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Suppose PA is consistent. Godel's theorem says that it cannot prove its own consistency. So, could there be a proof in PA that PA is inconsistent ? That wouldn't lead to a direct contradiction, so could that situation actually happen ?

Then what ? The theory would be lying to us even though it is consistent. How do we know it's not lying about other stuff ?

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  • $\begingroup$ Re your final question, if you believe the axioms of PA are true, then it can't lie to us because its rules of inference are truth-preserving. If you believe that not all the axioms of PA are true, then of course it is lying to us. $\endgroup$
    – Steven Landsburg
    Dec 10, 2016 at 0:35
  • $\begingroup$ @StevenLandsburg The last clause in your comment should probably read "then of course you believe that it is lying to us." $\endgroup$
    – Andreas Blass
    Dec 10, 2016 at 0:44
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    $\begingroup$ @AndreasBlass: Yes, I completely accept your correction. The same correction applies to my first sentence as well. $\endgroup$
    – Steven Landsburg
    Dec 10, 2016 at 1:00

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To answer the question in the title: Yes, there are consistent theories that prove their own inconsistency. One example is the theory obtained from PA by adding the axiom "PA is inconsistent." This is consistent, by Gödel's second incompleteness theorem, and it proves its own inconsistency because it proves the inconsistency of the (provably) weaker theory PA.

On the other hand, PA itself is not an example for the title question, because all its axioms are true of the standard natural numbers, logical deduction preserves truth, and "PA is inconsistent" isn't true.

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    $\begingroup$ I'm trying to wrap my mind around what you wrote. If I were to assert that a theory proved its own inconsistency I would mean "There exists a statement P for which the system proves P, and it also proves ~P." This is clearly impossible for an actual consistent system. By "proving its own inconsistency" are you only asserting that there is a provable statement Q which can merely be interpreted as asserting the system is inconsistent? $\endgroup$
    – Pace Nielsen
    Dec 10, 2016 at 4:17
  • $\begingroup$ @PaceNielsen To say that a theory $T$ proves its own consistency means that $\neg\text{Con}(T)$ is a theorem of $T$, where $\text{Con}(T)$ is the usual formalization of consistency of $T$, exactly as in the statement of Gödel's second incompleteness theorem. $\endgroup$
    – Andreas Blass
    Dec 10, 2016 at 17:35
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    $\begingroup$ @PaceNielsen The first two of your last 3 comments are wrong (or perhaps meaningless, since you use the notation $\text{Con}_X(X)$ for various theories $X$ without saying what it is supposed to mean or how it's supposed to differ from the customary $\text{Con}(X)$). The second incompleteness theorem says that a consistent sufficiently strong theory can't prove its own consistency (when the consistency is formulated in a natural way, satisfying the derivability conditions); it does not prevent such a theory from proving its own inconsistency. [continued in nest comment] $\endgroup$
    – Andreas Blass
    Dec 12, 2016 at 18:56
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    $\begingroup$ [continuation] Also, the implication from $\neg\text{Con}(PA)$ to $\neg\text{Con}(PA+S)$ is provable in PA (and in far weaker theories), for any sentence $S$. (To avoid weird formalizations, whatever formula $A(x)$ is used to formalize "$x$ is an axiom of PA", one should use "$A(x)\lor x=\bar s$ to formalize "$x$ is an axiom of PA + S", where $s$ is the Gödel number of $S$ and $\bar s$ is the numeral for $s$.) $\endgroup$
    – Andreas Blass
    Dec 12, 2016 at 18:59
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    $\begingroup$ @PaceNielsen Your third comment has a typo, a missing negation, but what you intended to cite from Feferman is correct (except that Feferman seems to carefully put "model" in quotation marks when he means interpretation rather than an actual model). PA + $\neg$Con(PA) is interpretable in PA, and PA + Con(PA) is not interpretable in PA. $\endgroup$
    – Andreas Blass
    Dec 12, 2016 at 20:34

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