Appendix B: proof of theorem 9
Using propositions 7 and 14 we get the result for i = 0. Let us prove the theorem by the principle of recurrence. We assume now that the result is true until rank i - 1. Computing  for all β ≥ (i + 1)α we get
For all 1 ≤ j ≤ iα we have β - j ≥ β - iα ≥ α so we get
For all iα + 1 ≤ j ≤ β - α we have j - 1 ≥ iα and β - j ≥ β - iα ≥ α and so, with the help of lemma 8 we get
thanks to the recurrence assumption, it is easy to see than B contribute with polynomial terms of degree i (as we have a sum on O(β) terms).
For all β - α + 1 ≤ j ≤ β we have j - 1 ≥ iα so we get
C contributing with polynomial terms of degree i - 1 (as we have a sum on α terms) Summing up all terms we get the result at rank i and the theorem is proved.