# Is arithmetic on the naturals $\omega$-consistent?

Is there any first order theory of arithmetic of the natural numbers that is known to be $\omega$-consistent if it is consistent?

If yes, then how?

• $\omega$-consistency implies consistency. PA is $\omega$-consistent as is any theory that has the standard model $\Bbb{N}$ as a model. – Rob Arthan Sep 16 '17 at 23:10
• I understand that. I'm asking if we can show that there exists a theory of arithmetic on $\mathbb{N}$ such that its consistency implies its $\omega$-consistency. Are you asserting that $\mathbb{N}$ exists and is consistent by fiat? If so, I'm asking basically is there any reason to believe arithmetic on $\mathbb{N}$ is indeed $\omega$-consistent. – oneq Sep 16 '17 at 23:13
• The definition of $\omega$-consistency is contingent on the existence of $\Bbb{N}$. If you are trying to make some more subtle point in your question, please edit your question to clarify. – Rob Arthan Sep 16 '17 at 23:19
• Ok, that answers my question. I was just wondering if there was any reason to believe any infinite theory is $\omega$-consistent, if one is not convinced of the existence of some true (intuitively $\omega$-consistent) arithmetic on $\mathbb{N}$. It appears then that the answer is no. Thanks for the help. – oneq Sep 16 '17 at 23:21
• So basically one way the world could be is that all infinite theories are $\omega$-inconsistent, even if not all are inconsistent. That's what I was trying to figure out. – oneq Sep 16 '17 at 23:26

The empty theory over the language of arithmetic is certainly $\omega$-consistent.

This theory has the property that whenever it proves some formula $\phi$, you can replace every atomic formula $t_1=t_2$ by "true" and ignore all of the quantifiers, and the resulting Boolean expression will then evaluate to true. (Intuitively this is because the one-element universe is a model, but you can verify it purely syntactically one inference rule at a time).

Therefore it is impossible for the theory to prove both $\neg\phi(0)$ and $(\exists x)\,\phi(x)$, since the two Boolean expressions they translate to are each other's negations.

(This same argument also works for full Peano Arithmetic minus the axiom stating that $0$ is not a successor).

Let me add a fact that doesn't answer your question literally as stated, but may be interesting to you and other readers of this question: if you accept the axioms of ZFC, then pretty much any theory of arithmetic is $$\omega$$-consistent. Namely,

Theorem (ZFC). For any theory of arithmetic $$\mathcal{T}$$ such that $$\mathbb{N} \vDash \mathcal{T}$$, $$\mathcal{T}$$ is $$\omega$$-consistent.$$^1$$

Here "theory" means "set of axioms", and $$\vDash$$ is the satisfaction relation from model theory: $$\mathcal{M} \vDash \mathcal{T}$$ means that the structure $$\mathcal{M}$$ (in this case, natural numbers with addition, multiplication, etc.) satisfies all of the axioms $$\mathcal{T}$$.

Since pretty much any theory of $$\mathbb{N}$$ is supposed to start from true axioms, that is axioms we believe are actually true about $$\mathbb{N}$$, most such theories will be $$\omega$$-consistent. In particular:

Corollary. Assuming the axioms of ZFC, Peano Arithmetic (PA) and Robinson Arithmetic (RA) are $$\omega$$-consistent.

This may be a bit surprising! So what is the basis for these results? The theorem requires a strong meta-theory, in this case ZFC, in which we can develop all of model theory (in particular, in which we can define what a formula is, what a theory is, and what satisfaction $$\vDash$$ means). In such a meta-theory, there is also an object called $$\mathbb{N}$$, the natural numbers (in ZFC the existence of $$\mathbb{N}$$ is given as an axiom). Because such an object exists, we can then show that any correct theory (one that consists of axioms which are true about $$\mathbb{N}$$) is $$\omega$$-consistent.

If we do not get to assume strong meta-theory axioms like ZFC, but instead are wondering about the $$\omega$$-consistency of a theory like PA in a much weaker theory, then we may no longer be able to show that PA is $$\omega$$-consistent.

$$^1$$ Proof of the theorem: We proceed by contradiction: let $$P(x)$$ be a formula with one free variable, such that $$\mathcal{T}$$ proves (implies) $$P(0)$$, $$P(1)$$, and so on, but $$\mathcal{T}$$ also proves (implies) $$\exists x. \lnot P(x)$$.

Now since $$\mathbb{N} \vDash \mathcal{T}$$, $$\mathbb{N} \vDash \varphi$$ for any logical consequence (implication) $$\varphi$$ of the axioms $$\mathcal{T}$$. Therefore, $$\mathbb{N} \vDash P(i)$$ for all natural numbers $$i \in \mathbb{N}$$, and moreover, $$\mathbb{N} \vDash \exists x. \lnot P(x)$$. But these statements are in direct contradiction: by definition, the latter statement means that there exists some natural number $$j$$ such that $$\mathbb{N} \vDash \lnot P(j)$$, which is the exact opposite of the former statement. So we are done.