Mixed Model Equations In this paper on page 1924 it is stated that 
\begin{equation}
\text{Var}(u \mid y) = \sigma^2[G - GZ^\top H^{-1}ZG]
\end{equation}
can be written as 
\begin{equation}
\text{Var}(u \mid y) = \sigma^2[G - (Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1}ZG],
\end{equation}
where $H = ZGZ^\top + R$. Can someone show how this step is achieved? 
Note that $H, G,$ and $R$ are symmetric matrices. 
Edit: How does 
\begin{equation}
\begin{aligned}
(Z^\top R^{-1}Z + G^{-1})GZ^\top &= Z^\top R^{-1}ZGZ^\top + Z^\top \\
&= Z^\top R^{-1}(ZGZ^\top + R) \\
&=Z^\top R^{-1}H. 
\end{aligned}
\end{equation}
Imply 
\begin{equation}
(Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1} = GZ^\top H^{-1}.
\end{equation}
 A: Noting that $(A + BCD)^{-1} = A^{-1} - A^{-1}B(C^{-1} + DA^{-1}B)^{-1}DA^{-1}$ we can write 
\begin{equation}
H^{-1} = (ZGZ^\top + R) = R^{-1} - R^{-1}Z(G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}.
\end{equation}
Using $H^{-1}$ we can write 
\begin{equation}
\begin{aligned}
GZ^\top H^{-1}ZG &= GZ^\top R^{-1}ZG - GZ^\top R^{-1}Z(G^{-1} + Z^\top R^{-1}Z)^{-1} Z^\top R^{-1} ZG \\
& = \Big[ G - GZ^\top R^{-1}Z(G^{-1} + Z^\top R^{-1}Z)^{-1}\Big]Z^\top R^{-1}ZG \\
& = \Big[G(G^{-1} + Z^\top R^{-1}Z) - GZ^\top R^{-1}Z\Big](G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}ZG \\
& = \Big[(I + GZ^\top R^{-1}Z) - GZ^\top R^{-1}Z\Big](G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}ZG \\
& = (G^{-1} + Z^\top R^{-1}Z)^{-1}Z^\top R^{-1}ZG.
\end{aligned}
\end{equation}
Thus $\text{Var}(u \mid y)$ can be written as 
\begin{equation}
\begin{aligned}
\text{Var}(u \mid y) & = \sigma^2(G - GZ^\top H^{-1}ZG) \\
&=  \sigma^2(G -  [G^{-1} + Z^\top R^{-1}Z]^{-1}Z^\top R^{-1}ZG) \\
&=  \sigma^2(G -  [G^{-1} + Z^\top R^{-1}Z]^{-1}[Z^\top R^{-1}Z + G^{-1} - G^{-1}]G) \\
& = \sigma^2(G^{-1} + Z^\top R^{-1}Z)^{-1}.
\end{aligned}
\end{equation}
Edit: Noting that 
\begin{equation}
(Z^\top R^{-1}Z + G^{-1})GZ^\top = Z^\top R^{-1}H. 
\end{equation}
Multiplying on the left by $(Z^\top R^{-1}Z + G^{-1})^{-1}$ and on the right by $H^{-1}$ we obtain
\begin{equation}
\begin{aligned}
(Z^\top R^{-1}Z + G^{-1})^{-1}(Z^\top R^{-1}Z + G^{-1})GZ^\top H^{-1} & = (Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1}HH^{-1} \\
\implies GZ^\top H^{-1} & = (Z^\top R^{-1}Z + G^{-1})^{-1}Z^\top R^{-1}.
\end{aligned}
\end{equation}
