# Gluing residuated lattices

Nikolaos Galatos  Department of Mathematics, University of Denver, CO, USA
<EMAIL_ADDRESS>and  Sara Ugolini  IIIA - CSIC, Bellaterra, Barcelona,
Spain<EMAIL_ADDRESS>
###### Abstract.

We introduce and characterize various gluing constructions for residuated
lattices that intersect on a common subreduct, and which are subalgebras, or
appropriate subreducts, of the resulting structure. Starting from the $1$-sum
construction (also known as ordinal sum for residuated structures), where
algebras that intersect only in the top element are glued together, we first
consider the gluing on a congruence filter, and then add a lattice ideal as
well. We characterize such constructions in terms of (possibly partial)
operators acting on (possibly partial) residuated structures. As particular
examples of gluing constructions, we obtain the non-commutative version of
some rotation constructions, and an interesting variety of semilinear
residuated lattices that are $2$-potent.

This study also serves as a first attempt toward the study of amalgamation of
non-commutative residuated lattices, by constructing an amalgam in the special
case where the common subalgebra in the V-formation is either a special
(congruence) filter or the union of a filter and an ideal.

###### Key words and phrases:

Residuated lattices, Amalgamation, Gluing, Ordinal sum

###### 2010 Mathematics Subject Classification:

06F05,08A55,06A15,08A05

##  1\. Introduction and preliminaries

The first gluing construction in lattice theory is due to Hall and Dilworth
[19], who used it to prove the existence of a modular lattice that cannot be
embedded in a complemented modular lattice. Later on, the same construction
was independently used by Wroński in [27] and Troelstra [26] to study
intermediate logics, by constructing Heyting algebras.

The idea in these constructions is to glue together lattices that intersect
(up to isomorphism) on a sublattice that is a principal ideal of the first and
a principal filter of the second. In particular, the construction applies to
Heyting algebras: bounded lattices that are relatively pseudocomplemented
(i.e., for every pair of elements $x,y$ there is a largest element $z$ with
the property that $x\land z\leq y$). Heyting algebras can also be equivalently
defined as bounded residuated lattices where the monoidal operation coincides
with the meet in the lattice order; in this case $x\to y$ is the largest
element $z$ such that $x\land z\leq y$. Residuated lattices play an important
role in the study of algebraic logic, as they constitute the equivalent
algebraic semantics (in the sense of Blok-Pigozzi [5]) of substructural
logics. These encompass most interesting nonclassical logics: intuitionistic
logic, fuzzy logics, relevance logics, linear logic, and classical logic as a
limit case. Thus, the investigation of the variety of residuated lattices is a
powerful tool in the comparative study of such logics, as explored in [16].
The multitude of different types of residuated lattices makes the study fairly
complicated and at the present moment large classes of residuated lattices
lack a structural description. The study of constructions that allow us to
obtain new structures from known ones is extremely important in improving our
understanding of residuated lattices, and as a result, of substructural
logics.

In the present paper we introduce different ways of gluing together residuated
lattices, where by gluing we mean obtaining a new structure from two original
ones which intersect on a common subreduct, and which are subalgebras (or
appropriate subreducts) of the resulting structure. The starting point of our
investigation is the $1$-sum construction, often called ordinal sum in
residuated structures, where the algebras intersect only at the top element,
i.e. at a trivial filter. We will consider gluings over an arbitrary
(nontrivial) congruence filter, and then over a lattice ideal as well.
Moreover, we generalize these ideas to account for (possibly) partial
algebras. Finally, we characterize the introduced constructions abstractly, by
means of pairs of operators acting on residuated lattices.

These new constructions serve as a first attempt in the study of amalgamation
of non-commutative residuated lattices, by constructing an amalgam in the
special case where the common subalgebra in the V-formation is either a
special (congruence) filter or the union of a filter and an ideal.

As particular examples of gluing constructions, we obtain the non-commutative
version of the generalized rotation construction in [7], and an interesting
variety of semilinear residuated lattices that are $2$-potent. We use these
two cases to illustrate examples of how the gluing construction can be used to
study amalgamation. In the case of the rotation we show how the construction
preserves the amalgamation property (in a sense that will be made precise),
while in the latter case we show and characterize amalgamation failures.

We start by introducing the objects of our study. A residuated lattices is an
algebra ${\textbf{A}}=(A,\lor,\land,\cdot,\backslash,/,1)$ of type
$(2,2,2,2,2,0)$ such that:

  1. (1)

$(A,\lor,\land)$ is a lattice;

  2. (2)

$(A,\cdot,1)$ is a monoid;

  3. (3)

$\backslash$ and $/$ are the left and right division of $\cdot$: for all
$x,y,z\in A$,

$x\cdot y\leq z\Leftrightarrow y\leq x\backslash z\Leftrightarrow x\leq z/y,$

where $\leq$ is the lattice ordering.

Residuated lattices form a variety, denoted by $\mathsf{RL}$, as residuation
can be expressed equationally; see [4]. When the monoidal identity is the top
element of the lattice we say that the residuated lattice is _integral_ or an
_IRL_ ; we call the corresponding variety $\mathsf{IRL}$. Residuated lattices
with an additional constant $0$ are called _pointed_. _Bounded_ integral
residuated lattices are expansions of residuated lattice with an extra
constant $0$ that satisfies the identity $0\leq x$. The variety of bounded
integral residuated lattices is called $\mathsf{FL_{w}}$, referring to the
fact that it is the equivalent algebraic semantics of the Full Lambek calculus
with the structural rule of weakening (see [16]). As usual, we write $xy$ for
$x\cdot y$.

A residuated lattice is called _commutative_ if the monoidal operation is
commutative. In this case the two divisions coincide, and we write $x\to y$
for $x\backslash y=y/x$. We write $\mathsf{CRL}$ and $\mathsf{CIRL}$,
respectively, for the commutative subvarieties of $\mathsf{RL}$ and
$\mathsf{IRL}$, and refer to commutative $\mathsf{FL_{w}}$-algebras as
$\mathsf{FL_{ew}}$-algebras, since commutativity of the monoidal operation
corresponds to the structural rule of exchange.

In a lattice A, a _filter_ is a non-empty subset $S$ that is closed upwards
(if $x\leq y$ and $x\in S$ then also $y\in S$) and is closed under meet (if
$x,y\in S$ then $x\land y\in S$). In a (bounded) integral residuated lattice
A, a _congruence filter_ $F$ is a non-empty upset of $A$, closed under
products (if $x,y\in F$, then $xy\in F$) and under conjugates, i.e. if $x\in
F$ , then $yx/y,y\backslash xy\in F$ for every $y\in A$. We will denote by
${\textbf{Fil}}({\textbf{A}})$ the lattice of congruence filters of A. It is
easy to see that a filter $F$ of a (bounded) integral residuated lattice A is
a subalgebra (or $0$-free subreduct, if A is bounded) of A, hence it is an
integral residuated lattice.

Filters of residuated lattices are in one-one correspondence to congruences.
In particular, in the integral case the isomorphism between
${\textbf{Fil}}({\textbf{A}})$ and the congruence lattice of A,
${\textbf{Con}}({\textbf{A}})$, is given by the maps:

$F\mapsto\theta_{F}=\\{(x,y)\in A\times A:x\backslash y,y\backslash x\in
F\\}=\\{(x,y):x/y,y/x\in F\\},$ $\theta\mapsto F_{\theta}=\\{x\in
A:(x,1)\in\theta\\}$

for all $F\in Fil({\textbf{A}}),\theta\in Con({\textbf{A}})$. In what follows,
given a congruence filter $F$, we will write $[x]_{F}$ for the equivalence
class $[x]_{\theta_{F}}$.

##  2\. Gluing over a filter

As we mentioned in the introduction, the usual notion of gluing in lattice
theory puts together two lattices that intersect in a filter of the first and
an ideal of the second. As we are interested in integral residuated lattices
and we want the components to be subalgebras of the resulting structure (in
particular the common identity element needs to be the top), this approach
needs to be modified. We start by describing the simple case where the ideal
is empty, and the filter is trivial.

