The sharp terminal leave in random triangle removal

Proof. We give the specialization of Joos–Kühn’s stopping estimates, including the parameters and rooted conventions. Their Sections 5–7 (pp. 9–11) set up the trajectories and stopping times; Lemma 10.1 (p. 50) controls those times, and Lemma 7.15 (pp. 16–17) supplies the two general extension bounds [JK25].

Public parameters and initial hypotheses. Denote their small parameter by \eta, keeping it distinct from \epsilon in (2.1). Choose \eta small enough for their hierarchy and so that

Choose their \delta>0 sufficiently small in terms of \eta, with \delta<1/2. These choices remain fixed as n grows. The removed graph is a triangle, with uniformity parameter k=2 and 2-density (3-1)/(3-2)=2; it is strictly 2-balanced. The initial density in their normalization is \theta=2e(K_n)/n^2=1-1/n. Their density trajectory is therefore exactly p_i in (2.2). Their horizon has density n^{-1/2+\eta}, strictly below p_{i_0} for large n.

Their Section 5 requires an (\eta^4,\delta,2)-pseudorandom initial graph. For completeness, K_n satisfies all its defining conditions (P1)–(P4) on p. 3. The public convention permits edges among distinguished vertices and ignores them in both counts and scalings. For a template with r distinguished vertices, a other vertices, and b edges not internal to the distinguished set, every fixed root injection has exactly (n-r)_a extensions in K_n; here (x)_a=x(x-1)\cdots(x-a+1) and (x)_0=1. For all templates with at most \eta^{-4} vertices,

Their initial error parameter is \zeta_{\mathrm{init}}=n^\delta/(\sqrt n\,\theta). The displayed relative error is smaller than both \zeta_{\mathrm{init}} and \zeta_{\mathrm{init}}^\delta, which verifies the relative-error requirements (P1) and (P2). If a\ge1, the scaling is asymptotic to n^a, while the upper scaling threshold in (P3) is at most n^{1/2}; thus neither small-scaling condition (P3) or (P4) applies. If a=0, count and scaling are both exactly one, and the remaining conditions hold directly.

Balanced counts and the stopping threshold. The dynamic error parameter is

The balanced-template controls in Section 7 give relative error \zeta_i^\delta until the relevant scaling reaches \zeta_i^{-\sqrt\delta}, simultaneously over the templates and root injections in their families. For a balanced rooted template with a\ge1 free vertices and density at most two, its scaling is at least x_i^a. Throughout i\le i_0, we have x_i\ge n^{2\epsilon} and

Thus each template used for parts (i) and (ii) belongs to the tracked family initially and never reaches its threshold before i_0.

A rooted edge counts degrees. A path of length two with its endpoints distinguished counts common neighbors, even when the prescribed endpoint images are adjacent. For a link cycle of length j, use the wheel whose center is distinguished and whose rim is the labeled C_j. This template has j free vertices and 2j edges. Any subset of s free vertices has at most s spokes and s rim edges, so the wheel is balanced with density two. Its extensions are exactly the labeled link-cycle embeddings and have scaling D^j. The largest wheel has r_0+1\le\eta^{-4} vertices. The degree and codegree templates are likewise balanced, with scalings np and D respectively.

At i_0,

Choose any fixed c with 0<c<\delta(\epsilon-\eta^2). The public relative errors then imply parts (i) and (ii), with slack for fixed constants.

General rooted extensions. On the same stopping controls, Lemma 7.15 applies to every template up to the public size cap, with no balance assumption on the target template. Its first case states that, if all same-root subtemplate scalings are at least one, the extension count is at most its scaling times (1+\log n)^{\alpha_{H,I}}. Its second case gives the bound (1+\log n)^{\alpha_{H,I}} when every completion scaling after enlarging the distinguished set is at most one. Here

In our independent-root convention these hypotheses are exactly those in (2.4) and (2.5). For the latter, the public convention ignores the edges internal to the enlarged root set J, giving precisely the exponent e(H)-e(H[J]). We may take C_0=2^{H_0+1}-2. Its size is immaterial: it is fixed independently of n.

