LemmaStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)

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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.

Every natural number is a transitive set and is not a member of itself

Statement

For every $n \in N$ : (a) $n$ is transitive, i.e. $x \in n \Rightarrow x \subseteq n$ ; and (b) $n \notin n$ .

Facts & Assumptions

Given: $N$ (The natural numbers $N$ (von Neumann)) with $0 = \emptyset$ , $σ (n) = n \cup {n}$ ; the induction principle (The principle of mathematical induction). Let $P (n)$ be the conjunction " $n$ is transitive and $n \notin n$ ".

[L1]

Induction: if $P (0)$ and $P (n) \Rightarrow P (σ (n))$ , then $P$ holds for all $n$ (The principle of mathematical induction).

Proof

technique · induction

1.1base

Base $P (0)$ : $0 = \emptyset$ has no members, so it is vacuously transitive and $\emptyset \notin \emptyset$ .

1.2ih

Inductive hypothesis: $P (n)$ holds, i.e. $n$ is transitive and $n \notin n$ .

2.1step 1.2

$σ (n) = n \cup {n}$ is transitive: take $x \in σ (n)$ ; if $x \in n$ then $x \subseteq n \subseteq σ (n)$ by the inductive hypothesis transitivity; if $x = n$ then $x = n \subseteq σ (n)$ since $n \subseteq n \cup {n}$ ; either way $x \subseteq σ (n)$ .

2.2step 1.2

$σ (n) \notin σ (n)$ : suppose $σ (n) \in σ (n) = n \cup {n}$ , so $σ (n) \in n$ or $σ (n) = n$ ; if $σ (n) = n$ then $n \in n$ (as $n \in {n} \subseteq σ (n) = n$ ), contradicting the inductive hypothesis $n \notin n$ ; if $σ (n) \in n$ then $σ (n) \subseteq n$ by transitivity of $n$ , and since $n \in n \cup {n} = σ (n) \subseteq n$ we get $n \in n$ , again contradicting $n \notin n$ ; hence $σ (n) \notin σ (n)$ .

3.1step 2.1step 2.2discharge-induction∎

Hence $P (σ (n))$ holds; by induction [L1], $P (n)$ for all $n \in N$ : every natural number is transitive and satisfies $n \notin n$ .

Every natural number is a transitive set and is not a member of itself

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources