A Discrete Dynamical Model of Signed Partitions

Abstract

We use a discrete dynamical model with three evolution rules in order to analyze the structure of a partially ordered set of signed integer partitions whose main properties are actually not known. This model is related to the study of some extremal combinatorial sum problems.

1. Introduction

A dynamical system is by definition a system whose state changes with time 𝑡. We have a discrete dynamical system when 𝑡 is an integer or a natural number, and the elements of the system can be obtained in the form 𝑢_𝑡+1=𝐹(𝑢_𝑡), where 𝐹 is some global function which describes the evolution rule of the system (see [1]). In a series of very recent works, the theory of discrete dynamical systems has been applied in several contexts. In [2], the theory of discrete dynamical systems is applied in order to analyze some models of concurrent computing systems. In [3], a dynamical model of parallel computation on bi-infinite time scale with an approach similar to two-sided symbolic dynamics is constructed. In [4], the authors analyze the orbit structure of parallel discrete dynamical systems over directed dependency graphs, with Boolean functions as global functions. In [5], the authors extend the manner of defining the evolution update of discrete dynamical systems on Boolean functions, without limiting the local functions to being dependent restrictions of a global one. Finally, in [6] is given a complete characterization of the orbit structure of parallel discrete dynamical systems with maxterm and minterm Boolean functions as global functions. In [7], the authors have introduced the poset (𝑆(𝑛,𝑑,𝑟),⊑) of all the signed integer partitions 𝑟≥𝑎_𝑟≥⋯≥𝑎₁≥0≥𝑏₁≥⋯≥𝑏_𝑛−𝑟≥−(𝑛−𝑟), where the positive parts and the negative parts are all distinct between them and the number of nonzero parts is exactly 𝑑. In this case, the partial order ⊑ is that on the components. The concept of signed integer partition has been introduced in [8] and studied in [9] from an arithmetical point of view. In [7, 10], it is shown that the structure of the lattice 𝑆(𝑛,𝑑,𝑟) is strictly related to the study of some combinatorial extremal sum problems. In this paper, we study some particular type of signed partitions (specifically, of the signed partitions of 𝑆(𝑛,𝑑,𝑟)) by adopting the interesting point of view of the discrete dynamical models. In particular, we analyze the lattice 𝑆(𝑛,𝑑,𝑟) as a discrete dynamical model with three local evolution rules whose dynamics is studied in a sequential mode. The way to study a lattice of classical partitions as a discrete dynamic model having some particular evolution rules is implicit in [11], where Brylawski proposed a dynamical approach to study the lattice 𝐿_𝐵(𝑛) of all the partitions of the fixed positive integer 𝑛 with the dominance ordering. In such a context, a configuration of the system is represented by an ordered partition of an integer 𝑛, that is, a decreasing sequence 𝑎=(𝑎₁,…,𝑎_𝑛) having sum 𝑛, whose blocks of the corresponding Young diagram are considered as mobile sand grain, and the movement of a sand grain, respects the following rules.

Rule 1 (vertical rule). One grain can move from a column to the next column if the difference of height of these two columns is greater than or equal to 2.

Rule 2 (horizontal rule). If a column containing 𝑝+1 grains is followed by a sequence of columns containing 𝑝 grains and then one column containing 𝑝−1 grains, then one grain of the first column can slip to the last column.

In this paper, we prove that the covering relation in the lattice 𝑆(𝑛,𝑑,𝑟) is uniquely determined by three evolution rules of our discrete dynamical model. We articulate the proof of this result by contradiction by using only the first-order logic. This method of proof is very laborious; however, its advantage is that it can be easily implemented on a proof assistant checker based on the first-order logic (e.g., Mizar). The paper is articulated as follows. In Section 2, we recall some basic definitions and preliminary results, for example, the definition of 𝑆(𝑛,𝑑,𝑟) and some of its properties. In Section 3, we explain how to see the signed partitions of 𝑆(𝑛,𝑑,𝑟) as configurations of our discrete dynamical model and we also describe its evolution rules. In Section 4, we assume by contradiction that there is a generic element 𝑤^′ of 𝑆(𝑛,𝑑,𝑟) that covers another generic element 𝑤∈𝑆(𝑛,𝑑,𝑟) but 𝑤^′ is not generated by 𝑤 with none of our evolution rules. From this assumption we deduce several propositions and conditions that we use in order to prove our principal result, that is, Theorem 4.

2. Definitions and Preliminary Results

If (𝑋,≤) is a poset and 𝑥,𝑦∈𝑋, we write 𝑦⋗𝑥 (or𝑥⋖𝑦) if 𝑦 covers 𝑥. Now we briefly recall the definition of the lattice 𝑆(𝑛,𝑟) that we have introduced in [7] in a more formal context. Let 𝑛 and 𝑟 be two nonnegative integers such that 𝑟≤𝑛.

We call (𝑛,𝑟)-string a 𝑛-tuplet of integers 𝑎_𝑟⋯𝑎₁∣𝑏₁⋯𝑏_𝑛−𝑟, (1) such that(i)𝑎₁,…,𝑎_𝑟∈{1,…,𝑟,0};(ii)𝑏₁,…,𝑏_𝑛−𝑟∈{−1,…,−(𝑛−𝑟),0};(iii)𝑎_𝑟≥⋯≥𝑎₁≥0≥𝑏₁≥⋯≥𝑏_𝑛−𝑟;(iv)the unique element in (1) which can be repeated is 0.If 𝑤 is a (𝑛,𝑟)-string, we call parts of 𝑤 the integers 𝑎_𝑟,…,𝑎₁,𝑏₁⋯𝑏_𝑛−𝑟, nonnegative parts of 𝑤 the integers 𝑎_𝑟,…,𝑎₁, and nonpositive parts of 𝑤 the integers 𝑏₁⋯𝑏_𝑛−𝑟. We set 𝑤₊=𝑎_𝑟⋯𝑎₁| and 𝑤₋=|𝑏₁⋯𝑏_𝑛−𝑟. In some cases, we do not distinguish between nonnegative and nonpositive parts of 𝑤, and we write more simply 𝑤=𝑙₁⋯𝑙_𝑛 instead of 𝑤=𝑎_𝑟⋯𝑎₁|𝑏₁⋯𝑏_𝑛−𝑟. We also denote by |𝑤|_> the number of parts of 𝑤 that are strictly positive, with |𝑤|_< the number of parts of 𝑤 that are strictly negative, and we set ‖𝑤‖=|𝑤|_>+|𝑤|_<. 𝑆(𝑛,𝑟) is the set of all the (𝑛,𝑟)-strings. If 𝑤=𝑎_𝑟⋯𝑎₁|𝑏₁⋯𝑏_𝑛−𝑟 and 𝑤^′=𝑎^′_𝑟⋯𝑎^′₁|𝑏^′₁⋯𝑏^′_𝑛−𝑟 are two (𝑛,𝑟)-strings, we set 𝑤₊=𝑤^′₊ if 𝑎^′_𝑖=𝑎_𝑖 for all 𝑖=𝑟,…,1,𝑤₋=𝑤^′₋ if 𝑏^′_𝑗=𝑏_𝑗 for all 𝑗=1,…,𝑛−𝑟, and 𝑤=𝑤^′ if 𝑤₊=𝑤^′₊ and 𝑤₋=𝑤^′₋. On 𝑆(𝑛,𝑟), we consider the partial order on the components that we denote by ⊑. To simplify the notations, in all the numerical examples the integers on the right of the vertical bar | will be written without minus sign. Since (𝑆(𝑛,𝑟),⊑) is a finite distributive lattice, it is also graded, with minimal element 0⋯0|12⋯(𝑛−𝑟) and maximal element 𝑟(𝑟−1)⋯21|0⋯0.

We recall now the concept of involution poset (see [12, 13] for some recent studies on such class of posets). An involution poset (IP) is a poset (𝑋,≤,𝑐) with a unary operation 𝑐∶𝑥∈𝑋↦𝑥^𝑐∈𝑋, such that (I1)(𝑥^𝑐)^𝑐=𝑥, for all 𝑥∈𝑋; (I2) if 𝑥,𝑦∈𝑋 and if 𝑥≤𝑦, then 𝑦^𝑐≤𝑥^𝑐.The map 𝑐 is called complementation of 𝑋 and 𝑥^𝑐 the complement of 𝑥. Let us observe that if 𝑋 is an involution poset, by (𝐼1) it follows that 𝑐 is bijective and by (𝐼1) and (𝐼2) it holds that if 𝑥,𝑦∈𝑋 are such that 𝑥<𝑦, then 𝑦^𝑐<𝑥^𝑐. If (𝑋,≤,𝑐) is an involution poset and if 𝑍⊆𝑋, we will set 𝑍^𝑐={𝑧^𝑐∶𝑧∈𝑍}. We note that if 𝑋 is an involution poset, then 𝑋 is a self-dual poset because from (𝐼1) and (𝐼2) it follows that if 𝑥,𝑦∈𝑋, we have that 𝑥≤𝑦, if and only if 𝑦^𝑐≤𝑥^𝑐, and this is equivalent to say that the complementation is an isomorphism between 𝑋 and its dual poset 𝑋^∗. In [7], it has been shown that (𝑆(𝑛,𝑟),⊑) is an involution poset and its complementation map 𝑐 is the following: 𝑎_𝑘⋯𝑎₁0⋯0∣0⋯0𝑏₁⋯𝑏_𝑙^𝑐=𝑎^′_𝑟−𝑘⋯𝑎^′₁0⋯0∣0⋯0𝑏^′₁⋯𝑏^′_{𝑛−𝑟−𝑙}, (2) where {𝑎^′₁,…,𝑎^′_𝑟−𝑘} is the usual complement of {𝑎₁,…,𝑎_𝑘} in {1,…,𝑟}, and {𝑏^′₁,…,𝑏^′_{𝑛−𝑟−𝑙}} is the usual complement of {𝑏₁,…,𝑏_𝑙} in {−1,…,−(𝑛−𝑟)} (e.g., in 𝑆(7,4), we have that (4310|001)^𝑐=2000|023). If 𝑑 is an integer such that 0≤𝑑≤𝑛, we set now 𝑆(𝑛,𝑑,𝑟)={𝑤∈𝑆(𝑛,𝑟)∶‖𝑤‖=𝑑}. Its easy to see that (𝑆(𝑛,𝑑,𝑟),⊑) is a sublattice of (𝑆(𝑛,𝑟),⊑) and obviously |𝑆(𝑛,𝑑,𝑟)|=(^𝑛_𝑑).

