Answer :
No, we cannot conclude that l2 is regular from the fact that l1 ∪ l2 and l1 are regular.
Does the regularity of l1 ∪ l2 and l1 imply the regularity of l2?
The regularity of a language means that there exists a finite automaton that recognizes that language. The union of two languages l1 and l2 is the set of all strings that are in either l1 or l2 or both.
Suppose that l1 ∪ l2 and l1 are regular. Then there exist finite automata A1 and A2 that recognize l1 ∪ l2 and l1, respectively. However, this does not imply that there exists a finite automaton that recognizes l2.
To see why, consider the example where l1 = {a^n b^n | n >= 0} and l2 = {a^n b^n c^n | n >= 0}. Both l1 and l1 ∪ l2 are regular languages, but l2 is not regular. This can be proven using the pumping lemma for regular languages.
Therefore, the regularity of l1 ∪ l2 and l1 does not necessarily imply the regularity of l2.
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