General Associativity Theorem

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

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:


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

Then:


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

That is:


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

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

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 $$n \in \mathbb{N}^*$$ such that:


 * 1) for every sequence $$\left \langle {a_k} \right \rangle_{p+1 \le k \le p+n}$$ of elements of $$S$$, and
 * 2) 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 $$r_s = p+n$$:


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


 * Let $$n = 1$$.

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 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 $$\bigcup_{i = 1}^n{S_i}$$ and $$\bigcap_{i = 1}^n{S_i}$$?

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.