# Are the ordered field axioms consistent?

Today in class a student asked to the professor

"Are the ordered field axioms consistent?"

And my prof replied something along the lines of "Yes, as we have a model of them: $\Bbb R$, this implies that the axioms are consistent."

I couldn't understand it, why does the existence of a model imply the consistency of the system? Could someone explain this?

• By the completeness theorem of FOL, a theory is consistent iff it has a model. If you remain skeptical, then perhaps you suspect that mathematics is inconsistent. – BrianO Jan 28 '16 at 4:34
• What would it mean for the axioms to be contradictory? It would mean that you could prove that the statement "False" was a consequence of them. But if you happen to know that the axioms are true in some particular structure, then you will then be able to prove that "False" is true for that structure, which is absurd. – David Jan 28 '16 at 4:40
• @David I don't understand why that is absurd, could you elaborate a bit? – YoTengoUnLCD Jan 28 '16 at 4:43
• $\exists x ~ | ~ P(x)$ and $\forall y, ~ P(y) \implies \bot$ seems quite absurd. – B. Freitas Jan 28 '16 at 4:45
• The statement "False" can never be true in any structure. – David Jan 28 '16 at 4:46

This is an instance of the soundness theorem (for first-order logic; there is an analogous soundness theorem for propositional logic, which it might be good at to look at first). The theorem states that if $T$ proves $\theta$, then $\theta$ is true in every model of $T$. The proof is by induction on the length of the proof, breaking into cases depending what the last inference rule used is. For instance, suppose the last rule used is "from $\varphi$ and $\psi$ deduce $\varphi\wedge\psi$" (I'm writing somewhat informally). Then if this is the last rule used to derive $\varphi\wedge\psi$, by the induction hypothesis $\varphi$ is true in any model of $T$, and $\psi$ is true in any model of $T$. So $\varphi\wedge\psi$ is true in any model of $T$.

As a consequence of this theorem, any theory with a model is consistent. This is because if a theory is inconsistent, it proves "$\exists x(x\not=x)$" (for example), but this sentence is not true of any structure - in particular, it's not true of the structure assumed to be a model of the theory.

So in particular, if $\mathbb{R}$ satisfies the ordered field axioms, then the ordered field axioms are consistent.