When proving the asociativity of some operation denoted by $\bullet$ one first proves that $a\bullet(b\bullet c)=(a\bullet b)\bullet c$ for any $a$, $b$ and $c$.
This automatically implies that $$a_1\bullet a_2\bullet\cdots\bullet a_n$$ is well defined for $a_1, a_2,\dots, a_n$ with $n$ arbitrarily large, but I would like to see a symbollic proof using the fact that all "parenthesifications" are equivalent, for example $$(a\bullet b)\bullet(c\bullet d)=a\bullet(b\bullet(c\bullet d))$$
So, given $a_1,a_2,\dots,a_n$, is there a notation to represent any such combination of parenthesis and $\bullet$s?
With this notation one could prove the equivalence of any two combinations by transforming any one of then by the rule $a\bullet(b\bullet c)=(a\bullet b)\bullet c$ until arriving at the "standard" $$a_1\bullet(a_2\bullet(\cdots(a_{n-1}\bullet a_n)\cdots))$$
A good notation should behave "mecanically" through the rule
Thanks in advance!
I have used numbers under the operation to specify the order of evaluation. Thus:
$$(ab)c=a\underset1\bullet b\underset2\bullet c,\qquad a(bc)=a\underset2\bullet b\underset1\bullet c,$$ $$a((bc)d)=a\underset3\bullet b\underset1\bullet c\underset2\bullet d,\qquad a(b(c(de)))=a\underset4\bullet b\underset3\bullet c\underset2\bullet d\underset1\bullet e.$$
However, this means we have two different ways of writing $(ab)(cd)$:
$$a\underset1\bullet b\underset3\bullet c\underset2\bullet d,\qquad a\underset2\bullet b\underset3\bullet c\underset1\bullet d.$$
An application of the operation to $n$ elements $a_1,a_2,\cdots,a_n$ is specified by a permutation of the numbers $1,2,\cdots,n-1$.
Three-term associativity means that any two consecutive (not necessarily adjacent) numbers in the permutation can be swapped. For example,
$$a\underset3\bullet b\underset2\bullet c\underset5\bullet d\underset1\bullet e\underset4\bullet f=(a(bc))((de)f)$$ $$=a\underset3\bullet b\underset2\bullet c\underset4\bullet d\underset1\bullet e\underset5\bullet f=((a(bc))(de))f$$
(let the three terms be $(a(bc))$ and $(de)$ and $(f)$). If there's a larger number between the two consecutive numbers, then they can be swapped without assuming associativity, as in the $(ab)(cd)$ example.