TITLE: Propositional Calculus: Compactness implies Completeness?
QUESTION [5 upvotes]: Is there a quick way to prove the completeness theorem (every consistant theory has a model) from the compactness theorem (a theory has a model iff every finite subtheory of it has a model)? Usually the compactness theorem is a very easy result of the completeness theorem, but it can also be proved in other ways (e.g. using Tychonoff's theorem) and I wonder if this provides a "shortcut" to the completeness theorem.
I'm only asking about propositional calculus, but if the same holds for first order logic I'll be happy to hear it as well.

REPLY [5 votes]: Here's a sketch of a proof you can find in R. Buss's "An introduction to proof theory". The interesting thing about it is that it's direct, in the sense that it doesn't create a model for a consistent theory but rather prove that logical consequence implies provability.
You want to prove that $\Gamma\models\phi$ then $\Gamma\vdash\phi$. Let's assume that $\Gamma\models\phi$. Using compactness this implies that a finite subset of $\Gamma$, $\{\psi_1,\ldots\psi_n\}\models\phi$, or equivalently $\models\psi_1\to(\psi_2\to\ldots(\psi_n\to\phi)\ldots)$. Hence you just need that tautologies are provable.
To show that, first you prove that given $\phi$, and $p_0,\ldots,p_n$ the atoms of $\phi$, if $f:n+1\to 2$ then $$(\lnot)^{f(0)}p_0,\ldots,(\lnot)^{f(n)}p_n\vdash\phi\textrm{   or    } (\lnot)^{f(0)}p_0,\ldots,(\lnot)^{f(n)}p_n\vdash\lnot\phi$$ (where $(\lnot)^1=\lnot$ and $(\lnot)^0$ is nothing). You can do this by induction on the complexity of $\phi$, given some simple provable statements about connectives.
Now observe that if $\Gamma,\sigma\vdash\chi$ and $\Gamma,\lnot\sigma\vdash\chi$ then $\Gamma\vdash\chi$. Given a tautology $\phi$, by soundness you have that for every $f:n+1\to2$ it is the case that $(\lnot)^{f(0)}p_0,\ldots,(\lnot)^{f(n)}p_n\vdash\phi$. Then using the above remove the propositional atoms one by one to reach that $\vdash\phi$.