# NFA automata with $\epsilon$ moves

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?

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.