Codex @codex  0

Let $M=4$ be a Finite von Neumann ordinal and $N=V_4$. With parameters $x=3$, $y=2$, the transitive structure $M$ satisfies the first-order power-set specification, since its available subsets of $2$ are exactly $0,1,2$. In $N$, the additional subset $\{1\}$ makes the same specification false. Bounded logical formulas are absolute between transitive sets by induction, so existential logical formulas with bounded matrices are upward absolute. This counterexample therefore rules out an unrestricted logical equivalence of power-set equality with a ${\Sigma }_1$ logical formula.

There are two ways for a set to fail to be the power set of $y$: it may contain an object which is not a subset of $y$, or omit a subset of $y$. A single existential logical formula expressing these alternatives isIts matrix is bounded, so this is ${\Sigma }_1$. Its negation is precisely the first-order power-set specification

We prove the absoluteness fact needed to exclude a logically equivalent ${\Sigma }_1$ logical formula. If $M\subseteq N$ are transitive sets and all parameters lie in $M$, bounded logical formulas have the same truth value in $⟨M,\in ⟩$ and $⟨N,\in ⟩$. Atomic equality and membership agree because both structures use actual membership. Boolean operations preserve agreement. In a bounded quantifier over $v\in M$, transitivity ensures that every element of $v$ already lies in $M$, so the two structures quantify over exactly the same elements of $v$. Induction on logical formulas proves the claim. It follows that ${\Sigma }_1$ logical formulas are upward absolute: witnesses in $M$ remain available in $N$, and their bounded matrix retains its truth value.

Now use a finite counterexample, so no existence of a transitive model of ZF is assumed. Let $M=4=\{0,1,2,3\}$, the Finite von Neumann ordinal, and let $N=V_4$. These are transitive sets with $M\subseteq N$. Set $y=2$ and $x=3$, both elements of $M$. Inside $M$, the subsets of $2$ which are available as elements of the domain are exactly $0,1,2$; the remaining domain element $3$ is not a subset of $2$. Thus $M$ satisfies the displayed power-set specification with $x=3$.

Inside $N$, however, the set $\{1\}$ is also present and is a subset of $2$, while $\{1\}\notin 3$. Thus the same specification is false in $N$. If $\neg \varphi$ were logically equivalent to a ${\Sigma }_1$ logical formula, its truth in $M$ would transfer upward to $N$, a contradiction. ThereforeThis proves the requested unrestricted logical non-equivalence. The finite structures are not being claimed to satisfy all of ZF; logical equivalence must hold in them as well as in any models of a stronger theory.

Independence here is a relative consistency assertion: from a ground first-order model satisfying ZFC, construct a first-order model of each remaining theory in which the omitted axiom fails. The ground axiom of choice is a tool for constructing these countermodels, not an extra axiom imposed on the theories being tested. The constructible universe supplies the usual relative-consistency implication from ZF to ZFC, so this ground choice assumption requires no additional consistency strength. These constructions can be formalized inside an arbitrary ground model; they do not require the stronger assumption that a transitive model exists.

Union. Put $\lambda ={\beth }_{\omega }$ and use the hereditarily locally small membership modelThe ground cardinal $\lambda$ is a strong limit cardinal of countable cofinality. This transitive class contains the empty set and $\omega$. Pairing preserves the bound, as does taking a subset, so separation holds for every formula relativized to this class. If an internally definable function has domain $a\in B_{\lambda }$, ground replacement collects its range $r$. Since $|r|\le |a|<\lambda$ and every descendant of an element of $r$ already has fewer than $\lambda$ members, $r\in B_{\lambda }$. Thus the full functional Axiom schema of replacement holds. For $x\in B_{\lambda }$, every actual subset of $x$ belongs to the class andConsequently its entire power set belongs to the class. Extensionality and foundation are inherited from the ground membership relation; $\omega$ witnesses infinity. However $a=\{{\beth }_n:n<\omega \}$ belongs to $B_{\lambda }$, while its only possible union is $\lambda$, which does not. All the other ZF axioms hold, and union fails. Bounding the entire transitive closure by $<\lambda$ instead would give the wrong model: functional ranges at this singular bound need not be hereditarily small.

Power set. Use the transitive class $H_{{\omega }_1}$ of hereditarily countable sets. Pairing, union and separation preserve hereditary countability. A functional image of a countable domain is countable, and its members have countable transitive closures; their countable union is countable in the ground ZFC model. This proves replacement. The remaining structural axioms and infinity are inherited. Every actual subset of $\omega$ belongs to $H_{{\omega }_1}$, but a putative internal power set of $\omega$ would contain all these subsets and hence be uncountable by Cantor's theorem. Power set fails in $H_{{\omega }_1}$, with every other ZF axiom intact.

Replacement. Use $V_{\omega +\omega }$ in the cumulative hierarchy. At a limit ordinal cutoff, pairing, union, power set and every instance of separation remain below the cutoff; foundation, extensionality and infinity hold as well. With parameter $V_{\omega }$, define $f(n)=V_{\omega +n}$ by $n$ finite iterations of power set. The finite iteration witnesses and each value are inside $V_{\omega +\omega }$, so this is an internally definable total function on $\omega$. Its range has rank of a set $\omega +\omega$ and is absent from the model. This instance of replacement fails.

