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Godel's incompleteness theorem tells us that there cannot be any complete consistent theory $T$ at least as strong as Peano arithmetic, because we can't prove $\text{Con}(T)$ within $T$.

But suppose that we have a theory $T_0$ that we intuitively trust, for example Peano arithimetic or ZFC. We won't be able to prove $\text{Con}(T_0)$, yet we know that it's "true", so we also trust $T_1 = T_0 + \text{Con}(T_0)$. Similarly, we trust $T_2 = T_1 + \text{Con}(T_1)$, despite not being able to prove it, and so on.

This means that if we trust $T_0$, we'll trust $T_0, T_1, T_2, ...$ We can keep going for every ordinal, letting $T_\lambda = \sum_{\alpha < \lambda} T_\alpha$ for limit ordinals $T_\lambda$. This eventually gives us $T_\infty = \sum_{\text{countable ordinal }\alpha} T_\alpha$. But this still isn't the strongest theory we trust. Even though we ran out of ordinals, there's still $T_\infty + \text{Con}(T_\infty)$ that we trust.

It seems to be impossible to define what the actual strongest theory we trust, since whenever we have a theory $T$ we trust, there will always be a stronger one $T + \text{Con}(T)$ that we trust. But intuitively, there is a "thing" $T$ which is the closure of $T_0$ under consistency.

What's the explanation for this? Godel's theorem tells us there is no complete consistent theory, but the sentence it builds is exactly the one that we intuitively know to be true.

  • Propositional calculus and predicate calculus are bout complete and consistent. – Doug Spoonwood Jun 12 '20 at 22:00
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    @DougSpoonwood Those are logics, not theories; the OP is talking about theories, and the incompleteness theorem kicks in there (note that the (in)completeness of a logic is totally different from the (in)completeness of a theory - there's annoying terminology overload). – Noah Schweber Jun 12 '20 at 22:02
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    "We can keep going for every ordinal" Actually, when we dig into the details we can't - you have to show how to express the relevant consistency principles in the relevant language, and eventually you run into ordinals which can't be defined in that language. Even with ordinals you can define the choice of definition matters: different expressions for the same ordinal can yield different theories. This has come up variously on MSE and MO: 1, 2, 3. – Noah Schweber Jun 12 '20 at 22:04
  • Interesting, I'll look at those. But my main question is the last part. I know the incompleteness theorem tells us that we can't prove $\text{Con}(T)$ within $T$, but intuitively, this doesn't matter to me because I know $\text{Con}(T)$ is true despite not being able to give you a proof within $T$. –  Jun 12 '20 at 22:07
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    @Stefan First of all, note that in general we should not be as confident in the consistency of $T+Con(T)$ as we are in the consistency of $T$: a theory can prove its own inconsistency while being consistent, and if $S$ is such a theory then $S$ is consistent but $S+Con(S)$ is inconsistent. – Noah Schweber Jun 12 '20 at 22:09
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    I have my doubt about the construction you describe. The language of $T$, if it is even half way reasonable, will have only so many formulas, but there are arbitrarily many ordinals, so you can't be adding a new formula at every stage. You would have to hit a stage in your recursion where the process simply stalls out somehow, wouldn't you? – Malice Vidrine Jun 12 '20 at 22:11
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    @MaliceVidrine Indeed, per my comment second comment above you'd hit a problem long before cardinality became an issue. – Noah Schweber Jun 12 '20 at 22:12
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    @Stefan The issue would be the following: what if we had a theory $A$ such that $A$ contains exactly the sentences we're certain are true? Well, first of all we might not be certain that $A$ itself has that property - this "global" fact might be quite subtle. If we are so certain, then of course we should additionally be certain that $A+Con(A)$ is also true (so $Con(A)\in A$), but that has another implicit hypothesis, namely that $A$ itself is arithmetically definable so that we can even express "$Con(A)$" in the relevant language. (continued) – Noah Schweber Jun 12 '20 at 22:14
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    But even granting that, the second incompleteness theorem turns out to break down once we shift from computably axiomatized theories to the broader context of arithmetically defined theories (see e.g. here). So really the only thing the second incompleteness theorem rules out is the possibility of finding a theory $A$ which $(i)$ is computably axiomatizable, $(ii)$ proves exactly the "certainly true" facts, and such that $(iii)$ property $(ii)$ is itself "certainly true." So it's not really that intricate a situation here. – Noah Schweber Jun 12 '20 at 22:16
  • (My previous comment should say "the only thing the second incompleteness theorem rules out in this context" - obviously IT2 has other applications!) – Noah Schweber Jun 12 '20 at 22:22

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