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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.

Formal Set-Theoretic Syntax, Structures, and Satisfaction: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These examples calculate satisfaction in small structures, expose variable capture and the failure of agreement at a single assignment under substitution, and evaluate empty-class guards. The final refutation concerns the explicitly stated truth-with-parameters scheme; it does not presume a sentence-only arithmetization theorem.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Parsing and evaluating a membership formula

Example

In the membership structure on M={0,1}, where 0=∅ and 1={0}, let ϕ=∃v1(v1∈v0). Then FV⁡(ϕ)={v0}. It is false when s(v0)=0 and true when s(v0)=1.

Facts & Assumptions

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement.

[F1]

Assignments agreeing on FV⁡(e) give equal denotations when e is a term and the same truth value when e is a formula. Also, passage to a reduct preserves denotations and satisfaction for expressions in the smaller signature. In particular sentence truth is independent of assignment; the truth of a formula can be specified by any tuple assigning all its free variables. (Coincidence for term values and satisfaction)

Verification

1.1F1

The outer code is (exists,v1,(rel,∈,(var,v1),(var,v0))). The body has both variables free; the outer binder removes v1. By coincidence all other assignment values are irrelevant.

2.1step 1.1given∎

The only membership pair in M2 is (0,1). For value 0 at v0, neither candidate witness is a member of 0. For value 1, witness 0 works. These are the two asserted truth values.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Variable capture changes satisfaction

Statement refuted

False claim: replacing free occurrences by a variable without first avoiding capture always has the intended substitution semantics. In ϕ=∃v2(v2∈v1), naively substitute v2 for v1.

Facts & Assumptions

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement refuted.

[F1]

There is a canonical total capture-avoiding substitution ϕ⟨t/x⟩ on formulas, obtained by renaming obstructing binders using least unused variable indices. It replaces the original free occurrences of x and satisfies M,s⊨ϕ⟨t/x⟩  ⟺  M,s[x:=⟦t⟧s]⊨ϕ. Canonicity refers to the specified coding and traversal, not to literal invariance under other fresh-variable conventions. (Canonical capture-avoiding substitution)

Counterexample

1.1F1

In M={∅,{∅}} with restricted membership, take s(v2)={∅}. The intended substitution truth, supplied by capture-avoiding substitution, is the truth of ϕ after setting v1 to s(v2). It is true, with witness ∅.

2.1step 1.1given∎

Naive replacement gives ∃v2(v2∈v2), false since neither carrier element belongs to itself. Renaming the binder to v3 first gives ∃v3(v3∈v2), true at the displayed assignment. Thus the naive operation changes the required truth value.

ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Relativization to the empty class

Example

For C={z:z≠z}, the relativization of ∃x(x=x) is false and that of ∀x(x=x) is true. The empty class is not an admitted structure.

Facts & Assumptions

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement.

[F1]

Fix a pure membership formula δ(z,p⃗) defining a class C={z:δ(z,p⃗)}. This is eliminable notation for a predicate, not a class object. For a fixed pure membership formula ϕ, first rename its binders away from the parameter variables p⃗. Define ϕC by keeping atoms, commuting with Boolean constructors, and setting (∃xψ)C=∃x(δ(x,p⃗)∧ψC),(∀xψ)C=∀x(δ(x,p⃗)→ψC). Copies of δ(x,p⃗) are inserted by capture-avoiding substitution and fresh internal bound variables. In the set case C=M use the predicate z∈M, with a fresh parameter variable for M. This operation is meaningful even for an empty class. It does not make the empty class an admitted structure: carriers of structures remain nonempty. For proper classes, evaluation of ϕC means a separate ambient formula for each fixed ϕ, not a uniform universe satisfaction relation. Conventions and prerequisites: prop-capture-avoiding-substitution. (Relativization to sets and definable classes)

Verification

1.1F1

The existential relativization is ∃x(x≠x∧x=x). Its matrix is false at every set, so it has no witness.

