1
$\begingroup$

$ \models \phi \to \forall x.\phi$ where $\phi$ is WFF and x not free in $\phi$

Does that render that $ \phi \to \forall x.\phi$ is true in any cases?

  • 0
    @Zev, it's not clear that they are really duplicates -- the OP here seemed to want to know what was being asked of him, whereas in the other question wanted help actually doing it. But in any case the question is now academic (!); the assignment was due yesterday. (Hmm, I see now that the asker was the same. Didn't notice that before).2011-11-18

1 Answers 1

1

Yes, $\vDash\psi$ means that $\psi$ is logically valid, that is, true in every interpretation of the language of $\psi$.

  • 0
    Also, don't use `->` to produce an arrow in (La)TeX; it looks horrible. Use `\rightarrow` or its synonym `\to` instead.2011-11-16