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The obvious answer is no, but this is not so easy to defend. One way to argue is that for a dfa to accept {a} ∗ , its initial state must be a final state. Removing any edge will not change this, so the resulting automaton still accepts λ.

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ankit-saha asked Mar 19, 2022
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Design an nfa with no more than five states for the set $\left \{ abab^n: n >0 \right \} \cup \left \{ ab{a}^n : n\geq 0 \right \}$