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<Paper uid="E93-1016">
  <Title>Parsing with polymorphism *</Title>
  <Section position="4" start_page="123" end_page="127" type="metho">
    <SectionTitle>
3 A partial decision procedure for
</SectionTitle>
    <Paragraph position="0"> L(/,\,v) While there are problems in the way of a general decision procedure for L (/'\'V), I claim a partial decision procedure for L (/'\'v) is possible. Partial in the sense of covering only a certain class of sequents, but one sufficiently large, I claim, to cover all linguistically relevant cases. The procedure will be a partial decision procedure for L (/,\,v) via being a partial decision procedure for L(0/'\'v).</Paragraph>
    <Paragraph position="1"> To describe the class of sequents that the procedure applies to I need definitions of the 'polarity' of an occurrence of a category. Let the category polarity of an occurrence of z in a category y (pol(z, y)) be: pol(x, z) = + if z occurs in y, pol(:~,y/z)</Paragraph>
    <Paragraph position="3"> of an occurrence of x in y in a sequent r is the same as the category polarity if y is an antecedent, and  otherwise it is opposite. I use 'polarity' as short for 'sequent polarity'. An example: (4) sk(V-X.X/(Xknp)) ::~ sk(V+X.X/(X\np)) 4I have found non-terminating consecutively bounded  depth first search to happen on the Prolog implementation of the calculus that these paragraphs suggest The decision procedure to be described is applicable to sequents whose negative occurrences of polymorphic categories are unlimited, but whose positive polymorphic categories are drawn from: (5) VX.X/(X\np), VX.X\(X/np), vx.x/x, VX.((cn\cn)/(s\X)/(X/np) vx.((x\x)/x), I will now make three observations concerning proofs in L (L\'v), leading up to the definition of the procedure.</Paragraph>
    <Paragraph position="4"> Observation One In the categories in (5) there is exactly one positive and one or two negative occurrence of the bound variable. This leads to the predictable occurrence of certain sequents. To help describe these I need to define some more terminology. An initial labelling of a proof is the assignment of unique integers to some of the categories in some sequent of the proof. A completed labelling is got from an initial labelling by a certain kind of propagation up the tree: a label is passed up when a labelled category is simply copied upward, and in a (VL) inference the label is distributed to the occurrences of the categories chosen for the variable. In other inferences where a labelled category is active, the label is not passed up. For example:  quent occurs in a labelled L (/'\'v) proof and the label on ai has been passed from a labelled occurrence of Vi. Correspondingly, call a sequent T ::~ ai 'negative for Vi'. Now note that in the above proof, the Vl in the root led to one V + and one V~&amp;quot; branch. This is no accident: one can predict the existence of such branches in any proof of a sequent with a positive occurrence of ViX.X/X. To see this, let me first define a notion reflecting how 'embedded' a category is: path(a, a) = O.</Paragraph>
    <Paragraph position="5"> Where a occurs in x, path(a, x/y) - (/,path(a, x)),</Paragraph>
    <Paragraph position="7"> With the exception of bound variable, if a category occurs with a path (C,p), and a polarity 6, in the conclusion of an inference, then it occurs in the premises of that inference with the same polarity, and with either the same path or with path p. Also, in leaves of a proof in L (/'\'v), categories only occur with zero path. Therefore, if we have  a proof of a sequent with a positive occurrence of ViX.X/X and with non-zero path, then there must occur higher in the proof, a sequent with V,.X.X/X occurring again positively and this time with with zero-path. In other words there must occur a node U, ViX.X/X, V =~ w. Then if there were no (VL) inference in this proof introducing the category ViX.X/X, the category ViX.X/X would be present in the leaves of the proof. Because the leaves can only feature hasic categories, there must be a (VL) inference, and therefore a node U ~, ai/ai, V ~ =~ w ~. Reasoning in a similar vein concerning the category ai/ai, we can be sure there must be a (/L) inference, with premises U ~,ai,V&amp;quot;=~w # and T ~=~al. These are V + and V~- sequents.</Paragraph>
    <Paragraph position="8"> Provable sequents having a positive occurrence of one of the polymorphic categories from (5), labelled with i, will generate an L~/'\'v) proof such that corresponding to each of the positive and negative occurrences of the bound variable, there are (distinct) V + and V~- branches.</Paragraph>
