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Here is the proof of a∗ being star-free:

$\Sigma* = \bar{\emptyset} $

$ A∗= \overline{Σ∗(Σ∖A)Σ∗} $

Would this be a proof for $a * b$? : $ A∗B= \overline{Σ∗(Σ∖A)Σ∗(Σ∖B)} $

For $(A * B )*$ it seems more complicated if not impossible, but is that proofable?

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$a^*b = (a^*)\cdot b$

$(a^*b)^* = \{\varepsilon\} \cup \{a,b\}^* b$

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  • $\begingroup$ Thank you! I forgot that star-free languages are closed regarding boolean operators / concatenation. I assume $\{A,B\}* = \overline{\Sigma*((\Sigma∖A)\cup(\Sigma∖B))\Sigma*}$? If you have time to spare, would you mind telling me whether (a+ b)∗ is non-star free or whether I could alter that grammar easily to obtain a non-star free grammar? More importan: do you know the dot-depth of those two grammars? Do I simply write out the extended form without the star and then count how often boolean expressions and concatenation alternate? I.e. for ab it would be 2 and for (ab)* its 2 or 2.5? $\endgroup$
    – Crea Teeth
    Oct 20, 2023 at 1:44
  • $\begingroup$ It seems that $(a^+b)^* = (a^*\cdot ab)^*$ is helpful here. Note that star-free languages are first-order definable, thus "every $b$ is preceded by $a$ and last letter is $b$". I am not familiar enough with dot-depth to answer. $\endgroup$ Oct 20, 2023 at 11:32

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