CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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.

Sets $a$ and $b$ with $\mathcal{P}(a) \cup \mathcal{P}(b) \subsetneq \mathcal{P}(a \cup b)$

Statement refuted

Refuted claim: $\mathcal{P}(a) \cup \mathcal{P}(b) = \mathcal{P}(a \cup b)$ for all sets $a$ and $b$ . The witness is $a := \{\varnothing\}$ and $b := \{\{\varnothing\}\}$ , for which the inclusion from left to right is proper: $a \cup b$ is a subset of $a \cup b$ , so it lies in the right-hand side, but it is a subset of neither $a$ nor $b$ .

Facts & Assumptions

Given: $a := \{\varnothing\}$ and $b := \{\{\varnothing\}\}$ .

[L1]

$\mathcal{P}(a) \cup \mathcal{P}(b) \subseteq \mathcal{P}(a \cup b)$ ( $\mathcal{P}(\varnothing) = \{\varnothing\}$ ; $\varnothing \in \mathcal{P}(a)$ and $a \in \mathcal{P}(a)$ ; $a \subseteq b$ if and only if $\mathcal{P}(a) \subseteq \mathcal{P}(b)$ ; $\bigcup \mathcal{P}(a) = a$ ; $\bigcap \mathcal{P}(a) = \varnothing$ ; and $\mathcal{P}(a \cap b) = \mathcal{P}(a) \cap \mathcal{P}(b)$ while only $\mathcal{P}(a) \cup \mathcal{P}(b) \subseteq \mathcal{P}(a \cup b)$ holds).

[L2]

The inclusion $\mathcal{P}(a) \cup \mathcal{P}(b) \subseteq \mathcal{P}(a \cup b)$ is an equality if and only if $a \subseteq b$ or $b \subseteq a$ ( $\mathcal{P}(\varnothing) = \{\varnothing\}$ ; $\varnothing \in \mathcal{P}(a)$ and $a \in \mathcal{P}(a)$ ; $a \subseteq b$ if and only if $\mathcal{P}(a) \subseteq \mathcal{P}(b)$ ; $\bigcup \mathcal{P}(a) = a$ ; $\bigcap \mathcal{P}(a) = \varnothing$ ; and $\mathcal{P}(a \cap b) = \mathcal{P}(a) \cap \mathcal{P}(b)$ while only $\mathcal{P}(a) \cup \mathcal{P}(b) \subseteq \mathcal{P}(a \cup b)$ holds).

[L3]

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

[L4]

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

[L5]

$\{x,y\}$ is the set whose elements are exactly $x$ and $y$ , and $\{x\} := \{x,x\}$ (The unordered pair $\{x,y\}$ and the singleton $\{x\} = \{x,x\}$ ).

[L6]

$z \in a \cup b$ holds if and only if $z \in a$ or $z \in b$ ( $\bigcup \varnothing = \varnothing$ , $\bigcup \{a\} = a$ , $\bigcup \{a,b\} = a \cup b$ , $\bigcap \{a\} = a$ , and $\bigcap \{a,b\} = a \cap b$ ).

[L7]

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

[L8]

$a \cup b := \bigcup\{a,b\}$ (The union $\bigcup x$ of a set, and the binary union $a \cup b := \bigcup \{a,b\}$ ).

Counterexample

technique · direct

1.1

Neither of the two sets is included in the other. The only element of $a$ is $\varnothing$ and the only element of $b$ is $\{\varnothing\}$ , and these differ because $\{\varnothing\}$ has an element while $\varnothing$ has none; so $\varnothing \in a$ with $\varnothing \notin b$ , and $\{\varnothing\} \in b$ with $\{\varnothing\} \notin a$ .

L4L5L7

2.1

The general inclusion holds, and it is an equality exactly when one of the two sets is included in the other; by step 1.1 that fails here, so the inclusion is proper.

L1L2step 1.1

3.1

The witnessing element is $a \cup b$ itself: it is a subset of $a \cup b$ , hence an element of $\mathcal{P}(a \cup b)$ , whereas $a \cup b \subseteq a$ would force $b \subseteq a$ and $a \cup b \subseteq b$ would force $a \subseteq b$ , both excluded by step 1.1; so $a \cup b$ is in neither $\mathcal{P}(a)$ nor $\mathcal{P}(b)$ .

L3L4L6L8step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 17 results over 7 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

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

Sets aaa and bbb with P(a)∪P(b)⊊P(a∪b)\mathcal{P}(a) \cup \mathcal{P}(b) \subsetneq \mathcal{P}(a \cup b)P(a)∪P(b)⊊P(a∪b)