Existence of proof of consistency of stronger axiomatic systems

Question

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.

Answer 1

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.