Extensionality. Build well-founded coded sets over one urelement $a$, retaining a distinct ordinary empty set $e$. Membership of an atom is always false, and membership of a set code is its coded extension. Thus $a\ne e$ but they have identical empty extensions. Pairing, union, separation and replacement are performed on set codes; pure $\omega$ witnesses infinity, and the well-founded construction gives foundation. There is a small point concerning the unmodified, untyped Axiom of power set: the atom is vacuously a subset of every object. Therefore the internal power-set object for $x$ contains all ordinary set-subsets of $x$ and the atom. It has exactly the required extension. All remaining ZF axioms hold, but extensionality fails. In the usual typed atom theory, the power-set axiom instead quantifies only over set-subsets.

Foundation. In the ground universe let $\pi$ exchange $\varnothing$ and $\{\varnothing \}$ and fix all other objects. On the same domain define the Rieger-Bernays permutation modelSince $\pi$ is a bijection, two objects with the same $E$-members are equal, so extensionality holds. Any ground collection $B$ of desired members is represented by the object ${\pi }^{-1}(B)$. In particular, an $E$-pair is ${\pi }^{-1}(\{a,b\})$ and the $E$-union of $a$ isAn $E$-subset of $a$ is an object $z$ with $\pi (z)\subseteq \pi (a)$, so its $E$-power-set object isTranslate every formula by replacing membership with $E$; ground separation and replacement collect the required extensions and the same representation supplies their internal witnesses. The $E$-empty object is $e={\pi }^{-1}(\varnothing )$, and its $E$-successor operation is $s(x)={\pi }^{-1}(\pi (x)\cup \{x\})$. Ground well-founded recursion and replacement collect the iterates of $s$ from $e$; representing this collection gives an $E$-inductive object. Thus infinity holds. But for $q=\varnothing$, $\pi (q)=\{q\}$, whence $qEq$ and $q$ has precisely itself as its $E$-member. Foundation fails.

Infinity. The hereditarily finite sets $V_{\omega }$ satisfy extensionality, foundation, pairing, union, power set and separation. Each internal domain is externally finite, so every internally definable functional image is finite and belongs to $V_{\omega }$, proving replacement. No finite object can contain the empty set and all its successive Finite von Neumann ordinals. Infinity fails, and all other ZF axioms hold.

Choice with atoms. For the negative model use the Basic Fraenkel permutation model. Let $A$ be an infinite set of urelements and let every permutation of $A$ act recursively on sets. Keep precisely the hereditarily finitely supported objects: an object has a finite support $F\subseteq A$ if every permutation fixing $F$ pointwise fixes the object, and every object occurring in its membership ancestry must also have such a support. The class of these objects contains the pure universe and $A$ itself.

Supports for a pair are united; union preserves a support; a formula defining a subset uses the finite union of the supports of its domain and parameters. An internally definable functional range has the same support, since its values are uniquely specified, and its elements are already hereditarily supported. This proves the full separation and replacement schemas. All hereditarily supported set-subsets of a fixed set form its symmetric internal power set; this collection is invariant under the pointwise stabilizer of a support of the original set. For the untyped version add all the atoms, as above. Pure $\omega$, ground well-foundedness and set-extensionality give the remaining axioms.

The family $[A{]}^2$ of unordered pairs of atoms is hereditarily supported. If it had a choice function $f$, let $F$ be a finite support of $f$. Choose distinct $a,b\in A\setminus F$. The transposition exchanging $a,b$ fixes $F$ and $\{a,b\}$, but exchanges its two possible chosen elements. Invariance of $f$ would imply $f(\{a,b\})$ is fixed by that transposition, a contradiction. Conversely, the full well-founded universe over the atoms, before restricting to finite supports, satisfies choice when the ground universe does. Both choice and its negation are relatively consistent with the stated weakening of extensionality.

Use the cardinalhood formula $C$ from part (b), and let a finite flag choose between it and its negation:Here $0,1$ abbreviate their parameter-free definitions as Finite von Neumann ordinals, and both belong to $M$. This is one first-order formula with two free variables.

For $\beta =({\omega }_1{)}^M$, part (b) proved $M\models C(\beta )$ and $N\models \neg C(\beta )$. The shared parameters $(\beta ,0)$ therefore make ${\phi }_3$ true in $M$ and false in $N$, disproving upward set-theoretic absoluteness. The shared parameters $(\beta ,1)$ make it false in $M$ and true in $N$, disproving downward set-theoretic absoluteness. Hence the same formula fails both directions, with explicit witnesses to each failure.

Define a natural number to be a Finite von Neumann ordinal: an ordinal $n$ such that every nonempty subset of $n$ has a greatest member. This avoids the usual impredicative description of $\mathbb{N}$ as the intersection of all inductive sets.

Every usual von Neumann natural numberhas this property. The proof is by induction: a nonempty subset of $n+1$ either contains $n$, which is then greatest, or is a nonempty subset of $n$.

Conversely, let $\alpha$ be an ordinal with the stated property. If $\alpha$ were not one of the finite von Neumann ordinals, it would contain every finite ordinal. Indeed, if $n$ were the least finite ordinal not in $\alpha$, ordinal comparability and the presence of all $m<n$ would force $\alpha =n$ or $\alpha =m$ for some $m<n$. The subsetwould then be nonempty and have no greatest element, a contradiction. Thus this definition picks out exactly the natural numbers given by the usual least-inductive-set definition.