# Can one have a theory that includes its own consistency as an axiom?

Consider the theory with the following axioms:

• The axioms of ZFC
• The "axiom of consistency": "This theory, including this axiom and all of the theory's other axioms, is consistent." Phrased differently (and equivalently):

• This theory does not prove $\text{False}$
• This theory states its own consistency as an axiom.

Can one have such a theory, one that includes its own consistency as an axiom? Can this theory be consistent?

• I seem to recall reading that Hilbert tried to get away with this sort of thing, but it's clear it can't work in general. If there's a contradiction in ZFC, there will still be a contradiction when you add "ZFC is consistent" to it. – Cheerful Parsnip Jul 4 '16 at 4:33
• A lot of things studied in Axiomatic Set Theory is the metamathematics. If you ever read Kunen and learn about basic model theory, you find that a lot of the deeper stuff is about showing how Con(ZFC) |-- Con(ZFC + CH) and things of the sort. What I just wrote there means that assuming ZFC is consistent, then so is ZFC with the continuum hypothesis (CH) as an additional axiom. You can also (through a much harder process called forcing) show how The consistence of ZFC implies the consistency of ZFC plus the negation of the continuum hypothesis as an additional axiom. – gorzardfu Jul 4 '16 at 5:10
• The crux of this Question is making sense of how "this axiom" refers to itself. There will be many Godel-like encodings of meta-properties of a set theory, such as Con(ZFC), so we cannot rely on an informal mention of such a property in formalizing axioms. – hardmath Jul 4 '16 at 18:28
• I'm not entirely sure what's unclear in this question. Yes, this "axiom of consistency" is unclear, but it is the crux of the issue here. This is the confusion... – Asaf Karagila Jul 17 '16 at 10:10
• Is there anything wrong with my answer? – Asaf Karagila Aug 4 '16 at 4:57

What is this "axiom of consistency"?

Is it the statement $\operatorname{Con}(T)$? Because $T+\operatorname{Con}(T)$ is not the same as $T$. Or do you mean that $T=T+\operatorname{Con}(T)$? Which is just to say that $T$ proves its own consistency.

And if $T$ proves its own consistency, then it must violate one of the conditions of Gödel's theorem:

1. So either $T$ is not recursively enumerable,
2. or it is not consistent,
3. or it is not strong enough to interpret arithmetic.

If by adding one axiom to $\sf ZFC$ you managed to violate any of these, it has to be the second condition. So either $\sf ZFC$ was inconsistent in the first place, or that $\sf ZFC$ proves the negation of your new axiom, and you've added a new contradiction to the system.

• Adding the axiom $Con(T)$ to $T$ is obviously not the same as adding to $T$ the axiom that we can prove $Con(T)$ inside $T$, so what did you mean ? And it seems for saying $T+Cont(T)$ is not consistent or not able to formalize the arithmetic, we need to prove $Cont(T+Con(T))$ inside $T+Con(T)$ – reuns Jul 18 '16 at 1:24
• Con (T) is not the same as con ([T]) ,where [T] is an encoding of T . In some systems, such as T=ZF or T=ZFC, there is no sentence Con (T) as any attempt to state it would require an infinitely long sentence. – DanielWainfleet Jul 18 '16 at 2:32
• @user254665: No, Con(T) is a statement about the natural numbers, after coding T via Goedel coding and internalizing the first-order logic. Since ZF and ZFC are recursively enumerable, Con(ZF) is a simple first-order statement in the language of arithmetic, which can be interpreted in ZF using the finite ordinals and ordinal arithmetic. So it is most certainly not an infinitely long sentence. – Asaf Karagila Jul 18 '16 at 4:19
• @user1952009: The OP said that the axiom of consistency for T is "T with this axiom is consistent". So this is not the situation that we add to T the axiom Con(T). – Asaf Karagila Jul 18 '16 at 4:20

Let me address a point which is implicit in the question but hasn't been squarely addressed by the other answers. Your description of the "axiom of consistency" is not precise: it is not at all obvious how to encode this axiom in the language of set theory. In particular, there is no direct way to encode the self-reference involved in saying "this axiom".

Here is one way you might try to make it precise. Let us say that a sentence $\varphi$ in the language of set theory can encode your "axiom of consistency" if $$ZFC\vdash \varphi\Leftrightarrow\text{"ZFC+\varphi is consistent"}.$$ Now unfortunately, this criterion is rather weak. For instance, if $\varphi$ is any sentence that ZFC proves to be false, then it satisfies this criterion, with both sides of the implication simply being false.

Nevertheless, there still is a reasonably "natural" candidate for such a $\varphi$: you can construct a sentence $\varphi$ that can naturally be interpreted as saying that ZFC$+\varphi$ is consistent, by essentially the same trick as is used in Gödel's incompleteness theorem. More generally, such self-referential statements can be constructed by the diagonal lemma.

So there is a reasonable candidate for the theory you are proposing. Unfortunately, this theory is inconsistent by Gödel, as discussed in Asaf's answer. In fact, Asaf's answer shows that the sentences $\varphi$ satisfying the criterion above are exactly the sentences which ZFC disproves!

Yes, one can have a theory that includes its own consistency as an axiom.

That theory is always (by Gödel's Second Incompleteness Theorem) inconsistent.

• That's not true (see Asaf's answer). – Stefan Mesken Jul 4 '16 at 10:59
• @Stefan The theory described in the question is meant to be recursive. – Andrés E. Caicedo Jul 4 '16 at 13:19