The name is absent

for i = 2, ∙ ∙ ∙, n + 1. Notice that

k^t, = max k^t-

JEN ¹

for any i ∈ N.

The following cases need to be addressed:

(1). If 1 ≤ Z ≤ n, we have that for any i ∈ N, it holds

ajx^l-bι = ɑɪx⁰ - bι ⅛ E⅛=ι k^l_hafq(hf

≤ — Цац — ∑^ⁿ_h=ι,h≠ι k^^l_h^aih + ⅛₊ι E⅛=ι ^aih

≤ -Цац - ∑ⁿ_h=₁,_h≠_l k}aι_h + ⅛₊ι ∑⅛=ι ^α⅛

≤ (⅛₊₁-⅛)Σ∑₁¾

≤ 0.

It implies that /(ж^г) Ц I for z = l, ∙∙∙,n⅛l.

(2). If I = n + 1, we have that for any i ∈ N, it holds

= af₊₁x⁰ - b_n+1 + E⅛=∣ ⅛⅛(⅛)

≤ EL⅜⅛(⅛)

≤ ^- E⅛=l ^⅛β(_ra+l)⅛ + k_n+1 E⅛=l ^a(n+l~)h

— ~ ∑⅛=l k_n+i^a(n+l)h + k_n+1 E⅛=l ^a(n+l)h

—  (⅛ + l ^- ⅛÷1) ∑⅛=l ^a(n+l)h

= 0.

It implies that /(ж^г) Ц n ⅛ 1 for all i = 1, ∙ ∙ ∙, n ⅛ 1. We conclude from (1) and
(2) with a contradiction. We are done.

Moreover, it is easy to derive the following lemma.

Lemma 3.6 Let a simplex P be given in standard form with the additional
condition that ∑_j≠i ∣¾∣ < <⅛ holds for all i = 1, ∙ ∙ -, n. If P contains an integral
point, then the Labeling Rule results in at least n ⅛ 1 completely labeled simplices of
type I and no completely labeled simplex of type II.

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