If $M$ and $N$ are a set then for pratical reason let's we put $$ \mathcal F(N,M):=M^N $$
Now if $(X,\bot)$ a semigroup then by the recursion theorem it is not hard to prove (see here for details) the existence of sequence $(\bot^n)_{n\in\omega_+}$ into $\bigcup_{n\in\omega_+}\mathcal F(X^n,X)$ such that $\bot^n$ is in $\mathcal F(X^n,X)$ and such that for any $x_1,\dots,x_n,x_{n+1}$ in $\omega_+$ in $X$ with $n$ in $\omega_+$ the equality $$ \tag{0}\label{0}\bot^{n+1}(x_1,\dots , x_n, x_{n+1})=\big(\bot^n(x_1,\dots , x_n)\big)\bot x_{n+1} $$ holds; moreover, it is possibile to prove that $\bot^2$ is just $\bot$ and for any $x_1,\dots, x_m,x_{m+1},\dots,x_{m+n}$ in $X$ with $m$ and $n$ in $\omega_+$ the equality $$ \begin{equation}\tag{1}\label{1} \bot^{m+n}(x_1,\dots,x_m,x_{m+1},\dots x_{m+n}) = \big(\bot^m(x_1, \ldots, x_m)\big) \bot\big(\bot^n(x_{m+1}, \ldots, x_{m+n})\big) \end{equation} $$
So, by this it is possibile to put $$ x_1\bot\dots\bot x_n:=\bot^n(x_1,\dots, x_n) $$ for any $x_1,\dots,x_n$ in $X$ with $n$ in $\omega_+$ so that let's we prove that for any permutations $\sigma$ of $\{1,\dots, n\}$ then the equality $$ \tag{2}\label{2}x_{\sigma(1)}\bot\dots\bot x_{\sigma(n)}=x_1\bot\dots\bot x_n $$ provided that $(X,\bot)$ is commutative.
First of all let be $\lambda$ the permutation which works as $\sigma$ except for $\sigma^{-1}(n+1)$ and for $n$ whereas let be $\tau$ the permutation with interchange only $\sigma^{-1}(n+1)$ with $n$ and vice versa: in matrix notation we can represent $\lambda$ with the matrix $$ \begin{pmatrix} 1 & \dots & \big(\sigma^{-1}(n+1)-1\big) & \sigma^{-1}(n+1) & \big(\sigma^{-1}(n+1)+1\big) & \dots & n & n+1\\ \sigma(1) & \dots & \sigma\big(\sigma^{-1}(n+1)-1\big) & \sigma^{-1}(n+1) & \sigma\big(\sigma^{-1}(n+1)+1\big) & \dots & \sigma(n) & n+1 \end{pmatrix} $$ whereas we can represent $\tau$ with the matrix $$ \begin{pmatrix} 1 & \dots & \big(\sigma^{-1}(n+1)-1\big) & \sigma^{-1}(n+1) & \big(\sigma^{-1}(n+1)+1\big) & \dots & n & n+1\\ 1 & \dots & \big(\sigma^{-1}(n+1)-1\big) & n+1 & \big(\sigma^{-1}(n+1)+1\big) & \dots & n & \sigma^{-1}(n+1) \end{pmatrix} $$
So, observing that $\sigma$ is just the composition $\tau\circ\lambda$ and observed that $\lambda$ does not changes the $n+1$-th element and $\tau$ does not changes the first element, let's we assume that \eqref{2} holds for any $n$ so that by associativity we observe that $$ x_{\sigma(1)}\bot\dots\bot x_{\sigma(n)}\bot x_{\sigma(n+1)}=x_{\tau(\lambda(1))}\bot\dots\bot x_{\tau(\lambda(n))}\bot x_{\tau(\lambda(n+1))}=\\ x_{\lambda(1)}\bot\dots\bot x_{\tau(\lambda(n))}\bot x_{\tau(\lambda(n+1))}= x_{\lambda(1)}\bot\Big(x_{\tau(\lambda(2))}\bot\dots\bot x_{\tau(\lambda(n))}\bot x_{\tau(\lambda(n+1))}\Big)=\\ x_{\lambda(1)}\bot\Big(x_{\lambda(2)}\bot\dots\bot x_{\lambda(n)}\bot x_{\lambda(n+1)}\Big)= x_{\lambda(1)}\bot\dots\bot x_{\lambda(n)}\bot x_{\lambda(n+1)}=\\ x_{\lambda(1)}\bot\dots\bot x_{\lambda(n)}\bot x_{n+1}=\Big(x_{\lambda(1)}\bot\dots\bot x_{\lambda(n)}\Big)\bot x_{n+1}=\\ \Big(x_{1}\bot\dots\bot x_{n}\Big)\bot x_{n+1}=x_{1}\bot\dots\bot x_{n}\bot x_{n+1} $$ which proves that it holds for $n+1$; moreover commutativity implies that \eqref{2} holds for $n$ equal to $2$: so by induction we conclude that \eqref{2} holds for any $n$ in $\omega_+$ bigger than $2$.
So, I would like to know if \eqref{2} holds and so if I well proved it. Moreover, as you can see, I used associativity not only to prove \eqref{2} but also to define $\bot^n$ with $n$ in $\omega_+$ which actually let to make sense to any string $x_1\bot\dots\bot x_n$ with $x_1,\dots ,x_n$ in $X$ and with $n$ in $\omega_+$: so I ask if the result can hold even without associativity. Could someone help me, please?