We call signed Young diagrams (briefly SYD) an ordered couple 𝐷=𝐷₁∶𝐷₂, where 𝐷₁ is a Young diagram of an integer partition built with decreasing columns instead with decreasing rows, and 𝐷₂ is a Young diagram of an integer partition built with increasing columns. In the sequel, we will call pile a column of a Young diagram and grain a square of a pile. For example, (3) is determined by the partitions 4331 and 113. If 𝑤=𝑎_𝑟⋯𝑎₁|𝑏₁⋯𝑏_𝑛−𝑟∈𝑆(𝑛,𝑟), we denote by 𝐷(𝑤) the signed Young diagram 𝐷(𝑤)=𝐷⁺(𝑤)∶𝐷⁻(𝑤), where 𝐷⁺(𝑤) is the Young diagram of the partition (𝑎_𝑟,…,𝑎₁) and 𝐷⁻(𝑤) is the Young diagram of the partition (−𝑏₁,…,−𝑏_𝑛−𝑟). Now, if 𝐷=𝐷₁∶𝐷₂ is a SYD, where (𝑎_𝑡,…,𝑎₁) is the decreasing partition having 𝐷₁ as Young diagram and ((−𝑏₁),…,(−𝑏_𝑠)) is the increasing partition having 𝐷₂ as Young diagram, we set 𝑝(𝐷)=𝑎_𝑡⋯𝑎₁|(−𝑏₁)⋯(−𝑏_𝑠) and we call 𝑝(𝐷) the signed partition of 𝐷. We say that a signed Young diagram 𝐷 is a (𝑛,𝑟)-SYD if it results that 𝑝(𝐷)∈𝑆(𝑛,𝑟) and we denote with 𝑌𝑆(𝑛,𝑟) the set of all the (𝑛,𝑟)-SYD. It is clear that the map 𝜙∶𝑤↦𝐷(𝑤) is a bijection between 𝑆(𝑛,𝑟) and 𝑌𝑆(𝑛,𝑟), whose inverse function is the map 𝜓∶𝐷↦𝑝(𝐷); therefore, in the sequel, we will identify the elements of 𝑆(𝑛,𝑟) with the (𝑛,𝑟)-SYDs of 𝑌𝑆(𝑛,𝑟). If 𝐴 and 𝐵 are two usual Young diagrams, we set 𝐴⊆𝐵 if 𝐴 is a subdiagram of 𝐵. If 𝐷=𝐷₁∶𝐷₂ and 𝐷^′=𝐷^′₁∶𝐷^′₂ are two SYDs, we set 𝐷⊑𝐷^′⟺𝐷₁⊆𝐷^′₁,𝐷₂⊇𝐷^′₂. (4) Therefore, it is clear that we can identify (𝑆(𝑛,𝑟),⊑) with (𝑌𝑆(𝑛,𝑟),⊑). Now we set 𝑌𝑆(𝑛,𝑑,𝑟)={𝐷(𝑤)∶𝑤∈𝑆(𝑛,𝑑,𝑟)}. Obviously we can identify (𝑌𝑆(𝑛,𝑑,𝑟),⊑) with (𝑆(𝑛,𝑑,𝑟),⊑). If (𝐻) is a statement, in the sequel we denote by ¬(𝐻) the negation of (𝐻).

3. Description of the Evolution Rules

In our dynamic model, the set of all configurations is exactly 𝑌𝑆(𝑛,𝑑,𝑟). Our goal is to define some rules of evolution that starting from the minimum of 𝑌𝑆(𝑛,𝑑,𝑟) allow us to reconstruct the Hasse diagram of 𝑌𝑆(𝑛,𝑑,𝑟) (and therefore to determine the covering relations in 𝑆(𝑛,𝑑,𝑟)).

Let 𝑤=𝑎_𝑟⋯𝑎₁|𝑏₁⋯𝑏_𝑛−𝑟∈𝑆(𝑛,𝑑,𝑟) and 𝐷=𝐷(𝑤)=𝐷⁺(𝑤)∶𝐷⁻(𝑤). If 1≤𝑖≤𝑟, we call 𝑖th-plus pile of 𝐷, and we denote it by 𝐶₊(𝑖,𝐷), the 𝑖th pile of 𝐷⁺(𝑤) that corresponds to part 𝑎_𝑖 of 𝑤 and, if 1≤𝑗≤𝑛−𝑟, we call 𝑗th-minus pile of 𝐷, and we denote it by 𝐶₋(𝑗,𝐷), the 𝑗th pile of 𝐷⁻(𝑤) that corresponds to part 𝑏_𝑗 of 𝑤. We call the pile 𝐶⁺(𝑖,𝐷)  plus  singleton  pile if 𝑎_𝑖=1 and 𝐶⁻(𝑗,𝐷)minus  singleton  pile if 𝑏_𝑗=1. Let us note that if there exists a plus singleton pile, then it is necessarily unique, analogously for a minus singleton pile. If 1<𝑖≤𝑟, we set Δ⁺_𝑖(𝑤)=𝑎_𝑖−𝑎_𝑖−1 and we call Δ⁺_𝑖(𝑤) the plus height difference of 𝑤 in 𝑖. If 1<𝑗≤𝑛−𝑟, we set Δ⁻_𝑗(𝑤)=|𝑏_𝑗|−|𝑏_𝑗−1| and we call Δ⁻_𝑗(𝑤) the minus height difference of 𝑤 in 𝑗. If 1<𝑖≤𝑟, we say that 𝑤 has a plus cliff at 𝑖 if Δ⁺_𝑖(𝑤)≥2. If 1<𝑗≤𝑛−𝑟, we say that 𝑤 has a minus cliff at 𝑗 if Δ⁻_𝑗(𝑤)≥2.

Remark 1. When we apply the following rules to one element 𝑤∈𝑆(𝑛,𝑟), we impose that there is an “invisible” extra pile in the imaginary place (𝑟+1) of 𝐷⁺(𝑤) having exactly 𝑟+1 grains. This is only a formal trick for decreas the number of rules necessary for our model; therefore, the (𝑟+1)th pile must be not considered as a part of 𝑤.

3.1. Evolution Rules

Rule 1. If the 𝑖th-plus pile has at least one grain and if 𝑤 has a plus cliff at 𝑖+1, then one grain must be added on the 𝑖th-plus pile: (5)

Rule 2. If there is not a plus singleton pile and there is a minus singleton pile, then the latter must be shifted to the side of the lowest non empty plus pile: (6)

Rule 3. One grain must be deleted from the 𝑗th-minus pile if 𝑤 has a minus cliff at 𝑗: (7)

Remark 2. (i) Under the hypothesis in Rule 3, the 𝑗th plus must have at least 2 grains.
(ii) In Rule 2, the lowest non empty plus pile can also be the invisible column in the place 𝑟+1. In this case, all the plus piles are empty and a possible minus singleton pile must be shifted in the place 𝑟.

4. Main Result

In the sequel, we write 𝑤→^𝑘𝑤^′ (or 𝑤^′=𝑤→^𝑘) to denote that 𝑤^′ is an 𝑛-tuplet of integers obtained from 𝑤 applying the Rule 𝑘, for 𝑘=1,2,3, and that ¬(𝑤→^𝑘𝑤^′) if the condition 𝑤→^𝑘𝑤^′ is false. We also set ∇(𝑤)=𝑤^′∶𝑤⟶^𝑘𝑤^′,𝑘=1,2,3. (8)

Proposition 3. If 𝑤∈𝑆(𝑛,𝑑,𝑟), then ∇(𝑤)⊆{𝑤^′∈𝑆(𝑛,𝑑,𝑟)∶𝑤^′⋗𝑤}.

