text:
\begin{document}
	\maketitle
\begin{abstract}
Let $\mathcal{A}_i: =(A_i,\le_i,\top_i,\odot_i,\to_i,\bot_i)$, $i=1,2$ be two  complete residuated multilattices, $G$ (set of objects) and $M$ (set of attributes) be two nonempty sets and $(\varphi, \psi)$ a Galois connection
between $A_1^G$ and $A_2^M$. In this work we prove that
 \[\mathcal{C}: =\{(h,f)\in A_1^G\times A_2^M \mid \varphi(h)=f \text{ and } \psi(f)=h \}\]
 is a complete residuated multilattice. This is a generalization of a result by Ruiz-Calvi{\~n}o and Medina
 \cite{RM12} saying that if the (reduct of the) algebras $\mathcal{A}_i$, $i=1,2$ are complete multilattices,  then $\mathcal{C}$ is a complete multilattice.
 \end{abstract}

\section{Introduction: FCA and Fuzzyfication}
In the theory of fuzzy concept analysis, the underlying set of truth values is generally a lattice. Medina and Ruiz-Calvi{\~n}o proposed in \cite{RM12} a new approach of fuzzy concept analysis by using multilattices as underlying set of truth values. Multilattices were introduced by Benado~\cite{Bm55}. The main idea is to relax the requirement on the existence of least upper bounds and greatest lower bounds, and to ask only that the set of minimal upper bounds and the set of maximal lower bounds of any pair of elements are non-empty. In this paper, we show that residuation on multilattices can be useful in fuzzy concept analysis to evaluate the attributes and objects. Especially, we will use the adjoint pair in a residuated multilattice to built a residuated concept  multilattice. We organize the present contribution as follows: in Section~\ref{s:CLmL} we recall some notions of Formal Concept Analysis and multilattices. In Section~\ref{s:TDS}, we apply the ordinary sum construction to create a new residuated multilattice from the old ones and prove that every bounded pure residuated multilattice has order equal or greater than $7$. We also briefly show how  Medina and Ruiz-Calvi{\~n}o used multilattices as truth degree sets. Section~\ref{s:RCML} is .

\section{Concept Lattices and Multilattices} \label{s:CLmL}
A \neu{formal context} is set up with the sets $G$ of objects, $M$  of attributes and a binary relation
$\I\subseteq G\times M$. We denote it  by $\KK:=(G,M,\I)$, and write $(g,m)\in\I$ or $g\I m$ to mean that the
object $g$ has the attribute $m$. In the whole paper we assume that $G$ and $M$ are non-empty. To extract knowledge from formal contexts, one can get clusters of objects/attributes  called concepts.

The Port-Royal Logic School considers a \neu{concept} as defined by two parts: an extent and an intent.
The \neu{extent} contains all entities belonging to the concept and the \neu{intent} is the set of all attributes common to all entities in the concept. To formalize the notion of concept we need the
\neu{derivation operator} $'$, defined on $A\subseteq G$ and $B\subseteq M$ by:
	\begin{align*}
	A':=\{m\in M \mid g\I m\text{ for all }g\in A\} \quad \text{ and } \quad
	B':=\{g\in G \mid g\I m\text{ for all }m\in B\}.
	\end{align*}
A \neu{formal concept} is a pair $(A,B)$ with $A'=B$ and $B'=A$. The set of formal concepts of $\KK$ is denoted by $\mathfrak{B}(\KK)$. It forms a complete lattice, called \neu{concept lattice} of the context $\KK$, when ordered by the \neu{concept hierarchy} below:
$$\left(A_{1},B_{1}\right)\leqslant\left(A_{2},B_{2}\right):\iff A_{1}\subseteq A_2.$$

Recall that a \neu{lattice} is a poset in which all finite subsets have a least upper bound and a greatest lower bound. A lattice is complete if every subset has a least upper bound and a greatest lower bound.
For any subset $X$ of $L$ we denote by $\bigvee X$ its least upper bound and by $\bigwedge X$ its greatest lower bound, whenever they exist. For $X=\{x,y\}$ we write $x\vee y:=\bigvee X$ and $x\wedge y:=\bigwedge X$.

For any context $(G,M,I)$ the pair $(',')$ forms a Galois connection between $\mathcal{P}(G)$ and $\mathcal{P}(M)$ and $c:X\mapsto X''$ a closure operator (on $\mathcal{P}(G)$ or $\mathcal{P}(M)$, ordered by the inclusion). Recall that a \neu{closure operator} on a poset $(P,\le)$ is a map $c:P\to P$ such that
\[x\le c(y) \iff c(x)\le c(y)\quad \text{ for all } x,y\in P.\]
A pair $(\varphi,\psi)$ is a \neu{Galois connection} between the two posets $(P,\le)$ and $(Q,\le)$ if
$\varphi: P\to Q$ and $\psi: Q\to P$ are maps such that
$$p\le \psi q \iff q\le \varphi p\quad \text{ for all } p\in P \text{ and  for all } q\in Q.$$
In general we will call any pair $(p,q)$ a \neu{concept} if $\varphi p=q$ and $\psi q=p$, whenever $(\varphi,\psi)$ is a Galois connection.
The maps $\varphi\circ\psi$ and $\psi\circ \varphi$ (denoted $\varphi\psi$ and $\psi\varphi$ for short) are the corresponding closure operators on $P$ and $Q$ respectively. An element $p\in P$ is \neu{closed} if $\psi\varphi(p)=p$. Similarly $q\in Q$ is closed if $\varphi\psi(q)=q$.

Objects, attributes or incidences can be of fuzzy nature. Several sets of truth degrees have been considered,
starting from the interval $[0,1]$ \cite{ZL65} to lattices \cite{GJ67}, multilattices \cite{MOR13} and residuations on these structures.

 In \cite{RM12} Ruiz-Calvi{\~n}o and Medina investigate the use of multilattices as underlying set of
 truth-values for attributes and objects, and prove that the set of all concepts is a multilattice.

 We will use  residuated multilattices as the set of truth-values to evaluate attributes and objects,
 and show that the set of all concepts in this case is a residuated multilattice.

 Multilattices generalize lattices and were introduced by Mihail Benado \cite{Bm55}. Let $(P,\le)$ be a
 poset and $x,y\in P$. We say that $x$ \neu{is below} $y$ or  $y$ \neu{is above} $x$, whenever
 $x\le y$.\footnote{We also accept that any element is below and above itself.}
\begin{defin}\cite{RM12}
 A poset $(P,\le)$ is called \neu{coherent} if every chain has supremum and infimum.
 \end{defin}
 \begin{defin}\cite{MOR07}
 A poset $(P,\le)$ is called a \neu{multilattice} or \neu{m-lattice} if for any finite subset $X$ of $P$,
 each upper bound of $X$ is above a minimal upper bound of $X$ and each lower bound of $X$ is below a maximal
 lower bound of $X$.
 \end{defin}
For any subset $X$ of $P$, we denote by $\sqcup X$  (resp. $\sqcap X$) the set of its minimal upper bounds
(resp. maximal lower bounds). In general, $\sqcup X$ and $\sqcap X$ could be empty (when they are not empty, the multilattice is said to be \neu{full}) . But if $(P,\le)$ is a bounded finite poset, then $\sqcup X$ and $\sqcap X$ are non-empty. For $X=\{x,y\}$ we write $x\sqcup y$  and $x\sqcap y$ instead of $\sqcup X$  and $\sqcap X$. When  $\sqcup X$  or $\sqcap X$ is a singleton it will be consider as an element of $P$, i.e., we write $\sqcup X\in P$  or $\sqcap X\in P$ instead of $\sqcup X\subseteq P$  or $\sqcap X\subseteq P$. If $(P,\le)$ is a lattice and $X$ a finite subset of $P$, then the set of upper (resp. lower) bounds of $X$ is non-empty and has exactly one minimal (resp. maximal) element, namely $\vee X$ (resp. $\wedge X$). Moreover, any upper (resp. lower) bound of $X$ is above (resp. below) $\vee X$ (resp. $\wedge X$). Thus any lattice is a multilattice with $\sqcup X=\{\vee X\}$ and $\sqcap X=\{\wedge X\}$. We call a multilattice \neu{pure} if it is not a lattice.

