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Question 1. Let $\operatorname{con}(\mathsf{ZFC})$ be a sentence in $\mathsf{ZFC}$ asserting that $\mathsf{ZFC}$ has a model.

Let $S$ be the theory $\mathsf{ZFC}+\operatorname{con}(\mathsf{ZFC})$. Let $\operatorname{con}(S)$ be a sentence in $S$ (or $\mathsf{ZFC}$) asserting that $S$ has a model. We assume throughout that $\mathsf{ZFC}$ is consistent.

Is it provable in $\mathsf{ZFC}$: $\operatorname{con}(\mathsf{ZFC})\leftrightarrow\operatorname{con}(S)$?

Question 2. Assume now S is inconsistent i.e. that is $S$, has no model.This means that there is no model of $\mathsf{ZFC}$ in which $\operatorname{con}(\mathsf{ZFC})$ is true, so the negation of $\operatorname{con}(\mathsf{ZFC})$ is true in any model of $\mathsf{ZFC}$ and therefore (by completeness theorem) is a theorem of $\mathsf{ZFC}$?

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migrated from mathoverflow.net Jan 5 '14 at 15:16

This question came from our site for professional mathematicians.

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    $\begingroup$ This could be a perfectly reasonable question. It also could be a perfectly reasonable homework problem after learning the incompleteness theorem. $\endgroup$ – Goldstern Jan 5 '14 at 12:57
  • $\begingroup$ Looking at the edited question, along with the comment by @Goldstern, I think that this is better migrated to math.SE instead. $\endgroup$ – Asaf Karagila Jan 5 '14 at 15:05
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No, not at all.

The consistency of $S$ is a strictly stronger assertion than the consistency of $\sf ZFC$. Note that $S$ proves $\operatorname{Con}\sf(ZFC)$ but not $\operatorname{Con}(S)$ (as per the usual incompleteness arguments).

If $\sf ZFC$ would prove the equivalence between the two consistency statements, then $S$ would prove that, in which case $S$ would prove its own consistency, because it can prove $\operatorname{Con}\sf(ZFC)$.


To the edit,

Note that you do have an additional assumption in the meta-theory. So now we actually have $\sf M=T+\operatorname{Con}(ZFC)+\lnot\operatorname{Con}(ZFC+\operatorname{Con}(ZFC))$ as our meta-theory (where $\sf T$ is some sufficient theory e.g. $\sf ZFC$).

This means that $M$ proves that $\sf ZFC$ proves that $\lnot\operatorname{Con}\sf(ZFC)$. It doesn't mean that $\sf ZFC$ proves that directly. The proof lies in the two additional consistency assumptions, rather than the part of $\sf ZFC$.

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