Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Finite graph Ramsey theorem: (s+t2s1)(s,t)2\binom{s+t-2}{s-1}\to(s,t)^2 for all positive s,ts,t

Statement

Facts & Assumptions

Given: Positive natural numbers s,ts,t.

[L1]

If m(s1,t)2m\to(s-1,t)^2 and n(s,t1)2n\to(s,t-1)^2, then m+n(s,t)2m+n\to(s,t)^2 for s,t2s,t\ge2 (If m(s1,t)2m\to(s-1,t)^2 and n(s,t1)2n\to(s,t-1)^2, then m+n(s,t)2m+n\to(s,t)^2 for s,t2s,t\ge2).

Proof

technique · induction
1.1

If s=1s=1 or t=1t=1, every nonempty vertex set contains the required one-vertex set in the corresponding colour convention, and the displayed binomial coefficient is 11.

base
1.2

Assume s,t2s,t\ge2 and that the formula holds whenever the sum of the two positive parameters is smaller than s+ts+t. Then (s+t3s2)(s1,t)2\binom{s+t-3}{s-2}\to(s-1,t)^2 and (s+t3s1)(s,t1)2\binom{s+t-3}{s-1}\to(s,t-1)^2 by the induction hypothesis.

ih
2.1

Apply [L1] to the two witnesses in step 1.2 and use [L2] to identify their sum as (s+t2s1)\binom{s+t-2}{s-1}. This gives the displayed arrow for (s,t)(s,t).

step 1.2L1L2
3.1

The base faces and the induction step cover all positive s,ts,t, so the explicit binomial witness works universally.

step 1.1step 2.1discharge-induction

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 69 results over 24 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