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How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Reflection, Absoluteness, and Elementary Submodels: Examples and Counterexamples

1 · Prerequisites

2 · Summary

Explicit graph and rank calculations illustrate the direction of absoluteness. A diagonal real demonstrates how an internal power set can omit an external subset, conditional on a countable transitive model. The elementary-submodel counterexample retains an uncountable ordinal as a parameter, forcing the submodel to fail transitivity. The final calculation tracks order positions through collapse embeddings.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-10Open item page →

Bounded formulas and the direction of absoluteness

Example

Bounded graph agreement can be checked explicitly on a={0}, b={0,1} and f={0,1}. Here f:ab is injective. Existence of such a graph is a separate existential assertion whose witness must be retained for upward transfer.

Facts & Assumptions

[F1]

Absolute basic set operations and relations: The graphs of empty set, subset, unordered pair, singleton, union, intersection (with =), difference, Kuratowski ordered pair, Cartesian product, relation domain/range, functionhood, evaluation and injection have Δ0 definitions. Thus their values agree between transitive membership structures whenever the input and output sets are in the smaller domain. This is graph agreement, not an assertion that an arbitrary transitive domain is closed under these operations.

[F2]

Bounded formulas are absolute for transitive sets: If MN are nonempty transitive sets, every Δ0 formula is absolute between them on parameter tuples from M. No internal set-theory axioms are required. The analogous assertion for definable transitive classes is a formula-by-formula scheme.

[F3]

Sigma-one truth goes upward: For nonempty transitive MN, a literal existential block over a Δ0 matrix transfers truth upward, and its universal dual transfers truth downward. For formulas classified only by ZF-provable equivalence, assume both structures satisfy ZF (or all axioms used in the equivalence proof).

Verification

Given: 0=, 1={0}, a={0}, b={0,1}, f={0,1}.

1.1

The formula for xy is ux(uy); for z={x,y} it is xzyzuz(u=xu=y). In the instance, the only element 0 of a belongs to b, and the only elements of b are 0,1, so both tests hold for ab and b={0,1}.

givenalgebra
2.1

Use the bounded K(p,x,y) graph of F1. A bounded injection test is: every pf has coordinates xa,yb with K(p,x,y); every xa occurs in such a p; and for p,qf and their coordinates in a,b, equal first coordinates imply equal second coordinates and conversely. Here 0,1={{0},{0,1}} is the sole pair, its first coordinate is 0, its value is 1b, and comparing the only pair with itself verifies both uniqueness and injection.

F1step 1.1
3.1

In transitive domains containing a,b,f, F2 preserves these bounded graph tests. If the smaller domain contains the displayed witness f, F3 transfers the sentence f(f:ab) upward by retaining it. If only a,b are present, graph absoluteness alone neither constructs f in that domain nor supplies downward transfer of its existence.

F2F3step 2.1
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Power sets need not agree across transitive models

Example

If M is a countable transitive model of ZF, then externally its internal power set of ω is countable and misses an actual subset of ω. This conditional example does not assert that ZF proves the existence of such an M.

Facts & Assumptions

[F1]

Ordinals and omega in transitive models: In ambient ZF, ordinalhood is absolute between transitive membership domains containing the parameter. A transitive set model of ZF contains precisely the real finite ordinals as its natural numbers and has ωM=ω. Its ordinals form an initial segment of the actual ordinals.

[F2]

Absolute basic set operations and relations: The graphs of empty set, subset, unordered pair, singleton, union, intersection (with =), difference, Kuratowski ordered pair, Cartesian product, relation domain/range, functionhood, evaluation and injection have Δ0 definitions. Thus their values agree between transitive membership structures whenever the input and output sets are in the smaller domain. This is graph agreement, not an assertion that an arbitrary transitive domain is closed under these operations.

[F3]

Ranks agree and hierarchy membership is absolute: If MN are transitive models of ZF and xM, then rankM(x)=rankN(x). For every αMOrd, (Vα)M=M(Vα)N. Equality of the two internal power sets or stage sets is not asserted.

Verification

Given: A countable transitive MZF and an external countability injection.

1.1

By F1, ωM=ω. Fix an external injection h:Mω. Define e(n) to be the unique xM with h(x)=n when there is one, and otherwise M. Thus e:ωM is onto, without any further choice. Put An=e(n) if e(n)ω, and An= otherwise.

