Alphabeta Math
CounterexampleConstruction: AI-adaptedVerification: 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.

Sets a and b with P(a)∪P(b)⊊P(a∪b)

Statement refuted

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

Facts & Assumptions

Given: a:={∅} and b:={{∅}}.

[L3]

z∈P(x) holds if and only if z⊆x (The power set P(x)={ z:z⊆x }).

[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}).

[L7]

There is exactly one set with no elements (There is exactly one set with no elements, written ∅).

Counterexample

technique · direct
1.1

Neither of the two sets is included in the other. The only element of a is ∅ and the only element of b is {∅}, and these differ because {∅} has an element while ∅ has none; so ∅∈a with ∅∉b, and {∅}∈b with {∅}∉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∪b itself: it is a subset of a∪b, hence an element of P(a∪b), whereas a∪b⊆a would force b⊆a and a∪b⊆b would force a⊆b, both excluded by step 1.1; so a∪b is in neither P(a) nor P(b).

L3L4L6L8step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

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