    <Paragraph position="9"> Observation Two We just argued that in any proof of a sequent with a positive occurrence of quantified category, there must occur a node at which the quantifier is introduced by a (VL) inference, and that for the categories in (5), V~ sequents must appear above this. For each of the V~ sequents, the minimum number of steps there can be between the conclusion of the (VL) step and the V~ sequent is the length of the paths to the associated occurrence of the bound variable in the quantified category. Proofs featuring such minimum intervals between the quantified category and the associated V~ sequents I will call orderly. One can ask the question whether whenever there is a proof of a sequent whose positive quantitiers are drawn from the list in (5), there is also an (equivalent) orderly proof. And the answer is that there is.</Paragraph>
    <Paragraph position="10"> Proof sketch We want to show that for any category x in (5), for each of the occurrence of a variable in it, that if there is a proof of U, x, V =~ w, then there is a proof in which the steps leading from the lowest occurrence of the relevant V~ sequent to the (VL) inference correspond to the path to the bound variable in x.</Paragraph>
    <Paragraph position="11"> Let me define the spine of a category as: sp(x/y) = (/, sp(x)), sp(VZ.x) = (V, sp(x)), sp(x) = O, where z is basic.</Paragraph>
    <Paragraph position="12"> We will show first for categories such that sp(x) = (V, slash), and sp(z) = (slashl, slash2), that when there is a proof such that the left inferences for the first two elements of the spine are separated by n steps there must be an equivalent proof where they are separated by n - 1 steps.</Paragraph>
    <Paragraph position="13"> One considers all the possibilities for the last intervening step, 1, and shows that the step associated with the first element of the spine could have been done before l, thus lowering by 1 the number of steps intervening between the first two elements of the spine. There is not the space to show all the cases. (7), (S) and (9, (10) are representative examples for sp(w) = (V, sp(x)). Note that in (9) and (10) there are side-conditions to the (VR) inferences. Satisfaction of these for (9) entails satisfaction for (10). (11), (12) and (13),(14) show representative examples for sp(w) = (slash1, slash2). In (14), X' is some variable chosen to be not free in U, x/y/z, T, V and w. The provability of the upper premise U, x/y, V w\[X'/X\] follows from that of U, z/y, V ~ w by substitution for the variable X throughout. 5 As to the equivalence of the proofs, one can confirm that in the term-associated versions, the same term is paired with the succedent category in each case.</Paragraph>
    <Paragraph position="15"> This is enough to show orderly proofs for VX.X/X and VX.(X\X)/X. For VX.X/(X\np) and VX.((cn\cn)/(s\X)/(X/np)) we must further show that if there is a proof of T =~ x/y whose last step is not a (/R) inference introducing x/y, then there is an equivalent proof whose last step is a (JR) inference introducing ~./y. One can show this by showing if there is a proof whose last two steps use (/R) followed by some rule *, then there is an equivalent proof reversing that order. (15) and (16) illustrate this.</Paragraph>
    <Paragraph position="16">  (15) U, a, V, y ~ x /R U,a,V ~x/y T~b /n U, a/b, T, V ~ z/y (16) U, a, V, y =~ x T =~ b /L  U, a, T, V, y ~ U, a/b, T, V =~ x/y/R So much by way of a sketch of a proof. I will put the fact that orderly proofs exist to the following use. For sequents whose positive quantifiers are drawn from the list in (5), one can be sure that if they have proofs at all, they have a proofs which instantiate quantifiers 'one at a time'. One at time in the sense that once a there is a (VL) inference, one can suppose there will be no more (VL) on the branches leading to the first occurrences of a V~ sequents. Observation Three Bearing in mind Observation One, the question whether a given choice, hi, for the value of the quantified variable is a good one will come to depend, sooner or later, on the derivability, of a certain set of V/6 sequents, containing one V~ sequent and one or two V~- sequents. In relation to this consider the following: Fact 1 (Unknown elimination) (i) and (ii) are  equivalent (i) There is an x such that L(/,\,v)\[-U,x,V ~ w,</Paragraph>
    <Paragraph position="18"> The proof of this, from left to right uses Cut and Cut-Elimination. For example, from L(/,\,v)~-U, x, V =C/. w, L(/,\,v)~-Ti =C/, x, we deduce L(/'\'V)+ Cut ~-U, Ti, V ~ w. Therefore by Cut elimination, L(I,\,v)~U, T1, V ~ w. For the right to left direction, let me say that (w\U)/V is a shorthand for (w\ui ...\us) /v,, .../vi. We choose the x to be (w\U)/Y. Clearly for this x, L(I,\,v)~-U,x,V ~ w. Also each of the claims L(/,\,v)~-T/ =~ x, follows from the assumed U, 7~,V~w, simply by sufficiently many slash Right inferences.</Paragraph>
