A Minimum Problem for Finite Sets of Real Numbers with Nonnegative Sum

Abstract

Let 𝑛 and 𝑟 be two integers such that 0<𝑟≤𝑛; we denote by 𝛾(𝑛,𝑟)[𝜂(𝑛,𝑟)] the minimum [maximum] number of the nonnegative partial sums of a sum ∑^𝑛1=1𝑎_𝑖≥0, where 𝑎₁,…,𝑎_𝑛 are 𝑛 real numbers arbitrarily chosen in such a way that 𝑟 of them are nonnegative and the remaining 𝑛−𝑟 are negative. We study the following two problems: (𝑃1) which are the values of 𝛾(𝑛,𝑟) and 𝜂(𝑛,𝑟) for each 𝑛 and 𝑟, 0<𝑟≤𝑛? (𝑃2) if 𝑞 is an integer such that 𝛾(𝑛,𝑟)≤𝑞≤𝜂(𝑛,𝑟) , can we find 𝑛 real numbers 𝑎₁,…,𝑎_𝑛, such that𝑟 of them are nonnegative and the remaining 𝑛−𝑟 are negative with ∑^𝑛1=1𝑎_𝑖≥0, such that the number of the nonnegative sums formed from these numbers is exactly 𝑞?

1. Introduction

In [1] Manickam and Miklós raised several interesting extremal combinatorial sum problems, two of which will be described below. Let 𝑛 and 𝑟 be two integers such that 0<𝑟≤𝑛; we denote by 𝛾(𝑛,𝑟)[𝜂(𝑛,𝑟)] the minimum [maximum] number of the nonnegative partial sums of a sum ∑^𝑛1=1𝑎_𝑖≥0, when 𝑎₁,…,𝑎_𝑛 are 𝑛 real numbers arbitrarily chosen in such a way that 𝑟 of them are nonnegative and the remaining 𝑛−𝑟 are negative. Put 𝐴(𝑛)=min{𝛾(𝑛,𝑟)∶0<𝑟≤𝑛}. In [1] the authors answered the following question.

(𝑄1) Which is the value of 𝐴(𝑛)?

In [1, Theorem 1] they found that 𝐴(𝑛)=2^𝑛−1. On the other side, from the proof of Theorem 1 of [1] also it follows that 𝛾(𝑛,𝑟)≥2^𝑛−1 for each 𝑟 and that 𝛾(𝑛,1)≤2^𝑛−1; therefore 𝛾(𝑛,1)=2^𝑛−1 (since 2^𝑛−1=𝐴(𝑛)≤𝛾(𝑛,1)≤2^𝑛−1). It is natural to set then the following problem which is a refinement of (𝑄1).

(𝑃1) Which are the values of 𝛾(𝑛,𝑟)and 𝜂(𝑛,𝑟) for each 𝑛 and 𝑟 with 0<𝑟≤𝑛?

In the first part of this paper we solve the problem (𝑃1), and we prove (see Theorem 4.1) that 𝛾(𝑛,𝑟)=2^𝑛−1 and 𝜂(𝑛,𝑟)=2^𝑛−2^𝑛−𝑟 for each positive integer 𝑟≤𝑛.

A further question that the authors raised in [1] is the following.

(𝑄2)  “We do not know what is the range of the possible numbers of the nonnegative partial sums of a nonnegative 𝑛-element sum. The minimum is 2^𝑛−1 as it was proven and the maximum is obviously 2^𝑛−1 but we do not know which are the numbers between them for which we can find reals 𝑎₁,…,𝑎_𝑛 with ∑^𝑛1=1𝑎_𝑖≥0 such that the number of the nonnegative sums formed from these numbers is equal to that number.’’

The following problem is a natural refinement of (Q2).

(𝑃2)  If 𝑞 is an integer such that 𝛾(𝑛,𝑟)≤𝑞≤𝜂(𝑛,𝑟), can we find 𝑛 real numbers 𝑎₁,…,𝑎_𝑛, such that 𝑟 of them are nonnegative and the remaining 𝑛−𝑟 are negative with ∑^𝑛1=1𝑎_𝑖≥0, such that the number of the nonnegative sums formed from these numbers is exactly 𝑞?

In the latter part of this paper (see Theorem 4.5) we give a partial solution to the problem (𝑃2).

To be more precise in the formulation of the problems that we study and to better underline the links with some interesting problems raised in [1], it will be convenient-identify a finite set of real-numbers with an appropriate real-valued function. Let then 𝑛 and 𝑟 be two fixed integers such that 0<𝑟≤𝑛 and let 𝐼_𝑛={1,2,…,𝑛} (we call 𝐼_𝑛 the index set). We denote by 𝑊(𝑛,𝑟), the set of all the functions 𝑓∶𝐼_𝑛→ℝ such that ∑_{𝑥∈𝐼𝑛}𝑓(𝑥)≥0 and |{𝑥∈𝐼_𝑛∶𝑓(𝑥)≥0}|=𝑟. If 𝑓∈𝑊(𝑛,𝑟), we set 𝛼(𝑓)=|{𝑌⊆𝐼_𝑛∶∑_𝑦∈𝑌𝑓(𝑦)≥0}|. It is easy to observe that 𝛾(𝑛,𝑟)=min{𝛼(𝑓)∶𝑓∈𝑊(𝑛,𝑟)} and 𝜂(𝑛,𝑟)=max{𝛼(𝑓)∶𝑓∈𝑊(𝑛,𝑟)}. We can reformulate the problem (𝑃2) in an equivalent way using the functions terminology instead of the sets terminology.

(𝑃2)  If 𝑞 is an integer such that 𝛾(𝑛,𝑟)≤𝑞≤𝜂(𝑛,𝑟), does there exist a function 𝑓∈𝑊(𝑛,𝑟) with the property that 𝛼(𝑓)=𝑞?

To solve the problem (𝑃1) and (partially) (𝑃2), we use some abstract results on a particular class of lattices introduced in [2, 3]. In this paper we substantially continue the research project started in [4], which is the attempt to solve some extremal sum problems as started in [1] and further studied in [3, 5–11].

