General Operation from Binary Operation

Theorem
Let $$\circ$$ be a binary operation on $$S$$.

Then there a unique sequence $$\left \langle {\circ_k} \right \rangle_{k \ge 1}$$ such that:

$$\forall n \in \N^*: \circ_n$$ is an $n$-ary operation on $$S$$ such that:

$$\forall \left({a_1, \ldots, a_k}\right) \in S^k: \circ_k \left({a_1, \ldots, a_k}\right) = \begin{cases} a: & k = 1 \\ \circ_n \left({a_1, \ldots, a_n}\right) \circ a_{n+1}: & k = n + 1 \end{cases} $$

In particular, $$\circ_2$$ is the same as the given binary operation $$\circ$$.

The $$n$$th term $$\circ_n$$ of the sequence $$\left \langle {\circ} \right \rangle$$ is called the $n$-ary operation defined by $$\circ$$.

Proof
Let $$\mathbb{S} = \left\{{\odot:}\right.$$ for some $$n \in \N^*$$, $$\odot$$ is an $n$-ary operation on $$\left.{S}\right\}$$.

Let $$s: \mathbb{S}\to \mathbb{S}$$ be the mapping defined as follows.

Let $$\odot$$ be any $n$-ary operation defined on $$\mathbb{S}$$.

Then $$s \left({\odot}\right)$$ is the $$\left({n+1}\right)$$-ary operation defined by:

$$\forall \left({a_1, \ldots, a_n, a_{n+1}}\right) \in S^{n+1}: s \left({\odot}\right) \left({a_1, \ldots, a_n, a_{n+1}}\right) = \odot \left({a_1, \ldots, a_n}\right) \circ a_{n+1}$$

By the Principle of Recursive Definition, there is a unique sequence $$\left \langle {\circ_k} \right \rangle_{k \ge 1}$$ such that $$\circ_1$$ is the unary operation defined as $$\circ_1 \left({a}\right) = a$$ and $$\circ_{n+1} = s \left({\circ_n}\right)$$ for each $$n \in \N^*$$.

By induction it can be shown that $$\circ_n$$ is an $n$-ary operation on $$S$$ for each $$n \in \N^*$$, and by the definition of $$s$$, the hypothesis holds for every $$n+1$$-tuple in $$S^{n+1}$$.