LemmaStatement: 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.

$\emptyset \subseteq x$ , $x \subseteq x$ , inclusion is transitive, and $x = y$ if and only if $x \subseteq y$ and $y \subseteq x$

Statement

For all sets $x$ , $y$ and $z$ :

(i) $\emptyset \subseteq x$ ;
(ii) $x \subseteq x$ ;
(iii) if $x \subseteq y$ and $y \subseteq z$ then $x \subseteq z$ ;
(iv) $x = y$ if and only if $x \subseteq y$ and $y \subseteq x$ .

Facts & Assumptions

Given: sets $x$ , $y$ and $z$ .

[L1]

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

[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 $\emptyset$ ).

Proof

technique · direct

1.1L1L3

Claim (i): no $t$ satisfies $t \in \emptyset$ , so the implication " $t \in \emptyset$ implies $t \in x$ " holds vacuously for every $t$ , which is $\emptyset \subseteq x$ .

1.2L1

Claim (ii): every $t$ with $t \in x$ satisfies $t \in x$ , which is $x \subseteq x$ .

1.3L1

Claim (iii): assume $x \subseteq y$ and $y \subseteq z$ , and let $t \in x$ ; then $t \in y$ by the first inclusion and $t \in z$ by the second, so every element of $x$ is an element of $z$ .

1.4L1L2

Claim (iv), from right to left: assume $x \subseteq y$ and $y \subseteq x$ ; for any $t$ , the first inclusion gives that $t \in x$ implies $t \in y$ and the second gives that $t \in y$ implies $t \in x$ , so $t \in x$ holds if and only if $t \in y$ , and therefore $x = y$ .

1.5L1

Claim (iv), from left to right: assume $x = y$ ; then $t \in x$ and $t \in y$ are the same statement for every $t$ , so each of $x \subseteq y$ and $y \subseteq x$ holds.

$\emptyset \subseteq x$ , $x \subseteq x$ , inclusion is transitive, and $x = y$ if and only if $x \subseteq y$ and $y \subseteq x$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

∅⊆x∅⊆x, x⊆xx⊆x, inclusion is transitive, and x=yx=y if and only if x⊆yx⊆y and y⊆xy⊆x

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$\emptyset \subseteq x$ , $x \subseteq x$ , inclusion is transitive, and $x = y$ if and only if $x \subseteq y$ and $y \subseteq x$