\fA^0$.

-   $ Z_R(\operatorname{{reg}}_*(\om_1\sha \om_0-\om_1*\om_0))=0$ for all $\om_1\in
       \fA^1$ and all $\om_0\in \fA^0$.

-   $ Z_R(\operatorname{{reg}}_\sha(b_0^m*w))=0$ for all $m\ge 1$ and all $w\in \fA^0$.

-   $ Z_R(\operatorname{{reg}}_*(b_0^m\sha w-b_0^m*w))=0$ for all $m\ge 1$ and all $w\in \fA^0$.

If $Z_R$ satisfies any one of these then we say that $Z_R$ has the *regularized double shuffle* (RDS) property.

Notice that RDS automatically implys FDS. The proof of the theorem is almost the same as that of [@IKZ Thm. 2] but for completeness we give the most important details in the following because there is some subtle difference for MPVs of arbitrary level.

Denote by $\caS$ the set of the $y_{s,j}$ ($s\in \Z_{>0},
j=0,\dots, N-1$). For convenience we write $\tau_z=\tau_j$ if $z=y_{s,j}\in\caS$. If $w=y_{s_1,i_1}\dots y_{s_n,i_n}\in \fA^1$ then we put $\tau_w=\tau_{i_1+\dots+i_n}$ and $\tau_{-w}=\tau_{-i_1-\dots-i_n}$. Then (cf. [@IKZ Prop. 2])

\[prop:gdz\] We have

1.  For $z\in \caS$ the map $\gd_z:\fA^1\to \fA^1$ defined by $$\gd_z(w):=z*w-z\tau_z(w)$$ is a “twisted derivation” in the sense that $$\gd_z(ww')=\gd_z(w)\tau_z(w')+w\tau_w\Big(\gd_z\big(\tau_{-w}(w')\big)\Big).$$ Moreover, all these twisted derivations commute.

2.  The above twisted derivations extend to a twisted derivation on all of $\fA$ after setting $\tau_a=\text{id}$, with values on the letters $a,
        b_j$ given by $$\gd_z(a)=0, \quad \gd_z(b_j)=(a+b_j)\tau_j(z)\quad (z\in \caS, j=0,\dots,N-1).$$ In particular, $\gd_z$ preserves $\fA^0$.

Easy 