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<Paper uid="E93-1005">
  <Title>Decidability and Undecidability in stand-alone Feature Logics</Title>
  <Section position="4" start_page="31" end_page="33" type="metho">
    <SectionTitle>
3 Decidability
</SectionTitle>
    <Paragraph position="0"> To begin our search for decidable fragments we will take our cue from Kasper and Rounds' original work.</Paragraph>
    <Paragraph position="1"> Kasper and Rounds' system was negation free, so the first question to ask is: what happens if we simply remove negation from L KR:~? Of course, if this is all we do we trivialise the satisfiability problem: it is immediate by induction on the structure of negation free wffs C/, that every negation free L KR~ wff is satisfied in the following model: M = ({w}, {Rt}tec, V) where Rz = {(w,w)} for all/E/~, and Y(p) = {w} for all propositional variables p. So we have regained decidability, but in a very uninteresting way.</Paragraph>
    <Paragraph position="2"> Now, what made the results of Kasper and Rounds interesting was that not only did they consider the negation free fragment (of LKR), they also imposed certain semantic restrictions. Only extensional models without constant-constant or constant-compound clashes were considered. 6 Will imposing any (or all) of these restrictions make it easier to find decidable fragments of L KR~ ? In fact demanding extensionality (that is, working only with models in which each atomic symbol is true at at most one node), does make it easy to find a decidable fragment.</Paragraph>
    <Paragraph position="3"> The fragment is the following. We consider wffs of the following form: Here C/ is a metavariable over L Kn wffs (that is, C/ contains no occurrences of =C/,); the ai (1 &lt; i &lt; n) SThese papers take the universal modality \[\] as primitive rather that =C/,, as it is somewhat easier to work with unary modalities. In the presence of full Boolean expressivity I:3 and :=~ axe interdefmable: Elf is True :=~ C/, and ::~ is 1:3(C/ --. ~b). However in what follows we will work with fragments without enough Boolean expressivity to interdefine these operators. As :C/, is the operator we are really interested in we have chosen it as our primitive here.</Paragraph>
    <Paragraph position="4"> SAs Kasper and Rounds showed, introducing this limited form of negation failure results in an NP complete satisfiability problem.</Paragraph>
    <Paragraph position="5">  are metavariables over combinations of sort symbols containing only V and A as logical operators; and the ai (1 &lt; i &lt; n) are metavariable over L KR wits.</Paragraph>
    <Paragraph position="6"> Note the general form of the wffs of this fragment.</Paragraph>
    <Paragraph position="7"> We have an L KR wff C/ conjoined with n general constraints o~i :2z tq. 7 The C/ can be thought of as the AVM associated with some particular natural language sentence, while the wffs of the form c~i ::~ ~i can be thought of as encoding the generalisations embodied in our grammatical theory. Looking for a satisfying model for a wff from this fragment is thus like asking whether the analysis of some particular string of symbols is compatible with a grammar.</Paragraph>
    <Paragraph position="8"> The proof that this fragment has a decidable satisfiability problem is straightforward. We're going to show that given any wff@ belonging to this fragment, there is an upper bound on the size of the models that need to be inspected to determine whether or not (I) is satisfiable. The fact that such an upper bound exists is a direct consequence of three lemmas which we will now prove.</Paragraph>
    <Paragraph position="9"> The first lemma we need is extremely obvious, but will play a vital role.</Paragraph>
    <Paragraph position="10"> Lemma 3.1 Let a be any wff containing no logical connectives apart from V and A. Then in any extensional model, c~ is satisfied at at most m nodes, where m is the number of distinct sort symbols in a.</Paragraph>
    <Paragraph position="11"> Proof: By induction on the construction of a. \[\] The importance of this lemma is that it gives us an upper bound on the number of nodes at which the antecedents ai of the constraints permitted in our fragment can be satisfied.</Paragraph>
    <Paragraph position="12"> Next we need a similar result for the L KR wffs of the fragment; that is, for the C/ and the consequents ~i of the constraints. As the next two lemmas establish, given any L KR wff C which is satisfiable at a node w in some model M, we can always manufacture a very small model MInodes(C , w) which also satisfies 4. How we go about defining MInodes(C, w) is suggested by the following observation: when evaluating a formula in some model, only certain of the model's nodes are relevant to the truth or falsity of the wff; all the irrelevant nodes can be thrown away.</Paragraph>
    <Paragraph position="13"> What the following two lemmas essentially tell us is that we can manufacture the small models we need by discarding nodes.</Paragraph>
    <Paragraph position="14"> The nodes that are relevant when evaluating an L KR wff C at a node w in a model M are the nodes selected by the function nodes : WFF x W , Pow(W) that satisfies the following conditions: ;'In what follows we refer to the cq as the antecedents of the constraints, and the ai as the consequents.</Paragraph>
    <Paragraph position="16"> One aspect of the definition of nodes may be bothering the reader: there is no clause for the path equations. In fact to give such a clause is rather messy, and it seems better to proceed as follows. Given a wff C of L Ks we define C* to be the result of replacing every subformula of the form</Paragraph>
