PropositionStatement: 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.

$P (\emptyset) = {\emptyset}$ ; $\emptyset \in P (a)$ and $a \in P (a)$ ; $a \subseteq b$ if and only if $P (a) \subseteq P (b)$ ; $⋃ P (a) = a$ ; $⋂ P (a) = \emptyset$ ; and $P (a \cap b) = P (a) \cap P (b)$ while only $P (a) \cup P (b) \subseteq P (a \cup b)$ holds

Statement

For all sets $a$ and $b$ :

(i) $P (\emptyset) = {\emptyset}$ ;
(ii) $\emptyset \in P (a)$ and $a \in P (a)$ ;
(iii) $a \subseteq b$ if and only if $P (a) \subseteq P (b)$ ;
(iv) $⋃ P (a) = a$ ;
(v) $⋂ P (a) = \emptyset$ ;
(vi) $P (a \cap b) = P (a) \cap P (b)$ ;
(vii) $P (a) \cup P (b) \subseteq P (a \cup b)$ ;
(viii) the inclusion in (vii) is an equality if and only if $a \subseteq b$ or $b \subseteq a$ .

Facts & Assumptions

Given: sets $a$ and $b$ .

[L1]

$z \in P (x)$ holds if and only if $z \subseteq x$ (The power set $P (x) = {z : z \subseteq x}$ ).

[L2]

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

[L3]

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

[L4]

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

[L5]

If $x \subseteq y$ and $y \subseteq z$ then $x \subseteq z$ ( $\emptyset \subseteq x$ , $x \subseteq x$ , inclusion is transitive, and $x = y$ if and only if $x \subseteq y$ and $y \subseteq x$ ).

[L6]

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

[L7]

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).

[L8]

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

[L9]

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

[L10]

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

[L11]

${x} : = {x, x}$ , the singleton of $x$ , is the set whose only element is $x$ (The unordered pair ${x, y}$ and the singleton ${x} = {x, x}$ ).

[L12]

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

Proof

technique · direct

1.1L1L2L10L11L12

Claim (i): $z \in P (\emptyset)$ holds exactly when $z \subseteq \emptyset$ , that is, when every element of $z$ lies in $\emptyset$ ; since $\emptyset$ has no elements this says $z$ has no elements, hence $z = \emptyset$ . So the elements of $P (\emptyset)$ are exactly $\emptyset$ , and $P (\emptyset) = {\emptyset}$ .

1.2L1L3L4

Claim (ii): $\emptyset \subseteq a$ and $a \subseteq a$ , so both $\emptyset$ and $a$ are elements of $P (a)$ .

1.3L1L2L4L5

Claim (iii): if $a \subseteq b$ and $z \in P (a)$ , then $z \subseteq a$ and $a \subseteq b$ give $z \subseteq b$ , so $z \in P (b)$ ; conversely if $P (a) \subseteq P (b)$ then $a \in P (a)$ , since $a \subseteq a$ , so $a \in P (b)$ , which says $a \subseteq b$ .

1.4L1L2L4L6L12

Claim (iv): $z \in ⋃ P (a)$ holds exactly when $z$ belongs to some element of $P (a)$ , that is, to some $s$ with $s \subseteq a$ , and any such $z$ lies in $a$ ; conversely if $z \in a$ then $z$ lies in the element $a$ of $P (a)$ . So the two sets have the same elements.

1.5L1L3L7L10

Claim (v): $\emptyset \in P (a)$ , so $P (a) \neq \emptyset$ and $⋂ P (a)$ is defined; an element of it would have to belong to every element of $P (a)$ , in particular to $\emptyset$ , which has no elements. So $⋂ P (a)$ has no elements and equals $\emptyset$ .

1.6L1L2L9L12

Claim (vi): $z \in P (a \cap b)$ holds exactly when every element of $z$ lies in $a \cap b$ , that is, lies in $a$ and in $b$ ; this is exactly $z \subseteq a$ together with $z \subseteq b$ , that is, $z \in P (a)$ and $z \in P (b)$ , which is $z \in P (a) \cap P (b)$ .

1.7L1L2L8

Claim (vii): if $z \in P (a) \cup P (b)$ then $z \subseteq a$ or $z \subseteq b$ ; in either case every element of $z$ lies in $a$ or in $b$ , hence in $a \cup b$ , so $z \in P (a \cup b)$ .

2.1L1L2L4L5L8step 1.7

Claim (viii): if $a \subseteq b$ then the elements of $a \cup b$ are exactly those of $b$ , so $z \subseteq a \cup b$ is equivalent to $z \subseteq b$ and $P (a \cup b) = P (b) \subseteq P (a) \cup P (b)$ , which with the reverse inclusion gives equality; the case $b \subseteq a$ is symmetric. Conversely, if equality holds then $a \cup b \subseteq a \cup b$ puts $a \cup b$ in $P (a) \cup P (b)$ , so $a \cup b \subseteq a$ or $a \cup b \subseteq b$ , and since $b \subseteq a \cup b$ and $a \subseteq a \cup b$ these give $b \subseteq a$ and $a \subseteq b$ respectively.

$P (\emptyset) = {\emptyset}$ ; $\emptyset \in P (a)$ and $a \in P (a)$ ; $a \subseteq b$ if and only if $P (a) \subseteq P (b)$ ; $⋃ P (a) = a$ ; $⋂ P (a) = \emptyset$ ; and $P (a \cap b) = P (a) \cap P (b)$ while only $P (a) \cup P (b) \subseteq P (a \cup b)$ holds

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

P(∅)={∅}P(∅)={∅}; ∅∈P(a)∅∈P(a) and a∈P(a)a∈P(a); a⊆ba⊆b if and only if P(a)⊆P(b)P(a)⊆P(b); ⋃P(a)=a⋃P(a)=a; ⋂P(a)=∅⋂P(a)=∅; and P(a∩b)=P(a)∩P(b)P(a∩b)=P(a)∩P(b) while only P(a)∪P(b)⊆P(a∪b)P(a)∪P(b)⊆P(a∪b) holds

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

$P (\emptyset) = {\emptyset}$ ; $\emptyset \in P (a)$ and $a \in P (a)$ ; $a \subseteq b$ if and only if $P (a) \subseteq P (b)$ ; $⋃ P (a) = a$ ; $⋂ P (a) = \emptyset$ ; and $P (a \cap b) = P (a) \cap P (b)$ while only $P (a) \cup P (b) \subseteq P (a \cup b)$ holds