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I've been reading about Gödel's Incompleteness Theorems and there's something that I don't quite understand. It's about adding new statements as axioms to a system.

I'm not sure if I'm understanding anything wrong so here is a brief summary of what I think I know:

  1. Gödel's First Incompleteness Theorem

Any consistent system $S$ strong enough to express arithmetic is incomplete. Gödel shows this by explicitly constructing an undecidable statement G that's reads something like:

G $\iff$ There is no Gödel number $x$ that corresponds to a proof of G in $S$

Assuming $S$ is consistent, if G is provable, then there exists $x$ that proves G, which is a contradiction. If not G is provable, then there is a proof of G, which leads to a contradiction. Therefore, G cannot be proven or disproven, so $S$ is incomplete.

  1. Gödel's Second Incompleteness Theorem

If $S$ is proven to be consistent, then $S$ is inconsistent. We get this by noting the implication 'if $S$ is consistent, then G is true'. So if there is a proof of $S$'s consistency, then we can prove $G$, which leads to a contradiction.

I understand that if a statement $A$ is independent of a set of consistent axioms, then there are models where $A$ or not $A$ are true, and I can add $A$ or not $A$ to the axioms while maintaining consistency.

The problem is that '$S$ is consistent' is undecidable, so does that mean I can add '$S$ is inconsistent' as an axiom? What are the implications of adding such an axiom? I understand that adding this as an axiom won't lead to contradictions, because I can't prove '$S$ is consistent' anyway, but I just find something very strange about adding this as an axiom.

ghost
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    It is "strange" but consistent... Compare with the well-known case of Euclidean geometry and Parallel postulate: the postulate is independent from the other axioms and thus we can add it to them to obtain the "obvioulys" consistent (if the other axioms are consistent) "classical" Euclidean geometry. OR we can add to the other axioms the negation of the Parallel postulate and what we get is the "not obviously" consistent (same as before) not-Euclidean geometry. – Mauro ALLEGRANZA Oct 27 '21 at 13:15
  • In the same way, as we can agree (or not?) that the "real" space is Euclidean and that two parallel lines will never intersect in it, we can agree that the "real" number system is consistent, and thus the formal statement $\text {Cons}(\mathsf {PA})$ is true in it (as well as the unprovable (in $\mathsf {PA}$) $G$ sentence), but the fact is that the formal system $\mathsf {PA}$ cannot prove that statement and thus we can "imagine" a different number system where all axioms of $\mathsf {PA}$ hold but also the previous sentence are (because we have "added" them as axioms). – Mauro ALLEGRANZA Oct 27 '21 at 13:24
  • For a good intro, see Gödel’s Theorem: An Incomplete Guide to Its Use and Abuse by Torkel Franzén – Mauro ALLEGRANZA Oct 27 '21 at 13:40
  • Yes, $\mathsf{PA}+\neg\mathsf{Con(PA)}$ is consistent (but not $\omega$-consistent), and is a very important theory in the study of formal systems. – Trebor Oct 27 '21 at 13:41
  • @Peter More precisely, the statement "S can prove everything" can be proved. But not everything can be proved. Can you see why? – Trebor Oct 27 '21 at 13:43
  • @Trebor To be honest, I do not get why this is consistent although it is , per axiom , not. – Peter Oct 27 '21 at 13:44
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    @Peter Well, if adding that axiom causes a contradiction, then by reduction ad absurdum, you can prove that the axiom is false. So you proved a theory's own consistency! – Trebor Oct 27 '21 at 13:46
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    @Peter Assuming S is consistent then adding the axiom that it is inconsistent results in a consistent theory. This is standard result in proof theory. – Trebor Oct 27 '21 at 13:55
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    @Peter “S + S is inconsistent” is inconsistent if and only if S proves that S is consistent. And per Godel, this is often not the case. – spaceisdarkgreen Oct 27 '21 at 14:30
  • @spaceisdarkgreen. I’m a bit confused. What if in addition to the “s is inconsistent axiom”, one also adds “s is inconsistent implies p, for all p”. Does this not cause the theory to imply all p and thus be inconsistent? In fact “s is inconsistent implies p, for all p” is true whether it is an axiom or not, so it still appears that all p is proven making the theory actually inconsistent. – David Okogbenin Oct 27 '21 at 14:46
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    @DavidOkogbenin It's very important to separate the actual statements from S's codes for them. The theory does prove "every statement of S is provable in S" and for each natural number $n$, it proves "the statement coded by $n$ is provable"... but still, it doesn't prove every p! The disconnect is that for a given natural number $n$ that encodes some statement p, S can't necessarily prove that "if the statement encoded by $n$ is provable, then p holds." In fact it can only do so if p is already provable in S. – spaceisdarkgreen Oct 27 '21 at 15:17
  • @DavidOkogbenin So if you added to "S + S is inconsistent" the scheme "S is inconsistent -> p" for each statement $p,$ the resulting theory would be inconsistent, but "S + S is inconsistent" does not necessarily prove every instance of that scheme. – spaceisdarkgreen Oct 27 '21 at 15:22
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    @Peter This is a common misconception. Various discussions on this site are relevant, e.g. here or here; see also Lob's theorem. Roughly, appropriate theories don't prove that the things they prove are true, and so can't get from "I prove $\varphi$" to "$\varphi$." – Noah Schweber Oct 27 '21 at 15:30
  • @NoahSchweber Thank you for this insight. – Peter Oct 27 '21 at 15:32
  • @DavidOkogbenin Building off of spaceisdarkgreen's most recent comment, see Lob's theorem - but also note (re: their emphasis on the word "scheme") that the expression "For all $p$, if $S$ proves $p$ then $p$ is true" isn't even expressible in the first place per Tarski. – Noah Schweber Oct 27 '21 at 15:32
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    I don't understand the votes to close as "unclear/needs details"; what details are missing here? – Noah Schweber Oct 29 '21 at 04:28

