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MyWikiBiz, Author Your Legacy — Tuesday September 24, 2024
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→‎The Cactus Language : Syntax: clear up notational problems
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^{\backprime\backprime}\operatorname{blank}^{\prime\prime}
 
^{\backprime\backprime}\operatorname{blank}^{\prime\prime}
 
& = &
 
& = &
\operatorname{blank} \, \cdot\,
+
\operatorname{blank} \, \cdot \,
 
^{\backprime\backprime}\operatorname{blank}^{\prime\prime} \\
 
^{\backprime\backprime}\operatorname{blank}^{\prime\prime} \\
 
\\
 
\\
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| The ''surcatenation'' of the <math>k\!</math> strings <math>(s_j)_{j = 1}^k</math> is the string of the form <math>^{\backprime\backprime} \, \operatorname{(} \, ^{\prime\prime} \, \cdot \, s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, \ldots \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, s_k \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math>
 
| The ''surcatenation'' of the <math>k\!</math> strings <math>(s_j)_{j = 1}^k</math> is the string of the form <math>^{\backprime\backprime} \, \operatorname{(} \, ^{\prime\prime} \, \cdot \, s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, \ldots \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, s_k \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math>
 
|}
 
|}
  −
(NB. I will need to figure out what I meant by the following definitions.)
      
These definitions can be made a little more succinct by defining the following sorts of generic operators on strings:
 
These definitions can be made a little more succinct by defining the following sorts of generic operators on strings:
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<ol style="list-style-type:decimal">
 
<ol style="list-style-type:decimal">
   −
<li>The ''concatenation'' <math>\operatorname{Conc}^k</math> of the <math>k\!</math> strings <math>s_j, j = 1 \ldots k,\!</math> is defined recursively as follows:</li>
+
<li>The ''concatenation'' <math>\operatorname{Conc}_{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>
    
<ol style="list-style-type:lower-alpha">
 
<ol style="list-style-type:lower-alpha">
   −
<li><math>\operatorname{Conc}^1_j s_j \ = \ s_1.</math></li>
+
<li><math>\operatorname{Conc}_{j = 1}^1 (s_j)_{j = 1}^k \ = \ s_1.</math></li>
   −
<li><p>For <math>k > 1,\!</math></p>
+
<li><p>For <math>\ell > 1,\!</math></p>
   −
     <p><math>\operatorname{Conc}^k_j s_j \ = \ (\operatorname{Conc}^{k-1}_j s_j) \, \cdot \, s_k.</math></p></li>
+
     <p><math>\operatorname{Conc}_{j=1}^\ell (s_j)_{j = 1}^k \ = \ (\operatorname{Conc}_{j=1}^{\ell - 1} (s_j)_{j = 1}^k) \, \cdot \, s_\ell.</math></p></li>
    
</ol>
 
</ol>
   −
<li>The ''surcatenation'' <math>\operatorname{Surc}^k</math> of the <math>k\!</math> strings <math>s_j, j = 1 \ldots k,\!</math> is defined recursively as follows:</li>
+
<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>
    
<ol style="list-style-type:lower-alpha">
 
<ol style="list-style-type:lower-alpha">
   −
<li><math>\operatorname{Surc}^1_j s_j \ = \ ^{\backprime\backprime} \, \operatorname{(} \, ^{\prime\prime} \, \cdot \, s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></li>
+
<li><math>\operatorname{Surc}_{j=1}^1 (s_j)_{j = 1}^k \ = \ ^{\backprime\backprime} \, \operatorname{(} \, ^{\prime\prime} \, \cdot \, s_1 \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></li>
   −
<li><p>For <math>k > 1,\!</math></p>
+
<li><p>For <math>\ell > 1,\!</math></p>
   −
     <p><math>\operatorname{Surc}^k_j s_j \ = \ (\operatorname{Surc}^{k-1}_j s_j) \, \cdot \, ( \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime} \, )^{-1} \, \cdot \, ^{\backprime\backprime} \, \operatorname{,} \, ^{\prime\prime} \, \cdot \, s_k \, \cdot \, ^{\backprime\backprime} \, \operatorname{)} \, ^{\prime\prime}.</math></p></li>
+
     <p><math>\operatorname{Surc}_{j=1}^\ell (s_j)_{j = 1}^k \ = \ (\operatorname{Surc}_{j=1}^{\ell - 1} (s_j)_{j = 1}^k) \, \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>
    
</ol></ol>
 
</ol></ol>
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