    <Paragraph position="18"> 4&amp;quot; is (all we've done is make the node existence demands encoded in the path equalities explicit). The usefulness of this transformation is simply that the two new conjuncts make available to the simple version of nodes defined above all the information hidden in the path equations. From now on we'll assume that all the L KR wffs we work with have been transformed in this fashion.</Paragraph>
    <Paragraph position="19"> With these preliminaries out of the way we are ready to proceed. Given a model M, an L KR wff C/ and a node w we form Mlnodes(C , w) in the obvious way: the nodes of the model are nodes(C, w), and the relations and valuation are the restriction of those of M to this subset. As the following simple lemma shows, nodes indeed picks out the correct nodes: Lemma 3.2 (Selection Lemma) For all models M, all nodes w of M and all L h'R~ wffs g'.</Paragraph>
    <Paragraph position="20"> M ~ C\[w\] iff Mlnodes(C/, w) ~ C/\[u,\].</Paragraph>
    <Paragraph position="21"> Proof: By induction on the structure of C/,. (Note that it follows from the definition of nodes that w E nodes(C,w). Once this is observed the induction is straightforward.) D The selection lemma is a completely general fact about modal languages. It doesn't depend on any special assumptions made in this paper, and in particular it doesn't make any use of the fact that we are only working with models in which each of the Rt is a partial function. Once this additional fact is taken into account, however, we see that M\[nodes(~,, w)is pleasingly small: there can only be one more node in M\[nodes(C/, w) than there are occurrences of modalities in C/. That is, we have:  Lemma 3.3 (Size Lemma) Let W be an L KR wff, and let mod(W) be the number of occurrences of modalities in W. Then for all models M and all nodes w in M we have that Inodes(W,w)\{w}l &lt; ,nod(W).</Paragraph>
    <Paragraph position="22"> Proof: By induction on the structure of ~b. Cl We now have all the pieces we need to establish the decidability result. Using these lemmas we can show that given any wff (I) of our fragment it is possible to place an upper bound on the size of models that need to be checked to determine whether or not (I) is satisfiable. So, suppose (b is a wff of the form</Paragraph>
    <Paragraph position="24"> that is satisfiable. That is, there is a model M and a node w in M such that M ~ ~\[w\]. Now, simply forming M\[nodes(q~,w) is not a process guaranteed to make a smaller model satisfying (I). The problem is that while this model certainly satisfies ~, in the course of selecting all the needed nodes we may be forced to select a node that verifies an antecedent ai of one of the general constraints, but we have no guarantee that we have selected all the nodes needed to make the matching consequent tC/i true.</Paragraph>
    <Paragraph position="25"> But this is easy to fix. We must not only form Mlnodes(e~, w), but in addition, for all i (1 &lt; i &lt; n) we must form Mlnodes(cq ^ ai, s), where s ranges over all the nodes in M that satisfy c~i. More precisely, we define a new model M' by taking as nodes all the nodes in all these models (that is, we take the union of all the nodes in all these models) and we define the M' relations and valuation to be the restriction of the relations and valuation in M to this subset.</Paragraph>
    <Paragraph position="26"> The new model M' has two nice properties.</Paragraph>
    <Paragraph position="27"> Firstly, it is clear that it makes ~ true at w and moreover, whenever it makes one of the ai true it makes the corresponding ~i true also. (This follows because of our choice of the nodes of M'; essentially we're making multiple use of the selection lemma here.) Secondly, it is clear that M' is finite, for its nodes were obtained as a finite union of finite sets. Indeed by making use of lemma 3.1 and the size lemma we can give an upper bound on the size of M' in terms of the number of symbols in (I). (This is just a matter of counting the number of general constraints in (I), the number of distinct propositional variables in the c~i, and the number of modal operators in the and tzi; we leave the details to the reader.) Thus the decidability result follows: given a wff if) of our fragment, bounded search through finite models suffices to determine whether or not (I) is satisfiable.</Paragraph>
    <Paragraph position="28"> Alas, this is not a very powerful result. The fragment simply is not expressive enough to function as a stand-alone formalism. Its Achilles heel lies in the strong condition imposed on the ai. There are two problems. First, because the ai cannot contain occurrences of features or path equations, many important constraints that stand-alone feature might have to impose cannot be expressed. Second, it is far from clear that the restriction to extensional models is realistic for stand alone formalisms. Certainly if we were trying to capture the leading ideas of HPSG it would not be; the freedom to decorate different nodes with the same sortal information plays an important role in HPSG.</Paragraph>
    <Paragraph position="29"> Can some of the restrictions on the ai be dropped? As the proof of the result shows, there is no obvious way to achieve this: as soon as we allow features or path equations in the (~i, the assumption of extensionality no longer helps us find an upper bound on the number of satisfying nodes, and the proof no longer goes through. Essentially what is needed is a way of strengthening lemma 3.1, but it is hard to find a useful way of doing this. Even imposing an acyclicity assumption on our models doesn't seem to help. As the results of the next section show, this is no accident. The combination of ~ and =* is intrinsically dangerous.</Paragraph>