F1given
2.1

Let d={nω:nAn}. For every k, kd iff kAk, so dAk. If dM, surjectivity gives e(k)=d for some k, whence Ak=d because dω, a contradiction. Therefore d is a subset of the actual omega outside M.

step 1.1algebra
3.1

Let p be the internal power set of omega. Transitivity gives pM. Bounded subset agreement F2 and the internal power-set axiom give p={xM:xω}. Restricting h to p proves countability, and step 2.1 proves dp. Thus the hierarchy intersection identity in F3 does not imply equality of internal and external power sets.

F2F3step 1.1step 2.1
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Reflecting a finite family with parameters

Example

For fixed formulas ϕ(x),ψ(x,y) and a set a, there is β with aVβ such that both formulas are absolute on every tuple from Vβ. For instance take a={}, ϕ(x)z(xz) and ψ(x,y)z(xzyz).

Facts & Assumptions

[F1]

Montague–Lévy reflection for a finite formula family: In ZF, for each fixed finite family Φ and every ordinal α, some β>α makes Φ absolute between Vβ and V, for all tuples in Vβ. More generally the same holds between Wβ and W for a definable increasing continuous hierarchy of sets exhausting a definable nonempty class W. For an empty class, the relativization statement is interpreted as a scheme rather than satisfaction in an empty structure.

[F2]

Transitive models of fixed finite axiom fragments: For each fixed external finite ΓZF, ZF proves that some transitive Vβ satisfies Γ, with β above any prescribed ordinal bound. In ZFC the analogous scheme holds for fixed finite ΓZFC. These are schemes indexed by external fragments, not a single internal assertion of models for all coded fragments.

Verification

Given: A fixed pair of formulas and a set parameter; the displayed instance uses von Neumann ranks.

1.1

In the instance, rank(a)=1 and the witnesses for ϕ(a) and ψ(a,a) can both be {a}. It has rank 2, so it lies in V3, while aV2. Thus the witness may require a later stage than the parameter.

givenalgebra
2.1

For the general pair, close both formulas under subformulas and apply F1 starting above rank(a)+1. The produced β contains a and reflects every formula of this finite closure. In particular it reflects the two original formulas at all tuples in Vβ, not merely at the named instance. Applying F2 is an additional option when the displayed formulas include a fixed axiom fragment.

F1F2step 1.1
3.1

Choosing only witnesses for the two formulas at a would not cover their subformula instances at the new witnesses and at all other parameters in Vβ. The construction in F1 bounds every existential subformula at every tuple from each stage and iterates those bounds. That is why its conclusion supplies the required all-tuple agreement.

F1step 2.1
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A countable elementary submodel need not be transitive

Statement

False statement: every countable elementary submodel of a transitive set is transitive. In ZFC there is a countable XVθ with ω1X that is not transitive.

Facts & Assumptions

[F1]

Hartogs: an ordinal that does not inject into a given set: For every set A there is an ordinal (def-ordinal) that does not inject into A, that is, admits no injective function into A. The least such ordinal is the Hartogs number (A), and it is exactly

(A)={ot(S,R):SA and R well-orders S},

the set of order types (thm-mostowski-collapse) of the well-ordered subsets of A.

The proof is choice free. That is the whole point of the theorem: in ZF alone, with no assumption that A can be well ordered, one still gets an ordinal too long to be laid inside A.

[F2]

Countable elementary submodels and their collapses: In ZFC, if an infinite set membership structure M satisfies Extensionality, then for every at most countable AM there is a countably infinite XM containing A, and X has a countable transitive collapse. To retain a set aM as one parameter, use A={a}.

[F3]

What the collapse fixes: Let π:XXˉ be a collapse of actual membership as above. It fixes every transitive subset AX pointwise. If αX is an actual ordinal, π(α) is the order type of Xα. In particular, if Xα is transitive, π(α)=Xα.

[F4]

The Axiom of Choice: Every family of nonempty sets has a choice function

Refutation

Given: Ambient ZFC and actual cumulative hierarchy stages.

1.1

By F1 let κ be the least ordinal not injecting into ω, that is ω1. Choose θ=κ+ω. Then κVθ, the stage is infinite and transitive, and it satisfies Extensionality: any actual distinguishing member of two elements remains in the stage by transitivity. With AC as in F4, apply F2 to the parameter set {κ} to get a countable XVθ containing κ.

