For reference I am studying via Mathematical Logic by Ebbinghaus, and am operating under the following definition of correctness: a sequent $\Gamma\phi$ is called correct if $\Gamma\models\phi$ (where $\Gamma$ is some other sequent)
Okay, now the question: Decide whether the following rule is correct.
$\frac{\phi \ \ \ \ \ \ \ \psi}{\exists x\phi \ \ \ \ \exists x\psi}$
I have come to the conclusion that it is correct via the following logic. Please tell me if my logic is incorrect, and also, if I profusely over complicated this question.
I broke the question down into four cases
(i) $x \in $ free$(\phi)$ $x \in $ free$(\psi)$
(ii) $x \notin $ free$(\phi)$ $x \notin $ free$(\psi)$
(iii) $x \notin $ free$(\phi)$ $x \in $ free$(\psi)$
(iv) $x \in $ free$(\phi)$ $x \notin $ free$(\psi)$
As for (i) it trivially holds as the witness for the result of the calculus can be the same in both the succeedant and antecedant
(ii) holds trivially as in this case the corresponding antecedants and succedants are logically equivalent.
(iii) In this case however, any thing can serve as the witness for the succeedant of the result of the calculus, as anything that models $\phi$ models $\psi$, and $\phi$ does not contain $x$ free, meaning the assignment of $x$ can vary, meaning $x$ can be anything
And in the case of (iv), $\psi$ is not dependant on the assignment of $x$, just of it's existence, as whenever $\phi$ is true, $\psi$ is true despite not having $x$ free, meaning that whenever $\exists x \phi$ is true $\psi$ is true by the fact that the witness of $\exists x\phi$ exists.
All four cases being true means the rule must be correct. (my explanation of (iv) is what is throwing me off) Thanks for any help.