Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (openai/gpt-5.4)audited 2026-07-25
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.

Order is preserved by adding a constant and by adding inequalities

Statement

Let FF be an ordered field (Ordered field) with positive cone PP, and let a,b,c,dFa, b, c, d \in F.

  1. Translation invariance. If a<ba < b then a+c<b+ca + c < b + c.
  2. Adding inequalities. If a<ba < b and c<dc < d then a+c<b+da + c < b + d.

Facts & Assumptions

Given: An ordered field FF with positive cone PP, and elements a,b,c,dFa, b, c, d \in F.

[L1]

For x,yFx, y \in F, the relation x<yx < y means yxPy - x \in P (Ordered field).

[L2]

PP is closed under addition: if u,vPu, v \in P then u+vPu + v \in P (axiom O2 of Ordered field).

Proof

technique · direct
1.1

Assume a<ba < b; by the definition of the order this means baPb - a \in P.

assume-hypL1
1.2

For every cFc \in F the field identities give (b+c)(a+c)=ba(b + c) - (a + c) = b - a.

algebra
1.3

Assume moreover c<dc < d; by the definition of the order this means dcPd - c \in P.

assume-hypL1
1.4

The field identities give (b+d)(a+c)=(ba)+(dc)(b + d) - (a + c) = (b - a) + (d - c).

algebra
2.1

Hence (b+c)(a+c)=baP(b + c) - (a + c) = b - a \in P, which is exactly a+c<b+ca + c < b + c, proving claim 1.

step 1.1step 1.2L1
2.2

Since baPb - a \in P and dcPd - c \in P, closure under addition gives (ba)+(dc)P(b - a) + (d - c) \in P.

step 1.1step 1.3L2
3.1

Therefore (b+d)(a+c)=(ba)+(dc)P(b + d) - (a + c) = (b - a) + (d - c) \in P, which is exactly a+c<b+da + c < b + d, proving claim 2.

step 1.4step 2.2L1

Depends on

Used by

…and 130 more results.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 2 results over 2 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