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MyWikiBiz, Author Your Legacy — Saturday June 29, 2024
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Returning to the case of the cactus language, the process of recognizing an iterative type or a recursive type can be illustrated in the following way.  The operative phrases in the simplest sort of recursive definition are its ''initial part'' and its ''generic part''.  For the cactus language <math>\mathfrak{C} (\mathfrak{P}),</math> one has the following definitions of concatenation as iterated precatenation and of surcatenation as iterated subcatenation, respectively:
 
Returning to the case of the cactus language, the process of recognizing an iterative type or a recursive type can be illustrated in the following way.  The operative phrases in the simplest sort of recursive definition are its ''initial part'' and its ''generic part''.  For the cactus language <math>\mathfrak{C} (\mathfrak{P}),</math> one has the following definitions of concatenation as iterated precatenation and of surcatenation as iterated subcatenation, respectively:
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{| align="center" cellpadding="8" width="90%"
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|
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<math>\begin{array}{llll}
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1.
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& \operatorname{Conc}^0
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& =
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& ^{\backprime\backprime\prime\prime}
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\\ \\
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& \operatorname{Conc}_{j=1}^k S_j
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& =
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& \operatorname{Prec} (\operatorname{Conc}_{j=1}^{k-1} S_j, S_k)
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\\ \\
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2.
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& \operatorname{Surc}^0
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& =
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& ^{\backprime\backprime} \, \operatorname{()} \, ^{\prime\prime}
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\\ \\
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& \operatorname{Surc}_{j=1}^k S_j
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& =
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& \operatorname{Subc} (\operatorname{Surc}_{j=1}^{k-1} S_j, S_k)
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\\ \\
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\end{array}</math>
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|}
    
<pre>
 
<pre>
1.  Conc^0        =  "".
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  −
    Conc^k_j S_j  =  Prec(Conc^(k-1)_j S_j, S_k).
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  −
2.  Surc^0        =  "-()-".
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  −
    Surc^k_j S_j  =  Subc(Surc^(k-1)_j S_j, S_k).
  −
   
In order to transform these recursive definitions into grammar rules,
 
In order to transform these recursive definitions into grammar rules,
 
one introduces a new pair of intermediate symbols, "Conc" and "Surc",
 
one introduces a new pair of intermediate symbols, "Conc" and "Surc",
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