Prove that a function is not PRF?

Question

Let $G$ $PRG$ and $G'(s)$ equal to the first $n$-bits of $G(s)$, where $|s| = n$. Show that $F_k(x) = G'(k) ⊕ x$ is not a $PRF$.

My attempt:
Lets assume a $k$ and 2 messages $x_1$ and $x_2$. So, for $x_1$ I have $F_k(x_1) = G'(k) ⊕ x_1$, for $x_2$ we have $F_k(x_2) = G'(k) ⊕ x_2$. If the attacker does $F_k(x_1) ⊕ F_k(x_2)$ he gets $x_1 ⊕ x_2$. And then he can simply $XOR$ that by $x_1$ or $x_2$ and then get the message from that.

I have no idea if this is the correct way to prove this. Can you give my any hint if this is the correct way to show that a function is not $PRF$, maybe I have to do a proof by contradiction ?

EDIT:

I also know that a $PRG$ must not be predictable, if it is then that's not good, so for the function $F$ to use $G'$ as the $PRG$ which is always the first $n$-bits of $G$ it's not good because it violates the pseudorandomness property. I don't know if this is correct or helpful at all.

Answer 1

For something to be a PRF, it must be impossible for a probabilistic polynomial time algorithm to distinguish it from a random function. The distinguisher you note in your attempt is correct, as a random function would not have that property (xor two calls with the same key but different x values and you get the xor of the two x values).

As a side note, these notes are very good when it comes to PRFs. Especially 3.3 and 3.4.