1
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
5
  • $\begingroup$ 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? $\endgroup$ – jonaprieto Jul 2 '16 at 5:02
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jul 2 '16 at 5:04
  • $\begingroup$ Yes. Thanks Eric $\endgroup$ – jonaprieto Jul 2 '16 at 5:07
  • $\begingroup$ Sorry, I've updated my question, adding the proof for the first part. Can you check it if is right? $\endgroup$ – jonaprieto Jul 2 '16 at 5:21
  • 1
    $\begingroup$ Looks good to me. $\endgroup$ – Eric Wofsey Jul 2 '16 at 5:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.