Non-regular context-free language and endless regular sublanguages

Question

I had a job for the university, which basically said:

"Demonstrates that the irregular language L = {0 ^ n 1 ^ n: n natural} did not have infinite regular sublanguages."

I demonstrated this by contradiction. I basically said that there is a language S, which is a sublanguage of L, and this is a common language. Since possible regular expressions for S are 0 *, 1 *, (1 + 0) * and (0o1) *. I check every grammar and demonstrate that none of them are part of L.

However, how could I prove that ANY non-regular contextual free language cannot contain any regular infinite sublanguages?

I don’t need to prove on my own, I just want to be directed in the right direction.

Answer 1

L = {0 ^ n 1 ^ n: n natural} is an irregular free context.

M = 2 * 3 * is infinitely regular.

N = L∪M is an irregular context for free. N contains M.

Answer 2

For the language 0 ^ n 1 ^ n it may be useful to study the pumping lemma. I think when I learned the pumping lemma, it was used in the language a ^ nb ^ n (the same thing.) Perhaps the pumping lecture can help in your proof.

You can also assume that regular languages ​​are closed under padding, combining, intersecting, and star wedges.

That is, if L1 and L2 are regular, then:

L1 L2 (concatenation) is also regular.
L1 n L2 is regular
L1 U L2 is regular
¬L1 is regular 
L1* is regular

, , , , , .

Answer 3

. .

-, , ", L / CF", . , ", X, ...", ( ) .

Answer 4

: , -

Answer 5

Since you just need some clues (and, fortunately, since I forgot how to make evidence since college), look at the definition of a common language and what properties it has. Just by looking, I had enough information to prove the claim.