generated by the relations ${\mathcal{R}}_0
  \equiv x_{\mu} \otimes 1 - 1 \otimes  x_{\mu} =0. $ The relations ${\mathcal{R}}_0$ generate the equivalence classes on ${\mathcal{A}} \otimes {\mathcal{A}}$. It can be shown that $\Delta_0 ({\mathcal{A}}) = {\mathcal{A}} \otimes {\mathcal{A}} / {\mathcal{R}}_0 $ is an algebra isomorphic to ${\mathcal{A}}$. Similarly, from the Leibniz rule for $p_{\mu }$ it follows: $$\begin{aligned}
    \Delta_0 p_{\mu} =p_{\mu} \otimes 1 + 1 \otimes p_{\mu}. \label{copp}\end{aligned}$$ Note that $\Delta_0 (T) $ is an algebra isomorphic to $T$.

Generally, $\Delta_0 h $ for any $h \in H, $ can be obtained by the homomorphism $\Delta_0 (h_1h_2) = (\Delta_0 h_1) (\Delta_0 h_2) $ that follows from Eq.(\[compositionrule\]). Note that $\left[p_{\mu} \otimes 1, x_{\nu} \otimes 1 - 1 \otimes  x_{\nu} \right] \neq 0, $ but $ \left[ \Delta_0 p_{\mu}, x_{\nu} \otimes 1 - 1 \otimes  x_{\nu}
  \right] = 0 $ and generally the consistency requirement $ \left[ \Delta_0 h, x_{\nu} \otimes 1 - 1 \otimes  x_{\nu}
  \right] = \left[ \Delta_0 h, {\mathcal{R}}_0 \right] = 0,$ for all $h \in H, $ is satisfied. Taking $h_1=p_\mu, \,h_2=x_\nu$, and using Eqs., and $\mathcal{R}_0$, one gets $$\begin{aligned}
\Delta_0(p_\mu x_\nu) &=& (\Delta_0p_\mu)(\Delta_0x_\nu) \nonumber \\
&=&  \left\{      \begin{array}{clrr}
& (p_\mu \otimes 1+1 \otimes p_\mu) (x_\nu \otimes 1) 