 FCA can be seen as applied theory of complete lattices. Below is how the completeness can be carried out to
 multilattices.

\begin{defin}\cite{MOR07}
A poset $(P, \le)$ is a \neu{complete multilattice}~\footnote{In \cite{RM12} the authors defined
complete multilattices as coherent posets $(P,\le)$ with no infinite antichain and
$\sqcup X\neq \emptyset\neq\sqcap X$ for all $X\subseteq P$.} if for all $X\subseteq P$, the
 sets $\sqcup X, \sqcap X$ are non empty and each upper bound of $X$ is above an element of $\sqcup X$
 and each lower bound is below an element of $\sqcap X$.
\end{defin}

\begin{figure}[!htbp]
	\begin{center}
	\begin{tikzpicture}[scale=0.7]
	\draw[fill] (2,0) circle (0.05);
	\draw[fill] (1,1) circle (0.05);
	\draw[fill] (3,1) circle (0.05);
	\draw[fill] (3,3) circle (0.05);
	\draw[fill] (1,3) circle (0.05);
	\draw[fill] (2,4) circle (0.05);
	\draw (2,0) -- (1,1) -- (1,3) -- (2,4)--(3,3)--(3,1)--(2,0);
	\draw (1,1) -- (3,3);
	\draw (1,3) -- (3,1);
	\draw (2,0) node[below]{$\bot$};
	\draw (1,1) node[left]{a};
	\draw (3,1) node[right]{b};
	\draw (3,3) node[right]{d};
	\draw (1,3) node[left]{c};
	\draw (2,4) node[above]{$\top$};
	\end{tikzpicture}
\hspace{1cm}
	\begin{tikzpicture}[scale=0.7]
	\draw[fill] (2,0) circle (0.05);
	\draw[fill] (1,1) circle (0.05);
	\draw[fill] (3,1) circle (0.05);
	\draw[fill] (3,3) circle (0.05);
	\draw[fill] (1,3) circle (0.05);

	\draw (3,1)--(2,0)--(1,1) -- (1,3) -- (3,1) -- (3,3) -- (1,1);
	\draw (2,0) node[below]{$\bot$};
	\draw (1,1) node[left]{a};
	\draw (3,1) node[right]{b};
	\draw (3,3) node[right]{d};
	\draw (1,3) node[left]{c};

	\end{tikzpicture}
\end{center}
\caption{A complete multilattice (left) and a non complete multilattice (right).} \label{fig:cmlat}
\end{figure}

On Figure~\ref{fig:cmlat} we can see the smallest bounded poset that is a complete multilattice but not a lattice. We will usually refer to it as the multilattice $\mathrm{ML_6}$. All finite lattices are complete lattices,
and all finite bounded posets are complete multilattices.

In the framework of multilattices, the concept of homomorphism has been originally introduced by M.
Benado \cite{Bm55}.

\begin{defin}\cite{Bm55}
A map $h: P \rightarrow P'$ between multilattices is said to be a \textbf{homomorphism} if
$h(x\sqcup y)\subseteq h(x)\sqcup h(y)$ and $h(x\sqcap y) \subseteq h(x) \sqcap h(y)$, for all $x, y\in P$.
\end{defin}
When the initial multilattice is full, the notion of homomorphism can be characterized in terms of equalities.
\begin{prop}\cite{CCMO14}
Let $h: P \rightarrow P'$ be a map between multilattices where $P$ is full. Then $h$ is a homomorphism if and only if, for all $x$, $y\in P$, \linebreak[4] $h(x\sqcup y)= (h(x)\sqcup h(y))\cap h(P)$ and
$h(x\sqcap y) = (h(x) \sqcap h(y))\cap h(P)$.
\end{prop}
A homomorphism $h$ will be called an \textbf{isomorphism} when it is one-to-one homomorphism.

In \cite{MOR07}, J.Medina, M. Ojeda-Aciego and J. Ruiz-Calvi{\~n}o defined two different types of submultilattices, namely : full submultilattice (or f-submultilattice) and restricted submultilattice
(or r-submultilattice).

\begin{defin}\cite{MOR07}
Let $(P,\le)$ be a multilattice and $X$ be a nonempty subset of $P$.

\begin{itemize}
\item[(i)] $X$ is called a \textbf{full submultilattice} ($f$-\textbf{submultilattice}) of $P$ if for all $x, y\in X$, $x\sqcup y\subseteq X$ and $x\sqcap y\subseteq X$ (SML-1).
     \item[(ii)] $X$ is called a \textbf{restricted submultilattice} (\textbf{$r$-submultilattice}) of $P$ if for all $x, y\in X$, $(x\sqcup y)\cap X\ne \emptyset$ and $(x\sqcap y)\cap X\ne \emptyset$ (SML-2).
\end{itemize}
\end{defin}
It was proved in \cite{MOR07} that if $X$ is a full submultilattice of a multilattice $P$, then equipped with
the restriction of the partial order from  $P$, $X$ is a multilattice on its own right. However, this is not
the case for restricted submultilattices.
\begin{prop}\label{containsM6}
Every pure and bounded multilattice contains a restricted submultilattice isomorphic to $\mathrm{ML_6}$.
\end{prop}
\begin{proof} Let $(P,\le)$ be a pure and bounded multilattice.
Then, there exists at least two elements $ x, y\in P$ such that $ x\sqcup y$ is not a singleton. Let
$a, b\in x\sqcup y$ with $a\ne b$. If $a\sqcup b$ and $x\sqcap y$ are singleton, it is enough to choose
$\{x\sqcap y, x, y, a, b, a\sqcup b\}$ as an isomorphic copy of $\mathrm{ML_6}$, which is indeed a full submultilattice of $P$. Otherwise, fix an element $z\in x\sqcap y$ and another element $c\in a\sqcup b$. Then
$\{z, x, y, a, b, c\}$ is a restricted submultilattice which is isomorphic to $\mathrm{ML_6}$.
\end{proof}
A pure and bounded multilattice does not always contain a full submultilattice isomorphic to $\mathrm{ML_6}$, as the following example shows.

Consider the multilattice depicted in the following Hasse diagram.

\begin{figure}[!htbp]
	\begin{center}
$\xymatrix@-1pc@M= 1.5pt{& &\top& &  \\
h\ar@{-}[urr]&&i\ar@{-}[u] &&j\ar@{-}[ull] \\
d\ar@{-}[u]\ar@{-}[urr]&& f\ar@{-}[u]\ar@{-}[urr]\ar@{-}[ull]&&g\ar@{-}[u]\ar@{-}[ull] \\
a\ar@{-}[u]\ar@{-}[urr]&& b\ar@{-}[u]\ar@{-}[urr]\ar@{-}[ull]&&c\ar@{-}[u]\ar@{-}[ull] \\
&& \bot\ar@{-}[ull]\ar@{-}[u]\ar@{-}[urr]& & \\}$
\end{center}
\caption{A pure bounded multilattice with no full  submultilattice .}\label{fig:pmlatnofsubmlat}
\end{figure}

All the restricted submultilattices of this multilattice isomorphic to $\mathrm{ML_6}$ are:
 $S_1=\{\bot,a, b, d, f, i\}; S_2=\{\bot,a, b, d, f, h\}; S_3=\{\bot, b, c, f, g, i\}; S_4=
 \{\bot, b, c, f, g, j\};\\
  S_5=\{a, h, i, d, f, \top\}; S_6=\{b, d, f, h, i, \top\}; S_7=\{b, f, g, i, j,  \top\};
  S_8=\{c, f, g,  i, j, \top\};$

$S_9=\{\bot, a, c, j, h, \top\}$. None of these is a full  submultilattice.

Now that we know what a complete multilattice is, we are ready to introduce residuated multilattices.
These will serve as truth degree sets to evaluate objects and attributes.

\section{Residuation and Truth Degree Sets} \label{s:TDS}

The structures of truth values in fuzzy logic and fuzzy set theory are usually lattices or residuated
lattices (see \cite{GJ67, HP98, HU96}). Particularly in fuzzy concept analysis, residuated lattices are used to evaluate the attributes and objets. But the necessity of the use of a more general structure arise in some
examples. In  \cite{RM12} multilattices are used as the underlying set of truth-values in FCA.
\begin{defin}\cite{JT02}
A structure $\mathcal{A}:= (A,\le, \top, \odot, \rightarrow)$ is called \neu{pocrim} (partially ordered
commutative residuated integral monoid) if  $(A,\leq,\top)$ is a poset with a maximum $\top$ and
$(A, \odot, \top)$ is a commutative monoid such that  for all  $a, b, c\in A$,
\[
(\ddagger)\quad a\odot c\leq b\iff c\leq a\to b.
\]
Any pair $(\odot,\to)$ satisfying $(\ddagger)$ is called a \neu{residuated couple} or an \neu{adjoint couple}
on $(A,\le)$.
\end{defin}

The following properties hold in any pocrim and for the properties of complete residuated lattices, we refer to \cite{JT02}$\mathcal{A}$.

