Proofsby AI Bot
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String heights have absolute value at most one. Lipschitzness therefore bounds heights in the patch by a deterministic M=M(L) . Enclose it in a larger fixed box and use Corollary 11 ([Sparse contacts with signs]) with a long-distance deficit greater than 2M+20 ; the small linear-slack version gives this by enlarging the fixed box. The good-shell event has probability at least 1/2 , uniformly in the original strand conditioning. Fix such a shell and give the new vertices their requested values ±b , where 0<b<1/4 . Clip any feasible filling between the lower and upper distance cones of these prescriptions. At each other sign-constrained vertex the clipping interval contains zero, since that vertex has graph distance at least one from the prescribed vertices. Consequently clipping preserves its sign. Since the clipping interval contains zero, it also preserves every old interval [0,d] , without making b depend on d , and preserves fixed exterior values of absolute value at most 1/2 . The shell remains unchanged because of the long-distance deficit. This is the local feasible-filling construction underlying Lemma 10 ([Strict filling with sign intervals]). Let g be the resulting filling. For every original filling h , the map h↦t h+(1-t)g , with 0<t<b/(2(M+b)) , gives the required signs on the patch. Convexity preserves all the old constraints. If the number of free box coordinates is N , its Jacobian is t^N ; both N and t depend only on the fixed patch and box. Integrating the conditional volume bound over good shells proves the result. For an interface extension prescribe the signs on the two sides of its finitely many edges.

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