< F(m(r) + 1) + Fm(r+1) -- by induction hypothesis
<= F(m(r+1) - 1) + F(m(r+1)) -- by the previous paragraph
= F(m(r+1) + 1)
This completes the proof of our claim.
Now to prove uniqueness, we show that if we have two sequences Fm1, Fm2, ..., Fm(r) and Fm'1, Fm'2, ... , Fm'(r') with equal sums satisfying the conditions of the theorem, them m=m' and r=r'.
To do this, we first argue that the largest elements of the two sequences are equal, i.e., m(r) = m'(r'). Suppose, for the sake of contradiction, that this is not the case. Then, without loss of generality, let m(r) < m' (r'). By the claim we just proved, it follows that the sum Fm1 + ... + Fm(r) < F(m(r) + 1). But since m(r) < m'(r'), m(r) + 1 <= m'(r'), so we also get Fm1 + ... + Fm(r) < F(m'(r')), which is impossible as the two sequences have equal sums. Thus the two largest elements of the sequences must be equal.
It follows immediately that Fm1 + ... Fm(r-1) = Fm'1 + ... + Fm'(r'-1). Therefore, we can now repeat the argument to show that the second largest elements must be equal. This can be continued and we have shown that the two sequences are equal. (Formally, this is an inductive proof, on the simultaneous order (r,r').)