<?xml version="1.0" standalone="yes"?>
<Paper uid="P03-1045">
  <Title>k-valued Non-Associative Lambek Categorial Grammars are not Learnable from Strings</Title>
  <Section position="4" start_page="0" end_page="0" type="metho">
    <SectionTitle>
3 Limit Point Construction
</SectionTitle>
    <Paragraph position="0"/>
    <Section position="1" start_page="0" end_page="0" type="sub_section">
      <SectionTitle>
3.1 Method overview and remarks
</SectionTitle>
      <Paragraph position="0"> Form of grammars. We define grammars Gn where A, B, Dn and En are complex types and S is the main type of each grammar: Gn = fa 7! A = B; b 7! Dn; c 7! En n Sg Some key points.</Paragraph>
      <Paragraph position="1"> We prove that fakbc j 0 k ng L(Gn) using the following properties:</Paragraph>
      <Paragraph position="3"> The condition A 6' B is crucial for strictness of language inclusion. In particular:</Paragraph>
      <Paragraph position="5"> This construction is in some sense more complex than those for the other systems (Foret and Le Nir, 2002a; Foret and Le Nir, 2002b) since they do not directly translate as limit points in the more restricted system NL.</Paragraph>
    </Section>
    <Section position="2" start_page="0" end_page="0" type="sub_section">
      <SectionTitle>
3.2 Definition and Main Results
</SectionTitle>
      <Paragraph position="0"> Definitions of Rigid grammars Gn and G Definition 1 Let p, q, S, three primitive types. We define:</Paragraph>
      <Paragraph position="2"> k) non-associative Lambek grammars (not allowing empty sequence and without product) admits a limit point ; the class of rigid (or k-valued for an arbitrary k) non-associative Lambek grammars (not allowing empty sequence and without product) is not learnable from strings.</Paragraph>
    </Section>
    <Section position="3" start_page="0" end_page="0" type="sub_section">
      <SectionTitle>
3.3 Details of proof for Gn
Lemma
</SectionTitle>
      <Paragraph position="0"> fakbc j 0 k ng L(Gn) Proof: It is relatively easy to see that for 0 k n, akbc 2 L(Gn). We have to consider</Paragraph>
      <Paragraph position="2"> For the converse, (for technical reasons and to ease proofs) we use both free group and free pregroup models of NL since a sequent is valid in NL only if its interpretation is valid in both models.</Paragraph>
      <Paragraph position="3"> Translation in free groups The free group translation for the types of Gn is:</Paragraph>
      <Paragraph position="5"> Translation in free pregroups The free pregroup translation for the types of Gn is:</Paragraph>
      <Paragraph position="7"> Proof: Let n denote the type assignment by the rigid grammar Gn. Suppose n(w) ' S, using free groups [[ n(w)]] = S;  - This entails that w has exactly one occurrence of c (since [[ n(c)]] = p 1S and the other type images are either 1 or p) - Then, this entails that w has exactly one occurrence of b on the left of the occurrence of c (since</Paragraph>
      <Paragraph position="9"> We now discuss possible deductions (note that pplppl ppl = ppl): if k0 and k00 6= 0: ppplprSppl S impossible. if k0 6= 0 and k00 = 0: ppplprS S impossible. if k0 = 0 and k00 6= 0: pprSppl S impossible. if k0 = k00 = 0: w 2 fakbc j 0 kg  (Final) Lemma L(Gn) fakbc j 0 k ng Proof: Suppose n(w) ' S, using pregroups [ n(w)] S. We can write w = akbc for some k, such that :</Paragraph>
      <Paragraph position="11"> We use the following property (its proof is in Ap- null pendix A) that entails that 0 k n. (Auxiliary) Lemma:</Paragraph>
      <Paragraph position="13"> where nbalt counts the alternations of p's and q's sequences (forgetting/dropping their exponents). null</Paragraph>
    </Section>
    <Section position="4" start_page="0" end_page="0" type="sub_section">
      <SectionTitle>
3.4 Details of proof for G
Lemma
</SectionTitle>
      <Paragraph position="0"> fakbc j 0 kg L(G ) Proof: As with Gn, it is relatively easy to see that for k 0, akbc 2 L(G ). We have to consider  An interest point of this construction: It provides a limit point for the whole hierarchy of Lambek grammars, and pregroup grammars.</Paragraph>
      <Paragraph position="1"> Limit point for pregroups The translation [ ] of Gn gives a limit point for the simple free pregroup since for i 2 f ;0;1;2;::: g: i(w) 'NL S iff w 2 LNL(Gi) by definition ; i(w) 'NL S implies [ i(w)] S by models ; [ i(w)] S implies w 2 LNL(Gi) from above.</Paragraph>
      <Paragraph position="2"> Limit point for NL; The same grammars and languages work since for i 2 f ;0;1;2;::: g: i(w) 'NL S iff [ i(w)] S from above ; i(w) 'NL S implies i(w) 'NL; S by hierarchy ; i(w) 'NL; S implies [ i(w)] S by models.</Paragraph>
      <Paragraph position="3"> Limit point for L and L; The same grammars and languages work since for i 2 f ;0;1;2;::: g : i(w) 'NL S iff [ i(w)] S from above ; i(w) 'NL S implies i(w) 'L S using hierarchy ; i(w) 'L S implies i(w) 'L; S using hierarchy ; i(w) 'L; S implies [ i(w)] S by models.</Paragraph>
      <Paragraph position="4"> To summarize : w 2 LNL(Gi) iff [ i(w)] S iff w 2 LNL;(Gi) iff w 2 LL(Gi) iff w 2 LL;(Gi)</Paragraph>
    </Section>
  </Section>
class="xml-element"></Paper>