I'm having trouble figuring out the proof to the proposition: for any $a_1,a_2,\ldots,a_n \in \mathbb{G}$ the value of $a_1~R~a_2~R~a_3~R\cdots R~a_n$ is independent of how the expression is bracketed (this is called the generalized associative law).
If $R$ was an equivalence relation this would be quite easy by using the symmetry of the relation to show commutativity and associativity.
I don't really know which definitions/assumptions to start with to prove this. I want to prove this locally for a few elements first and then by induction extend that to all elements in $G$. How may I do this?
Best Answer
The induction starts with three elements - that's just the associativity axiom!
If we have $n$ elements, $a_1, \ldots, a_n$, then we either have $a_1 (a_2 \ldots a_n)$ or $(a_1 \ldots a_{n-1}) a_n$. Notice that inside the bracket we have $n-1$ elements, so by induction, it doesn't matter how we bracket them.
In particular, for the first case we can choose the bracketing to be $a_1 ((a_2 a_3) \ldots a_n)$, which is the same as $(a_1 ((a_2 a_3) \ldots a_{n-1}))a_n$. Again, we have $n-1$ elements in the left-hand bracket so it doesn't matter how we bracket them, so this is the same as $(a_1 \ldots a_{n-1})a_n$.
So all expressions of the second form are equal to each other, and all expressions of the first form are equal to an expression of the second form. Therefore it doesn't matter where you put the brackets!
Edit: as pointed out in the comments, I had missed an important case. We might have $(a_1 \ldots a_r) (a_{r+1} \ldots a_n)$. However, this can be dealt with in the same way: both of these brackets have fewer than $n$ elements, so it doesn't matter how we bracket inside them. In particular, it's the same as $(a_1 (a_2 \ldots a_r)) (a_{r+1} \ldots a_n)$, which is the same as $a_1 ((a_2 \ldots a_r)(a_{r+1} \ldots a_n))$, which is the same as $a_1 (a_2 \ldots a_n)$ by the induction hypothesis. Therefore bracketings of this new form agree with the two cases previously considered.