2. A Partial Order on the Subsets of 𝐼_𝑛

Of course if we take two functions 𝑓,𝑔∈𝑊(𝑛,𝑟) such that 𝑓(𝐼_𝑛)=𝑔(𝐼_𝑛), then 𝛼(𝑓)=𝛼(𝑔). This implies that if we define on 𝑊(𝑛,𝑟) the equivalence relation 𝑓∼𝑔 iff 𝑓(𝐼_𝑛)=𝑔(𝐼_𝑛) and we denote by [𝑓] the equivalence class of a function 𝑓∈𝑊(𝑛,𝑟), then the definition 𝛽([𝑓])=𝛼(𝑓) is well placed. It is also clear that it holds 𝛾(𝑛,𝑟)=min{𝛽([𝑓])∶[𝑓]∈𝑊(𝑛,𝑟)/∼} and 𝜂(𝑛,𝑟)=max{𝛽([𝑓])∶[𝑓]∈𝑊(𝑛,𝑟)/∼}. Now, when we take an equivalence class [𝑓]∈𝑊(𝑛,𝑟)/~, there is a unique 𝑓^∗∈[𝑓] such that𝑓^∗(𝑟)≥⋯≥𝑓^∗(1)≥0>𝑓^∗(𝑟+1)≥⋯≥𝑓^∗(𝑛).(2.1) We can then identify the quotient set 𝑊(𝑛,𝑟)/~  with the subset of all the functions 𝑓^∗∈𝑊(𝑛,𝑟) that satisfy the condition (2.1). This simple remark conducts us to rename the indexes of 𝐼_𝑛 as follows: 𝑟̃ instead of 𝑟,…,1̃ instead of 1, 1 instead of 𝑟+1,…,𝑛−𝑟 instead of 𝑛. Therefore, if we set 𝐼(𝑛,𝑟)={1̃,…,𝑟,̃1,…,𝑛−𝑟}, we can identify the quotient set 𝑊(𝑛,𝑟)/∼  with the set of all the functions 𝑓∶𝐼(𝑛,𝑟)→ℝ that satisfy the following two conditions:𝑓𝑓(𝑟()̃𝑟)̃≥+⋯⋯≥+𝑓𝑓11̃̃≥+0𝑓>1𝑓+⋯+𝑓𝑛−𝑟≥0,1≥⋯≥𝑓𝑛−𝑟.(2.2) Now, if a generic function 𝑓∶𝐼(𝑛,𝑟)→ℝ that satisfies (2.2) is given, we are interested to find all the subsets 𝑌⊆𝐼(𝑛,𝑟) such that ∑_𝑦∈𝑌𝑓(𝑦)≥0. This goal becomes then easier if we can have an appropriate partial order⊑ on the power set 𝒫(𝐼(𝑛,𝑟)) “compatible” with the total order of the partial sums inducted by 𝑓, that is, a partial order ⊑ that satisfies the following monotonicity property: if 𝑌,𝑍∈𝒫(𝐼(𝑛,𝑟)), then ∑_𝑧∈𝑍𝑓(𝑥)≤∑_𝑦∈𝑌𝑓(𝑦) whenever 𝑍⊑𝑌. To have such a partial order ⊑on 𝒫(𝐼(𝑛,𝑟)) that has the monotonicity property we must introduce a new formal symbol that we denote by0^§. We add this new symbol to the index set 𝐼(𝑛,𝑟), so we set 𝐴(𝑛,𝑟)=𝐼(𝑛,𝑟)∪{0^§}. We introduce on 𝐴(𝑛,𝑟) the following total order:𝑛−𝑟≺⋯≺2≺1≺0^§≺1̃≺2̃≺⋯≺𝑟.̃(2.3)If 𝑖,𝑗∈𝐴(𝑛,𝑟), then we write: 𝑖⪯𝑗 for 𝑖=𝑗 or 𝑖≺𝑗. We denote by 𝑆(𝑛,𝑟) the set of all the formal expressions 𝑖₁⋯𝑖_𝑟∣𝑗₁⋯𝑗_𝑛−𝑟 (hereafter called strings) that satisfy the following properties:(i)𝑖₁,…,𝑖_𝑟∈{1̃,…,𝑟,̃0^§}, (ii)𝑗₁,…,𝑗_𝑛−𝑟∈{1,…,𝑛−𝑟,0^§}, (iii)𝑖₁≽⋯≽𝑖_𝑟≽0^§≽𝑗₁≽⋯≽𝑗_𝑛−𝑟, (iv)the unique element which can be repeated is 0^§.

In the sequel we often use the lowercase letters 𝑢,𝑤,𝑧,… to denote a generic string in 𝑆(𝑛,𝑟). Moreover to make smoother reading, in the numerical examples the formal symbols which appear in a string will be written without ^~  − and ^§; in such way the vertical bar ∣ will indicate that the symbols on the left of | are in {1̃,…,𝑟,̃0^§} and the symbols on the right of ∣ are elements in {0^§,1,…,𝑛−𝑟}. For example, if 𝑛=3 and 𝑟=2, then 𝐴(3,2)={2̃≻1̃≻0^§≻1} and 𝑆(3,2)={21|0,21|1,10|0,20|0,10|1,20|1,00|1,00|0}.

Note that there is a natural bijective setcorrespondence ∗∶𝑤∈𝑆(𝑛,𝑟)↦𝑤^∗∈𝒫(𝐼(𝑛,𝑟))between 𝑆(𝑛,𝑟) and 𝒫(𝐼(𝑛,𝑟)) defined as follows: if 𝑤=𝑖₁⋯𝑖_𝑟∣𝑗₁⋯𝑗_𝑛−𝑟∈𝑆(𝑛,𝑟), then 𝑤^∗ is the subset of 𝐼(𝑛,𝑟) made with the elements 𝑖_𝑘 and 𝑗_𝑙 such that 𝑖_𝑘≠0^§ and 𝑗_𝑙≠0^§. For example, if 𝑤=4310∣013∈𝑆(7,4), then 𝑤^∗={1̃,3̃,4̃,1,3}. In particular, if 𝑤=0⋯0∣0⋯0, then 𝑤^∗=∅.

Now, if 𝑣=𝑖₁⋯𝑖_𝑟∣𝑗₁⋯𝑗_𝑛−𝑟 and 𝑤=𝑖^1⋯𝑖^𝑟∣𝑗^1⋯𝑗^{𝑛−𝑟} are two strings in 𝑆(𝑛,𝑟), we define: 𝑣⊑𝑤 iff 𝑖₁⪯𝑖^1,…,𝑖_𝑟⪯𝑖^𝑟,𝑗₁⪯𝑗^𝑟,…,𝑗_𝑛−𝑟⪯𝑗^{𝑛−𝑟}.

It is easily seen that (1)(𝑆(𝑛,𝑟),⊑) is a finite distributive (hence also graded) lattice with minimum element 0⋯0∣12⋯(𝑛−𝑟) and maximum element 𝑟(𝑟−1)⋯21∣0⋯0; (2)(𝑆(𝑛,𝑟),⊑)has the following unary complementary operation 𝑐: 𝑝₁⋯𝑝_𝑘0⋯0∣0⋯0𝑞₁⋯𝑞_𝑙^𝑐=𝑝^1⋯𝑝^{𝑟−𝑘}0⋯0∣0⋯0𝑞^1⋯𝑞^{𝑛−𝑟−𝑙},(2.4)where {𝑝^1,…,𝑝^{𝑟−𝑘}} is the usual complement of {𝑝₁,…,𝑝_𝑘} in {1̃,…,𝑟}̃ and {𝑞^1,…,𝑞^{𝑛−𝑟−𝑙}} is the usual complement of {𝑞₁,…,𝑞_𝑙} in {1,…,𝑛−𝑟} (e.g., in 𝑆(7,4), we have that (4310∣001)^𝑐=2000∣023).

Since we have the formal necessity to consider functions 𝑓 defined on the extended set 𝐴(𝑛,𝑟) instead of on the indexes set 𝐼(𝑛,𝑟), then we will put 𝑓(0^§)=0. Precisely we can identify the quotient set 𝑊(𝑛,𝑟)/∼ with the set 𝑊𝐹(𝑛,𝑟), defined by 𝑊𝐹(𝑛,𝑟)=𝑓=∶0𝐴>(𝑛𝑓,𝑟)→ℝ∶𝑓(𝑟)̃≥⋯≥𝑓1̃≥𝑓0^§1≥⋯≥𝑓𝑛−𝑟,𝑓(𝑟)̃+⋯+𝑓1̃+𝑓1+⋯+𝑓𝑛−𝑟≥0.(2.5)

We call an element of 𝑊𝐹(𝑛,𝑟) a (𝑛,𝑟)-weight function, and if 𝑓∈𝑊𝐹(𝑛,𝑟), we will continue to set 𝛼(𝑓)∶=|{𝑌⊆𝐼(𝑛,𝑟)∶∑_𝑦∈𝑌𝑓(𝑦)≥0}|. Therefore, with these last notations we have that 𝛾(𝑛,𝑟)=min{𝛼(𝑓)∶𝑓∈𝑊𝐹(𝑛,𝑟)}, 𝜂(𝑛,𝑟)=max{𝛼(𝑓)∶𝑓∈𝑊𝐹(𝑛,𝑟)} and the question (𝑃2) becomes equivalent to the following.

(𝑃2)  If 𝑞 is an integer such that 𝛾(𝑛,𝑟)≤𝑞≤𝜂(𝑛,𝑟), does there exist a function 𝑓∈𝑊𝐹(𝑛,𝑟)with the property that 𝛼(𝑓)=𝑞?

3. Boolean Maps Induct by Weight Functions

We denote by 𝟐 the boolean lattice composed of a chain with 2 elements that we denote 𝑁 (the minimum element) and 𝑃 (the maximum element). A Boolean map (briefly BM) on 𝑆(𝑛,𝑟) is a map 𝐴∶dom(𝐴)⊆𝑆(𝑛,𝑟)→𝟐; in particular if dom(𝐴)=𝑆(𝑛,𝑟),we also say that 𝐴 is a Boolean total map (briefly BTM) on 𝑆(𝑛,𝑟). If 𝐴 is BM on 𝑆(𝑛,𝑟), we set 𝑆^+𝐴(𝑛,𝑟)={𝑤∈dom(𝐴)∶𝐴(𝑤)=𝑃}.

