# prove the intersection of a positive closure of a non-regular language and a finite language is a regular language

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:

1. $$n = 0$$, There no words in $$L^+ \cap F$$ and the empty set is a regular language so the statement stands
2. Let's assume that for $$n\in\mathbb{N}$$, the number of words in $$L^+ \cap F$$, the intersection is a regular language
3. 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

• Hint: A subset of a finite set is finite. Mar 20, 2022 at 21:58
• ... And every finite language is regular
– rici
Mar 20, 2022 at 22:12
• oh I see it now, I just went full retard Mar 20, 2022 at 22:39