Given $L$ a non-regular language and $F$ a finite language I need to prove, or disprove, that $L^+ \cap F$ is a regular language. I tried to prove this using induction on the number of words in $L^+ \cap F$, as from $F$ finite nature I get that the intersection is also finite, hence:
- $n = 0$, There no words in $L^+ \cap F$ and the empty set is a regular language so the statement stands
- Let's assume that for $n\in\mathbb{N}$, the number of words in $L^+ \cap F$, the intersection is a regular language
- for $k = n + 1, k\in\mathbb{N}$ I know from the assumption of the induction that there is an automaton for the first $n$ words in the intersection so we can add a new chain of states (or a path) to that automaton with the final state the one accepting the kth word, therefore $L^+ \cap F$ is a regular language
In my head, this seems to be a sufficient proof, but I still feels as if the last part for $n + 1$ is lacking something.
I then tried a different approach, I know $F$ is finite and therefore regular and has an automaton, and as $L^+ \cap F$ is a subset of $F$ I can take the automaton for $F$ and remove the paths for the words in $F - L^+ \cap F$ leaving an automaton for every word in $L^+ \cap F$, hence the intersection is a regular language. But while the second explanation seems better, to me at least, it doesn't feel like a formal proof
