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The lattice lemma in the certificate note can combine elements at very different scales: write $z=q_zg+\rho_z$ and require distinct 2-adic valuations of the nonzero residues, at most one zero residue, and $\sum_z|\rho_z|<g$. The following observations, by Plasma AI and author-recorded, delimit a proposed averaging proof of this condition. They do not prove its existence for an arbitrary set.

Virtual points and repeated sources

Let $g=z_0/q_0>0$, with $q_0$ odd, and use odd positive multipliers $q_z$ for other elements. The virtual point $v_z=z/q_z$ satisfies

$$\rho_z=z-q_zg=q_z(v_z-g), \qquad v_2(\rho_z)=v_2(v_z-g).$$

Here $v_2$ is extended to nonzero rationals by subtracting the valuations of numerator and denominator. The lattice proof still works for rational $g$: multiplying all residues and $g$ by a common denominator preserves the distinctness of their valuations and the strict budget. A companion's cost is $q_z|v_z-g|$, so different sources carry different weights.

Under the strict total budget, a selected collection cannot use two different representations of the same element $z$. Indeed, if the multipliers are $p\ne q$, then

$$|z-pg|+|z-qg|\ge |p-q|g\ge g.$$

For two odd multipliers the lower bound is $2g$. Thus repeated-source exclusion is automatic for this strict-budget construction. The statement does not cover relaxed graded budgets whose total may exceed $g$.

The real center can also be chosen independently of the virtual pivot. Let $v_i=z_i/q_i$ for positive odd integers $q_i$, and suppose the nonzero differences $v_i-v_0$, $1\le i\le k$, have distinct 2-adic valuations. If some real $c>0$ satisfies

$$\sum_{i=0}^k |z_i-q_ic|<c, \tag{F}$$

then $\{z_0,\ldots,z_k\}$ is dissociated. Repeated sources are excluded by the same strict-budget argument. In a ternary relation write $Q=\sum_i\varepsilon_iq_i$. If $Q\ne0$, rewriting around $c$ contradicts (F). If $Q=0$, rewriting around $v_0$ gives $\sum_{i\ge1}\varepsilon_iq_i(v_i-v_0)=0$; the least valuation is unique since every $q_i$ is odd. All these coefficients vanish, and $Q=0$ then forces the remaining coefficient to vanish. This is a weighted version of the free-center lemma. Its averaging problem is still open, but the center need not be constrained to one of the virtual points.

A generic weighted packing inequality is false

Let $T$ be a full binary tree whose leaves are labeled by a finite set of intervals $I_i=[x_i-r_i,x_i+r_i]$ in $[0,1]$. Set $S_i=\{j:x_i\in I_j\}$, which contains $i$. Let $m_i$ be the depth of leaf $i$ in the tree induced on $S_i$, after vertices with just one child are suppressed. There is no absolute constant $K$ such that

$$\sum_i r_i2^{-m_i}\le K \tag{W}$$

for every such interval family and binary tree.

Here is an explicit family with linear growth in the number of scales. For an integer $H\ge0$ and each $s=0,\ldots,H$, take the $2^s$ dyadic cell centers and radii

$$x_{s,j}=\frac{2j+1}{2^{s+1}},\qquad r_{s,j}=2^{-s-2},\qquad 0\le j<2^s.$$

The centers are pairwise distinct and each interval lies strictly inside its cell. At a center of level $s$, no interval of a finer level can contain that center: it is a cell boundary at every finer level. No other interval of level $s$ contains it either. Thus $S_{s,j}$ consists of $(s,j)$ itself and labels of coarser levels.

Let $B_s$ be any full binary tree on the level-$s$ labels. Put $T_0=B_0$ and recursively make $T_s$ the tree with children $T_{s-1}$ and $B_s$. Use $T_H$ as the tree on all labels. In its restriction to $S_{s,j}$, all finer levels disappear, and $(s,j)$ is the sole surviving leaf of $B_s$. Every other surviving leaf is in $T_{s-1}$. Therefore its induced depth $m_{s,j}$ is either zero or one, and

$$\sum_i r_i2^{-m_i} \ge \frac12\sum_{s=0}^H2^s2^{-s-2} =\frac{H+1}{8}.$$

This is unbounded, proving the failure of (W). The construction is explicit for every $H$ and has no computational premise.

What remains for folding

The virtual points of a fixed source $z$ are not arbitrary interval centers: as the odd multiplier varies, they form the specific family $z/q$, and a threshold on weighted distance gives intervals of radius proportional to $1/q$. The obstruction above does not assert that its interval family has this form with the required arithmetic relations. Consequently it refutes a proof using arbitrary weighted intervals and only the Kraft equality, not the folding route itself.

A successful averaging argument must use the compatibility of the different multipliers of each source, or another arithmetic property absent from (W), to produce a modulus with sufficiently many cheap residues at distinct levels. This is the unresolved interface in the constant-loss route.