$(\forall x.B(x) \leftrightarrow \exists y. R(xy)) \leftrightarrow (\forall x \exists y.B(x)\rightarrow R(xy)) \land \forall xy. (R(x,y) \rightarrow B(x))$ is a valid formula.
My attempt to prove the validity of RHS→LHS
RHS: $B(x)$ is true iff there exists a $R$-successor to $x$.
LHS: if B(x) is true and there exists an $R$-successor to $x$, then the both conjuncts seem to make sense and are right. If $B(x)$ and there exists no R-successor to $x$, then both conjuncts are trivially satisfied.
The other direction is not so clear. Is there a simple-intuitive way to prove it?