On the 2nd edition of Artin's Algebra, the writer uses proposition 2.1.4 to imply that generalised associative law works for associative binary operation:
Proposition 2.1.4 Let an associative law of composition be given on a set $S$. There is a unique way to define, for every integer $n$, a product of $n$ elements $a_1...a_n$ of $S$, denoted temporarily by $[a_1...a_n]$, with the following properties:
(i) The product $[a_1]$ of one element is the element itself.
(ii) The product $[a_1a_2]$ of two elements is given by the law of composition.
(iii) For any integer $i$ in the range $1\le i<n$, $[a_1...a_n]=[a_1...a_i][a_{i+1}...a_n]$
I have no problem understanding the proof of this proposition, but I would like to ask how this proposition implies the generalised associative law.
My understanding:
Artin's notation $[a_1...a_n]$ should be equivalent to a function $f_n(a_1,...,a_n):S^n\to S$.
Let $b(a_1,...,a_n)$ denotes a product of $a_1...a_n$ with some well-defined bracketing.
One could observe that, for any $b(a_1,...,a_n)$, we have $f_n(a_1,...,a_n)=b(a_1,...,a_n)$ by (i) and (iii). So, any well-defined bracketing on a product evaluates to the same value (i.e generalised associative law).
May I ask, if my understanding is correct, why would we even need to prove for uniqueness? We can just show the existence of $f_1$, then show the existence of $f_n$ for $n>1$ by defining $f_n$ recursively using $f_1$ (for each $n$ check (i) (ii) (iii) using induction). Once we have $f_i$ for all $1\le i\le n$, we could check that $f_n(a_1,...,a_n)=b(a_1,...,a_n)$ for any two bracketings.
If my understanding is incorrect, may I ask how Proposition 2.1.4 implies generalised associative law?
Let p(a_1,...,a_n) be any product of n elements in S. We can use strong induction to prove that p(a_1,...,a_n) = [a_1...a_n] as follows. Suppose this is true for all k < n. Now p(a_1, ..., a_n) = p'(a_1,...,a_k) p"(a_(k+1),...,a_n) for some products p' on k elements and p" on n-k elements. By the strong inductive hypothesis, p' and p" are identical to the product defined in Proposition 2.1.4, and so the rhs is equal to [a_1...a_k] [a_(k+1)...a_n] which, by (iii) is equal to [a_1...a_n]