Knowing that $G(s)$ is a PRG, is the following construction $G'(s) = G(s||0)$ a PRG? I know there already is a very similar question. However, I can't understand why $G'(s)$ is not a PRG. If $G(s)$ is a PRG, then why $G(s||0)$ can't also be a PRG? How can the distribution over $G(s||0)$ be distinguished from a truly uniform distribution of the same output range?
Knowing that $G(s)$ is a PRG, is the following construction $G'(s) = G(s||0)$ a PRG?


Maybe.


  If $G(s)$ is a PRG, then why $G(s||0)$ can't also be a PRG?


$G'$ can "also be a PRG". (It's just not necessarily also a PRG.)


  How can the distribution over $G(s||0)$ be distinguished from a truly uniform distribution of the same output range?"


$G$ can be such that the last bit of its output is always equal to the last bit of its input.