Common Contracts
1 similar null contracts
PDF hosted at the Radboud Repository of the Radboud University NijmegenNovember 10th, 2010
we have 0 = (r • t)(z, y)= x t(z, x) · r(x, y)= t(z, 2y) · 1+ t(z, 2y + 1) · —1. Hence t(z, 2y)= t(z, 2y + 1), so that t solves the equations r(—, y) = 0. Since t is bifinite, there are for a fixed z ∈ Z, only finitely many y with t(z, y) /= 0. Hence we can express t(z, ) R(N) in terms of the base vectors in B, say as:
