1
$\begingroup$

In any formal system S that is susceptible to Godel's proof, we can make a formula G which is undecidable. That should mean that we can add either $G$ or $\neg G$ as an axiom to S and still end up with a consistent system, but I'm not sure exactly how $S + \neg G$ can be consistent.

So, $G$ says: "There does not exist a number $n$, which is the Godel number for a proof of $G$." If G were provable, it would be false, and the system would be inconsistent (because $G$ implies $\neg G$). If $\neg G$ were provable, however, the system does not have to be blatantly inconsistent, but only omega-inconsistent. $\neg G$ says "There does exist a number $n$, which is the Godel number for a proof of $G$." But it doesn't specify what that number is. So S can still have theorems: "1 does not prove $G$", "2 does not prove $G$", "3 does not prove $G$" and so on, and it would only be omega-inconsistent (since there is no provable statement that is blatantly the negation of another). But we assume that $S$ is omega-consistent, so we are forced to conclude that $G$ is undecidable. Okay. (as a side-note, I thought that Godel's proof didn't need the assumption of omega-consistency, but it seems required here ... ?)

Now, I understand what happens when you add $G$ to S. In this new $S+G$ system, $G$ says, "There is no number which is the Godel number of a proof for $G$ in S", which is, of course, true (and presumably consistent).

But in $S + \neg G$, $\neg G$ says "There is a number which is the Godel number of a proof for $G$ in S"; yet, since every provable statement of $G$ is also provable in $S + \neg G$, this new system will also say "1 does not prove $G$ in S", "2 does not prove $G$ in S", ... and so on. So isn't $S + \neg G$ omega-inconsistent? And doesn't that go against the idea that we should be able to add either $G$ or $\neg G$ to S and still end up with a consistent system? Does that "consistency promise" not extend to omega-consistency? If not, isn't there a "logically better" way to extend S, even though both extensions are possible?

  • 0
    This is an old post but there was something in the comments that was not cleared up properly. You claimed that consistency is$a$very well-defined concept, but the definition you proceeded to give used "provable", which is not well-defined! If you unpack what you 'mean', you will see why. You would need to say "there exists a proof of ...", but what **exactly** is a proof? Ok, you say, "there exists a (finite) string of symbols such that ...". Now we've hit the infinite regress, since you can't possibly define "(finite) string" non-circularly!2016-05-02

2 Answers 2

5

Your conclusion that $S+\neg G$ is not $\omega$-consistent is right. It is (assuming that $S$ is consistent) consistent, but fails to be $\omega$-consistent. Being $\omega$-consistent is a stronger condition than consistency, satisfied by fewer systems.

Would it be "logically better" to extend $S$ in an $\omega$-consistent way than in one that isn't? I don't think so. As a sometimes Platonist, I would certainly favor the extension $S+G$ over $S+\neg G$, perhaps even claiming that the former is true whereas the latter isn't, but that preference is not based on logical properties (being true is not a logical property).

Part of the confusion is that "$\omega$-consistency" is something of a misnormer -- because the name contains "consistency" one is tempted to think that like ordinary consistency it is an intrinsic property of the theory. But really it isn't; saying that a theory is $\omega$-consistent is a statement between the relation between what the theory proves and arithmetic at the meta-level. Being $\omega$-consistent is a necessary criterion for the theorems of the theory to be truths about the intuitive naturals, but failure to express arithmetic truth is not a logical problem for the theory -- it is innocent of our ambitions about what we might like it to model or not.

One might consider modifications of the concept of $\omega$-consistency such that it looks more intrinsic to the theory. For example we could define that a theory $T$ is $\omega'$-consistent iff, whenever $T\vdash \exists x. \phi(x)$ there is some closed term $t$ such that $T\vdash \phi(t)$. (The difference is that in ordinary $\omega$-consistency we require $t$ to be a "numeral"; the generalized definition allows arbitrary closed terms). However, under this generalization it is still not clear why one would consider $\omega'$-consistency to be a desirable property of theories in general, even if we don't intend the theory to model arithmetic.

  • 0
    @HenningMakholm: I agree with your comment, but I interpreted your statement that "under this **generalization** it is **still** not clear why ... $ω'$-consistency [is] **desirable**" to imply that you felt that it is not clear why $ω$-consistency is desirable. Clearly we do desire it (as long as we believe natural numbers exist).2016-05-02
1

Omega-consistency is certainly a syntactical property; in that sense, it is an intrinsic property of systems, not a property they have under a particular interpretation.

However, if consistency is desirable on simple logical grounds, the desirability of omega-consistency is arithmetical, for only if we take our universe of discourse to be the natural numbers (or some other items with the same structure), we will consider desirable for a system denying a property P of each numeral (0, S0, SS0, ...) to not prove the existence of an object with P.

If we wish to admit objects other than the naturals in our universe, omega-consistency need not be desirable.