# Proof of the Completeness Theorem in Predicate Calculus

I recently learned the Completeness Theorem for First Order Logic in class.

Completeness of FOL (Godel, 1930) If $\Gamma\models\varphi$, then $\Gamma\vdash\varphi$. Equivalently, any consistent set of formulas is satisfiable.

It seems that most proofs of the above theorem rely on extending $\Gamma$ to a maximal consistent set $\Delta$. Moreover, in this procedure one often relies on a Skolemisation trick that adds instances/existential wffs to $\Gamma$, such as $\exists x_1\phi_1\rightarrow(\phi_1)^{x_1}_{c_1}$.

I don't quite get the significance of this Skolemisation step. It seems to me that no part of this Skolemisation trick is used in the later part of the proof (in my case, doing the standard procedure of considering the quotient of a structure $\mathfrak{M}$). Indeed, why can't we simply skip this step and define (i) $\Phi=$ set of all formulas, which is countable if our underlying language $\mathcal{L}$ is countable; (ii) $\mathscr{A}=\{\Sigma\subseteq\Phi:\Gamma\subset\Sigma~~\wedge~\Sigma\text{ is consistent}\}$, and then use Zorn's Lemma to construct a maximal set $\Delta\in\mathscr{A}$ directly?

Could someone point out where I must have been mistaken? Thanks in advance!

In order to build the term model is needed the Skolemisation to provide for every statement $\exists \phi(x)$ that holds in the theory an element $c$ of the term model which satisfy $\phi(c)$ (making the statement $\exists \phi(x)$ true in the model).
About the second part, your proof proves that if $\Sigma$ is a consisten theory then there is a consistent maximal theory (in the same languange) which contains it. But you wanted to prove that your theory is satisfiable, so you should prove that a maximal consistent theory is satisfiable.