Proof. Let 𝑤=𝑎_𝑟⋯𝑎₁|𝑏₁⋯𝑏_𝑛−𝑟∈𝑆(𝑛,𝑑,𝑟), 𝑎_𝑟+1=𝑟+1 (the invisible pile in the place 𝑟+1), and 𝐷=𝐷(𝑤). We distinguish the three possible cases related to the previous rules.
Case  1. Let us assume that 𝑟≥𝑖≥1, 𝑎_𝑖≠0, and that 𝑤 has a plus cliff at 𝑖+1. If 𝑤^′=𝑤→¹, then 𝑤^′=𝑎_𝑟⋯𝑎_𝑖+1(𝑎_𝑖+1)𝑎_𝑖−1⋯𝑎₁|𝑏₁⋯𝑏_𝑛−𝑟. It is clear that ‖𝑤^′‖=𝑑 because 𝑎_𝑖≠0. Since there is a plus cliff at 𝑖+1, we have 𝑎_𝑖+1−𝑎_𝑖≥2; hence, 𝑎_𝑖+1≥𝑎_𝑖+2>𝑎_𝑖+1>𝑎_𝑖>𝑎_𝑖−1, and this implies that 𝑤^′∈𝑆(𝑛,𝑑,𝑟). We must show now that 𝑤^′ covers 𝑤 in 𝑆(𝑛,𝑑,𝑟). Since 𝑤 and 𝑤^′ differ between them only in the place 𝑖 for 𝑎_𝑖 and 𝑎_𝑖+1, respectively, it is clear that there does not exist an element 𝑧∈𝑆(𝑛,𝑑,𝑟) such that 𝑤⊏𝑧⊏𝑤^′. Hence, 𝑤^′⋗𝑤.
Case  2. Let us assume that in 𝐷(𝑤) there is not a plus singleton pile and that there is a minus singleton pile (we say 𝐶⁻(𝑗,𝐷), for some 1≤𝑗≤𝑛−𝑟). Since 𝑎_𝑟+1=𝑟+1, we can assume that 𝑎_𝑖+1>0,𝑎_𝑖=0, for some 1≤𝑖≤𝑟. This means that 𝑤 has the following form: 𝑤=𝑎_𝑟⋯𝑎_𝑖+100⋯0|0⋯0(−1)𝑏_𝑗+1⋯𝑏_𝑛−𝑟, where 𝑎_𝑖+1>1 (otherwise 𝐷(𝑤) has a plus singleton pile). Applying Rule 2 to 𝑤, we obtain 𝑤^′=𝑤→², where 𝑤^′=𝑎_𝑟⋯𝑎_𝑖+110⋯0|0⋯00𝑏_𝑗+1⋯𝑏_𝑛−𝑟. It is clear then that 𝑤^′∈𝑆(𝑛,𝑟) and ‖𝑤^′‖=𝑑 since 𝑤^′ is obtained from 𝑤 with only a shift of the pile-1 to the left in the place 𝑖. Let us note that the only elements 𝑧₁,𝑧₂∈𝑆(𝑛,𝑟) such that 𝑤⊏𝑧₁⊏𝑤^′ and 𝑤⊏𝑧₂⊏𝑤^′ are 𝑧₁=𝑎_𝑟⋯𝑎_𝑖+110⋯0|0⋯0(−1)𝑏_𝑗+1⋯𝑏_𝑛−𝑟 and 𝑧₂=𝑎_𝑟⋯𝑎_𝑖+100⋯0|0⋯00𝑏_𝑗+1⋯𝑏_𝑛−𝑟, but ‖𝑧₁‖=𝑑+1 and ‖𝑧₂‖=𝑑−1; hence, 𝑧₁,𝑧₂ are not elements of 𝑆(𝑛,𝑑,𝑟). This implies that 𝑤^′ covers 𝑤 in 𝑆(𝑛,𝑑,𝑟).
Case  3. If 1<𝑗≤𝑛−𝑟 and 𝑤 has a minus cliff at 𝑗, we apply Rule 3 to 𝑤 on the pile 𝐶⁻(𝑗,𝐷) and we obtain 𝑤^′=𝑤→³, where 𝑤^′=𝑎_𝑟⋯𝑎₁|𝑏₁⋯𝑏_𝑗−1(𝑏_𝑗+1)𝑏_𝑗+1⋯𝑏_𝑛−𝑟. Since 𝑤 has a minus cliff at 𝑗, we have −𝑏_𝑗+𝑏_𝑗−1=|𝑏_𝑗|−|𝑏_𝑗−1|≥2; therefore, 𝑤^′∈𝑆(𝑛,𝑟) because 0≥𝑏_𝑗−1≥𝑏_𝑗+2>𝑏_𝑗+1>𝑏_𝑗>𝑏_𝑗+1 and ‖𝑤^′‖=𝑑 since 𝑏_𝑗≤−2 implies 𝑏_𝑗+1<0. As in Case 1, we note that 𝑤^′ covers 𝑤 in 𝑆(𝑛,𝑑,𝑟) because they differ between them only for a grain in the place 𝑗.

The main result of this paper is to prove that if 𝑤,𝑤^′∈𝑆(𝑛,𝑑,𝑟), then 𝑤^′ covers 𝑤 in 𝑆(𝑛,𝑑,𝑟) if and only if 𝑤→^𝑘𝑤^′ for some 𝑘∈{1,2,3}. Obviously this result can be reformulated as follows.

Theorem 4. If 𝑤∈𝑆(𝑛,𝑑,𝑟), then ∇(𝑤)={𝑤^′∈𝑆(𝑛,𝑑,𝑟)∶𝑤^′⋗𝑤}.

By Proposition 3, to prove Theorem 4, it suffices to show that {𝑤^′∈𝑆(𝑛,𝑑,𝑟)∶𝑤^′⋗𝑤}⊆∇(𝑤). To show this inclusion, we proceed by contradiction as follows. We start with the following hypotheses: 𝑤,𝑤^′∈𝑆(𝑛,𝑑,𝑟),𝑤=𝑎_𝑟⋯𝑎₁∣𝑏₁⋯𝑏_𝑛−𝑟,𝑤^′=𝑎^′_𝑟⋯𝑎^′₁∣𝑏^′₁⋯𝑏^′_𝑛−𝑟,𝑤^′⋗𝑤 (9) and we also assume that ¬𝑤⟶^𝑘𝑤^′foreach𝑘=1,2,3. (10) Then the proof of Theorem 4 will consist into show that the hypotheses (9) and the conditions (10) always lead to contradictory cases, so that it must be necessarily verified the inclusion {𝑤^′∈𝑆(𝑛,𝑑,𝑟)∶𝑤^′⋗𝑤}⊆∇(𝑤).

We will articulate the proof of Theorem 4 in several cases. Each possible case will be placed in the form of a proposition and we will obtain a contradiction in all cases, showing, hence, the thesis of the theorem. Before proceeding, let us note that in some cases we also use the alternative notations 𝑤=𝑙₁⋯𝑙_𝑛 and 𝑤^′=𝑙^′₁⋯𝑙^′_𝑛.

We set now Ω₁(𝑤)=𝑖∈{1,…,𝑟}∶𝑎_𝑖>0,𝑎_𝑖+1−𝑎_𝑖,Ω≥2₃(𝑤)=𝑗∈{2,…,𝑛−𝑟}∶𝑏_𝑗−1−𝑏_𝑗,Λ≥2₊𝑤,𝑤^′=ℎ∈{1,…,𝑟}∶𝑎^′_ℎ>𝑎_ℎ,Λ⁻𝑤,𝑤^′=𝑙∈{1,…,𝑛−𝑟}∶𝑏^′_𝑙>𝑏_𝑙. (11)Formalization of Rules 1, 2, and 3.

The description of Rules 1, 2, and 3 in terms of parts of 𝑤 and 𝑤^′ is the following: (1)𝑤→¹𝑤^′ is equivalent to Ω₁(𝑤)≠∅ AND (∃𝑖∈Ω₁(𝑤)∶(𝑎^′_𝑖=𝑎_𝑖+1) AND (Λ⁺(𝑤,𝑤^′)⧵{𝑖}=Λ⁻(𝑤,𝑤^′)=∅)); (2)𝑤→²𝑤^′ is equivalent to (forall𝑘∈{1,…,𝑟}𝑎_𝑘≠1) AND (∃𝑗∈{1,…,𝑛−𝑟} such that 𝑏_𝑗=−1) AND (∃𝑖∈{1,…,𝑟} such that 𝑎_𝑖+1>0, 𝑎_𝑖=0) AND (Λ⁺(𝑤,𝑤^′)⧵{𝑖}=Λ⁻(𝑤,𝑤^′)⧵{𝑗}=∅, 𝑎^′_𝑖=1, 𝑏^′_𝑗=0); (3)𝑤→³𝑤^′ is equivalent to Ω₃(𝑤)≠∅ AND (∃𝑗∈Ω₃(𝑤)∶(𝑏^′_𝑗=𝑏_𝑗+1) AND (Λ⁺(𝑤,𝑤^′)=Λ⁻(𝑤,𝑤^′)⧵{𝑗}=∅)).