F1F2F4given
2.1

If X were transitive, κX would imply κX. Composing this inclusion with a countability injection Xω would inject κ into ω, contradicting its definition. Thus this X satisfies the hypotheses of the proposed assertion and fails its conclusion.

step 1.1algebra
3.1

F3 computes the collapse value of κ as ot(Xκ), a countable ordinal because the trace is countable. It cannot equal the uncountable κ. The transitive collapse is therefore a different membership presentation, as required.

F3step 1.1step 2.1
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Transporting an elementary-chain map through collapse

Example

For collapses πα:MαMˉα of an elementary membership chain, the maps are jαβ=πβιαβπα1. Their action on an ordinal is an order-type embedding, and equality with inclusion is an additional condition. In ZFC the two-stage chain XVκ+ω constructed below gives an explicit failure of inclusion.

Facts & Assumptions

[F1]

Elementary chains and compatible collapses: A nonempty set-ordinal elementary chain of actual membership structures satisfying Extensionality has a union elementary over every stage, and the union has a transitive collapse. Conjugating the inclusions by stage and union collapses gives coherent elementary embeddings; these are not asserted to be inclusions of the transitive images.

[F2]

What the collapse fixes: Let π:XXˉ be a collapse of actual membership as above. It fixes every transitive subset AX pointwise. If αX is an actual ordinal, π(α) is the order type of Xα. In particular, if Xα is transitive, π(α)=Xα.

[F3]

Countable elementary submodels and their collapses: In ZFC, if an infinite set membership structure M satisfies Extensionality, then for every at most countable AM there is a countably infinite XM containing A, and X has a countable transitive collapse. To retain a set aM as one parameter, use A={a}.

[F4]

Hartogs: an ordinal that does not inject into a given set: For every set A there is an ordinal (def-ordinal) that does not inject into A, that is, admits no injective function into A. The least such ordinal is the Hartogs number (A), and it is exactly

(A)={ot(S,R):SA and R well-orders S},

the set of order types (thm-mostowski-collapse) of the well-ordered subsets of A.

The proof is choice free. That is the whole point of the theorem: in ZF alone, with no assumption that A can be well ordered, one still gets an ordinal too long to be laid inside A.

[F5]

The Axiom of Choice: The Axiom of Choice (AC) is the following statement.

Every family of nonempty sets has a choice function (def-choice-function).

Written out: for every set F all of whose members are nonempty, there exists a function g with domain F satisfying g(S)S for all SF.

An equivalent formulation is that a product of nonempty sets is nonempty: if Xi for every iI, then iIXi. Here iIXi is the set of functions f with domain I such that f(i)Xi for every iI; when a family of nonempty sets is indexed by itself, such an f is precisely a choice function for it.

Verification

Given: A chain and its actual collapse maps, with a named ordinal at an earlier stage; ambient ZFC for the concrete witness.

1.1

For a named actual ordinal ξMα, put τα=ot(Mαξ). F2 gives πα(ξ)=τα, so direct substitution in F1 yields jαβ(τα)=πβ(ξ)=ot(Mβξ)=τβ. On a predecessor ηMαξ, its order position ot(Mαη) is sent to ot(Mβη). These equalities specify the induced order embedding.

F1F2given
2.1

For three stages the calculation is jβγ(jαβ(u))=πγ(πβ1(πβ(πα1(u))))=πγ(πα1(u))=jαγ(u), with inclusions understood at the displayed domain changes. For two identical stages this computes the identity. If the named ordinal has τατβ, step 1.1 moves it, whereas literal inclusion would fix it. Thus inclusion requires additional agreement of collapse values and does not follow from the conjugation formula.

step 1.1algebra
3.1

Here is a chain for which the values differ. In ZFC let κ=(ω) from F4, and put θ=κ+ω. The transitive infinite set Vθ contains κ and satisfies Extensionality: all members of each of its elements remain in its domain, so internal agreement of members is actual agreement. F3, with the singleton parameter set {κ} and the AC assumption F5, supplies a countable XVθ containing κ. Take the two-stage chain M0=X, M1=Vθ. F2 makes π1 the identity, and gives τ=π0(κ)=ot(Xκ). This ordinal injects into ω: compose the inverse order isomorphism with a countable enumeration inverse for X. F4 says κ does not inject into ω, so τκ. Step 1.1 now calculates j01(τ)=κτ. This elementary transported map moves an element of its domain and therefore is not literal inclusion.

F2F3F4F5step 1.1

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