For all $a, b, c \in A$,
\begin{itemize}
	\itemsep4pt
	\item[\textbf{P1}] $a\odot b \leq a$ and $a\odot b \leq b$;
	\item[\textbf{P2}]
	$\left\{\begin{array}{rcccl}
	a\odot(a\rightarrow b)&\leq& a &\leq& b\rightarrow (a\odot b) \\
	b\odot(a\rightarrow b)&\leq& b &\leq& a\rightarrow (a\odot b);
	\end{array}\right.$
	\item[\textbf{P3}] $a\leq b$ $\Leftrightarrow$ $a\rightarrow b=\top$;
    \item[\textbf{P4}] $a\leq b\implies\left\{\begin{array}{rcl}
	a\odot c &\le& b \odot c,\\
	c\rightarrow a &\le& c\rightarrow b,\\
	b\rightarrow c &\le & a\rightarrow	c;
	\end{array}  \right. $

    \item[\textbf{P5}] $a\rightarrow (b\rightarrow c)\ =\ b\rightarrow (a\rightarrow c)$;
    \item[\textbf{P6}] $(a\rightarrow b)\odot (b\rightarrow c)\ \leq\ a\rightarrow c$;
    \item[\textbf{P7}]  $\left\{\begin{array}{rcl}
    a\rightarrow b &\leq & (a\odot c)\rightarrow (b\odot c) \\
	a\rightarrow b &\leq & (c\rightarrow a)\rightarrow (c\rightarrow b) \\
	a\rightarrow b &\leq & (b\rightarrow c)\rightarrow (a\rightarrow c)
	\end{array}\right.$
\end{itemize}
\vspace{0.5cm}
A pocrim is bounded if it also has a lower bound, usually denoted by $\bot$.

Residuated multilattices were introduced in \cite{CCMO14}.
\begin{defin}\cite{CCMO14}
A (bounded) \neu{residuated multilattice} (write $\mathcal{RML}$ for short)  $\mathcal{A}$ is a (bounded)
 pocrim whose underlying poset $(A,\le)$ is a multilattice.
A \neu{complete residuated multilattice}  is a pocrim whose underlying poset is a complete multilattice.
\end{defin}

To obtain a general procedure to construct new residuated multilattices, we will extend the ordinal sum
construction of pocrims to residuated multilattices.

First, we recall the construction of the ordinal sum of pocrims. Let $(A,\leq_A, \odot_A, \to_A, \top_A)$
and $(B, \leq_B, \odot_B,\to_B, \top_B)$ be two pocrims. Consider the set
$C:=(A\amalg B) / \{\top_A\equiv\top_B\}$ (that is the disjoint union of $A$ and $B$ with $\top_A$ and $\top_B$
identified) and define $\leq $ on $C$ by $x\leq y$ if $x, y\in A$ and $x\leq_A y$, or $x, y\in B$ and $x\leq_B y$
or $x\in A\setminus \{\top_A\}$ and $y\in B$.
Define $\odot$ and $\to$ on $C$ by:
\begin{align*}
	x\odot y=
	\begin{cases}
	x\odot_A y &\text{ if }\ x, y\in A \\
	x\odot_B y &\text{ if }\ x, y\in B \\
	x &\text{ if }\ x\in A\setminus \{\top_A\}\ \text{ and }\ y\in B  \\
	y &\text{ if }\ y\in A\setminus \{\top_A\}\ \text{ and }\ x\in B
	\end{cases}
\end{align*}
\begin{align*}
	x\rightarrow y=
	\begin{cases}
		x\to_A y &\text{ if }\ x, y\in A  \\
	x\to_B y &\text{ if }\ x, y\in B  \\
	\top_B & \text{ if }\ x\in A\setminus \{\top_A\}\ \text{ and }\ y\in B  \\
	y & \text{ if }\ y\in A\setminus \{\top_A\}\ \text{ and }\  x\in B
	\end{cases}
\end{align*}

\vspace{0.5cm}
Then $(C, \leq, \odot, \to, \bot_A, \top_B)$ is a pocrim, called the ordinal sum of $A$ and $B$ and denoted
by $A\oplus B$. Note that in $A\oplus B$, $A\setminus\{\top_A\}$ is below every element of $B$. Our goal is to apply this construction to create new $\mathcal{RML}$ from old ones. The first step in this process consists in identifying the smallest pure $\mathcal{RML}$.

In what follows in this part, our aim is to prove by contradiction that there is no pocrim structure on $\mathrm{ML_6}$ and to give an example of pocrim structure on a pure complete multilattice with $7$ elements.

\begin{lem}\label{fleche}
	Suppose that the multilattice $\mathrm{ML_6}$ is endowed with a pocrim structure
$(\mathrm{ML}_6, \leq, \odot, \to, \bot, \top)$. Then the implication $\to$ is given by Table $1$ below.
\end{lem}
\begin{proof}
	We shall use repeatedly the fact that in any pocrim $y\leq x\to y$ and $x\to (y\to z)=y\to (x\to z)$ hold.
	\begin{itemize}
		\item[(i)] Since $c\leq d\to c\ne \top$, then $d\to c=c$. Similarly $c\to d=d$.
		\item[(ii)] Note that since $b\leq c\to b$ and $b\leq d\to b$, then $c\to b\in \{b, c, d\}$ and
$d\to b\in \{b, c, d\}$. We narrow down the possibilities further. Note that \linebreak[4]
$c\to b\leq c\to d=d$, hence $c\to b\ne c$. Likewise, $d\to b\ne d$. Thus $c\to b\in \{b, d\}$ and
$d\to b\in \{b, c\}$. We show that $c\to b=b$ and $d\to b=b$ by showing that the other combinations lead to
contradiction.
		\begin{itemize}
			\item If $c\to b=b$ and $d\to b=c$, then

 $c=d\to b=d\to (c\to b)=c\to (d\to b)=c\to c=\top$, which is a contradiction.
			\item If $c\to b=d$ and $d\to b=b$, then

$\top=d\to d=d\to (c\to b)=c\to (d\to b)=c\to b$, which implies that $c\leq b$. This is impossible.
			\item If $c\to b=d$ and $d\to b=c$, then

$\top =a\to c=a\to (d\to b)=d\to (a\to b)$. So $d\leq a\to b\ne \top$, hence $d=a\to b$. Now,
$\top=a\to d=a\to (c\to b)=c\to (a\to b)=c\to d$, which is again a contradiction.\\
		\end{itemize}
		Since $a$ and $b$ play symmetrical roles, we deduce $c\to a=a$ and $d\to a=a$.
		\item[(iii)] Since $b\leq a\to b$, then $a\to b \in \{b, c, d\}$. Again, we show that
$a\to b \in \{c, d\}$ is impossible. Indeed, if $a\to b=c$, then

$\top=c\to c=c\to (a\to b)=a\to (c\to b)=a\to b$, which is impossible. Similarly, if $a\to b=d$, then
$\top=d\to d=d\to (a\to b)=a\to (d\to b)=a\to b$, which is the same contradiction. Hence, $a\to b=b$. The proof
that $b\to a=a$ is analogous.
		\item[(iv)] It remains to show that $a\to \bot=b, b\to \bot =a$ and $c\to \bot=d\to \bot=\bot$.\\
		Note that $a\odot b\leq a, b$, so $a\odot b=\bot$. Thus, $a\leq b\to \bot$ and $b\leq a\to \bot$. On the
other hand, $b\to \bot\leq b\to a=a$. Hence, $b\to \bot =a$ and a similar argument shows that $a\to \bot =b$.
Finally, observe that $a, b\leq c$, so $c\to \bot\leq a\to \bot, b\to \bot$. Hence, $c\to \bot\leq b, a$ and
consequently $c\to \bot=\bot$. The verification that $d\to \bot =\bot$ is similar.
	\end{itemize}
	We have justified all the entries of the following table.
\end{proof}
\begin{table}[h]
\begin{center}
	\begin{tabular}{|c|c|c|c|c|c|c|}
		\hline

		$\rightarrow$  & $\bot$ & a & b & c & d & $\top$ \\
		\hline
		$\bot$& $\top$ & $\top$ & $\top$ & $\top$ & $\top$ & $\top$ \\
		\hline
		a & b & $\top$ & b & $\top$ &$\top$ & $\top$ \\
		\hline
		b & a & a & $\top$ &  $\top$ &$\top$ & $\top$ \\
		\hline
		c & $\bot$ & a & b & $\top$ & d & $\top$\\
		\hline
		d & $\bot$ & a & b & c  & $\top$ & $\top$\\
		\hline
		$\top$ &$\bot$ & a & b & c & d & $\top$ \\
		\hline
	\end{tabular}
\end{center}
\caption{}\label{tableML_6}
\end{table}

\begin{prop}\label{noprml}
	There does not exist a residuated multilattice structure on $\mathrm{ML_6}$ extending its existing partial order.
\end{prop}
\begin{proof}
	By contradiction suppose that there exist $\odot$ and $\to$ such that
$(\mathrm{ML}_6,\leq, \odot, \to, \bot, \top)$ is a residuated multilattice. Then by Lemma \ref{fleche}, $\to$
is given by table 1 above.

