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I'm trying to understand the NFA and $\epsilon$ moves logic. Let's say there is an NFA $A$ with $\epsilon$ moves, if $\epsilon ∈ L(A)$ then $q_0 \in F$? and vice versa? If $\epsilon \in L(A)$ then $q_0 \in F$? How can I prove it?

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It is true that if $q_0 \in F$ then $\epsilon \in L(A)$. The converse holds for DFAs and for NFAs without $\epsilon$ moves, but doesn't hold for NFAs which are allowed to have $\epsilon$ moves. That is, there is an automaton $A$ whose initial state is not accepting, yet $\epsilon \in L(A)$. I'll let you find an example.

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  • $\begingroup$ Thank you! You mean something like δ(q0,ε)=q1 when q1 is the accepting state? $\endgroup$ – Liana Jan 6 '17 at 15:21
  • $\begingroup$ Yes, for example. $\endgroup$ – Yuval Filmus Jan 6 '17 at 15:22

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