= 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 $\mathit{idle}.0,\mathit{idle}.1,\mathit{idle}.2$, so all nodes are idle. Then take this <CSP trace>:
$$
\begin{aligned}
&\mathit{ring}.1.\mathrm{true},\quad\mathit{active}.2,\quad\mathit{ring}.2.\mathrm{true},\quad\mathit{ring}.0.\mathrm{false},\\
&\mathit{ring}.1.\mathrm{false},\quad\mathit{ring}.2.\mathrm{false},\quad\mathit{ring}.0.\mathrm{false},\quad\mathit{idle}.2.
\end{aligned}
$$
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 $\mathit{idle}.2$, all nodes are idle again. The coordinator may then perform $\mathit{active}.0,\mathit{idle}.0$ forever without starting another survey. This is an infinite execution with no shutdown and no assumption of fairness that would exclude it.
\Image[/past-exam-of-the-computer-science-course-of-the-university-of-oxford/2015/concurrency-2015-ring-counterexample.png]
{title=A node can activate after the survey starts but before the token reaches it}
Therefore
$$
\boxed{\text{The proposed eventual-shutdown property fails in general, already for }N=3.}
$$
This 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 $\mathit{ring}.1.\mathrm{true}$ 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.
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