  </Section>
  <Section position="5" start_page="33" end_page="33" type="metho">
    <SectionTitle>
4 Undecidability
</SectionTitle>
    <Paragraph position="0"> The starting point for this section is the undecidability result for the full L KR=~ language (see \[Blackburn and Spaan 1991, 1992\]) which was proved using reduction from a certain undecidable tiling problem.</Paragraph>
    <Paragraph position="1"> We're going to strengthen this undecidability result, and we're going to do so by using further tiling arguments. As the use of tiling arguments seem to be something of a novelty in the computational linguistics literature, we include a little background discussion of the method.</Paragraph>
    <Paragraph position="2"> Tiling arguments are a well known proof technique in computer science for establishing computability and complexity results. (In fact, tiling arguments are used to introduce the basic concepts of complexity, decidability and undecidability in \[Lewis and Papadimitriou 1981\], one of the standard introductions to theoretical computer science.) They are also a popular method for analysing the complexity of logics; both \[Harel 1983\] and \[Hard 1986\] are excellent guides to the versatility of the method for this application.</Paragraph>
    <Paragraph position="3"> One of the most attractive aspects of tiling problems is that they are extremely simple to visualise. A tile T is just a 1 x 1 square, fixed in orientation, that has coloured edges right(T), left(T), up(T), and down(T) taken from some denumerable set. A tiling problem takes the following form: given a finite set 7&amp;quot; .-.I 34 of tile types, can we cover a certain part of Z x Z (Z denotes the integers) using only tiles of this type, in such a way that adjacent tiles have the same colour on the common edge, and such that the tiling obeys certain constraints? For example, consider the following problem. Suppose 7- consists of the following four types of tile: Can an 8 by 4 rectangle be tiled with the fourth type of tile placed in the left hand corner? The answer is 'yes' -- but we'll leave it to the reader to work out how to do it.</Paragraph>
    <Paragraph position="4"> There exist complete tiling problems for many complexity classes. In the proof that follows we make use of a certain II deg complete tiling problem, namely the problem of tiling the entire positive quadrant of the plane, that is, the problem of tiling N x N where N is the set of natural numbers.</Paragraph>
    <Paragraph position="5"> We begin with the following remark: by inspection of the undecidability proof for L KR~ in \[Blackburn and Spaan 1991, 1992\], it is immediate that we still have undecidability if we restrict the language to formulas that consist of a conjunction of formulas of the form C/1 ::~ ~b2, where C/1 and C/2 are L KR formulas with negations applied to atoms only, and C/2 is satisfiable. (The stipulation that C/2 must be satisfiable prevents it from playing the role of False and thus smuggling in illicit negations.) Call this language L-. Let's see if we can strengthen this result further. null So, suppose we look at L- formulas with V as the only binary boolean connective in C/1 and C/2. In this case, we show that the corresponding satisfiability problem is still undecidable by constructing another reduction from N x N tiling.</Paragraph>
    <Paragraph position="6"> Let 7- = {7&amp;quot;1,...,Tk} be a set of tiles. We construct a formula C/ such that: 7&amp;quot; tiles N x N iff C/ is satisfiable.</Paragraph>
    <Paragraph position="7"> First of all we will ensure that, if C/ is satisfiable in a model M, then M contains a gridlike structure.</Paragraph>
    <Paragraph position="8"> The nodes of M (henceforth W), play the role of points in a grid, R, is the right successor relation, and Ru is the upward successor relation. Define: C/9ri d = (TrUe ~ (,)(U) ~,~ (u)(r)).</Paragraph>
    <Paragraph position="9"> Clearly C/arid forces gridlike models.</Paragraph>
    <Paragraph position="10"> Next we must tile the model. To do this we use propositional variables tl,.-., tk, such that ti is true at some node w, iff tile Ti is placed at w. To force a proper tiling, we need to satisfy the following three  Let C/ be earid A C/1 A C/2 A C/3. It is not too difficult to prove that C/ is satisfiable iff T tiles N x N, which implies that the satisfiability problem for our fragment of L- is undecidable.</Paragraph>
    <Paragraph position="11"> Are there weaker undecidable fragments? Yes: we can remove True from C/. We do this by using a new propositional variable PT which plays the role of True. Insisting that</Paragraph>
    <Paragraph position="13"> ensures that PT behaves like True.</Paragraph>
    <Paragraph position="14"> Are even weaker fragments undecidable? Yes: we can ensure that V occurs at most once in each clause. In fact we only have to rewrite part of C/1 (namely, k True =~ Vi=l ti), for this is the only place in C/ where V occurs. We use new variables b2 ..... bk-1 for this purpose and we ensure that bi is true iff \[j is true for some j _&lt; i. We do this as follows:</Paragraph>
    <Paragraph position="16"> Clearly this has the desired effect.</Paragraph>
  </Section>
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