“…from the identity (6) in the samy way as (18). Let us fix 0 ≤ j < k − r. By (14) and Proposition 4.6, if j + r is odd, then ζ ⋆ A 3 ({1} k−r−j ) is divisible by p and T ⋆ j+r,r is divisible by p 2 and if j + r is even, then ζ ⋆ A 3 ({1} k−r−j ) is divisible by p 2 and T ⋆ j+r,r is divisible by p. Therefore, ζ ⋆ A 3 ({1} k−r−j )T ⋆ j+r,r = 0 and we see that the left hand side of (23) is equal to T ⋆ k,r . On the other hand, by using Proposition 4.1, Proposition 4.2 and (22), we see that the right hand side of (23) is equal to Let us fix 1 ≤ j ≤ r − 1 and j ≤ l ≤ k − r + j.…”