Hybrid Sequence:
The starting point is $\lib{prg-real}^H$.
We can apply a three-hop maneuver to the first call to $G$, replacing ``$A\|B := G(\seed)$'' with ``$A\|B \gets \bits^{2\secpar}$.''
Uniformly sampling $2\secpar$ bits is the same as uniformly (and independently) sampling its two halves.
Now we can apply a three-hop maneuver to the remaining call to $G$.
Concatenating $\secpar$ uniformly sampled bits with $2\secpar$ \emph{independent,} uniformly sampled bits is the same as sampling $3\secpar$ uniform bits. The result of this change is the $\lib{prg-rand}^H$ library, which completes the proof.
$\lib{prg-real}^H$
$\prgsamp_H$( ):
$\seed \gets \bits^\secpar$
// $H(\seed)$:
$A $
$\| B $
${}:= {}$
$G(\seed)$
$\prgsamp_G()$
${}\gets \bits^{2\secpar}$
${}\gets \bits^\secpar$
$C \| D $
${}:= {}$
$G(B)$
$\prgsamp_G()$
${}\gets \bits^{2\secpar}$
return
$A \| C \| D$
$Y$
$\link$
$\lib{prg-real}^G$
$\prgsamp_G$( ):
$\seed \gets \bits^\secpar$
return $G(\seed)$
$\lib{prg-rand}^G$
$\prgsamp_G$( ):
$Y \gets \bits^{2\secpar}$
return $Y$