Why Korean Is Not a Regular Language: A Proof

Natural language string sets are known to require a grammar with a generative capacity slightly beyond that of Context Free Grammars. Proofs regarding complexity of natural language have involved particular properties of languages like English, Swiss German and Bambara. While it is not very difficult to prove that Korean is more complex than the simplest of the many infinite sets, no proof has been given of this in the literature. I identify two types of center embedding in Korean and use them in proving that Korean is not a regular set, i.e. that no FSA's can recognize its string set. The regular language i salam i (i salam ul$)^j$ michi (key ha)^k$ essta is intersected with Korean, to give {i salam i (i salam ul$)^j$ michi (key ha$)^k$ essta i $$\mid$$ j, k $\geq$ 0 and j $\leq$ k}. This latter language is proved to be nonregular. As the class of regular sets is closed under intersection, Korean cannot be regular.