Negation of Rules 1, 2, and 3.  (1)¬(𝑤→¹𝑤^′)⇔ [(1A^′) OR (1B)], where (1B) is (Ω₁(𝑤)=∅), (1A^′) is [Ω₁(𝑤)≠∅ AND (forall𝑖∈Ω₁(𝑤) (𝑎^′_𝑖≠𝑎_𝑖+1) OR (Λ⁺(𝑤,𝑤^′)⧵{𝑖}≠∅ OR Λ⁻(𝑤,𝑤^′)≠∅))]. (2)¬(𝑤→²𝑤^′)⇔ [(2A^′) OR (2B) OR (2C) OR (2D)], where (2A^′) is (forall𝑘∈{1,…,𝑟}𝑎_𝑘≠1) AND  (∃𝑗∈{1,…,𝑛−𝑟} such that 𝑏_𝑗=−1) AND  (∃𝑖∈{1,…,𝑟} such that 𝑎_𝑖+1>0,𝑎_𝑖=0) AND ((Λ⁺(𝑤,𝑤^′)⧵{𝑖}≠∅ OR Λ⁻(𝑤,𝑤^′)⧵{𝑗}≠∅) OR 𝑎^′_𝑖≠1 OR 𝑏^′_𝑗≠0); (2B) is (∃𝑘∈{1,…,𝑟} such that 𝑎_𝑘=1); (2C) is (forall𝑗∈{1,…,𝑛−𝑟}𝑏_𝑗≠−1); (2D) is (forall𝑖∈{1,…,𝑟}𝑎_𝑖+1=0 OR 𝑎_𝑖>0). (3)¬(𝑤→³𝑤^′)⇔ [(3A^′) OR (3B)], where (3B) is (Ω₃(𝑤)=∅); (3A^′) is [Ω₃(𝑤)≠∅ AND (forall𝑗∈Ω₃(𝑤) (𝑏^′_𝑗≠𝑏_𝑗+1) OR (Λ⁺(𝑤,𝑤^′)≠∅ OR Λ⁻(𝑤,𝑤^′)⧵{𝑗}≠∅))]. we set (1.1b)∶𝑤=00⋯0∣𝑏₁⋯𝑏_𝑛−𝑟,(1.2b)∶𝑤=𝑟(𝑟−1)⋯(𝑘+1)𝑘⋯0∣𝑏₁⋯𝑏_𝑛−𝑟forsome𝑘∈{1,…,𝑟}. (12)(13)

Proposition 5. Consider (1B) ⇒ ((1.1b) OR (1.2b)).

Proof. The condition Ω₁(𝑤)=∅ means that for all 𝑖∈{1,…,𝑟},𝑎_𝑖=0 or 0≤𝑎_𝑖+1−𝑎_𝑖≤1 (i.e., 𝑎_𝑖+1−𝑎_𝑖=0 or 𝑎_𝑖+1−𝑎_𝑖=1). We have then the following possibilities: (j)𝑎_𝑖=0; (jj)𝑎_𝑖+1−𝑎_𝑖=0 (and therefore 𝑎_𝑖+1=𝑎_𝑖=0 since 𝑤∈𝑆(𝑛,𝑟)); (jjj)𝑎_𝑖+1=𝑎_𝑖+1. If 𝑎_𝑖=0 for all 𝑖∈{1,…,𝑟} then (12) holds. Let us assume therefore that 𝑘∈{1,…,𝑟} is the smallest index such that 𝑎_𝑘>0; hence, 𝑎_𝑟>𝑎_𝑟−1>⋯>𝑎_𝑘+1>𝑎_𝑘>0 and 𝑎_𝑘−1=⋯=𝑎₁=0. This means that for all indexes 𝑖∈{𝑟,𝑟−1,…,𝑘+1,𝑘} holds necessarily (jjj). In particular, for 𝑖=𝑟 we have 𝑟+1=𝑎_𝑟+1=𝑎_𝑟+1; hence, 𝑎_𝑟=𝑟. Now, if 𝑖=𝑟−1, 𝑟=𝑎_𝑟=𝑎_𝑟−1+1, that is, 𝑎_𝑟−1=𝑟−1. Iterating (13) follows.

We set (2.1d)∶𝑤=𝑟(𝑟−1)⋯21∣𝑏₁⋯𝑏_𝑛−𝑟. (14)

Proposition 6. Consider (2D) ⇔ (2.1d).

Proof. By (2D), if we take 𝑖=𝑟, we have 𝑎_𝑟+1=0 or 𝑎_𝑟>0; therefore, it is necessarily 𝑎_𝑟>0 because by hypothesis, 𝑎_𝑟+1=𝑟+1>0. Thus, still by (2D), for 𝑖=𝑟−1, we have 𝑎_(𝑟−1)+1=𝑎_𝑟=0 or 𝑎_𝑟−1>0; therefore, it must be 𝑎_𝑟−1>0. By iterating, we can deduce that 𝑎_𝑟>0,𝑎_𝑟−1>0,…, 𝑎₁>0; hence, 𝑎₁=1, 𝑎₂=2,…, 𝑎_𝑟=𝑟 because 𝑤 is an element of 𝑆(𝑛,𝑟), that is, exactly (14). On the other side, if 𝑤=𝑟(𝑟−1)⋯21|𝑏₁⋯𝑏_𝑛−𝑟, then it is immediate to note that (2D) is verified.

We set (3.1b)∶𝑤=𝑎_𝑟⋯𝑎₁()∣00⋯0,3.2b∶𝑤=𝑎_𝑟⋯𝑎₁()⋯(){}.∣00⋯0−1)(−2−𝑙forsome𝑙∈1,…,𝑛−𝑟 (15)(16)

Proposition 7. Consider (3B) ⇒ ((3.1b) OR (3.2b)).

Proof. The condition Ω₃(𝑤)=∅ means that for all 𝑗∈{2,…,𝑛−𝑟},0≤𝑏_𝑗−1−𝑏_𝑗≤1 (i.e., 𝑏_𝑗−1−𝑏_𝑗=0 or 𝑏_𝑗−1−𝑏_𝑖=1). We have then the following possibilities:(i)𝑏_𝑗−1−𝑏_𝑗=0 (and therefore 𝑏_𝑗−1=𝑏_𝑖=0 since 𝑤∈𝑆(𝑛,𝑟));(ii)𝑏_𝑗−1=𝑏_𝑗+1.
If (i) holds for all 𝑗∈{2,…,𝑛−𝑟}, then (15) holds. We can assume therefore that 𝑘∈{2,…,𝑛−𝑟} is the smallest index such that (ii) holds. Let us observe then that if in a place 𝑗∈{2,…,𝑛−𝑟}, (ii) holds; then also in the places 𝑗+1,…,𝑛−𝑟, (ii) holds. In fact, if 𝑏_𝑗=𝑏_𝑗−1−1 and by absurd 𝑏_𝑗=𝑏_𝑗+1=0, then 𝑏_𝑗−1=1 which is a contradiction because 0≥𝑏_𝑗−1. Therefore, if (ii) holds in 𝑗, then it also holds in 𝑗+1, and therefore, by iteration, it holds for 𝑗+1,𝑗+2,…,𝑛−𝑟. Now we assume at first that 𝑘=2. In this case, (ii) holds for all indexes in {2,…,𝑛−𝑟}; therefore, 𝑏₁=0 or 𝑏₁=−1 and 𝑏₂=𝑏₁−1, 𝑏₃=𝑏₂−1⋯. If 𝑏₁=0, then 𝑤=𝑎_𝑟⋯𝑎₁|0(−1)(−2)⋯(−(𝑛−𝑟)+1) and if 𝑏₁=−1, then 𝑤=𝑎_𝑟⋯𝑎₁|(−1)(−2)⋯(−(𝑛−𝑟)); thus in both cases (ii) holds. Let now 𝑘≥3. Then 𝑏_𝑘−1=𝑏_𝑘−2=0 (because in the place 𝑘−1 (i) holds) and hence 0=𝑏_𝑘−1=𝑏_𝑘−2=⋯=𝑏₂=𝑏₁. Moreover, since (i) holds in the places 𝑘,𝑘+1,…,𝑛−𝑟, we have 𝑏_𝑘=𝑏_𝑘−1−1=0−1=−1; therefore, 𝑏_𝑘+1=𝑏_𝑘−1=−2,…, 𝑏_𝑛−𝑟=𝑏_{𝑛−𝑟−1}−1=−(𝑛−𝑟−𝑘)−1=−𝑙, where 𝑙=𝑛−𝑟−𝑘+1. This completes the proof.

Proposition 8. Consider (3.1b) ⇒ (𝑤₋=𝑤^′₋=∣00⋯0).

Proof. Since 𝑤⊏𝑤^′, it must be 0≥𝑏^′_𝑗≥𝑏_𝑗, where 𝑏_𝑗=0 for all 𝑗∈{1,…,𝑛−𝑟} because (15) implies 𝑤=𝑎_𝑟⋯𝑎₁|00⋯0. Therefore, 𝑏_𝑗=𝑏^′_𝑗=0 for all 𝑗∈{1,…,𝑛−𝑟}; that is, 𝑤^′=𝑎^′_𝑟⋯𝑎^′₁∣00⋯0.

