General Associativity Theorem/Formulation 1

Theorem
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:


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

Proof

 * Let $T$ be the set of all $n \in \N_{>0}$ 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, $T = \N^*$.