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

1.1

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 $\emptyset \subseteq x$ .

L1L3

1.2

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

1.3

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 $z$ .

1.4

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 $x = y$ .

L1L2

1.5

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.

2.1

Claims (i) to (iv) are established, which is the statement.

step 1.1step 1.2step 1.3step 1.4step 1.5∎

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

B. Kaya, MATH 320 Set Theory (METU), §1.2 (standard reference, not scraped)
Subset (Wikipedia) (standard reference, not scraped)
C. Wilson, A Brief Introduction to ZFC (Chicago REU 2016), §2.1 (standard reference, not scraped)

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