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<Paper uid="C04-1012">
  <Title>Restrictions on Monadic Context-Free Tree Grammars</Title>
  <Section position="4" start_page="0" end_page="0" type="metho">
    <SectionTitle>
3 Linearity and Nondeletion on Monadic
</SectionTitle>
    <Paragraph position="0"> CFTGs Because linear, nondeleting, monadic CFTGs generate the same class of string languages as TAGs, the question is whether the restrictions of linearity and nondeletion on monadic CFTGs are necessary to generate the same class of languages. First, it will be shown that nondeletion is unnecessary.</Paragraph>
    <Paragraph position="1"> Theorem 3.1 For any linear, monadic CFTG G, there exists an equivalent linear, nondeleting, monadic CFTG G0.</Paragraph>
    <Paragraph position="2"> Proof. Let G = (N; ;P;S) be a linear, monadic CFTG. An equivalent linear, nondeleting, monadic CFTG G0 = (N0; ;P0;S) can be constructed as follows.</Paragraph>
    <Paragraph position="3"> The set of nonterminal is N0 = N00 [ N01 such that N00 = N0 [ fAjA 2 N1g and N01 = N1.</Paragraph>
    <Paragraph position="4"> For the preparation of the definition of P 0, for 2 TN[ (X1) we define ( ) TN0[ (X1) as the smallest set satisfying the following conditions:</Paragraph>
    <Paragraph position="6"> Because of the construction of N0 and P0, G0 is monadic and nondeleting.</Paragraph>
    <Paragraph position="7"> To show the equivalence of G and G0, we prove the following statement holds for any 2 TN[ and 2 T by induction on the length of derivations: null G ) if and only if there exitsts ^ 2 ( ) such that ^ G0 ) .</Paragraph>
    <Paragraph position="8"> We start with proving &amp;quot;only-if&amp;quot; part. Let Gk) . If k = 0, then clearly = , 2 ( ) and G0 ) . For k 1, assume that the statement holds for any derivation of length less than k. If a rule of the form A ! with A 2 N0 is used at the first step, the proof is rather simple, so we only prove the other case. Suppose that a rule A(x) ! with A 2 N1 is used at the first step and = 0[A( 00)] G) 0[ [ 00]] G ) for some 0 2 TN[ (dX1e) \ TN[ (bX1c) and 00 2 TN[ . By the induction hypothesis, there exist 2 ( 0[ [ 00]]) such that G0 ) . Here, we have to think of the three different cases: (1) 2 ( 0), (2) can be written as 0[^ ] for some 0 2 ( 0) and  ^ 2 ( ), and (3) can be written as 0[^ [ 00]] for some 0 2 ( 0), ^ 2 ( ) and 00 2 ( 00). See  In the case (2), A ! ^ is in P 0 and therefore, 0[ A ] 2 ( ) and 0[ A ] G0) 0[^ ] G0 ) . And in the case (3), A(x) ! ^ is in P 0 and therefore,</Paragraph>
    <Paragraph position="10"> sume that the statement holds for any derivation of length less than k. The rule used at the first step is one of the following forms: (1) A ! ^ with  Next, consideration will be given to whether the restriction of linearity can be removed from monadic CFTGs to generate the same class of languages. The answer is negative. The following example is a non-linear, monadic CFTG that generates a string language that no linear, monadic CFTG can generate.</Paragraph>
    <Paragraph position="11"> Example 3.2 The following is an example of a monadic CFTG that generates the string language Lw4 = fwwww j w 2 fa;bg+g. G = (N; ;P;S) where N = fS;Ag, the ranks of S and A are 0 and 1, respectively, = fa;b;c;dg, the ranks of a, b, c and d are 0, 0, 2 and 4, respectively, and P consists of the following rules:</Paragraph>
    <Paragraph position="13"> Because G has the rule A(x) ! d(xxxx), G is not linear.</Paragraph>
    <Paragraph position="14"> Theorem 3.3 There exists a monadic CFTG which is not weakly equivalent to any linear, monadic CFTG.</Paragraph>
    <Paragraph position="15"> Proof. It is known that the string language Lw4 in Example 3.2 cannot be generated by any TAG. It cannot be generated by any linear, monadic CFTG, neither.</Paragraph>
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