Proposition 9. Consider(i)(1.1b) AND (2B) ⇒ absurd,(ii)(1.1b) AND (2.1d) ⇒ absurd,(iii)(2C) AND (3.2b) ⇒ absurd,(iv)(1.1b) AND (3.1b) ⇒ absurd,(v)(1.2b) AND (3.1b) ⇒ absurd.

Proof. (i), (ii), and (iii) are obvious.(iv) By (12) and (15), it follows that 𝑤=0⋯0|0⋯0, absurd.(v) By Proposition 8, we have 𝑤^′₋=𝑤₋. Then, since 𝑤⊏𝑤^′, ‖𝑤^′‖=‖𝑤‖ and 𝑤^′₋=𝑤₋, by (13) we deduce that 𝑤^′₊=𝑤₊. Hence, the absurd 𝑤^′=𝑤.

Proposition 10. Consider (1.2b) AND (2B) ⇒ (2.1d).

Proof. By (2B), it follows that in (13) it must be 𝑘=1; therefore, 𝑤=𝑟(𝑟−1)⋯21|𝑏₁⋯𝑏_𝑛−𝑟.

Proposition 11. Consider (2.1d) AND (3.2b) ⇒ absurd.

Proof. By (14), it is immediate to observe that 𝑤^′₊=𝑤₊ because 𝑤⊏𝑤^′. Then by (16), since 𝑤^′₊=𝑤₊, ‖𝑤^′₋‖=‖𝑤₋‖ and 𝑤⊏𝑤^′, it must also be 𝑤^′₋=|00⋯0(−1)(−2)⋯(−𝑙). Hence, we obtain 𝑤=𝑤^′, absurd.

Proposition 12. Consider (2𝐴^′) ⇒ absurd.

Proof. We set(F) [(forall𝑘∈{1,…,𝑟}𝑎_𝑘≠1) AND (∃𝑗∈{1,…,𝑛−𝑟} such that 𝑏_𝑗=−1) AND (∃𝑖∈{1,…,𝑟} such that 𝑎_𝑖+1>0,𝑎_𝑖=0)]; (G) (Λ⁺(𝑤,𝑤^′)⧵{𝑖}≠∅ OR Λ⁻(𝑤,𝑤^′)⧵{𝑗}≠∅);(H) (𝑎^′_𝑖≠1 OR 𝑏^′_𝑗≠0).
Then (2A^′) is equivalent to ((F) AND (G)) OR ((F) AND (H)). We show at first that ((F) AND (H)) leads to a contradiction. Next we show that ((F) AND (G) AND ¬(H)) also leads to an absurd. This will prove the thesis;
((F) AND (H)) ⇒ absurd.
By (F), it follows that 𝑤 satisfies the hypothesis of Rule 2; therefore, by Proposition 3 we can take an element 𝑧∈𝑆(𝑛,𝑑,𝑟) such that 𝑧=𝑤→². As in the proof of the Case 2 in Proposition 3, it results that 𝑤 and 𝑧 must have the following form: 𝑤=𝑎_𝑟⋯𝑎_𝑖+100⋯0∣0⋯0(−1)𝑏_𝑗+1⋯𝑏_𝑛−𝑟,𝑧=𝑎_𝑟⋯𝑎_𝑖+110⋯0∣0⋯00𝑏_𝑗+1⋯𝑏_𝑛−𝑟, (17)(18) where 𝑎_𝑖+1>1 and 𝑏_𝑗+1<−1. We distinguish now several cases.
(I) 𝑏^′_𝑗≠0. Then 𝑏^′_𝑗≤−1, and also 𝑏^′_𝑗≥𝑏_𝑗=−1 because 𝑤⊏𝑤^′; hence 𝑏^′_𝑗=−1.
We note now that 𝑎^′_𝑟>0,…,𝑎^′_𝑖+1>0 (19) because 𝑎^′_𝑟≥𝑎_𝑟>0,…,𝑎^′_𝑖+1≥𝑎_𝑖+1>0. Moreover, since 𝑏^′_𝑗=−1, it follows that −1=𝑏^′_𝑗>𝑏^′_𝑗+1>⋯>𝑏^′_𝑛−𝑟 (20) because 𝑤^′∈𝑆(𝑛,𝑟). Now, since 𝑤∈𝑆(𝑛,𝑑,𝑟), from the form of 𝑤 it results that 𝑑=(𝑟−𝑖)+(𝑛−𝑟−𝑗+1). Moreover, since also 𝑤^′∈𝑆(𝑛,𝑑,𝑟), we have ‖𝑤^′‖=𝑑=(𝑟−𝑖)+(𝑛−𝑟−𝑗+1). Therefore, since in (19), there are exactly 𝑟−𝑖 inequalities and in (20) there are exactly 𝑛−𝑟−𝑗+1 inequalities; it follows that 𝑤^′=𝑎^′_𝑟⋯𝑎^′_𝑖+100⋯0∣0⋯0(−1)𝑏^′_𝑗+1⋯𝑏^′_𝑛−𝑟. (21)
(II) 𝑎^′_𝑖=0. Since 𝑤⊏𝑤^′, by (17) we have two possible cases: 𝑤^′ has the form (21) or 𝑤^′=𝑎^′_𝑟⋯𝑎^′_𝑖+100⋯0∣0⋯00𝑏^′_𝑗+1⋯𝑏^′_𝑛−𝑟 (22) with 𝑎^′_𝑖+1≥𝑎_𝑖+1>1 and 0≥𝑏^′_𝑗+1. If (22) holds, then ‖𝑤^′‖≤(𝑟−𝑖)+(𝑛−𝑟−𝑗)=𝑑−1, which is absurd because 𝑤^′∈𝑆(𝑛,𝑑,𝑟); hence it must hold necessarily (21).
(III) 𝑎^′_𝑖>1. Since 𝑤⊏𝑤^′ and 𝑑=‖𝑤‖=‖𝑤^′‖, by (17) it follows that 𝑤^′=𝑎^′_𝑟⋯𝑎^′_𝑖+1𝑎^′_𝑖0⋯0∣0⋯00𝑏^′_𝑗+1⋯𝑏^′_𝑛−𝑟. (23) Then from the hypothesis 𝑎^′_𝑖≥2 and by (18), we have 𝑤⊏𝑧⊏𝑤^′, which is a contradiction.
Hence the unique possibility is that 𝑤^′ has the form (21). Since 𝑤^′⋗𝑤, we have 𝑎^′_𝑘≥𝑎_𝑘 for 𝑘=𝑟,…,𝑖+2,𝑖+1 and 𝑏^′_𝑙≥𝑏_𝑙 for 𝑙=𝑗+1,𝑗+2,…,𝑛−𝑟. Moreover, at least one inequality of the type 𝑎^′_𝑘>𝑎_𝑘 or the type 𝑏^′_𝑙>𝑏_𝑙 must hold. We distinguish now the several cases.
(Δ1) We assume that a unique inequality of the type 𝑎^′_𝑝>𝑎_𝑝 holds for some 𝑝=𝑟,…,𝑖+2,𝑖+1. Therefore, we have 𝑎^′_𝑘=𝑎_𝑘 for all 𝑘≠𝑝 and 𝑏^′_𝑙=𝑏_𝑙 for all 𝑙=𝑗+1,𝑗+2,…,𝑛−𝑟. Then, since 𝑎_𝑝+1=𝑎^′_𝑝+1>𝑎^′_𝑝>𝑎_𝑝>0 (let us note that the previous inequalities also hold if 𝑝=𝑟 because 𝑎_𝑟+1=𝑟+1=𝑎^′_𝑟+1), we have that 𝑎_𝑝+1−𝑎_𝑝≥2 and 𝑎_𝑝>0. Hence 𝑤 satisfies the hypotheses of Rule 1 in the place 𝑝 and so 𝑤→¹𝑢, where 𝑢=𝑎_𝑟⋯𝑎_𝑝+1𝑎_𝑝+1𝑎_𝑝−1⋯𝑎_𝑖+1()𝑏0⋯0∣0⋯0−1_𝑗+1⋯𝑏_𝑛−𝑟=𝑎^′_𝑟⋯𝑎^′_𝑝+1𝑎_𝑝+1𝑎^′_𝑝−1⋯𝑎^′_𝑖+1()𝑏0⋯0∣0⋯0−1^′_𝑗+1⋯𝑏^′_𝑛−𝑟, (24) with 𝑢∈𝑆(𝑛,𝑑,𝑟). Now, if 𝑎^′_𝑝=𝑎_𝑝+1, it results that 𝑤^′=𝑢; therefore, 𝑤→¹𝑤^′, which is a contradiction with the hypothesis ¬(𝑤→¹𝑤^′). Otherwise, if 𝑎_𝑝+1<𝑎^′_𝑝, then we have 𝑤⊏𝑢⊏𝑤^′, which is absurd because 𝑤^′⋗𝑤.
(Δ2) If a unique inequality of the type 𝑏^′_𝑞>𝑏_𝑞 holds for some 𝑞=𝑗+1,𝑗+2,…,𝑛−𝑟, we deduce a contradiction as in the proof of the previous case by using Rule 3 instead of Rule 1.
(Δ3) We assume now that at least two strict inequalities among the 𝑎^′_𝑘≥𝑎_𝑘 and 𝑏^′_𝑙≥𝑏_𝑙 hold, and let us also assume that the first (by left) of such strict inequalities is 𝑎^′_𝑝>𝑎_𝑝 for some 𝑝∈{𝑟,…,𝑖+1}. We consider then the following (𝑛,𝑟)-string: 𝑣=𝑎_𝑟⋯𝑎_𝑝+1𝑎_𝑝+1𝑎_𝑝−1⋯𝑎_𝑖+1()𝑏0⋯0∣0⋯0−1_𝑗+1⋯𝑏_𝑛−𝑟. (25) Since 𝑎^′_𝑝>𝑎_𝑝 is the first strict inequality by left, we have 𝑎^′_𝑝+1=𝑎_𝑝+1 (also if 𝑝=𝑟); hence 𝑎_𝑝+1=𝑎^′_𝑝+1>𝑎^′_𝑝≥𝑎_𝑝+1>𝑎_𝑝>𝑎_𝑝−1 because 𝑎_𝑝>0; this implies that 𝑣∈𝑆(𝑛,𝑟); therefore 𝑣∈𝑆(𝑛,𝑑,𝑟) because ‖𝑣‖=‖𝑤‖=𝑑. Moreover, it is clear that 𝑤⊏𝑣. Let us observe now that 𝑣⊏𝑤^′; in fact, by hypothesis, 𝑤^′ must have at least another place 𝑞≠𝑝 where 𝑎^′_𝑞>𝑎_𝑞 or 𝑏^′_𝑞>𝑏_𝑞 and 𝑣 differs by 𝑤 only in the place 𝑝. This shows that 𝑤⊏𝑣⊏𝑤^′, with 𝑣∈𝑆(𝑛,𝑑,𝑟), which is absurd because 𝑤^′⋗𝑤.
(Δ4) If at least two strict inequalities among the 𝑎^′_𝑘≥𝑎_𝑘 and 𝑏^′_𝑙≥𝑏_𝑙 hold and the first (by left) of such strict inequalities is 𝑏^′_𝑝>𝑏_𝑝 for some 𝑝∈{𝑗+1,𝑗+2,…,𝑛−𝑟}, then we deduce a contradiction in the same way as in the previous case.
This concludes the proof that ((F) AND (H)) ⇒ absurd. To complete the proof we show now that ((F) AND (G) AND ¬(H)) ⇒ absurd.
Since (H) is false, we have 𝑎^′_𝑖=1 and 𝑏^′_𝑗=0; therefore, 𝑤 has the form (17) and 𝑤^′=𝑎^′_𝑟⋯𝑎^′_𝑖+110⋯0|0⋯00𝑏^′_𝑗+1⋯𝑏^′_𝑛−𝑟. We take 𝑧∈𝑆(𝑛,𝑑,𝑟) as in (18), so that 𝑤 and 𝑧 differ between them only in the places 𝑖,𝑗 and 𝑤⊏𝑧. Moreover, since 𝑤^′=𝑎^′_𝑟⋯𝑎^′_𝑖+110⋯0|0⋯00𝑏^′_𝑗+1⋯𝑏^′_𝑛−𝑟, by (G), (18), and (17) it follows that 𝑧⊏𝑤^′. Therefore, we obtain 𝑤⊏𝑧⊏𝑤^′ that is in contrast with the hypothesis 𝑤^′⋗𝑤.