If 𝑓∈𝑊𝐹(𝑛,𝑟), the sum function Σ_𝑓∶𝑆(𝑛,𝑟)→ℝ induced by 𝑓 on 𝑆(𝑛,𝑟) is the function that associates to 𝑤=𝑖_𝑟⋯𝑖₁∣𝑗₁⋯𝑗_𝑛−𝑟∈𝑆(𝑛,𝑟) the real number Σ_𝑓(𝑤)=𝑓(𝑖₁)+⋯+𝑓(𝑖_𝑟)+𝑓(𝑗₁)+⋯+𝑓(𝑗_𝑛−𝑟), and therefore we can associate to 𝑓∈𝑊𝐹(𝑛,𝑟) the map 𝐴_𝑓∶𝑆(𝑛,𝑟)→𝟐 setting:𝐴_𝑓(𝑤)=⎧⎪⎨⎪⎩𝑃ifΣ^𝑓𝑁𝑁iiff𝑤Σ_𝑓=(𝑤0)⋯≥00,∣0𝑤⋯≠00,⋯0∣0⋯0,(𝑤)<0.(3.1)

Let us note that |𝑆^+𝐴_𝑓(𝑛,𝑟)|=|{𝑤∈𝑆(𝑛,𝑟)∶𝐴_𝑓(𝑤)=𝑃}|=𝛼(𝑓), and so 𝛾(𝑛,𝑟)=min{|𝑆^+𝐴_𝑓(𝑛,𝑟)|∶𝑓∈𝑊𝐹(𝑛,𝑟)},𝜂(𝑛,𝑟)=max{|𝑆^+𝐴_𝑓(𝑛,𝑟)|∶𝑓∈𝑊𝐹(𝑛,𝑟)}.

Our goal is now to underline that some properties of such maps simplify the study of our problems. It is easy to observe that the map 𝐴_𝑓 has the following properties.(i)𝐴_𝑓 is orderpreserving.(ii)If 𝑤∈𝑆(𝑛,𝑟) is such that 𝐴_𝑓(𝑤)=𝑁, then 𝐴_𝑓(𝑤^𝑐)=𝑃,(iii)𝐴_𝑓(10⋯0∣0⋯0)=𝑃, 𝐴(0⋯0∣0⋯0)=𝑁, and 𝐴_𝑓(𝑟⋯21∣12⋯(𝑛−𝑟))=𝑃.

Example 3.1. Let 𝑓 be the following (5,3)-weight function: 𝑓∶3̃2̃1̃↓1↓10↓.9−↓01.8−↓22.1(3.2) We represent the map 𝐴_𝑓 defined on 𝑆(5,3) by using the Hasse diagram of this lattice, on which we color green the nodes where the map 𝐴_𝑓 assumes value 𝑃 and red the nodes where it assumes value 𝑁 (See Figure 1).
Note that if we have a generic Boolean total map 𝐴∶𝑆(𝑛,𝑟)→𝟐 which has the properties (i), (ii), and (iii) of 𝐴_𝑓, that is, the following:(BM1)𝐴 is orderpreserving,(BM2)if 𝑤∈𝑆(𝑛,𝑟) is such that 𝐴(𝑤)=𝑁, then 𝐴(𝑤^𝑐)=𝑃,(BM3)𝐴(10⋯0∣0⋯0)=𝑃, 𝐴(0⋯0∣0⋯0)=𝑁, and𝐴(𝑟⋯21∣12⋯(𝑛−𝑟))=𝑃,in general there does not exist a function 𝑓∈𝑊𝐹(𝑛,𝑟) such that 𝐴_𝑓=𝐴 (see [3] for a counterexample).
We denote by 𝒲₊(𝑆(𝑛,𝑟),𝟐) the set of all the maps 𝐴∶𝑆(𝑛,𝑟)→𝟐 which satisfy (BM1) and (BM2) and by 𝒲₊(𝑛,𝑟) the subset of all the maps in 𝒲₊(𝑆(𝑛,𝑟),𝟐) which satisfy also (BM3). We also set 𝛾^∗(𝑛,𝑟)∶=min{|𝑆^+𝐴(𝑛,𝑟)∶𝐴∈𝒲₊(𝑛,𝑟)} and 𝜂^∗(𝑛,𝑟)∶=max{|𝑆^+𝐴(𝑛,𝑟)|∶𝐴∈𝒲₊(𝑛,𝑟)}. Let us observe that 𝛾^∗(𝑛,𝑟)≤𝛾(𝑛,𝑟)≤𝜂(𝑛,𝑟)≤𝜂^∗(𝑛,𝑟). A natural question raises at this point.
(𝑄):if 𝑞 is an integer such that 𝛾^∗(𝑛,𝑟)≤𝑞≤𝜂^∗(𝑛,𝑟), does there exist a map 𝐴∈𝒲₊(𝑛,𝑟) with the property that |𝑆^+𝐴(𝑛,𝑟)|=𝑞?
The question (𝑄) is the analogue of (𝑃2) expressed in terms of Boolean total maps on 𝑆(𝑛,𝑟) instead of (𝑛,𝑟)-weight functions, and if we are able to respond to (𝑄), we provide also a partial answer to (𝑃2). In Section 4 we give an affirmative answer to the question (𝑄), and also we give a constructive method to build the map 𝐴.

Figure 1 

4. Main Results

In the sequel of this paper we adopt the classical terminology and notations usually used in the context of the partially ordered sets (see [12–14] for the general aspects on this subject). If 𝑍⊆𝑆(𝑛,𝑟), we will set ↓𝑍={𝑥∈𝑆(𝑛,𝑟)∶∃𝑧∈𝑍suchthat𝑥⊑𝑧},↑𝑍={𝑥∈𝑆(𝑛,𝑟)∶∃𝑧∈𝑍suchthat𝑧⊑𝑥}. In particular, if 𝑧∈𝑆(𝑛,𝑟), we will set ↓𝑧=↓{𝑧}={𝑥∈𝑆(𝑛,𝑟)∶𝑧⊒𝑥}, ↑𝑧=↑{𝑧}={𝑥∈𝑆(𝑛,𝑟)∶𝑧⊑𝑥}. 𝑍 is called a downset of 𝑆(𝑛,𝑟) if for 𝑧∈𝑍 and 𝑥∈𝑆(𝑛,𝑟) with 𝑧⊒𝑥, then 𝑥∈𝑍. 𝑍 is called an upset of 𝑆(𝑛,𝑟) if for 𝑧∈𝑍 and 𝑥∈𝑆(𝑛,𝑟) with 𝑧⊑𝑥, then 𝑥∈𝑍. ↓𝑍 is the smallest down-set of 𝑆(𝑛,𝑟) which contains 𝑍, and 𝑍 is a downset of 𝑆(𝑛,𝑟) if and only if 𝑍=↓𝑍. Similarly, ↑𝑍 is the smallest up-set of 𝑆(𝑛,𝑟) which contains 𝑍, and 𝑍 is an upset in 𝑆(𝑛,𝑟) if and only if 𝑍=↑𝑍.

Theorem 4.1. If 𝑛 and 𝑟 are two integers such that 0<𝑟<𝑛, then 𝛾(𝑛,𝑟)=𝛾^∗(𝑛,𝑟)=2^𝑛−1 and 𝜂(𝑛,𝑟)=𝜂^∗(𝑛,𝑟)=2^𝑛−2^𝑛−𝑟.

