1
$\begingroup$

Working in first order logic with equality and membership $``\sf FOL(=,\in)"$

Let $\phi x$ be a formula in which only $x$ occur free, and never bound.

Let $\pi_i x \vec{z}$ be the formula $\forall y (y \in x \leftrightarrow \psi_i y\vec{z})$ where $\psi_i y \vec{z}$ is a formula in which only symbols $``y,z_1,..,z_n"$ occur free, and never bound; such that:

$\sf FOL(=,\in)$ $ \vdash \forall \vec{z} \forall x: \pi_i x \vec{z} \to \phi x $

Let $\sf T$ be a theory that extends $\sf FOL(=, \in)$, with only the following axioms:

  1. $\exists x. \phi x$

  2. $\forall \vec{z} (\exists x. \pi_i x \vec{z}), _{ i=1,..,m} $

The idea is that $\sf T$ only says that there exists an object that fulfills $\phi$, and stipulate $m$-many naive comprehension axioms each assuring the existence of a set of all objects satisfying a formula among $\psi_1,..,\psi_m$ formulas, and all those sets in turn are provable to satisfy $\phi$ in just the background language of $\sf T$.

My question is that given the above conditions, is there a known set theory that is provably consistent relative to some extension of ZF in which the following is provable? $$\sf Con(\forall x. \phi x) \land Con( T) \\\to Con(T+ \forall x. \phi x)$$.

My guess is to the negative, but I don't know of a counter-example.

$\endgroup$
3
  • $\begingroup$ I must be missing something. If you take $\phi$ to be $x=x$, and $\psi$ to be $(x=x)\land (z_1=z_1)\land \cdots \land (z_n=z_n)$, can't you just take H=T=FOL? Or, if you really want to prove the consistency statement inside H (rather than metatheoretically), you could take H to be any (supposedly consistent) system where it proves the consistency you want? $\endgroup$ Mar 4, 2021 at 19:15
  • $\begingroup$ @PaceNielsen, you are asking about a particular theory $\sf T$ that only has axioms of $\exists x: x=x$ and $\forall z_1,..,\forall z_n \exists x: x=x \land z_1=z_1 \land ...\land z_n=z_n$. Now this theory clearly fulfills the rule I've given since there clearly exists a consistent theory H that proves $\forall x. x=x$ and also we have that if $\sf T$ is consistent, then $\sf Con(T + \forall x. x=x )$. That's clear you can take H to be simply $\sf FOL(=)$, and clearly $T$ is consistent with $\forall x. x=x$ since $\sf T + \forall x. x=x$ is a fragment of $\sf FOL(=)$ $\endgroup$ Mar 4, 2021 at 19:28
  • $\begingroup$ @PaceNielsen, in the last line of the above comment I mean a 'fragment of FOL(=,$\in$)'. Back to your question, of course for this particular choice of $\phi$ and $\psi$ you can take H=T=FOL(=,$\in$). But generally speaking I'm thinking of H to be a simple theory, much simpler than T+$\forall x. \phi x$, the latter theory can be much stronger than T and H, that's the benefit of this rule. $\endgroup$ Mar 4, 2021 at 19:58

1 Answer 1

2
$\begingroup$

Take $\psi(y)$ to be $(y=y)\land \exists u\, \forall v\, \neg(v\in u)$, so that $\psi$ expresses the statement that there is some $\in$-minimal object. Thus, $\pi(x)$ is just false if there are no $\in$-minimal objects, but if there are $\in$-minimal objects then it says $x$ has all objects as elements.

Let $\phi(x)$ be the statement $\pi(x)\lor \forall u\, \exists v\, (v\in u)$. So, this says either (1) $x$ has every object as a member and there is an $\in$-minimal object, or (2) there are no $\in$-minimal objects.

Clearly $\forall x\, \phi(x)$ is consistent. FOL proves $\pi(x)\rightarrow \phi(x)$ (and really all you need is propositional logic). Also $T$ is consistent. But $T+\forall x\, \phi(x)$ is not consistent.

$\endgroup$
6
  • $\begingroup$ Thanks a lot. You are right. This calls for requiring every axiom of $\sf T$ to be consistent with $\forall x. \phi(x)$, but I still think this would be hopeless. $\endgroup$ Mar 5, 2021 at 7:10
  • $\begingroup$ You’re not allowing for the existence of the empty set. $\endgroup$ Mar 5, 2021 at 8:21
  • $\begingroup$ @MonroeEskew, your comment is addressed to whom? $\endgroup$ Mar 5, 2021 at 10:21
  • $\begingroup$ Pace. . . . . . $\endgroup$ Mar 5, 2021 at 10:29
  • $\begingroup$ @MonroeEskew That's right. However, if you want to allow for the existence of the empty set, or all of ZFC, you could do something like the following. First, assume ZFC holds. Let $\psi(y)$ express "$y$ is empty and $V=L$". Thus $\pi(x)$ is never satisfied if $\neg(V=L)$, and is satisfied only for $x=\{\emptyset\}$ if $V=L$. Let $\phi(x)$ be $\pi(x) \lor \neg(V=L)$. etc... $\endgroup$ Mar 5, 2021 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.