Is $G$ a secure PRG? Is the following function a secure PRG?


  Given $F$ is a secure PRG and $k$ is choosen random from key space $K$.
  
  $$G(x) = F(k,x) \oplus F(k,x \oplus 1^s)$$


My solution is $x \oplus 1^s = x'$ so $G(x)$ becomes $f \oplus f'$ with $f = F(k,x)$ and $f' = F(k,x')$. Clearly $f$ and $f'$ are PRG, i.e. generate random strings; and $\oplus$ of these will be random.

Correct me if I am wrong.
To give this question its deserved answer, I’ll repeat what Ricky Demer noted in his comment:

$$G(x \oplus 1^s) = F(k,x \oplus 1^s) \oplus F(k,x \oplus 1^s \oplus 1^s)  \\
\downarrow  \\
F(k,x \oplus 1^s) \oplus F(k,x \oplus 1^s \oplus 1^s) = F(k,x \oplus 1^s) \oplus F(k,x \oplus 0^s)  \\
\downarrow  \\
F(k,x \oplus 1^s) \oplus F(k,x \oplus 0^s) = F(k,x \oplus 1^s) \oplus F(k,x)  \\
\downarrow  \\
F(k,x \oplus 1^s) \oplus F(k,x) = F(k,x) \oplus F(k,x \oplus 1^s)  \\
\downarrow  \\
F(k,x) \oplus F(k,x \oplus 1^s) = G(x)$$

As tylo commented: This shows G is not a secure PRG.