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

1.1

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 $P (\emptyset) = {\emptyset}$ .

L1L2L10L11L12

1.2

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

L1L3L4

1.3

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 $a \subseteq b$ .

L1L2L4L5

1.4

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.

L1L2L4L6L12

1.5

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 $\emptyset$ .

L1L3L7L10

1.6

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 $z \in P (a) \cap P (b)$ .

L1L2L9L12

1.7

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 $z \in P (a \cup b)$ .

L1L2L8

2.1

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.

L1L2L4L5L8step 1.7

3.1

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

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6step 1.7step 2.1∎

Depends on

Used by

Sets a and b with P(a) ∪ P(b) ⊊ P(a ∪ b) Counterexample
P(∅) = {∅}, P({∅}) = {∅,{∅}}, and P({∅,{∅}}) listed in full Example

Dependency tree · two levels

15 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

Power set (Wikipedia) (standard reference, not scraped)
B. Kaya, MATH 320 Set Theory (METU), §1.2 (standard reference, not scraped)
Algebra of sets (Wikipedia) (standard reference, not scraped)

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