Proof. Assume that 0<𝑟<𝑛. We denote by 𝑆₁(𝑛,𝑟) the sublattice of 𝑆(𝑛,𝑟) of all the strings 𝑤 of the form 𝑤=𝑖₁⋯𝑖_𝑟∣𝑗₁⋯𝑗_{𝑛−𝑟−1}(𝑛−𝑟), with 𝑗₁⋯𝑗_{𝑛−𝑟−1}∈{0^§,1,…,𝑛−𝑟−1} and by 𝑆₂(𝑛,𝑟) the sublattice of 𝑆(𝑛,𝑟) of all the strings 𝑤 of the form 𝑤=𝑖₁⋯𝑖_𝑟∣0𝑗₂⋯𝑗_𝑛−𝑟, with 𝑗₂⋯𝑗_𝑛−𝑟∈{0^§,1,…,𝑛−𝑟−1}. It is clear that 𝑆(𝑛,𝑟)=𝑆₁(𝑛,𝑟)⋃̇𝑆₂(𝑛,𝑟)and 𝑆₁(𝑛,𝑟)≅𝑆₂(𝑛,𝑟)≅𝑆(𝑛−1,𝑟). We consider now the following further sublattices of 𝑆(𝑛,𝑟): 𝑆𝑆⁺¹(𝑛,𝑟)∶=𝑤∈𝑆₁(𝑛,𝑟)∶𝑤=𝑟(𝑟−1)⋯21∣𝑗₁⋯𝑗_{𝑛−𝑟−1}(𝑛−𝑟)_±1,𝑆₋₁(𝑛,𝑟)∶=𝑤∈𝑆₁(𝑛,𝑟)∶𝑤=𝑖₁⋯𝑖_𝑟−10∣𝑗₁⋯𝑗_{𝑛−𝑟−1}(𝑛−𝑟),𝑖₁≻0^§,𝑆(𝑛,𝑟)∶=𝑤∈𝑆₁(𝑛,𝑟)∶𝑤=0⋯0∣𝑗₁⋯𝑗_{𝑛−𝑟−1}(𝑛−𝑟)₊₂,𝑆_±2(𝑛,𝑟)∶=𝑤∈𝑆₂(𝑛,𝑟)∶𝑤=𝑟(𝑟−1)⋯21∣0𝑗₂⋯𝑗_𝑛−𝑟,𝑆₋₂(𝑛,𝑟)∶=𝑤∈𝑆₂(𝑛,𝑟)∶𝑤=𝑖₁⋯𝑖_𝑟−10∣0𝑗₂⋯𝑗_𝑛−𝑟,𝑖₁≻0^§,(𝑛,𝑟)∶=𝑤∈𝑆₂(𝑛,𝑟)∶𝑤=0⋯0∣0𝑗₂⋯𝑗_𝑛−𝑟.(4.1)It occurs immediately that 𝑆_𝑖(𝑛,𝑟)=𝑆^+𝑖(𝑛,𝑟)⋃̇𝑆^±𝑖(𝑛,𝑟)⋃̇𝑆^−𝑖(𝑛,𝑟), for 𝑖=1,2 and 𝑆^±𝑖(𝑛,𝑟) is a distributive sublattice of 𝑆_𝑖(𝑛,𝑟) with 2^𝑛−1−2⋅2^{𝑛−𝑟−1}=2^𝑛−1−2^𝑛−𝑟 elements, for 𝑖=1,2.
Now we consider the following (𝑛,𝑟)-weight function 𝑓∶𝐴(𝑛,𝑟)→ℝ:𝑓∶𝑟̃⋯1̃0^§(𝑛↓−𝑟)⋯↓(𝑛↓−𝑟)↓0−↓11⋯⋯↓(𝑛−−↓𝑟1−1)(𝑛−𝑟()𝑛(1↓−−𝑟)𝑟)−1(4.2) Then it follows that Σ_𝑓∶𝑆(𝑛,𝑟)→ℝ is such that Σ_𝑓(𝑤)≥0if𝑤∈𝑆^±2<0if𝑤∈𝑆_±1(𝑛,𝑟),(𝑛,𝑟).(4.3) It means that the boolean map 𝐴_𝑓∈𝒲₊(𝑛,𝑟) is such that 𝐴_𝑓(𝑤)𝑃if𝑤∈𝑆^±2𝑁if𝑤∈𝑆_±1(𝑛,𝑟),(𝑛,𝑟).(4.4) This shows that |𝑆^+𝐴_𝑓(𝑛,𝑟)|=|𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟)⋃̇𝑆^±2(𝑛,𝑟)|=2^{𝑛−𝑟−1}+2^{𝑛−𝑟−1}+2^𝑛−1−2⋅2^{𝑛−𝑟−1}=2^𝑛−1.
In [1, Theorem 1] it has been proved that 𝛾(𝑛,1)=2^𝑛−1 and 𝛾(𝑛,𝑟)≥2^𝑛−1. Since 𝛾(𝑛,𝑟)≥𝛾^∗(𝑛,𝑟), using a technique similar to that used in the proof Theorem 1 of [1], it easily follows that 𝛾^∗(𝑛,𝑟)≥2^𝑛−1. As shown above, it results that |𝑆^+𝐴_𝑓(𝑛,𝑟)|=2^𝑛−1. Hence 2^𝑛−1≥𝛾(𝑛,𝑟)≥𝛾^∗(𝑛,𝑟)≥2^𝑛−1, that is, 𝛾(𝑛,𝑟)=𝛾^∗(𝑛,𝑟)=2^𝑛−1. This proves; the first part of theorem; it remains to prove the latter part.
We consider now the following (𝑛,𝑟)-positive weight function 𝑔∶𝐴(𝑛,𝑟)→ℝ: 𝑔∶↓1𝑟̃⋯⋯↓↓11̃0↓0^§−↓11⋯↓(𝑛−↓𝑟−1)(𝑛↓−𝑟)𝑛−𝑟⋯𝑛−−1𝑟𝑛−−1𝑟(4.5) It results then that the sum Σ_𝑔∶𝑆(𝑛,𝑟)→ℝ is such that Σ_𝑔(𝑤)≥0 if 𝑤∈𝑆^±1(𝑛,𝑟)⋃̇𝑆^±2(𝑛,𝑟); that is, the boolean map 𝐴_𝑔∈𝒲₊(𝑛,𝑟) is such that 𝐴_𝑔(𝑤)=𝑃if 𝑤∈𝑆^±1(𝑛,𝑟)⋃̇𝑆^±2(𝑛,𝑟). This shows that |𝑆^+𝐴_𝑔(𝑛,𝑟)|=|𝑆^±1(𝑛,𝑟)⋃̇𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟)|=(2^𝑛−1−2^𝑛−𝑟)+(2^𝑛−1−2^𝑛−𝑟)+2^{𝑛−𝑟−1}+2^{𝑛−𝑟−1} = 2^𝑛−2^𝑛−𝑟.
On the other hand, it is clear that for any 𝐴∈𝒲₊(𝑛,𝑟) it results 𝐴(𝑤)=𝑃 for each 𝑤∈𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟)and 𝐴(𝑤)=𝑁 for each 𝑤∈𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟).The number (2^𝑛−2^𝑛−𝑟) is the biggest number of values 𝑃 that a boolean map 𝐴∈𝒲₊(𝑛,𝑟) can assume. Hence, since 𝜂(𝑛,𝑟)≤𝜂^∗(𝑛,𝑟), we have 2^𝑛−2^𝑛−𝑟=|𝑆^+𝐴_𝑔(𝑛,𝑟)|≤𝜂(𝑛,𝑟)≤𝜂^∗(𝑛,𝑟)≤2^𝑛−2^𝑛−𝑟; that is be; 𝜂(𝑛,𝑟)=𝜂^∗(𝑛,𝑟)=2^𝑛−2^𝑛−𝑟. This concludes the proof of Theorem 4.1.

To better visualize the previous result, we give a numerical example on a specific Hasse diagram. Let 𝑛=6 and 𝑟=2 and let 𝑓 be as given in previous theorem; that is 𝑓∶2̃1̃0^§4↓↓4↓0−↓11−↓21−↓31−↓45(4.6) In Figure 2 we have shown the Hasse diagram of the lattice 𝑆(6,2), where 𝑆⁺¹(𝑛,𝑟) is black, 𝑆^±1(𝑛,𝑟) is violet, 𝑆⁻¹(𝑛,𝑟) is red, 𝑆⁺²(𝑛,𝑟) is blue, 𝑆^±2(𝑛,𝑟) is brown, 𝑆⁻²(𝑛,𝑟) is green. Therefore 𝐴_𝑓 assume the following values.(i)The blue, black, and brown nodes correspond to values 𝑃 of 𝐴_𝑓,(ii)The violet, red, and green nodes correspond to values 𝑁 of 𝐴_𝑓.