    <Paragraph position="19"> On the basis of these observations, I suggest the following decision procedure: 6 Definition 1 (Decision procedure) Where A, r vary over possibly empty sequences of sequents, let a rewrite procedure 7~ be defined as follows  1. A, z =t, x, r .,~ A, r, where x is atomic 2. A, T :=~ w, r .,., A, O, r, if T &amp;quot;=~ w follows from 0 by some rule of L(/'\'v) other than O/L) 3. A, U, VZ.z, V =~ w, r ~ A, z\[x/Z\], V =~ w, r, where X is an unknown, and there are no other unknowns in A, U, VZ.z, V ::~ w, r 4. A, U,X,V =~ w, Tx =~ X .... , T, ~X, r ..~ A U, T1, V =C/, w, ..., U, Tn, V ~ w, r  A sequent T ~ w is accepted iff the sequence consisting of just this sequence can be rewritten to the empty sequence by 7C/.</Paragraph>
    <Paragraph position="20"> The fourth clause slightly oversimplifies what I intend in the two respects that (i) the rewrite can apply when the U, X, V =C/, w, T1 =C/, X, ..., T, =C/, X occur dispersed in any order through the sequence, and (it) it can only apply if the unknown X does not occur in sequents other than those mentioned. Note because of clause 3, there will only ever be one unknown in the state of the procedure. This corresponds to Observation Two above. I will show that this procedure is terminating and correct when applied to sequents whose positive quantifiers are drawn from (5). By correctness of the procedure, I mean that the procedure accepts riff L(/,\,V)\]--r. The implication left to right I will call soundness, and from right to left completeness.</Paragraph>
    <Paragraph position="21"> There is a term associated version of this decision procedure, rewriting a pair consisting of a set of equations, and a sequence of term-associated sequents. On the basis of the discussion earlier, for the most part the the reader should be able to easily imagine what embellishments are required to the clauses of the rewrite. I will just give the full version of the Clause 4 rewrite. The input will be:</Paragraph>
    <Section position="1" start_page="125" end_page="126" type="sub_section">
      <SectionTitle>
3.1 Termination
</SectionTitle>
      <Paragraph position="0"> If there are any rewrites possible for a sequence there at most finitely many. So we require that no rewrite series can be infinitely long. Call the sequents featuring an unknown a linked set. At any one time nSince writing this paper, I have discovered that the above observation concerning unknown elimination have been made before \[Moortgat, 1988\], \[Benthem, 1990\].</Paragraph>
      <Paragraph position="1"> This will be further discussed at the end of the paper  there is at most one linked set. Let the degree, d, of a sequence be the total number of connectives. All rewrites on a sequence that has no linked set lower the degree. So rewriting can only go on finitely long before it stops or a linked set is introduced. A linked set is introduced by a clause 3 rewrite, introducing an unknown into some particular sequent. Call this the input sequent. While the sequence contains a linked set, either the degree of the whole sequence goes down, and the sequence remains one containing a linked set (clause 1, clause 2), or the sequence becomes one no longer containing a linked set (clause 4). So a rewrite can only go on finitely long before it either stops, or has a phase where a linked set is introduced and then eliminated. Call the sequents which result from the elimination of the unknown in a clause 4 rewrite, the oulpul sequents. Now considering any such phase of unknown introduction followed by elimination, one can say that the count of positive quantifiers in the input sequent must be strictly greater than the count of positive quantifiers in any of the outputs. This, taken together with the fact that the maximum count of positive quantifiers is never increased outside of such phases, means that there can only by finitely many such phases in a rewrite.</Paragraph>
    </Section>
    <Section position="2" start_page="126" end_page="126" type="sub_section">
      <SectionTitle>
3.2 Soundness
</SectionTitle>
      <Paragraph position="0"> We show that if the procedure accepts a sequence of n sequents (n &gt; 1), then there is substitution for the unknowns such that there are n proofs of the n substituted for sequents. This subsumes soundness, which is where n = 1 and there are no unknowns. I shall use sub(A) to refer to the sequence of sequents got from A by some substitution for the unknowns in A, and L(/,\,v)~-A for the claim that there are proofs of each of the sequents in A The proof is by induction on the length of the shortest accepting rewrite. When the shortest accepting rewrite is of length 1, the sequence must consist simply of an axiom, and so there is a proof. Now suppose the statement is true for all sequences whose shortest accepting rewrite is less than 1. Then for sequences whose shortest accepting rewrite is of length l, we consider case-wise what the first rewrite might be.</Paragraph>
