If \(L_1 = \{a^n | n \geq 0 \}\) and \(L_2 = \{b^n | n \geq 0 \}\), consider…

Key insight: L1 = a* and L2 = b*, so L1⋅L2 = a*b*.

• Regularity: a* and b* are regular, and regular languages are closed under concatenation, so a*b* is regular.

• Equality check: a*b* = {a^i b^j | i, j ≥ 0}, which is not the same as {a^n b^n | n ≥ 0}. For example, aab (a^2 b^1) belongs to a*b* but not to {a^n b^n}.

Conclusion: The statement that the concatenation is a regular language is true, while the statement that the concatenation equals {a^n b^n | n ≥ 0} is false. Therefore, only the claim that the concatenation is regular is correct.