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Directory:Jon Awbrey/Papers/Cactus Language (view source)
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, 23:26, 27 December 2008→The Cactus Language : Syntax: markup
<p><math>\operatorname{Conc}^k_j s_j \ = \ \operatorname{Prec}(\operatorname{Conc}^{k-1}_j s_j, s_k).</math></p></li>
<p><math>\operatorname{Conc}^k_j s_j \ = \ \operatorname{Prec}(\operatorname{Conc}^{k-1}_j s_j, s_k).</math></p></li>
</ol></ol></ol>
</ol></ol>
<li>
<p>The conception of the <math>k\!</math>-place surcatenation operation can be extended to include its natural "prequel":</p>
<p><math>\operatorname{Surc}^0 \ = \ ^{\backprime\backprime} \, \operatorname{()} \, ^{\prime\prime}.</math></p>
<p>Finally, the construction of the <math>k\!</math>-place surcatenation can be broken into stages by means of the following conceptions:</p>
<pre>
<ol style="list-style-type:lower-alpha">
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<p>A ''subclause'' in <math>\mathfrak{A}^*</math> is a string that ends with a <math>^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></p></li>
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<p>The ''subcatenation'' <math>\operatorname{Subc} (s_1, s_2)</math> of a subclause <math>s_1\!</math> by a string <math>s_2\!</math> is the string that is defined as follows:</p>
<p><math>\operatorname{Subc} (s_1, s_2) \ = \ s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime} \, )^{-1} \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, s_2 \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></p>
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<p>The ''surcatenation'' of the <math>k\!</math> strings <math>s_1, \ldots, s_k\!</math> can now be defined as an iterated subcatenation over the sequence of <math>k + 1\!</math> strings that starts with the string <math>s_0 \ = \ \operatorname{Surc}^0 \ = \ ^{\backprime\backprime} \, \operatorname{()} \, ^{\prime\prime}</math> and then continues on through the other <math>k\!</math> strings:</p></li>
<ol style="list-style-type:lower-roman">
<li>
<p><math>\operatorname{Surc}^0_j s_j \ = \ \operatorname{Surc}^0 \ = \ ^{\backprime\backprime} \, \operatorname{()} \, ^{\prime\prime}.</math></p></li>
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<p>For <math>k > 0,\!</math></p>
<p><math>\operatorname{Surc}^k_j s_j \ = \ \operatorname{Subc}(\operatorname{Surc}^{k-1}_j s_j, s_k).</math></p></li>
</ol></ol></ol>
<pre>
Notice that the expressions Conc^0_j z_j and Surc^0_j z_j
Notice that the expressions Conc^0_j z_j and Surc^0_j z_j
are defined in such a way that the respective operators
are defined in such a way that the respective operators