Lemma 22 ([Relative cost of the soft tests]). Under the probability measure obtained by normalizing a row's hard birth weights, the probability that at least one of the tests fails is O(ε_s/h_s) , with an absolute implied constant. In …

Fix a recognized model n_0/q and the static anchor choices. For p q , reducedness gives p n_0 , so that prime does not divide qz+n_0 . For p q above the threshold H_v^ ε_s , the model's own hard exclusion already makes its valuation zero. At a remaining prime, at most the other model can impose a nonconstant exclusion. Conditioning on it inflates each geometric valuation tail by at most an absolute factor. The scalar factors and the tag factors do not change this conclusion. Therefore 𝔼_ hard (qz+n_0,V) ∑_ p≤H_v^ ε_s ( p)/(p-1) ε_s H_v. For the last inequality, if H_v^ ε_s <2 the prime sum is empty; otherwise (2H_v^ ε_s ) ε_s H_v . At p=2 conditioning on the one possible other exclusion costs at most a factor two, so the same absolute tail bound applies. If n_0=0 , reducedness gives q=1 ; the valuation tails are still the same geometric tails, truncated at the exponents in V , including the numerator residue zero. Markov's inequality gives failure probability O(ε_s/h_s) for this model. There are at most two recognized models, so the union bound proves the assertion. For each fixed anchor choice this bounds failed hard mass by O(ε_s/h_s) times its hard mass. Integrate this inequality over anchor choices before normalizing. The normalized average is a mixture of the conditioned laws, and has the same relative bound, without an additional factor c_ hard ^ -1 .