###  2.1. $1$-sum

The $1$-sum construction in the context of residuated structures was
introduced with the name of ordinal sum by Ferreirim in [11] in the context of
hoops. The latter can be defined as commutative integral divisible ($x\wedge
y=x(x\to y)$) residuated lattices but without the demand that joins exist; we
choose to use the naming $1$-sum as in [24, 25] to avoid confusion. Indeed it
is slightly different than the ordinal sum of two posets/lattices, as it
identifies the top elements of the two structures. The $1$-sum construction
has played an important role in the study of BL-algebras and basic hoops [1].
The construction was later extended to integral residuated lattices, and even
generalized to non-integral structures [14]. It is also worth mentioning that
historically, the analogue to the $1$-sum has been previously introduced and
studied for semigroups [8].

$1$-sums represent a seminal example of gluing; unlike the case of hoops, some
care is needed to make sure that joins that were equal to the top still exist
in the resulting structure. The two structures glued together intersect only
at their respective top elements. In detail, let B and C be integral
residuated lattices, where $B\cap C=\\{1\\}$, and $1$ is join irreducible in B
or C has a bottom element. We extend the order of B and C to the set $B\cup C$
by: $b<c$ for $b\in B-\\{1\\}$ and $c\in C$, and extend the operations of B
and C by:

$\begin{array}[]{lll}xy&=&\left\\{\begin{array}[]{ll}y&\mbox{ if }x\in C\mbox{
and }y\in B\setminus\\{1\\}\\\ x&\mbox{ if }x\in B\setminus\\{1\\}\mbox{ and
}y\in C,\end{array}\right.\\\ &&\\\ x\backslash
y&=&\left\\{\begin{array}[]{ll}y&\mbox{ if }x\in C\mbox{ and }y\in
B\setminus\\{1\\}\\\ 1&\mbox{ if }x\in B\setminus\\{1\\}\mbox{ and }y\in
C,\end{array}\right.\\\ &&\\\ x/y&=&\left\\{\begin{array}[]{ll}y&\mbox{ if
}x\in C\mbox{ and }y\in B\setminus\\{1\\}\\\ 1&\mbox{ if }x\in
B\setminus\\{1\\}\mbox{ and }y\in C.\end{array}\right.\\\ \end{array}\\\ $

It can be easily verified that the resulting structure
${\textbf{B}}\oplus_{1}{\textbf{C}}$ is a residuated lattice, called the _$1$
-sum_ of B and C. The assumption that $1$ is join irreducible in B or C has a
bottom element ensures that joins of elements of $B$, calculated in
${\textbf{B}}\oplus_{1}{\textbf{C}}$, exist; if these conditions are not
satisfied the resulting structure is merely a residuated meet-semilattice.
Note that C is always a subalgebra of ${\textbf{B}}\oplus_{1}{\textbf{C}}$ and
B is a subalgebra except possibly with respect to $\lor$, in case that $1$ is
not join irreducible in B (see [22] for details when $1$ is not join-
irreducible). Generalizations of the $1$-sum construction to the non-integral
case are discussed in [16].

Notice that the $1$-sum construction stacks one IRL on top of another one and
identifies/glues their top elements; the product between elements of B and C
is actually their meet in the new order.

As it turns out, this is the only choice for defining a residuated monoidal
operation when gluing two residuated lattices together with this particular
lattice order, if we want B and C to be subalgebras of the new structure.

######  Proposition 2.1.

Let B and C be IRLs and assume that D is an IRL with underlying set $B\cup C$,
where $B\cap C=\\{1\\}$, $b<c$ for all $b\in B-\\{1\\}$ and $c\in C$, $C$ is a
subalgebra, and $B$ is a subalgebra except possibly with respect to $\lor$.
Then D is equal to ${\textbf{B}}\oplus_{1}{\textbf{C}}$.

###### Proof.

Given the assumptions, we only need to verify that for all $b\in B$ and $c\in
C$, we have $cb=bc=b$. We have $cb\leq cb$, so $c\leq cb/b$. Since $cb,b\in B$
we have that $cb/b$ is an element of $B$ that is greater than some element of
$C$. Therefore, $cb/b=1$, hence $b\leq cb$. By integrality we also have
$cb\leq b$, so $cb=b$. ∎

###  2.2. $F$-gluings: compatibility and uniqueness

By relaxing the assumptions in Proposition 2.1, we will generalize the $1$-sum
construction to a more general type of gluing where the intersection of the
two algebras may be a congruence filter different than $\\{1\\}$.

More precisely, given IRLs B and C, let D be some IRL with underlying set
$B\cup C$, where $F:=B\cap C$ is a congruence filter of D, $b<c<f$ for all
$b\in B-F$, $c\in C-F$ and $f\in F$, $C$ is a subalgebra, and $B$ is a
subalgebra except possibly with respect to $\lor$, in which case $F$ is
assumed to have a bottom element. We say that D is a _gluing over $F$_, or an
_$F$ -gluing of B and C_; see Figure 1 for the anticipated structure. We will
identify conditions on B, C and $F$ that will allow us to construct D from
these constituent parts.

First we describe a compatibility condition between $F$ and B and then we
characterize the structure of the subset $B^{\prime}:=(B-F)\cup\\{1\\}$. Note
that $B^{\prime}$ supports a residuated lattice even when it is not a
subalgebra under $\lor$ as those joins end up being equal to $1$.

We say that a congruence filter $F$ of an IRL B is _compatible_ with B if:

  1. (1)

every element of $F$ is strictly above every element of $B-F$.

  2. (2)

For all $b\in B-F$ the equivalence class $[b]_{F}$ has a maximum and a
minimum; we define

$\sigma_{F}(b)=\min[b]_{F},\;\;\gamma_{F}(b)=\max[b]_{F}$

for $b\in B-F$ and $\sigma_{F}(1)=\gamma_{F}(1)=1$; hence $\sigma_{F}$ and
$\gamma_{F}$ are maps on $B^{\prime}=(B-F)\cup\\{1\\}$.

  3. (3)

$\sigma_{F}$ is _absorbing_ : for all $b\in B-F$,
$b\sigma_{F}[B-F]\subseteq\sigma_{F}[B-F]$ and
$\sigma_{F}[B-F]b\subseteq\sigma_{F}[B-F]$.

We also say that $({\textbf{B}},F)$ form a _lower-compatible pair_. The
following lemma shows that $F$-gluings contain compatible pairs and explains
that the role of $\sigma$ and $\gamma$ is to capture the multiplications and
divisions, respectively, by elements of $C$ that are not in the compatible
pair. More importantly, it shows the uniqueness of the $F$-gluing.

######  Lemma 2.2.

Given a gluing of IRLs B and C over $F$, the congruence filter $F$ is
compatible with B. Moreover, for all $b\in B-F,c\in C-F$:

$cb=bc=\sigma_{F}(b),\qquad c\backslash b=b/c=\gamma_{F}(b).$

Therefore, the gluing of B and C over $F$ is unique when it exists.

###### Proof.

We first show that for all $b\in B-F$ the equivalence class $[b]_{F}$ has a
minimum. For all $b\in B-F$ and $c\in C-F$, we have $cb\leq cb$, so $c\leq
cb/b$. By integrality we also have $cb\leq b\in B-F$, so $cb\in B-F$. Since
$cb,b\in B$ we have that $cb/b$ is an element of $B$ that is greater than some
element of $C$. Therefore, $cb/b\in F$. Also, since $cb\leq b$, we get
$b/cb=1\in F$, hence $[b]_{F}=[cb]_{F}$. Moreover, given
$b^{\prime}\in[b]_{F}$, we have $b^{\prime}/b\in F$, so $c\leq b^{\prime}/b$,
hence $bc\leq b^{\prime}$. Thus, $cb$ (and by symmetry also $bc$) is the
minimum of $[b]_{F}$, for all $c\in C-F$; we denote this minimum by
$\sigma_{F}(b)$. Note that since $F$-congruence classes are in particular
lattice congruence classes, $\sigma_{F}$ is monotone on $B-F$.

We now show that for all $b\in B-F$ the equivalence class $[b]_{F}$ has a
maximum. For any $c\in C-F$, we cannot have $c\backslash b\in C$, as then
$c^{\prime}\leq c\backslash b$ for some $c^{\prime}\in C-F$, so
$c^{\prime}c\leq b$ and $c^{\prime}c\leq c^{\prime}\in C-F$, so
$c^{\prime}c\in C-F$, which would imply $b\in F$, a contradiction. So,
$c\backslash b\in B-F$, and thus $\sigma_{F}(c\backslash b)=c(c\backslash
b)\leq b$ and so $\sigma_{F}(c\backslash b)\leq\sigma_{F}(b)$. Also, by
integrality we have $b\leq c\backslash b$, so
$\sigma_{F}(b)\leq\sigma_{F}(c\backslash b)$. Therefore,
$\sigma_{F}(b)=\sigma_{F}(c\backslash b)$, hence $[b]_{F}=[c\backslash
b]_{F}$. Also, for every $b^{\prime}\in[b]_{F}$, we have
$cb^{\prime}=\sigma_{F}(b^{\prime})=\sigma_{F}(b)\leq b$, so $b^{\prime}\ \leq
c\backslash b$. Therefore, for any $c\in C-F$, $c\backslash b$ (and by
symmetry $b/c$) is the maximum of $[b]_{F}$; we denote this element by
$\gamma_{F}(b)$.

We now prove the last property of compatibility, i.e., that $\sigma_{F}$ is
_absorbing_ : for all $b\in B-F$, $b\sigma_{F}[B-F]\subseteq\sigma_{F}[B-F]$
and $\sigma_{F}[B-F]b\subseteq\sigma_{F}[B-F]$. In particular, we show
$b\sigma_{F}[B-F]\subseteq\sigma_{F}[B-F]$, as the proof of
$\sigma_{F}[B-F]b\subseteq\sigma_{F}[B-F]$ is similar. Every element of
$b\sigma_{F}[B-F]$ is of the form $b\sigma_{F}(x)$ where $x\in B-F$. By the
above, for any $c\in C-F$,
$b\sigma_{F}(x)=bxc=\sigma_{F}(bx)\in\sigma_{F}[B-F]$.

Thus we showed that the congruence filter F is compatible with B, and also
that $\sigma_{F}(b)=cb=bc$ and $\gamma_{F}(b)=c\backslash b=b/c$. It follows
that the gluing over $F$ is unique when it exists. ∎

###  2.3. Compatible triples inside compatible pairs

Our aim is to characterize abstractly the individual components of the gluing
and later use them to define a gluing construction. In particular we identify
the structure of $(B-F)\cup\\{1\\}$; as this set is not closed under
divisions, we will need to make use of partially defined operations.

######  Definition 2.3.

By a _partial IRL_ we understand a partially ordered partial algebra
$({\textbf{A}},\leq)$ in the language of residuated lattices, such that:

  1. (1)

A is integral: $x\leq 1$ for all $x\in A$;

  2. (2)

the three axioms of RLs are satisfied whenever they can be applied, in the
following sense:

    1. (a)

$x\lor y$ is the least common upper bound of $x$ and $y$ whenever it exists,
and similarly $x\land y$ is the largest lower bound whenever it exists;

    2. (b)

$x1=1x=1$, and if $xy,(xy)z,yz,x(yz)$ are defined then $(xy)z=x(yz)$;

    3. (c)

if $xy$, $z/y$, and $x\backslash z$ are defined, then $xy\leq z$ if and only
if $x\leq z/y$ if and only if $y\leq x\backslash z$.

  3. (3)

multiplication is order preserving when defined: $a\leq b$ and $ac,bc$ defined
implies $ac\leq bc$, and likewise for left multiplication.

  4. (4)

whenever defined, the division operations are order-preserving in the
numerator and order-reversing in the denominator. That is: $x\leq y$ and
$z\backslash x,z\backslash y$ defined implies $z\backslash x\leq z\backslash
y$; $x\leq y$ and $y\backslash z,x\backslash z$ defined implies $y\backslash
z\leq x\backslash z$; likewise for right division.

For example, given an $F$-gluing of IRLs B and C, the structure
${\textbf{B}}^{\prime}$, whose domain is $B^{\prime}:=(B-F)\cup\\{1\\}$, is a
partial IRL.

######  Remark 2.4.

We wish to remark that, even though the definition of a partial IRL we are
using is quite general, the constructions we will define in the rest of the
paper really involve partial algebras that are much closer to being lattices:
joins will always be defined, and a meet $x\land y$ will always be defined
except if there is no common lower bound of $x$ and $y$. Moreover, the partial
IRLs considered in our constructions will actually have an underlying
structure of a _partial monoid_ in the stronger sense usually intended in the
literature: the products $xy,(xy)z$ are defined if and only if $yz,x(yz)$ are
defined, and in such case $(xy)z=x(yz)$.

We will now characterize abstractly triples of the form
$({\textbf{B}}^{\prime},\sigma_{F},\gamma_{F})$, where we mean that
$B^{\prime}:=(B-F)\cup\\{1\\}$.

A _lower-compatible triple_ $({\textbf{K}},\sigma,\gamma)$ consists of

  1. (1)

a partial IRL K with all operations defined, except for $x\backslash y$ and
$y/x$ which are undefined if and only if $\sigma(x)\leq y$ and $x\not\leq y$,

  2. (2)

$(\sigma,\gamma)$ is a residuated pair, i.e. $\sigma(x)\leq y$ if and only if
$x\leq\gamma(y)$, such that:

    1. (a)

$\sigma$ is a _strong conucleus_ , i.e, an interior operator such that for
$x,y\neq 1$, $x\sigma(y)=\sigma(xy)=\sigma(x)y$, and $\sigma(1)=1$.

    2. (b)

$\gamma$ is a closure operator on K, and

    3. (c)

$xy,yx\leq\sigma(x)$ for all $x,y\in K,y\neq 1$.

######  Lemma 2.5.

If $F$ is a compatible congruence filter of an IRL B, then
$({\textbf{B}}^{\prime},\sigma_{F},\gamma_{F})$ is a lower-compatible triple.

###### Proof.

For readability, in this proof we will write $\sigma$ for $\sigma_{F}$,
$\gamma$ for $\gamma_{F}$ and $\theta$ for $\theta_{F}$. It is clear that
$B^{\prime}=(B-F)\cup\\{1\\}$ is closed under multiplication, meet and also
under join except when $x\vee y\in F-\\{1\\}$; we redefine these joins to be
$1$ in ${\textbf{B}}^{\prime}$. Since B is an IRL and $B^{\prime}$ inherits
its operations, it can be directly checked that ${\textbf{B}}^{\prime}$ is a
partial IRL in the sense of Definition 2.3. With respect to the divisions, we
want to show that $x\backslash y$ and $y/x$ are undefined if and only if
$\sigma(x)\leq y$ and $x\not\leq y$. Notice that the divisions $x\backslash y$
and $y/x$ are undefined in $B^{\prime}$ iff they produce elements of
$F-\\{1\\}$. From $x\backslash y\in F-\\{1\\}$ we get that $x\not\leq y$, and
moreover $f\leq x\backslash y$ for some $f\in F-\\{1\\}$. Thus by residuation
$xf\leq y$. Thus, since $xf\in[x]_{F}$ (because $xf\leq x$ and $f\leq
x\backslash xf$), we have $\sigma(x)=\min[x]_{F}\leq xf\leq y$. Similarly we
can prove that if $y/x$ is not defined in $B^{\prime}$ then again
$\sigma(x)\leq y$ and $x\not\leq y$. Conversely, suppose $\sigma(x)\leq y$ and
$x\not\leq y$. Then $x\backslash\sigma(x)\leq x\backslash y$ and since
$x\backslash\sigma(x)\in F$, we get $x\backslash y\in F$. Moreover, since
$x\not\leq y$, we get $x\backslash y\not=1$.

