How to formalize "It doesn't matter how places the brackets"

Question

Let $X$ be a set and $X\times X\to X: (a, b)\mapsto a\cdot b$ an associative operation on $X$. Now one can prove by induction that it doesn't matter how one places the brackets in a product $x_1\cdot x_2\cdot\text{ }\dots\text{ } \cdot x_n$, the product always evaluates to the same value. This result justifies the notation of a product $x_1\cdot x_2\cdot\text{ }\dots\text{ } \cdot x_n$ without any brackets. But how can one formalize it?

Answer 1

One nice way to formalize the general associative law is via binary trees. For us, a tree of operations will be a finite tree, with a root node, where each node is either a leaf (no successors) or has exactly two successors (a left successor and a right successor), and each leaf is labelled with an element of $X$.

The idea is that each leaf of the tree represents an input, and each non-leaf node is an instance of the operation $*$ applied to the values associated to the $*$-product of the left and right sub-trees of that node. So, for instance, $(a*b)*c$ is represented by a tree with one split at the root, and then the left side splits once more; and the leaves are labelled $a, b, c$, respectively.

A vauation of a tree of operations is a map $m$ from the nodes to $X$, satisfying $m(\sigma)$ is the label already on $\sigma$ - if $\sigma$ is a leaf - and $m(\sigma)=m(S_{left}(\sigma))*m(S_{right}(\sigma)),$ where $S_{left}$ and $S_{right}$ are the left and right successors of $\sigma$, if $\sigma$ is not a leaf. We can prove by induction that each tree $T$ of operations has exactly one valuation; call this "$v(T)$."

Note that the leaves of a tree of operations are ordered in a natural way: one leaf $\sigma$ is before another leaf $\tau$ if it is "to the left" of $\tau$, that is, when $\rho$ is the furthest node of the tree above both $\sigma$ and $\tau$ we have that $\sigma$ goes left from $\rho$ while $\tau$ goes right. The general associative law is then:

If $T_1, T_2$ are two trees of operations with the same leaf-labels occurring in the same left-right order, then $v(T_1)$ and $v(T_2)$ assign the same element of $X$ to the root of $T_1$ and $T_2$, respectively.

Answer 2

To each rooted planar binary tree $\tau$ we can associate an operation $X^{\times n}\to X$, where $n$ is the arity of the tree (i.e. the number of leaves): if $x_1,\ldots,x_n\in X$ then $\tau(x_1,\ldots,x_n)$ is obtained by putting $x_i$ at the $i$th leaf of the tree, and then "moving down towards the root" by applying the multiplication at the vertices. This shows that planar binary trees are in bijective correspondence with bracketings. Then the statement you are looking for is:

Lemma: Any two planar binary trees of the same arity induce the same operation.

This is easily proven by induction.

Remark: These ideas come quite naturally once you are acquainted with operad theory.

Answer 3

Assume your operation is associative, i.e. $\forall a,b,c\in X$: $$a\cdot (b\cdot c) =(a\cdot b)\cdot c$$ Now, we want to prove that brackets don't matter for any general expression using induction.

Starting with a product of three items, the claim is true due to the associative property. Now assume the claim is true for some expression involving $n>3$ terms, which we'll denote by $x_n$. Any expression involving $n+1$ terms, is of the forms: $$x_{n+1}=a\cdot (x_n)\ \text{or}\ x_{n+1}=(a\cdot x_{n_1})\cdot x_{n_2}$$ Where $n_1 + n_2 = n$, and we want to prove these two forms are identical. Due again to the associative property, and the induction hypotheses we have that: $$(a\cdot x_{n_1})\cdot x_{n_2} = a\cdot(x_{n_1}\cdot x_{n_2}) = a \cdot (x_n)$$ Q.E.D

Answer 4

This is called the associative law which says precisely: $$\forall x,y,z [(x*(y*z))=((x*y)*z))] $$

That it extends to arbitrary finite products can then proven using induction.

Some more details: you will prove that every expresention is equivalent to the one with brackets on left e.g. to $(((x*y)*z)*w))$ (every expresion uses each variable only once). Then you can define complexity $C:\mathrm{EXP}\to\mathbb{N}$ of an expression recursively as : $C(x)=0$, and for composed expressions $C(e*f)=C(e)+C(f)+\mathrm{length}(f)-1$, where $e,f$ are expressions and $\mathrm{length}(f)$ is a number of variables in $f$.

Examples: $C(x*y)=C(x)+C(y)+1-1=0$,

$C((x*y)*z))=C(x*y)+C(z)+1-1=0$,

but $C(x*(y*z))=C(x)+C(y*z)+2-1=1$,

and use induction on the complexity. Calculate that any one application of associativity decreases the complexity.