Changes

357 bytes removed ,  20:30, 19 January 2009
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<li>
 
<li>
<p>For <math>k > 1,\!</math></p>
+
<p>For <math>k > 0,\!</math></p>
    
<p><math>\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
 
<p><math>\operatorname{Parse} (\operatorname{Conc}_{j=1}^k s_j) ~=~ \operatorname{Node}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
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</ol>
 
</ol>
   −
<li>The ''surcatenation'' <math>\operatorname{Surc}_{j=1}^k</math> of the sequence of <math>k\!</math> strings <math>(s_j)_{j=1}^k</math> is defined recursively as follows:</li>
+
<li>The parse of the surcatenation <math>\operatorname{Surc}_{j=1}^k</math> of the <math>k\!</math> sentences <math>(s_j)_{j=1}^k</math> is defined recursively as follows:</li>
    
<ol style="list-style-type:lower-alpha">
 
<ol style="list-style-type:lower-alpha">
   −
<li><math>\operatorname{Surc}_{j=1}^1 s_j \ = \ ^{\backprime\backprime} \, \operatorname{(} \, ^{\prime\prime} \, \cdot \, s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></li>
+
<li><math>\operatorname{Parse} (\operatorname{Surc}^0) ~=~ \operatorname{Lobe}^0.</math>
    
<li>
 
<li>
<p>For <math>\ell > 1,\!</math></p>
+
<p>For <math>k > 0,\!</math></p>
   −
<p><math>\operatorname{Surc}_{j=1}^\ell s_j \ = \ \operatorname{Surc}_{j=1}^{\ell - 1} s_j \, \cdot \, ( \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime} \, )^{-1} \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, s_\ell \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></p></li>
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<p><math>\operatorname{Parse} (\operatorname{Surc}_{j=1}^k s_j) ~=~ \operatorname{Lobe}_{j=1}^k \operatorname{Parse} (s_j).</math></p></li>
    
</ol></ol>
 
</ol></ol>
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