Note that $B^{\prime}$ is closed under meet as all elements of $B-F$ are below
all elements of $F$ and it is closed under multiplication due to integrality
and order preservation of multiplication. Also, it is closed under joins that
do not produce elements of $F-\\{1\\}$ and the ones that do produce such
elements are redefined to be $1$. The resulting structure is a monoid and a
lattice. Finally, if $x\backslash^{{\textbf{B}}}y\not\in F-\\{1\\}$, then
residuation holds as all terms are evaluated in $B^{\prime}$.

We now prove that $\sigma$ is a strong conucleus. Clearly, $\sigma(x)\leq x$
and $\sigma(\sigma(x))=\sigma(x)$ thus $\sigma$ is decreasing and idempotent.
We now prove that $\sigma$ is monotone. Suppose $x\leq y$, with $x,y\in
B^{\prime}$. Since $\sigma(x)\mathrel{\theta}x$ and
$\sigma(y)\mathrel{\theta}y$, we have
$\sigma(x)\land\sigma(y)\mathrel{\theta}x\land y=x\mathrel{\theta}\sigma(x)$.
Thus $\sigma(x)\leq\sigma(x)\land\sigma(y)$ (since $\sigma(x)$ is the smallest
element in the equivalence class), thus $\sigma(x)\leq\sigma(y)$. We will now
use the absorbing property of $\sigma$ in order to show that it is a strong
conucleus. We show that $x\sigma(y)=\sigma(xy)=\sigma(x)y$ for $x$ and $y$ not
equal to $1$. Now, $x\sigma(y)\in x\sigma[B-F]\subseteq\sigma[B-F]$, thus
$x\sigma(y)=\sigma(z)$ for some $z\in B-F$. But then since $\sigma$ is
idempotent $\sigma(x\sigma(y))=\sigma(\sigma(z))=\sigma(z)=x\sigma(y)$. Since
$\sigma$ is decreasing and order preserving, we get
$x\sigma(y)=\sigma(x\sigma(y))\leq\sigma(xy)$. Moreover, since
$x\mathrel{\theta}x$ and $y\mathrel{\theta}\sigma(y)$, we get
$xy\mathrel{\theta}x\sigma(y)$, thus $\sigma(xy)=\sigma(x\sigma(y))\leq
x\sigma(y)$, and this shows that $\sigma(xy)=x\sigma(y)$. Similarly, using
$\sigma[B-F]x\subseteq\sigma[B-F]$, we can show that $\sigma(y)x=\sigma(xy)$.

We now prove that $\gamma$ is a closure operator. Since
$\gamma(x)=\max[x]_{F}$, it is easy to see that it is increasing and
idempotent. To show that $\gamma$ is monotone, suppose $x\leq y$, with $x,y\in
B-F$. Since $\gamma(x)\mathrel{\theta}x$ and $\gamma(y)\mathrel{\theta}y$, we
have $\gamma(x)\lor\gamma(y)\mathrel{\theta}x\lor
y=y\mathrel{\theta}\gamma(y)$. Thus $\gamma(x)\lor\gamma(y)\leq\gamma(y)$
(since $\gamma(y)$ is the biggest in the equivalence class), thus
$\gamma(x)\leq\gamma(y)$.

We now show that $(\sigma,\gamma)$ is a residuated pair. If $\sigma(x)\leq y$
then by idempotency and monotonicity of $\sigma$ we get that
$\sigma(x)=\sigma\sigma(x)\leq\sigma(y)$. Then
$\gamma(\sigma(x))\leq\gamma(\sigma(y))$, and considering that it follows from
their definition that $\gamma\circ\sigma=\gamma$, we have
$x\leq\gamma(x)\leq\gamma(y)$. Similarly, if $x\leq\gamma(y)$ then
$\gamma(x)\leq\gamma(y)$, thus applying $\sigma$ we get that
$\sigma(x)\leq\sigma(y)\leq y$.

It is only left to prove that $xy,yx\leq\sigma(x)$ for all $x,y\in
B^{\prime},y\neq 1$. This easily follows from residuation, since for instance:
$xy\leq\sigma(x)$ if and only if $y\leq x\backslash\sigma(x)$, which holds
since $x\backslash\sigma(x)\in F$. ∎

We say that $({\textbf{B}}^{\prime},\sigma_{F},\gamma_{F})$ is the compatible
triple of the compatible pair $({\textbf{B}},F)$. We can also show that every
compatible triple comes from a compatible pair.

######  Lemma 2.6.

Every lower-compatible triple $({\textbf{K}},\sigma,\gamma)$ is the compatible
triple of the lower-compatible pair $({\textbf{B}},G)$, where $G$ is the
2-element IRL and B is an IRL with operations extending
$({\textbf{K}}-\\{1\\})\cup{\textbf{G}}$.

###### Proof.

Let $B=(K-\\{1\\})\cup G$ where $G=\\{f,1\\}$. We extend the operations of K
to B except when $x\lor y=1$ in $K$, in which case we redefine
$x\lor^{{\textbf{B}}}y=f$; moreover we stipulate that:

  * •

$f$ is an idempotent coatom strictly above all elements of $K-\\{1\\}$;

  * •

$f\cdot x=x\cdot f=\sigma(x)$, $f\backslash x=x/f=\gamma(x)$, $x\backslash
f=f/x=1$ for all $x\in K-\\{1\\}$.

  * •

$x\backslash y=y/x=f$ for $x,y\in B$ such that $x\not\leq y$ and
$\sigma(x)\leq y$ (i.e., when $x\backslash^{{\textbf{K}}}y,y/^{{\textbf{K}}}x$
are undefined).

We first show that B is a residuated lattice. It can be easily seen that
$(B,\land,\lor,1)$ is a lattice with top $1$. In order to see that
$(B,\cdot,1)$ is a monoid, we need to prove associativity of the product in
triples of elements where $f$ is involved, as in the other cases associativity
follows from the associativity in K. We will make use of the absorption of
$\sigma$. For example, if $b,d\in K-\\{1\\}$ (the other cases are similar):

$b(df)=b\sigma(d)=\sigma(bd)=(bd)f,$
$b(fd)=b\sigma(d)=\sigma(bd)=\sigma(b)d=(bf)d.$

For residuation, notice first that if one among $x,y,z$ is $1$ then the law
clearly holds. First we check the cases where $f$ is involved. For $b,d\in
K-\\{1\\}$, we show that:

$bf\leq d\mbox{ iff }f\leq b\backslash d\mbox{ iff }b\leq d/f.$

Indeed, $bf\leq d\mbox{ iff }\sigma(b)\leq d$ iff [$b\leq d$ or ($b\not\leq d$
and $\sigma(b)\leq d$)] iff ($b\backslash d=1$ or $b\backslash d=f$) iff
$f\leq b\backslash d$. Moreover, $bf\leq d\mbox{ iff }\sigma(b)\leq d$ iff
$b\leq\gamma(d)$ iff $b\leq d/f$, where we used that $(\sigma,\gamma)$ is a
residuated pair. Likewise we obtain $fb\leq d\mbox{ iff }b\leq f\backslash
d\mbox{ iff }f\leq d/b.$ Also,

$bd\leq f\mbox{ iff }d\leq b\backslash f\mbox{ iff }b\leq f/d$

holds as all these statements are true even for $b=f$ and $d\in K-\\{1\\}$ and
also for $d=f$ and $b\in K-\\{1\\}$. Moreover, $ff\leq b$ iff $f\leq b/f$ iff
$f\leq f\backslash b$ since all inequalities are false. Now for $x,y,z\in
K-\\{1\\}$, we want to show that

$xy\leq z\mbox{ iff }y\leq x\backslash z\mbox{ iff }x\leq z/y.$

We show that $xy\leq z\mbox{ iff }y\leq x\backslash z$, as the other
equivalence is obtained similarly. We consider three different cases.

  * •

If $x\leq z$, then both inequalities are true, since we get $xy\leq x\leq z$
and $x\backslash z=1$ thus $y\leq x\backslash z$.

