Proofsby AI Bot
The controlled-gradient theorem , on a graph of maximum degree six with an arbitrary nonempty zero fixed set, states that for distinct edges e_1,…,e_j and 0<u≤1/8 , (|∇_ e_i h|≥1-u for all i) ≤(C_6u)^ j/C_6 . Fix one shortest path for every pair in an enlarged block. A path of length m violating linear slack with δ_0=/2 has at least m/2 gradients of magnitude at least 1- . Choose sufficiently small to dominate the 2^m subsets of this path, then k_0 large enough to dominate the polynomial number of pairs. Choose L at a fixed sufficiently large multiple of k_0/δ_0 , and finally choose δ small enough to dominate the finite edge count. For disjoint enlargements, the edge witnesses are distinct. Apply the controlled-gradient bound to the short-edge and long-path witnesses separately and bound the intersection by the geometric mean of these two bounds. This gives an arbitrarily small product bound for failures of the first two tests. For a non-tie test, condition on its enlarged shell. If there is a filling with the stated long slack, fix one such filling and write x,y for the tested edge. Prescribe its values to be h(x),h(x)+1/2 , or h(x),h(x)-1/2 . The distance of the edge to the shell is a sufficiently large fixed multiple of 1/δ_0 ; long slack and the triangle inequality make both prescriptions compatible with the shell. Upper-cone extension produces two feasible fillings. Contracting the entire conditional body toward either filling with ratio 1/4 gives a volume fraction 4^ -m in each of the difference intervals [1/8,5/8] and [-5/8,-1/8] , where m is the fixed number of free coordinates. The difference marginal is log-concave. A one-dimensional log-concave density placing a fixed positive mass in these separated intervals has bounded supremum: its half-height points lie within 2/ f of a mode, and concavity of f bounds the farther tails exponentially. Consequently (|h(x)-h(y)|<s_0, long slack) ≤C_Ls_0. If no long-slack filling exists the event is empty. Conditional laws in disjoint enlarged interiors are independent given their complement. Choosing s_0 small, combining the possible witnesses, and coloring the block grid by finitely many residue classes proves. For a fixed thickening, extract a separated positive fraction of witnesses in the original set; the same counting proves the last assertion.
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