- #1

evinda

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MHB

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I want to prove that if $L$ is regular then $L^R=\{ w | w^R \in L \}$ is regular.

I have thought the following:

We suppose that $L$ is regular. Then there is a dfa that recognizes $L$.

Assume that $q_0$ is the starting state and $q_n$ is an accepting state, where $n \in \mathbb{N}$.

Taking as $q_n$ the starting state and as $q_0$ the accepting state, we get a dfa that recognizes $L^R=\{ w | w^R \in L \}$. So $L^R=\{ w | w^R \in L \}$ is also regular.

Is my idea right?