The name is absent

23

It means that l(yⁿ⁺¹) ψ n ⅛ 1. It is also a contradiction.                        □

We still have to prove Theorem 3.7. It suffices to consider the case in which P
has an integral point x⁰, i.e., Ax⁰ ≤ b. Following the same line of the proof of the
above theorem, we can show that there is at most one completely labeled simplex
of type II in this case. It is clear we only need to demonstrate that the procedure
will produce at least two different completely labeled simp lices. Let us suppose to
the contrary that the procedure only generates a completely labeled simplex, say,
σ(z¹, ∙ ∙ ∙ , æⁱ⁺¹), of type II. It is easy to see that for each к ∈ N, the starting point
v^k ∈ C_k can be expressed as

v^k = x⁰ - ∑ ʌ^(ʌ)

⅛∈^^r-⅛

where λ^k are positive integers for all h E N-к- Notice that in Step (2) the procedure
operates by only using q(Jι) for h ∈ N~k for each к ∈ N. Hence, starting from
v^k ∈ Ck, the vertices x^kj generated by the procedure can be written as

jΛ = ,V+ £ i⅛)

⅛∈^^r-⅛

where λ^k' are integers for all h ∈ N_k- Since starting from the n ⅛ 1 starting
points c^l, ∙ ∙ ∙, uⁿ+¹, the procedure generates a unique completely labeled simplex
σ(x¹, ∙ ∙ ∙ , æⁿ⁺¹) of type II, it implies that for each к ∈ N,

xⁱ = ∙^, + Σ⅛_t,⅛ⁱ⅛(4

= ∙^, + Σ⅛Mfr(J)

(4.2)

= 4'° + Σ½V._l.₊ ,, W’"*’ 4W

where λ^k_h^3k are integers for all h ∈ .∖~k∙ It is not difficult to derive that

ʌ ⅛⅛ _ ʌ 4ι

^λh - ^λh

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