 Solution

For a general ring, no. Being idle at the instant the coordinator starts is not the same as remaining idle until sampled. A downstream node that has not yet received the first-round message is still in its $\mathrm{Free}$ state and may become active.

For a concrete $N=3$ example, first perform $idle.0,idle.1,idle.2$, so all nodes are idle. Then take this CSP trace:At the first displayed event the coordinator has started and all nodes are idle. Node $1$ has received true and is frozen, but node $2$ can still activate before receiving the token. It forwards false, so the entire second round carries false and nobody shuts down. After the final $idle.2$, all nodes are idle again. The coordinator may then perform $active.0,idle.0$ forever without starting another survey. This is an infinite execution with no shutdown and no assumption of fairness that would exclude it.

ThereforeThis does not contradict part (e): its stronger trigger, the true first-round return, never occurs in the failed survey. For the special $N=2$ ring, starting by $ring.1.\mathrm{true}$ already samples and freezes the only non-coordinator. If both nodes are idle at that event, this particular counterexample is impossible and the successful-survey progress argument applies. The general claim in the question must cover larger rings, where the unsampled downstream node exists.