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<Paper uid="E93-1053">
  <Title>Localising Barriers Theory</Title>
  <Section position="5" start_page="446" end_page="447" type="concl">
    <SectionTitle>
5 Conclusion
</SectionTitle>
    <Paragraph position="0"> We have described a mechanism that handles global constraints on long movement from a local basis. The device has been derived from a logical formulation of Chomsky's \[1986\] theory so that equivalence to this theory is easily proved. We have sketched methods to use the logic for early determination of ungrammatical readings in a parser. In my thesis (\[Schiehlen, 1992\]) the technique has been implemented in an Earley parser that generates all readings in parallel. In this system local conditions are couched into feature terms. Feature clashes lead to creation and abolition of dependencies modelling the GB notion of failed feature assignment and last resource. The barriers logic restricts rule choice for the predictor (descending ancestor lines) and discards analyses in the completer (ascending ancestor lines). Ongoing work is centred around an application of the barriers framework to the generation of semantic structure (Discourse Representation Structure). Kraeht's \[1992\] approach to analysing barriers theory is related to the one presented here. However, Kracht's emphasis is not so much on parsing.</Paragraph>
    <Paragraph position="1">  A Proofs Proof of (6) is trivial.</Paragraph>
    <Paragraph position="2"> The theorem (7) is symmetric for al and a2. Suppose alRb A &amp;quot;~a2Rb. a2 and b are unconnected. So there exist kl barriers not dominating al (kl &lt; n) and k2 barriers not dominating a2 (k2 &gt; n). Suppose c is a barrier not dominating a2 but dominating al (there are at least k2-kl &gt; 1 such barriers), c&gt;b and y&gt;b, hence c and y are connected. But y&gt;_c entails y&gt;al.</Paragraph>
    <Paragraph position="3"> Ifc&gt;y then either x&gt;c&gt;y or c&gt;x. But c&gt;x implies c &gt; a2.</Paragraph>
    <Paragraph position="4"> To prove (9) note that all barriers for y dominate y by (5). Hence they also dominate a e \[y\].</Paragraph>
    <Paragraph position="5"> We now turn to (10). Take al E \[x\] and a2 E \[y\]. a2 and y are not connected. We show that if --c &gt; a2 and c &gt; b then -~c &gt; al. Assume c &gt; b and c &gt; ax.</Paragraph>
    <Paragraph position="6"> Then x and c are connected both dominating b. We know that -~x _&gt; c &gt; ax. Hence c &gt; x &gt; y. Suppose y! is y's father. Then c &gt; x &gt;_ y! ~.- y and equally c&gt; x &gt; y! ~- a2. We obtain that {c I B(c,b) A -~c&gt; el} D {c I B(c,b) ^ -~c&gt;a2}. Hence -~\[x\]Rb.</Paragraph>
    <Paragraph position="7"> We prove (24). Suppose c is a barrier for x. Then by (23) there is a sequence of nodes xl = c and xn &gt; xn+l = x. But xn &gt; x &gt; b, so c is a barrier for b as well. a and y are unconnected. Suppose c is a barrier for y but not x. Then xl = c and xn&gt;x~+l = y. xn and x are connected both dominating y. We know that -~x &gt; xn &gt; y and ~xn &gt; x else c would be a barrier for x. Hence Xn = x and we get x, = x &gt; b.</Paragraph>
    <Paragraph position="8"> There are at least as many barriers for b as there are for y, so -~aRb.</Paragraph>
    <Paragraph position="9"> To prove (25) we adopt the argumentation of the foregoing proof and infer that x is a barrier for y.</Paragraph>
    <Paragraph position="10"> bILz shows that b, x are unconnected, hence -~x &gt; b and -~bRy.</Paragraph>
    <Paragraph position="11"> (26) follows if we prove B(c, bl) ~ B(c, b2) by induction. The theorem is symmetric. Assume a c such that B(c, bl). Then either scheme (16) holds: L(c) A c&gt;bx hence c&gt;b2. Or (17) and L(c) A 3d : B(d, bl) A c&gt; d A Ve : c&gt; e &gt; d ---* ~L(e). By induction B(d, b2) as well. For the negative scheme (18) we use symmetry to extend the implication I(c, bx, B, L) ---, I(c, b2, B, L) to an equivalence. For (28) we give a proof by cases. Either B(y, b) --. I(z, b, B, L). y is the barrier node d referred to in the consequent. Or I(y, b, B, L) A -~L(y) --* I(x, b, B, L). We set the barrier node d of the first inheritance clause equal to the one of the second. Does a node e between x and d satisfy L? y does not, nor do the nodes between y and d, and there is no node between x and y. But y and e must be connected, both dominating d. We show I(x, b, B, L) --* B(y, b) V I(y, b, B, L). The barrier node d of the antecedent clause and y are connected, both dominating b (see 5). d cannot sit between x and y. If d - y the first disjunct holds. If y &gt; d we set d equal to the barrier node of the second disjunct. No e between y and d satisfies L.</Paragraph>
    <Paragraph position="12"> We reduce (29) to (10). If a &gt; y &gt; b we make use of (6). Otherwise let x! be the smallest node that dominates both y and a and let x be such that x! ~- x &gt;__ y. Then by (10) &amp;quot;~\[x\] Rb, meaning --~aRb (see 8).</Paragraph>
  </Section>
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