Pseudorandom generator: Let $($PRG$)$ G : $\{0,1\}"\{$O$,1\}2$m be a pseudorandom generator and let's denote

G $($x$)=$ y$1||$ yz$,$ where y$1,$ Yz are both n$-$bit strings, and we use G$1($x$)$ to represent y$1,$ and G$2($x$)$ to represent y$2-$

Question: Is G$1($x$)$ is still a PRG$?($Hint: Explain in detail why if G$1()$ is not a PRG$,$ then G$(-)$ will not be a PRG