Large filters of quasiorder lattices can be generated by few elements

TL;DR

This paper proves that for small posets, the filter of quasiorders extending the poset order can be generated by a few elements, revealing a structural simplicity in such lattices.

Contribution

It establishes that quasiorder filters of small posets are finitely generated, providing new insights into the structure of these lattices.

Findings

Filters can be generated by few elements for small posets

Structural simplicity of quasiorder lattices in small posets

Advances understanding of lattice generation in order theory

Abstract

For a poset $(P;\leq)$, the quasiorders (AKA preorders) extending the poset order "$\leq$" form a complete lattice $F$, which is a filter in the lattice of all quasiorders of the set $P$. We prove that if the poset order "$\leq$" is small, then $F$ can be generated by few elements.

Equations

$$\left.\parbox{284.52756pt}{there is a $(k-1)$-element subset $Y(T)$ of $\textup{Quo}^{\kern-1.0pt\leq}(T;\mu_{T})$ such that $Y(T)\cup\{\textup{qu}_{\mu}(y,x):x\prec y\text{ in }T\}$ generates $\textup{Quo}^{\kern-1.0pt\leq}(T;\mu_{T})$.}\,\,\right\}$$

$$n\le \left(\genfrac{}{}{0ex}{}{k}{\lfloor k/2\rfloor }\right)$$

$$\Bigl{(}n_{\textup{cextr}}\cdot f(n_{\textup{cmp}})+{n_{\textup{cedge}}}+n_{\textup{corr}}\Bigr{)}\text{-generated}$$

$$\Bigl{(}n_{\textup{cextr}}\cdot f(n_{\textup{cmp}})+\textup{LASp}({n_{\textup{cedge}}})+n_{\textup{corr}}\Bigr{)}\text{-generated}$$

$$\bigvee \{{\rho }_j:j\in J\}=\text{tr}\left(\bigcup \{{\rho }_j:j\in J\}\right)\text{ and }\bigvee {}^{\mu }\{{\tau }_j:j\in J\}=\text{tr}\left(\bigcup \{{\tau }_j:j\in J\}\right);$$

$$T\in \text{Comp}(P)$$

$$\left.\parbox{312.9803pt}{if $\emptyset\neq X\subseteq P$ such that $X$ has an at most one-element intersection with each component of $(P;\leq)$, then $\varphi_{X}\colon(\textup{Quo}(X);\subseteq)\to\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ defined by $\rho\mapsto\textup{tr}(\mu\cup\rho)$ is a complete lattice embedding and $\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{X};\leq)=\varphi_{X}(\textup{Quo}(X))$; in particular, $\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{X};\leq)$ is a complete sublattice of $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ and this sublattice is isomorphic to $(\textup{Quo}(X);\leq)$.}\,\,\right\}$$

$${\gamma }^{+}={\text{qu}}_{\mu }(\gamma )=\mu \vee \text{quo}(\gamma )$$

$${\mathbf{\Xi }}_{\text{stp}}$$

$$n_{\text{cextr}}$$

$$\left.\parbox{275.99164pt}{if $P$ is a forest of finite component size, then $\forall(u,v)\in\textup{Edge}(P)$, \\ $\textup{qu}_{\mu}(v,u)\nleq\bigvee\{\textup{qu}_{\mu}(y,x):(x,y)\in\textup{Edge}(P)\setminus\{(u,v)\}\}$.}\right\}$$

$$\left.\parbox{199.16928pt}{if $(y,x)\in S\in\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$ and $x,y\in T\in\textup{Comp}(P)$, then $\textup{qu}_{\mu}(y,x)=\textup{qu}_{\mu}(S)\cap\nabla^{+}_{{\kern-1.5ptT}}$.}\,\,\right\}$$

$$\left.\parbox{270.30118pt}{if $P$ is a forest of finite component size, then $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ contains an at most $\textup{LASp}({n_{\textup{cedge}}})$-element subset $G$ such that for each $S\in\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$, $\textup{qu}_{\mu}(S)$ is the meet of some members of $G$.}\,\,\right\}$$

$$\{S\}=\bigcap \{X:S\in X\in G_0\}\text{ for every }S\in {\mathbf{\Xi }}_{\text{stp}}.$$

$$\displaystyle\textup{qu}_{\mu}(X)=\bigvee\{\textup{qu}_{\mu}(S):S\in X\}=\textup{qu}_{\mu}\bigl{(}\bigcup\{S:S\in X\}\bigr{)},$$

$$\text{and we define }G:=\{{\text{qu}}_{\mu }(X):X\in G_0\}.$$

$${\text{qu}}_{\mu }(S)=\bigwedge \{{\text{qu}}_{\mu }(X):X\in G_0\text{ and }S\in X\}.$$

$$p\ne q\text{ and }(p,q)\in {\text{qu}}_{\mu }(X)\text{ for all }X\in G_0\text{ containing }S.$$

$$\vec{r}(X):r_0(X)=p,r_1(X),\dots ,r_{m(X)-1}(X),r_{m(X)}(X)=q$$

$$\vec{r}(X)\text{ does not depend on }X\text{ for }S\in X\in G_0.$$

$$\left.\parbox{256.0748pt}{For $a<b\in P$ and $\rho\in\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$, $(b,a)\in\rho$ if and only if $(y,x)\in\rho$ for all edges $[x,y]$ such that $a\leq x\prec y\leq b$.}\,\,\right\}$$

$$\left.\parbox{261.76526pt}{$|G|={n_{\textup{cedge}}}$ for Part (A), $|G|=\textup{LASp}({n_{\textup{cedge}}})$ for Part (B), and (for both of these two parts of the theorem) for each $S\in\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$, $\textup{qu}_{\mu}(S)$ is the meet of some members of $G$. }\,\,\right\}$$

$$H:=\bigcup \{H_D:D\in {\mathbf{\Theta }}_{\text{thd}}\}\text{; note that }|H|=n_{\text{cextr}}\cdot f(n_{\text{cmp}}).$$

$$X\in \text{Comp}(P)$$

$$i\in \{1,2\}$$

$$\text{and }|F(M_1^{▶◀})|+|F(M_2^{▶◀})|\le n_{\text{corr}}.$$

$$E:=F(M_1^{▶◀})\cup F(M_2^{▶◀})\cup G\cup H.$$

$$\text{for all }D\in {\mathbf{\Theta }}_{\text{thd}},{\text{Quo}}^{\le }(P\rceil {}_D;\le )\subseteq K.$$

$$i\in \{1,2\}$$

$$M_i^{▶◀}\subseteq K$$

$$\left.\parbox{199.16928pt}{for any $T\in\textup{Comp}(P)$ and $x_{\ast},y^{\ast}\in\textup{Extr}(T)$, if $x_{\ast}\leq y^{\ast}$, then $\textup{qu}_{\mu}(y^{\ast},x_{\ast})\in K$.}\,\,\right\}$$

Full text

This research was supported by the National Research, Development and Innovation Fund of Hungary, under funding scheme (i.e., grant number) K 138892.

Our goal is to prove Theorem 3.3, which is motivated by earlier results; for a mini-survey of these results, see Section 1 with emphasis on its last paragraph. The secondary goal is to point out that the lattices in the paper could be applied in our 2021 secret key authentication and cryptographic protocol and, furthermore, the underlying mathematical problem of this protocol is likely to be hard; see Section 5.

1. Introduction

First, we survey the previously known results on generation of quasiorder lattices. Reflexive and transitive relations of a set $X$ are called quasiorders or (in a part of the literature) preorders (of $X$). For a set $X$, we will denote by $\textup{Quo}(X)=(\textup{Quo}(X);\subseteq)$ the complete lattice of quasiorders of $X$. We say that a complete lattice $L$ is $n$-generated as complete lattice if $L$ has an at most $n$-element subset $Y$ (called a * set generating* $L$ as a complete lattice) such that no proper complete sublattice of $L$ includes $Y$. For a finite lattice, this concept is equivalent to the existence of an at most $n$-element generating set in the usual sense. Note that our convention in this paper:“generated” always means that “generated as a complete lattice” even when this is not emphasized. Furthermore, “$k$-generated” means the existence of an at most $k$-element generating set.

Six-element generating sets:  Letting $\kappa_{0}:=\aleph_{0}$ and, for any integer $n\in\mathbb{N}_{0}=\{0,1,2,\dots\}$, $\,\kappa_{n+1}:=2^{\kappa_{n}}$, we proved in Chajda and Czédli [3] that whenever $X$ is a set with $|X|\leq\kappa_{n}$ for some $n\in\mathbb{N}_{0}$, then $\textup{Quo}(X)$ is 6-generated (but note that the six-element generating set given in [3] has an additional property). A cardinal number $\eta>\aleph_{0}$ is called inaccessible if it satisfies the following two conditions:

• •

for any cardinal $\lambda$,  $\lambda<\eta\Rightarrow 2^{\lambda}<\eta$, and

• •

for any set $I$ of cardinals smaller than $\eta$,  $|I|<\eta\Rightarrow\sup I<\eta$.

We call $\eta$ accessible111Note that “accessible” and “inaccessible” are not the opposites of each others. if there exists no inaccessible $\eta^{\prime}$ such that $\eta^{\prime}\leq\eta$. In particular, the finite cardinals, the previously mentioned cardinals $\kappa_{n}$, and $\sup\{\kappa_{n}:n\in\mathbb{N}_{0}\}$ are accessible as well as (almost) all cardinals that algebraists meet in their work.

Takách [23] extended the result of [3] to every set $X$ of accessible cardinality.

Five-element generating sets:  Dolgos [14] proved that $\textup{Quo}(X)$ is 5-generated for any set $X$ with size (=cardinality) $|X|\leq\aleph_{0}$, and Kulin [16] extended this result to all sets of accessible cardinalities.

Four-element generating sets:  With the exception of $|X|\in\{4,6,8,10\}$, Czédli [7] proved that if $X$ is a set with $|X|\leq\aleph_{0}$, then $\textup{Quo}(X)$ is 4-generated. Furthermore, [7] also proves that for any set $X$ with at least three elements, $\textup{Quo}(X)$ is not 3-generated. Soon thereafter, Czédli and Kulin [11, Theorem 3.5] reduced the number of exceptions and permitted all infinite sets $X$ with accessible cardinalities. Thus, we know that

Lemma 1.1** (Czédli [7], Czédli and Kulin [11]).**

If $X$ is an at least two-element set of an accessible cardinality and this cardinality is different from $4$, then the complete lattice $\textup{Quo}(X)$ is $4$-generated. For $|X|\geq 3$, $\textup{Quo}(X)$ is not $3$-generated. For $|X|=4$, $\textup{Quo}(X)$ is $5$-generated but we do not know whether it is $4$-generated or not.

