Reverseaction reverses the link (p, q) to (q, p). Note that whenever this hap- pens, p is staying or p is leaving and p < q. The changes (that is, cases 2 and 3) to a link (p, q) over timeform asequence of links (p, q) = (p0, q0), (p1, q1), (p2, q2), . . . that we call the trace of (p, q). The caseslisted above implythe followingMonotonicitylemma.Lemma 3. (Monotonicity) For every (pr, qr) in the trace of (p, q), pr, qr ∈ [p, q]. This and the fact that we have a bounded number ofprocesses may seem to imply that every trace is bounded, but for now we cannot exclude the case that a link is reversed an unbounded number of times between two processes. It will only be implied later when we know that eventually all leaving processes will exit the network.Consider an arbitraryfixedcomputation of £ÐA. A link that does not change is stable. A steady chain of processes xk, . . . , x0 is a sequence of leaving and not yet gone processes of increasingorder with stable links (xi, xi—1). A steady chain is maximal if it cannot be extended to the left or right. See Figure 8 for an illustration. Note thatat every state of the computation, every leaving process is part of at least one maximal steady chain (which might just be a chain consisting of itself). Also, the following lemma holds.Lemma 4. A maximal steady chain can only change in two ways: either (1) process xk exits the network, or (2) the chain is extended to the left or right due to new stable edges. Since the number of processes is bounded, this means that eventually a maximal steady chain is stable, that is, it does not change for the rest of the computation. This is a stable chain.Lemma 5. In every computation of £ÐA, the only stable chain is the empty chain. Proof. Consider the contrary that we have a non-empty stable chain xk, . . . , x0. Our goal will be to prove that eventually there is no incoming link from non-gone processes in PG to xk. This implies that eventually xk has no more messagesto process, so NIÐSC will eventually be true. Therefore, xk can exit the network, which contradicts our assumption that the chain is stable.First, suppose there is an incoming link (p, xk) with p < xk. If there is a reversal in the trace of that link, then we end up with a link (xk, pr) with p ≤ pr < xk. If this causes xk to delegate pr away, then due to the Monotonicity Lemma that link will never include xk again. Otherwise, xk stores pr in left , and since a leaving process never reverses its link to left , xk eithe...