# $\vDash \phi \Rightarrow\, \vDash \psi$

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).

The statement $\vDash \phi \Rightarrow\, \vDash \psi$ tells you that if $\phi$ is true in all structures, then $\psi$ is true for all structures. But if $\phi$ is not true in all structures, it tells you nothing at all. That is, if $\phi$ is any statement that is not true in all structures, then then the implication $\vDash \phi \Rightarrow\, \vDash \psi$ automatically holds no matter what $\psi$ is.
So for instance, if $\phi$ is a statement that is true in some structures but not all structures, and $\psi$ is a statement that is false in all structures, then $\vDash \phi \Rightarrow\, \vDash \psi$ is true but there exist structures $R$ such that $R\vDash \phi \Rightarrow\, R\vDash \psi$ is false (namely, any structure in which $\phi$ is true).
• It's a little confusing. I think my problem is that I'm assuming that condition as I already have $\vDash \phi$ but doesn't not good. Right? – jonaprieto Jul 2 '16 at 5:02
• Right. If $\vDash \phi$ is true, then $\vDash \phi \Rightarrow\, \vDash \psi$ implies $R\vDash \phi \Rightarrow\, R\vDash \psi$ for any $R$. But if $\vDash\phi$ is not true, you can't say anything of the sort. – Eric Wofsey Jul 2 '16 at 5:04