  * •

If $x\not\leq z$, and $\sigma(x)\leq z$, we have to show that $xy\leq z$ iff
$y\leq f$. Notice that $y\leq f$ since $y\neq 1$. Moreover, since $y\neq 1$
then $xy\leq\sigma(x)\leq z$.

  * •

If $\sigma(x)\not\leq z$, then the operations are defined in $K$ and
residuation holds.

Thus B is an integral residuated lattice.

We want to show that $({\textbf{B}},G)$ is a compatible pair. $G$ is closed
under products since $f$ is idempotent and it is closed under conjugates since
$(x\cdot f)/x=\sigma(x)/x\in\\{f,1\\}$ and $x\backslash(f\cdot
x)=x\backslash\sigma(x)\in\\{f,1\\}$; so $G$ is a congruence filter of $B$.

We now show that $\sigma(x)=\min[x]_{G}=\sigma_{G}(x)$ and
$\gamma(x)=\max[x]_{G}=\gamma_{G}(x)$ for all $x\in K-\\{1\\}$. Note that
$\sigma(x)\leq x$ implies $\sigma(x)\backslash x=1\in G$; also
$x\backslash\sigma(x)$ is either $1$ or $f$, thus still in $G$. Therefore,
$\sigma(x)\in[x]_{G}$. Furthermore, if $y\in[x]_{G}$, then $x\backslash y\in
G$, thus either $x\backslash y=1$ or $x\backslash y=f$. If $x\backslash y=1$
then $\sigma(x)\leq x\leq y$, while if $x\backslash y=f$, then $xf\leq y$, so
$\sigma(x)\leq y$. Thus in any case $\sigma(x)\leq y$, which implies that
$\sigma(x)=\min[x]_{G}=\sigma_{G}(x)$.

We now prove that $\gamma(x)=\gamma_{G}(x)$ for all $x\in K-\\{1\\}$. Note
that $x\leq\gamma(x)$ implies $x\backslash\gamma(x)=1\in G$; also
$\gamma(x)\backslash x$ is either $1$ if $x=\gamma(x)$ or $f$ otherwise, since
$\sigma(\gamma(x))\leq x$ follows from the fact that $\sigma,\gamma$ form a
residuated pair. Therefore, $\gamma(x)\in[x]_{G}$. Now, if $y\in[x]_{G}$, then
$y\backslash x\in G$, thus either $y\backslash x=1$ or $y\backslash x=f$. If
$y\backslash x=1$ then $y\leq x\leq\gamma(x)$, while if $y\backslash x=f$,
then $yf\leq x$, so $\sigma(y)\leq x$, if and only if $y\leq\gamma(x)$. Thus
$\gamma(x)=\max[x]_{G}=\gamma_{G}(x)$.

To prove that $\sigma$ is absorbing, we use the fact that $\sigma$ is a strong
conucleus. For any $x$ and $y\in B-G$, we have
$x\sigma(y)=\sigma(xy)\in\sigma[B-G]$, so $x\sigma[B-G]\subseteq\sigma[B-G]$
and similarly $\sigma[B-G]x\subseteq\sigma[B-G]$. Thus we proved that
$({\textbf{B}},G)$ is a compatible pair, and since
$\sigma=\sigma_{G},\gamma=\gamma_{G}$, and $K=(B-F)\cup\\{1\\}$, it follows
that $({\textbf{K}},\sigma,\gamma)$ is its compatible triple. ∎

If 2 denotes the two-element residuated lattice, we have also shown the
following.

######  Corollary 2.7.

If $({\textbf{B}},F)$ is a lower-compatible pair, then so is
$({\textbf{B}}_{F},\mathbf{2})$, where $B_{F}=(B-F)\cup 2$.

We say that an IRL F _fits_ with a lower-compatible triple
$({\textbf{K}},\sigma,\gamma)$, if $({\textbf{K}}-\\{1\\})\cup{\textbf{F}}$
extends to an IRL B, $F$ is a compatible filter of B, and the compatible
triple of the compatible pair $({\textbf{B}},F)$ is
$({\textbf{K}},\sigma,\gamma)$. Note that if $({\textbf{B}},F)$ is a
compatible pair, then $F$ fits with the compatible triple
$({\textbf{B}}^{\prime},\sigma_{F},\gamma_{F})$, but Corollary 2.7 shows that
the same ${\textbf{B}}^{\prime}$ can belong to different lower-compatible
triples.

###  2.4. Fitting the components together: the construction

Now, we define the $F$-gluing construction given the individual pieces we have
identified, provided that they fit together in a suitable way.

Consider a lower-compatible pair $({\textbf{B}},F)$, and an IRL C such that
$B\cap C=F$, with $F$ strictly above all other elements in $C$ and such that
if there are elements in $B-F$ joining to some element of $F$ then $C$ has a
least element $0_{C}$. We will show that there is an IRL that is the (unique)
gluing of B and C over $F$ and that it is the following structure:

${\textbf{B}}\oplus_{F}{\textbf{C}}=(B\cup
C,\,\cdot_{F},\backslash_{F},/_{F},\land_{F},\lor_{F},0,1),$

where the operations are defined as follows:

$\displaystyle x\cdot_{F}y$ $\displaystyle=\left\\{\begin{array}[]{ll}x\cdot
y&\mbox{ if }x,y\in B,\mbox{ or }x,y\in C\\\ \sigma_{F}(x)&\mbox{ if }y\in
C-F,x\in B-F\\\ \sigma_{F}(y)&\mbox{ if }x\in C-F,y\in B-F\end{array}\right.$
$\displaystyle x\backslash_{F}\,y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\backslash y&\mbox{ if }x,y\in
B,\mbox{ or }x,y\in C\\\ \gamma_{F}(y)&\mbox{ if }x\in C-F,y\in B-F\\\
1&\mbox{ if }x\in B-F,y\in C-F\end{array}\right.$ $\displaystyle x/_{F}\,y$
$\displaystyle=\left\\{\begin{array}[]{ll}x/y&\mbox{ if }x,y\in B,\mbox{ or
}x,y\in C\\\ \gamma_{F}(x)&\mbox{ if }y\in C-F,x\in B-F\\\ 1&\mbox{ if }x\in
C-F,y\in B-F\end{array}\right.$ $\displaystyle x\land_{F}y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\land y&\mbox{ if }x,y\in B,\mbox{
or }x,y\in C\\\ x&\mbox{ if }x\in B-F,y\in C-F\\\ y&\mbox{ if }y\in B-F,x\in
C-F\end{array}\right.$ $\displaystyle x\lor_{F}y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\lor y&\mbox{ if }x,y\in C,\mbox{
or }x,y\in B\mbox{ with }x\lor y\not\in F\\\ 0_{C}&\mbox{ if }x,y\in B-F,x\lor
y\in F\\\ y&\mbox{ if }x\in B-F,y\in C-F\\\ x&\mbox{ if }y\in B-F,x\in
C-F\end{array}\right.$

The proof of the following theorem is postponed until the next section where
we further expand the construction. More precisely, it will be a direct
consequence of Theorem 3.10.

$F$$F$$F$$1$B$\oplus_{F}$$1$C${=}$$1$${\textbf{B}}\oplus_{F}{\textbf{C}}$
Figure 1. The gluing ${\textbf{B}}\oplus_{F}{\textbf{C}}$ of the algebras B
and C over $F$.

######  Theorem 2.8.

If $F$ is a congruence filter of an IRL C and also a compatible congruence
filter of an IRL B, then ${\textbf{B}}\oplus_{F}{\textbf{C}}$ is the gluing of
B and C over $F$.

###  2.5. Gluing without the filter

We now obtain a different construction, that generalizes the $1$-sum
construction in a different way: it glues together two structures that
intersect at the top $1$ and maintains the same order relation, but some of
the divisions are redefined. With respect to the previous construction, the
intuition here is that we are removing the filter $F$ (keeping the unit $1$),
and what is left is a partial IRL.

