Is it possible to prove a mathematical statement by proving that a proof exists?

Question

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?

Answer 1

There is a disappointing way of answering your question affirmatively: If $\phi$ is a statement such that First order Peano Arithmetic PA proves "$\phi$ is provable", then in fact PA also proves $\phi$. You can replace here PA with ZF (Zermelo Fraenkel set theory) or your usual or favorite first order formalization of mathematics. In a sense, this is exactly what you were asking: If we can prove that there is a proof, then there is a proof. On the other hand, this is actually unsatisfactory because there are no known natural examples of statements $\phi$ for which it is actually easier to prove that there is a proof rather than actually finding it.

(The above has a neat formal counterpart, Löb's theorem, that states that if PA can prove "If $\phi$ is provable, then $\phi$", then in fact PA can prove $\phi$.)

There are other ways of answering affirmatively your question. For example, it is a theorem of ZF that if $\phi$ is a $\Pi^0_1$ statement and PA does not prove its negation, then $\phi$ is true. To be $\Pi^0_1$ means that $\phi$ is of the form "For all natural numbers $n$, $R(n)$", where $R$ is a recursive statement (that is, there is an algorithm that, for each input $n$, returns in a finite amount of time whether $R(n)$ is true or false). Many natural and interesting statements are $\Pi^0_1$: The Riemann hypothesis, the Goldbach conjecture, etc. It would be fantastic to verify some such $\phi$ this way. On the other hand, there is no scenario for achieving anything like this.

Answer 2

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 $k$.

1. 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.
2. 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).
3. 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 $p > 0$.
4. 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.