We set now (1A)∶Ω₁(𝑤)≠∅,∀𝑖∈Ω₁(𝑤)𝑎^′_𝑖≠𝑎_𝑖()+1,3A∶Ω₃(𝑤)≠∅,∀𝑗∈Ω₁(𝑤)𝑏^′_𝑗≠𝑏_𝑗+1. (26)(27)

Proposition 13. Consider(i)(1𝐴^′) ⇔ (1A),(ii)(3𝐴^′) ⇔ (3A).

Proof. (i) The implication ⇐ is obvious. We assume now that (1A^′) holds and there exists 𝑖∈Ω₁(𝑤) such that 𝑎^′_𝑖=𝑎_𝑖+1 and also that Λ⁺(𝑤,𝑤^′)⧵{𝑖}≠∅ (i.e., there exists ℎ≠𝑖 such that 𝑎^′_ℎ>𝑎_ℎ) or Λ⁻(𝑤,𝑤^′)≠∅ (i.e., there exists 𝑙 such that 𝑏^′_𝑙>𝑏_𝑙). Since 𝑖∈Ω₁(𝑤), we can apply Rule 1 to 𝑤 and we can take 𝑧=𝑤→¹. By Proposition 3, we know that 𝑧∈𝑆(𝑛,𝑑,𝑟) and 𝑧⋗𝑤. Moreover, since 𝑎^′_ℎ>𝑎_ℎ (ℎ≠𝑖) or 𝑏^′_𝑙>𝑏_𝑙, we also have 𝑧⊏𝑤^′; therefore, 𝑤⊏𝑧⊏𝑤^′, which is in contrast with the hypothesis 𝑤^′⋗𝑤. Hence this proves that the condition 𝑎^′_𝑖≠𝑎_𝑖+1 must be verified for all 𝑖∈Ω₁(𝑤), which is exactly (26).
(ii) The proof is similar to (i) applying Rule 3 to 𝑤 instead of Rule 1.

We set (1a)∶Ω₁(𝑤)≠∅,∀𝑖∈Ω₁(𝑤)𝑎^′_𝑖=𝑎_𝑖,(3a)∶Ω₃(𝑤)≠∅,∀𝑗∈Ω₃(𝑤)𝑏^′_𝑗=𝑏_𝑗. (28)(29)

Proposition 14. Consider(i)(1A) ⇒ (1a),(ii)(3A) ⇒ (3a).

Proof. (i) By (1A), we have Ω₁(𝑤)≠∅, and by the definition of Ω₁(𝑤) this is equivalent to say that 𝑤 satisfies the hypotheses of Rule 1 for some 𝑖∈{1,…,𝑟}. Therefore, by Proposition 3, there exists an element 𝑧_𝑖∈𝑆(𝑛,𝑑,𝑟) such that 𝑤→¹𝑧_𝑖, where 𝑧_𝑖=𝑎_𝑟⋯𝑎_𝑖+1(𝑎_𝑖+1)𝑎_𝑖−1⋯𝑎₁|𝑏₁⋯𝑏_𝑛−𝑟 and obviously 𝑤⊏𝑧_𝑖. Since 𝑎^′_𝑖≠𝑎_𝑖+1, we distinguish two cases.(a)𝑎^′_𝑖>𝑎_𝑖+1. In this case 𝑤⊏𝑧_𝑖⊏𝑤^′, that is in contrast with the hypothesis 𝑤^′⋗𝑤.(b)𝑎^′_𝑖<𝑎_𝑖+1. In this case 𝑎^′_𝑖≤𝑎_𝑖; hence 𝑎^′_𝑖=𝑎_𝑖 because 𝑎^′_𝑖≥𝑎_𝑖 from the hypothesis 𝑤^′⋗𝑤.
(ii) The proof of this case is similar to (i).

Lemma 15. Assume that (1a) holds. If 𝑞∈{2,3,…,𝑟,𝑟+1}, 𝑎^′_𝑞=𝑎_𝑞, and 𝑎_𝑞−1>0, then 𝑎^′_𝑞−1=𝑎_𝑞−1.

Proof. If 𝑞−1∈Ω₁(𝑤), then the thesis holds by the definition of (1a) in (28). We assume now that 𝑞−1∉Ω₁(𝑤). Then 𝑎_𝑞−1=0 or 0≤𝑎_(𝑞−1)+1−𝑎_𝑞−1=𝑎_𝑞−𝑎_𝑞−1≤1, and since 𝑎_𝑞−1>0, we necessarily have 0≤𝑎_𝑞−𝑎_𝑞−1≤1. Now, if 𝑎_𝑞−𝑎_𝑞−1=0, we have 𝑎_𝑞=𝑎_𝑞−1=0 since 𝑤∈𝑆(𝑛,𝑟), which is in contradiction with 𝑎_𝑞−1>0. Hence, it must be 𝑎_𝑞−𝑎_𝑞−1=1; that is, 𝑎_𝑞=𝑎_𝑞−1+1. Moreover, 𝑎^′_𝑞>𝑎^′_𝑞−1 because 𝑎^′_𝑞−1≥𝑎_𝑞−1>0 and 𝑤^′∈𝑆(𝑛,𝑟). Therefore, since 𝑎^′_𝑞=𝑎_𝑞, we have 𝑎_𝑞−1+1=𝑎_𝑞=𝑎^′_𝑞>𝑎^′_𝑞−1, which implies 𝑎_𝑞−1+1>𝑎^′_𝑞−1≥𝑎_𝑞−1 because 𝑤⊏𝑤^′. By subtracting 𝑎_𝑞−1 from each term of the previous inequalities, we obtain 1>𝑎^′_𝑞−1−𝑎_𝑞−1≥0; therefore, it must be 𝑎^′_𝑞−1=𝑎_𝑞−1.