Cryptography, authentication, and further results:   As Ahmed and Czédli [1], Czédli [5], [6], [8], [9], Czédli and Oluoch [13], Dolgos [14], Strietz [21]–[22], and Zádori [24] indicate, there are several results related or analogous to the aforementioned ones. To save space, we do not recall these results; we only make a few comments on them. Strietz’s papers [21] and [22] were the first to prove results on small generating sets of complicated and large lattices (namely, equivalence lattices). However, it was Zádori’s excellent approach in [24] that motivated almost all subsequent results in this topic. Working with equivalence lattices is usually more feasible than working with quasiorder lattices. Surprisingly, it was quasiorder lattices that showed the way how to extend results from finite sets to some infinite sets; see Chajda and Czédli [3]. Finally, note that Czédli [9] and [10] outline (without elaborating all details) how large and (in case of [9]) complicated finite lattices with small generating sets and easy-to perform lattice operations can be used for secret key crypto-systems and authentication222Just to say a bit more about cryptography and authentication, we recall an idea of Czédli [9] and complement it with “$g_{\textup{str}}$” (see later). Consider the equation $x^{y}=z$ in a large finite prime field $\mathbb{Z}_{p}$; any two of the three unknowns determine the third. However, the computational difficulty heavily depends on which of the three is to be found; cryptographic protocols are based on this fact. Now let $L$ be a very large finite lattice with a complicated structure in which, however, it is easy to describe the elements and to perform the lattice operations. (Equivalence lattices and quasiorder lattices are such.) Let $b$ be a positive integer like $100$ or of a similar magnitude. Let $\vec{h}$ be a $k$-dimensional vector formed from the elements of (a superset of) a generating set of $L$ (we can think of $k\approx 120$ or of a similar magnitude), let $\vec{p}$ be a $b$-dimensional vector of $k$-ary lattice terms, and let $\vec{u}\in L^{b}$. In the equation $\vec{p}(\vec{h})=\vec{u}$, any two of $\vec{h}$, $\vec{p}$, and $\vec{u}$ “determine” the remaining one (in some sense but not always uniquely), but the computational difficulty depends on which one of $\vec{h}$, $\vec{p}$, and $\vec{u}$ is what we are looking for. Keeping the hardness of the (classical) discrete logarithm in our mind, the idea is, perhaps, more visible if we denote $\vec{p}(\vec{h})$ by $\vec{p}^{\kern 1.0pt\vec{h}}$. Then (provided that some wrong cases are excluded) it is difficult to obtain $\vec{h}$ from $\vec{p}$ and $\vec{u}:=\vec{p}^{\kern 1.0pt\vec{h}}$. In the practice (but not in the theory as we can always use equations like the first two in (5.7)), it is even more difficult to obtain $\vec{h}$ from $\vec{p}$ and $\vec{p}^{\kern 1.0pt\vec{h}}=\vec{p}(\vec{h})$ if we allow that the components $p_{1},\dots,p_{b}$ of $\vec{p}$ are algebraic functions (built from lattice operations, $k$ many variables, and a few constant elements of $L$). Finally, let $g_{\textup{str}}$ be a function with domain $L^{b}$. (Even the simplest case where $g_{\textup{str}}$ is the omitted identity function is interesting. Note that $g_{\textup{str}}$ need not be a lattice polynomial.) Assume that Kati (the Hungarian first name corresponding to Kate and Cathy) wants to communicate with her Bank via an open channel; they both know and only they know $\vec{h}$ and $g_{\textup{str}}$. The Bank can verify Kati’s identity in the following way. The Bank sends Kati a random $\vec{p}$. (Some trivial $\vec{p}$’s should be avoided, of course.) Then Kati sends $\vec{u}:=g_{\textup{str}}(\vec{p}^{\kern 1.0pt\vec{h}})$ to the Bank. Even if an Adversary has intercepted the latest and all earlier messages, neither he can extract $\vec{h}$ from earlier pairs (\vec{p}\,^{\prime},\vec{u}\,^{\prime}):=\bigl{(}\vec{p}\,^{\prime},\,g_{\textup{str}}(\vec{p}\,^{\prime}(\vec{h}))\bigr{)} nor he can use earlier messages as the Bank always uses a new random $\vec{p}$. Thus, the Adversary cannot cause harm by pretending to be Kati or the Bank, and he cannot break the authentication protocol outlined above. In addition to authentication, Kati and the Bank can use $\vec{u}$ as the key (frequently to be renewed) of Vernam Cipher. Even though the underlying mathematical problem is hard in the sense of Section 5, this does not imply the security of our protocol in itself by various well-known reasons (some of them are mentioned in [10]). However, the hardness of the underlying problem gives hope and we believe in the security of the crypto- and authentication system outlined above as quasiorder lattices are quite large and involved. .

In [9], equivalence lattices are used; for a set $X$, let $\textup{Equ}(X)=(\textup{Equ}(X),\subseteq)$ stand for the (complete) lattice of equivalences of $X$. For an $n$-element finite set $X$, $|\textup{Quo}(X)|$ is much larger than $|\textup{Equ}(X)|$. For example, for a 7-element set $X$, $|\textup{Equ}(X)|=877$ while $|\textup{Quo}(X)|=9\,535\,241$; see Ahmed and Czédli [1, Table 1] or Sloane [20, A000110 and A000798]. The present paper indicates that, in addition to $\textup{Equ}(X)$, $\textup{Quo}(X)$, and their direct products or powers (see Czédli [9] and Czédli and Oluoch [13]), many large filters of $\textup{Quo}(X)$ also offer reasonable options to design crypto-systems and authentication protocols.

2. What do we mean by a large filter of $\textup{Quo}(X)$?

For an element $u$ of a lattice $L$, the principal filter $\{x\in L:u\leq x\}$ will be denoted by $\mathord{\uparrow}\kern 0.5ptu$ or, if confusion threatens, by $\mathord{\uparrow}_{\kern-1.0ptL}\kern 0.5ptu$. Similarly, $\mathord{\downarrow}\kern 0.5ptu=\mathord{\downarrow}_{\kern-1.0ptL}\kern 0.5ptu$ stands for the the principal ideal $\{x\in L:x\leq u\}$. The height $u$ is the supremum of the lengths of chains in $\mathord{\downarrow}\kern 0.5ptu$. By a large filter we mean a principal filter $\mathord{\uparrow}\kern 0.5ptu$ such that the height of $u$ is a (small) finite number. So, for a set $A$, a large filter of $\textup{Quo}(A)$ is of the form $\mathord{\uparrow}\kern 0.5pt\mu=\mathord{\uparrow}_{\kern-1.0pt\textup{Quo}(A)}\kern 0.5pt\mu$ such that $\mu\in\textup{Quo}(A)$ is of a small finite height. Let $\Theta(\mu)$ denote $\mu\cap\mu^{-1}=\{(x,y)\in A^{2}:(x,y)\in\mu\text{ and }(y,x)\in\mu\}$. Then $\Theta(\mu)\in\textup{Equ}(A)$. The $\Theta(\mu)$-block of an element $x\in A$ is denoted by $x/\Theta(\mu):=\{y\in A:(x,y)\in\Theta(\mu)\}$. Letting $\mu/\Theta(\mu):=\{(x/\Theta(\mu),y/\Theta(\mu)):(x,y)\in\mu\}$, it is easy to see, and it is explicitly stated in (the last sentence of) Czédli and Lenkehegyi [12, Theorem 1.6] that $\mathord{\uparrow}_{\kern-1.0pt\textup{Quo}(A)}\kern 0.5pt\mu$ is isomorphic to $\mathord{\uparrow}_{\kern-1.0pt\textup{Quo}(A/\Theta(\mu))}\kern 0.5pt{\mu/\Theta(\mu)}$. Therefore, we can assume that $\Theta(\mu)$ equals $\Delta_{A}=\{(x,x):x\in A\}$ since otherwise the study of the filter $\mathord{\uparrow}_{\kern-1.0pt\textup{Quo}(A)}\kern 0.5pt\mu$ reduces to that of another filter over a “smaller” base set, $A/\Theta(\mu)$. (The quotient marks indicate that “smaller” if $A$ is finite and “not larger” otherwise.) However, for $\mu\in\textup{Quo}(A)$, we have that $\Theta(\mu)=\Delta_{A}$ if and only if $\mu$ is a partial order. Consequently, we do not loose anything by restricting our attention to principal filters $\mathord{\uparrow}_{\kern-1.0pt\textup{Quo}(A)}\kern 0.5pt\mu$ such that $\mu$ is a partial order. To deal with this case, we introduce the following terminology. Let $(A;\mu)$ be a poset (AKA partially ordered set); note that \boxed{\text{we often write \leq$instead of$\mu}}. The lattice of quasiorders extending $\mu$ will be denoted by $\textup{Quo}^{\kern-1.0pt\leq}(A;\mu)=(\textup{Quo}^{\kern-1.0pt\leq}(A;\mu);\subseteq)$, it is the same as the filter $\mathord{\uparrow}_{\kern-1.0pt\textup{Quo}(A)}\kern 0.5pt\mu$.

3. Stating the result

First, we recall or introduce the concepts and notations that are necessary to formulate and understand the result. Posets (AKA partially ordered sets) with more than one element are said to be nontrivial while the singleton poset is trivial. The length of a finite $n$-element chain is $n-1$. For a non-negative integer number $n\in\mathbb{N}_{0}$, we say that a poset is of length $n$ if it has a chain of length $n$ but it has no chain of length $n+1$. If $P=(P;\leq)$ is a poset and it is of length $n$ for some $n\in\mathbb{N}_{0}$, the we say that $P$ is of finite length. In this case, $\textup{len}(P)$ will stand for the length of $P$. One of the reasons we always assume this is that then the covering relation $\prec$ of $P$ determines the (partial) order $\leq$ of $P$. While the Hasse diagram of $P$ is a directed graph333Indeed, every edge is directed upwards., this diagram can also be considered an undirected graph, the graph of the poset $P$. The edges of this graph are the two-element subsets $\{x,y\}$ of $P$ such that $x\prec y$ or $y\prec x$. We say that the poset $P$ is a forest if its graph contains no circle of positive length. The $($connectivity$)$ components444That is, the blocks of the least equivalence of $P$ that collapses every edge. of $P$ are called trees. By an extremal element of $P$ we mean an element that is maximal or minimal. The set of maximal elements, that of minimal elements, and that of extremal elements of a poset $Q$ are denoted by $\textup{Max}(Q)$, $\textup{Min}(Q)$, and $\textup{Extr}(Q)=\textup{Max}(Q)\cup\textup{Min}(Q)$, respectively.

In order to extract a lot from our ideas how to find small generating sets, we need to define several functions and parameters; we present them in the following two definitions. As these functions and parameters occur555We could only state a weaker result without these parameters and functions. in Theorem 3.3 and make it look complicated, we note in advance that the functions and parameters are illustrated by Figures666In the figures, which give lots of information, the only purpose of the colors, fill patterns, and gradients is to make them easier to read. There has been an attempt to choose the colors so that they should remain different after grey-scaled printing. 1 and 2. Furthermore, some particular cases of the theorem will be formulated in an easy-to-read way in Corollary 4.1 (and even in Examples 4.2 and 4.3) in Section 4.

Definition 3.1** (Parameters of our poset).**

Let $P=(P;\leq)=(P;\mu)$ be a poset777Let us repeat that, in the whole paper, we often use $\mu$ to denote the order relation of $P$. of finite length.

(A) Let $\textup{Comp}(P)$ stand for the set of (connectivity) components of $P$. We say that $P$ is of finite component size if there is a $k\in\mathbb{N}^{+}=\{1,2,3,\dots\}=\mathbb{N}_{0}\setminus\{0\}$ such that for every $T\in\textup{Comp}(P)$, $T$ has at most $k$ elements. In this case, the least such $k$ is called the component size of $P$ and it is denoted by $n_{\textup{csize}}$.

(B) The cardinality of the set of the components is denoted by $n_{\textup{cmp}}$. So $n_{\textup{cmp}}=|\textup{Comp}(P)|$.

(C) Let $\textup{Edge}(P)$ stand for the set of edges of $P$; by an edge, we mean a pair $(x,y)$ belonging to the covering relation $\prec$. Similarly, for $T\in\textup{Comp}(P)$, $\textup{Edge}(T)$ stands for the set of edges of $T$.

(D) Assuming that $P$ is of a finite component size, $n_{\textup{csize}}^{2}$ a finite upper bound on $\{|\textup{Edge}(T)|:T\in\textup{Comp}(P)\}$ and so we can define the component edge number of $P$ as ${n_{\textup{cedge}}}:=\max\{|\textup{Edge}(T)|:T\in\textup{Comp}(P)\}$. (Note that even if ${n_{\textup{cedge}}}$ is finite, $P$ can have infinitely many edges.)

(E) Assuming that $P$ is of finite component size, $n_{\textup{csize}}$ is a finite upper bound on $\{|\textup{Extr}(Q)|:Q\in\textup{Comp}(P)\}$ and so we can define the component extremal number by $n_{\textup{cextr}}:=\max\{|\textup{Extr}(Q)|:Q\in\textup{Comp}(P)\}$, which is a positive integer even if $\textup{Extr}(P)$ is infinite.

(F) For a finite tree $T=(T;\mu_{T})=(T;\leq)$, we define the tree parameter $n_{\textup{tp}}(T)$ as follows. If $T$ is a singleton, then $n_{\textup{tp}}(T)=0$. If $T$ is a chain with more than one element, then $n_{\textup{tp}}(T)=1$. Otherwise, $n_{\textup{tp}}(T)$ is the smallest number $k\in\mathbb{N}^{+}$ such that, denoting the least member of $\textup{Quo}^{\kern-1.0pt\leq}(T;\mu_{T})$ containing $(y,x)$ by $\textup{qu}_{\mu}(y,x)$,

[TABLE]

Note that, in spite of its notation, we do not claim that $Y(T)$ above is unique888We have not investigated (and do not plan to investigate) whether $Y(T)$ is unique or not..

(G) Assuming that $n_{\textup{cmp}}\geq 3$, the correction number $n_{\textup{corr}}$ is defined and the two selector components $M^{\mathord{\bowtie}}_{1}$ and $M^{\mathord{\bowtie}}_{2}$ of $P$ are designated as follows. Pick a connected component $T$ of $P$ with the least tree parameter; this component is $M^{\mathord{\bowtie}}_{1}$. Then pick another component $T^{\prime}$ with the least tree parameter among the components distinct from $M^{\mathord{\bowtie}}_{1}$; this $T^{\prime}$ is $M^{\mathord{\bowtie}}_{2}$. Finally, let $n_{\textup{corr}}:=n_{\textup{tp}}(M^{\mathord{\bowtie}}_{1})+n_{\textup{tp}}(M^{\mathord{\bowtie}}_{2})$.

Next, we define two functions.

Definition 3.2** (Two functions).**

(A) For a positive integer $n$, let $\textup{LASp}(n)$ denote the smallest $k\in\mathbb{N}^{+}=\{1,2,3,\dots\}$ such that

[TABLE]

where $\lfloor k/2\rfloor$ denotes the (lower) integer part of $k/2$, and let $\textup{LASp}(0):=0$. For example, $\textup{LASp}(1)=1$, $\textup{LASp}(2)=2$, $\textup{LASp}(3)=3$, $\textup{LASp}(4)=\textup{LASp}(5)=\textup{LASp}(6)=4$, …, $\textup{LASp}(1000)=13$, $\textup{LASp}(10^{20})=70$, and $\textup{LASp}(10^{100})=337$. (The origin of the acronym LASp is explained in Czédli [10].)

(B) We define a function $f\colon\{\text{accessible cardinals}\}\setminus\{0\}\to\mathbb{N}_{0}$ by the rules $f(1):=0$, $f(2):=2$, $f(3):=4$, $f(4):=5$ and999If the answer to the question in Lemma 1.1 is affirmative, then we could let $f(4)=4$., for any accessible cardinal $x\geq 5$, $f(x):=4$.

Based on Definitions 3.1 and 3.2, we are in the position to formulate our result on the lattice $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ of all of quasiorders of $P$ that extend $\mu$ = “$\leq$”.

Theorem 3.3**.**

Let $(P;\leq)=(P;\mu)$ be a poset of finite component size $n_{\textup{csize}}$ such that $|P|$ is an accessible cardinal. Then the following three assertions101010Let us repeat: some particular cases are formulated in an easy-to-read way in Corollary 4.1 and also in Examples 4.2 and 4.3. Note that, for any $k\in\mathbb{N}^{+}$, $k$-generated lattices are also $k^{\prime}$-generated for every $k^{\prime}>k$. So a warning is appropriate here: we do not claim that (3.3) and (3.4) are optimal in general; see also Remark 4.4. hold.

(A)* If $n_{\textup{cmp}}\geq 3$, then the lattice $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)=(\textup{Quo}^{\kern-1.0pt\leq}(P;\mu);\subseteq)$ is*

[TABLE]

as a complete lattice.

(B)* If $P$ is a forest and $n_{\textup{cmp}}\geq 3$, then the lattice $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ is*

[TABLE]

as a complete lattice.

(C)* If $P$ is a finite chain, then $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ is $\textup{LASp}({n_{\textup{cedge}}})$-generated but not $\left(\textup{LASp}({n_{\textup{cedge}}})-1\right)$-generated.*

Proof of Theorem 3.3.

While the meet operation in $\textup{Quo}(P)$ and $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ is the set theoretic intersection, the join in these complete lattices are described with the help of the reflexive and transitive closure tr as follows:

[TABLE]

here and later, tr is understood over $P$. Now it is clear that $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ is a complete sublattice of $\textup{Quo}(P)$, whereby we will use $\vee$ and $\bigvee$ for the elements of $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$. For a subset $X$ of $P$, we define $\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{X};\leq)$ as $\{\textup{tr}(\mu\cup\rho):\rho\in\textup{Quo}(X)\}$; it is a subposet of $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$.

For later use, we formulate the following “independence principle”.

[TABLE]

For a subset $X$ of $P$, the “full relation” $X^{2}$ of $X$ is often denoted by $\nabla_{{\kern-1.5ptX}}$. Furthermore, we let $\nabla^{+}_{{\kern-1.5ptX}}:=\textup{qu}_{\mu}(\nabla_{{\kern-1.5ptX}})\in\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$. Although (3.6) can easily be derived from (3.5), there is an even simpler way. Namely, $\textup{qu}_{\mu}(p,q)\subseteq\nabla^{+}_{{\kern-1.5ptT}}=\nabla_{{\kern-1.5ptT}}\cup\mu\in\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ and even $\nabla_{{\kern-1.5ptT}}\cup\mu$ fails to contain $(x,y)\in(P\setminus T)^{2}$ if $x\nleq y$. This implies (3.6).