Since $a\odot a\leq a$, we have $a\odot a\in \{\bot, a\}$.

If $a\odot a=\bot$, then $a\odot a\leq b$ and $a\leq a\to b=b$ (Table 1), which is a contradiction.

Suppose $a\odot a=a$. Since $a\leq c$, we have $a=a\odot a\leq a\odot c\leq a$. Hence, $a\odot c=a$. From
$a\leq d$, we have $a=c\odot a\leq c\odot d\leq c, d$. Because $c$ and $d$ are incomparable, $c\odot d=a$. It
follows that $d\leq c\to a=a$ (Table 1), which is again a contradiction.
\end{proof}
Combining Proposition \ref{containsM6} and Proposition \ref{noprml} we obtain that every bounded pure
$\mathcal{RML}$ has order greater than or equal to $7$. The next example shows that there is indeed a bounded
pure $\mathcal{RML}$ of order $7$, which shall subsequently be denoted by $\mathrm{RML}_7$,
\begin{figure}[!htbp]
	\begin{center}
		\begin{tikzpicture}[scale=0.5]
		\draw[fill] (2,0) circle (0.05);
		\draw[fill] (1,1) circle (0.05);
		\draw[fill] (3,1) circle (0.05);
		\draw[fill] (3,3) circle (0.05);
		\draw[fill] (1,3) circle (0.05);
		\draw[fill] (2,4) circle (0.05);
		\draw[fill] (2,5) circle (0.05);

		\draw (2,0) -- (1,1) -- (1,3) -- (2,4) -- (2,5) -- (2,4) --(3,3)--(3,1)--(2,0);
		\draw (1,1) -- (3,3);
		\draw (1,3) -- (3,1);

		\draw (2,0) node[below]{$\bot$};
		\draw (1,1) node[left]{a};
		\draw (3,1) node[right]{b};
		\draw (3,3) node[right]{d};
		\draw (1,3) node[left]{c};
		\draw (2,4) node[below]{e};
		\draw (2,5) node[above]{$\top$};
		\end{tikzpicture}
	\end{center}
	\caption{A complete residuated multilattice.}\label{fig:rcmlat}
\end{figure}

with the operations $\odot$ and $\rightarrow$ defined in the following tables.

\begin{table}[h]
\begin{center}
	\begin{tabular}{|c|c|c|c|c|c|c|c|}
		\hline

		$\odot$ & $\bot$ &    a   &    b   &    c   &    d   &    e   & $\top$ \\
		\hline
		$\bot$  & $\bot$ & $\bot$ & $\bot$ & $\bot$ & $\bot$ & $\bot$ & $\bot$ \\
		\hline
		   a    & $\bot$ &    a   & $\bot$ & $\bot$ &    a   & $\bot$ &  a \\
		\hline
		   b    & $\bot$ & $\bot$ &    b   &    b   & $\bot$ & $\bot$ &  b \\
		\hline
		   c    & $\bot$ & $\bot$ &    b   &    c   & $\bot$ & $\bot$ &  c \\
		\hline
		   d    & $\bot$ &    a   & $\bot$ & $\bot$ &    d   & $\bot$ &  d \\
		\hline
		   e    & $\bot$ & $\bot$ & $\bot$ & $\bot$ & $\bot$ &    a   &  e \\
		\hline
		$\top$  & $\bot$ &    a   &    b   &    c   &    d   &    e   & $\top$ \\
		\hline
	\end{tabular}
	\hspace{2cm}
	\begin{tabular}{|c|c|c|c|c|c|c|c|}
		\hline

		$\rightarrow$  & $\bot$ &    a   &    b   &    c   &    d   & e & $\top$ \\
		\hline
		     $\bot$    & $\top$ & $\top$ & $\top$ & $\top$ & $\top$ & $\top$ & $\top$ \\
		\hline
		        a      &    b   & $\top$ &    b   & $\top$ & $\top$ & $\top$ & $\top$ \\
		\hline
		        b      &    a   &    a   & $\top$ & $\top$ & $\top$ & $\top$ & $\top$ \\
		\hline
		        c      & $\bot$ &    a   &    b   & $\top$ &    d   & $\top$ & $\top$\\
		\hline
		        d      & $\bot$ &    a   &    b   &    c   & $\top$ & $\top$ & $\top$\\
		\hline
		        e      &    d   &    e   &    d   &    e   &    d   & $\top$ & $\top$\\
		\hline
		     $\top$    & $\bot$ &    a   &    b   &    c   &    d   &    e   & $\top$ \\
		\hline
	\end{tabular}
\end{center}
\caption{Of $\odot$ (left) and $\rightarrow$ (right) on $\mathrm{RML}_7$}\label{tableML_7}
\end{table}
Now that we have setup a smaller pure $\mathcal{RML}$, we can give the structure on the set of all mappings from
a non-empty set to a residuated multilattice.

Given a residuated multilattice $\mathcal{A}$ and a nonempty set $X$, we will denote by $A^X$ the set of all
mappings from $X$ to $A$. An order
on $A^X$  as well as the operations $\odot$, $\to$ are defined pointwise: i.e., for $f_1,f_2\in A^X$ we have
\begin{itemize}
	\item $f_1\leq f_2:\iff f_1(x)\leq f_2(x)$ for all $x\in X$,
	\item $\left(f_1\odot f_2\right)(x):=f_1(x)\odot f_2(x)$, for any $x\in X$,
	\item $\left(f_1\to f_2\right)(x):= f_1(x)\to f_2(x)$, for any $x\in X$.
\end{itemize}
\begin{lem}
	Let $\mathcal{A}$ be a complete residuated multilattice and $X$ a nonempty set. Then
$\left(A^X,\le,\odot,\to,\top,\bot \right)$ is a complete residuated multilattice.
\end{lem}
\begin{proof}
Trivially, $\le$ is an order relation on $A^X$. The map $\uuh:x\mapsto \top$ is the greatest element of $A^X$ and
the map $\ddh :x\mapsto \bot$ is the smallest element of $A^X$. For $f_1,f_2,f_3\in A^X$, we have
\begin{align*}
	f_1\odot f_2 \le f_3 &\iff (f_1\odot f_2)(x) \le f_3(x),\ \forall x\in X \\
	&\iff f_1(x)\odot f_2(x) \le f_3(x),\ \forall x\in X \\
	&\iff f_2(x) \le f_1(x) \to f_3(x),\ \forall x\in X \\
	&\iff f_2 \le f_1 \to f_3
\end{align*}
We still need to define $\sqcup$ and $\sqcap$ on $A^X$. For all $x\in X$, the sets $f_1(x)\sqcup f_2(x)$ and
$f_1(x)\sqcap f_2(x)$ are non-empty. By the axiom of choice, define mappings
$f_\sqcup:x\mapsto f_\sqcup(x)\in f_1(x)\sqcup f_2(x)$ and
$f_\sqcap:x\mapsto f_\sqcap(x)\in f_1(x)\sqcap f_2(x)$.  Each map $f_\sqcup$ (resp. $f_\sqcap$) is a
minimal upper (resp. maximal lower) bound of $\{f_1,f_2\}$.

