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Nota bene - Wikipedia
http://ssthjj.leshan.gov.cn/shbj/gggs/202404/234b22ff943a4f48a19303d9da916d0c/files/deca05ebd85d48a2b765826d4318cd86.xls WebMar 16, 2024 · L 2 = {b n n ≥ 0} Therefore, L 2 is a regular language. L 1 ⋅ L 2 = {ϵ, a, b, aa, bb, ab, ba, … } L 1 ⋅ L 2 = a *.b * L 1 ⋅ L 2 = {a n b m n ≥ 0, m ≥ 0} Statement I: CORRECT. Therefore L 1 ⋅ L 2 is a regular language. Statement II: INCORRECT. L 1 ⋅ L 2 = {a n b n n ≥ 0} It doesn’t accept string: {a, b, aa, bb ... suthahar relative fat mass
How to identify if a language is regular or not - GeeksforGeeks
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