We set (){}𝑎1.1a∶∃𝑙∈𝑟,…,2suchthat_𝑙>0,𝑤=𝑎_𝑟𝑎_𝑟−1⋯𝑎_𝑙+1𝑎_𝑙0⋯0∣𝑏₁⋯𝑏_𝑛−𝑟,𝑤^′=𝑎_𝑟𝑎_𝑟−1⋯𝑎_𝑙+1𝑎_𝑙𝑎^′_𝑙−1⋯𝑎^′₁∣𝑏^′₁⋯𝑏^′_𝑛−𝑟. (30)

Proposition 16. Consider (1a) ⇒ (1.1a).

Proof. By (28), there exists at least one element 𝑖∈Ω₁(𝑤). In this place we have 𝑎_𝑖>0 and moreover 𝑎_𝑖+1−𝑎_𝑖≥2. Since 𝑤∈𝑆(𝑛,𝑟), it must be 𝑎₁=0 or 𝑎₁=1. If 𝑎₁=1, then 𝑎₂=2,…,𝑎_𝑟=𝑟, which is in contrast with 𝑎_𝑖+1−𝑎_𝑖≥2. Hence 𝑎₁=0 and therefore there exists 𝑙∈2,…,𝑟 such that 𝑎_𝑙>0 and 𝑎_𝑙−1=⋯=𝑎₁=0. We take 𝑞=𝑟+1; that is, 𝑞−1=𝑟. Since 𝑎^′_𝑞=𝑎_𝑞=𝑟+1 and 𝑎_𝑞−1=𝑎_𝑟≥𝑎_𝑙>0, by Lemma 15 it follows that 𝑎^′_𝑞−1=𝑎_𝑞−1; that is, 𝑎^′_𝑟=𝑎_𝑟. Let now 𝑞=𝑟. If 𝑟=𝑙, then we have the thesis otherwise 𝑞=𝑟>𝑙; that is, 𝑞−1≥𝑙; therefore 𝑎_𝑞−1≥𝑎_𝑙>0 because 𝑤∈𝑆(𝑛,𝑟). Then since 𝑎^′_𝑞=𝑎^′_𝑟=𝑎_𝑟=𝑎_𝑞, we can apply Lemma 15 and we deduce 𝑎^′_𝑞−1=𝑎_𝑞−1; that is, 𝑎^′_𝑟−1=𝑎_𝑟−1. By iteration the thesis follows.

Lemma 17. Assume that (3a) holds. If 𝑞∈{2,3,…,𝑛−𝑟−1}, 𝑏^′_𝑞=𝑏_𝑞<0, then 𝑏^′_𝑞+1=𝑏_𝑞+1.

Proof. If 𝑞+1∈Ω₃(𝑤), then the thesis holds by definition of (3a) in (29). We assume now that 𝑞+1∉Ω₃(𝑤). Now, if 𝑏_𝑞−𝑏_𝑞+1=0, we have 𝑏_𝑞=𝑏_𝑞+1=0 since 𝑤∈𝑆(𝑛,𝑟), which is in contradiction with the hypothesis 𝑏_𝑞<0. Hence it must be 𝑏_𝑞−𝑏_𝑞+1=1; that is, 0>𝑏_𝑞=𝑏_𝑞+1+1; therefore 0>𝑏_𝑞+1+1=𝑏_𝑞=𝑏^′_𝑞>𝑏^′_𝑞+1≥𝑏_𝑞+1 since 𝑤^′∈𝑆(𝑛,𝑟) and 𝑤⊏𝑤^′. The last inequalities provide 𝑏_𝑞+1+1>𝑏^′_𝑞+1≥𝑏_𝑞+1. By subtracting 𝑏_𝑞+1 from each term of the previous inequalities, we obtain 1>𝑏^′_𝑞+1−𝑏_𝑞+1≥0; therefore it must be 𝑏^′_𝑞+1=𝑏_𝑞+1.

Proposition 18. If (3a) holds, then 𝑤 has the following form: 𝑤=𝑎_𝑟⋯𝑎₁|0⋯0𝑏_𝑙𝑏_𝑙+1⋯𝑏_𝑛−𝑟, with 𝑏_𝑙<0 and 𝑙≥2. Moreover, if𝑏_𝑙≤−2, we also have 𝑤^′₋=𝑤₋.

Proof. By (3a) in (29), we know that Ω₃(𝑤)≠∅. Moreover, if 𝑗∈Ω₃(𝑤), then 𝑗≥2 and 𝑏_𝑗<0 because 𝑏_𝑗−1−𝑏_𝑗≥2. Therefore, we can take the minimum 𝑙∈2,…,𝑛−𝑟 such that 𝑏_𝑙<0; hence 𝑏_𝑙−1=⋯=𝑏₁=0 and this means that 𝑤=𝑎_𝑟⋯𝑎₁|0⋯0𝑏_𝑙𝑏_𝑙+1⋯𝑏_𝑛−𝑟. Assume now that 𝑏_𝑙≤−2. Since 𝑏_𝑙−1=0, we have 𝑏_𝑙−1−𝑏_𝑙≤2; therefore 𝑙∈Ω₃(𝑤). By (3a), follows then that 𝑏^′_𝑙=𝑏_𝑙<0. Now, if 𝑙=𝑛−𝑟, the thesis is true, otherwise, if 𝑙∈{2,…,𝑛−𝑟−1}, by Lemma 17 we have 𝑏^′_𝑙+1=𝑏_𝑙+1<0. Again, if 𝑙+1=𝑛−𝑟, the thesis is true; otherwise, if 𝑙+1∈{2,…,𝑛−𝑟−1}, by Lemma 17 we have 𝑏^′_𝑙+2=𝑏_𝑙+2<0. Proceeding repeatedly in this way we obtain 𝑏^′_𝑙=𝑏_𝑙, 𝑏^′_𝑙+1=𝑏_𝑙+1,…,𝑏^′_𝑛−𝑟=𝑏_𝑛−𝑟. Finally, since 𝑏₁=⋯=𝑏_𝑙−1=0 and 𝑤⊏𝑤^′, we also have 𝑏^′₁=⋯=𝑏^′_𝑙−1=0. This proves that 𝑤^′₋=𝑤₋.

Proposition 19. Consider ((2C) AND (3a)) ⇒𝑤^′₋=𝑤₋.

Proof. (2C) implies that 𝑤 has the form 𝑤=𝑎_𝑟⋯𝑎₁|0⋯0𝑏_𝑙𝑏_𝑙+1⋯𝑏_𝑛−𝑟, with 𝑏_𝑙≤−2. Since (3a) holds, by Proposition 18, the thesis follows.

Proposition 20. Consider ((1.1a) AND (2C) AND (3a)) ⇒ absurd.

Proof. By Proposition 19, we have that 𝑤^′₋=𝑤₋. Moreover, by (1.1a) in (30), we also have 𝑤^′₊=𝑤₊ because ‖𝑤^′‖=‖𝑤‖=𝑑 and 𝑤^′₋=𝑤₋; therefore, 𝑤=𝑤^′. This is absurd because 𝑤^′⋗𝑤 by hypothesis.

Proposition 21. Consider ((1.1b) AND (2C) AND (3a)) ⇒ absurd.

Proof. By (1.1b) in (12), we know that 𝑤=0⋯0|𝑏₁⋯𝑏_𝑛−𝑟, and this implies ‖𝑤₊‖=0 and ‖𝑤₋‖=𝑑. Moreover, by Proposition 19, 𝑤^′₋=𝑤₋; therefore it follows that ‖𝑤^′₋‖=𝑑; hence ‖𝑤^′₊‖=0 since 𝑤^′∈𝑆(𝑛,𝑑,𝑟). This shows that 𝑤=𝑤^′ and this is absurd because 𝑤⋗𝑤^′.

Proposition 22. Consider ((1.1a) AND (2B)) ⇒𝑤^′₊=𝑤₊.

Proof. By (1.1a) in (30) and (2B), it follows that 𝑤=𝑎_𝑟⋯𝑎_𝑙+110⋯0|𝑏₁⋯𝑏_𝑛−𝑟 and 𝑤^′=𝑎_𝑟⋯𝑎_𝑙+11𝑎^′_𝑙−1⋯𝑎^′₁|𝑏₁⋯𝑏_𝑛−𝑟. Now, since 𝑤^′∈𝑆(𝑛,𝑟) and 𝑎_𝑙=𝑎^′_𝑙=1, we have that 1=𝑎^′_𝑙>𝑎^′_𝑙−1≥⋯≥𝑎^′₁≥0; that is, 𝑎^′_𝑙−1=⋯=𝑎^′₁=0. Therefore 𝑤^′₊=𝑤₊.

Proposition 23. Consider ((1.1a) AND (2B) AND (3.1b)) ⇒ absurd.

Proof. By Proposition 22𝑤^′₊=𝑤₊ and by Proposition 8  𝑤=𝑤^′₋; hence 𝑤=𝑤^′ and this is a contradiction.

Proposition 24. Consider ((1.1a) AND (3.1b)) ⇒ absurd.

