What is the proof that First-order logic is complete?

Question

According to Wikipedia, first order logic is complete. What is the proof of this?

(Also, in the same paragraph, it says that its undecidable. Couldn't you just enumerate all possible proofs and disproofs to decide it though?)

Answer 1

I won't cover why first order logic is complete - an adequate textbook on mathematical logic will provide the proof. I recommend Mathematical Logic by Kunen, but it may be a bit more extensive than your needs.

As for decidability: Every necessarily true statement is provable, and every necessarily false statement has a proof of its negation (that's what it means to be "complete") but there are statements which are neither. For example, $(\forall x)P(x)$ may be either true or false depending on the universe of discourse and the definition of $P$. On such a statement, your proposed algorithm would never halt, because it would never find a proof or a disproof; but at no point could we be sure that no proof or disproof exists.

Answer 2

I think there's a bit of confusion about the meaning of the theorem. The phrase "First-order logic is complete" means exactly "If a sentence $\varphi$ is true in every model of $\Gamma$, then $\Gamma\vdash\varphi$" (so it's saying something about how the semantics and a specific deduction system interact; note that this means that the phrase isn't totally appropriate, and should really be along the lines of e.g. "Natural deduction is complete," or better yet "Natural deduction is complete for the usual semantics of first-order logic").

As to the proof, it's given in many easily-findable sources so I won't give it in detail, but the key observation is the following. The completeness theorem is equivalent to the statement "If $\Gamma$ is consistent (that is, $\Gamma\not\vdash\perp$) then $\Gamma$ is satisfiable (that is, there is some $\mathcal{M}\models\Gamma$)." So the proof proceeds by constructing a model: we give a way to associate a structure $\mathcal{M}_\Gamma$ to any theory $\Gamma$, and then want to argue that $\mathcal{M}_\Gamma\models\Gamma$ if $\Gamma$ is consistent.

Since our only hypothesis on $\Gamma$ is syntactic (namely, its consistency), it's reasonable that $\mathcal{M}_\Gamma$ should itself be related to syntactic ideas; the most obvious first guess is that $\mathcal{M}_\Gamma$ should consist of terms in the language modulo $\Gamma$-provable equality (with the obvious structure imposed). The problem is that depending on $\Gamma$ this might not actually work, either because the language doesn't have enough terms or because $\Gamma$ itself is too weak. Instead, we have to first construct a possibly-larger theory in a possibly-larger language, $\hat{\Gamma}\supseteq\Gamma$, such that if $\Gamma$ is consistent then $(i)$ $\hat{\Gamma}$ is consistent if $\Gamma$ is and $(ii)$ $\mathcal{M}_{\hat{\Gamma}}\models \hat{\Gamma}$ (note that $(ii)$ implies $(i)$ so this is a bit redundant).

It's worth noting that this is not Godel's original proof, but rather Henkin's. Godel's (quite neat) original proof was more proof-theoretic; it's not as easily findable, but this article by Avigad has an excellent outline of it (section 4).

Especially in light of the undecidability of first-order logic and the syntactic incompleteness of many first-order theories like PA (note that this notion of incompleteness is different from that of the completeness theorem; saying that a theory, rather than a logic, is incomplete is to say that there is some sentence the theory neither proves nor disproves), the completeness theorem may appear quite surprising; see this old question for some discussion of this.

Answer 3

I would recommend Lou Van den dries' lecture notes in first order logic. He does it in three chapters. Here's a link https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.math.uiuc.edu/~henson/Math570/Fall2009/Math570notes.pdf&ved=0ahUKEwiR--aI5KnPAhVS0GMKHdYADEAQFggbMAA&usg=AFQjCNFRgSAJkPtYyywzaa5YWygqpRZAhQ