let $L\subseteq\Sigma^*$ be a regular language.

for $\sigma \in \Sigma$ prove that $L'=\{w_1\sigma w_2:w_1w_2\in L\}$ is a regular languge.

I tried induction on the length of the regular expression of L.

there are 3 cases in the induction:

$r=r_1\cup r_2, r =r_1\cdot r_2 , r=(r_1)^*$.

the first one is pretty easy but now I stuck on the 2nd and the 3rd.

in the 2nd case:

$w_1\sigma w_2\in L' \leftrightarrow w_1w_2\in L=L(r)=L(r_1\cdot r_2 )=L(r_1)\cdot L(r_2 )$

but at this point I stuck cause I can't say that $w_1\in L(r_1)$ or something like that.

in the 3rd case:

$w_1\sigma w_2\in L' \leftrightarrow w_1w_2\in L=L(r)=L((r_1)^*)=L(r_1)^* \leftrightarrow w_1 w_2=\epsilon$ or $w_1w_2=w_1\cdot w_2...w_k, k>0, \forall i =1...k, w_i \in L(r_1)$

and I have no idea how to finish that case.

  • $\begingroup$ Can you create a finite state machine for $L^\prime$? You already know you can create a FSM for $L$, and you can use that as a starting point. $\endgroup$ – Larry B. Nov 1 '18 at 17:01
  • $\begingroup$ @FabioSomenzi any clue? $\endgroup$ – UltimateMath Nov 1 '18 at 18:52
  • $\begingroup$ Sorry for the misleading comment. I misread the definition of $L'$. I put a comma between $w_1$ and $w_2$, which made the problem trivial to solve with regular expressions. The actual problem is better tackled with automata as suggested by Larry B.. $\endgroup$ – Fabio Somenzi Nov 1 '18 at 22:07

The problem can be solved using automata, but your suggested approach is very interesting. However, you have to be very careful since there is nothing like "the regular expression of $L$" (a regular language admits infinitely many regular expressions that represent it). To avoid this potential problem, you can argue directly on regular languages as follows.

Let $a$ be a fixed letter of the alphabet $A$. For each language $L$ of $A^*$, let $$ L' = \{uav \mid u,v \in A^*,\ uv \in L\} $$ Let $\mathcal{C}$ be the class of all regular languages $L$ of $A^*$ such that $L'$ is regular. Our aim is to show that $\mathcal{C}$ contains all regular languages.

Step 1. The class $\mathcal{C}$ contains the empty language, the language $\{1\}$ and the languages $\{b\}$ for each letter $b$. Trivial (actually you could say directly that any finite language belongs to $\mathcal{C}$).

Step 2. The class $\mathcal{C}$ is closed under finite union. This follows from the formula $(L_1 \cup L_2)' = L_1' \cup L_2'$, which you already observed.

Step 3. The class $\mathcal{C}$ is closed under product. This follows from the formula $(L_1L_2)' = L_1L_2' \cup L_1'L_2$, which essentially says that if you want to insert an $a$ in $L_1L_2$, you have to insert it either in a word of $L_1$ or in a word of $L_2$.

Step 4. The class $\mathcal{C}$ is closed under star. This follows from the formula $(L^*)' = L^*L'L^* \cup \{a\}$ that I let you verify.

Thus $\mathcal{C}$ contains the finite languages and is closed under union, product and star: thus it contains all regular languages.


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