Concept
Why inherited precision fits the tapered bound along a shrinking path
Assuming the wire-shrinkage invariant, at depth d inherited precision is at most s lambda^{O(1)} 2^(a_(d-1)). Previous increments are sums of 8hD_t+8M(a+1) across O(log lambda) depths. For an active t, i is less than u and a_d=a_(d-1)+2u, giving D_t>=2^(a_(d-1)+u). Since M>H and u>=lambda^(4/5), the extra factor 2^u dominates every fixed power of lambda. Both inherited p and the evaluation-loss term are therefore O_C(MD_t/u); the new 8hD_t term has that order directly. Hence p_t=O_C(MD_t/u). All precisions remain polynomial in root s because a_d<=4epsilon lambda.
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Updated 2026-11-08
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Math