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

Let $f_1\!:\,S \to 2^S$ be the mapping $$f_1(a_1) := \{a_1\}\;\; \forall a_1 \in S.$$ We define recursively the mapping $$f_n\!:\,\underbrace{S\!\times\!S\!\times\!\ldots\!\times\!S}_n \to 2^S$$ such that

(1)

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 $$|f_1(a_1)| = 1,\quad |f_2(a_1,\,a_2)| = 1,\quad |f_3(a_1,\,a_2,\,a_3)| = 1.$$

We shall show by induction that

(2)

We make the induction hypothesis, that (2) is true for all $n < k.$

Now let $z,\,z'$ be arbitrary elements of $f_k(a_1,\ldots,\,a_k)$ . Then there exist the elements $x,\,y,\,x',\,y'$ of $S$ and the integers $r,\,s \in \{1,\,\ldots,\,k\!-\!1\}$ such that $$z = x\cdot y,\quad x \in f_r(a_1,\ldots,\,a_r),\quad y \in f_{k-r}(a_{r+1},\ldots,\, a_k),$$ $$z' = x'\cdot y',\quad x' \in f_s(a_1,\ldots,\,a_s),\quad y' \in f_{k-s}(a_{s+1},\ldots,\, a_k).$$ If specially $r = s$ , then, by the induction hypothesis, $x = x'$ and $y = y'$ , whence $z = x\cdot y = x'\cdot y' = z'$ . If on the contrary, $r \neq s$ , e.g. $r < s$ , then the induction hypothesis guarantees the existence of an element $v$ of $S$ such that $$f_{s-r}(a_{r+1},\ldots,\,a_s) = \{v\}$$ and $$x' = x\cdot v,\quad y = v\cdot y'.$$ Since ``$\cdot$ '' is associative, we have $$z = x\cdot y = x\cdot(v\cdot y') = (x\cdot v)\cdot y' = x'\cdot y' = z'.$$ Thus the equation (2) is in force for $n = k$ .