Proof of A Finite Sum of Vectors with Vanishing Tail

Write (O$k$) for claim $k$ of Properties of the Order on the Natural Numbers.

Let $A$ be the set of those $i\in\mathbb{N}$ with the property that if $i\in[n]$ and $j\le i$, then $\sum_{k=1}^{i}u_{k}=\sum_{k=1}^{j}u_{k}$.

Base case. Suppose $1\in[n]$ and $j\le 1$. By (O4) we also have $1\le j$, so $j=1$ by (O2) and the assertion is an identity. Hence $1\in A$.

Induction step. Let $i\in A$, and suppose $i+1\in[n]$ and $j\le i+1$. If $j=i+1$ the assertion is an identity, so assume $j\ne i+1$; then $j\le i$ by (O5), and $j<i+1$ by (O3). By (O4) we have $1\le i$, and by (O6) and (O1) we have $i\le i+1\le n$, so $i\in[n]$, and therefore $\sum_{k=1}^{i}u_{k}=\sum_{k=1}^{j}u_{k}$ because $i\in A$.

Since $j<i+1$ and $i+1\in[n]$, the hypothesis gives $u_{i+1}=0_{V}$. The recursion part of claim 1 of Properties of Finite Sums of Vectors, together with the defining property of the zero vector in claim 1 of Elementary Identities in a Vector Space, therefore gives

Hence $i+1\in A$.

By Principle of Induction for the Natural Numbers, $A=\mathbb{N}$. Since $j\in[n]$ we have $j\le n$, and $n\in[n]$ by (O1) and (O4), so applying the property of $A$ to $i=n$ gives the assertion.