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Proof of Lemma 3, p. 14 — with K₂ = M³(M + 1), H₂ = M(M + 1) as in Lemma 2, every plan with q₂ ≠ M costs more than the threshold

Proved
JRPHard.RangeDiv.item2_pinned

by mikedeng1 · Oct 15, 2026 · Mathlib 0df444a (Lean v4.33.1)

eoqjoint-replenishmentp2o-batch-pfp2ap2o-gran-per-chapterp2o-plan-paperp2o-v1range-divisor

Let M≥U≥L>0M \ge U \ge L > 0M≥U≥L>0 be integers with L+U<22LUL + U < 2\sqrt{2LU}L+U<22LU​, and consider the periodic instance JRPM,U,L\mathrm{JRP}_{M,U,L}JRPM,U,L​ with K0=LUK_0 = LUK0​=LU, K1=0K_1 = 0K1​=0, H1=1H_1 = 1H1​=1, K2=M3(M+1)K_2 = M^3(M+1)K2​=M3(M+1), H2=M(M+1)H_2 = M(M+1)H2​=M(M+1). For all positive integers q1,q2q_1, q_2q1​,q2​ with q2≠Mq_2 \neq Mq2​=M,

JRPM,U,L(q1,q2)>L+U+2M2(M+1).\mathrm{JRP}_{M,U,L}(q_1, q_2) > L + U + 2M^2(M+1).JRPM,U,L​(q1​,q2​)>L+U+2M2(M+1).

So no plan with q2≠Mq_2 \ne Mq2​=M reaches the threshold zM,L,Uz_{M,L,U}zM,L,U​: item 2's heavy EOQ term pins q2=Mq_2 = Mq2​=M, exactly as in the proof of Lemma 2.

Formalization Note The paper says only that K2,H2K_2, H_2K2​,H2​ are "sufficiently large … as in the proof … in Lemma 2". The statement uses Lemma 2's values, so the claim is about one explicit instance.

Preamble
import Mathlib
import Definitions.Def_JRPHard_RangeDiv_Setting
Formal statement
namespace JRPHard.RangeDiv

theorem item2_pinned (M L U : ℕ) (hI : IsRangeDivisorInstance M L U) (q₁ q₂ : ℕ)
    (hq₁ : 0 < q₁) (hq₂ : 0 < q₂) (hne : q₂ ≠ M) :
    (threshold M L U : ℝ) < rangeCost M L U q₁ q₂ := by sorry

end JRPHard.RangeDiv
Source
Schulz and Telha, Integer factorization and Riemann's hypothesis: Why two-item joint replenishment is hard, arXiv:2007.09045v1, p. 14, §3.1, proof of Lemma 3, first paragraph (with p. 11, proof of Lemma 2, "Optimal replenishment cycle for item 2")
Human review
  • Endorsed by Shuze Chen · Oct 15, 2026

    Confirmed by the moderator at approval.

  • Endorsed by mikedeng1 · Oct 15, 2026

    Confirmed by the mission captain (proposal self-audit).

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