CorollaryStatement: 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.

There is no set $y$ with $x \in y \leftrightarrow \forall s (s \in \emptyset \to x \in s)$ , so $⋂ \emptyset$ is undefined

Statement

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

$x \in y \leftrightarrow \forall s (s \in \emptyset \to x \in s) .$

The defining condition for $⋂ x$ therefore determines no set when $x = \emptyset$ , which is why $⋂ \emptyset$ is left undefined.

Facts & Assumptions

Given: nothing beyond the results cited below.

[L1]

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

[L2]

There is no set $U$ such that $y \in U$ for every set $y$ (There is no set $U$ with $y \in U$ for every set $y$ ).

[L3]

For $x \neq \emptyset$ , $⋂ x$ is the set whose elements are exactly the sets belonging to every element of $x$ (The intersection $⋂ x$ of a nonempty set, the binary intersection $a \cap b : = ⋂ {a, b}$ , and disjointness).

Proof

technique · contradiction

1.1assume-contra

Suppose there is a set $y$ such that, for every set $x$ , $x \in y$ holds if and only if $x$ belongs to every element of $\emptyset$ .

1.2L1

$\emptyset$ has no elements, so for every set $x$ the condition " $x$ belongs to every element of $\emptyset$ " is satisfied vacuously.

2.1step 1.1step 1.2

Combining, every set $x$ satisfies $x \in y$ .

3.1L2L3step 2.1discharge-contradiction∎

A set with every set as an element does not exist, so the supposition fails; the condition defining $⋂$ therefore determines no set at $\emptyset$ , and $⋂ \emptyset$ is undefined.

There is no set $y$ with $x \in y \leftrightarrow \forall s (s \in \emptyset \to x \in s)$ , so $⋂ \emptyset$ is undefined

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

There is no set yy with x∈y↔∀s (s∈∅→x∈s)x∈y↔∀s(s∈∅→x∈s), so ⋂∅⋂∅ is undefined

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

There is no set $y$ with $x \in y \leftrightarrow \forall s (s \in \emptyset \to x \in s)$ , so $⋂ \emptyset$ is undefined