Generally, given $\{f_i| i\in I\}$ a non-empty subset of $A^X$, for all $x\in X$,
\linebreak[4]$\sqcup\{f_i(x)| i\in I\}$ and $\sqcap\{f_i(x)| i\in I\}$ exist and they are non-empty. Hence, by
the axiom of choice, we can define the mappings $f_\sqcup:x\mapsto f_\sqcup(x)\in \sqcup\{f_i(x)| i\in I\}$ and
$f_\sqcap:x\mapsto f_\sqcap(x)\in \sqcap\{f_i(x)| i\in I\}$.
\end{proof}
Let $(A_1, \leq_1)$ and $(A_2, \leq_2)$ be two complete
multilattices, $G$ and $M$ two non-empty sets (of objects and attributes), and $(\varphi,
\psi)$ a Galois connection between $(A_1^G,\le_1)$ and $(A_2^M,\le_2)$. Recall that a concept is a pair
$(h,f)\in A_1^G\times A_2^M$ such that $\psi(f)=h$ and $\varphi(h)=f$. Concepts are ordered by
\[(h_1, f_1)\le (h_2, f_2):\iff h_1\leq_1 h_2 (\texttt{or equivalently }f_1\geq_2 f_2)\].
\begin{thm}\label{thm:1}
Let $(A_i, \leq_i)$, $i=1,2$ be two complete
multilattices, $G$ and $M$ be two non-empty sets  and $(\varphi,\psi)$ be a Galois connection between
$(A_1^G,\le_1)$ and $(A_2^M,\le_2)$. Let $H\subseteq A_1^G$, $F\subseteq A_2^M$ and $\mathcal{C}$ the set of
concepts of $A_1^G\times A_2^M$. Then
\begin{itemize}
	\item[\textup{(i)}] $\sqcap\psi(F)\subseteq \psi(\sqcup F)$  and
	$\sqcap\varphi(H)\subseteq \varphi(\sqcup H)$.
	\item[\textup{(ii)}] The poset $(\mathcal{C},\le)$ of all concepts of $A_1^G\times A_2^M$ is a complete
multilattice, with:

$\bigsqcap\limits_{j\in J}(h_j, f_j):=\{(h,\varphi(h)); h\in \sqcap\{h_j\mid j \in J\}\}$ and

$\bigsqcup\limits_{j\in J}(h_j, f_j):=\{(\psi(f), f); f\in \sqcap\{f_j;\mid j \in J\}\}$.

\end{itemize}
\end{thm}
\begin{proof}
\begin{itemize}
	\item[\textup{(i)}] Let $h\in\sqcap\psi(F)$; i.e. $h$ is a maximal lower bound of $\psi(F)$. For all
$f\in F$, we have  $h\le\psi(f)$, and $\varphi h\ge \varphi\psi (f)\ge f$. Thus $\varphi h$ is an upper bound
of $F$, and is by then above a minimal upper bound of $F$. Thus there is $f_0\in \sqcup F$ such that
$\varphi h \ge f_0$.

	For any $f\in F$ we have $f_0\ge f$ and $ h\le  \psi\varphi(h) \le \psi(f_0)\le \psi(f)$. Thus
$h, \psi\varphi h$ and $\psi f_0$ are all lower bounds of $\psi(F)$. From the maximality of $h$ we get
$h=\psi\varphi h =\psi f_0$, and $h\in \psi(\sqcup F)$. Thus $\sqcap\psi(F)\subseteq \psi(\sqcup F)$. The
inclusion $\sqcap\varphi(H)\subseteq \varphi(\sqcup H)$ follows immediately.
	\item[\textup{(ii)}] Let $C_J:= \{(h_j,f_j)\mid j\in J\}$ be a set of concepts.	We need to show that $C_J$
has a maximal lower bound, and that any lower bound of $C_J$ is below a maximal lower  bound of $C_J$. First,
because $\sqcap\{h_j\mid j \in J\}$ and $\sqcap\{f_j;\mid j \in J\}$ are non-empty, the sets
$\{(h,\varphi(h)); h\in \sqcap\{h_j\mid j \in J\}\}$ and $\{(\psi(f), f); f\in \sqcap\{f_j;\mid j \in J\}\}$
are non-empty.
Let $h\in \sqcap\{h_j\mid j \in J\}$. We have $h\leq h_j$ and $f_j=\varphi(h_j)\leq \varphi(h)$, for all
$j \in J$. Therefore, $(h, \varphi(h))\leq(h_j,f_j)$, for all $j \in J$. That is $(h, \varphi(h))$ is a lower
bound of $C_J$ and by the maximality of $h$, it is a maximal lower bound of $C_J$.
And because $h\leq \psi\varphi(h)\leq \psi\varphi(h_j)$, for all $j \in J$, by the maximality of $h$, we
have $\psi\varphi(h)=h$. Hence, $(h, \varphi(h))$ is a concept.

Similarly, we prove that, for any $f\in \sqcap\{f_j;\mid j \in J\}$, $(\psi(f), f)$ is a concept and it is a
minimal upper bound of $C_J$.

\end{itemize}

\end{proof}
We are going to show that starting with two complete residuated multilattices $\mathcal{A}_1$ and $\mathcal{A}_2$,
the set of concepts again forms a complete residuated multilattice.

\section{\textbf{Residuated concept multilattices}} \label{s:RCML}
From now on, $\mathcal{A}_i:=(A_i,\le_i,\top_i,\odot_i,\to_i,\bot_i)$, $i=1,2$ are complete residuated
multilattices, $G$ is a set of objects (to be evaluated in $\mathcal{A}_1$), $M$ is a set of attributes
(to be evaluated in $\mathcal{A}_2$), and $(\varphi,\psi)$ is a Galois connection between $A_1^G$ and $A_2^M$.
Further we  denote by $\mathcal{C}$ the set of concepts,  $\Ext(\mathcal{C})$ the set of extents and
$\Int(\mathcal{C})$ the set of intents. i.e
\begin{align*}
	\mathcal{C}&:=\{(h,f)\in A_1^G\times A_2^M \mid \varphi(h)=f \text{ and } \psi(f)=h \}, \\

	\Ext(\mathcal{C})&:=\{h\in A_1^G\mid (h,\varphi(h))\in \mathcal{C}\} =\{h\in A_1^G\mid \psi\varphi(h)=h\},\\

	\Int(\mathcal{C})&:=\{f\in A_2^M\mid (\psi(f),f)\in \mathcal{C}\} =\{f\in A_2^M\mid \varphi\psi(f)=f\}.

\end{align*}
The operations $\odot_i$ and $\to_i$ are defined componentwise on $A_1^G$ and $A_2^M$. Let $\ddh_i$ and $\uuh_i$
be the constant maps with values $\bot_i$ and $\top_i$: i.e. for $g\in G$ and $m\in M$,
 \[
 \ddh_1(g)=\bot_1,\quad \ddh_2(m)=\bot_2,\quad \uuh_1(g)=\top_1 \quad \text{ and }\quad  \uuh_2(m)=\top_2.
 \]
 For any $h\in A_1^G$ and $f\in A_2^M$ we have
\[
h\odot_1\ddh_1\ =\ \ddh_1,\quad  h\odot_1\uuh_1\ =\ h,\quad
h\to_1\uuh_1\ =\ \uuh_1\quad \text{ and } \]
\[f\odot_2\ddh_2\ =\ \ddh_2,\quad  f\odot_2\uuh_2\ =\ f,\quad f\to_2\uuh_2\ =\ \uuh_2.
\phantom{\quad \text{ and }}\]
We are interested in constructing a residuated couple on $(\mathcal{C},\le)$; We are looking for a suitable
product ($\odot$) and implication ($\to$) such that \[(h_1,f_1)\odot (h_2,f_2)\le (h_3,f_3) \iff (h_1,f_1)\le
(h_2,f_2)\to (h_3,f_3).\]

\begin{lem} Let $h_1,h_2\in A_1^G$,  $f_1,f_2\in A_2^M$, and $(\varphi,\psi)$ a Galois connection. Then
\begin{enumerate}
	\itemsep4pt
	\item $\psi\varphi(h_1\odot_1 h_2) = \min\{\psi(f)\mid f\in A_2^M$   and $h_1\odot_1 h_2\leq_1 \psi(f)\}$
	\item $\varphi\psi(f_1\odot_2 f_2) = \min\{\varphi(h)\mid h\in A_1^G \text{ and } f_1\odot_2 f_2\leq_2
\varphi(h)\}$
	\item $\psi\varphi(h_1\to_1 h_2) =\max\{\psi\varphi(h)\mid h\in A_1^G \text{ and } h\odot_1  h_1\le_1 h_2 \}$
	\item $\varphi\psi(f_1\to_2 f_2)=\max\{\varphi\psi(f)\mid f\in A_2^M \text{ and } f\odot_2 f_1\leq_2 f_2 \}$.
\end{enumerate}

\end{lem}

\begin{proof}
	 Let $(\varphi,\psi)$ be a Galois connection.
\begin{enumerate}
\itemsep4pt
\item Let $h_1,h_2\in A_1^G$.  We set
$H:= \{\psi(f)\mid f\in A_2^M \text{ and } h_1\odot_1 h_2\leq_1 \psi(f) \}.$ and
$F=\{f\in A_2^M \mid h_1\odot_1 h_2\leq_1 \psi(f)\}$ and get $H=\psi(F)$.

