Let $F_n$ be the free group generated by $x_1,\cdots,x_n$ and for each $1\leq i\leq n$, let a homomorphism $d_i:F_n\to F_{n-1}$ be defined as follows:

 - $d_i(x_r)=x_r$, if $i>r$;
 - $d_i(x_r)=1$, if $i=r$;
 - $d_i(x_r)=x_{r-1}$, if $i<r$.

The equalizer $E_n\leq F_n$ defined by $$E_n=\{x\in F_n\mid d_1(x)=d_2(x)=\cdots=d_n(x)\}$$ is clearly a subgroup of $F_n$, and thus a free group (Every subgroup of a free group is free). 

**Question**: What is a minimal generating set of $E_n$?


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My guess is the set consists of the following elements may be a (**Is it minimal?**) generating set; but I am not sure if it is correct:

 1. $x_{\sigma(1)}x_{\sigma(2)}\cdots x_{\sigma(n)}$;

 2. $[x_{\sigma(1)},x_{\sigma(2)}][x_{\sigma(1)},x_{\sigma(3)}]\cdots[x_{\sigma(1)},x_{\sigma(n)}][x_{\sigma(2)},x_{\sigma(3)}]\cdots[x_{\sigma(2)},x_{\sigma(n)}]\cdots[x_{\sigma(n-1)},x_{\sigma(n)}]$;

 3. $[[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(3)}][[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(4)}]\cdots[[x_{\sigma(n-2)},x_{\sigma(n-1)}],x_{\sigma(n)}]$;

 4. $\cdots$;

 5. $[\cdots[[x_{\sigma(1)},x_{\sigma(2)}],x_{\sigma(3)}],\cdots x_{\sigma(n)}]$,

where $[x,y]=x^{-1}y^{-1}xy$ and $\sigma\in\Sigma_n$, the symmetric group acts on $\{1,2,\cdots,n\}$.