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I know that if a language $L_1$ and a language $L_2$ are regular, then $L1/L2$ is regular.

When we construct a DFA $M'=(Q, \Sigma, \delta, q_0, F)$, for each state $i$ we can make $i$ the start state (we'll represent is using $L'i$) and if $L'i \cap L_2 \neq 0$ then put $q_i$ in $F'$ in $M'$.

I understand the process of constructing an automaton M if both languages are regular, but what if L1 is regular but L2 is any language? How would I construct an automaton for this?

Thank you!

user3295674
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  • What does $L'i$ denote? – DylanSp Feb 15 '16 at 14:22
  • Also, any finite language is regular, so the question as it stands is a finite regular language and another, not necessarily regular language. (Just checking this is what you mean) – Luke Mathieson Feb 15 '16 at 14:24
  • @LukeMathieson you are right, correct with those two, let me fix it up. So if L1 is a regular language and L2 is any language, not necessarily regular – user3295674 Feb 15 '16 at 14:31
  • @DylanSp sorry, let me specify that. When I make i the start state, that's representing Li'. – user3295674 Feb 15 '16 at 14:35
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    Actually, I did some better searching and I am guessing the answer to this question (they asked more questions in one post, but just look at the top paragraph) would answer what I am asking http://cs.stackexchange.com/questions/34006/are-regular-languages-closed-under-division . Correct? Should I delete my question then? (Wanting to check on the proper etiquette :) ) – user3295674 Feb 15 '16 at 14:42
  • @user3295674 It will be closed as a duplicate. – Yuval Filmus Feb 15 '16 at 14:53

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