 Since $h_1\odot_1 h_2\le_1\psi\varphi(\uuh_1) =\uuh_1$, it follows that $\uuh_1\in H$ and $\varphi(\uuh_1)\in F$.
 We claim that $H$ has a least element. In fact, for all $\psi(f)\in H$, $h_1\odot_1 h_2\le_1\psi(f)$.
 Therefore there is a maximal lower bound $h$ of $H$ such that  $h_1\odot_1 h_2\le_1 h$. Thus
 $h\in\sqcap H = \sqcap \psi(F)\subseteq \psi(\sqcup F)$, and there is $f_0\in \sqcup F$ such that $h=\psi(f_0)$.
Hence, $h\in H$ and $h\in\sqcap H$. Henceforth  $h$ is the smallest element of $H$.
Moreover, from $h_1\odot_1 h_2 \le_1\psi\varphi(h_1\odot_1 h_2)$ we get $\varphi(h_1\odot_1 h_2)\in F$ and
$\psi\varphi(h_1\odot_1 h_2)\in H$. It follows that
\[
h_1\odot_1 h_2 \le_1\psi\varphi(h_1\odot_1 h_2)\le_1 \psi\varphi\psi(f_0)=\psi(f_0)=h,
\]
and $\psi\varphi(h_1\odot_1 h_2)=h=\min H$.

\item Similarly, the set $\{\varphi(h)\mid h\in A_1^G \text{ and } f_1\odot_2 f_2\leq_2 \varphi(h) \}$ is
non-empty and has a least element, namely $\varphi\psi(f_1\odot_2 f_2)$, for any $f_1,f_2\in A_2^M$.
\item  Let $h_1,h_2\in A_1^G$. We set $H= \{\psi\varphi(h)\mid h\in A_1^G \text{ and } h\odot_1 h_1\le_1 h_2\}$.
Then
\begin{align*}

H &=\{\psi\varphi(h)\mid h\in A_1^G \text{ and }  h\odot_1 h_1\leq_1  h_2 \} \\
  &=\{\psi\varphi(h)\mid h\in A_1^G \text{ and }  h\le_1 h_1\to_1  h_2 \} \\
&\subseteq\{\psi\varphi(h)\mid h\in A_1^G \text{ and }
\psi\varphi(h)\le_1\psi\varphi(h_1\to_1  h_2) \}  , \quad \text{ $\psi\varphi$ isotone operator}
\end{align*}
Thus $\psi\varphi(h_1\to_1  h_2)$ is an upper bound of $H$. Moreover, $h_1\to_1  h_2\le_1 h_1\to_1  h_2$,
then $\psi\varphi(h_1\to_1  h_2)$ is in $H$.

Thus, $\psi\varphi(h_1\to_1  h_2)=max\{\psi\varphi(h)\mid h\in A_1^G \text{ and } h\le_1 h_1\to_1  h_2\}$.
\item The proof is similar to 3.
\end{enumerate}

\end{proof}

\noindent
Let $h_1,h_2\in A_1^G$ and  $f_1,f_2\in A_2^M$. We define the operations $\otimes_1$, $\otimes_2$,
$\rightrightarrows_1$ and $\rightrightarrows_2$ as follows:
\begin{align*}
h_1\otimes_1 h_2:= \psi\varphi(h_1\odot_1 h_2) && h_1\rightrightarrows_1 h_2:= \psi\varphi(h_1\to_1 h_2) \\
f_1\otimes_2 f_2:= \varphi\psi(f_1\odot_2 f_2) &&
f_1\rightrightarrows_2 f_2:= \varphi\psi(f_1\to_2 f_2).
\end{align*}
\noindent
Observe that the operations $\otimes_1$ and $\rightrightarrows_1$ are defined on all pairs $h_1, h_2\in A_1^G$
but their results are in  $\Ext(\mathcal{C})$.

Similarly, the operations $\otimes_2$ and $\rightrightarrows_2$ are defined on all pairs $f_1, f_2\in A_2^M$
but their results are in  $\Int(\mathcal{C})$. We can then infer that  $(h_1\otimes_1 h_2,
\varphi(h_1\otimes_1 h_2))=(\psi\varphi(h_1\odot_1 h_2), \varphi(h_1\odot_1 h_2))$ is a concept and  also that
$(h_1\rightrightarrows_1 h_2, \varphi(h_1\rightrightarrows_1 h_2)) = (\psi\varphi( h_1\to_1 h_2),
\varphi(h_1\to_1 h_2))$ is a concept. Similarly,
$(\psi(f_1\otimes_2 f_2), f_1\otimes_2 f_2)$ and

$(\psi(f_1\rightrightarrows_2 f_2), f_1\rightrightarrows_2 f_2)$

are concepts.

It is obvious that the operations $\otimes_i$, $i\in \{1, 2\}$ are commutative. Now, let us look about their
associativity and the residuated couple needed.

\begin{lem}\label{lemme 1} Let $(\varphi,\psi)$ be a Galois connection. If $\Ext(\mathcal{C})$ and
$\Int(\mathcal{C})$ are closed under $\to_i$ then $(\otimes_i, \rightrightarrows_i)$, $i=1,2$
are residuated couples and $\otimes_1$ and $\otimes_2$ are associative.
\end{lem}

\begin{proof} Let $h_1, h_2, h_3 \in A_1^G$ such that $h_1\otimes_1 h_2\leq_1 h_3$.
\begin{align*}
h_1\otimes_1 h_2\leq_1 h_3 &\implies h_1\leq_1 \psi\varphi(h_1)\in
 \{\psi\varphi(h)\mid h\in A_1^G \text{ and } h\otimes_1 h_2\leq_1  h_3 \}\\
&\implies 	h_1\leq_1 \max\{\psi\varphi(h)\mid h\in A_1^G \text{ and } h\otimes_1 h_2\leq_1  h_3 \} \\
&\implies 	h_1\leq_1 \max\{\psi\varphi(h)\mid h\in A_1^G \text{ and }\psi\varphi(h\odot_1 h_2)\leq_1  h_3 \}\\
&\implies 	h_1\leq_1 \max\{\psi\varphi(h)\mid h\in A_1^G \text{ and } h\odot_1 h_2\leq_1  h_3 \} \\
&\implies 	h_1\leq_1 \max\{\psi\varphi(h)\mid h\in A_1^G \text{ and } h\le_1 h_2\to_1  h_3 \} \\
&\implies h_1\leq_1 \varphi\psi(h_2\to_2 h_3) = h_2\rightrightarrows_2 h_3.
\end{align*}
Henceforth,  for all $h_1, h_2, h_3$ in $A_1^G$ we get \[
h_1\otimes_1 h_2\leq_1 h_3\implies h_1\leq_1 h_2\rightrightarrows_1 h_3.\]
The converse holds if $h_3$ and $h_2\to_1 h_3$ are closed. In fact,
\begin{align*}
h_1\leq_1 h_2\rightrightarrows_1 h_3 &\iff h_1\leq_1 \psi\varphi(h_2\to_1 h_3)\\
&\implies h_1\leq_1 h_2\to_1 h_3,\quad \text{assuming $h_2\to_1 h_3$ closed}\\
&\iff h_1\odot_1 h_2\leq_1 h_3 \\
&\implies \psi\varphi(h_1\odot_1 h_2)\leq_1 h_3, \quad \text{ assuming $h_3$ closed} \\
&\iff h_1\otimes_1 h_2\leq_1 h_3.
\end{align*}
Thus $(\otimes_1, \rightrightarrows_1)$ is an adjoint couple on $\Ext(\mathcal{C})$, if $\Ext(\mathcal{C})$
is closed under $\to_1$.

Similarly, we can proved that $(\otimes_2, \rightrightarrows_2)$ is an adjoint couple on $\Int(\mathcal{C})$,
if it is closed under $\to_2$.