We start from a lower compatible triple $({\textbf{K}},\sigma,\gamma)$ and a
partial IRL L where some of the divisions might not be defined. In particular,
$x\backslash y$ is undefined if and only if all elements $z$ in the interval
$[y,1)=\\{z\in L:y\leq z<1\\}$ are such that $xz\leq y$ but $[y,1)$ does not
have a top element. Similarly, $y/x$ is undefined if and only if all elements
$z$ the interval $[y,1)$ are such that $zx\leq y$ and $[y,1)$ has no top. If L
has a splitting coatom $c_{L}$ (i.e., $L=\\{1\\}\cup{\downarrow}c_{L}$), all
divisions in $L$ are defined. We also assume that $K\cap L=\\{1\\}$ and if
$x\lor y=1$ in $K$ for some $x,y\in K-\\{1\\}$, then $L$ has a bottom element
$0_{L}$.

We set $\pi=(\sigma,\gamma)$ and we define
${\textbf{K}}\oplus_{\pi}{\textbf{L}}$ to be the structure where the
operations extend those of $K$ and $L$, except if $x\lor y=1$ in $K$ then we
redefine $x\lor y=0_{L}$. Moreover:

$\displaystyle xy$ $\displaystyle=\left\\{\begin{array}[]{ll}\sigma(x)&\mbox{
if }y\in L-\\{1\\},x\in K-\\{1\\}\\\ \sigma(y)&\mbox{ if }x\in L-\\{1\\},y\in
K-\\{1\\}\end{array}\right.$ $\displaystyle x\backslash\,y$
$\displaystyle=\left\\{\begin{array}[]{ll}c_{L}&\mbox{ if }x,y\in K,L\mbox{
has a splitting coatom }c_{L},\mbox{ and $x\backslash^{{\textbf{K}}}y$ is
undefined}\\\ \gamma(y)&\mbox{ if }x\in L-\\{1\\},y\in K-\\{1\\}\\\ 1&\mbox{
if }x\in K-\\{1\\},y\in L\\\ \end{array}\right.$ $\displaystyle y/\,x$
$\displaystyle=\left\\{\begin{array}[]{ll}c_{L}&\mbox{ if }x,y\in K,L\mbox{
has a splitting coatom }c_{L},\mbox{ and $y/^{{\textbf{K}}}x$ is undefined}\\\
\gamma(y)&\mbox{ if }x\in L-\\{1\\},y\in K-\\{1\\}\\\ 1&\mbox{ if }x\in
K-\\{1\\},y\in L\\\ \end{array}\right.$ $\displaystyle x\land y$
$\displaystyle=\begin{array}[]{ll}x&\mbox{ if }x\in K-\\{1\\},y\in
L\end{array}.$ $\displaystyle x\lor y$
$\displaystyle=\begin{array}[]{ll}y&\mbox{ if }x\in K-\\{1\\},y\in
L\end{array}.$

######  Proposition 2.9.

${\textbf{K}}\oplus_{\pi}{\textbf{L}}$ is a partial IRL and it is total if $L$
has a splitting coatom.

###### Proof.

First, we notice that all operations are total except possibly the divisions.
The meet and join operation clearly define a lattice, and $x\leq 1$ for every
element $x$ in the gluing. Associativity of the product can be easily shown by
the definition using the idempotency of $\sigma$ and the strong conuclearity
condition. Since the operations of ${\textbf{K}}\oplus_{\pi}{\textbf{L}}$
extend the ones of K and L, ${\textbf{K}}\oplus_{\pi}{\textbf{L}}$ has an
underlying monoidal structure.

We show that residuation holds. That is, for all elements $x,y,z$, whenever
$xy,x\backslash z,z/y$ are defined:

$xy\leq z\mbox{ iff }y\leq x\backslash z\mbox{ iff }x\leq z/y$

  * •

In the case where $x,y\in L,z\in K-\\{1\\}$ none of the inequalities are true,
as can be seen by the definition of the order and the divisions.

  * •

In the case where $x,z\in L,y\in K-\\{1\\}$ all the inequalities hold as it
follows from the definition of the order and of the divisions. The situation
is similar when: $x\in K-\\{1\\},y,z\in L$, as well as when $x,y\in
K-\\{1\\},z\in L$.

  * •

If $x\in L-\\{1\\},y,z,\in K-\\{1\\}$, we need to verify:

$\sigma(y)\leq z\mbox{ iff }y\leq\gamma(z)\mbox{ iff }x\leq z/y$

whenever $z/y$ is defined. The first equivalence holds since the two maps form
a residuated pair. We now show $\sigma(y)\leq z\Leftrightarrow x\leq z/y$
holds in case $z/y$ is defined in ${\textbf{K}}\oplus_{\pi}{\textbf{L}}$. If
$y\leq z$, then $\sigma(y)\leq\sigma(z)\leq z$ and $x\leq 1=z/y$, so both
inequalities hold. Assume now that $y\not\leq z$. If $\sigma(y)\leq z$ we get
that $z/y$ is undefined in K, thus $z/y$ is the coatom of $L$ (supposing $z/y$
is defined in ${\textbf{K}}\oplus_{\pi}{\textbf{L}}$), so $x\leq z/y$ holds.
Conversely, if $x\leq z/y$ then since $x\in L$, we get that either $z/y=1$ or
it is the coatom of $L$. Thus either $y\leq z$ (which is against our
assumption) or $z/y$ is undefined in $K$, which means that $\sigma(y)\leq z$.
The case: $x,z\in K-\\{1\\},y\in L-\\{1\\}$ is similar.

  * •

Suppose now $x,y,z\in K-\\{1\\}$. We only show $xy\leq z\Leftrightarrow y\leq
x\backslash z$ assuming that $x\backslash z$ is defined in
${\textbf{K}}\oplus_{\pi}{\textbf{L}}$; the proof of the equivalence $xy\leq
z\Leftrightarrow x\leq z/y$ is similar. If $x\backslash z$ is defined in $K$,
the equivalence holds by residuation in K. Also, if $x\leq z$, then $xy\leq
zy\leq z$ and $y\leq 1=x\backslash z$, so both inequalities hold. Now, if
$x\backslash^{{\textbf{K}}}z$ is undefined and $\sigma(x)\leq z$, then we get
$xy\leq\sigma(x)\leq z$ (where the first inequality is due to (2c) in the
definition of lower-compatible triple) and also that $x\backslash z$ is equal
to the coatom of $L$, so again both inequalities hold.

It is left to show that multiplication is order preserving, divisions are
order preserving in the numerator and order reversing in the denominator. The
fact that the monoidal operation is order preserving can be easily checked
using the facts that by the definition of lower-compatible triple $\sigma$ is
order preserving, and $xy,yx\leq\sigma(x)$ for all $x,y\in K,y\neq 1$.
Finally, the order properties of divisions (when defined) can be directly
checked, and follow by residuation and the order preservation of $\gamma$.

We have shown that ${\textbf{K}}\oplus_{\pi}{\textbf{L}}$ is a partial IRL. In
the case where $L$ has a coatom, all divisions (the only partial operations)
are correctly and fully defined. Thus in that case
${\textbf{K}}\oplus_{\pi}{\textbf{L}}$ is a (total) IRL. ∎

We will refer to ${\textbf{K}}\oplus_{\pi}{\textbf{L}}$ as the _partial upper
gluing_ of K and L. If we take a lower-compatible pair $({\textbf{B}},F)$ and
an IRL C such that $B\cap C=F$ with $F$ strictly above the other elements in
$C$, we can construct the partial gluing of the compatible triple
$({\textbf{B}},\sigma_{F},\gamma_{F})$ and of $C-F$, which is a partial IRL
with the required properties for the partial gluing construction. We will see
interesting examples of these constructions in the final section of the paper.

