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Comparison nodes of the fourth term 3b−2a3b-2a3b−2a

Proved
Erdos195.fourth_term_node

by Ethan Cheung Wang · Oct 23, 2026 · Mathlib 0df444a (Lean v4.33.1)

arithmetic-progressionscombinatoricserdos-problems

Let a≠ba\ne ba=b be integers, c=3b−2ac=3b-2ac=3b−2a, k=ν2(b−a)k=\nu_2(b-a)k=ν2​(b−a). Then: (i) ν2(c−a)=k\nu_2(c-a)=kν2​(c−a)=k and bitk(c)=bitk(b)\mathrm{bit}_k(c)=\mathrm{bit}_k(b)bitk​(c)=bitk​(b); (ii) ν2(c−b)=k+1\nu_2(c-b)=k+1ν2​(c−b)=k+1 and c≡b(mod2k+1)c\equiv b\pmod{2^{k+1}}c≡b(mod2k+1); (iii) for x≠bx\ne bx=b with p=ν2(b−x)p=\nu_2(b-x)p=ν2​(b−x) and 2p∣a−x2^p\mid a-x2p∣a−x (i.e. p≤ν2(a−x)p\le\nu_2(a-x)p≤ν2​(a−x) or x=ax=ax=a):

ν2(c−x)=p,bitp(c)=bitp(b);\nu_2(c-x)=p,\qquad \mathrm{bit}_p(c)=\mathrm{bit}_p(b);ν2​(c−x)=p,bitp​(c)=bitp​(b);

(iv) for x≠ax\ne ax=a with ν2(a−x)+1<ν2(b−x)\nu_2(a-x)+1<\nu_2(b-x)ν2​(a−x)+1<ν2​(b−x): ν2(c−x)=ν2(a−x)+1\nu_2(c-x)=\nu_2(a-x)+1ν2​(c−x)=ν2​(a−x)+1. Thus the comparison node of the fourth term ccc with xxx coincides with that of bbb whenever p≤qp\le qp≤q. Used for completing pairs in tree-order constructions (Erdős Problem 195).

Formalization note: treeLt is from Definitions.Def_Erdos195_tree; Safe, bitOne from Definitions.Def_Erdos195_defs; positions are List.idxOf, ν2\nu_2ν2​ is padicValInt 2.

Preamble
import Mathlib
import Definitions.Def_Erdos195_tree
Formal statement
namespace Erdos195

theorem fourth_term_node (a b : ℤ) (hab : a ≠ b) :
    (padicValInt 2 ((3 * b - 2 * a) - a) = padicValInt 2 (b - a) ∧
      (bitOne (3 * b - 2 * a) (padicValInt 2 (b - a)) ↔ bitOne b (padicValInt 2 (b - a)))) ∧
    (padicValInt 2 ((3 * b - 2 * a) - b) = padicValInt 2 (b - a) + 1 ∧
      (3 * b - 2 * a) % 2 ^ (padicValInt 2 (b - a) + 1) = b % 2 ^ (padicValInt 2 (b - a) + 1)) ∧
    (∀ x : ℤ, x ≠ b → (2 : ℤ) ^ padicValInt 2 (b - x) ∣ a - x →
      padicValInt 2 ((3 * b - 2 * a) - x) = padicValInt 2 (b - x) ∧
      (bitOne (3 * b - 2 * a) (padicValInt 2 (b - x)) ↔ bitOne b (padicValInt 2 (b - x)))) ∧
    (∀ x : ℤ, x ≠ a → padicValInt 2 (a - x) + 1 < padicValInt 2 (b - x) →
      padicValInt 2 ((3 * b - 2 * a) - x) = padicValInt 2 (a - x) + 1) := by sorry

end Erdos195
Source
Erdős Problem 195 (https://www.erdosproblems.com/195); statement from GPT 6 Astra (Oct 23 2026 tree answers), Lean-verified by Grok Bot for Ethan Cheung

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