Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-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.

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FALSE: every list of mnmn pairwise distinct reals has a strictly increasing sublist of length m+1m+1 or a strictly decreasing sublist of length n+1n+1

Statement

FALSE. The statement

for all m,nNm, n \in \mathbb{N}, every pairwise distinct finite list of reals of length mnmn has a strictly increasing sublist of length m+1m+1 or a strictly decreasing sublist of length n+1n+1 (A finite list of reals, and its strictly increasing and strictly decreasing sublists).

This is Every list of mn+1mn+1 pairwise distinct reals has a strictly increasing sublist of length m+1m+1 or a strictly decreasing sublist of length n+1n+1 with the length lowered from mn+1mn+1 to mnmn. The true theorem is sharp, so lowering the length by one destroys it, and it does so at every pair (m,n)(m,n) rather than at some exceptional pair.

Facts & Assumptions

Given: The naturals mm and nn, and lists of reals with their sublists as in A finite list of reals, and its strictly increasing and strictly decreasing sublists.

[L1]

For all m,nNm, n \in \mathbb{N} there is a pairwise distinct list of reals of length mnmn with no strictly increasing sublist of length m+1m+1 and no strictly decreasing sublist of length n+1n+1 (For all mm and nn there is a list of mnmn pairwise distinct reals with no strictly increasing sublist of length m+1m+1 and no strictly decreasing sublist of length n+1n+1).

[L2]

Every pairwise distinct list of reals of length mn+1mn+1 has a strictly increasing sublist of length m+1m+1 or a strictly decreasing sublist of length n+1n+1 (Every list of mn+1mn+1 pairwise distinct reals has a strictly increasing sublist of length m+1m+1 or a strictly decreasing sublist of length n+1n+1).

[L3]

22=42 \cdot 2 = 4 (Multiplication of natural numbers, Order on the natural numbers), and the terms of a list are elements of the ordered field R\mathbb{R} (Ordered field).

Refutation

technique · constructive
1.1

Read the displayed claim at m=n=2m = n = 2: every pairwise distinct list of reals of length 44 would have a strictly increasing sublist of length 33 or a strictly decreasing sublist of length 33.

givenL3
1.2

By [L1] at m=n=2m = n = 2 there is a pairwise distinct list of reals of length 22=42\cdot 2 = 4 with no strictly increasing sublist of length 33 and no strictly decreasing sublist of length 33.

L1L3construct
2.1

That list refutes the reading of step 1.1, so the displayed claim is false. The same argument runs at every pair (m,n)(m,n), since [L1] produces a witness for each of them; the claim therefore fails everywhere, not at an exceptional pair.

step 1.1step 1.2L1
3.1

What survives is [L2]: the conclusion holds once the length is raised to mn+1mn+1. So the least length at which the alternative becomes unavoidable is mn+1mn+1, and the claim above is exactly the assertion that it is mnmn.

step 2.1L1L2discharge-construct

Remarks

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 55 results over 20 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