= Solution
Because $D$ is minimal, every state is <accessible state of a deterministic finite automaton>[accessible]. Choose a word $u$ taking the initial state to $p$, and let the nonempty word $v$ label the closed path $\gamma$. If a word $w$ took $p$ to an accept state, then
$$
uv^rw\in\mathcal L(D)\qquad(r=0,1,2,\ldots).
$$
These words have strictly increasing lengths because $v\ne\epsilon$, contradicting the assumption that $\mathcal L(D)$ is finite. Hence \b[no accept state is reachable from $p$.]
Back to article page
