TheoremStatement: AI-adaptedProof: 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.

Under Foundation, $x \notin x$ for every set $x$ , there are no sets with $x \in y \in x$ , and there are no sets with $x \in y \in z \in x$

Statement

Assume the Axiom of Foundation. Then:

(i) $x \notin x$ for every set $x$ ;
(ii) there are no sets $x$ and $y$ with $x \in y$ and $y \in x$ ;
(iii) there are no sets $x$ , $y$ and $z$ with $x \in y$ , $y \in z$ and $z \in x$ .

Cycles of length greater than three are not treated here: stating "there is no membership cycle of any finite length" needs a finite sequence, that is a function on a natural number, and neither notion is available at this point in the reading order.

Facts & Assumptions

Given: the Axiom of Foundation.

[L1]

Every set with at least one member has a member $s$ that shares no member with it (The Axiom of Foundation: $\forall S (\exists t (t \in S) \to \exists s (s \in S \land \neg \exists u (u \in s \land u \in S)))$ ).

[L2]

${x, y}$ is the set whose elements are exactly $x$ and $y$ , and ${x} : = {x, x}$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L3]

$z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ( $⋃ \emptyset = \emptyset$ , $⋃ {a} = a$ , $⋃ {a, b} = a \cup b$ , $⋂ {a} = a$ , and $⋂ {a, b} = a \cap b$ ).

[L4]

$a \cup b : = ⋃ {a, b}$ (The union $⋃ x$ of a set, and the binary union $a \cup b : = ⋃ {a, b}$ ).

Proof

technique · contradiction

1.1assume-contra

Suppose the statement fails, so that at least one of the following holds: there is a set $x$ with $x \in x$ ; there are sets $x$ and $y$ with $x \in y$ and $y \in x$ ; there are sets $x$ , $y$ and $z$ with $x \in y$ , $y \in z$ and $z \in x$ .

2.1L1L2step 1.1

Suppose $x \in x$ and put $S : = {x}$ . Its only member is $x$ , so $S$ has a member and Foundation applies; the member it supplies must be $x$ , so $x$ shares no member with $S$ . But $x \in x$ and $x \in S$ , so $x$ is a member of both.

2.2L1L2step 1.1

Suppose $x \in y$ and $y \in x$ , and put $S : = {x, y}$ . Foundation supplies a member of $S$ sharing no member with $S$ , and that member is $x$ or $y$ . If it is $x$ , then $y \in x$ and $y \in S$ make $y$ a member of both; if it is $y$ , then $x \in y$ and $x \in S$ make $x$ a member of both.

2.3L1L2L3L4step 1.1

Suppose $x \in y$ , $y \in z$ and $z \in x$ , and put $S : = {x, y} \cup {z}$ , whose members are exactly $x$ , $y$ and $z$ . Foundation supplies a member of $S$ sharing no member with $S$ . If it is $x$ , then $z \in x$ and $z \in S$ ; if it is $y$ , then $x \in y$ and $x \in S$ ; if it is $z$ , then $y \in z$ and $y \in S$ . In each case that member shares a member with $S$ .

3.1step 2.1step 2.2step 2.3discharge-contradiction∎

Each of the three suppositions contradicts the member Foundation supplies, so none of them holds: $x \notin x$ for every $x$ , no sets satisfy $x \in y \in x$ , and no sets satisfy $x \in y \in z \in x$ .

Under Foundation, $x \notin x$ for every set $x$ , there are no sets with $x \in y \in x$ , and there are no sets with $x \in y \in z \in x$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Under Foundation, x∉xx∈/x for every set xx, there are no sets with x∈y∈xx∈y∈x, and there are no sets with x∈y∈z∈xx∈y∈z∈x

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Under Foundation, $x \notin x$ for every set $x$ , there are no sets with $x \in y \in x$ , and there are no sets with $x \in y \in z \in x$