Part (of the proof) 1**.**

We claim that

[TABLE]

To prove this, consider the functions $\varphi^{\prime}_{X}\colon(\textup{Quo}(X);\subseteq)\to(\textup{Quo}(P);\subseteq)$ defined by the equality $\varphi^{\prime}_{X}(\rho):=\rho\cup\Delta_{P}$ and $\varphi^{\flat}_{X}\colon\varphi^{\prime}_{X}((\textup{Quo}(X);\subseteq))\to(\textup{Quo}(P);\subseteq)$ defined by $\varphi^{\flat}_{X}(\rho):=\rho\vee\mu$. It follows from (3.5) and $\Delta_{P}\cup\mu=\mu$ that $\varphi_{X}=\varphi_{X}^{\flat}\circ\varphi^{\prime}_{X}$. Thus, as $\varphi^{\prime}_{X}$ is clearly an embedding, it suffices to show that $\varphi^{\flat}_{X}$ is also such. Clearly, $\varphi^{\flat}_{X}$ commutes with joins. For $\rho\in\textup{Quo}(P)$, let $\rho^{-}$ denote the irreflexive version $\rho\setminus\Delta_{P}$ of $\rho$. For $\rho\in\varphi^{\prime}_{X}(\textup{Quo}(X))$, we have that $\rho^{-}\subseteq X^{2}$. Assume that $\rho_{j}\in\varphi^{\prime}_{X}(\textup{Quo}(X))$ for all $j$ in an index set $J$ and $(x,y)\in\bigwedge_{j\in J}\varphi^{\flat}_{X}(\rho_{j})=\bigcap_{j\in J}\varphi^{\flat}_{X}(\rho_{j})$; we need to show that (x,y)\in\varphi^{\flat}_{X}\bigl{(}\bigcap_{j\in J}\rho_{j}\bigr{)}. Since this is trivial if $(x,y)\in\mu$, we assume that $(x,y)\notin\mu$, that is, $x\nleq y$. As $(x,y)\in\bigcap_{j\in J}\varphi^{\flat}_{X}(\rho_{j})$, we have that $(x,y)\in\varphi^{\flat}_{X}(\rho_{j})=\rho_{j}\vee\mu$ for all $j\in J$. By the first line of (3.7), the transitivity of $\rho_{j}$, that of $\mu$, $\rho_{j}^{-}\subseteq X^{2}$, $x\nleq y$, and (3.5), there are elements $z_{0}=x$, $z_{1}$, $z_{2}$, and $z_{3}=y$ such that $\{z_{1},z_{2}\}\subseteq X$, $z_{1}\neq z_{2}$, $(z_{0},z_{1})\in\mu$, $(z_{1},z_{2})\in\rho_{j}$, and $(z_{2},z_{3})\in\mu$. The first two memberships give that $z_{1}$ is the only element in the intersection of $X$ and the component of $z_{0}=x$. Similarly, the last two memberships give that $z_{2}$ is the only element in the intersection of $X$ and the component of $z_{3}=y$. Thus, $z_{1}$ and $z_{2}$ do not depend on $j$, whereby $(z_{1},z_{2})\in\bigcap_{j\in J}\rho_{j}$ and we obtain that (x,y)\in\Bigl{(}\bigcap_{j\in J}\rho_{j}\Bigr{)}\vee\mu=\varphi^{\flat}_{X}\Bigl{(}\bigcap_{j\in J}\rho_{j}\Bigr{)}. So \bigcap_{j\in J}\varphi^{\flat}_{X}(\rho_{j})\subseteq\varphi^{\flat}_{X}\Bigl{(}\bigcap_{j\in J}\rho_{j}\Bigr{)}. This yields that $\varphi^{\flat}_{X}$ commutes with meets as it is isotone. So $\varphi^{\flat}_{X}$ is a complete lattice homomorphism. For $\rho\in\textup{Quo}(P)$, let $\rho\rceil_{\kern-1.5ptX}$ denote the restriction of $\rho$ to $X$. That is, $\rho\rceil_{\kern-1.5ptX}=\rho\cap(X\times X)$. The first line of (3.7) together with (3.5) imply easily that for $\rho\in\varphi_{X}^{\prime}(\textup{Quo}(X))$, we have that $\rho=\varphi^{\flat}_{X}(\rho)\rceil_{\kern-1.5ptX}\cup\Delta_{P}$. This yields the injectivity of $\varphi^{\flat}_{X}$ and we have shown (3.7).

Let us repeat and (motivated by $\nabla^{+}_{{\kern-1.5ptX}}$) introduce that for a pair $(x,y)\in P^{2}$ and a set $\gamma\subseteq P^{2}$ of such pairs, let $\textup{quo}(x,y)$ and $\textup{quo}(\gamma)$ stand for the least quasiorder in $\textup{Quo}(P)$ containing $(x,y)$ and including $\gamma$, respectively. Similarly, $\textup{qu}_{\mu}(x,y)$ and $\textup{qu}_{\mu}(\gamma)$ or $\gamma^{+}$ are the least members of $(\textup{Quo}^{\kern-1.0pt\leq}(P);\mu)$ containing $(x,y)$ and including $\gamma$, respectively. Note that

[TABLE]

In addition to $\textup{Edge}(P)=\{(x,y)\in P^{2}:x\prec y\}$ and, for $T\in\textup{Comp}(P)$, $\textup{Edge}(T)=\{(x,y)\in T^{2}:x\prec y\}$, we define $\textup{Edge}^{-1}(P):=\{(x,y):(y,x)\in\textup{Edge}(P)\}$ and $\textup{Edge}^{-1}(T):=\{(x,y):(y,x)\in\textup{Edge}(T)\}=\{(x,y):\{x,y\}\subseteq T\text{ and }y\prec x\}$.

Next, we introduce two crucial concepts for the proof. By a strip of $P$, we will mean a nonempty subset $S$ of $\textup{Edge}^{-1}(P)$ such that for all $T\in\textup{Comp}(P)$, we have that $|S\cap\textup{Edge}^{-1}(T)|\leq 1$. Similarly, a thread111111Some strips and threads are visualized in Figures 1 and 2. Namely, a strip $S$ looks like a colorful ribbon through the edges that $S$ contains; threads are thinner and go through the elements they contain.

is a subset $D$ of $\textup{Extr}(P)$ such that for each $T\in\textup{Comp}(P)$,  $|D\cap\textup{Extr}(T)|=1$. (Note the difference: “$\leq 1$” for strips but “$=1$” for threads.) As $P$ is of finite component size for all the three parts of the theorem, ${n_{\textup{cedge}}}$ and $n_{\textup{cextr}}$ are also finite. Clearly,

[TABLE]

Similarly,

[TABLE]

Part (of the proof) 2**.**

This part is only needed for Parts (B) and (C) of the theorem. We claim that

[TABLE]

To verify this, suppose the contrary. Then, combining (3.5) and (3.8) with $\mu=\textup{tr}(\textup{Edge}(P))$, we can find a shortest sequence $z_{0}=v$, $z_{1}$,…, $z_{n-1}$, $z_{n}=u$ of elements of $P$ such that for each $s\in\{1,\dots,n\}$, either $z_{s-1}\prec z_{s}$, or $z_{s-1}\succ z_{s}$ but $(z_{s-1},z_{s})\neq(v,u)$. Our sequence, being the a shortest one, is repetition-free. Hence, in particular, none of the $z_{1}$,…, $z_{n-1}$ is $u$ or $v$. But then $z_{0}=v$, $z_{1}$,…, $z_{n-1}$, $z_{n}=u$, $z_{n+1}=z_{0}$ is a non-singleton circle in the graph of $P$, which is a contradiction proving (3.11).

For a strip $S\in\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$, see (3.9), $S$ is a set of reversed edges and so a subset of $P^{2}$, whence $\textup{qu}_{\mu}(S)$ makes sense. For $T\in\textup{Comp}(P)$, $|\textup{Edge}^{-1}(T)\cap S|\leq 1$. Therefore, using a shortest sequence like in the previous paragraph and observing that this sequence cannot “jump” from a component to another component, we obtain easily that

[TABLE]

We claim that

[TABLE]

To prove this, consider the powerset lattice \textup{Pow}(\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}})=(\textup{Pow}(\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}});\subseteq)=\bigl{(}\{S:S\in\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}\};\subseteq\bigr{)}. Since $|\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}|={n_{\textup{cedge}}}$, we know from Czédli [10] that $\textup{Pow}(\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}})$ has a $\textup{LASp}({n_{\textup{cedge}}})$-element generating set $G_{0}$. The members of $G_{0}$ are subsets of $\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$. For each strip $S$, the singleton $\{S\}\in\textup{Pow}(\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}})$ belongs to the sublattice $[G_{0}]$ generated by $G_{0}$. Since $\textup{Pow}(\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}})$ is distributive, $\{S\}$ is obtained by applying a disjunctive normal form to appropriate elements of $G_{0}$. That is, $\{S\}$ is the union of intersections of elements of $\textup{Pow}(\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}})$. However, $\{S\}$ is join-irreducible (i.e., union-irreducible), whereby no union is needed and $\{S\}$ is the intersection of some members of $G_{0}$. So

[TABLE]

For $X\in G_{0}$, we let

[TABLE]

We are going to show that $G$ satisfies the requirements of (3.13). As $|G_{0}|=\textup{LASp}({n_{\textup{cedge}}})$, we have that $|G|\leq\textup{LASp}({n_{\textup{cedge}}})$, as required. Next, we are going to show that for $S$ occurring in (3.14), we have that

[TABLE]

The “$\subseteq$” part of (3.17) is clear since $S\in X$ implies, by (3.15), that $\textup{qu}_{\mu}(S)\leq\textup{qu}_{\mu}(X)$. To verify the converse inclusion, assume that $p,q\in P$ such that

[TABLE]

For an $X$ such that $S\in X\in G_{0}$, as in (3.18), the containment $(p,q)\in\textup{qu}_{\mu}(X)$, $\mu=\textup{tr}{(\textup{Edge}(P)})$, and (3.5) yield a shortest sequence121212written in the form of a vector

[TABLE]

of elements of $P$ such that for each $s\in\{1,\dots,m(X)\}$,

• $(\bullet)$

$r_{s-1}(X)\prec r_{s}(X)$ or

• $(\flat)$

$(r_{s-1}(X),r_{s}(X))\in\textup{qu}_{\mu}(S^{\prime})$ for some $S^{\prime}\in X$.

By the same reason, $(\flat)$ also has an expansion to a sequence of elements of $P$. Therefore, assuming that this expansion has already been made (and resetting the notation), we can replace $(\flat)$ by the following condition:

• $(\sharp_{X})$

$(r_{s-1}(X),r_{s}(X))\in S^{\prime}\subseteq\textup{Edge}^{-1}(P)$ for some $S^{\prime}\in X$.

So, for each $s\in\{1,\dots,m(X)\}$, either $(\bullet)$ or $(\sharp_{X})$ holds. In both cases, $r_{s-1}(X)$ and $r_{s}(X)$ are the two endpoints of an edge of the (undirected) graph of $P$. Hence, $\vec{r}(X)$ is a path in the graph of $P$. In a forest, the shortest path connecting two vertices is unique. (Indeed, if we had two different shortest paths, then we would also have a circle of positive length, which is impossible.) Therefore, the sequence

[TABLE]

This allows us to write $m$, $\vec{r}$, and $r_{i}$ instead of $m(X)$, $\vec{r}(X)$, and $r_{i}(X)$, respectively. As distinct strips are disjoint, $S^{\prime}$ in $(\sharp_{X})$ is uniquely determined. Keeping in mind that $X$ has contained $S$ since (3.18), (3.20) and $(\sharp_{X})$ yield that for each $s\in\{1,\dots,m\}$, either $(\bullet)$, i.e. $r_{s-1}\prec r_{s}$, or

• $(\sharp_{\forall})$

$(r_{s-1},r_{s})\in S^{\prime}\subseteq\textup{Edge}^{-1}(P)$ for some $S^{\prime}$ belonging to all the $X\in G_{0}$ that contain $S$.

