1},x)=Y^{*}({\bf 1},x)=1$. Then it remains to prove the Jacobi identity.

Let $u,v\in V,\; \alpha\in {\cal{D}}_{P(z)}(W)$. By Proposition \[pbasic1\], there exists $k_{0}\in {\N}$ such that for [*all*]{} $m\in {\Z}$, $$\begin{aligned}
\label{ey*v*m}
(x_{1}-z)^{k_{0}}Y^{*}(u,x_{1})v^{*}_{m}\alpha\in W^{*}((x_{1})),\end{aligned}$$ hence (recall Lemma \[cbasic\]) $$\begin{aligned}
\label{e3.27}
(x_{1}-z)^{k_{0}}Y^{R}(u,x_{1})Y^{*}(v,x_{2})\alpha
=(x_{1}-z)^{k_{0}}Y^{*}(u,x_{1})Y^{*}(v,x_{2})\alpha.\end{aligned}$$ Let $k'\ge k_{0}$ be such that $$(x_{2}-z)^{k'}Y^{R}(v,x_{2})\alpha=
(x_{2}-z)^{k'}Y^{*}(v,x_{2})\alpha\in W^{*}((x_{2})).$$ Then by (\[e3.27\]), $$\begin{aligned}
\label{eyruv}
& &(x_{2}-z)^{k'}(x_{1}-z)^{k'}
Y^{R}(u,x_{1})Y^{R}(v,x_{2})\alpha\nonumber\\
&=&(x_{2}-z)^{k'}(x_{1}-z)^{k'}Y^{*}(u,x_{1})Y^{*}(v,x_{2})\alpha.\end{aligned}$$ Similarly, there exists $k''\in {\N}$ such that $$\begin{aligned}
\label{eyrvu}
& &(x_{1}-z)^{k''}(x_{2}-z)^{k''}Y^{R}(v,x_{2})Y^{R}(u,x_{1})\alpha
\nonumber\\
&=&(x_{1}-z)^{k''}(x_{2}-z)^{k''}Y^{*}(v,x_{2})Y^{*}(u,x_{1})\alpha.\end{aligned}$$ Set $k=k'+k''$. For any $w\in W$, using (\[eyruv\]), (\[eyrvu\]) and (\[ejacobiy\*m\]) we obtain $$\begin{aligned}
\label{e3.30}
& &x_{0}^{-1}\delta\left(\frac{x_{1}-x_{2}}{x_{0}}\right)
(x_{1}-z)^{k}(x_{2}-z)^{k}\<Y^{R}(u,x_{1})Y^{R}(v,x_{2})\alpha,w\>\nonumber\\
& &-x_{0}^{-1}\delta\left(\frac{x_{2}-x_{1}}{-x_{0}}\right)
(x_{1}-z)^{k}(x_{2}-z)^{k}\<Y^{R}(v,x_{2})Y^{R}(u,x_{1})\alpha,w\>\nonumber\\
&=&x_{0}^{-1}\delta\left(\frac{x_{1}-x_{2}}{x_{0}}\right)
(x_{1}-z)^{k}(x_{2}-z)^{k}\<Y^{*}(u,x_{1})Y^{*}(v,x_{2})\alpha,w\>
\nonumber\\
& &-x_{0}^{-1}\delta\left(\frac{x_{2}-x_{1}}{-x_{0}}\right)
(x_{1}-z)^{k}(x_{2}-z)^{k}\<Y^{*}(v,x_{2})Y^{*}(u,x_{1})\alpha,w\>
\nonumber\\
&=&x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)
(x_{1}-z)^{k}(x_{2}-z)^{k}\<Y^{*}(Y(u,x_{0})v,x_{2})\alpha,w\>.\end{aligned}$$ Let $m$ be [*any fixed*]{} integer. Since $u_{n}v\ne 0$ for only finitely many $n\ge m$, there is a nonnegative integer $k_{1}$ ([*depending*]{} on $m$) such that for [*all*]{} $n\ge m$, $$\begin{aligned}
\label{eunv}
(x_{2}-z)^{k_{1}}Y^{*}(u_{n}v,x_{2})\alpha\in {\cal{D}}_{P(z)}(W)((x_{2})),\end{aligned}$$ hence (recall Lemma \[cbasic\]) $$\begin{aligned}
(x_{2}-z)^{k_{1}}Y^{R}(u_{n}v,x_{2})\alpha=
(x_{2}-z)^{k_{1}}Y^{*}(u_{n}v,x_{2})\alpha.\end{aligned}$$ Then $$\begin{aligned}
\label{econn1}
& &{\rm Res}_{x_{0}}x_{0}^{m}
x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)
(x_{2}-z)^{k_{1}}\<Y^{R}(Y(u,x_{0})v,x_{2})\alpha,w\>\nonumber\\
&=&{\rm Res}_{x_{0}}x_{0}^{m}
x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)
(x_{2}-z)^{k_{1}}\<Y^{*}(Y(u,x_{0})v,x_{2})\alpha,w\>.\end{aligned}$$ Choosing $k_{1}\ge k$ and then combining (\[e3.30\]) with (\[econn1\]) we obtain $$\begin{aligned}
\label{ejac}
& &{\rm Res}_{x_{0}}x_{0}^{m}
x_{0}^{-1}\delta\left(\frac{x_{1}-x_{2}}{x_{0}}\right)
(x_{1}-z)^{k_{1}}(x_{2}-z)^{k_{1}}\<Y^{R}(u,x_{1})
Y^{R}(v,x_{2})\alpha,w\>\nonumber\\
& &-{\rm Res}_{x_{0}}x_{0}^{m}
x_{0}^{-1}\delta\left(\frac{x_{2}-x_{1}}{-x_{0}}\right)
(x_{1}-z)^{k_{1}}(x_{2}-z)^{k_{1}}\<Y^{R}(v,x_{2})Y^{R}(u,x_{1})\alpha,w\>
\nonumber\\
&=&{\rm Res}_{x_{0}}x_{0}^{m}
x_{2}^{-1}\delta\left(\frac{x_{1}-x_{0}}{x_{2}}\right)
(x_{1}-z)^{k_{1}}(x_{2}-z)^{k_{1}}\<Y^{R}(Y(u,x_{0})v,x_{2})\alpha,w\>.\end{aligned}$$ In view of Remark \[rbim\], we may multiply both sides of (\[ejac\]) by $(-z+x_{1})^{-k_{1}}(-z+x_{2})^{-k_{1}}$ to get $$\begin{aligned}
