Under what circumstances are one-way functions reducible to their hard-core bits?

Question

I've seen competing claims about hard-core predicates. On the one hand, there's the basic definition

Let $f : \{0,1\}^n \rightarrow \{0,1\}^k$ and $b : \{0,1\}^n \rightarrow \{0,1\}$ be computable in polynomial time. We say that $f$ is a one-way function with hard-core bit $b$ if, for all randomized algorithms $B$ and all constants $c$, $$ \mathbb{P}(B(f(x)) = b(x)) = \frac{1}{2}+o(n^{-c})\,. $$

But I've also read the statement

We want the problem of finding $x$ to be reducible, in polynomial time, to the problem of finding $b(x)$, so that finding $b(x)$ is just as hard as finding all of $x$. In that case $b$ is called a hard-core bit for $f$.

How are these requirements related? Is one stronger than the other? Or do they imply each other?