Alphabeta Math
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.

A×(B∪C)=(A×B)∪(A×C), A×(B∩C)=(A×B)∩(A×C), A×(B∖C)=(A×B)∖(A×C), (A∩B)×(C∩D)=(A×C)∩(B×D); A×B=∅ if and only if A=∅ or B=∅; and for nonempty A and B, A×B⊆C×D if and only if A⊆C and B⊆D

Statement

For all sets A, B, C, D:

  • (i) A×(B∪C)=(A×B)∪(A×C);
  • (ii) A×(B∩C)=(A×B)∩(A×C);
  • (iii) A×(B∖C)=(A×B)∖(A×C);
  • (iv) (A∩B)×(C∩D)=(A×C)∩(B×D);
  • (v) A×B=∅ if and only if A=∅ or B=∅;
  • (vi) if A≠∅ and B≠∅, then A×B⊆C×D if and only if A⊆C and B⊆D.

Facts & Assumptions

Given: sets A, B, C, D.

[L1]

z∈A×B holds if and only if z=(a,b) for some a∈A and some b∈B (The Cartesian product A×B:={ z∈P(P(A∪B)):∃a∈A ∃b∈B z=(a,b) }).

[L2]

(a,b)=(c,d) if and only if a=c and b=d ((a,b)=(c,d) if and only if a=c and b=d).

[L7]

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

[L8]

If every z satisfies z∈x if and only if z∈y, then x=y (The Axiom of Extensionality: ∀x ∀y (∀z (z∈x↔z∈y)→x=y)).

[L9]
[L10]

a∪b:=⋃{a,b}, and ⋃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∪b:=⋃{a,b}).

[L11]

For x≠∅, ⋂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∩b:=⋂{a,b}, and disjointness).

Proof

technique · direct
1.1

Membership criterion: for all sets a and b, (a,b)∈A×B holds if and only if a∈A and b∈B. Indeed, an element of A×B is a pair (a′,b′) with a′∈A and b′∈B, and (a,b)=(a′,b′) forces a=a′ and b=b′; the converse is immediate from the description of A×B. Every element of a product is a pair, so it suffices in each identity below to compare pairs.

L1L2L9
2.1

Claim (i): (a,t)∈A×(B∪C) holds exactly when a∈A and t∈B or t∈C, that is, exactly when (a,t)∈A×B or (a,t)∈A×C.

L3L8L10step 1.1
2.2

Claim (ii): (a,t)∈A×(B∩C) holds exactly when a∈A, t∈B and t∈C, that is, exactly when (a,t)∈A×B and (a,t)∈A×C.

L4L8L11step 1.1
2.3

Claim (iii): (a,t)∈A×(B∖C) holds exactly when a∈A, t∈B and t∉C. On the other side, (a,t)∈(A×B)∖(A×C) holds exactly when a∈A, t∈B, and it is not the case that a∈A and t∈C; given a∈A, that last condition is t∉C.

L5L8step 1.1
2.4

Claim (iv): (u,v)∈(A∩B)×(C∩D) holds exactly when u∈A, u∈B, v∈C and v∈D, that is, exactly when (u,v)∈A×C and (u,v)∈B×D.

L4L8L11step 1.1
2.5

Claim (v): if A=∅ or B=∅ then no pair satisfies the membership criterion, so A×B has no elements and equals ∅; conversely if both are nonempty, fix a∈A and b∈B, and then (a,b)∈A×B.

L7step 1.1
2.6

Claim (vi): assume A≠∅ and B≠∅. If A×B⊆C×D, fix b0∈B; for any a∈A the pair (a,b0) lies in A×B, hence in C×D, so a∈C, and A⊆C follows; fixing a0∈A and running the same argument on the second coordinate gives B⊆D. Conversely, if A⊆C and B⊆D, then any (a,b)∈A×B has a∈C and b∈D, so it lies in C×D.

L6L7step 1.1
3.1

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

step 2.1step 2.2step 2.3step 2.4step 2.5step 2.6∎

Depends on

Used by

Dependency tree · two levels

19 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