Talk:General Associativity Theorem

Isn't this technically limited to proving associativity for any countable number of operations? Of course, you probably will never have an uncountable number of associative operations, but we should still be clear. --Cynic (talk) 04:51, 10 May 2010 (UTC)

Probably is, actually. It's a fine point with (it seems to me) limited relevance ... I need to tidy this page up some time, but it won't be immediate. --Matt Westwood 05:37, 10 May 2010 (UTC)

Proof
There is a (the?) proof at http://groupprops.subwiki.org/wiki/Associative_implies_generalized_associative with ref. to Dummit/Foote. I would like to tersify it for ProofWiki to better get my head around it. As this is my first transcription here, I'll append it here before moving to the article page.

Theorem
Let $\left({S, \circ}\right)$ be a semigroup.

Let $a_i$ denote elements of $S$.

Let $\circ$ be associative.

Let $n$ be any positive integer: $n>3, n\in\N$.

Then every possible parenthesization of the expression:


 * $\displaystyle a_1 \circ a_2 \circ \dots \circ a_n$

is equivalent.

Lemma
Let $\circ$ be associative.

Then any parenthesization of $a_1 \circ a_2 \circ \dots \circ a_n$ is equal to the left-associated expression:
 * $\displaystyle ((\dots ( a_1 \circ a_2) \circ a_3) \circ \dots ) \circ a_n)$

Proof
...

It follows that all parenthesizations of $a_1 \circ a_2 \circ \dots \circ a_n$ yield identical results.

So the theorem holds.


 * Is it good, that Dummit & Foote? It's not a work I've come across. --prime mover (talk) 18:39, 1 October 2014 (UTC)