Prove
Lemma 49 ([Finite realization and product labels]). Fix finitely many old…Theorem 8.1 ([Address lemma]). Suppose the binary symbols under consideration…Proposition
The one-edge assertion is precisely the arithmetic contraction rule, with product replacements justified by Lemma 49 ([Finite realization and product labels]). Induction proves the tree assertion: a split--split merger is split, a split--simple merger is simple, and the first merger of two components containing simple vertices has zero derivative. Along a tree there is a unique choice of relative orientations once one root orientation is fixed. Its product template depends on support types and external bits, not on the internal bits being differentiated. For clarity, consider first the two variables on a split pair p,q . Differentiation in the inertia bit at one place over p gives the symbol formed with the entire rational prime p omitted. Its elementary extension is unramified at both places over p . Differentiating in the other oriented bit is therefore zero. Repetition of the same variable is zero by Δ_z^2=0 . This argument would not follow from omission of only one oriented prime ideal. For a longer cycle, first contract a spanning path. By the product-label rule the closing variable has become internal to the merged cluster. Its change affects neither the product's support data nor any incidence to a remaining vertex. Equivalently it is an inertia change at an omitted endpoint, covered by the preceding unramifiedness argument. Thus its derivative vanishes. Possible conjugation of the product generator only changes the induced orientation and does not reintroduce that endpoint. Since there is no self-residue among the inputs to the replacement symbol, there is no extra closing-edge datum to retain. Additional derivatives commute with this zero derivative. For example, contract the aligned edge 12 in a triangle. With π_ 12 =π_1π_2 , the two bits toward vertex 3 are t_0=z_ 13 ^0+z_ 23 ^0 and t_1=z_ 13 ^1+z_ 23 ^1 , up to fixed type corrections. An aligned edge 23 then differentiates t_0 ; a conjugately aligned closing edge 31 differentiates t_1 . Their mixed derivative is the omitted-endpoint zero just proved. If the closing edge is aligned instead, both become the same difference and its square is zero. Contracting a longer spanning path reduces to exactly these two alternatives.