Assume $\mathsf{ZFC}$ is consistent. Are there axioms $\mathsf{P}$ and $\mathsf{Q}$ independent from $\mathsf{ZFC}$ such that?:

  • $\mathsf{Q}$ is a stronger axiom than $\mathsf{P}$

  • There is a proof under $\mathsf{ZFC}$ that $\mathsf{ZFC} + \mathsf{P}$ is consistent

  • There is no proof under $\mathsf{ZFC}$ that $\mathsf{ZFC} + \mathsf{Q}$ is consistent

  • There is a proof under $\mathsf{ZFC} + \mathsf{P}$ that $\mathsf{ZFC} + \mathsf{Q}$ is consistent

This question arose while developing a new logic system. This logic system deals with undecidable statements by assigning them a truth value other than true and false. They are the elements of a $\sigma$-algebra on $\mathsf{ZFC}$'s house.

If such $\mathsf{P}$ and $\mathsf{Q}$ don't exist, it would imply that all nonempty subhouses (which all have uncountably many residents) have the trivial topology, which is a very disappointing result. Well, it turns out I asked the wrong question. This question is essentially equivalent to verifying that $X = \overline{Y} \land Y = \overline{Z} → X = \overline{Z}$, where overlines mean topological closure. I'll try to formulate the original question I had, but that should be another post.

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    $\begingroup$ When you say "proof" and "undecidable" what proof system are these with respect to? $\endgroup$ Dec 26 '20 at 1:07
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    $\begingroup$ Just the second and third of your four requirements are already impossible to satisfy. The second requirement implies, in particular, that ZFC proves its own consistency. So, by Gödel's second incompleteness theorem, it would follow that ZFC is inconsistent and therefore proves everything, contrary to the third requirement. (If you know that ZFC is consistent, then it follows, by the first part of the same argument, that the second requirement can't be satisfied.) $\endgroup$ Feb 20 at 23:48

Well right off the bat, there's a problem: if $\mathsf{ZFC}$ proves that $\mathsf{ZFC}+P$ is consistent, then $\mathsf{ZFC}$ is inconsistent right off the bat per Godel's second incompleteness theorem! I was silly and missed this initially.

We can rephrase this by using conditional consistency claims, as follows:

Can there be $P,Q$ such that:

  1. $\mathsf{ZFC}\vdash Q\rightarrow P$,

  2. $\mathsf{ZFC}\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC}+P)$,

  3. $\mathsf{ZFC}+P\vdash Con(\mathsf{ZFC}+P)\rightarrow Con(\mathsf{ZFC}+Q)$, but

  4. $\mathsf{ZFC}\not\vdash Con(\mathsf{ZFC})\rightarrow Con(\mathsf{ZFC}+Q)$?

However, this changes the picture drastically. Take $P=\neg Con(\mathsf{ZFC})$. Then bulletpoint $(3)$ holds trivially for every $Q$, because $\mathsf{ZFC}+P\vdash\neg Con(\mathsf{ZFC}+P)$. So just pick any $Q$ such that $(1)$ and $(4)$ hold.

  • $\begingroup$ You're ignoring the obvious: if ZFC proves that ZFC + P is consistent, then it proves that ZFC is consistent. $\endgroup$
    – Asaf Karagila
    Feb 21 at 0:35
  • $\begingroup$ @AsafKaragila ... Oh wow, yes, d'oy. Technically I wasn't wrong though! $\endgroup$ Feb 21 at 0:42

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