Proof. By Proposition 8, 𝑤=𝑤^′₋=|00⋯0; hence ‖𝑤^′₊‖=‖𝑤₊‖=𝑑 because 𝑤,𝑤^′∈𝑆(𝑛,𝑑,𝑟). Moreover, by (1.1a) in (30), it follows that 𝑤^′₊=𝑤₊=𝑎_𝑟⋯𝑎_𝑙+1𝑎_𝑙0⋯0. (31) Therefore, 𝑤=𝑤^′, which is absurd.

Proposition 25. Consider ((1.1a) AND (2B) AND (3a)) ⇒ absurd.

Proof. By Proposition 18, we have 𝑤=𝑎_𝑟⋯𝑎₁|0⋯0𝑏_𝑙𝑏_𝑙+1⋯𝑏_𝑛−𝑟 with 𝑏_𝑙<0 and 𝑙≥2. By Proposition 22,𝑤^′₊=𝑤₊. Now, if 𝑏_𝑙≤−2, by Proposition 18 we have 𝑤^′₋=𝑤₋ and therefore 𝑤^′=𝑤, which is absurd. So we can assume that 𝑏_𝑙=−1. Since 𝑤^′₊=𝑤₊, ‖𝑤^′‖=‖𝑤‖=𝑑 and 𝑤⊏𝑤^′, it follows that 𝑏^′_𝑙 cannot be 0; otherwise, it is ‖𝑤^′‖<𝑑. This implies that it must be 𝑏^′_𝑙=−1. We distinguish now two cases.
Case  1. If 𝑙∈{2,…,𝑛−𝑟−1}, since 𝑏_𝑙=𝑏^′_𝑙=−1, by Lemma 17 (with 𝑞=𝑙) we obtain 𝑏_𝑙+1=𝑏^′_𝑙+1<0. Now, if 𝑙+1=𝑛−𝑟, then 𝑤₋=𝑤^′₋, and as before this provides a contradiction. If 𝑙+1<𝑛−𝑟, we can still apply Lemma 17 (with 𝑞=𝑙+1) and therefore have 𝑏_𝑙+2=𝑏^′_𝑙+2<0. Proceeding in this way in all cases we always obtain 𝑤₋=𝑤^′₋ and hence is absurd.
Case  2. If 𝑙=𝑛−𝑟, then 𝑤=𝑎_𝑟⋯𝑎₁∣0⋯0(−1), and since ‖𝑤^′‖=‖𝑤‖=𝑑, 𝑤⊏𝑤^′, and 𝑤^′₊=𝑤₊, it must also be 𝑤^′=𝑎_𝑟⋯𝑎₁|0⋯0(−1); that is, 𝑤=𝑤^′, absurd.

Proposition 26. Consider ((1.1a) AND (2B) AND (3.2b)) ⇒ absurd.

Proof. By Proposition 22, 𝑤^′₊=𝑤₊. Then by (3.2b) in (16), since 𝑤^′₊=𝑤₊, ‖𝑤^′₋‖=‖𝑤₋‖, and 𝑤⊏𝑤^′, it must also be 𝑤^′₋=∣00⋯0(−1)(−2)⋯(−𝑙). (32) Hence we obtain 𝑤=𝑤^′, absurd.

Proposition 27. Consider ((1.2b) AND (2C) AND (3a)) ⇒ absurd.

Proof. By Proposition 19, 𝑤^′₋=𝑤₋. Then, since 𝑤⊏𝑤^′, ‖𝑤^′‖=‖𝑤‖, and 𝑤^′₋=𝑤₋, by (1.2b) in (13) we deduce that 𝑤^′₊=𝑤₊. Hence the absurd 𝑤^′=𝑤.

Proposition 28. Consider ((2.1d) AND (3a)) ⇒ absurd.

Proof. Since 𝑤⊏𝑤^′ by (2.1d) in (14), it follows that 𝑤^′₊=𝑤₊. Then, since (3a) in (29) holds, the same proof of Proposition 25 leads to the absurd 𝑤=𝑤^′.

We complete now the proof of our main result; that is, we provide three syntactic trees where each possible path from the roots to the leaves leads to a contradiction. In the following we draw each tree and next we also write all the implication chains for each path from the roots to the leafs.

We recall at first some useful implications and equivalences already proved in the previous section:(Θ1A) (1A^′) ⇔ [by Proposition 13 (i)] (1A) ⇒ [by Proposition 14] (1a) ⇒ [by Proposition 16] (1.1a)(Θ2A) (2A^′) ⇒ [by Proposition 12] absurd(Θ2D) (2D) ⇔ [by Proposition 6] (2.1d)(Θ3A) (3A^′) ⇔ [by Proposition 13 (ii)] (3A) ⇒ [by Proposition 14] (3a) ⇒ [see statement of Proposition 18] Proposition 18.

The first tree that we consider is that having the condition (1A^′) as root (we denote such tree with TREE(1A^′)).

TREE (1A^′): (33)

We observe now that by virtue of Proposition 5 when we start from the condition (1B), we must consider two subtrees, the first with root (1.1b) (which we denote by TREE(1.1b)) that is

TREE (1.1b): (34)

and the second (which we denote by TREE(1.2b)) that is

TREE (1.2b): (35)

Let us note that the ramification (36)

in each of the previous trees follows by Proposition 7.

We have then exactly 27 possible paths from the roots to the leafs in all the previous trees (we exclude the paths that lead to (2A^′) because we already know that they lead to a contradiction by Θ2A). In the following we write all these 27 paths and, for each, the related proposition that leads to a contradiction together with some implication among the Θ1A, Θ2D, and Θ3A. In the sequel, the node (3B) is omitted for brevity because it is only an intermediate node and in the paths that involve (3B) we use only the leafs (3.1b) and (3.2b) to deduce the related contradictions: (1A^′) AND (2B) AND (3A^′) ⇒ absurd by Θ1A, Θ3A, and Proposition 25 (1A^′) AND (2B) AND (3.1b) ⇒ absurd by Θ1A and Proposition 23 (1A^′) AND (2B) AND (3.2b) ⇒ absurd by Θ1A and Proposition 26 (1A^′) AND (2C) AND (3A^′) ⇒ absurd by Θ1A, Θ3A, and Proposition 20 (1A^′) AND (2C) AND (3.1b) ⇒ absurd by Θ1A and Proposition 24 (1A^′) AND (2C) AND (3.2b) ⇒ absurd by Θ1A and Proposition 9 (iii) (1A^′) AND (2D) AND (3A^′) ⇒ absurd by Θ2D, Θ3A, and Proposition 28 (1A^′) AND (2D) AND (3.1b) ⇒ absurd by Θ1A and Proposition 24 (1A^′) AND (2D) AND (3.2b) ⇒ absurd by Θ2D and Proposition 11

 (1.1b) AND (2B) AND (3A^′) ⇒ absurd by Proposition 9 (i) (1.1b) AND (2B) AND (3.1b) ⇒ absurd by Proposition 9 (i) (1.1b) AND (2B) AND (3.2b) ⇒ absurd by Proposition 9 (i) (1.1b) AND (2C) AND (3A^′) ⇒ absurd by Θ3A and by Proposition 21 (1.1b) AND (2C) AND (3.1b) ⇒ absurd by Proposition 9 (iv) (1.1b) AND (2C) AND (3.2b) ⇒ absurd by Proposition 9 (iii) (1.1b) AND (2D) AND (3A^′) ⇒ absurd by Θ2D and by Proposition 9 (ii) (1.1b) AND (2D) AND (3.1b) ⇒ absurd by Θ2D and by Proposition 9 (ii) (1.1b) AND (2D) AND (3.2b) ⇒ absurd by Θ2D and by Proposition 9 (ii)

 (1.2b) AND (2B) AND (3A^′) ⇒ (1.2b) AND (2D) AND (3A^′) (that we examine below) by Θ2D and by Proposition 10 (1.2b) AND (2B) AND (3.1b) ⇒ (1.2b) AND (2D) AND (3.1b) (that we examine below) by Θ2D and by Proposition 10 (1.2b) AND (2B) AND (3.2b) ⇒ (1.2b) AND (2D) AND (3.2b) (that we examine below) by Θ2D and by Proposition 10 (1.2b) AND (2C) AND (3A^′) ⇒ absurd by Θ3A and by Proposition 27 (1.2b) AND (2C) AND (3.1b) ⇒ absurd by Proposition 9 (v) (1.2b) AND (2C) AND (3.2b) ⇒ absurd by Proposition 9 (iii) (1.2b) AND (2D) AND (3A^′) ⇒ absurd by Θ2D, Θ3A, and Proposition 28  (1.2b) AND (2D) AND (3.1b) ⇒ absurd by Proposition 9 (v) (1.2b) AND (2D) AND (3.2b) ⇒ absurd by Θ2D and by Proposition 11

This concludes the proof of Theorem 4. Below we draw the Hasse diagram of the lattice 𝑆(5,3,2) by using the evolution Rules 1, 2, and 3 starting with the minimum element of this lattice, which is 00|123. We label a generic edge of the next diagram with the symbol 𝑅𝑘 if it leads to a production that uses the Rule 𝑘, for 𝑘∈{1,2,3}: (37)