2.6.12: Summary
The completeness theorem is the converse of the soundness theorem . In one form it states that if \(\Gamma \Entails A\) then \(\Gamma \Proves A\) , in another that if \(\Gamma\) is consistent then it is satisfiable. We proved the second form (and derived the first from the second). The proof is involved and requires a number of steps. We start with a consistent set \(\Gamma\) . First we add infinitely many new constant symbols \(c_i\) as well as formulas of the form \(\lexists{x}{A(x)} \lif A(c)\) where each formula \(A(x)\) with a free variable in the expanded language is paired with one of the new constants. This results in a saturated consistent set of sentences containing \(\Gamma\) . It is still consistent. Now we take that set and extend it to a complete consistent set . A complete consistent set has the nice property that for any sentence \(A\) , either \(A\) or \(\lnot A\) is in the set (but never both). Since we started from a saturated set, we now have a saturated, complete, consistent set of sentences \(\Gamma^*\) that includes \(\Gamma\) . From this set it is now possible to define a structure \(\Struct{M}\) such that \(\Sat{M(\Gamma^*)}{A}\) iff \(A \in \Gamma^*\) . In particular, \(\Sat{M(\Gamma^*)}{\Gamma}\) , i.e., \(\Gamma\) is satisfiable. If \(=\) is present, the construction is slightly more complex.
Two important corollaries follow from the completeness theorem. The compactness theorem states that \(\Gamma \Entails A\) iff \(\Gamma_0 \Entails A\) for some finite \(\Gamma_0 \subseteq \Gamma\) . An equivalent formulation is that \(\Gamma\) is satisfiable iff every finite \(\Gamma_0 \subseteq \Gamma\) is satisfiable. The compactness theorem is useful to prove the existence of structures with certain properties. For instance, we can use it to show that there are infinite models for every theory which has arbitrarily large finite models. This means in particular that finitude cannot be expressed in first-order logic. The second corollary, the Löwenheim-Skolem Theorem , states that every satisfiable \(\Gamma\) has a countable model. It in turn shows that uncountability cannot be expressed in first-order logic.