Consider \(𝐿 = 𝐿_1∩𝐿_2\) where \(L_1 = \{ 0^m 1^m 20^n 1^n \mid m,n \geq 0…

Key idea: use the forms of the two languages to simplify the intersection.

• Strings in the first language have the form 0^m1^m2 0^n1^n (with m,n ≥ 0).

• Strings in the second language have the form 0^p1^q2^r (with p,q,r ≥ 0), so all 1s occur before any 2s and any 2s (if present) are at the end.

• For a string to be in the intersection, the suffix 0^n1^n (which occurs after the 2 in the first form) must be compatible with the second form. Because the second form does not allow 0s or 1s after the 2, that suffix must be empty, so n = 0.

• Therefore the intersection equals {0^m1^m2 | m ≥ 0}.

Why this language is context-free:

• A context-free grammar generating {0^m1^m2 | m ≥ 0} is: S -> A 2 ; A -> 0 A 1 | ε.

Why this language is not regular:

• The substring before the 2 is {0^m1^m | m ≥ 0}, which is a standard non-regular language. One can use the pumping lemma or closure properties to show non-regularity; hence {0^m1^m2 | m ≥ 0} is not regular.

Conclusion: The intersection language is context-free but not regular.