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