Probability and conditioning. Lemma 10.1 bounds the probability of failure of the required stopping controls before the public horizon by \exp(- (\log n)^{4/3}). In particular the triangle count is controlled through i_0 around the positive trajectory n^3p_i^3/6, with relative error tending to zero. Termination at or before i_0 would contradict that control, so i_0 is reached on this event. Define \mathcal G_n to be the event of reaching i_0 and satisfying the properties in the proposition at that time. It is determined by the prefix alone. The later public success event is a subset of \mathcal G_n and proves its probability estimate. We do not condition the continuation on any future stopping control.□

Proof. The priority order is a uniform random permutation of \cT. Conditional on any scanned prefix, the unscanned triangles have uniform relative order. Every currently present triangle is unscanned: an accepted triangle has lost its edges, and a rejected triangle already had a deleted edge and can never become present again. Until the next acceptance the graph stays fixed. Thus the next accepted triangle, if there is one, is the first currently present triangle in the remaining permutation, and is uniform among those triangles. This also identifies the absorbing state when none remains.□

Proof. Along a recursive path the thresholds strictly decrease through the priorities of triangle types. No type can occur twice on that path, so its length is at most |\cT|. The branching is finite, proving termination.

We prove correctness by induction on the number of triangle priorities strictly below the call’s threshold. Every child has a smaller threshold, so the induction hypothesis applies to it. Write \mathcal F for the focus of the present call. We claim that some edge of \mathcal F is deleted before t if and only if some candidate S with U_S<t has both edges of S\setminus\{e(S)\} untouched before U_S.

For the forward implication, take an accepted triangle that deletes a focus edge before t. It is a candidate and all its edges are untouched just before its acceptance. The omitted parent, when present, is at the threshold itself and cannot be this triangle. Conversely, suppose such a candidate exists and every focus edge is untouched before t. Its focus edge is then present at U_S, as are its two other edges. The scan accepts it, contradicting the assumed survival of the focus. This proves the claim. By induction the candidate condition is precisely that its child returns true, and the test returns false exactly when at least one such child exists.□

Proof. Put d_{\max}=\max_{e\in\cE}d_e and M=2d_{\max}; every candidate list has at most M entries. For large n, the good-graph estimates give M\le4D. From any fixed possible node there are at most M^j descendant paths of length j. On each path the j occurrence priorities are independent uniforms, and the probability that they strictly decrease is 1/j!. Hence

The events on the left decrease with j. Almost surely their depths are therefore bounded; finite branching makes the entire decreasing dependency tree finite. This argument uses threshold 1, and thus works simultaneously for every smaller threshold. There are only countably many nodes in the forests for all possible root types, so a countable intersection gives the asserted simultaneous conclusion.

Truncate the recursion at depth j, using any fixed Boolean value at the boundary. Its output is a measurable function of the threshold and finitely many priorities. On the probability-one event just proved, these outputs stabilize as j\to\infty, for all thresholds, to the output of the full evaluation. Defining an arbitrary value on the null exceptional set proves joint measurability.□

Proof. After omitting T, the candidate sets at f and g are disjoint: a triangle containing both edges would equal T. Their occurrence variables and all descendant variables are independent. The pair returns true precisely when every candidate in both sets fails to give a successful child below t. The two separate events have the laws of the singleton roots with T omitted, proving (3.2). The ordering of the joint list changes which children are visited, but not this event that all relevant children fail.

Consider a candidate S=\{e,f,g\} at a singleton root e. Conditional on its occurrence priority being v, its child is a pair call at threshold v, with S omitted and independent descendant priorities. Equation (3.2) gives its success probability as q_{f,S}(v)q_{g,S}(v). Thus I_{e,S}(t) is exactly the probability that this candidate has priority below t and a successful child. Different root candidates use disjoint sets of occurrence variables, including their descendants. Taking the probability that none succeeds proves both products in (3.3).

Initially the probabilities are measurable and bounded between 0 and 1. Equation (3.1) therefore makes each I_{e,T} Lipschitz. The finite products imply continuity of all q_e and q_{e,T}. Substitution back into (3.1) now gives a continuous integrand, so the integrals, and then the products, are continuously differentiable. Derivatives at the endpoints have their one-sided interpretation. The initial values follow directly from the formulas.□

Proof. We first approximate the adjacency matrix on each star by an averaging operator. We then combine these averages and estimate the effect of the error. All constants in the proof are uniform in G.

