General Associativity Theorem

Theorem
If an operation is associative on 3 entities, then it is associative on any number of them.

Also known as the general (or generalized) associative law.

Formal Statement
Let $\left({S, \circ}\right)$ be a semigroup.

Let $\left \langle {a_k} \right \rangle_{p+1 \le k \le p+n}$ be a sequence of elements of $S$.

Let $\left \langle {r_k} \right \rangle_{0 \le k \le s}$ be a strictly increasing sequence of natural numbers such that $r_0 = p$ and $r_s = p+n$.

Suppose:


 * $\displaystyle \forall k \in \left[{1 .. s}\right]: b_k = \prod_{j=r_{k-1}+1}^{r_k} {a_j}$

Then:


 * $\displaystyle \prod_{k=1}^s {b_k} = \prod_{k = p + 1}^{p + n} {a_k}$

That is:


 * $\displaystyle \prod_{k=1}^s \left({a_{r_{k-1}+1} \circ a_{r_{k-1}+2} \circ \ldots \circ a_{r_k}}\right) = a_{p+1} \circ \ldots \circ a_{p+n}$

Alternative Formulation
Let $a_1, a_2, \ldots$ be elements of a set $S$ and let $n \in \N^*$.

Let $\circ$ be an associative operation.

Let the set $P_n \left({a_1, a_2, \ldots, a_n}\right)$ be defined inductively by:


 * $P_1 \left({a_1}\right) = \left\{{a_1}\right\}$
 * $P_2 \left({a_1, a_2}\right) = \left\{{a_1 \circ a_2}\right\}$
 * $P_n \left({a_1, a_2, \ldots, a_n}\right) = \left\{{x \circ y: x \in P_r \left({a_1, a_2, \ldots, a_r}\right) \land y \in P_s \left({a_{r+1}, a_{r+2}, \ldots, a_{r+s}}\right), n = r+s}\right\}$

Then $P_n \left({a_1, a_2, \ldots, a_n}\right)$ consists of a unique entity which we can denote $a_1 \circ a_2 \circ \ldots \circ a_n$.

Formal Statement

 * Let $T$ be the set of all $n \in \N^*$ such that:


 * $(1): \quad$ for every sequence $\left \langle {a_k} \right \rangle_{p+1 \le k \le p+n}$ of elements of $S$

and:
 * $(2): \quad$ for every strictly increasing sequence $\left \langle {r_k} \right \rangle_{0 \le k \le s}$ of natural numbers such that $r_0 = p$ and $r_s = p+n$:


 * $\displaystyle b_k = \prod_{j=r_{k-1}+1}^{r_k} {a_j}$


 * Let $n = 1$.

Then:

So $1 \in T$.


 * Now suppose $n \in T$.

Let $\left \langle {a_k} \right \rangle_{p+1 \le k \le p+n+1}$ be a sequence of elements of $S$.

Let $\left \langle {r_k} \right \rangle_{0 \le k \le s}$ be a strictly increasing sequence of natural numbers such that $r_0 = p$ and $r_s = p+n+1$.

Then $r_{s-1} \le p + n$.

There are two cases:


 * 1) $r_{s-1} = p + n$;
 * 2) $r_{s-1} < p + n$.


 * First, suppose $r_{s-1} = p + n$.

Then $b_s = a_{p + n + 1}$.

Thus:


 * Secondly, suppose $r_{s-1} < p + n$.

Let $b'_s = a_{r_{s-1}+1} \circ \ldots \circ a_{r_s+1}$.

Then $b_s = b'_s \circ a_{p+n+1}$ by definition of composite.

Now $n \in T \implies a_{p+1} \circ \ldots \circ a_{p+n} = b_1 \circ \ldots \circ b_{s-1} \circ b'_s$.

Thus:

Thus in both cases $n + 1 \in T$.

So by the Principle of Finite Induction, $T = \N^*$.

Alternative Formulation
The cases where $n = 1$ and $n = 2$ are clear.

Let $a = x \circ y \in P_n: x \in P_r, y \in P_s$.

If $r > 1$ then we write $x = a_1 \circ z$ where $z = a_2 \circ a_3 \circ \ldots \circ a_r$ by induction.

Then $x \circ y = \left({a_1 \circ z}\right) \circ y = a_1 \circ \left({z \circ y}\right) = a_1 \circ \left({a_2 \circ a_3 \circ \ldots \circ a_n}\right)$ (again by induction).

If $r=1$, then by induction $x \circ y = a_1 \circ y = a_1 \circ \left({a_2 \circ a_3 \circ \ldots \circ a_n}\right)$.

Thus in either case, $x \circ y = a_1 \circ \left({a_2 \circ a_3 \circ \ldots \circ a_n}\right)$ which is a single element of $P_n$.

Hence we see that $P_n \left({a_1, a_2, \ldots, a_n}\right)$ consists of a single element.

Comment
This theorem answers the following question:

It has been proved that, for example, union and intersection are associative in Union is Associative and Intersection is Associative.

That is: $R \cup \left({S \cup T}\right) = \left({R \cup S}\right) \cup T$ and the same with intersection.

However, are we sure that there is only one possible answer to $\displaystyle \bigcup_{i = 1}^n{S_i}$ and $\displaystyle \bigcap_{i = 1}^n{S_i}$?

That is, is it completely immaterial where we put the brackets in an expression containing an arbitrary number of multiple instances of one of these operations?

The question is a larger one than that - given any associative operation, is it completely associative?

This result shows that it is. Always.