In the OLS model, we assume that $E(X'U)=0$ (with $u$ being the error term), which comes from $E(U|X=x)=0$, providing us that $E(U)=0$ and $cov(x_i, u)=0$ $\forall x_i$. I understand this argument intuitively, as $E(U|X=x)=0$ would be violated if $x$ and $u$ were correlated, e.g. $u$ increases with $x$ would cause $E(U|X=x)>0$ for large $x$. But I also know that this is not sufficient for a formal proof, and one is not presented in Woolridge (2010).
I am also wondering if this proof goes the other way as well, i.e. does $E(x'u)=0 \Rightarrow E(U|X=x)=0$ given that $E(u)=0$.
I'm guessing these are both fairly straightforward (thus why they were omitted in the text) but an explanation or a hint in the right direction would be appreciated.