Closed walks in a link. For u\in[n], let \cE_u=\{uv:v\in N_G(u)\} and N_u=|\cE_u|. Identify uv\in\cE_u with v\in N_G(u), and let M_u be the adjacency matrix of the link G[N_G(u)] on this index set. We have N_u=(1\pm n^{-c})np, and every row sum of M_u is (1\pm n^{-c})D, by the codegree assumption. Put r=r_0=8000.

The trace \operatorname{tr}(M_u^r) counts closed walks of length r with a specified starting position. Walks whose r positions are distinct are precisely embeddings of the labeled cycle C_r; their number is (1\pm n^{-c})D^r. For all other walks, classify the patterns of identifications among the r positions. There are only finitely many patterns, since r is fixed. The support of any realized pattern is a connected simple graph; patterns forcing a loop have no realizations.

If the support is a tree, every support edge is traversed at least twice: a walk crossing the cut obtained by deleting that edge must cross back. Thus the support has at most r/2 edges. Choose the image of one vertex and then expose the other vertices along this tree. The link has maximum degree at most 2D, so this gives at most O(np)(2D)^{r/2} realizations of each pattern.

If the support contains a cycle, let it have j<r vertices and choose one of its simple cycles, of length b\ge3. The link-cycle condition gives O(D^b) choices for this labeled cycle. A forest rooted at its vertices attaches all the other vertices of the connected support, using at most (2D)^{j-b} choices. Ignoring its remaining edges only increases the count. The resulting bound is O(D^j), and hence O(D^{r-1}) for each such pattern. These cases exhaust the non-injective walks. Dividing their total contribution by D^r shows that

Indeed, D\sim n^{1/1000} and npD^{-4000}\sim n^{-6999/2000}.

The leading eigendirection. Write B_u=M_u/D, and order its real eigenvalues as \lambda_1\ge\lambda_2\ge\cdots. The row-sum bound gives \lambda_1\le1+n^{-c}; the Rayleigh quotient of the normalized all-ones vector gives \lambda_1\ge1-n^{-c}. Because r is even,

Consequently, with the fixed positive constant \gamma=c_*/r, all eigenvalues other than \lambda_1 have absolute value O(n^{-\gamma}). In particular, \lambda_1 is simple for large n.

Let \mathbf j_u=N_u^{-1/2}\mathbf1 and let v_u be a unit eigenvector for \lambda_1. The row sums also give \lVert(B_u-\mathrm{Id})\mathbf j_u\rVert_2\le n^{-c}. Write \mathbf j_u=\alpha v_u+w, where w\perp v_u. Every eigenvalue of B_u-\mathrm{Id} on v_u^\perp has absolute value at least 1/2 for large n, so \lVert w\rVert_2\le2n^{-c}. The orthogonal projectors onto the lines spanned by v_u and \mathbf j_u therefore differ in operator norm by at most 2n^{-c}. Combining this with the eigenvalue estimates gives

Summing the star averages. Extend every star matrix by zero to the full edge space \mathbb R^{\cE}, and let P be the sum of the averaging matrices \mathbf1\mathbf1^{\mathsf T}/N_u on the stars. Distinct edges lying in a common triangle have a unique common endpoint, so A is the sum of the extended matrices M_u. Define

The matrix P is symmetric and positive semidefinite. For every x\in\mathbb R^{\cE}, estimate (4.9) gives

Each edge lies in exactly two stars, which explains the last equality and prevents a loss proportional to the number of vertices. Since R is symmetric, it follows that \lVert R\rVert_2=O(n^{-\gamma}). Every row of P has sum two, while a row of A/D has sum 2d_e/D. Both matrices are entrywise nonnegative, so their absolute row sums also give

We have thus obtained a small Euclidean error, whereas the evolution in (4.6) requires the maximum norm. The special form of P supplies this stronger control. Define

Then P=UV, \lVert U\rVert_\infty=2, and \lVert V\rVert_\infty=1. If W is the neighbor random-walk matrix, defined by (Wy)_u=N_u^{-1}\sum_{v\in N_G(u)}y_v, then VU=\mathrm{Id}+W and \lVert W\rVert_\infty=1. The identity (UV)^j=U(VU)^{j-1}V for j\ge1, applied to the exponential power series, yields

The exponential inside the integral satisfies \lVert e^{-v(\mathrm{Id}+W)}\rVert_\infty \le e^{-v}e^{v\lVert W\rVert_\infty}=1. Hence, for every s\ge0,

