Proof of Concatenation of Finite Sums

All sums are finite sums in a field, and by the definition of a tuple a $p$-tuple in $K$ is a map from $[p]$ to $K$, so those sums apply to tuples without change. We use claim 1 of Properties of Finite Sums in two ways: the recursion

and the restriction property, which says that the value of a finite sum is unchanged when the tuple of summands is replaced by a restriction of it defined on the indices being summed. We also use the identities $S(p)=p+1$ and $p+S(q)=S(p+q)$ of Natural Numbers, the order properties of Properties of the Order on the Natural Numbers, and associativity of addition in the field $K$.

Fix $n\in\mathbb{N}$. For $m\in\mathbb{N}$ let $P(m)$ be the assertion of the statement for that value of $m$: for every $a\in K^{n+m}$, writing $a'\in K^{n}$ for the restriction of $a$ to $[n]$ and $b\in K^{m}$ for the $m$-tuple with $b_{k}=a_{n+k}$,

We prove $P(m)$ for every $m$ by induction on $m$.

Base case. Let $m=1$, so $n+m=n+1=S(n)$. By the recursion,

By the restriction property, $\sum_{k=1}^{n}a_{k}=\sum_{k=1}^{n}a'_{k}$. Moreover $\sum_{k=1}^{1}b_{k}=b_{1}=a_{n+1}=a_{S(n)}$. Hence $P(1)$ holds.

Induction step. Assume $P(m)$ and let $a\in K^{n+S(m)}$. By Natural Numbers, $n+S(m)=S(n+m)$. Since $m\le S(m)$ (claims 1 and 5 of Properties of the Order on the Natural Numbers) we get $n+m\le n+S(m)$ by claim 6 of that lemma, so $[n+m]\subseteq[n+S(m)]$ and the restriction $\tilde a\in K^{n+m}$ of $a$ to $[n+m]$ is defined. By the recursion and the restriction property,

Let $b\in K^{S(m)}$ be given by $b_{k}=a_{n+k}$ and let $\tilde b\in K^{m}$ be its restriction to $[m]$, so that $\tilde b_{k}=\tilde a_{n+k}$ for $k\in[m]$. The restriction of $\tilde a$ to $[n]$ is the restriction $a'$ of $a$ to $[n]$, so $P(m)$ applied to $\tilde a$ gives

the last equality again by the restriction property. Substituting and using associativity of addition in $K$, together with $b_{S(m)}=a_{n+S(m)}$ and the recursion applied to $b$,

This is $P(S(m))$. By induction, $P(m)$ holds for every $m\in\mathbb{N}$, and since $n$ was arbitrary the lemma follows.