Let M1 = (Q1, {q1}, ∆1, F1), where ∆1 ⊆ Q1 × (Σ ∪ {ϵ}) × Q1, be a non deterministic
finite automaton (NFA) accepting a language L1 ⊆ {0, 1}∗
.Let ϵ denote the null string.
We construct a new NFA M2 = (Q2, {q2}, ∆2, F2), where ∆2 ⊆ Q2 × (Σ ∪ {ϵ}) × Q2,
as follows.
• Q2 = Q1.• q2 = q1.• F2 = F1 ∪ {q1}.• (p, a, p′) ∈ ∆2 iff either (p, a, p′) ∈ ∆1 or (p ∈ F1 and a = ϵ and p′ = q1)
Prove or disprove: The language L2 accepted by M2 is L1*.
I tried and got that yes indeed l2=L1*