General Associativity Theorem/Formulation 1

Theorem
Let $\struct {S, \circ}$ be a semigroup.

Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ be a sequence of elements of $S$.

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

Suppose:


 * $\ds \forall k \in \closedint 1 s: b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} {a_j}$

Then:


 * $\ds \forall n \in \N_{>0}: \prod_{k \mathop = 1}^s {b_k} = \prod_{k \mathop = p \mathop + 1}^{p \mathop + n} {a_k}$

That is:


 * $\ds \forall n \in \N_{>0}: \prod_{k \mathop = 1}^s \paren {a_{r_{k - 1} + 1} \circ a_{r_{k - 1} + 2} \circ \ldots \circ a_{r_k} } = a_{p + 1} \circ \ldots \circ a_{p + n}$

Proof
The proof will proceed by the Principle of Mathematical Induction on $\N$.

Let $T$ be the set of all $n \in \N_{>0}$ such that:


 * $(1): \quad$ for every sequence $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + n}$ of elements of $S$

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

the statement:
 * $\ds b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} a_j$

holds.

Basis for the Induction
Let $n = 1$.

Then:

So $1 \in T$.

This is our basis for the induction.

Induction Hypothesis
It is to be shown that, if $m \in T$ where $m \ge 1$, then it follows that $m + 1 \in T$.

This is the induction hypothesis:


 * $(1): \quad$ for every sequence $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m}$ of elements of $S$

and:
 * $(2): \quad$ for every strictly increasing sequence $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of natural numbers such that $r_0 = p$ and $r_s = p + m$:

the statement:
 * $\ds b_k = \prod_{j \mathop = r_{k-1} \mathop + 1}^{r_k} a_j$

holds.

It is to be demonstrated that it follows that:


 * $(1): \quad$ for every sequence $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m + 1}$ of elements of $S$

and:
 * $(2): \quad$ for every strictly increasing sequence $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ of natural numbers such that $r_0 = p$ and $r_s = p + m + 1$:

the statement:
 * $\ds b_k = \prod_{j \mathop = r_{k - 1} \mathop + 1}^{r_k} a_j$

holds.

Induction Step
This is our induction step:

Let $\sequence {a_k}_{p + 1 \mathop \le k \mathop \le p + m + 1}$ be a sequence of elements of $S$.

Let $\sequence {r_k}_{0 \mathop \le k \mathop \le s}$ be a strictly increasing sequence of natural numbers such that $r_0 = p$ and $r_s = p + m + 1$.

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

There are two cases:


 * $(1): \quad r_{s - 1} = p + m$
 * $(2): \quad r_{s - 1} < p + m$

First, suppose:
 * $r_{s - 1} = p + m$

Then:
 * $b_s = a_{p + m + 1}$

Thus:

Secondly, suppose:
 * $r_{s - 1} < p + m$

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

Then by definition of composite:
 * $b_s = b\,'_s \circ a_{p + m + 1}$

Now by the Induction Hypothesis:
 * $m \in T \implies a_{p + 1} \circ \ldots \circ a_{p + m} = b_1 \circ \ldots \circ b_{s - 1} \circ b\,'_s$

Thus:

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

So by the Principle of Mathematical Induction:
 * $T = \N_{>0}$

Hence the result.