Figure 2 

First to give the proof of Theorem 4.5 we need to introduce some useful results and the concept of basis in 𝑆(𝑛,𝑟). In the following first lemma we show some properties of the sublattices of 𝑆(𝑛,𝑟).

Lemma 4.2. Here the following properties hold, where 𝜃=00⋯0∣0⋯0 and Θ=𝑟⋯21∣12⋯(𝑛−𝑟). (i)↑Θ=𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟).(ii)↓𝜃=𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟).(iii)↑𝑆^±2(𝑛,𝑟)⊆𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟).(iv)↓𝑆^±1(𝑛,𝑟)⊆𝑆^±1(𝑛,𝑟)⋃̇𝑆⁻¹(𝑛,𝑟).(v)((𝑆^±1(𝑛,𝑟))^𝑐=𝑆^±2(𝑛,𝑟).

Proof. (i) If 𝑤∈(↑Θ), then Θ⊑𝑤; that is, it has the form 𝑤=𝑟⋯1∣𝑗₁⋯𝑗_𝑛−𝑟, where 𝑗₁⋯𝑗_𝑛−𝑟∈{0^§,1,…,𝑛−𝑟}; therefore 𝑤∈𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟). If 𝑤∈𝑆⁺¹(𝑛,𝑟), it has the form 𝑤=𝑟⋯1∣𝑗₁⋯𝑗_{𝑛−𝑟−1}(𝑛−𝑟), where 𝑗₁⋯𝑗_{𝑛−𝑟−1}∈{0^§,1,…,𝑛−𝑟−1}; if 𝑤∈𝑆⁺²(𝑛,𝑟), it has the form 𝑤=𝑟⋯1∣0𝑗₂⋯𝑗_𝑛−𝑟, where 𝑗₂⋯𝑗_𝑛−𝑟∈{0^§,1,…,𝑛−𝑟−1}. In both cases it results that Θ⊑𝑤, that is, 𝑤∈(↑Θ).(ii)It is analogue to (i).(iii)The minimum of the sublattice 𝑆^±2(𝑛,𝑟)is 𝛼=10⋯0∣01⋯(𝑛−𝑟−1); since ↑𝑆^±2(𝑛,𝑟)⊆↑𝛼, it is sufficient to show that ↑𝛼=𝑆^±2(𝑛,𝑟)⋃𝑆⁺²(𝑛,𝑟). The inclusion 𝑆^±2(𝑛,𝑟)⋃𝑆⁺²(𝑛,𝑟)⊆↑𝛼follows by the definition of 𝑆^±2(𝑛,𝑟)and of 𝑆⁺²(𝑛,𝑟). On the other side, if 𝑤∈↑𝛼, it follows that 𝛼⊑𝑤; that is, 𝑤=𝑖₁⋯𝑖_𝑟∣0𝑗₂⋯𝑗_𝑛−𝑟, with 𝑖₁≻0^§and 𝑗₂⋯𝑗_𝑛−𝑟∈{0^§,1,…,𝑛−𝑟−1}. Therefore 𝑤∈𝑆^±2(𝑛,𝑟)⋃𝑆⁺²(𝑛,𝑟), and this proves the other inclusion.(iv)Let us consider the maximum of the sublattice 𝑆^±1(𝑛,𝑟); that is 𝑡₁=𝑟(𝑟−1)⋯20∣0⋯0(𝑛−𝑟). Since ↓𝑆^±1(𝑛,𝑟)⊆↓𝛽, it is sufficient to show that ↓𝑡₁=𝑆^±1(𝑛,𝑟)⋃𝑆⁻¹(𝑛,𝑟); this proof is similar to (iii).(v)If 𝑤∈𝑆^±1(𝑛,𝑟); it has the form 𝑤=𝑖₁⋯𝑖_𝑟−10∣𝑗₁⋯𝑗_{𝑛−𝑟−1}(𝑛−𝑟), with 𝑖₁≻0^§ and 𝑗₁⋯𝑗_{𝑛−𝑟−1}∈{0^§,1,…,𝑛−𝑟−1}; therefore its complement has the form 𝑤^𝑐=𝑖^1⋯𝑖^𝑟−10∣0𝑗^2⋯𝑗^{𝑛−𝑟}, with 𝑖₁′≻0^§ and 𝑗^2⋯𝑗^{𝑛−𝑟}∈{0^§,1,…,𝑛−𝑟−1}; hence 𝑤^𝑐∈𝑆^±2(𝑛,𝑟). This shows that (𝑆^±1(𝑛,𝑟))^𝑐⊆𝑆^±2(𝑛,𝑟).
Now, if 𝑤∈𝑆^±2(𝑛,𝑟), we write 𝑤 in the form (𝑤^𝑐)^𝑐; then 𝑤^𝑐∈(𝑆^±2(𝑛,𝑟))^𝑐⊆𝑆^±1(𝑛,𝑟); therefore 𝑤∈(𝑆^±1(𝑛,𝑟))^𝑐. This shows that 𝑆^±2(𝑛,𝑟)⊆(𝑆^±1(𝑛,𝑟))^𝑐.

At this point let us recall the definition of basis for 𝑆(𝑛,𝑟), as given in [3] in a more general context. In the same way as an antichain uniquely determines a Boolean order-preserving map, a basis uniquely determines a Boolean map that has the properties (BM1) and (BM2) (see [3] for details). Hence the concept of basis will be fundamental in the sequel of this proof.

Definition 4.3. A basis for 𝑆(𝑛,𝑟) is an ordered couple ⟨𝑌₊∣𝑌₋⟩, where 𝑌₊ and 𝑌₋ are two disjoint antichains of 𝑆(𝑛,𝑟) such that(B1)(↓𝑌₊)⋂(𝑌^𝑐−)=∅, (B2)((↑𝑌₊)⋃↑(𝑌^𝑐−))⋂↓𝑌₋=∅, (B3)𝑆(𝑛,𝑟)=((↑𝑌₊)⋃↑(𝑌^𝑐−))⋃↓𝑌₋. In the proof of Theorem 4.5 we will construct explicitly such a basis.
We will also use the following result that was proved in [3].

Lemma 4.4. Let ⟨𝑊₊∣𝑊₋⟩ be a basis for 𝑆(𝑛,𝑟). Then the map 𝐴(𝑥)𝑃if𝑥∈↑𝑊⁺𝑁if𝑥∈↓𝑊₋↑𝑊^𝑐−,(4.7) is such that 𝐴∈𝒲₊(𝑆(𝑛,𝑟),𝟐).

Theorem 4.5. If 𝑞 is a fixed integer with 2^𝑛−1≤𝑞≤2^𝑛−2^𝑛−𝑟, then there exists a boolean map 𝐴_𝑞∈𝒲₊(𝑛,𝑟) such that |𝑆^+𝐴_𝑞(𝑛,𝑟)|=𝑞.