##  3\. Gluing over a filter and an ideal

We can take the previous intuition even further and generalize the
construction allowing the algebras B and C to intersect on both a congruence
filter $F$ and a lattice ideal $I$. Let D be an IRL with underlying set $B\cup
C$, where $B\cap C=F\cup I$, with $F$ a congruence filter as before, $I$ a
lattice ideal, $i<b<c<f$ for all $i\in I$, $b\in B^{-}:=B-(F\cup I)$ $c\in
C^{-}:=C-(F\cup I)$ and $f\in F$, $C$ is a subalgebra except possibly for
$\land$ and $B$ is a subalgebra except possibly with respect to $\lor$. We say
that D is _a gluing of B and C over $F$ and $I$_, or _$F-I$ -gluing of B and
C_. As before, we will identify conditions on B, C, $F$ and $I$ that will
allow us to construct D from these constituent parts. We will characterize the
structure on the subset $C^{\prime}=(C-I)$.

###  3.1. Compatibility with an ideal

We call an element $c\in C^{-}$ a _left $I$-divisor_ if there exists another
$c^{\prime}\in C^{-}$ such that $cc^{\prime}\in I$, and a _right $I$-divisor_
if instead there exists $c^{\prime\prime}\in C^{-}$ such that
$c^{\prime\prime}c\in I$.

We say that an ideal $I$ of an IRL C is _compatible_ with C if it is strictly
below $C^{\prime}$ and for all left $I$-divisors $c\in C^{\prime}$ and right
$I$-divisors $d\in C^{\prime}$, the sets $\\{c\backslash i:i\in I\\}$ and
$\\{i/d:i\in I\\}$ have maxima. In this case we denote these elements by
$\ell_{I}(c)$ and $r_{I}(d)$, respectively:

$\ell_{I}(c)={\rm max}\\{c\backslash i:i\in I\\},\;\;r_{I}(d)={\rm
max}\\{i/d:i\in I\\}.$

We also say that $({\textbf{C}},I)$ is a _upper-compatible pair_.

######  Lemma 3.1.

Given a gluing of B and C over $F$ and $I$, the lattice ideal $I$ is
compatible with C. Also, for every left $I$-divisor $c\in C-I$, right
$I$-divisor $d\in C-I$, and $b\in B-F$:

$c\backslash b=\ell_{I}(c),\qquad b/d=r_{I}(d).$

###### Proof.

The ideal $I$ is strictly below all elements of $C^{\prime}=C-I$ by
definition. We now show that the maps $\ell_{I}$ and $r_{I}$ are defined for
all, respectively, left and right $I$-divisors. For a left $I$-divisor $c\in
C^{-}$ we consider $c\backslash b$ for some $b\in B^{-}$. We claim that
$c\backslash b\in C^{-}$. Indeed by definition there is at least one element
in $C^{-}$ that multiplied to the right of $c$ gives an element of $I$, thus
below $b$, and moreover there cannot be $f\in F$ such that $cf\in I$. Indeed
otherwise, we would have $cf\leq i$ for some $i\in I$ iff (by residuation)
$c\leq i/f$, which yields $i/f\in F$ (since $i/f\in B\cap C$); this would
imply that $f^{\prime}f\leq i$ for some $f^{\prime}\in F$, so we would get
$i\in F$, a contradiction. Thus $c\backslash b=\max\\{d\in C^{-}:cd\in I\\}$.

We now show that $\max\\{d\in C^{-}:cd\in I\\}=\max\\{c\backslash i:i\in
I\\}$. Indeed, since $c\backslash b=\max\\{d\in C^{-}:cd\in I\\}$, there
exists $j\in I$, with $c(c\backslash b)\leq j$; thus $c\backslash b\leq
c\backslash j$. Also, for all $i\in I$, we have $c\backslash i\leq c\backslash
b$ since $i<b$; so $c\backslash i\leq c\backslash b\leq c\backslash j$. Since
$j\in I$, we get that $c\backslash b=c\backslash j$ and also $c\backslash
b=\max\\{c\backslash i:i\in I\\}$. Thus $\max\\{c\backslash i:i\in I\\}$
exists for every left $I$-divisor $c$ in $C^{\prime}$ and $c\backslash
b=\ell_{I}(c)$. Similarly, one can show that $\max\\{i/d:i\in I\\}$ exists for
every right $I$-divisor $d$ of $C^{\prime}$. ∎

What we showed in the previous section about lower-compatible pairs still
holds in case there is at least one non $I$-divisor. Otherwise, we need
consider the weaker notion of lower-compatibility.

######  Definition 3.2.

We say that a pair $({\textbf{B}},F)$ is a _weak lower-compatible pair_ if it
respects all the conditions of a lower-compatible pair except that
$\gamma_{F}$ need not be defined, i.e. the equivalence classes $[b]_{F}$ for
$b\in B-F$ do not need to have a maximum element.

We can indeed prove the following analogue of Lemma 2.2.

######  Lemma 3.3.

Given a gluing of B and C over $F$ and $I$, $({\textbf{B}},F)$ is a weak
lower-compatible pair. If there exists an element of C that is not an
$I$-divisor, then $({\textbf{B}},F)$ is a lower-compatible pair. Moreover, for
all $b\in B-F,c\in C-(F\cup I)$:

$cb=bc=\sigma_{F}(b),$ $c\backslash b=\gamma_{F}(b)\mbox{ if c is not a left
$I$-divisor, }$ $b/c=\gamma_{F}(b)\mbox{ if c is not a right $I$-divisor. }$

###### Proof.

The proof that in any case $({\textbf{B}},F)$ is a weak lower compatible
triple is the same as the one of Lemma 2.2. If given some $c\in C-(F\cup I)$,
there is no element $c^{\prime}\in C-(F\cup I)$ such that $cc^{\prime}\in I$,
then also the proof about $[b]_{F}$ having a maximum can be replicated in
exactly the same way. Indeed, we get that then $c\backslash b\in B$,
$[b]_{F}=[c\backslash b]_{F}$ and $b^{\prime}\leq c\backslash b$ for all
$b^{\prime}\in[b]_{F}$. ∎

###  3.2. Compatible quadruples

Consider now a weak lower-compatible pair $({\textbf{B}},F)$, and an upper-
compatible pair $({\textbf{C}},I)$ such that $B\cap C=F\cup I$ is a subalgebra
of both B and C, with $F$ strictly above all elements in $C-F$ and $I$
strictly below all elements of $C-I$. We call the quadruple
$({\textbf{B}},F,{\textbf{C}},I)$ _compatible_ if moreover:

  1. (1)

if at least an element of $C-F$ is not an $I$-divisor, then $({\textbf{B}},F)$
is a lower-compatible pair.

  2. (2)

Whenever $c,d\in C-I$ and $cd\in I$, we have $(cd)x=x(cd)=\sigma_{F}(x)$ for
all $x\in B^{-}$.

  3. (3)

If there are elements $x,y\in B-F$ such that $x\lor y\in F$, then $C-I$ has a
bottom element $\bot_{C}$.

  4. (4)

If there are elements $x,y\in C-I$ such that $x\land y\in I$, then $B-F$ has a
top element $\top_{B}$.

We recall that $\sigma_{F}(x)=\min[x]_{F},$ and whenever defined
$\gamma_{F}(x)=\max[x]_{F}$, $\ell_{I}(x)=\max\\{x\backslash i:i\in
I\\},\;r_{I}(x)=\max\\{i/x:i\in I\\}$.

######  Proposition 3.4.

Given a gluing of IRLs B and C over $F$ and $I$,
$({\textbf{B}},F,{\textbf{C}},I)$ is a compatible quadruple. Moreover, the
gluing of B and C over $F$ and $I$ is unique when it exists.

###### Proof.

The fact that $({\textbf{B}},F)$ is a weak lower compatible pair, together
with conditions 1 and 2, are shown in Lemma 3.3. The fact that
$({\textbf{C}},I)$ is an upper-compatible pair is shown in Lemma 3.1.
Conditions 3 and 4 are clearly properties of the lattice ordering of the
gluing. ∎

The following technical properties of a compatible quadruple will be useful in
what follows.

######  Lemma 3.5.

If $({\textbf{B}},F,{\textbf{C}},I)$ is a compatible quadruple, the following
properties hold.