It follows from (3.14) that $S^{\prime}$ in $(\sharp_{\forall})$ is $S$. Hence, if $(\sharp_{\forall})$ holds for a subscript $s\in\{1,\dots,m\}$, then $(r_{s-1},r_{s})\in S\subseteq\textup{qu}_{\mu}(S)$. If $(\bullet)$ holds for this $s$, then $(r_{s-1},r_{s})\in\textup{qu}_{\mu}(S)$ again since $\mu\subseteq\textup{qu}_{\mu}(S)$. Therefore, we obtain that $(p,q)=(r_{0},r_{m})\in\textup{qu}_{\mu}(S)$ by transitivity. We have obtained the converse inclusion for (3.17). Hence, we have proved (3.17) and, thus, (3.13).

Part (of the proof) 3**.**

The following “convexity property” is valid for any poset $P$.

[TABLE]

The “if part” of (3.21) is clear by transitivity. To see the “only if part”, assume that $(b,a)\in\rho$ and $[x,y]$ is an edge in the interval $[a,b]$. Then $(y,b)\in\mu\subseteq\rho$, $(b,a)\in\rho$, and $(a,x)\in\mu\subseteq\rho$ imply $(y,x)\in\rho$ by transitivity, proving (3.21).

Now we are in the position to prove Part (C) of the theorem. So let $P$ be a chain of length ${n_{\textup{cedge}}}$ for a while. By Kim, Kwon, and Lee [15], $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ is a Boolean lattice with ${n_{\textup{cedge}}}$ atoms. Alternatively, since all the necessary tools occur in the present paper, we can argue shortly as follows. For each member $\rho$ of $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$, (3.21) gives that $\rho=\bigvee\{\textup{qu}_{\mu}(y,x):(y,x)\in\textup{Edge}^{-1}(P)\cap\rho\}$. By (3.11), none of the joinands can be removed, and it follows that $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ is (order) isomorphic to the powerset lattice of $\textup{Edge}^{-1}(P)$, whereby it is isomorphic to the Boolean lattice with ${n_{\textup{cedge}}}$ atoms. Therefore, Part (C) of Theorem 3.3 follows from Czédli [10].

Part (of the proof) 4**.**

Now we turn our attention to Parts (A) and (B) of the theorem. For Part (B), we choose $G$ according to (3.13). For Part (A), let $G:=\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$; see (3.9). It is trivial for Part (A) and it follows from (3.13) for Part (B) that

[TABLE]

For $D\in\boldsymbol{\boldsymbol{\Theta}}_{\textup{\kern-0.5ptthd}}$, see (3.10), $|D|=n_{\textup{cmp}}$ by definitions. So Lemma 1.1 allows us to pick an $f(n_{\textup{cmp}})$-element generating set $H_{D}^{(0)}$ of $(\textup{Quo}(D);\subseteq)$. By (3.7), $\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{D};\leq)$, which is a complete sublattice of $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$, also has an $f(n_{\textup{cmp}})$-element generating set $H_{D}:=\varphi_{D}(H_{D}^{(0)})$. So we can take

[TABLE]

Note that

[TABLE]

Next, we define the third ingredient needed by (3.3) and (3.4). Let $i\in\{1,2\}$. If $n_{\textup{tp}}(M^{\mathord{\bowtie}}_{i})=0$ (that is, if $|M^{\mathord{\bowtie}}_{i}|=1$), then $F^{(0)}(M^{\mathord{\bowtie}}_{i}):=\emptyset$. If $M^{\mathord{\bowtie}}_{i}$ is a finite chain with more than one element131313Note that $n_{\textup{tp}}(M^{\mathord{\bowtie}}_{i})=1$ in this case., then let $F^{(0)}(M^{\mathord{\bowtie}}_{i}):=\{\nabla_{{\kern-1.5ptM^{\mathord{\bowtie}}_{i}}}\}$. If $M^{\mathord{\bowtie}}_{i}$ is not a chain, then let $F^{(0)}(M^{\mathord{\bowtie}}_{i})$ be the union of $\{\nabla_{{\kern-1.5ptM^{\mathord{\bowtie}}_{i}}}\}$ and the set $Y(M^{\mathord{\bowtie}}_{i})$ defined in (3.1). In all cases, $F^{(0)}(M^{\mathord{\bowtie}}_{i})\subseteq\textup{Quo}^{\kern-1.0pt\leq}(M^{\mathord{\bowtie}}_{i};\leq_{M^{\mathord{\bowtie}}i})$. Note that

[TABLE]

For $\rho\in F^{(0)}(M^{\mathord{\bowtie}}_{i})$, we can take $\textup{qu}_{\mu}(\rho)$ since $\rho\subseteq{M^{\mathord{\bowtie}}_{i}}^{2}\subseteq P^{2}$. Using (3.5) or (3.6), it follows easily that the function $\psi_{i}\colon\textup{Quo}^{\kern-1.0pt\leq}(M^{\mathord{\bowtie}}_{i};\leq_{M^{\mathord{\bowtie}}_{i}})\to\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ defined by $\rho\mapsto\textup{qu}_{\mu}(\rho)$ is a complete141414As the domain of $\psi_{i}$ is a finite lattice, the adjective “complete” can be dropped. lattice embedding. Now we pass from $M^{\mathord{\bowtie}}_{i}$ to $P$. Let M^{\scriptscriptstyle{\mathord{\blacktriangleright}\kern-3.0pt\mathord{\blacktriangleleft}}}_{i}:=\psi_{i}\Bigl{(}\textup{Quo}^{\kern-1.0pt\leq}(M^{\mathord{\bowtie}}_{i};\leq_{M^{\mathord{\bowtie}}i})\Bigr{)} and $F(M^{\scriptscriptstyle{\mathord{\blacktriangleright}\kern-3.0pt\mathord{\blacktriangleleft}}}_{i}):=\psi_{i}(F^{(0)}(M^{\mathord{\bowtie}}_{i}))$. Note the difference: $M^{\mathord{\bowtie}}_{i}$ is in $\textup{Comp}(P)$ and so it consists of some elements of $P$ while $M^{\scriptscriptstyle{\mathord{\blacktriangleright}\kern-3.0pt\mathord{\blacktriangleleft}}}_{i}$ is a sublattice of $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ and consists of some quasiorders of $P$. It is clear by (3.1) and (3.25) that

[TABLE]

Now (3.22), (3.23), and (3.26) allow us to define

[TABLE]

It follows from (3.22), (3.23), and (3.27) that $|E|$ is (at most) what (3.3) and (3.4) require. So we only need to show that $E$ generates $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ as a complete lattice. Assume that $K$ is a complete sublattice of $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ such that $E\subseteq K$. To complete the proof, we need to show that $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)\subseteq K$.

It follows from $H_{D}\subseteq H\subseteq E$ that, with the notation used in (3.7),

[TABLE]

Next, we claim that

[TABLE]

To see this, take the unique $S\in\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$ that contains the reversed edge $(y,x)$ of $M^{\mathord{\bowtie}}_{i}$. Since $G\subseteq E$, (3.22) yields that $\textup{qu}_{\mu}(S)$ is in $K$. So is $\nabla^{+}_{{\kern-1.5ptM^{\mathord{\bowtie}}_{i}}}$ by $F(M^{\scriptscriptstyle{\mathord{\blacktriangleright}\kern-3.0pt\mathord{\blacktriangleleft}}}_{i})\subseteq E$ and (3.26). Hence (3.12) gives that $\textup{qu}_{\mu}(y,x)=\textup{qu}_{\mu}(S)\wedge\nabla^{+}_{{\kern-1.5ptM^{\mathord{\bowtie}}_{i}}}\in K$, proving (3.30). Now it follows from (3.30), $F(M^{\scriptscriptstyle{\mathord{\blacktriangleright}\kern-3.0pt\mathord{\blacktriangleleft}}}_{i})\subseteq E$, and (3.26) that for all $i\in\{1,2\}$,

[TABLE]

Our next task151515This task will be completed after (3.38), where (3.32) will appear to be a particular case of (3.37). is to show that

[TABLE]

To show this, we can assume that $x_{\ast}<y^{\ast}$. (Then $x_{\ast}\in\textup{Min}(T)$ and $y^{\ast}\in\textup{Max}(T)$.) By (3.31), we can also assume that $T\notin\{M^{\mathord{\bowtie}}_{1},M^{\mathord{\bowtie}}_{2}\}$. Let $D^{\prime},D^{\prime\prime}\in\boldsymbol{\boldsymbol{\Theta}}_{\textup{\kern-0.5ptthd}}$ be threads containing $x_{\ast}$ and $y^{\ast}$, respectively; they are distinct but they need not be unique. For $i\in\{1,2\}$, let $x_{\ast i}$ and $y^{\ast}_{i}$ be the unique element of $D^{\prime}\cap M^{\mathord{\bowtie}}_{i}$ and that of $D^{\prime\prime}\cap M^{\mathord{\bowtie}}_{i}$, respectively. So $\{x_{\ast i},x_{\ast}\}\subseteq D^{\prime}$ and $\{y^{\ast},y^{\ast}_{i}\}\subseteq D^{\prime\prime}$ for $i\in\{1,2\}$. By the choice of $H_{D}$ (with $D=D^{\prime}$ or $D=D^{\prime\prime}$), (3.7), and (3.23),

[TABLE]

We know from (3.31) that

[TABLE]

We claim that (3.35) and (3.36) below hold; note in advance that their proofs will use only the assumption that $x_{\ast}\in\textup{Min}(P)$, $y^{\ast}\in\textup{Max}(P)$, and $x_{\ast}<y^{\ast}$.

