proof of general associativity

We suppose that “$\cdot$” is an associative binary operation of the set $S$.

Let  $f_1:S\to 2^S$  be the mapping

$$f_1(a_1):=\{a_1\}\mathit{\hspace{1em}}\forall a_1\in S.$$

We define recursively the mapping

$$f_n:\underset{n}{\underbrace{S\times S\times \mathrm{\dots }\times S}}\to 2^S$$

such that

$f_n(a_1,\mathrm{\dots },a_n):={\displaystyle \bigcup _{r=1}^{n-1}}{f}_r(a_1,\mathrm{\dots },a_r)\cdot f_{n-r}(a_{r+1},\mathrm{\dots },a_n)$(1)

for  $n=2,\mathrm{\hspace{0.17em}3},\mathrm{\hspace{0.17em}4},\mathrm{\dots }$

For example,

$$f_2(a_1,a_2)=\{a_1\}\cdot \{a_2\}=\{a_1\cdot a_2\},$$

$$f_3(a_1,a_2,a_3)=\{a_1\cdot (a_2\cdot a_3)\}\cup \{(a_1\cdot a_2)\cdot a_3\}=\{a_1\cdot (a_2\cdot a_3),(a_1\cdot a_2)\cdot a_3\}=\{(a_1\cdot a_2)\cdot a_3\};$$

the last equality due to the associativity.  It’s clear that always

$$|f_1(a_1)|=\mathrm{\hspace{0.33em}1},|f_2(a_1,a_2)|=\mathrm{\hspace{0.33em}1},|f_3(a_1,a_2,a_3)|=\mathrm{\hspace{0.33em}1}.$$

We shall show by induction that

$|f_n(a_1,a_2,\mathrm{\dots },a_n)|=\mathrm{\hspace{0.33em}1}$(2)

for each positive integer $n$.  This means that all groupings of the $n$ fixed elements using parentheses in forming the products with “$\cdot$” yield one single element.

We make the induction hypothesis, that (2) is true for all 

Now let $z$ and $z^{\prime }$ be arbitrary elements of  $f_k(a_1,\mathrm{\dots },a_k)$. Then there exist the elements $x,y,x^{\prime },y^{\prime }$ of $S$ and theintegers  $r,s\in \{1,\mathrm{\dots },k-1\}$  such that

$$z=x\cdot y,x\in f_r(a_1,\mathrm{\dots },a_r),y\in f_{k-r}(a_{r+1},\mathrm{\dots },a_k),$$

$$z^{\prime }=x^{\prime }\cdot y^{\prime },x^{\prime }\in f_s(a_1,\mathrm{\dots },a_s),y^{\prime }\in f_{k-s}(a_{s+1},\mathrm{\dots },a_k).$$

If specially  $r=s$,  then, by the induction hypothesis,  $x=x^{\prime }$  and  $y=y^{\prime }$,  whence  $z=x\cdot y=x^{\prime }\cdot y^{\prime }=z^{\prime }$.  If on the contrary,  $r\ne s$,  e.g.  ,  then the induction hypothesis guarantees the existence of an element $v$ of $S$ such that

$$f_{s-r}(a_{r+1},\mathrm{\dots },a_s)=\{v\}$$

and

$$x^{\prime }=x\cdot v,y=v\cdot y^{\prime }.$$

Since “$\cdot$” is associative, we have

$$z=x\cdot y=x\cdot (v\cdot y^{\prime })=(x\cdot v)\cdot y^{\prime }=x^{\prime }\cdot y^{\prime }=z^{\prime }.$$

Thus the equation (2) is in force for  $n=k$.