We still need to check the associativity of $\otimes_1$ and $\otimes_2$. Let $h_1,h_2,h_3$ in $A_1^G$. Then

\begin{align*}
(h_1\otimes_1 h_2)\otimes_1 h_3 &= \psi\varphi((h_1\otimes_1 h_2)\odot_1 h_3) \\ &=\min\{\psi(f)\mid
 f\in A_2^M \text{ and } (h_1\otimes_1
h_2)\odot_1 h_3\leq_1 \psi(f)\},
\end{align*}
and
\begin{align*}
h_1\otimes_1 (h_2\otimes_1 h_3) &= \psi\varphi(h_1\odot_1 (h_2\otimes_1 h_3)) \\ &=\min\{\psi(f)\mid
 f\in A_2^M \text{ and } h_1\odot_1 (h_2\otimes_1 h_3)\leq_1 \psi(f)\}.
\end{align*}
It is enough to prove that the sets
\[H_1=\{\psi(f)\mid f\in A_2^M \text{ and } (h_1\otimes_1
h_2)\odot_1 h_3\leq_1 \psi(f)\}\]
and
\[
H_2=\{\psi(f)\mid f\in A_2^M \text{ and } h_1\odot_1 (h_2\otimes_1 h_3)\leq_1 \psi(f)\}
\]
are equal. Let $f\in A_2^M$ such that $(h_1\otimes_1 h_2)\odot_1 h_3\leq_1\psi(f)$.

\begin{align*}
(h_1\otimes_1 h_2)\odot_1 h_3\leq_1 \psi(f) &\implies  (h_1\otimes_1 h_2)\le_1 h_3\to_1 \psi(f) \\
&\implies \psi\varphi(h_1\odot_1 h_2)\le_1 h_3\to_1 \psi(f) \\
&\implies h_1\odot_1 h_2\le_1 h_3\to_1 \psi(f) \\
&\implies (h_1\odot_1 h_2)\odot_1 h_3 \le_1 \psi(f) \\
&\implies \psi\varphi((h_1\odot_1 h_2)\odot_1 h_3) \le_1 \psi(f) \\
&\implies \psi\varphi(h_1\odot_1 (h_2\odot_1 h_3)) \le_1 \psi(f) \\
&\implies h_1\otimes_1 (h_2\odot_1 h_3)) \le_1 \psi(f) \\
&\implies h_2\odot_1 h_3 \le_1 h_1\rightrightarrows_1 \psi(f) \\
&\implies \psi\varphi(h_2\odot_1 h_3) \le_1 h_1\rightrightarrows_1 \psi(f) \\
&\implies h_2\otimes_1 h_3 \le_1 h_1\rightrightarrows_1 \psi(f) \\
&\implies h_1\otimes_1 (h_2\otimes_1 h_3) \le_1  \psi(f)\\
&\implies h_1\odot_1 (h_2\otimes_1 h_3) \le_1  \psi(f).
\end{align*}
Thus $\psi(f)\in H_1 \implies \psi(f)\in H_2$.

Conversely, let $f\in A_2^M$ such that $h_1\odot_1 (h_2\otimes_1 h_3)\leq_1\psi(f)$.

\begin{align*}
h_1\odot_1 (h_2\otimes_1 h_3)\leq_1 \psi(f) &\implies  h_2\otimes_1 h_3 \le_1 h_1\to_1 \psi(f) \\
&\implies \psi\varphi(h_2\odot_1 h_3)\le_1 h_1\to_1 \psi(f) \\
&\implies h_2\odot_1 h_3\le_1 h_1\to_1 \psi(f) \\
&\implies h_1 \odot_1 (h_2\odot_1 h_3) \le_1 \psi(f) \\
&\implies \psi\varphi(h_1\odot_1 (h_2\odot_1 h_3)) \le_1 \psi(f) \\
&\implies \psi\varphi((h_1\odot_1 h_2)\odot_1 h_3) \le_1 \psi(f) \\
&\implies (h_1\odot_1 h_2)\otimes_1 h_3 \le_1 \psi(f) \\
&\implies h_1\odot_1 h_2 \le_1 h_3\rightrightarrows_1 \psi(f) \\
&\implies \psi\varphi(h_1\odot_1 h_2) \le_1 h_3\rightrightarrows_1 \psi(f) \\
&\implies h_1\otimes_1 h_2 \le_1 h_3\rightrightarrows_1 \psi(f) \\
&\implies (h_1\otimes_1 h_2)\otimes_1 h_3 \le_1  \psi(f)\\
&\implies (h_1\otimes_1 h_2)\odot_1 h_3 \le_1  \psi(f).
\end{align*}
Thus $\psi(f)\in H_2 \implies \psi(f)\in H_1$. Therefore $H_1=H_2$, and
\[
h_1\otimes_1 (h_2\otimes_1 h_3) = (h_1\otimes_1 h_2)\otimes_1 h_3.
\]
\end{proof}

We can now define the product $\odot$ and the implication $\rightarrow$
on the set of all concepts.
\begin{thm}\label{thm:main}
Let $\mathcal{A}_1$  and $\mathcal{A}_2$ be two complete residuated multilattices, $G$ and $M$ two non-empty sets
(of objects and attributes) and
$(\varphi, \psi)$ a Galois connection between $A_1^G$ and $A_2^M$. Then
\[\mathcal{C}=\{(h,f)\in A_1^G\times A_2^M \mid \varphi(h)=f \text{ and } \psi(f)=g \})\]
 is a complete residuated multilattice, if $\Ext(\mathcal{C})$ and
$\Int(\mathcal{C})$ are closed under $\to_1$ and $\to_1$ respectively, with
	\begin{align*}\label{e1}
	(h_1, f_1)\odot(h_2, f_2) &= (h_1\otimes_2 h_2, \varphi(h_1\otimes_2 h_2)) \text{ and } \\
	(h_1, f_1)\rightarrow(h_2, f_2) &= (h_1\rightrightarrows_2 h_2, \varphi(h_1\rightrightarrows_2 h_2)).
	\end{align*}
	for all concepts $(h_i, f_i)$, $i=1,2$.
\end{thm}
\begin{proof}
	In \cite{MOR13} it is shown that $(\mathcal{C}, \preceq)$ is a complete multilattice.

	Let $(h_1, f_1)$, $(h_2, f_2)$ and $(h_3, f_3)$ in $\mathcal{C}$. From Lemma \ref{lemme 1} we know that
$(h_1\otimes_1
	h_2, \varphi(h_1\otimes_1 h_2))$ and $(h_1\rightrightarrows_2 h_2,
	\varphi(h_1\rightrightarrows_1 h_2))$ are concepts. It remains to
	prove that $\odot$ and $\rightarrow$ satisfy the adjointness
	condition that is,
	\begin{align*}
	(h_1, f_1)\odot(h_3, f_3)\le (h_2, f_2)&\iff (h_3, f_3)\le (h_1, f_1)\rightarrow(h_2, f_2).
\end{align*}
This is equivalent to
\begin{align*}
(h_1\otimes_1 h_3, \varphi(h_1\otimes_1 h_3))\le (h_2, f_2) &\iff
(h_3, f_3)\le (h_1\rightrightarrows_1 h_2, \varphi(h_1\rightrightarrows_1 h_2)),
\end{align*}

which is again equivalent to prove that
\[
h_1\otimes_1 h_3\leq_1
	h_2\Leftrightarrow h_3\leq_1 h_1\rightrightarrows_1 h_2.
\]
This is true since $\otimes_1$ and $\rightrightarrows_1$ satisfy the adjointness condition,  by
Lemma \ref{lemme 1}.
\end{proof}

We have thus proved that with residuated multilattices we obtain a residuated concept  multilattice.
Replacing one of the residuated multilattices $\mathcal{A}_1$ and $\mathcal{A}_2$ by a residuated lattice, we
claim that the set of all concepts is a residuated lattice in the following corollary.
\begin{cor} Under the assumption of the hypothesis of Theorem~\ref{thm:main}, if $\mathcal{A}_1$ or
$\mathcal{A}_2$ is a residuated lattice, then $(\mathcal{C}, \preceq)$ is a residuated lattice.

\end{cor}
\begin{proof}
If $A_1$ is a residuated lattice the multi-infimum becomes a singleton and is indeed an infimum. Hence,
every set has an infimum and so $(\mathcal{C}, \le)$ is a complete residuated lower semilattice with the
greatest element element $(\uuh_1, \varphi(\uuh_1))$.

The proof with $A_2$ is similar.
\end{proof}
\section{Conclusion} \label{s:conclusion}
By choosing residuated multilattices as set of truth values, we have proved that the set of concepts forms a
complete residuated multilattice.