[TABLE]

note that $h^{\bullet}(y^{\ast},x_{\ast})$ is the term in the middle of (3.35). Observe that (3.36) is clear by (3.33) and (3.34). The first inequality in (3.35) is trivial by transitivity. To show the second inequality, assume that $(u,v)\in P^{2}$ belongs to the meet in (3.35), that is, it belongs to $h^{\bullet}(y^{\ast},x_{\ast})$. We also assume that $u\nleq v$ since otherwise $(u,v)\in\nabla^{+}_{{\kern-1.5ptT}}$ is trivial. Let $i\in\{1,2\}$. Using the description of joins, see (3.5), (3.8), and (3.24), we obtain a shortest sequence $\vec{w}:w_{0}=u,w_{1},w_{2},\dots,w_{t-1},w_{t}=v$ of pairwise different elements of $P$ such that for each $j\in\{1,\dots,t\}$, one of the following four alternatives holds:

• $(\spadesuit)$

$w_{j-1}<w_{j}$  (and so $w_{j-1}$ and $w_{j}$ belong to the same component of $P$);

• $(\clubsuit)$

$(w_{j-1},w_{j})=(y^{\ast},y^{\ast}_{i})$;

• (

$\spadesuit$

)

$(w_{j-1},w_{j})=(y^{\ast}_{i},x_{\ast i})$

• (

$\clubsuit$

)

$(w_{j-1},w_{j})=(x_{\ast i},x_{\ast})$

Since $u\nleq v$, at least one of $(\clubsuit)$, (

$\spadesuit$

), and (

$\clubsuit$

) holds some $j\in\{1,\dots,t\}$. Hence, at least one of the elements $w_{j}$ belongs to $T\cup M^{\mathord{\bowtie}}_{i}$. This fact and the parenthesized comment in $(\spadesuit)$ yield that all the $w_{j}$ belong to $T\cup M^{\mathord{\bowtie}}_{i}$. In particular, $\{u,v\}=\{w_{0},w_{t}\}\subseteq T\cup M^{\mathord{\bowtie}}_{i}$. As opposed to the elements $w_{j}$ for $j\in\{1,\dots,t-1\}$, we know that $\{u,v\}$ does not depend on $i\in\{1,2\}$. Hence, $\{u,v\}\subseteq(T\cup M^{\mathord{\bowtie}}_{i})\cap(T\cup M^{\mathord{\bowtie}}_{3-i})=T\cup(M^{\mathord{\bowtie}}_{i}\cap M^{\mathord{\bowtie}}_{3-i})=T\cup\emptyset=T$. Thus, $(u,v)\in\nabla^{+}_{{\kern-1.5ptT}}$, proving the second inequality of (3.35) and (3.35) itself.

Based on (3.35) and (3.36), now we show that

[TABLE]

First, we deal with the particular case, that is, let $(v,u)\in\textup{Edge}^{-1}(P)$. Let $S\in\boldsymbol{\boldsymbol{\Xi}}_{\textup{\kern-0.1ptstp}}$ be the unique strip that contains $(v,u)$. Since $G\subseteq E$, we know from (3.22) that $\textup{qu}_{\mu}(S)$ is in $K$. Pick $u_{\ast},v^{\ast}\in\textup{Extr}(P)$ such that $u_{\ast}\leq u\prec v\leq v^{\ast}$ and let $T$ be the unique component of $P$ that contains these four elements. The sentence preceding (3.35) makes it clear that (3.35) and (3.36) remain valid if we substitute $(x,y,x_{\ast},y^{\ast})$ with $(u,v,u_{\ast},v^{\ast})$. Hence, combining (3.12), (3.21), and (3.36), we have that

[TABLE]

These inequalities and (3.36) yield that $\textup{qu}_{\mu}(v,u)=\textup{qu}_{\mu}(S)\wedge h^{\bullet}(v^{\ast},u_{\ast})\in K$. That is, the particular case of (3.37) holds. It follows from transitivity and (3.21) that

[TABLE]

Therefore, the particular case implies the general case, and we have shown (3.37).

We have also shown (3.32) since it is a particular case of (3.37).

Now we are ready to show that

[TABLE]

Here, of course, $\mu_{T}$ stands for the restriction $\mu\rceil_{\kern-1.5ptT}$ of $\mu$ to $T$. We can assume that $|T|>1$. It suffices to show that $\nabla^{+}_{{\kern-1.5ptT}}=\bigvee\{\textup{qu}_{\mu}(y,x):(y,x)\in\textup{Edge}^{-1}(T)\}$, that is, $\nabla^{+}_{{\kern-1.5ptT}}=\textup{qu}_{\mu}(\textup{Edge}^{-1}(T))$, as all the $\textup{qu}_{\mu}(y,x)$ are in $K$ by (3.37) and so $\nabla^{+}_{{\kern-1.5ptT}}\in K$ will prompt follow. To see the “$\leq$” part, assume that $(x,y)\in\nabla^{+}_{{\kern-1.5ptT}}$. As $T$ is a connected component, there are elements $z_{0}=x,z_{1},\dots,z_{t-1},z_{t}=y$ in $T$ such $(z_{j-1},z_{j})\in\textup{Edge}(T)\cup\textup{Edge}^{-1}(T)$ for every $j\in\{1,\dots,t\}$. Regardless whether $(z_{j-1},z_{j})\in\textup{Edge}(T)$ or $(z_{j-1},z_{j})\in\textup{Edge}^{-1}(T)$, we have that $(z_{j-1},z_{j})\in\textup{qu}_{\mu}(z_{j-1},z_{j})\subseteq\textup{qu}_{\mu}(\textup{Edge}^{-1}(T))$. Hence, the required “$\leq$” part follows by transitivity. As the converse inequality is trivial by (3.6), we have shown (3.39).

The following easy assertion will be useful.

[TABLE]

The “$\geq$” part is clear by transitivity. To show the converse inequality, assume that $(p,q)\in\textup{qu}_{\mu}(a,b)\setminus\mu$. By (3.5) and (3.8), there is a sequence $x_{0}=p,x_{1},\dots,x_{t-1},x_{t}=q$ of elements in $P$ such that, for each $i\in\{0,\dots,t-1\}$, $(x_{i},x_{i+1})\in\mu$ or $(x_{i},x_{i+1})=(a,b)$. As $p\nleq q$, there is at least one $i$ such that $(x_{i},x_{i+1})=(a,b)$. The least such $i$ shows that $p=x_{0}\leq x_{1}\dots\leq x_{i}=a$, whence $p\in\mathord{\downarrow}\kern 0.5pta$, while the largest such $i$ yields that $b=x_{i+1}\leq\dots\leq x_{t}=q$, whereby $q\in\mathord{\uparrow}\kern 0.5ptb$. Hence, $(p,q)\in\mathord{\downarrow}\kern 0.5pta\times\mathord{\uparrow}\kern 0.5ptb\subseteq\mu\cup(\mathord{\downarrow}\kern 0.5pta\times\mathord{\uparrow}\kern 0.5ptb)$, proving (3.40).

Next, we claim that for any $T_{1}$ and $T_{2}$,

[TABLE]

To show this, first we note that $T_{1}\cap T_{2}=\emptyset$. Pick $a_{\ast},a^{\ast}\in\textup{Extr}(T_{1})$ and $b_{\ast},b^{\ast}\in\textup{Extr}(T_{2})$ such that $a_{\ast}\leq a\leq a^{\ast}$ and $b_{\ast}\leq b\leq b^{\ast}$. We let

[TABLE]

Observe that $H\subseteq E$ and (3.37) imply that $\beta^{\textup{up}}$, $\gamma_{\textup{dn}}$, and $\beta^{\textup{up}}\wedge\gamma_{\textup{dn}}$ are in $K$. Hence, to prove (3.43) and, consequently, (3.41), it suffices to prove the equality in (3.43). Actually, it suffices to show that $\textup{qu}_{\mu}(a,b)\geq\beta^{\textup{up}}\wedge\gamma_{\textup{dn}}$, as the converse inequality is trivial. To do so, assume that $(p,q)\in\beta^{\textup{up}}\wedge\gamma_{\textup{dn}}$. Pick a shortest sequence $\vec{x}:x_{0}=p$, $x_{1},\dots,x_{h}=q$ and a shortest sequence $\vec{y}:y_{0}=p$, $y_{1},\dots,$ $y_{k}=q$ of elements of $P$ witnessing that $(p,q)\in\beta^{\textup{up}}=\mu\vee\textup{quo}(a^{\ast},b^{\ast})\vee\textup{quo}(b^{\ast},b)$ and $(p,q)\in\gamma_{\textup{dn}}=\mu\vee\textup{quo}(a,a_{\ast})\vee\textup{quo}(a_{\ast},b_{\ast})$, respectively, according (3.5). So, for example, for every $j\in\{1,\dots,h\}$, exactly one of $x_{j-1}<x_{j}$, $(x_{j-1},x_{j})=(a^{\ast},b^{\ast})$, and $(x_{j-1},x_{j})=(b^{\ast},b)$ holds. First, assume that $q\in T_{1}$ and $(p,q)\in\beta^{\textup{up}}\wedge\gamma_{\textup{dn}}\subseteq\beta^{\textup{up}}$. If an element of the sequence $\vec{x}$ is in $T_{2}$, then all the subsequent elements of $\vec{x}$ are in $T_{2}$, contradicting that $q\in T_{1}$. Hence, no element of $\vec{x}$ is in $T_{2}$. So, for all $j\in\{1,\dots,h\}$,  $(x_{j-1},x_{j})$ is neither $(a^{\ast},b^{\ast})$ nor $(b^{\ast},b)$. Hence, $x_{j-1}<x_{j}$ for all $j\in\{1,\dots,h\}$, and so $(p,q)=(x_{0},x_{h})\in\mu\subseteq\textup{qu}_{\mu}(a,b)$. Let us summarize:

[TABLE]

Second, observe that if an element of $\vec{y}$ belongs to $T_{1}$, then all the preceding elements of $\vec{y}$ do so. Then, similarly to the argument for (3.44), we obtain that

[TABLE]

By the “$\in\mu$” part of (3.44) and (3.45), each of the sequences $\vec{x}$ and $\vec{y}$ implies that

[TABLE]

