Compactness theorem equivalences

Question

i have this equivalence to compactness theorem that i have problems to prove:

For every first-order theory $T$, every tuple $x̄$ of distinct variables and all sets $\Phi(x̄),\Psi(x̄)$ of first-order formulas, if $T\vDash\forall\bar{x}(\bigwedge\Phi(x̄)\leftrightarrow\bigvee\Psi(x̄))$ then there are finite sets $\Phi_{0}(x̄)\subseteq\Phi(x̄)$ and $,\Psi_{0}(x̄)\subseteq\Psi(x̄)$ such that $T\vDash\forall\bar{x}(\bigwedge\Phi_{0}(x̄)\leftrightarrow\bigvee\Psi_{0}(x̄))$

Answer 1

Write $\neg\Psi$ for the set containing the negation of formulas in $\Psi$. Consider only one implication for the moment. From

(0) $T\vDash\forall x(\bigwedge\Phi(x)\rightarrow\bigvee\Psi(x))$

we obtain

$T \cup \Phi(x) \cup \neg\Psi(x)\models\bot$

By compactness, for some finite set

$T \cup \Phi_0(x) \cup \neg\Psi(x)\models\bot$

Therefore

(1) $T \models \forall x [\bigwedge \Phi_0(x) \rightarrow \bigvee\Psi(x)]$

The converse of (0) is equivalent to $T\vDash\forall x(\bigwedge\neg\Psi(x)\rightarrow\bigvee\neg\Phi(x))$. Therefore the same argument yield $T \models \forall x [\bigwedge\neg\Psi_0(x) \rightarrow \bigvee\neg\Phi(x)]$ which is equivalent to

(2) $T \models \forall x [\bigwedge\Phi(x) \rightarrow \bigvee\Psi_0(x)]$

If you put (0), the converse of (0), (1) and (2) together you obtain

$T \models \forall x [\bigwedge\Phi(x)\rightarrow \bigwedge\Phi_0(x) \rightarrow \bigvee\Psi(x) \rightarrow\bigwedge\Phi(x) \rightarrow \bigvee\Psi_0(x)\rightarrow \bigvee\Psi(x)\rightarrow \bigwedge\Phi(x) ]$

EDIT. Note that we obtain a stronger claim than required. Namely, $\bigwedge\Phi(x)\leftrightarrow \bigwedge\Phi_0(x)$ and $\bigvee\Psi(x)\leftrightarrow \bigvee\Psi_0(x)$.