General Associativity Theorem/Formulation 2/Proof 2

Theorem
Let $n \in \N_{>0}$ and let $a_1, \ldots, a_n$ be elements of a set $S$.

Let $\circ$ be an associative operation on $S$.

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 $a_1 \circ a_2 \circ \ldots \circ a_n$.

Proof
Proof by strong induction:

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le n$.

$P \left({1}\right)$ is trivially true, as this just says $a_1 = a_1$.

$P \left({2}\right)$ is the case:
 * $a_1 \circ a_2 = a_1 \circ a_2$

for which there is also nothing to prove.

Basis for the Induction
$P \left({3}\right)$ is the case:
 * $\left({a_1 \circ a_2}\right) \circ a_3 = a_1 \circ \left({a_2 \circ a_3}\right)$

which is the definition of associativity.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k$.

Then we need to show:
 * The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_r$ such that $r \le k+1$.

Induction Step
This is our induction step:

Consider the expressions:
 * $(1): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$
 * $(2): \quad \left({a_1 \circ a_2 \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$

for some $i, j \in \Z: 1 \le i, j \le k$.

We require to show that whatever the values of $i$ and $j$, these expressions are equal.

Suppose WLOG that $i < j$.

Then the above expressions can be written:
 * $\left({a_1 \circ a_2 \circ \cdots \circ a_i}\right) \circ \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right) \circ \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$

By the basis for the induction, the General Associativity Theorem holds for each of the parts in parenthesis.

Let:
 * $a = \left({a_1 \circ a_2 \circ \cdots \circ a_i}\right)$
 * $b = \left({a_{i+1} \circ a_{i+2} \circ \cdots \circ a_j}\right)$
 * $c = \left({a_{j+1} \circ a_{i+2} \circ \cdots \circ a_{k+1}}\right)$

By definition of associative operation:
 * $\left({a \circ b}\right) \circ c = a \circ \left({b \circ c}\right)$

This demonstrates the equality of the expressions $(1)$ and $(2)$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

Therefore:
 * The General Associativity Theorem holds for all composites $a_1 \circ a_2 \circ \cdots \circ a_n$ for $n \in \N$.