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Let me define the following "grammar":

$$A_0 \leftarrow 1$$

$$A_{i+1} \leftarrow A_i \mid A_i \ K_{i+1} \mid A_i \ K_{i+1} \ A_i$$

where $1$ and $K_i$ are terminals (infinite amount of them: $K_1, K_2, \dots$ ), $A_0, A_1, \dots$ are non-terminals, $\mid$ means alternative, $\leftarrow$ defines a non-terminal, $i$ is an integer such that $i \geq 0$.

That is, it looks like a grammar composed of an infinite number of rules and terminals.

My questions are:

  1. can I call such a construction a "grammar"?
  2. in which grammar category can I include it?
  3. have such structures been studied before, can anyone give links or bibliographic references?

Thanks a lot.

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  • $\begingroup$ Please, if someone downvotes, it is welcome a comment to allow improve the question or to withdraw it in case there are something completly incorrect. $\endgroup$ – pasaba por aqui Dec 7 '15 at 15:10
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    $\begingroup$ I'm not sure why it was downvoted, maybe someone found the notation confusing. I tried to latexify it, and make it a bit clearer, hopefully the changed version will show up soon. $\endgroup$ – László Kozma Dec 10 '15 at 17:23
  • $\begingroup$ @Laszlo Kozma: Thanks a lot for the edits. I didn't know I could use $$ in my text. $\endgroup$ – pasaba por aqui Dec 11 '15 at 12:07
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The simple answer would be that, having an infinite set of rules, this is not a grammar in the usual sense. Languages over infinite alphabets have been investigated, but usually using register automata rather than grammars.

That being said, you could treat the language as a limit (i.e. infinite union) of finitary languages defined by the "truncations" $G_i$ which you obtain by omitting all indices greater than $i$. I am unfortunately not familiar with research into such beasts.

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