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If $L_1,\,L_1L_2$ are regular languages with $L_1\neq \emptyset,\,\lambda\notin L_1,\,\lambda\notin L_2,$ is $L_2$ regular?

I think I found a correct proof for this question but my professor says it cannot work. Her issue is with the right quotient step but I cannot see why it is incorrect. Can anyone shed some light on this?

$$ L_1L_2 = \{ w_1w_2 : w_1 \in L_1,\, w_2 \in L_2 \} \qquad \vert w_1 \vert \ge 1,\, \vert w_2 \vert \ge1 $$

$$ \left( L_1L_2 \right)^R = \{ w_2^Rw_1^R : w_1 \in L_1,\, w_2 \in L_2 \} $$

$$ \left( L_1L_2 \right)^R / L_1^R = \{ w_2^R : w_2 \in L_2 \} \qquad (w_1^R \in L_1^R)(\forall w_2^Rw_1^R) \implies w_2^R \in \left( L_1L_2 \right)^R / L_1^R $$

$$ \left( \left( L_1L_2 \right)^R / L_1^R \right)^R = \{ w_2 : w_2 \in L_2 \} = L_2 $$

$$ L_2 \, \text{is regular by closure of reversal and right quotient on regular languages.} $$

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How about this? $L_1=\{a^n\mid n>0\}$, $L_2 = \{a^{m^2}\mid m>0\}$. Certainly $L_1$ and $L_1L_2$ are regular, but $L_2$ certainly isn't.

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  • $\begingroup$ Took me a minute to think of how your $L_1L_2$ is regular but I see it now, nice counter example. I have a feeling my right quotient captures $L_2^R$, but also other strings. $\endgroup$ – Brady Dean Feb 22 at 0:13
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Hint #1:

Try to find a counterexample.

Hint #2:

What happens in the special case where $L_1 = \Sigma^+$? Can you solve that special case?

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