. (ii) every atom that has a strictly positive occurrence in G belongs to Q, and

3. (iii) every strongly connected component of $G^{pnn}(F\land G)$ is contained in P or in Q,

then any set of atoms is a stable model of $F\land G$ if and only if it is a stable model of each of the formulas:

\begin{equation*}\begin{array} cF\land\bigwedge_{A\in Q}(A\lor\neg A),\ G\land\bigwedge_{A\in P}(A\lor\neg A).\end{array}\end{equation*}

This assertion will become incorrect, however, if we replace $G^{pnn}(F\land G)$ in condition (iii) by $G^{sp}(F\land G)$ . A counterexample is given by formula (7) as $F\land G$ , $\{q\}$ as P, and $\{p\}$ as Q. Indeed, $\{p,q\}$ is a stable model of each of the formulas:

\begin{equation*}\begin{array} l(p\rightarrow q)\land(p\lor\neg p),\\(((q\rightarrow p)\rightarrow p)\rightarrow p)\land(q\lor\neg q),\end{array}\end{equation*}

but not a stable model of (7).

5 Proofs of theorems

It is convenient to prove Theorem 2 first.

For any formula F, ${\rm SPos}(F)$ stands for the set of atoms that have at least one strictly positive occurrence in F. For any propositional theory T, ${\rm SPos}(T)$ is the union of the sets ${\rm SPos}(F)$ over all formulas F in T.

Lemma 2 Let F be a propositional formula, let I,J be interpretations such that $J \subset I$ , and let M be an atom in $I\setminus J$ such that

(8) $${\rm{for \, every \, edge }}\,(M,A)\,{\rm{of}}\,{G^{sp}}({F^I}),\,A \in J{\rm{.}}$$

If M belongs to ${\rm SPos}(F^I)$ and J satisfies $F^I$ , then $I \setminus \{ M \}$ satisfies $F^I$ as well.

The proof is by structural induction. Formula F is neither an atom nor $\bot$ . Indeed, otherwise $F^I$ would be an atom or $\bot$ too; since $M\in{\rm SPos}(F^I)$ , $F^I=M$ . Since $M\in I\setminus J$ , this contradicts the assumption that J satisfies $F^I$ .

Let F be $F_1\land F_2$ so that $F^I$ is $F_1^I\land F_2^I$ . Since J satisfies $F^I$ , J satisfies $F^I_i$ $(i=1,2)$ . We need to show that $I\setminus\{M\}$ satisfies $F^I_i$ as well. Case 1: ${M \in {\rm SPos}(F_i^I)}$ . Since every rule of $F_i^I$ is a rule of $F^I$ , $G^{sp}(F_i^I)$ is a subgraph of $G^{sp}(F^I)$ ; from (2), we can conclude that

$${\rm{for \, every \, edge }}\,(M,A)\,{\rm{of}}\,{G^{sp}}({\rm{F}}_{\rm{i}}^{\rm{I}}),\,A \in J{\rm{.}}$$

Then $I \setminus\{M\}$ satisfies $F^I_i$ by the induction hypothesis. Case 2: ${M \not\in {\rm SPos}(F_i^I)}$ . Since ${\rm SPos}(F_i^I)$ is a subset of I, it follows that ${\rm SPos}(F_i^I) \subseteq I \setminus \{ M \}$ . On the other hand, I satisfies $F_i$ , because J satisfies $F_i^I$ . By Lemma 1, these two facts imply that $I \setminus \{ M \}$ satisfies $F_i^I$ .

If F is $F_1\lor F_2$ , then the proof is similar.

Let F be $F_1\to F_2$ . Then $F^I$ is $F_1^I \to F_2^I$ and ${{\rm SPos}(F^I) = {\rm SPos}(F_2^I)}$ so that ${M \in {\rm SPos}(F_2^I)}$ . It follows that for every atom A in ${\rm SPos}(F_1^I)$ , the graph $G^{sp}(F^I)$ has an edge from M to A. Hence, by assumption (8), every such atom A belongs to J. Thus,

(9) \begin{equation}{\rm SPos}(F_1^I) \subseteq J.\end{equation}

Case 1: J satisfies $F_2^I$ . Since every rule of $F_2^I$ is a rule of $F^I$ , $G^{sp}(F_2^I)$ is a subgraph of $G^{sp}(F^I)$ ; from (8) we can conclude that

$${\rm{for \, every \, edge }}\,(M,A)\,{\rm{of}}\,{G^{sp}}(F_2^I),\,A \in J.$$

By the induction hypothesis, it follows that $I\setminus\{M\}$ satisfies $F^I_2$ , and consequently satisfies $F^I$ .

Case 2: J does not satisfy $F_2^I$ . Then I does not satisfy $F_1$ . Indeed, otherwise we would be able to conclude by (9) and Lemma 1 that J satisfies $F_1^I$ , which contradicts the assumption that J satisfies $F^I$ . Hence ${F_1^I = \bot}$ , and $F^I$ is a tautology.

We will show first that the set $I\setminus J$ contains an atom M satisfying condition (8).

Case 1: $I\setminus J$ contains an atom that is not a vertex of $G^{sp}(T^I)$ . Then condition (8) holds for that atom trivially.

Case 2: all atoms in $I\setminus J$ are vertices of $G^{sp}(T^I)$ . Assume that condition (8) is not satisfied for any of the vertices M in $I\setminus J$ , so that

Since the set $I\setminus J$ is non-empty, it follows that the graph $G^{sp}(T^I)$ has an infinite path. But this is impossible, because $G^{sp}(T^I)$ is a subgraph of $G^{sp}(T)$ .

Take an atom M in $I\setminus J$ that satisfies condition (8), and any formula F from T. If $M \in {\rm SPos}(F^I)$ , then we conclude that $I\setminus\{M\}$ satisfies $F^I$ by Lemma 2. Otherwise, ${\rm SPos}(F^I)\subseteq I\setminus\{M\}$ , and $I\setminus\{M\}$ satisfies $F^I$ by Lemma 1.

Take any atom A in I; we need to show that $I\setminus\{A\}$ is not a model of $T^I$ . Since I is supported, T contains a nondisjunctive rule ${Body}\rightarrow A$ such that I satisfies Body. The atom A has no strictly positive occurrences in Body; otherwise, $A,A,\dots$ would be an infinite path in $G^{sp}(T)$ . Consequently,

\begin{equation*}{\rm SPos}({Body}^I) \subseteq{{\rm SPos}({Body}) \subseteq I \setminus \{ A \}}.\end{equation*}

By Lemma 1, it follows that $I \setminus \{ A \}$ satisfies ${Body}^I$ . Therefore, $I \setminus\{ A \}$ does not satisfy the formula ${Body}^I \to A$ , which belongs to $T^I$ .