The second inequality follows from the positive semidefiniteness of P. Likewise, A/D=P+R is symmetric with smallest eigenvalue at least -\lVert R\rVert_2, so

A finite perturbation expansion. It remains to transfer the maximum-norm estimate from E_0 to E. We use a finite form of the bounded-perturbation expansion for semigroups [EN00, Theorem III.1.10], estimating its initial terms and remainder in different norms. Choose a fixed integer K with K\gamma>2. Iterating the variation-of-constants identity

gives terms with j=0,\ldots,K-1 copies of R, followed by a remainder with K copies. The term with j copies is integrated over an ordered simplex of volume s^j/j! and contains j+1 factors E_0, all at nonnegative times at most s. By (4.11), its maximum norm is at most

The remainder has the exact form

Its Euclidean norm is at most

For an m\times m matrix H, Cauchy–Schwarz gives \lVert H\rVert_\infty\le\sqrt m\,\lVert H\rVert_2. Here m=|\cE|\le n^2/2, so, uniformly for s\le\log n,

The other K terms have maximum norm bounded by a fixed power of 1+\log n, by (4.10). This proves (4.7). The order K may be large, but is fixed independently of n and G.□

Proof. Use a and \delta_n from (4.3). All I_{e,T} vanish and all probabilities equal one at zero, so by continuity the conditions (4.4) hold on some interval starting at zero. On every such interval, (4.5) improves the second condition to I_{e,T}\le1/4 for large n.

Write h(s)=z'(s)+(A/D)z(s). As z(0)=0, variation of constants and Proposition 4.1 give, on the same interval,

The last equality follows from a>0, c-a>0 and 2\epsilon-a>0, together with D\sim n^{2\epsilon}: every fixed power of \log n is dominated by each of these positive powers of n. In particular, for large n this improves the first condition to \lVert z\rVert_\infty\le\delta_n/2. The spectral exponent \gamma only fixes the integer K in the semigroup estimate.

If either bootstrap condition had a first exit before s=S, these strict improvements would hold up to that exit. All product factors would then remain at least 3/4, so the finitely many functions and their logarithms would be continuous on a neighborhood of the exit. Both conditions would persist beyond it, a contradiction. Thus the bounds hold for s<S, and continuity gives them also at s=S; the endpoint factors remain positive. This proves z_e(s)=o(n^{-a}) throughout [0,S] and the asserted bound on I_{e,T}. Finally,

which establishes both probability estimates.□

Proof. Assign independent uniform priorities U_T, T\in\cT, for the finite experiment, and an independent supply of fresh uniform variables. Run the unfolding tests as follows. At the first exposure of a type T, use U_T; at each later exposure of that type, use a fresh variable instead.

This procedure has the independent-unfolding law. Indeed, the type queried next is determined by G, the root labels and the execution history. If Q is the set of types exposed so far, that history depends only on (U_T:T\in Q) and on the fresh variables already used. Every recursive computation contributing to a returned answer is part of this history, with its candidate lists exposed explicitly. Conditional on the history, the variables (U_T:T\notin Q) are still independent uniforms. A fresh variable used for a repeated type is independent of the same history. Induction over exposures therefore gives independent uniform occurrence priorities. The unused occurrence variables may be filled in independently afterward.

The evaluations terminate almost surely by Section 3. Until a repeated type is exposed, the unfolding and finite executions use the same priority at every candidate. They consequently have the same sorted lists, recursive calls, thresholds, answers and stopping decisions. On \mathcal C_k^c this agreement holds for the whole execution of all roots. The stated inequality follows from this coupling of the answer vectors.□

Proof. First suppose that \lambda is injective on all of \Gamma. If an edge of G between two labels is absent from the formal graph, its two formal endpoints give the required witness. It remains to show that a collision is impossible when there is no such extra edge.

Suppose that a triangle type T is exposed at two calls. It cannot occur twice in one list, so the exposing calls are distinct. Their responsible focus edges are distinct by property (ii), and their images are distinct by injectivity. Those two edges already represent all three vertices of T, even if neither candidate occurrence is traversed. Since there are no extra edges between labels, the three formal vertices form a triangle of \Gamma. By property (iv), it is a gluing triangle. Its two new edges are focused only at that gluing’s child call, which omits the triangle as its parent. Its attachment edge is focused at only one call. Hence the triangle can appear in at most one candidate list, a contradiction.

