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Let's assume that we have a consistent first-order theory, which was derived from a second order theory by replacing universal quantification over second order variables by axiom schemes for first-order definable predicates. Now let's compare this theory to two different second-order theories using Henkin semantics with suitable comprehension axiom schemes.

  1. If the comprehension axiom scheme allows quantification only over the first-order variables, can the resulting second-order theory be inconsistent? I guess the answer is no, and the resulting theory will be equivalent to the first-order theory with respect to provability of first-order formulas.
  2. If the comprehension axiom scheme allows quantification over both first and second-order variables, we can no longer be sure that the resulting second order theory is consistent. Is there a simple example, where the first-order theory is "provably consistent", and the second-order theory is "provably inconsistent"?

As already clarified in the comments, "provably (in)consistent" just means that assuming ZFC (or any other foundation) is fine, there is no need to restrict answers to syntactic derivations.

Thomas Klimpel
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  • If "the second-order theory is" inconsistent then "the second-order theory is 'provably inconsistent'". $\hspace{.49 in}$ –  May 11 '14 at 21:15
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    @RickyDemer By "provably inconsistent", I just mean that I don't insist on a possibly extremely long syntactic derivation of $\phi\land\lnot\phi$, but that a "short" proof (possibly assuming ZFC) that such a derivation exist is also sufficient. – Thomas Klimpel May 11 '14 at 21:21
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    While I don't know the answer to your question, here's a nice example. Consider that we work in the theory $\sf ZFC$ augmented by the assumptions of $\rm Con\sf (ZFC)$ and $\lnot\rm Con\sf (ZFC+\rm Con\sf (ZFC))$. Then there is a model of $\sf ZFC$, therefore there is a model of $\sf NBG$; but there is no model of $\sf MK$ since that would imply much more in terms of consistency strength (and there's also no model of $\sf ZFC_2$, since that would require an inaccessible cardinal to exist). – Asaf Karagila May 16 '14 at 05:21
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    @AsafKaragila Thanks, this really helped. I see now that the main obstacle to the second part of my question is the word "simple". If I drop it, I can just fix a Gödel numbering, and add $\lnot\mathsf{Con(PA)}$ as a single axiom to $\mathsf{PA}$. My best guess for an answer is then probably that either a "simple" independent statement for PA like Goodstein's theorem can be expressed as a single formula (so that I can add the negation of that formula as an axiom), or to see whether Robinson arithmetic $\mathsf{Q}$ allows for a simple formula (provably) independent of $\mathsf{Q}$. – Thomas Klimpel May 17 '14 at 12:27
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    $(\forall y)(\exists x)(x+x = y : \lor : S(x+x) = y) ;;$ is a simple formula that is provably independent of Q. $\hspace{1.01 in}$ –  Aug 11 '14 at 06:27

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