As next step, we will initiate and study residuated concept  multilattices with hedges. Also, from what was done
by Medina et al. \cite{MOR07} on the use of ordered multilattices as underlying sets of truth-values for a
generalized framework of logic programming, we are planing to extend their finding in the context of residuated
multilattices.

config: arxiv


TITLE: Huggins Displacement Theory and Retrocausality
QUESTION [2 upvotes]: I was looking at the Wikipedia entries on Time Travel and the Grandfather paradox and noticed a paragraph on the so-called Huggins Displacement Theory. I haven't been able to find the source although I assume it's originator was the Victorian astronomer Sir William Huggins. Does anyone know of the precise reference?
Normally sending information into one's past can lead to contradictions. For example based on some condition now I could send a message into my past that prevents that very condition from occurring in my present. Thus sending information into the past, and therefore arbitrarily changing it, seems to be forbidden physically.
However the Huggins Displacement Theory says that there is a loop-hole in this argument (or more accurately a "corner case"!) I could send information along my past lightcone so that if I influence something a year in my past then it must be a light-year away from me. Thus no change in the past has a enough time to travel back to me to cause a contradiction in my present.
For example imagine that I see a stock price going up today. The Huggins Theory says that I can beam knowledge of this stock price rise back in time provided that the message is also displaced by a comparable distance. Thus if I send it back a year in time then it must also travel away from me by a light-year in distance. If my twin gets the message then he might try to profit from the information by sending back a message to me to buy the stock. Unfortunately the message will only get back to me at the instant that the price rises so that neither myself nor my twin will be able to profit from this situation.
Maybe physics does allow information to be sent back in time along a past lightcone so that this effect could be measured. My hunch is that perhaps if one charged up an electrode then electrostatic influences from it might travel back along its lightcone to cause measureable repulsion or attraction on distant electrodes.
I don't think electromagnetic radiation itself could go backwards in time as in principle it could be reflected back onto an observer's past worldline and therefore trigger contradictions in his/her present.

REPLY [2 votes]: My name is Sarah Huggins, about three years ago, my dad wrote an article on Wikipedia to demonstrate how nothing on the site is reliable. That article was The Huggins Displacement Theory.  A couple years later, it's in three books, multiple movie and book reviews, and on physics blogs like this. He took the article down once it got flagged, but I thought I should let you know that it's not a legitimate theory and was made up to teach 12 year old me a lesson.


TITLE: Understanding if the following subsets of a vector space satisfy addition and scalar multiplication
QUESTION [0 upvotes]: Here, I'm supposed to justify if the following subsets are closed under scalar multiplication, addition, both or neither.
a) $W_1=\mathcal{S},Z_1=\{s \in\mathcal{S}|s \text{ is monotone}\}.$ Here, monotone sequences are defined as $a_{n+1} \geq a_n$ for all $n\geq 0$ or $a_{n+1} \leq a_n$ for all $n\geq 0$ for some sequence $(a_n)_n$.
b) $W_2=M_3(\mathbb{R}),Z_2=\{A \in M_3(\mathbb{R})|Tr(A)^3=Det(A) \}$.
a) I could pick $a_n=(0,1,1,...)$ and $b_n=(0,0,-1,-1,...)$ for some $a_n$, $b_n$ in $Z_1$ since this gives $(0,1,0,0,...)$ which isn't monotonic. Addition fails. I think multiplication will work though but I am not sure how to show this.
b) I'm almost certain addition fails but I am not so sure about scalar multiplication.
If someone could help me out, that would be great!

REPLY [1 votes]: For part $a)$ scalar multiplication, let $(a_n)$ be a monotone sequence and $s$ be a scalar. Then $s(a_n) = (s\cdot a_n)$ is also montone, since you can always just cancel from both sides of the $\leq$ sign.
For $b)$, consider
$$
A = \begin{bmatrix}
2&0&0\\
0&2&0\\
0&0&0
\end{bmatrix},B = \begin{bmatrix}
0&0&0\\
0&0&0\\
0&0&2
\end{bmatrix}
$$
Note that $A$ and $B$ have determinant and trace $0$ so they are in $Z_2$, but $A+B$ has trace $8$ and determinant $8$, but obviously $8^3 \neq 8$.
For scalar multiplication we don't have closure either. Consider the identity matrix $I$. It has cubed trace and determinant $1$, so it is in $Z_2$, however if for example we multiply the matrix by $2$, we get a trace of $8$ and a determinant of $8$ and as noted about $8^3 \neq 8$.


TITLE: Modifying the Hamiltonian when there is a presence of the Coulomb interaction
QUESTION [2 upvotes]: Referring to the Hamiltonian of a system of free electrons,
$$
 H_0= \sum_{\sigma} \int d^3rd^3r' \psi_{\sigma}^{\dagger}(\mathbf{r})\left(- \frac{\hbar^2}{2m}\nabla^2\right)\delta(\mathbf{r}-\mathbf{r'})\psi_\sigma (\mathbf{r'})
$$
When the Coulomb interaction is turn on, we can modify this Hamiltonian by imposing
$$
\partial^\mu\rightarrow\partial^\mu + i \frac{q}{\hbar c} A^\mu
$$
I expected the new Hamiltonian to contain the term like
$$
H_{\mathrm{Coulomb}} = \frac{1}{2}\sum_{\sigma,\sigma'}\int d^3rd^3r' \psi_\sigma^\dagger(\mathbf{r})\psi_\sigma(\mathbf{r})\left( \frac{q^2}{4\pi\epsilon_0|\mathbf{r}-\mathbf{r'}|}\right)\psi_{\sigma'}^\dagger(\mathbf{r'})\psi_{\sigma'}(\mathbf{r'})
$$
However, when I substituted this to the free Hamiltonian, I could not see anyway to obtain this result at all.

REPLY [3 votes]: There are a couple of issues with the phrasing of the question:
First: Coulomb interaction only affects the temporal component of the derivative operator (i.e. you need to focus on only $\mu = 0$). The general form you wrote is useful if you consider the full electromagnetic field (i.e. in the presence of both $A_0$ and $\mathbf A$).
Second: The Peierls substitution for $\partial^0$ leads to a minimal coupling between the electric potential ($A_0$) and electron density, which is unavailable in the hamiltonian formalism. To see this the action is required.
Within the action formalism the $A_0$ field can be integrated out, which will lead to the effective long-range interaction you have quoted. To demonstrate, I use the imaginary time ($\tau$) formalism and slightly modify the parameters in your equations to write the action,
$$S = \sum_\sigma \int d\tau  d\mathbf{r}  ~\Psi_\sigma^\dagger(\tau, \mathbf r) \left[\partial_\tau - i g A_0(\tau, \mathbf r) \right] \Psi_\sigma(\tau, \mathbf r) - \int d\tau H_0 + \frac{1}{2}\int d\tau  d\mathbf{r} |\mathbf \nabla A_0(\tau, \mathbf r)|^2.$$
Now from within the partition function $$ Z = \int DA_0 ~ D\Psi ~ D\Psi^{\dagger} ~ e^{-S},$$ one may integrate out $A_0$ (easier to do that if you Fourier transform and write $S$ in momentum space) using properties of Gaussian integration. This will result in $$S =  \sum_\sigma \int d\tau  d\mathbf{r}  ~\Psi_\sigma^\dagger(\tau, \mathbf r) \partial_\tau \Psi_\sigma(\tau, \mathbf r) - \int d\tau (H_{Coulomb} + H_0),$$ where $H_{Coulomb}$ is the equation noted in the question. Therefore, the total hamiltonian is $H = H_{Coulomb} + H_0$.


TITLE: Products of representables are regular on a regular skeletal Reedy category?
QUESTION [5 upvotes]: The definition of a (pre)regular skeletal Reedy category in the sense of Cisinski generalizes the intuition behind the Eilenberg-Zilber factorization for simplicial sets, specifically the property that the image of any section of a representable is again representable, or equivalently that all nondegenerate sections of representables are monic.
Of course, the classical Eilenberg-Zilber property for simplicial sets is stronger, namely that this regularity property holds not only for representables but also for products of representables (which yields the shuffle (i.e. Eilenberg-Zilber) decomposition of simplicial prisms that we all know and love).
Can the regularity of products of representables be deduced from just the axioms of regular skeletal Reedy categories, or is it an additional axiom that must be included? If it is an additional property, are there any easier-to-check equivalent properties?

REPLY [1 votes]: I asked Cisinski by e-mail if this was true, and he said that it isn't.  A counterexample is the category of planar trees ($\Omega_{\mathbf{pl}}$, not $\Omega$, which is not even normal) used in the theory of Dendroidal sets.  He said that this property can be axiomatized by asking that the class of regular presheaves is closed under the Cartesian product.