Two consequences of completeness

Theorem 3..16 (Compactness theorem)  

Let $S \subseteq L$, $t \in L: S \models t$. Then some finite $S' \subseteq S$ has $S' \models t$.

Proof. If $S \models t$ then $S \vdash t$. But proofs are finite so some finite $S' \subseteq S$ has $S' \vdash t$. Then $S' \models t$.

$\qedsymbol$

Corollary 3..17 (Equivalent formulation of compactness)  

If every finite subset of $S$ has a model, then $S$ is consistent.

Proof. There is no finite subset of $S$, such that $S \vdash \bot$. So $S \not \vdash \bot$.

$\qedsymbol$

Theorem 3..18 (Decidability theorem)  

There is an algorithm to determine, for any $S \subseteq L$ and $t \in L$ whether or not $S \vdash t$.

Note that this is not obvious at all.

Proof. Trivial by replacing $\vdash$ with $\models$. To decide if $S \models t$ just write down a truth table.

$\qedsymbol$