Next suppose that the labels are not injective. Choose the first repetition in a construction order with roots first and parents before children. The distinct-root assumption means that this repetition occurs when a new vertex z is attached through an edge xy. Write w for the earlier vertex with \lambda(w)=\lambda(z). All earlier labels are distinct. Property (i) shows that x,y,w are distinct and that \lambda(x)\lambda(w) and \lambda(y)\lambda(w) are edges of G.

If either xw or yw is absent formally, it gives an extra edge among the earlier vertices. The ancestral paths of its endpoints are entirely earlier in the construction, so their labels are injective. Property (iii) ensures that the pair remains nonadjacent in the completed graph. This is the desired witness.

Otherwise xyw is a formal triangle and hence an earlier gluing triangle. There are two possibilities for its latest vertex. If it is x or y, then xy was born at this gluing. The current call focusing xy is exactly its birth call, and omits the triangle xyw as its parent. It therefore cannot have the proposed child with the same triangle type. If the latest vertex is w, then w was attached through xy. The children introducing w and z would belong to the same unique call focusing xy, and would have the same triangle type. That call contains this candidate only once and traverses it at most once, again a contradiction. Thus a first repeated label always produces an earlier extra edge.

Finally, either endpoint of a uniform oriented root has distribution

by the degree estimate and 2m=n^2p. For two independent roots, any fixed pair of their endpoints coincides with probability \sum_v\pi(v)^2\le\max_v\pi(v). There are four pairs to consider; the two endpoints within each root are distinct. A union bound therefore gives the term O(n^{-1}) in (5.1).□

Proof. Write \kappa_n=(1+\log n)^{C_0} for the common factor in the two template bounds of Proposition 2.2. Recall that L=100, B=50, p\sim n^{-1/2+\epsilon}, and D=np^2\sim n^{2\epsilon}, where \epsilon=1/2000.

Fewer than L calls. Suppose \ell<L. Regard J^+ as a template with no distinguished vertices. Order all root vertices first, ordering the endpoints of each root edge, and then add the other vertices in any parent-first order of their birth calls. Before adding ab, each vertex has at most two neighbors preceding it. Moreover, the first vertex of any nonempty vertex subset has no preceding neighbor within that subset. An induced subgraph on v>0 vertices therefore has at most 2v-2 edges from J, and at most 2v-1\leq 2v from J^+. Every subgraph satisfies the same weaker bound e\leq 2v. Since p<1 and D>1 for large n, its scaling is at least

The empty subgraph has scaling one. The first template bound gives

Here the template has at most 4+(L-1)=103\leq H_0 vertices. The inequality

shows that \kappa_np\leq D^{-B} for large n. This proves (6.2) in the short case.

The last L births. Suppose \ell\geq L. Choose any parent-first ordering of the non-root calls, with all roots placed first, and let Z be the set of the last L vertices born. The labels of the preceding vertices can be assigned in at most

ways. Indeed, there are 2m choices for each oriented root, and at most 2D common neighbors of the two endpoints of each subsequent attachment edge. Discarding restrictions against repeated labels only increases this bound. We now fix an injective assignment of these preceding labels.

The birth call of every vertex in Z is an ancestor of the birth call of a marked vertex in Z. To see this, every call in the pattern lies on a path to a marked birth call. Once that path enters the final suffix, all its subsequent calls lie in the suffix, by the parent-first order. In particular, at least one endpoint of ab belongs to Z. Figure 2 illustrates how such a suffix can meet both arms.

Form a template (H,I) as follows. Its free vertices are Z. Retain the two backward edges of each vertex of Z, meaning its edges to the endpoints of its attachment edge, and retain ab. The distinguished set I consists of all preceding vertices needed as endpoints of these edges. Include no edges with both endpoints in I. We may discard such constraints because we seek an upper bound on extensions. Thus I is independent in the template, whether or not its prescribed images have edges between them in G.

