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Are regular languages closed under the following construction?

$f(L) = \{w \mid w \in L$ and for all prefixes $x$ of $w$ it holds that $x \notin L$ $\}$

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  • $\begingroup$ @Magdiragdag both formalisms are equivalent i think $\endgroup$ – collapsar Apr 3 '14 at 12:00
  • $\begingroup$ @Magdiragdag i think it was. the implication excluding certain elements in $L$ only applies if $|w| > 1$, so $\epsilon \in L$ is neither eliminated nor introduced by $f$. $\endgroup$ – collapsar Apr 3 '14 at 12:11
  • $\begingroup$ @Magdiragdag ok, so you regard $\epsilon$ as a prefix of any $w \ in L, |w| > 1$. i interpreted the op's phrasing meaning as 'non-empty prefix'. $\endgroup$ – collapsar Apr 3 '14 at 12:17
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Yes, $f(L)$ is regular if $L$ is.

Hint. Take a deterministic finite automaton whose language is $L$ and remove all the outgoing transitions from the accepting states.

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If $A$ is the alphabet, then $f(L) = L - LA^+$.

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