1 Answers1

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Your understanding is correct, apart from the caveat that there are a lot of details you glossed over. Most significantly, in the part about the second incompleteness theorem, while the first incompleteness theorem shows, 'if S is consistent, then G is true', what we really need is to show that S can prove this... and this is a major undertaking, even on top of the detailed work to show the first incompleteness theorem, and it requires new technical assumptions about S, more detailed than the conditions needed for the first theorem.

But let's put that aside, and address your question. Yes, it certainly feels odd that it is consistent to add "S is inconsistent" as an axiom (sometimes I think that this should be a "paradox" with a name). The reason it feels odd is that you are adding an axiom that is false.

The crucial thing to grasp here is that a theory doesn't need to be correct (according to some standard interpretation) in order to be consistent. In terms of interpretations, the standard interpretation (i.e. the natural numbers) is no longer a model of "S + S is inconsistent", but some of the nonstandard interpretations of S are still models. These nonstandard models have nonstandard natural numbers (larger than any number their initial segment that looks like $\mathbb N$) that they think encode proofs of "0=1" in S.

In "S + S is inconsistent", we also have an odd syntactic situation where for each natural number $n,$ it is provable that "$\mathbf n$ does not code a proof of 0=1 in S" and yet it is also provable that "there is a proof of 0=1 in S". This is called $\omega$-inconsistency. Of course, as in the previous paragraph, we can see that the resolution is that not every element of the domain of an interpretation needs to be represented by a numeral $\mathbf n.$

Perhaps we would like it if it were inconsistent to be wrong, but this would require that our theory decide every sentence (correctly), since, as you've reasoned, if a sentence isn't decidable, you can add either it or its negation and get a consistent system, and both can't be right. And unfortunately this is exactly what Godel's theorem rules out (for sufficiently strong, recursively axiomatizable theories).

spaceisdarkgreen
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