# Splitting up bracket terms

I found a statement saying: Let $\circledast$ be an associative binary operation on a set $\mathbb{X}$. A bracket term of length n, consisting of n elements $a_1, ..., a_n$ and arbitrary brackets, e.g. $K_4 = a_1 \circledast (a_2 \circledast a_3) \circledast a_4$, can be written as $K_n= (...(a_1 \circledast a_2)\circledast a_3) \circledast ...) \circledast a_{n-1})\circledast a_n$.

In other words: You can leave out the brackets.

Induction hypothesis: $K_k = (...(a_1 \circledast a_2)\circledast a_3) \circledast ...) \circledast a_{k-1})\circledast a_k$ for every bracket term of length $k, 3\leq k\leq n$.

The proof is simple, but in the induction step it makes use of the assumption that there are always $l, m \in \mathbb{N} \backslash \{0\}$ so that $l+m=n+1$ and $K_{n+1} = K_l \circledast K_m$

To my mind, this means that you can always split up a bracket expression at a certain position l (from left) or m (from right). Is this just a trivial fact or can/must this be further explained?

Edit: I have no problems with understanding the proof for the actual statement, but what's the reason for being able to split up arbitrary bracket terms?

Yes. That is what is usually called "strong induction," prove $p(n + 1)$ using $p(1)$ up to $p(n)$ true as the induction step.