We can assume that $p\nleq q$ as otherwise $(p,q)\in\textup{qu}_{\mu}(a,b)$ is obvious. But then each of the sequences $\vec{x}$ and $\vec{y}$ implies easily that $\{p,q\}\subseteq T_{1}\cup T_{2}$. Thus, it follows from (3.44), (3.45), and (3.46) that there is only one case left. Namely, from now on in the proof, we assume that $p\in T_{1}$ and $q\in T_{2}$. In the sequence $\vec{x}$, which begins in $T_{1}$ and terminates in $T_{2}$, let $x_{u}$ be the first element that is in $T_{2}$. As we mentioned in the argument for (3.44), the subsequent elements $x_{u+1}$, $x_{u+2}$, …, $x_{h}$ are all in $T_{2}$. Clearly, $x_{u}=b^{\ast}\geq b$. Using that, for $j\in\{u,\dots,h-1\}$, either $(x_{j},x_{j+1})=(b^{\ast},b)$ (and so $x_{j+1}=b\geq b$) or $(x_{j},x_{j+1})\in\mu$ (and so $x_{j}\geq b\Rightarrow x_{j+1}\geq b$), it follows by induction161616At most two inductive steps are needed as $\vec{x}$ is a shortest sequence. on $j$ that $q=x_{h}\geq b$, that is, $q\in\mathord{\uparrow}\kern 0.5ptb$. Similarly, letting $y_{v}$ denote the last member of $\vec{y}$ that belongs to $T_{1}$, the elements $y_{v},y_{v-1},\dots,y_{0}=p$ are all in $T_{1}$. Then $y_{v}=a_{\ast}\in\mathord{\downarrow}\kern 0.5pta$. Using that each of the pairs $(y_{j-1},y_{j})$, $j\in\{v,v-1,\dots,1\}$, is in $\mu$ or is of the form $(a,a_{\ast})$, we obtain171717By a trivial induction backwards on $j$ in at most two steps. that $p=y_{0}\in\mathord{\downarrow}\kern 0.5pta$. Hence $(a,b)\in\mathord{\downarrow}\kern 0.5pta\times\mathord{\uparrow}\kern 0.5ptb$. So $(a,b)\in\textup{qu}_{\mu}(a,b)$ by (3.40). We have proved (3.43) and (3.41).

Next, we claim that

[TABLE]

We can assume that $a\parallel b$ and $T\notin\{M^{\mathord{\bowtie}}_{1},M^{\mathord{\bowtie}}_{2}\}$ since otherwise (3.47) is trivial by (3.26) and (3.37). For $i\in\{1,2\}$, pick an element $c_{i}\in M^{\mathord{\bowtie}}_{i}$. It suffices to show that

[TABLE]

as the term on the right is in $K$ by (3.41). By transitivity, the “$\leq$” in place of the equality sign in (3.48) is clear. For $i\in\{1,2\}$, (3.40) yields that

[TABLE]

Using (3.5), the inclusions above imply that

[TABLE]

As $T$, $M^{\mathord{\bowtie}}_{1}$, and $M^{\mathord{\bowtie}}_{2}$ are pairwise disjoint, (3.49) implies that

[TABLE]

the only role of underbracing is to help reading. This shows the “$\geq$” inequality and the whole (3.48). Thus, (3.47) holds.

Finally, each elements of $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ is the join of (possibly infinitely many) elements of the form $\textup{qu}_{\mu}(a,b)$ where $a,b\in P$. Therefore, as we already know from (3.41) and (3.47) that all the $\textup{qu}_{\mu}(a,b)$ are in $K$, we conclude that $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)\subseteq K$, as required. The proof of Theorem 3.3 is complete.

∎

4. Notes on the main theorem

In this section, we present some particular cases of the (main) Theorem 3.3 and a remark. Let $\{P_{i}=(P_{i};\leq_{i})=(P_{i};\mu_{i})$ :  $i\in I\}$ be a set of posets that are assumed to be pairwise disjoint181818If they are not disjoint, then we have to replace them by isomorphic copies that are pairwise disjoint.. The cardinal sum   $\sum^{\textup{card}}_{i\in I}(P_{i};\leq_{i})$ of these posets is $(\bigcup_{i\in I}P_{i};\bigcup_{i\in I}\mu_{i})$. E.g., for $P$ in Theorem 3.3, $(P;\leq)=\sum^{\textup{card}}_{T\in\textup{Comp}(P)}(T;\leq_{T})$. If $I=\{1,2\}$, then we prefer to write $(P_{1};\leq_{1})\mathrel{+^{\textup{card}}}(P_{2};\leq_{2})$. The Y-poset is the 4-element poset $\{0,a,b,c\}$ such that $0\prec a\prec b$, $a\prec c$, but $b$ and $c$ are incomparable. The following statement is a straightforward consequence of Theorem 3.3.

Corollary 4.1**.**

Assume that $n_{\textup{cmp}}\geq 5$ is an accessible cardinal. Let $A_{2}$ denote the $2$-element antichain. The lattices below are complete lattices and lattice generation is understood accordingly.

(a)* If $(P;\leq)=(P;\mu)$ is the cardinal sum of $n_{\textup{cmp}}$ many singleton posets (that is, if $(P;\leq)$ is an antichain of size $n_{\textup{cmp}}$), then $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)=\textup{Quo}(P)$ is $4$-generated. So Theorem 3.3 implies the earlier result for antichains of size at least $5$, see Lemma 1.1.*

(b)* If $(P;\leq)=(P;\mu)$ is the cardinal sum of $n_{\textup{cmp}}$ many chains of lengths at most ${n_{\textup{cedge}}}\in\mathbb{N}^{+}$ and at least one of these chains is of length ${n_{\textup{cedge}}}$ (that is, the longest chain consists of ${n_{\textup{cedge}}}+1\geq 2$ many elements), then $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ is \bigl{(}10+\textup{LASp}({n_{\textup{cedge}}})\bigr{)}-generated while $\textup{Quo}^{\kern-1.0pt\leq}(P\mathrel{+^{\textup{card}}}A_{2})$ is \bigl{(}8+\textup{LASp}({n_{\textup{cedge}}})\bigr{)}-generated.*

(c)* For example, if $(P;\leq)$ is the cardinal sum of $n_{\textup{cmp}}$ many chains of length $10^{20}$ each, then $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ is $80$-generated while $\textup{Quo}^{\kern-1.0pt\leq}(P\mathrel{+^{\textup{card}}}A_{2})$ is $78$-generated.*

(d)* For example again, if $(P;\leq)$ is the cardinal sum of  $n_{\textup{cmp}}$ many Y-posets, then $n_{\textup{tp}}(M^{\mathord{\bowtie}}_{1})=n_{\textup{tp}}(M^{\mathord{\bowtie}}_{2})=3$, $n_{\textup{corr}}=6$, and $\textup{Quo}^{\kern-1.0pt\leq}(P;\mu)$ is $(3\cdot 4+3+6)$-generated (that is, $21$-generated) while $\textup{Quo}^{\kern-1.0pt\leq}(P\mathrel{+^{\textup{card}}}A_{2})$ is $15$-generated.*

Note that, in Corollary 4.1, the lattices $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ and $\textup{Quo}^{\kern-1.0pt\leq}(P\mathrel{+^{\textup{card}}}A_{2})$ in Part (d) with, say, $n_{\textup{cmp}}=50$ are (more than) large enough but still feasible for the purpose outlined in Footnote 2. For a subset $X$ of $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$, we write $[X]_{\textup{cgen}}$ to denote the complete sublattice of $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ generated by $X$. (If $P$ happens to be finite, then the adjective “complete” can be dropped.)

Example 4.2**.**

For $P=(P;\leq)$ drawn in Figure 1, Theorem 3.3 asserts that $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ is $23$-generated. Note that $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ has more than $9\cdot 10^{13}$ elements191919Since $[\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{D_{1}};\leq)\cup\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{D_{4}};\leq)]_{\textup{cgen}}\cong\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{D_{1}};\leq)\times\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{D_{4}};\leq)\cong(\textup{Quo}(D_{1});\leq)\times(\textup{Quo}(D_{4});\leq)$ is easy to verify and $|\textup{Quo}(D_{1})|=|\textup{Quo}(D_{4})|=9\,535\,241$ by Sloane [20, A000798], we have that $|\textup{Quo}(P)|>9\,535\,241^{2}\approx 9\cdot 10^{13}$. .

Example 4.3**.**

For $P=(P;\leq)$ drawn in Figure 2, Theorem 3.3 asserts that $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ is $12$-generated. Note that $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ has much more than $10^{23}$ elements202020Because $|[\textup{Quo}^{\kern-1.0pt\leq}(P\kern-0.2pt\boldsymbol{\rceil}\kern-1.1pt_{D_{1}};\leq)]_{\textup{cgen}}|=115\,617\,051\,977\,054\,267\,807\,460\approx 1.16\cdot 10^{23}$; see Sloane [20, A000798]. .

Remark 4.4**.**

With few exceptions, we do not know what the smallest possible size of a generating set of $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ for a poset$(P;\leq)$ in the scope of Theorem 3.3 is. We only know that the theorem is sharp for every antichain of size $n_{\textup{csize}}\neq 4$, see Lemma 1.1, and for every finite chain.

Remark 4.5**.**

The proof of our theorem heavily uses Lemma 1.1. For infinite sets, this lemma is taken from Czédli and Kulin [11]. We could only deal with accessible cardinalities in [11] and in the earlier paper Czédli [5], on which [11] relies. This explains that Theorem 3.3 is also restricted to (some posets of) accessible cardinalities. It is worth noting that by Kuratowski’s result, see [17] and see also Levy [18] for a secondary source, ZFC has a model in which all cardinal numbers are accessible.

5. Connection with NP-completeness and secret key authentication

In the first half of this section, we present a (presumably new) NP-complete problem. In the second half, we point out some connections among cryptographic authentication, NP-completeness, and Theorem 3.3.