2.1F1∎

The universal relativization is ∀x(x≠x→x=x). Its antecedent is always false, so it is true. These computations concern guarded formulas; the nonempty-carrier convention remains in force.

CounterexampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Agreement at one assignment is insufficient

Statement refuted

False claim: if two formulas agree at one assignment, they still agree there after the same free-for substitution. Use the two-element equality structure M={0,1} with constant cM=1, distinct variables x,y, and s(x)=s(y)=0.

Facts & Assumptions

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement refuted.

[F1]

If t is free for x in ϕ, then for every structure M and assignment s, M,s⊨ϕ[t/x]  ⟺  M,s[x:=⟦t⟧s]⊨ϕ. (Free-for substitution commutes with satisfaction)

Counterexample

1.1F1

At s, the formulas x=y and y=y are both true. The constant c has no variables, so is free for x in each formula. The free-for identity evaluates substitution by changing s(x) to 1.

2.1F1step 1.1∎

After substitution the formulas are c=y and y=y. Their truth values are respectively false (1≠0) and true (0=0). The original agreement at s imposes no agreement at the updated assignment, which is exactly the missing hypothesis.

False statementConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

V has a definable truth predicate

Statement

False statement, in the truth-with-parameters sense: there is a pure membership formula T(w,x,y,z) and a set t such that for every pure membership formula ψ with free variables among x,y and every pair of sets r,s,

ψ(r,s)  ⟺  T(⌜ψ⌝,r,s,t).

The displayed universal demand is a scheme of biconditionals for the proposed T,t.

Facts & Assumptions

Given: Work in ZF unless the statement explicitly weakens or supplements it; fix the objects and hypotheses of the statement.

[F1]

Set satisfaction and relativization must be distinguished. Set satisfaction is uniform in the code of a formula and the data of a set structure. Relativization to a definable proper class supplies an ambient formula separately for each fixed input formula. Writing V⊨ϕ(a⃗) in this latter sense merely abbreviates ϕ(a⃗). For the truth-with-parameters interface, a proposed pure membership formula T(w,x,y,z) and a set parameter t would have to satisfy, for every pure membership formula ψ with free variables among x,y and all sets r,s, the biconditional ψ(r,s)↔T(⌜ψ⌝,r,s,t). Here ⌜ψ⌝ is its finite set code. This is a scheme of requirements, not a single first-order assertion quantifying over ambient truths. The companion refutation tests this exact scheme by one formula built from the proposed T. No sentence-only arithmetized diagonal lemma or representability theorem is asserted here. Conventions and prerequisites: thm-relativization-and-set-satisfaction. (Set truth and the Tarski interface)

[F2]

There is a canonical total capture-avoiding substitution ϕ⟨t/x⟩ on formulas, obtained by renaming obstructing binders using least unused variable indices. It replaces the original free occurrences of x and satisfies M,s⊨ϕ⟨t/x⟩  ⟺  M,s[x:=⟦t⟧s]⊨ϕ. Canonicity refers to the specified coding and traversal, not to literal invariance under other fresh-variable conventions. (Canonical capture-avoiding substitution)

Refutation

1.1F1F2

Fix a proposed T,t. Rename bound variables and use capture-avoiding simultaneous replacement of its four free slots to form the pure membership formula ψ(x,y)=¬T(x,x,y,y). This notation means slot replacement in T, not an added predicate symbol. Its only possible free variables are x,y, and its finite code q=⌜ψ⌝ is a set.

2.1step 1.1F1∎

The demanded instance for this ψ, r=q and s=t gives ψ(q,t)  ⟺  T(q,q,t,t). The definition of ψ gives ψ(q,t)  ⟺  ¬T(q,q,t,t). Therefore that one instance equates a proposition with its negation, which is impossible in classical logic. This refutes every proposed T,t without a sentence-only arithmetization theorem.

Sources