# Is an extension of a consistent theory that assumes its own finite consistency consistent?

Fix a proof system for first-order arithmetic and let $\text{Con}(T,k)$ be an encoding of the statement that there is no proof of a contradiction in theory $T$ shorter than $k$ bits.

Now define $U_k = \text{PA} + \text{Con}(U_k, k)$ using Gödel's fixed point theorem. So each $U_k$ is the theory of PA with an additional axiom claiming that $U_k$ is consistent up to $k$-bit proofs.

Let's use ZFC as a meta-theory, so we know that PA is consistent. Are all the $U_k$ also consistent?

• Maybe I'm missing something, but if PA is consistent, then for each individual $k\in\mathbb{N}$ PA proves "there is no contradiction in me of length $<k$" (since there are only finitely many proofs of length $<k$ in the first place, and PA can enumerate them and check that each one isn't a proof of $0=1$). So isn't it provable in ZFC that PA and $U_k$ are equivalent? Commented Apr 11, 2017 at 18:28
• Noah, the theory $PA + Con(PA, k)$ is consistent and equivalent to PA, but I'm not so sure about $U_k = PA + Con(U_k, k)$. Commented Apr 11, 2017 at 19:54
• Rob, $\text{Con}(U_k,k)$ is a sentence that encodes "there is no proof of a contradiction shorter than $k$ in the theory whose axioms are the axioms of PA plus this sentence", I think it is possible to construct this, although not in a unique way. I don't see an issue with PA being infinite, all we need is a way to tell if a sentence is an axiom. Commented Apr 14, 2017 at 16:41
• @RobArthan The fixed point theorem is only being applied to a single formula here. Specifically, there's a formula $\varphi(x)$ asserting that $x$ is the Goedel number of a sentence which, when appended to PA, yields a $k$-consistent theory. By the fixed point theorem, we get a sentence $\psi$ with Goedel number $n$ such that PA proves $\psi\iff\varphi(n)$; the theory $U_k$ is just PA+$\psi$. Commented Apr 14, 2017 at 16:51
• @NoahSchweber: &@Dan: thanks for the clarifications. I think it would have been clearer if Dan had written $U_k = Con(PA + U_k, k)$ rather than making it look as if $U_k$ were a sentence that implies each axiom of $PA$. Commented Apr 14, 2017 at 19:08

It depends on the particular fixed-point that you use.

First, on the negative side, I claim that for some fixed points and large enough $k$, the answer is no, they can be inconsistent.

For example, let $\psi$ be any statement that can be refuted by PA in many fewer than $k$ bits, such as the statement $1\neq 1$. In this case, PA proves that $\psi$ is false, and it also proves that $\text{PA}+\psi$ proves a contradiction, and for large enough $k$ this proof will use fewer than $k$ bits. So PA proves that $\psi$ and $\text{Con}(\text{PA}+\psi,k)$ are equivalent, both being false. So it is one of the fixed points asserted to exist by the fixed-point lemma. But it is not consistent.

Meanwhile, on the positive side, let us assume that PA is consistent. Let $\psi$ be the sentence $1=1$, which is provable in PA and consistent with PA. For any fixed $k$, it follows that there will be no proof of a contradiction from $\text{PA}+\psi$ in fewer than $k$ bits (since we assumed PA is consistent), and furthermore PA will prove this for each particular $k$, since it can note the fact specifically about each of the finitely many proofs. So PA proves that $\psi$ is true and also that $\text{Con}(\text{PA}+\psi,k)$ is true, and so it is a fixed point. And for this fixed point, we have consistency.

Thinking about it a little more, I've come to realize that what is going on is the following:

Theorem. If PA proves or refutes $\psi$, then for any sufficiently large $k$, PA proves that $\psi$ is equivalent to $\text{Con}(\text{PA}+\psi,k)$, and so all such statements are fixed points.

Proof. If PA refutes $\psi$, then that proof has some length, and so for $k$ large enough, PA also proves that PA+$\psi$ is inconsistent, and thus PA proves $\psi\leftrightarrow \text{Con}(\text{PA}+\psi,k)$, since both sides are proven false.

If PA proves $\psi$, and without loss is consistent, then for any $k$ there will be no proof of a contradiction from PA+$\psi$ using fewer than $k$ bits, and PA will prove that by verifying each of those proofs. So in this case, PA proves $\psi\leftrightarrow \text{Con}(\text{PA}+\psi,k)$.

So in either case, $\psi$ is a fixed point. QED

Update. Although one might be disappointed by the fixed-points provided by the previous theorem, I've now realized that all fixed points are exactly like that.

Theorem. The following are equivalent for any sentence $\psi$:

1. $\text{PA}\vdash\psi\leftrightarrow\text{Con}(\text{PA}+\psi,k)$ for some $k$.
2. $\text{PA}\vdash\psi\leftrightarrow\text{Con}(\text{PA}+\psi,k)$ for all sufficiently large $k$.
3. $\psi$ is provable or refutable in PA.

Proof. I've already proved that if $\psi$ is provable or refutable, then $2$ holds, and this implies $1$. Conversely, suppose that $1$ holds. Since there are only finitely many possible proofs using $k$ at most $k$ bits, it follows for any particular $k$ that PA either proves or refutes $\text{Con}(\text{PA}+\psi,k)$, simply by inspecting the finitely many proofs. So PA proves or refutes $\psi$, since $\psi$ is equivalent to that statement. So all three statements are equivalent. QED

In particular, there are no fixed points that are themselves independent of PA.