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<Paper uid="C90-3034">
  <Title>A Quantifier Scoping Algorithm without A Free Variable Constraint</Title>
  <Section position="3" start_page="190" end_page="190" type="metho">
    <SectionTitle>
2 Quantification in Logic
</SectionTitle>
    <Paragraph position="0"> Semantic theories generally recurse over the syntax of the object language. For example, following the procedure and notation of \[Tennant 1978\], '~ we say that g satisfies&amp;quot;(Vxf(x))&amp;quot; iff for every 0, g(x --+ o) satisfies &amp;quot;f(x)&amp;quot; Thus, the satisfaction of &amp;quot;(Y x f(x))&amp;quot; is given in terms of the satisfaction of formulae of the form &amp;quot;f(x)&amp;quot;. Truth is defined as satisfaction by the null assignment, N. Given the following axiom g(x -* a) satisfies &amp;quot;f(x)&amp;quot; iff f'(a) then we can produce the following proof &amp;quot;( V x f(x))&amp;quot; is true iff N satisfies &amp;quot;(V x f(x))&amp;quot; for every o, N(x o) satisfies &amp;quot;fix)&amp;quot; iff \[or every o, f'(a) Finally~ formalising our meta.language gives &amp;quot;(V z f(x))&amp;quot; is true iff (V c~ f'(a)) This idea can be extended to structurally ambiguous sentences of English. Suppose C is some environment containing a complex term such as &amp;quot;&lt;a y woman(y)&gt;&amp;quot;, then g satisties C(&lt; all y woman (y) &gt;) if (All a g(y --~ a) satisfies &amp;quot;woman(y)&amp;quot; g(y a) satisfies c(y)) Here, C(y) indicates the environment C(&lt;a y woman(y) &gt;) with y replacing the complex term.</Paragraph>
    <Paragraph position="1"> The extension involves two key changes. First, we employ a four part notation in the metalanguage. Let us say that (All x f(x) g(x)) abbreviates the English: for every object x such ~We assume g is an assignment from variables to objects dealing with all variables required, g(x --~ a) is g modified so that x is assigned to ~. Greek letters are reserved for meta-language variables.</Paragraph>
    <Paragraph position="2"> that f(x) holds, g(x) alsoholds. Secondly, we use a simple conditional rather than a bi-conditional in the rule. The reason for this is simply that an ambiguous sentence such as 1) is true in either of two conditions. The theory will predict  &amp;quot;(loves &lt;a x woman(x)&gt; &lt;every y man(y)&gt;)&amp;quot; is true if (a a woman'(a) (every /9 man'(f/) loves'(c%fl)) and also that &amp;quot;(loves &lt;a x woman(x)&gt; &lt;every y man(y)&gt;)&amp;quot; is true if (every 0 man'(a) (a fl woman'(f~) loves'(a,/9))  We ensure 1) is not true in any other conditions by adopting a general exclusion clause that a sentence is not true except in virtue of the clauses of the given theory.</Paragraph>
  </Section>
  <Section position="4" start_page="190" end_page="191" type="metho">
    <SectionTitle>
3 Comparison and Illustration
</SectionTitle>
    <Paragraph position="0"> The primitive operation of our algorithm will be to apply a complex term to a formula containing it, e.g. to apply &lt;q x r(x)&gt; to p(&lt;q x r(x)&gt;). The result of application is a new four part quantifier expression whose first two parts are q and x, whose third part is the result of recursing on r(x) and whose fourth part is the result of recursing on p(x) (the formula itself with the complex term replaced by the variable it restricts).</Paragraph>
    <Paragraph position="1"> For example, by choosing &lt;a x woman(x)&gt; first in 1), the algorithm will construct a new expression derived from &amp;quot;a&amp;quot;, &amp;quot;x&amp;quot; and recursions on &amp;quot;woman(x)&amp;quot; and &amp;quot;loves(x &lt;every y man(y)&gt;)&amp;quot;. The first recursion will result in woman(x). The second will build yet another term from &amp;quot;every&amp;quot;, &amp;quot;y&amp;quot; and further recursion on &amp;quot;man(y)&amp;quot; and &amp;quot;loves(x,y)&amp;quot;. The final result will be</Paragraph>
    <Paragraph position="3"> Clearly, by choosing &lt;every y man(y)&gt; first, the alternative reading of the sentence would have been produced. Quantifiers chosen earlier receive wider scope. We work our way through the formula outside-in. \[Woods 1968\] explained the advantages of a top-down treatment of quantified noun phrases.</Paragraph>
