If $G$ is a PRG, is $F(s)=G(S)||G(\bar{s})$ necessarily a PRG? Let $G$ be a pseudorandom generator, and $F(s)=G(s)||G(\bar{s})$ (where $\bar{s}$ denotes the bitwise complement of $s$), is $F$ necessarily a PRG?

My intuition says that it is a PRG, as clearly $G(s)$ and $G(\bar{s})$ are PRGs on their own, and concatenating them shouldn't affect the randomness, but rather it will simply increase the expansion factor (length of output). However, I feel like there exists a counterexample where $G(s)$ and $G(\bar{s})$ are secure on their own but concatenating them will lose some randomness, i.e. make them predictable (because $F(s)$ and $F(\bar{s})$ should look pretty similar). I've thought about some examples where $G(s)=H(s) \oplus H(\bar{s})$, where $H(s)$ is a PRG (and $G(s)$ would also still be a PRG here), but I'm not sure if this leads to $F(s)$ not being a PRG. Any thoughts would be appreciated!
$F(s)=G(s)\mathbin\|G(\bar{s})$ is not necessarily a PRG. We'll construct a counterexample.

Assume $H(t)$ is a PRG.

Now define $G(t\mathbin\|0)=H(t)$ and $G(t\mathbin\|1)=H(\bar t)$. The $G(s)$ thus defined is a PRG with input width $|t|+1$. Yet $F(s)=G(s)\mathbin\|G(\bar s)$ is not a PRG, because its two halves are always equal.