Proof. Let 𝑞 be a fixed integer such that 2^𝑛−1≤𝑞≤2^𝑛−2^𝑛−𝑟. We determine a specific Boolean total map 𝐴∈𝒲₊(𝑛,𝑟) such that |𝑆^+𝐴(𝑛,𝑟)|=𝑞. We proceed as follows.
The case 𝑟=1 is proved in the previous Theorem 4.1. Let us assume then 𝑟>1. Since 𝑆^±1(𝑛,𝑟) is a finite distributive sublattice of graded lattice 𝑆(𝑛,𝑟), also 𝑆^±1(𝑛,𝑟) is a graded lattice. We denote by 𝑅 the rank of 𝑆^±1(𝑛,𝑟) and with 𝜌₁ its rank function. Note that the bottom of 𝑆^±1(𝑛,𝑟) is 𝑏₁=10⋯0∣1⋯(𝑛−𝑟−1)(𝑛−𝑟) and the top is 𝑡₁=𝑟(𝑟−1)⋯20∣0⋯0(𝑛−𝑟).
We write 𝑞 in the form 𝑞=2^𝑛−1+𝑝 with 0≤𝑝≤2^𝑛−2^𝑛−𝑟−2^𝑛−1=2ⁿ⁻¹−2^𝑛−𝑟=|𝑆^±1(𝑛,𝑟)|. We will build a map 𝐴∈𝑊₊(𝑛,𝑟) such that |𝑆^+𝐴(𝑛,𝑟)|=2^𝑛−1+𝑝.
If 𝑝=0, we take 𝐴=𝐴_𝑓, with 𝑓 as in Theorem 4.1, and hence we have |𝑆^+𝐴(𝑛,𝑟)|=2^𝑛−1.
If 𝑝=2^𝑛−1−2^𝑛−𝑟, we take 𝐴=𝐴_𝑔, with 𝑔 as in Theorem 4.1, and we have |𝑆^+𝐴(𝑛,𝑟)|=2^𝑛−2^𝑛−𝑟.
Therefore we can assume that 0<𝑝<2^𝑛−1−2^𝑛−𝑟.
If 0≤𝑖≤𝑅, we denote by ℜ_𝑖 the set of elements 𝑤∈𝑆^±1(𝑛,𝑟) such that 𝜌₁(𝑤)=𝑅−𝑖, and we also set 𝛽_𝑖∶=|ℜ_𝑖| to simplify the notation. We write each ℜ_𝑖 in the following form: ℜ_𝑖={𝑣_𝑖1,…,𝑣_𝑖𝛽𝑖}. If 0≤𝑙≤𝑅, we set 𝔘_l∶=⋃̇_{𝑖=0,…,𝑙}ℜ_𝑖.
If 0≤𝑙≤𝑅−2, we set 𝔅_𝑙∶=⋃̇_{𝑖=𝑙+2,…,𝑅}ℜ_𝑖 and 𝔅_𝑅−1∶=𝔅_𝑅∶=∅. We can then write 𝑝 in the form 𝑝=∑^𝑘𝑖=0|ℜ_𝑖|+𝑠=|𝔘_𝑘|+𝑠, for some 0≤𝑠<|ℜ_𝑘+1| and some 0≤𝑘≤𝑅−1. Depending on the previous number 𝑠 we partition ℜ_𝑘+1 into the following two disjoint subsets: ℜ_𝑘+1={𝑣_(𝑘+1)1,…,𝑣_(𝑘+1)𝑠}⋃̇{𝑣_{(𝑘+1)(𝑠+1)},…,𝑣_{(𝑘+1)𝛽𝑘+1}}, where the first subset is considered empty if 𝑠=0.
In the sequel, to avoid an overload of notations, we write simply 𝑣_𝑖 instead of 𝑣_(𝑘+1)𝑖, for 𝑖=1,…,𝛽_𝑘+1.
Let us note that 𝑆^±1(𝑛,𝑟)=𝔘_𝑘⋃̇ℜ_𝑘+1⋃̇𝔅_𝑘.
We define now the map 𝐴∶𝑆(𝑛,𝑟)→𝟐: 𝐴(𝑤)=⎧⎪⎪⎪⎨⎪⎪⎪⎩𝑃if𝑤∈𝑆^±2𝑃if𝑤∈𝔘_𝑘(𝑛,𝑟)̇𝑆⁺¹(𝑛,𝑟)̇𝑆⁺²(𝑛,𝑟),̇𝑣₁𝑁if𝑤∈𝔅_𝑘,…,𝑣_𝑠,̇𝑣_𝑠+1𝑁if𝑤∈𝑆₋₁,…,𝑣_𝛽𝑘+1,(𝑛,𝑟)̇𝑆⁻²(𝑛,𝑟).(4.8) Let us observe that |𝑆^+𝐴(𝑛,𝑟)| = |𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟)|+|𝔘_𝑘⋃̇{𝑣₁,…,𝑣_𝑠}| = (2^𝑛−1−2^𝑛−𝑟)+2^{𝑛−𝑟−1}+2^{𝑛−𝑟−1}+|𝔘_𝑘|+𝑠 = 2^𝑛−1+𝑝 = 𝑞.
Therefore, if we show that 𝐴∈𝒲₊(𝑛,𝑟), the theorem is proved.
We write ℜ_𝑘 in the following way: ℜ_𝑘={𝑤₁,…,𝑤_𝑡}⋃̇{𝑤_𝑡+1,…,𝑤_𝛽𝑘},where {𝑤₁,…,𝑤_𝑡} = ℜ_𝑘⋂↑{𝑣₁,…,𝑣_𝑠} and {𝑤_𝑡+1,…,𝑤_𝛽𝑘} = ℜ_𝑘⧵{𝑤₁,…,𝑤_𝑡}. Analogously ℜ_𝑘+2 = {𝑧₁,…,𝑧_𝑞}⋃̇{𝑧_𝑞+1,…,𝑧_𝛽𝑘+2}, where {𝑧_𝑞+1,…,𝑧_𝛽𝑘+2} = ℜ_𝑘+2⋂↓{𝑣_𝑠+1,…,𝑣_𝛽𝑘+1} and {𝑧₁,…,𝑧_𝑞} = ℜ_𝑘+2⧵{𝑧_𝑞+1,…,𝑧_𝛽𝑘+2}.
We can see a picture of this partition of the sublattice 𝑆^±1(𝑛,𝑟) in Figure 3.
Depending on 𝑠 and 𝑘, we build now a particular basis for 𝑆(𝑛,𝑟).
To such aim, let us consider the minimum 𝛼=10⋯0∣01⋯(𝑛−𝑟−1) of 𝑆^±2(𝑛,𝑟) and the subsets 𝑇₊∶={𝑣₁,…,𝑣_𝑠,𝑤_𝑡+1,…,𝑤_𝛽𝑘} and 𝑇₋∶={𝑣_𝑠+1,…,𝑣_𝛽𝑘+1,𝑧₁,…,𝑧_𝑞}. Let us distinguish two cases:(a1)𝛼∈↑𝑇₊,(a2)𝛼∉↑𝑇₊. We define two different couples of subsets as follows:
In the case (a1) we set 𝑌₊∶=𝑇₊and𝑌₋∶=𝑇₋⋃̇{𝜃};in the case (a2) we set 𝑌₊∶=𝑇₊⋃{𝛼} and 𝑌₋∶=𝑇₋⋃̇{𝜃}.
Step 1. ⟨𝑌₊∣𝑌₋⟩ is a couple of two disjoint antichains of 𝑆(𝑛,𝑟).
In both cases (a1) and (a2) it is obvious that 𝑌₊⋂𝑌₋=∅.
Case (a1)
The elements {𝑣₁,…,𝑣_𝑠} are not comparable between them because they have all rank 𝑅−(𝑘+1) and analogously for the elements {𝑤_𝑡+1,…,𝑤_𝛽𝑘} that have all rank 𝑅−𝑘. Let now 𝑣∈{𝑣₁,…,𝑣_𝑠} and let 𝑤∈{𝑤_𝑡+1,…,𝑤_𝛽𝑘}; then 𝑤∉↓𝑣 because 𝜌₁(𝑣)<𝜌₁(𝑤), and 𝑤∉↑𝑣 because {𝑤_𝑡+1,…,𝑤_𝛽𝑘}⋂↑{𝑣₁,…,𝑣_𝑠}=∅ by construction. For the elements in 𝑌₋ different from 𝜃, we can proceed as for 𝑌₊. On the other side, we can observe that 𝜃 is not comparable with none of the elements 𝑣_𝑠+1,…,𝑣_𝛽𝑘+1,𝑧₁,…,𝑧_𝑞 since these are all in 𝑆^±1(𝑛,𝑟) while 𝜃∈𝑆⁻²(𝑛,𝑟). Thus 𝑌₊ is an antichain. Case (a2)
In this case we only must show that 𝛼 is not comparable with none of the elements 𝑣₁,…,𝑣_𝑠,𝑤_𝑡+1,…,𝑤_𝛽𝑘. At first from the fact that 𝛼∉↑𝑇₊ it follows 𝑣_𝑖∉↓𝛼 for each 𝑖=1,…,𝑠, and 𝑤_𝑗∉↓𝛼 for each 𝑗=𝑡+1,…,𝛽_𝑘. Moreover, the elements 𝑣₁,…,𝑣_𝑠 and 𝑤_𝑡+1,…,𝑤_𝛽𝑘 are all in 𝑆^±1(𝑛,𝑟); hence they have the form 𝑖₁⋯𝑖_𝑟−10∣𝑗₁⋯𝑗_{𝑛−𝑟−1}(𝑛−𝑟), with 𝑖₁≻0^§, while 𝛼=10⋯0∣01⋯(𝑛−𝑟−1), so 𝛼∉↓𝑣_𝑖 and 𝛼∉↓𝑤_𝑗 for each 𝑖=1,…,𝑠 and each 𝑗=𝑡+1,…,𝛽_𝑘.Step 2. ⟨𝑌₊∣𝑌₋⟩ is a basis for 𝑆(𝑛,𝑟).
