Proof that in El Gamal DDH implies CPA
A key point is that for group $G$ and any arbitrary $m \in G$, choosing a uniform $k \in G$ and setting $\hat{k} = k \cdot m$ gives the same distribution for $\hat{k}$ as just choosing it uniformly from $G$. 
This is because:
$Pr[\hat{k}=k \cdot m] =Pr[k=\hat{k} \cdot m^{-1}] = {1 \over {|G|}}$ 
So in the reduction, when $A$ interacts with the challenge $&lt;u=g^y,v=g^z \cdot m&gt;$ for random $z$, $u$ is completely independent of $v$, and so they reveal no information over each other.
Notice that even though it's not a valid encryption scheme (because it's impossible to decrypt), the experiment is still well defined.