How can I prove the last part of the following exercise.
Show that $R\vDash \phi \Rightarrow R\vDash\psi$ for all structure $R$ implies $\vDash \phi \Rightarrow\, \vDash \psi$, but not vice versa.
I have the first part. Ans: If $\vDash \phi$ then for all structures $R$, $\phi$ is true, by hypothesis, it implies that $R\vDash \psi$ for all hypothesis, but that's by definition is $\vDash \psi$. Then, we conclude $\vDash \phi \Rightarrow \vDash \psi$.
But my concern is with the last one part of the question. "but no vice versa". That is,
$\vDash \phi \Rightarrow\, \vDash \psi$ doesn't implies $R\vDash \phi \Rightarrow R\vDash\psi$ for all structure $R$.
Why? I can't see even why it is actually true, I am using the fact that $\vDash \phi$ says for all structure and for all interpretation, we have $\phi$, right? Then, why not, $R\vDash \phi \Rightarrow R\vDash\psi$ for all structure $R$.
The exercise was taken from the semantics section of book's VanDalen. (3.4.7).