# Question about the consistency of assuming (via axiom) that $\kappa < \nu$ for certain pairs of cardinal numbers provably satisfying $\kappa \leq \nu$

Call an ordered pair of formulae $\langle P(\kappa,\tilde{a}), Q(\nu,\tilde{a})\rangle$ in the language of $\{\in\}$ unproblematic iff

1. ZFC proves that for all $\tilde{a}$ and all cardinal numbers $\kappa$ and $\nu$, if $P(\kappa,\tilde{a})$ and $Q(\nu,\tilde{a})$ hold, then $\kappa \leq \nu$ as cardinal numbers.
2. ZFC does not prove that for some $\tilde{a},$ there exists a cardinal number $\kappa$ such that $P(\kappa,\tilde{a})$ and $Q(\kappa,\tilde{a})$ hold.

Then clearly, if $\langle P(\kappa,\tilde{a}), Q(\nu,\tilde{a})\rangle$ is an unproblematic formula pair, then we may adjoin the following axiom to ZFC and be confident that the resulting theory will be consistent.

Axiom. For all $\tilde{a}$ and all cardinal numbers $\kappa$ and $\nu$, if $P(\kappa,\tilde{a})$ and $Q(\nu,\tilde{a})$ hold, then $\kappa < \nu$ as cardinal numbers.

My question is this. Suppose we have a family of ordered pairs of formulae $$\Phi=\{\langle P_i(\kappa,\tilde{a}), Q_i(\nu,\tilde{a})\rangle\}_{i \in I}$$ such that $\Phi_i$ is unproblematic for every $i \in I$. If we adjoin the corresponding family of axioms to ZFC, is the resulting theory necessarily consistent? And if so, is this extension necessarily conservative over ZFC for sentences in the language of first-order arithmetic?

• I'm not sure what you mean by "conservative over first-order arithmetic," since ZFC is already wildly more powerful than any theory I'd consider "first-order arithmetic" (except for true arithmetic, $TA$; but that's a complete theory, so anything extending it but not conservative over it (for sentences of arithmetic) must be inconsistent). Do you maybe mean, "Conservative over $ZFC$ for sentences in the language of first-order arithmetic?" – Noah Schweber May 2 '14 at 5:18
• @NoahS, thanks for your help, I've edited out my phrase and used yours in its stead. – goblin May 2 '14 at 5:22

The answer to each of your questions is "no," in a very strong way (and we don't even need to use the parameter $\tilde{\alpha}$):
Let $P(\kappa)=$"$\kappa=0$," and for a sentence $\varphi$ let $Q_{\varphi}(\nu)=$"$[\varphi\implies (\nu=1)]\wedge [\neg\varphi\implies (\nu=0)]"$. Then $\langle P, Q_\varphi\rangle$ is unproblematic as long as $\varphi$ is consistent with $ZFC$, and the axiom $A(P,Q_\varphi)=$"$P(\kappa), Q_\varphi(\nu)\implies \kappa<\nu$" associated to the pair is equivalent to $\varphi$.
• Any sentence $\varphi$ (consistent with $ZFC$) is a consequence of (equivalent to, in fact) a single unproblematic pair (so there is absolutely zero guaranteed conservativity);
• For any $\varphi$ independent of $ZFC$, the set of new axioms $\{A(P, Q_\varphi)$, $A(P, Q_{\neg\varphi})\}$ are contradictory even though it consists only of (axioms induced by) unproblematic pairs.