Let $H$ be a bialgebra and assume that $M$ is left $H$-module and left $H$-comodule.
$M$ is called Yetter-Drinfeld module iff $\forall h\in H$ and $\forall m \in M$, $$\left(h_{(1)}m\right)_{(-1)}h_{(2)}\otimes \left(h_{(1)}m\right)_{(0)} = h_{(1)}m_{(-1)}\otimes h_{(2)}m_{(0)}\,.$$
I'm trying to prove that if $H$ is Hopf algebra (with antipode $S$) then this definition is equivalent to the following statement:
$\forall h\in H$, and $\forall m \in M$ $$\delta (hm)=h_{(1)}m_{(-1)}S\left(h_{(3)}\right) \otimes h_{(2)}m_{(0)}\,.$$
where $\delta$ is a coaction given by structure of comodule.
First of all $\delta(hm)= h_{(1)}m_{(-1)}\otimes h_{(2)}m_{(0)}$, so we should prove that $$\left(h_{(1)}m\right)_{(-1)}h_{(2)}\otimes \left(h_{(1)}m\right)_{(0)}=h_{(1)}m_{(-1)}S\left(h_{(3)}\right) \otimes h_{(2)}m_{(0)}\,,$$ but we know that $\left(h_{(1)}m\right)_{(-1)}=h_{(1)}m_{(-1)}$ and that $\left(h_{(1)}m\right)_{(0)}=h_{(2)}m_{(0)}$. From the definition of a Hopf algebra we know that $S\in\operatorname{End}(H)$, so we can always write $S(x)$ instead of some $y$. Using the above properties we obtain what we want.
Is my argumentation correct ?
You can show that the two conditions are equivalent using the antipode condition $$S(h_{(1)}) h_{(2)}=h_{(1)} S(h_{(2)})=\varepsilon(h)1. $$