The backward edges use at most 2L boundary vertices, and the extra edge uses at most one further boundary vertex. Hence

The 2L backward edges are distinct, and ab is not one of them. Every retained edge touches Z, so

We verify every completion condition needed for the second template bound. If U\subsetneq Z, then

Indeed, assign each backward edge to its newer endpoint in Z. In H[I\cup U] there are at most two such edges per vertex of U. If (6.5) failed, the induced graph would retain ab and both backward edges of every vertex of U. Retaining ab puts every free mark in U. Retaining both backward edges of a vertex also forces the vertex born at its immediate parent call into U, whenever that vertex belongs to Z. Indeed, both focus edges at a non-root parent contain its newly born vertex, so the child’s attachment edge contains that vertex. Repeating this argument backward along each marked path includes every vertex of Z, by the ancestor property just proved. This would give U=Z, a contradiction.

For a proper U, (6.5) implies

Every intermediate set of distinguished vertices has the form J_0=I\cup U, with U\subseteq Z. Its completion scaling is

For proper U the numerator is less than one by (6.4), while the denominator is at least one. For U=Z the ratio equals one. All the required completion scalings are therefore at most one.

The second template bound now gives at most \kappa_n extensions for the labels of Z. In applying it we may ignore avoidance of previously assigned vertices outside I, again only enlarging the count. Since L-B=50>0 and D\sim n^{1/1000},

for all sufficiently large n. Multiplying this by (6.3) proves (6.2). The threshold for n is uniform: the template order and its polylogarithmic exponent are fixed independently of the path lengths.□

Proof. Condition only on the selected priority coordinates (u_y), not on actual visitation, return values, or previously exposed lists. For all required calls to be visited, these priorities must decrease along every required path.

Consider a required non-root call x that has one or two required children. Its focus is a pair \{f,g\}, and its omitted parent triangle is T. Let t_x be the largest priority of its required children. Define O_x prospectively to be the event that every off-pattern candidate at x with priority below t_x has a child query returning false, when that subtree is evaluated hypothetically. Reaching all the required children necessitates O_x. In particular,

If a required call has two required children, reaching the later child also requires the earlier selected child to return false. We discard this additional condition in (6.8).

Each O_x depends only on subtrees rooted at off-pattern children of x. These subtrees contain none of the selected priority coordinates and are disjoint for different required calls. In particular, when one required call is an ancestor of another, the path between them starts through a selected child, so it does not enter an off-pattern subtree. The events O_x are therefore independent conditional on (u_y). This assertion concerns the prospective independent unfolding.

For an off-pattern candidate S attached through focus edge e, the probability that it is not a successful child before t_x is exactly 1-I_{e,S}(t_x). By the product identities of Proposition 3.4,

where e(S) is the unique focus edge contained in S. The denominator removes the one or two selected candidate factors from the complete product. The product estimate (4.14) therefore gives, uniformly in the types and in t_x\in[0,1],

for all sufficiently large n.

We next distribute these factors among the selected child priorities. At a call with one required child of priority u, the bound is 2b(u). At a call with two required children of priorities u,v, we use

There is at most one such non-root fork. We ignore the off-pattern requirements at roots entirely. There are at most two required child steps leaving roots in total, and inserting their missing factors b(u)\geq C_D^{-1} costs at most C_D^2. The number of non-root calls with required children is at most \ell. Combining independence, (6.8), and (6.10), we obtain the conditional bound

when the priorities decrease along the required paths, and zero otherwise. Every selected priority now carries one factor b. This also covers root marks and empty arms, which introduce no new priority coordinates.

The selected priorities have joint Lebesgue density one. In integrating (6.11), retain the order constraints within the trunk and within each arm, but discard constraints joining different segments. For a segment of length j, symmetry of the product integrand gives

The three segments use disjoint coordinates. Their contributions multiply, proving (6.7). Priority ties have probability zero.□

Proof. Lemma 5.2 reduces a collision to initial root overlap, which has probability O(n^{-1}), or to an injectively labeled marked path union with an extra edge. For each fixed length triple, multiply the pattern count (6.1), the embedding bound (6.2), the root-matching probability (2m)^{-r}, and the visitation bound (6.7). The result is at most an absolute constant times

For two-tree patterns h=0; allowing all triples of nonnegative integers only enlarges the sum. A union bound over patterns and embeddings therefore bounds the witness probability by

All terms are nonnegative; the countable union bound applies to the almost surely finite evaluations established in Section 3. The identity 24D\mu=12\log C_D accounts for the exponent 15. Finally, B=50 and D\sim n^{1/1000} imply D^{-35}+n^{-1}=o(D^{-1}), proving (6.12).□