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**Example text**

Removal of two opposite lumped gates lumped with { , + 1} in In , where is either w or n − 1. Then we simply remove half of each gate, say w + 1 and + 1 using Operation III, gain λ2 | log η|2 and we extend the set of core indices to include w and extend the permutation σ by adding σ (w) = . Therefore we effectively gained λ| log η| from each such gate. 2) and we thus collect at least Cλ6−(8d+1)κ−O(δ) . 38). 4. Nest. The procedure is very similar to the analysis of the last gate, so we just outline the steps.

Let Pν be the lump of the vertex v − 1 right before to v in the circular ordering and assume v − 1 = 0, 0∗ (otherwise we consider v + 1 and the proof is slightly modified). 4) (uniformly in wev− ) to obtain the necessary decay between the two newly consecutive momenta. The same bound holds if some of the B(·) on the left-hand side is replaced with · −2d due to the set G. Now we integrate wev− to obtain a new delta function from dμ(wev− )δ wev+ − wev− − u σ δ wev− + ±we − u ν e∈L ± (Pν ) : e=ev− ⎛ ⎞ ≤ δ ⎝wev+ + ±we − (u σ + u ν )⎠ e∈L ± (Pν ) : e=ev− 52 L.

25) from [10] can be used only once. We just indicate that the set of A and B momenta are as follows: A := { p1 , p2 , . . , pb−1 , pb+1 , . . , pk+1 } , and we leave the details to the reader. B := { p˜ 1 , p˜ 2 , . . , p˜ k−1 , pb }, 32 L. Erd˝os, M. -T. Yau 2 < C λ | log λ | E*g 2 E*g+2 6 4 < C λ | log λ | E*g Eg+2 Fig. 9. 36), can be easily reduced to the case of a recollision (Fig. 9). 4). Clearly g = 0 in the case of a triple collision. 3) and we collect a factor λ2 | log λ|2 . The resulting Feynman graph has either a recollision or a gate at the end.