The proof of Theorem 2.
By Definition 2 we have
U(p,n,m,v) = ...(1),
U(p,n,m,v) = ...(2)
and
U(p,n,m,v) = ...(3),
where ...(4),
...(5)
and ...(6).
It is clear that for z = 0, 1, 2, ..., t2
...(7).
By (4), (5) and (6) t₁ = t₂ or t₁ = t₂+1. We are going to deal with these two cases separately.
(a) If t₁ = t₂ , by (1), (2), (3), (4), (5), (6) and (7) we have
U(p,n,m,v) + U(p,n,m+1,v) = U(p,n+1,m+1,v).
(b) If t₁ = t₂+1, then we have only to prove that the t₁ -th term of (1) is equal to the t₁ -th term of (3), since (2) does not have the t₁ -th term.
By the fact that t₁ = t₂+1 , we know that n-m+p-v+1 is a multiple of p, and hence we have
n-m+p-v+1=pt₁
From this we have
n-v-p(t₁-1) = m-1
and
n+1-v-p(t₃-1) = m
which imply that
By (1), (2), (3), (4), (5), (6), (7) and (8) we have
U(p,n,m,v) + U(p,n,m+1,v) = U(p,n+1,m+1,v).
Remark. Theorem 2 was proved by the authors and published in Reference [9].