Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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.

The Five Lemma for modules

Statement

In a commutative diagram with exact rows

ABCDEA0B0C0D0E0uavbwcxdeu0v0w0x0

the middle map c is injective if a is surjective and b,d are injective, and it is surjective if b,d are surjective and e is injective. In particular, if a,b,d,e are isomorphisms, then c is an isomorphism.

Facts & Assumptions

Given: The commutative diagram in the statement, with exact rows.

Diagram: u ⁣:A→B, v ⁣:B→C, w ⁣:C→D, x ⁣:D→E, u′ ⁣:A′→B′, v′ ⁣:B′→C′, w′ ⁣:C′→D′, x′ ⁣:D′→E′, a ⁣:A→A′, b ⁣:B→B′, c ⁣:C→C′, d ⁣:D→D′, e ⁣:E→E′.

[C1]

b∘u=u′∘a (given).

[C2]

c∘v=v′∘b (given).

[C3]

d∘w=w′∘c (given).

[C4]

e∘x=x′∘d (given).

[L1]

In such a diagram, a surjective with b,d injective implies c injective, while b,d surjective with e injective implies c surjective (The injective and surjective Four Lemmas).

Proof

technique · direct
1.1

Under the first set of hypotheses, the injective Four Lemma [L1] applied to the diagram [C1] to [C4] gives that c is injective.

C1C2C3C4L1
1.2

Under the second set of hypotheses, the surjective Four Lemma [L1] applied to the same diagram gives that c is surjective.

C1C2C3C4L1
2.1

If a,b,d,e are isomorphisms, then a,b,d meet the first hypotheses and b,d,e meet the second; steps 1.1 and 1.2 make c both injective and surjective, hence an isomorphism.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

3 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