Let $\Vdash \phi$ denote that $\phi$ is a tautology, i.e. that for any valuation $f$, $[\phi]_f = 1$. (Informally, $\phi$ is true no matter what truth value we give to the atoms composing it.)
Let $A \Vdash \phi$ denote that $\phi$ is a consequence of $A$, where $A$ is a set of propositional formulas. In other words, let it denote that whenever every $a \in A$ is true (or has a valuation $f$ such that $[a]_f = 1$), it follows that $\phi$ is true as well (or $[\phi]_f =1$).
I was requested to prove
$$ (\Vdash \psi \to \phi) \iff(\{ \psi \} \Vdash \phi) $$
I am confused with respect to the interpretation of $\Vdash$ as denoting a tautology. Precisely,
$$ \Vdash \psi \to \phi $$
may mean $a.$ ($\psi \to \phi$) is a tautology; $b$. $\psi$ is a tautology. It seems to me that the second case is correct. But assuming this, a new question arise. Namely, consider the proposition $\Vdash \psi \to \phi$, which reads "if $\psi$ is a tautology then $\phi$ follows". When considering the possible truth values of this proposition, should I
$1.$ Assume $\psi$ is always true (because it is a tautology) and only see how the proposition varies with respect to the truth values of $\phi$; or
$2.$ Consider $\Vdash \psi$ as a formula on itself, which could be true or false?
In short, my doubt boils down to not knowing whether $\Vdash$ is a notation declaring that something is the case, or an operator (such as $\neg$) which may (along with a proposition) be the case or not.
Thanks.
"$\Vdash \psi \to \phi$" should be parsed as $\Vdash (\psi \to \phi)$. It wouldn't make sense to parse it as $(\Vdash \psi) \to \phi$, because $\phi$ is an object (a propositional formula) in our metatheory, not an actual sentence.