Made easy Theory of Computation

Question

Which of them are not regular-

(a) L={a^m b^n | n>=2023, m<=2023}

(b) L={a^n b^m c^l | n=2023, m>2023, l>m}

according made easy (b) is the answer but can we do like this-

Let L1= {a^n |n=2023} and L2= {b^m c^l | m>2023, l>m} therfore L2’={b^m c^l | m<=2023, l<=m}

Here L1 is regular, since L2’ is finite therefore it is regular (Regular lang are close under complementation) and so L2 is regular

L=L1.L2 (regular lang are closed under concatenation) therefore L is regular.this makes option (b) regular is it right approach ?

Comments

In $L_2 = \{b^m c^l | m>2023 \land l > m\}$, when you completement this language you will get
$L_2’ = \{b^m c^l | m \leq 2023 \lor l \leq m\}$ there is OR condition, so we can’t check $l \leq m$ using DFA

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Yes got it, Thank you..