We must see that (B1), (B2), and (B3) hold in both cases (a1) and (a2).
Case (a1)
(B1)Let us begin to observe that 𝑇^𝑐−={𝑣^𝑐𝑠+1,…,𝑣^𝑐𝛽_𝑘+1,𝑧^𝑐1,…,𝑧^𝑐𝑞}⊆(𝑆^±1(𝑛,𝑟))^𝑐=𝑆^±2(𝑛,𝑟). Since 𝜃^𝑐=Θ, we have then 𝑌^𝑐−=𝑇^𝑐−⋃̇{𝜃^𝑐}⊆𝑆^±2(𝑛,𝑟)⋃̇{Θ}⊆𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟).On the other hand, since ↓𝑌₊⊆↓𝑆^±1(𝑛,𝑟), by Lemma 4.2(iv) we have also that ↓𝑌₊⊆𝑆^±1(𝑛,𝑟)⋃^.𝑆⁻¹(𝑛,𝑟). Hence (↓𝑌₊)⋂(𝑌^𝑐−)=∅. This proves (B1).(B2)We show at first that ↑𝑌^𝑐−⋂↓𝑌₋=∅. Since 𝑌^𝑐−⊆𝑆^±2(𝑛,𝑟)⋃̇{Θ}, we have that ↑𝑌^𝑐−⊆↑(𝑆^±2(𝑛,𝑟))⋃↑Θ. By Lemma 4.2(i) and (iii) we have then ↑𝑌^𝑐−⊆𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟). On the other side, since ↓𝑇₋⊆↓𝑆^±1(𝑛,𝑟), by Lemma 4.2(iv) it follows that ↓𝑇₋⊆𝑆^±1(𝑛,𝑟)⋃^.𝑆⁻¹(𝑛,𝑟). By Lemma 4.2(ii), we have ↓𝜃=𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟). Hence it holds ↓𝑌₋⊆𝑆^±1(𝑛,𝑟)⋃̇𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟). This proves that ↑𝑌^𝑐−⋂↓𝑌₋=∅. We show now that also ↑𝑌₊⋂↓𝑌₋=∅; we proceed by contradiction. Let us suppose that there exists an element 𝑧∈↑𝑌₊⋂↓𝑌₋, then there are two elements 𝑤₊∈𝑌₊ and 𝑤₋∈𝑌₋ such that 𝑤₊⊑𝑧⊑𝑤₋, hence 𝑤₊⊑𝑤₋. We will distinguish the following five cases, and in each of them we will find a contradiction.(a)𝑤₊∈{𝑣₁,…,𝑣_𝑠} and 𝑤₋∈{𝑣_𝑠+1,…,𝑣_𝛽𝑘+1}. In this case 𝑤₊ and 𝑤₋ are two distinct elements having both rank 𝑅−(𝑘+1) and such that 𝑤₊⊑𝑤₋; it is not possible.(b)𝑤₊∈{𝑣₁,…,𝑣_𝑠} and 𝑤₋∈{𝑧₁,…,𝑧_𝑞}. In this case 𝑤₊ has rank 𝑅−(𝑘+1) while 𝑤₋ has rank 𝑅−(𝑘+2)<𝑅−(𝑘+1), and this contradicts the condition 𝑤₊⊑𝑤₋.(c)𝑤₊∈{𝑤_𝑡+1,…,𝑤_𝛽𝑘} and 𝑤₋∈{𝑣_𝑠+1,…,𝑣_𝛽𝑘+1}.This case is similar to the previous because 𝑤₊ has rank 𝑅−𝑘 while 𝑤₋ has rank𝑅−(𝑘+1).(d)𝑤₊∈{𝑤_𝑡+1,…,𝑤_𝛽𝑘} and 𝑤₋∈{𝑧₁,…,𝑧_𝑞}.This is similar to the previous because 𝑤₊ has rank 𝑅−𝑘 while 𝑤₋ has rank 𝑅−(𝑘+2).(e)𝑤₊∈{𝑣₁,…,𝑣_𝑠,𝑤_𝑡+1,…,𝑤_𝛽𝑘} and 𝑤₋=𝜃. In this case the condition 𝑤₊⊑𝑤₋ implies that 𝑤₊∈↓𝜃; since ↓𝜃=𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟) this is not possible since 𝑤₊∈𝑆^±1(𝑛,𝑟).(B3)Since 𝛼 is the minimum of 𝑆^±2(𝑛,𝑟) we have that 𝑆^±2(𝑛,𝑟)⊆↑𝛼, moreover, 𝛼∈↑𝑌₊; therefore 𝑆^±2(𝑛,𝑟)⊆↑𝛼⊆↑𝑌₊. Since Θ=𝜃^𝑐∈𝑌^𝑐−, it follows that ↑Θ⊆↑𝑌^𝑐−. By Lemma 4.2(i) we have then that (𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟))=↑Θ⊆↑𝑌^𝑐−. By Lemma 4.2(ii) we also have that (𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟))=↓𝜃⊆↓𝑌₋.
To complete the proof of (B3) let us observe that 𝑆^±1(𝑛,𝑟)=(⋃^𝑘𝑖=0ℜ_𝑖)⋃̇ℜ_𝑘+1⋃̇(⋃^{𝑅𝑗=𝑘+2}ℜ_𝑗), where (⋃^𝑘𝑖=0ℜ_𝑖)⊆↑ℜ_𝑘⊆↑𝑌₊, (⋃^{𝑅𝑗=𝑘+2}ℜ_𝑗)⊆↓ℜ_𝑘+2⊆↓𝑌₋.
Moreover, since ℜ_𝑘+1={𝑣₁,…,𝑣_𝑠}⋃̇{𝑣_𝑠+1,…,𝑣_𝛽𝑘+1},with {𝑣₁,…,𝑣_𝑠}⊆𝑌₊⊆↑𝑌₊ and {𝑣_𝑠+1,…,𝑣_𝛽𝑘+1}⊆𝑌₋⊆↓𝑌₋, then ℜ_𝑘+1⊆(↑𝑌₊⋃↓𝑌₋). This shows that 𝑆^±1(𝑛,𝑟)⊆(↑𝑌₊⋃↓𝑌₋). Since the six sublattices 𝑆^+𝑖(𝑛,𝑟), 𝑆^±𝑖(𝑛,𝑟) and 𝑆^−𝑖(𝑛,𝑟) for 𝑖=1,2 are a partition of 𝑆(𝑛,𝑟), the property (B3) is proved.Case (a2)
(B1)We begin to show that 𝛼∉𝑌^𝑐−. In fact, it holds that 𝛼=(𝑟(𝑟−1)⋯20∣0⋯0(𝑛−𝑟))^𝑐; suppose that 𝛼∈𝑌^𝑐−, and this shows that we obtain a contradiction. By 𝛼∈𝑌^𝑐− it follows 𝛼^𝑐=𝑡₁∈𝑌₋. But 𝑡₁ is the top of 𝑆^±1(𝑛,𝑟), so 𝑌₋=𝑡₁, since 𝑌₋ is antichain. This means that ℜ_𝑘+1=ℜ₀ and this is not possible, of course. Since 𝑌₊=𝑇₊⋃̇{𝛼}, we have (↓𝑌₊)⋂𝑌^𝑐−=((↓𝑇₊)⋂𝑌^𝑐−)⋃((↓𝛼)⋂𝑌^𝑐−), and, moreover, as in the proof of (B1) in the case (a1), we also have that (↓𝑇₊)⋂𝑌^𝑐−=∅; therefore, to prove (B1) it is sufficient to show that (↓𝛼)⋂𝑌^𝑐−=∅; As in the proof of (B1) in the case (a1), we have 𝑌^𝑐−⊆𝑆^±2(𝑛,𝑟)⋃̇{Θ}. Since 𝛼 is the minimum of 𝑆^±2(𝑛,𝑟) and Θ∉↓𝛼, it follows that the unique element of ↓𝛼 that can belong to 𝑌^𝑐− is 𝛼, but we have shown before that this is not true.(B2)As in the previous case we have (↑𝑌^𝑐−)⋂↓𝑌₋=∅; moreover (↑𝑌₊)⋂(↓𝑌₋)=((↑𝑇₊)⋂(↑𝛼))⋂(↓𝑌₋)((↑𝑇₊)⋂(↓𝑌₋))⋃((↑𝛼)⋂(↓𝑌₋)). As in the case (a1) we have ((↑𝑇₊)⋂(↓𝑌₋))=∅; therefore, to prove (B2) also in the case (a2), it is sufficient to show that ((↑𝛼)⋂(↓𝑌₋))=∅. As in the proof of (B2) in the case (a1) it results that ↓𝑌₋⊆𝑆^±2(𝑛,𝑟)⋃̇𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟), and, by definition of 𝛼, it is easy to observe that (↑𝛼)⋂(𝑆^±1(𝑛,𝑟)⋃̇𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟))=∅. Hence ((↑𝛼)⋂(↓𝑌₋))=∅.(B3)It is identical to that of case (a1).Step 3. The map A defined in (4.8) is such that 𝐴∈𝒲₊(𝑛,𝑟).
