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.
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 2$: The Axioms of Group Theory: $(1.4)$
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.5$
Category:General Associativity Theorem