    <Paragraph position="4">  The basic operation of H&amp;S is similar. An application builds a four part term whose first two parts are q and x, whose third part is r(x) and whose fourth part is the formula with x replacing &lt;q x r(x)&gt;). The result is then recursed upon in order to deal with other complex terms in the formula.</Paragraph>
    <Paragraph position="5"> Now consider complex noun phrases such as &amp;quot;every representative of a company&amp;quot;. These are success cases for H&amp;S. The new algorithm deals with them without alteration. For example ~</Paragraph>
    <Paragraph position="7"> We allow &amp;quot;every&amp;quot; to take wide scope as follows.</Paragraph>
    <Paragraph position="8"> First, we construct a new term from &amp;quot;every&amp;quot; ,&amp;quot;x&amp;quot; and recursions on &amp;quot;arrived(x)&amp;quot; and &amp;quot;and(rep(x), &gt; &amp;quot; of(x, &lt;a y company(y) )) . The recursion on &amp;quot;arrived(x)&amp;quot; simply produces &amp;quot;arrived(x)&amp;quot;. The recursion on &amp;quot;and(rep(x), of(x, &lt;a y company(v)&gt;))&amp;quot; will lead us to construct a new term from &amp;quot;a', &amp;quot;y&amp;quot; and the results of recursions on &amp;quot;company (y)&amp;quot; and &amp;quot;and(rep(x),of(x,y))&amp;quot;. These last two recursions are again simple cases, 4 resulting in</Paragraph>
    <Paragraph position="10"> for &amp;quot;and(rep(z), of(x, &lt;a y company(y)&gt;))&amp;quot;.</Paragraph>
    <Paragraph position="11"> With this result, we can complete our analysis of 3 itself.</Paragraph>
    <Paragraph position="13"> for the whole input.</Paragraph>
    <Paragraph position="14"> In comparison, H&amp;S use a much more complex mechanism. They do this because otherwise deal- null on us by H&amp;S's syntactic analysis. The issue is whether quantifiers can be extracted across conjunctions. For present purposes, I assume they can - indeed, that the recursive rule for &amp;quot;and&amp;quot; only applies when the environments C and D in &amp;quot;and((,D)&amp;quot; contain no complex terms.</Paragraph>
    <Paragraph position="16"> which is not the required reading of the sentence. It also contains a free variable. H&amp;S therefore forbid the algorithm to apply complex terms which are embedded within other complex terms. Also, the restrictions of complex terms are recursively scoped with a flag set so that this call of the procedure returns partial results (still containing complex terms), as well as full results.</Paragraph>
  </Section>
  <Section position="5" start_page="191" end_page="192" type="metho">
    <SectionTitle>
4 Negation
</SectionTitle>
    <Paragraph position="0"> There are two readings of the sentence 4. Everyone isn't here depending on whether &amp;quot;not&amp;quot; or &amp;quot;every&amp;quot; takes wider scope. In ordinary logic we have &amp;quot;not(p)&amp;quot; is true iff it is not the case that &amp;quot;p&amp;quot; is true Suppose C is an environment containing an occurrence of &amp;quot;not&amp;quot;, then g satisfies C(..not..) if it is not the case that g satisfies C( .... ) Here the formula on the right-hand-side is just that on the left, with the occurrence of &amp;quot;not&amp;quot; removed. The ambiguity in 4) arises in exactly the same manner as quantifier scope ambiguities. Using one rule (negation) before another (quantification) leads to wider scope for the first application.</Paragraph>
    <Paragraph position="1"> In contrast, H&amp;S analyse 4 syntactically as not(here(&lt;every x person(x)&gt;)) and mark &amp;quot;not&amp;quot; as being opaque in its only argument. The rule for opaque arguments allows them to be scoped first thus giving H&amp;S the narrow scope &amp;quot;every&amp;quot; reading.</Paragraph>
    <Paragraph position="2">  This use of the terrn &amp;quot;opaque&amp;quot; is somewhat non-standard since &amp;quot;not&amp;quot; is not usually considered to be opaque.</Paragraph>
  </Section>
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