Theorem: General Associativity Theorem/Formulation 2

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 $\map {P_n} {a_1, a_2, \ldots, a_n}$ be defined inductively by:
:$\map {P_1} {a_1} = \set {a_1}$
:$\map {P_2} {a_1, a_2} = \set {a_1 \circ a_2}$
:$\map {P_n} {a_1, a_2, \ldots, a_n} = \set {x \circ y: x \in \map {P_r} {a_1, a_2, \ldots, a_r} \land y \in \map {P_s} {a_{r + 1}, a_{r + 2}, \ldots, a_{r + s} }, n = r + s}$
Then $\map {P_n} {a_1, a_2, \ldots, a_n}$ 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.
{{WLOG}}, suppose 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 $n \in \N$.
{{qed}}