Since we have proved that⟨𝑌₊∣𝑌₋⟩ is a basis for 𝑆(𝑛,𝑟), by Lemma 4.4 it follows that the map 𝐴∈𝒲₊(𝑆(𝑛,𝑟),𝟐) if the two following identities hold: ↑𝑌₊↑𝑌^𝑐−=𝑆^±2(𝑛,𝑟)̇S⁺¹(𝑛,𝑟)̇𝑆⁺²(𝑛,𝑟)̇𝔘_𝑘̇𝑣₁↓𝑌₋,…,𝑣_𝑠,(4.9)=𝔅_𝑘̇𝑣_𝑠+1,…,𝑣_𝛽𝑘+1̇𝑆⁻¹(𝑛,𝑟)̇𝑆⁻²(𝑛,𝑟).(4.10)We prove at first (4.10). By definition of 𝔅_𝑘 and 𝑇₋ it easy to observe that 𝔅_𝑘⋃̇{𝑣_𝑠+1,…,𝑣_𝛽𝑘+1}⊆↓𝑇₋, and, moreover, by Lemma 4.2(ii) we also have that ↓𝜃=𝑆⁻¹(𝑛,𝑟)⋃𝑆⁻²(𝑛,𝑟); hence, since ↓𝑌₋=↓𝜃⋃↓𝑇₋, it results that 𝔅_𝑘⋃̇{𝑣_𝑠+1,…,𝑣_𝛽𝑘+1}⋃𝑆⁻¹(𝑛,𝑟)⋃̇𝑆⁻²(𝑛,𝑟)⊆↓𝑌₋. On the other hand, by Lemma 4.2(iv) we have that ↓𝑇₋⊆𝑆^±1(𝑛,𝑟)⋃̇𝑆⁻¹(𝑛,𝑟) because 𝑇₋ is a subset of 𝑆^±1(𝑛,𝑟). At this point let us note that the elements of ↓𝑇₋ that are also in 𝑆^±1(𝑛,𝑟) must belong necessarily to the subset 𝔅_𝑘⋃̇{𝑣_𝑠+1,…,𝑣_𝛽𝑘+1}. This proves the other inclusion and hence (4.10).
To prove now (4.9) we must distinguish the cases (a1) and (a2). We set Δ∶=𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟)⋃𝔘_𝑘⋃̇{𝑣₁,…,𝑣_𝑠}. Let us first examine the case (a1). Since 𝛼 is the minimum of 𝑆^±2(𝑛,𝑟), we have 𝑆^±2(𝑛,𝑟)⊆↑𝛼⊆↑𝑌₊. Moreover, since 𝑌₋=𝑇₋⋃{𝜃}, it follows that ↑𝑌^𝑐−⊇↑𝜃^𝑐=↑(Θ)=𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟) by lemma 3.2(i). Finally, since 𝔘_𝑘⋃̇{𝑣₁,…,𝑣_𝑠}⊆↑𝑇₊=↑𝑌₊, the inclusion ⊇ in (4.9) is proved. To prove the other inclusion ⊆ we begin to observe that ↑𝑌₊⋂(𝑆⁻¹(𝑛,𝑟)⋃𝑆⁻¹(𝑛,𝑟))=∅; therefore the elements of ↑𝑌₊ that are not in 𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟) must be necessarily in 𝑆^±1(𝑛,𝑟), and such elements, by definition of 𝑇₊, must be necessarily in 𝔘_𝑘⋃̇{𝑣₁,…,𝑣_𝑠}.This proves that ↑𝑌₊⊆Δ, For ↑(𝑌^𝑐−), we have ↑(𝑌^𝑐−)=↑(𝑇^𝑐−⋃{𝜃}^𝑐)=(↑𝑇^𝑐−)⋃(↑Θ), where ↑(Θ)=𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟) by Lemma 4.2(i), and, since 𝑇₋⊆𝑆^±1(𝑛,𝑟), also 𝑇^𝑐−⊆(𝑆^±1(𝑛,𝑟))^𝑐=𝑆^±2(𝑛,𝑟), by Lemma 4.2(v). Therefore ↑𝑇^𝑐−⊆↑𝑆^±2(𝑛,𝑟)⊆𝑆^±2(𝑛,𝑟)⋃𝑆⁺²(𝑛,𝑟) by Lemma 4.2(iii). This shows that ↑(𝑌^𝑐−)⊆𝑆^±2(𝑛,𝑟)⋃𝑆⁺²(𝑛,𝑟)⋃𝑆⁺¹(𝑛,𝑟)⊆Δ, hence, the inclusion ⊆. The proof of (4.9) in the case (a1) is therefore complete.
Finally, to prove (4.9) in the case (a2), it easy to observe that the only difference with respect to case (a1) is when we must show that ↑𝑌₊⊆Δ. In fact, in the case (a2) it results that 𝛼∉↑𝑇₊ and 𝑌₊=𝑇₊⋃{𝛼}, while 𝑌₋ is the same in both cases (a1) and (a2). Therefore, in the case (a2), the elements of ↑𝑌₊ that are not in 𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟) must be in (↑𝑇₊)⋂𝑆^±1(𝑛,𝑟) or in (↑𝛼)⋂𝑆^±1(𝑛,𝑟). As in the case (a1) we have (↑𝑇₊)⋂𝑆^±1(𝑛,𝑟)=𝔘_𝑘⋃̇{𝑣₁,…,𝑣_𝑠}, and, since 𝛼 is the minimum of 𝑆^±2(𝑛,𝑟), it results that ↑𝛼=↑𝑆^±2(𝑛,𝑟)⊆𝑆^±2(𝑛,𝑟)⋃𝑆⁺²(𝑛,𝑟) by Lemma 4.2(iii); hence (↑𝛼)⋂𝑆^±1(𝑛,𝑟)=∅. Therefore, also in the case (a2), the elements of ↑𝑌₊ that are not in 𝑆^±2(𝑛,𝑟)⋃̇𝑆⁺¹(𝑛,𝑟)⋃̇𝑆⁺²(𝑛,𝑟) must be necessarily in 𝔘_𝑘⋃̇{𝑣₁,…,𝑣_𝑠}. The other parts of the proof are same as in case (a1). Hence we have proved the identities (4.9) and (4.10). By Lemma 4.4 it follows then that the map 𝐴∈𝒲₊(𝑆(𝑛,𝑟),𝟐). Finally, by definition of 𝐴, we have obviously 𝐴(𝜃)=𝑁, 𝐴(𝜉₁)=𝑃, and 𝐴(Θ)=𝑃. This shows that 𝐴∈𝒲₊(𝑛,𝑟). The proof is complete.

Figure 3 

To conclude we emphasize the elegant symmetry of the induced partitions on 𝑆(𝑛,𝑟) from the boolean total maps 𝐴_𝑞’s constructed in the proof of the Theorem 4.5.