If $G'(s)=G(s0^{|s|})$ and $G$ is a PRNG, is $G'$ necessarily a PRNG? I argue it is because there should be no more correlation between the outputs $G(s0^{|s|})$ than between the outputs $G(s)$. Thus we can say the $G'(s)$ is a PRNG that satisfies $|G'(s)| \geq 4|s|$. If we let $s' = s0^{|s|}$ and denote the set of all $s'$ by $S'$, then $s'$ should be chosen uniformly at random from $S'$. Since, as I stated there should be no correlation between $G'(s')$ for any $s'$, then effectively we are only increasing the expansion factor $l$.

Disagree? Please tell me why.
Suppose g is a PRNG. $\:$ Define G by the following pseudocode:

def G(w):
 m = length(w)
 n = ceiling(m/2)
 set x equal to the n leftmost bits of w
 set y equal to the m-n rightmost bits of w
 z = concatenate(g(x),y)
 output z


In this case, G is a PRNG and G' is not a PRNG.