can be written down by symmetry. Summing up all the four terms and applying induction, we obtain $$\begin{array}{r c l}
RHS_{u v} 
    & = & {\hat \Delta}(\hat G(u)) v + u {\hat \Delta}(\hat G(v)) + 
          \hat G(u) {\hat \Delta}(v) + {\hat \Delta}(u) \hat G(v) \\ 
    &   & \: \: \: - {\hat \Delta}(u) {{\bf c}}v - u {{\bf c}}{\hat \Delta}(v) -
           2 u {\otimes}v\\
    & = & {\hat \Delta}(\hat G(u) v) + {\hat \Delta}(u \hat G(v)) -
           {\hat \Delta}(u {{\bf c}}v) \\
    & = & {\hat \Delta}(\hat G(u v)) \\
    & = & LHS_{u v}.
\end{array}$$

\[l:coderivation\_identity\] Let $\hat G$, $\mu$ and ${\hat \Delta}$ be as defined before. Then $2 \hat G = \mu \circ {\hat \Delta}.$

The proof follows the same pattern as that of the previous theorem. The induction step is as follows. $$\begin{array}{r c l}
RHS_{u v} 
    & = & \mu({\hat \Delta}(u) v + u {\hat \Delta}(v)) \\
    & = & \mu({\hat \Delta}(u)) v + u \mu({\hat \Delta}(v)) -
          2 u {{\bf c}}v \\
    & = & 2 (\hat G(u) v + u \hat G(v) 
        - u {{\bf c}}v) \\
    & = & 2 (\hat G(u v)) \\
    & = & LHS_{u v}.
\end{array}$$

One may check 