\tilde{\Delta}(H).$$ From the property $\tau^{2}=1\otimes 1$ and the definitions of $\tau$ and $R$, one gets $$\label{rdr}
R\,\Delta h\, R^{-1}=\tilde{\Delta}h,$$ for every $h$ in $H$. For the family of twists , the corresponding $R$-matrix is given by $$\label{r}
R=\exp{\bigl(i(A\otimes x_kp_k -x_kp_k\otimes A)\bigr)},$$ and it satisfies $$\tau_0 \,R\,\tau_0=R^{-1},  \quad R^{-1}: \  \tilde{\Delta}(H)
 \longrightarrow  \Delta(H).$$ We note that if the $\star$-product is defined by $(f\star g)_{\mathcal{F}} =m_0\mathcal{F}^{-1}\triangleright (f\otimes g)_{\mathcal{R}}$, where $\mathcal{R}$ in the subscript denotes that $(f\otimes g)_{\mathcal{R}} \in \Delta ({\mathcal{A}})$, and we define $(f\star g)_{\widetilde{\mathcal{F}}} =m_0\widetilde{\mathcal{F}}^{-1}\triangleright (f\otimes g)_{\widetilde{\mathcal{R}}}$, then it follows that $$\label{sfstf}
(f\star g)_{\mathcal{F}}=(g\star f)_{\widetilde{\mathcal{F}}} \ \Rightarrow \ \mathcal{F}^{-1}\triangleright (f\otimes g)_{\mathcal{R}} = \widetilde{\mathcal{F}}^{-1}\triangleright (g\otimes f)_{\widetilde{\mathcal{R}}},$$ where $(g\otimes f)_{\widetilde{\mathcal{R}}} \in \tilde{\Delta}({\mathcal{A}}). $ Multiplying the second equation in by $\widetilde{\mathcal{F}}$ from the left, one finds $$R\triangleright (f\otimes g)_{\mathcal{R}}= (g\otimes f)_{\widetilde{\mathcal{R}}}.$$

By direct calculation (inserting and - into the LHS of ), it is easy to check that is the exact $R$-matrix for the generators of the Heisenberg algebra $H$ $$\begin{aligned}
&& R\Delta x_iR^{-1}= x_i\otimes Z^{\lambda -1} =Z^{\lambda}\otimes x_i =\tilde{\Delta}x_i \label{dtxi} \\
&& R\Delta x_0R^{-1}=x_0\otimes 1-a_0\lambda\cdot 1\otimes x_k p_k = 1\otimes x_0 +a_0(1-\lambda) x_kp_k\otimes 1= \tilde{\Delta}x_0 \label{dtx0} \\
&& R\Delta p_i R^{-1}=p_i\otimes Z^{1-\lambda}+Z^{-\lambda}\otimes p_i =\tilde{\Delta}p_i \\
&& R\Delta p_0 R^{-1}=p_0\otimes 1+1\otimes p_0 =\tilde{\Delta} p_0,\label{dtp_0}\end{aligned}$$ 