For $b,k\in\mathbb{N}^{+}$, let $\vec{p}=(p_{1},\dots,p_{b})$ stand for a vector (namely, a $b$-dimensional vector) of $k$-ary lattice terms. If $L$ is a nontrivial (that is, at least 2-element) lattice and $\vec{u}\in L^{b}$, then $\vec{p}(\vec{x})=\vec{u}$ denotes the system

[TABLE]

of equations, in which $\vec{x}=(x_{1},\dots,x_{k})$ is the vector of the unknowns and we are looking for a vector $\vec{g}\in L^{k}$ such that $\vec{p}(\vec{g})=\vec{u}$ holds in $L$. If there is such a $\vec{g}$, then we say that $\vec{p}(\vec{x})=\vec{u}$ is solvable in $L$ or $\vec{p}(\vec{x})=\vec{u}$ has a solution in $L^{k}$. For a fixed $b\in\mathbb{N}^{+}$ and a nontrivial lattice $L$, let

[TABLE]

While there are fast methods to solve a classical system of linear equations over a field, the situation with $\textup{EqsLat}(L,b)$ is quite different. The following statement generalizes Czédli [10, Remark 5.2] from finite Boolean lattices to all finite nontrivial lattices at the cost of making the condition on $b$ more restrictive. Since our approach here borrows several ideas from [10], which is quite detailed, we are going to prove the following statement in a concise way by omitting some widely-known details.

Proposition 5.1**.**

For any fixed $4\leq b\in\mathbb{N}^{+}$ and for any nontrivial finite lattice $L$, $\textup{EqsLat}(L,b)$, defined in (5.2), is an NP-complete problem.

Proof.

It is well known from Cook [4, Theorem 2] and from many other sources including Wikipedia that 3SAT is NP-complete. In a similar but a bit stronger statement, a conjunctive normal form will be called a CNF${}^{3}_{2}$ if each of its clauses is either the disjunction of three non-negated variables or the disjunction of two negated variables. For example,

[TABLE]

is a CNF${}^{3}_{2}$. Let SAT${}^{3}_{2}$ stand for the (Boolean) satisfiability problem of CNF${}^{3}_{2}$’s in the 2-element lattice $\{0,1\}$. Since SAT${}^{3}_{2}$ is NP-complete by Czédli [10, Observation 6.1]212121 The extended version of the paper., it suffices to reduce SAT${}^{3}_{2}$ to $\textup{EqsLat}(L,b)$ in polynomial time. Fix $a_{0},a_{1}\in L$ such that $a_{0}\prec a_{1}$, and let $h$ be an arbitrary CNF${}^{3}_{2}$. Let $x_{1},\dots,x_{m}$ be the variables of $h$. Denote by $\vec{x}=(x_{-1},\dots,x_{m})$ the vector formed from these variables and two additional variables, $x_{-1}$ and $x_{0}$. We define two lattice terms,

[TABLE]

By de Morgan’s laws, for any $\vec{g}\in\{0,1\}^{m}$,

[TABLE]

Denote the two-element interval $[a_{0},a_{1}]$ by $I$ and, for a $\vec{g}\in\{0,1\}^{m}$, let $\vec{g}^{I}=(g_{1}^{I},\dots,g_{m}^{I})$ stand for the vector $(a_{g_{1}},\dots,a_{g_{m}})$. Since $I$ is isomorphic to the (Boolean) sublattice $\{0,1\}$, (5.4) implies that for any $\vec{g}\in\{0,1\}^{m}$

[TABLE]

For $i\in\{1,2\}$, we define a $(2+m)$-ary lattice term $p_{i}$ by

[TABLE]

and we also define two very simple lattice terms, $p_{-1}(\vec{x}):=x_{-1}$ and $p_{0}(\vec{x}):=x_{0}$. The time we need to construct the system

[TABLE]

of equations from $h$ is (bounded from above by) a polynomial function of the length of $h$. Hence, to complete our argument on reducing SAT${}^{3}_{2}$ to $\textup{EqsLat}(L,b)$ in polynomial time, it suffices to show that (5.7) is solvable in $L$ if and only if $h$ is satisfiable in $\{0,1\}$.

Assume that $h$ is satisfiable, that is, we have a $\vec{g}\in\{0,1\}^{m}$ such that $h(\vec{g})=1$. Let $w_{-1}:=a_{0}$, $w_{0}:=a_{1}$, and for $j\in\{1,\dots,m\}$, $w_{j}:=g^{I}_{j}$. By its definition, $\vec{w}:=(w_{-1},\dots,w_{m})$ is a solution of the first two equations of (5.7). For $j\in\{1,\dots,m\}$, $g^{I}_{j}=a_{g_{j}}\in\{a_{0},a_{1}\}$. Hence, when substituting $\vec{x}$ with $\vec{w}$, $(x_{j}\vee x_{-1})\wedge x_{0}$ in (5.6) turns into $(g_{j}^{I}\vee a_{0})\wedge a_{1}=g_{j}^{I}$. Thus, (5.5) implies that $p_{i}(\vec{w})=p_{i}^{(0)}(\vec{g}^{I})=a_{2-i}$ for $i\in\{1,2\}$, showing that $\vec{w}$ is also a solution of the last two equations of (5.7). Therefore, if $h$ is satisfiable then (5.7) is solvable.

Conversely, assume that (5.7) is solvable and $\vec{w}=(w_{-1},\dots,w_{m})$ is a solution of it. For $j\in\{1,\dots,m\}$, as $(w_{j}\vee w_{-1})\wedge w_{0}=(w_{j}\vee a_{0})\wedge a_{1}\in[a_{0},a_{1}]=I=\{a_{0},a_{1}\}$, we can defined a vector $\vec{g}=(g_{1},\dots,g_{m})\in\{0,1\}^{m}$ by the rule $(w_{j}\vee w_{-1})\wedge w_{0}=g^{I}_{j}$. Thus, (5.6) and the last two equations in (5.7) show that $p_{1}^{(0)}(\vec{g}^{I})=a_{1}$ and $p_{2}^{(0)}(\vec{g}^{I})=a_{0}$. Hence, $h(\vec{g})=1$ by (5.5) and so $h$ is satisfiable, as required. The proof of Proposition 5.1 is complete. ∎

Large lattices generated by few elements are usually appropriate for the authentication protocol outlined in Footnote 2, provided they are easy to described (and stored) and the lattice operations can be performed fast enough. The results on 4-element generating sets mentioned in the Introduction, Theorem 3.3, our earlier experience with computer programs presented in Czédli [9, 10] and Czédli and Oluoch [13], and Zádori [25] indicate that finite Boolean lattices, equivalence lattices, quasiorder lattices, $\textup{Quo}^{\kern-1.0pt\leq}(P;\leq)$ from the scope of Theorem 3.3, and subspace lattices of finite vector spaces (or, equivalently, subspace lattices of coordinatizable finite projective geometries) are all good candidates for $L$.

The underlying mathematical problem of the authentication protocol presented in Footnote 2 is the following.

[TABLE]

As opposed to the decision problem $\textup{EqsLat}(L,b)$, (5.8) is a functional problem.

Definition 5.2**.**

We say that problem (5.8) belongs to the complexity class FP if there exist an algorithm $\mathcal{A}$ and a polynomial $f$ such that for every input $\vec{p}(\vec{x})=\vec{u}$ of (5.8) (that is, for any system (5.1) of equations), if the input $\vec{p}(\vec{x})=\vec{u}$ is solvable in $L$, then $\mathcal{A}$ finds one of its solutions in $f(\text{input-size})$ time. (There is no condition on a non-solvable input. So, for example, we allow even that the algorithm can give a false solution or it cannot ever halt where an input is not solvable or syntactically meaningless.)

Following Rich [19], we write FP (where “F” comes from “functional”) since P is reserved for decision problems. To indicate that $\textup{EqsLat}(L,b)$ is hard, we have the following statement; note that most of the experts222222At least, so Wikipedia says. conjecture that P $\neq$ NP.

Observation 5.3**.**

Assuming that $b\in\mathbb{N}^{+}$ and a finite lattice $L$ are both fixed such that $4\leq b$ and $2\leq|L|$, if problem (5.8) belongs to FP, then P equals NP.

Proof.

As the proof is almost the same as the analogous proof of [10, Proposition 5.1], here we present only an outline of the argument. Assuming that (5.8) belongs to FP and using the notation of Definition 5.2, we have the following new algorithm to solve problem $\textup{EqsLat}(L,b)$ defined in (5.2): compute the size $\xi$ of the input $\vec{p}(\vec{x})=\vec{u}$ and put $f(\xi)$ into a “counter” variable $c$, then perform the steps of $\mathcal{A}$ and decrease $c$ by 1 alternately. If $\mathcal{A}$ finds a solution, then output “yes” and halt. Otherwise, if $\mathcal{A}$ halts without finding a solution or $c$ reaches 0, output “no” and halt. This new algorithm solves $\textup{EqsLat}(L,b)$ in polynomial time. But $\textup{EqsLat}(L,b)$ is NP-complete by Proposition 4.1, whereby P = NP, as required. ∎

6. More about the dedication

The reader may wonder what happens if we drop the assumption ”$|P|$ is an accessible cardinal” from Theorem 3.3. Because of this natural question, this is the right place to make some comments on the dedication included in the paper.

Remark 6.1**.**

There are several sources, including https://en.wikipedia.org/wiki/Vilmos_Totik and https://mta.hu/koztestuleti_tagok?PersonId=19540 where one can read about Professor Vilmos Totik. As a witness, here I add three stories; the third story sheds some light on the question mentioned above. First, Vilmos was one of my roommates in Loránd Eötvös Students’ Hostel for (almost) three academic years from 1974 to 1977; his excellence was clear for all of us, fellow students, even that time. Second, he received the Researcher of the Year award from the University of Szeged in the field of physical sciences on November 9, 2019; this was the first occasion when this new award was delivered; see https://u-szeged.hu/news-and-events/2019/honorary-titles-and . As this link shows, that day was splendid for us also for another reason: Doctor Honoris Causa title was awarded to Professor Ralph McKenzie232323Professor Ralph McKenzie needs no introduction to the targeted readership.. I still remember the dinner we organized for that evening, where the two celebrated professors of quite different fields of mathematics discussed, among other things, their fishing experiences. Third, when (around 1999) I mentioned the basic idea of my proofs242424In spite of many changes in subsequent proofs, see Czédli [9], Czédli and Kulin [11], and Czédli and Oluoch [13], the basic idea of dealing with sets of large cardinalities is, unfortunately, still the same. in [5] and [8], it took Professor Totik less than a minute to come out with a proof showing that my method cannot work for all infinite cardinals. (Later, when I forgot his proof, I could not find it again and I needed a professor of set theory to find the proof again.) To conclude the stories, I wish Professor Vilmos Totik further successes and all the best.