TITLE: Is it possible to prove a mathematical statement by proving that a proof exists?
QUESTION [57 upvotes]: I'm sure there are easy ways of proving things using, well... any other method besides this!
But still, I'm curious to know whether it would be acceptable/if it has been done before?

REPLY [13 votes]: I'd say the model-theoretic proof of the Ax-Grothendieck theorem falls into this category.  There may be other ways of proving it, but this is the only proof I saw in grad school, and it's pretty natural if you know model theory.
The theorem states that for any polynomial map $f:\mathbb{C}^n \to\mathbb{C}^n$, if $f$ is injective (one-to-one), then it is surjective (onto).  The theorem uses several results in model theory, and the argument goes roughly as follows. 
Let $ACL_p$ denote the theory of algebraically closed fields of characteristic $p$.  $ACL_0$ is axiomatized by the axioms of an algebraically closed field and the axiom scheme $\psi_2, \psi_3, \psi_4,\ldots$, where $\psi_k$ is the statement "for all $x \neq 0$, $k x \neq 0$".  Note that all $\psi_k$ are also proved by $ACL_p$, if $p$ does not divide $k$.

The theorem is true in $ACL_p$, $p>0$.  This can be easily shown by contradiction: assume a counter example, then take the finite field generated by the elements in the counter-example, call that finite field $F_0$.  Since $F_0^n\subseteq F^n$ is finite, and the map is injective, it must be surjective as well.
The theory of algebraically closed fields in characteristic $p$ is complete (i.e. the standard axioms prove or disprove all statements expressible in the first order language of rings).
For each degree $d$ and dimension $n$, restrict Ax-Grothendieck to a statement $\phi_{d,n}$, which is expressible as a statement in the first order language of rings. Then $\phi_{d,n}$ is provable in $ACL_p$ for all characteristics $p > 0$.
Assume the $\phi_{d,n}$ is false for $p=0$.  Then by completeness, there is a proof $P$ of $\neg \phi_{d,n}$ in $ALC_0$.  By the finiteness of proofs, there exists a finite subset of axioms for $ACL_0$ which are used in this proof.  If none of the $\psi_k$ are used $P$, then $\neg \phi_{d,n}$ is true of all algebraically closed fields, which cannot be the case by (2).  Let $k_0,\ldots, k_m$ be the collection of indices of $\psi_k$ used in $P$.  Pick a prime $p_0$ which does not divide any of $k_0,\ldots,k_m$.  Then all of the axioms used in $P$ are also true of $ACL_{p_0}$.  Thus $ACL_{p_0}$ also proves $\neg \phi_{d,n}$, also contradicting (2).  Contradiction.  Therefore there is a proof of $\phi_{d,n}$ in $ACL_0$.

So the proof is actually along the lines of "for each degree $d$ and dimension $n$ there is a proof of the Ax-Grothendieck theorem restricted to that degree and dimension."  What any of those proofs are, I have no clue.