LemmaStatement: 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.

For a set $x \neq \emptyset$ the collection ${z : \forall s (s \in x \to z \in s)}$ is a set, and it does not depend on the member of $x$ used to separate it

Statement

Let $x$ be a set with $x \neq \emptyset$ . Then there is a set whose elements are exactly the sets belonging to every member of $x$ ; that is, the class ${z : \forall s (s \in x \to z \in s)}$ is a set. Moreover, for every $B \in x$ the separated set ${z \in B : \forall s (s \in x \to z \in s)}$ is that same set, so the construction does not depend on which member of $x$ is used.

Facts & Assumptions

Given: a set $x$ with $x \neq \emptyset$ .

[L1]

For any parameters $\overset{ˉ}{p}$ and any set $x$ , there is a set $y$ whose elements are exactly the elements $z$ of $x$ for which $φ (z, \overset{ˉ}{p})$ holds (The Axiom Schema of Separation: for each formula $φ$ , $\forall \overset{ˉ}{p} \forall x \exists y \forall z (z \in y \leftrightarrow (z \in x \land φ (z, \overset{ˉ}{p})))$ ).

[L2]

If every $z$ satisfies $z \in x$ if and only if $z \in y$ , then $x = y$ (The Axiom of Extensionality: $\forall x \forall y (\forall z (z \in x \leftrightarrow z \in y) \to x = y)$ ).

[L3]

There is exactly one set with no elements, written $\emptyset$ (There is exactly one set with no elements, written $\emptyset$ ).

[L4]

$x \subseteq y$ means that every element of $x$ is an element of $y$ (Subset $x \subseteq y$ , proper subset $x ⊊ y$ , and the separation notation ${z \in x : φ (z)}$ ).

Proof

technique · direct

1.1L3givenchoose

If $x$ had no members it would be a set with no elements and hence equal to $\emptyset$ , contrary to hypothesis; so $x$ has a member, and we fix one, $B \in x$ .

2.1L1L4step 1.1

Apply Separation to $B$ with the formula $φ (z, x) : = \forall s (s \in x \to z \in s)$ and the parameter $x$ : the collection $c_{B} : = {z \in B : \forall s (s \in x \to z \in s)}$ is a set, and $c_{B} \subseteq B$ .

3.1step 2.1step 1.1

For every $z$ , $z \in c_{B}$ holds exactly when $z \in B$ and $z$ belongs to every member of $x$ ; since $B$ is itself a member of $x$ , the second condition already forces $z \in B$ , so $z \in c_{B}$ holds exactly when $z$ belongs to every member of $x$ .

4.1L2step 3.1∎

The condition characterising the elements of $c_{B}$ in step 3.1 does not mention $B$ , so for any other member $B^{'}$ of $x$ the set $c_{B^{'}}$ has exactly the same elements as $c_{B}$ and equals it; the class is therefore a set and is independent of the member used to separate it.

For a set $x \neq \emptyset$ the collection ${z : \forall s (s \in x \to z \in s)}$ is a set, and it does not depend on the member of $x$ used to separate it

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For a set x≠∅x=∅ the collection { z:∀s (s∈x→z∈s) }{z:∀s(s∈x→z∈s)} is a set, and it does not depend on the member of xx used to separate it

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

For a set $x \neq \emptyset$ the collection ${z : \forall s (s \in x \to z \in s)}$ is a set, and it does not depend on the member of $x$ used to separate it