Prove
Corollary 23 ([Colours and actual leaf weights]). For a law satisfying, discard…Lemma 22 ([First peeling and restriction]). Take the constants in, and choose…Proposition
Under the raw primal law, each P_e is uniform injective. Two independent such images have nonzero intersection with probability at most (2^d-1)^2/(2^N-1) , by a union bound over nonzero coefficient pairs. The primal marginals under θ have density at most L relative to those laws. Union over components therefore bounds the probability of any plus-span intersection by O_g(L^22^ 2d-N )=2^ -Ω(N) . If the plus spans are disjoint in every component, a tensor common to the two endpoint primal tensor spaces has every component zero: each of its column spaces lies in the corresponding plus-span intersection. Injectivity of the primal frame maps then makes the sharing witnesses zero, contrary to the parity sum one in . A hole is thus included in the exceptional event just bounded. Before deleting internal holes the pair law is at most L^2π^2 . For a rank- t exact-image event, and bound its probability by L^2 2^ ε_ ch N-(1-ε)tN . Take ε<ζ/4 . Since ε_ ch <ζ/4 and t≥1 , division by the nonhole probability 1-2^ -Ω(N) leaves this at most 2^ -(1-ζ)tN for large N , uniformly for all ranks. The same normalization gives the endpoint caps. Empty pin spaces have zero primal projections, which gives the assertion about C_i .