Now evaluate T=t+alpha zeta, where zeta lies in O and alpha=2^(a+1). In a packet with unequal symbols at coordinate i, crossing rows and columns have EQUAL slot bits at i. Write each column as y=y0+2^(i+1)j with 0<=y0<2^(i+1). The columns in one tile have the same group. For a row crossing to that tile, v_2(t-y0+x)<=i: a larger valuation would force the tile bit to be opposite to the row bit. Adding alpha zeta does not change this valuation. These are the denominators expanded in the next step.