TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-08-06 (claude-opus-5)

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.

There is no $R$ with $x \in R \leftrightarrow x \notin x$ for every $x$

Statement

There is no set $R$ such that, for every set $x$ ,

$x \in R \leftrightarrow x \notin x .$

Facts & Assumptions

Given: the language of set theory, in which $x \in R$ and $x \notin x$ are formulas (The first-order language of set theory: $\in$ , $=$ , formulas with parameters, and class abbreviations).

Proof

technique · contradiction

1.1assume-contra

Suppose there is a set $R$ such that $x \in R$ holds if and only if $x \notin x$ , for every set $x$ .

2.1step 1.1

The hypothesis holds for every set $x$ , and $R$ is a set, so it holds for $x : = R$ : $R \in R$ if and only if $R \notin R$ .

3.1step 2.1discharge-contradiction∎

A statement equivalent to its own negation is contradictory: if $R \in R$ then $R \notin R$ , and if $R \notin R$ then $R \in R$ , so each alternative refutes itself. There is therefore no such $R$ .

Remarks

What fails is unrestricted comprehension, not a particular formula. The argument uses only that $x \notin x$ is a formula of the language and that the supposed $R$ is itself a set, so it refutes any principle asserting that every formula has a set of all its instances. Separation and Replacement build subsets of sets already in hand, which is exactly what blocks Russell's construction records how The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ avoids it.

Depends on

The first-order language of set theory: $\in$ , $=$ , formulas with parameters, and class abbreviations

There is no $R$ with $x \in R \leftrightarrow x \notin x$ for every $x$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 1 level

Sources

There is no RR with x∈R↔x∉xx∈R↔x∈/x for every xx

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 1 level

Sources

There is no $R$ with $x \in R \leftrightarrow x \notin x$ for every $x$