Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 12G / b / ii

Suppose a path took $p$ to a state $q$. Part (i) implies that no accept state is reachable from either $p$ or $q$. Thus every possible continuation is rejected from both states, so $p$ and $q$ are indistinguishable states of a deterministic finite automaton. Distinct indistinguishable states cannot occur in a minimal deterministic finite automaton. Therefore $q=p$, and there is no path from $p$ to any other state.

Equivalently, a cycle state in a minimal deterministic finite automaton for a finite language is the unique rejecting sink.