      <Paragraph position="1"> * clause 2 rewrite, for example: A, U, z/y, T, V ~ w, F .,.* A, U,x, V =~. w, T ::~ y, F. A, U,x, V ~ w, T ::~ y, r must have a shortest accepting rewrite of length &lt; l, so by induction there is a substitution such that L(/,\,v)~-sub(A), sub(U,x,V =~ w), sub(T ::V y), sub(r). From this it follows that L(/,\,V)Fsub(A), sub(U,z/y,T, V ~ ~), sub(r).</Paragraph>
      <Paragraph position="2"> The other possibilities for clause 2 rewrites work in a similar way * clause 3 rewrite: A, U, VZ.x,V=~w, F ~.~ A, U,x\[X/Z\], V =~ w, A. By induction there is a substitution such that L(l'\'v)~-sub(A), sub(U,.x\[X/Z\], Y ::V w, sub(A). Let sub' be the substitution that differs from sub simply by substituting nothing for X. sub'(VZ.x) -- VZ(sub'(x)), and sub(x\[X/Z\]) = subt(x)\[sub(X)/Z\]. It follows that L(/,\,v)~-sub'(~), sub'(U, VZ.~, V ~ ~), sub'(F) * clause 4 rewrite. A, U,X,V::~w, T1 ::~X, ..., Tn ~ X, r ..~ A U, T1, V =v w, ..., U, Tn, V =V w r. By induction: L(/,\,v)~-sub(A), sub(U, T1, V =~ w,..., U, T,, V :, w), sub(r). Let sub' be the substitution that differs from sub simply by substituting for X, sub(w\U/V). Clearly L(/,\,v)~ - sub'(U,X,V=~w). Also for each T~, it follows from L(/,\,v)~-sub(U, Ti, V :=0 w) that L(/,\'v)~-sub'(Ti =~ X). Hence L(/,\,v)~-subl(A), sub'(U, X, V =~ w), sub'(T1 =~ X), ..., sub'(T, ::~ X), sub'(r) \[\]</Paragraph>
    </Section>
    <Section position="3" start_page="126" end_page="127" type="sub_section">
      <SectionTitle>
3.3 Completeness
</SectionTitle>
      <Paragraph position="0"> I will now show completeness for sequents whose positive polymorphic categories are drawn from (5).</Paragraph>
      <Paragraph position="1"> By a frontier, f, in a proof, I will mean either the leaves of that proof or the leaves of a subtree having the same root. Given a frontier f in a proof p, which has some completed labelling, the procedure will be said to be in a state s that corresponds to f, if the state and the frontier are identical except that (i) s may have some axioms deleted as compared with f, and (ii) the occurrences of labelled, non-quantified ai in f, are transformed to occurrences of some unknown in s. Given a state s, I will say that a frontier, f, is accessible if there is a state corresponding to f that the procedure may reach from s.</Paragraph>
      <Paragraph position="2"> I assume the procedure is complete for unknown-free sequents whose positive quantifier count is zero. 7 Now suppose the procedure is complete for unknown-free sequents whose positive quantifier count is less than some particular n, and consider a sequent r, of positive quantifier count n, with some proof, p, and one of the form remarked upon in Observation Two.</Paragraph>
      <Paragraph position="3"> There will be (VL) inferences in this proof, amongst which is a set lower than any others. Take the conclusion of one such (VL) inference, U, VX.y, V ==~ w and from all other branches pick a point not above a (VL) inference. This set of points forms a frontier, f, which is accessible if the procedure starts at r. Call the corresponding state s. The sequents in the state other than U, VX.y, V =~z w are unknown-free, have a positive quantifier count of less than n, and have a proof, and so by induction the procedure is complete for them. So there is a possible later state s I which consists solely of the sequent U, VX.y, V ~ w.</Paragraph>
      <Paragraph position="4"> We now focus on the subproof of p that is rooted in U, VX.y, V =~ w. Consider VX.y as labelled with i, and labelling to have been propagated up the tree. I want to define a certain accessible frontier, if, in this tree. There are a certain finite number of branches ending in U, VX.y, V ::~ w. A certain subset of those 7I am of course assuming that all these positive quantified categories are drawn from the list in (5)  branches lead to V~ sequents, and without any intervening (VL) inferences. Select for the frontier f' tile lowest occurrences for the V~ sequents. From the other branches simply select a set of nodes, P, which is not preceded by a (VL). This frontier is accessible, and the corresponding state is: U, Xi, V =2,, w, T1 z=~ Xi, ..., Tn ~ Xi. By a clause 4 rewrite this leads to: U, T1, V =~ w, ..., U, T,, V ~ w. This state is unknown free, each of the sequents has positive quantifier count less than n, and each has a proof. So by induction, the procedure is complete for each of the sequents, and the state may be rewritten to O&amp;quot; \[\]</Paragraph>
    </Section>
  </Section>
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