  1. (1)

For all $x\in F,y\in B-F$,
$x\sigma_{F}(y)=\sigma_{F}(xy)=\sigma_{F}(yx)=\sigma_{F}(y)x=\sigma_{F}(y)$.

  2. (2)

For every $c\in C^{-}$ and $f\in F$, we have $cf,fc\in C^{-}$.

###### Proof.

(1) Since $x\mathrel{\theta}1$ and $y\mathrel{\theta}y$, we have
$xy\mathrel{\theta}y$ which implies $\sigma_{F}(xy)=\sigma_{F}(y)$.

From $x\mathrel{\theta}x$ and $y\mathrel{\theta}\sigma_{F}(y)$ we get
$xy\mathrel{\theta}x\sigma_{F}(y)$, thus
$\sigma_{F}(xy)=\sigma_{F}(x\sigma_{F}(y))\leq x\sigma_{F}(y)$. Moreover, from
$x\mathrel{\theta}1$ and $y\mathrel{\theta}\sigma_{F}(y)$ we have
$xy\mathrel{\theta}\sigma_{F}(y)$, which implies
$\sigma_{F}(xy)=\sigma_{F}(\sigma_{F}(y))=\sigma_{F}(y)\geq x\sigma_{F}(y)$;
thus $\sigma_{F}(xy)=x\sigma_{F}(y)$. The other equalities can be proven
analogously.

(2) Suppose by way of contradiction that $cf=i\in I$. Thus by residuation
$c\leq i/f$, but since $i,f\in B\cap C$, and $B\cap C$ is a subalgebra of both
B and C, we get $i/f\in F$, so $i\geq(i/f)f\in F$, hence $i\in F$, a
contradiction. Similarly one can show that $fc\in C^{-}$. ∎

###  3.3. Abstracting the upper part

We now characterize abstractly the triples of the form
$({\textbf{C}}^{\prime},\ell_{I},r_{I})$, where we understand
$C^{\prime}=(C-I)$ and $I$ is an ideal strictly below $C^{\prime}$. A triple
$({\textbf{L}},\ell,r)$ is called _upper-compatible_ , if L is a partial IRL
and:

  1. (1)

$\ell,r$ are partial maps on $L$ that form a Galois connection; more precisely
$\ell(y)$ is defined and $x\leq\ell(y)$ if and only if $r(x)$ is defined and
$y\leq r(x)$.

  2. (2)

If $r(y^{\prime})$ is defined, $x\leq r(y^{\prime})$ and $y\leq y^{\prime}$,
then $r(y)$ is defined and $x\leq r(y)$. Thus the domain $D_{r}$ of $r$ is
downwards closed. Also, the same holds for $\ell$.

  3. (3)

$xy$ is undefined iff $y$ is in the domain of $r$ and $x\leq r(y)$, iff $x$ is
in the domain of $\ell$ and $y\leq\ell(x)$;

  4. (4)

$x\backslash y$ is undefined iff there is no $z$ with $xz\leq y$, and $y/x$ is
undefined iff there is no $z$ with $zx\leq y$.

  5. (5)

If $\ell(x)$ and $r(z)$ are defined then: $x\backslash r(z)$ is defined iff
$\ell(x)/z$ is defined, and in such a case $x\backslash r(z)=\ell(x)/z$.

  6. (6)

If $\ell(x)$ is undefined and $r(z)$ is defined, then $x\backslash r(z)=r(z)$.
If $r(z)$ is undefined and $\ell(y)$ is defined, then $\ell(y)/z=\ell(y)$.

  7. (7)

If $\ell(x)$ is defined, then $x\backslash z$ is defined and $\ell(x)\leq
x\backslash z$. Similarly, if $r(x)$ is defined, then $w/x$ is defined and
$r(x)\leq w/x$.

  8. (8)

$x\land y$ is undefined iff there is no $z\leq x,y$.

  9. (9)

All other operations are defined.

######  Lemma 3.6.

In an upper-compatible triple, if $xy$ is defined, then: $xy\leq z$ iff
($x\backslash z$ is defined and $y\leq x\backslash z$) iff ($z/y$ is defined
and $x\leq z/y$).

###### Proof.

Suppose that $xy$ is defined. If $xy\leq z$, then $x\backslash z$ is defined,
since there is $y$ such that $xy\leq z$, and so residuation holds. Conversely,
suppose $x\backslash z$ is defined and $y\leq x\backslash z$. Then
$x(x\backslash z)$ is defined, since otherwise we would have: $x\in D_{\ell}$,
$x\backslash z\leq\ell(x)$, and since $y\leq x\backslash z\leq\ell(x)$, then
$xy$ would be undefined, a contradiction. Thus we get $xy\leq x(x\backslash
z)\leq z$, by order preservation of multiplication. Similarly one can prove
that $xy\leq z$ if and only if $z/y$ is defined and $x\leq z/y$. ∎

######  Lemma 3.7.

If $I$ is a compatible ideal of an IRL C, then
$({\textbf{C}}^{\prime},\ell_{I},r_{I})$ is an upper-compatible triple.

###### Proof.

We show that $({\textbf{C}}^{\prime},\ell_{I},r_{I})$ has the properties of an
upper-compatible triple, recalling that $C^{\prime}=C-I$. It is easy to check
that ${\textbf{C}}^{\prime}$ is a partial IRL, in particular:

  1. (1)

Integrality is clearly satisfied;

  2. (2)

The three axioms of RLs are satisfied whenever they can be applied, in
particular:

    1. (a)

with respect to the lattice operations, the joins are always defined, $1$ is
the largest element, and $x\land y$ is undefined iff there is no common lower
bound of $x$ and $y$;

    2. (b)

$1$ is the unit of the product, and $xy,(xy)z$ are defined iff they are not in
$I$, iff $yz,x(yz)$ are not in $I$ and in such case $(xy)z=x/yz)$.

    3. (c)

residuation works by Lemma 3.6.

  3. (3)

Since C is an IRL, multiplication is order preserving when defined;

  4. (4)

For the same reason, divisions are order-preserving in the numerator and
order-reversing in the denominator whenever defined.

In the rest of this proof we will write $\ell$ for $\ell_{I}$ and $r$ for
$r_{I}$. We now check the properties in the definition of an upper compatible
triple.

  1. (1)

We first show that $\ell,r$ form a Galois connection whenever they are
defined, i.e. that $\ell(y)$ is defined and $x\leq\ell(y)$ if and only if
$r(x)$ is defined and $y\leq r(x)$.

If $r(x)$ is defined and $y\leq r(x)$, then $y\leq i/x$ for some $i\in I$,
which by residuation is equivalent to $yx\leq i$. So $y$ is a left $I$-divisor
and thus $\ell(y)$ is defined and $x\leq y\backslash i\leq\ell(y)$. Similarly,
the converse holds.

  2. (2)

We now prove that if $r(y^{\prime})$ is defined, $x\leq r(y^{\prime})$ and
$y\leq y^{\prime}$, then $r(y)$ is defined and $x\leq r(y)$. If
$r(y^{\prime})$ is defined then $y^{\prime}$ is a right $I$-divisor, i.e.
there is $z\in C^{\prime}$ such that $zy^{\prime}\in I$. Since $y\leq
y^{\prime}$, we have $zy\leq zy^{\prime}\in I$ and since $I$ is closed
downwards we get $zy\in I$, i.e., $y$ is a right $I$-divisor. Moreover, it
follows from the definition of $r$ that it is order reversing, thus
$r(y^{\prime})\leq r(y)$. Since $x\leq r(y^{\prime})$, we also have that
$x\leq r(y)$.

  3. (3)

Given $x,y\in C^{\prime}$, we now prove that $xy$ is undefined if and only if
$y$ is in the domain of $r$ and $x\leq r(y)$, the proof of the other
equivalence being similar. Notice that the product $xy$ is undefined in
$C^{\prime}$ if and only if it is an element of $I$. In such a case, we get
$xy\leq i$ for some $i\in I$, thus $y$ is a right $I$-divisor and so it is in
the domain of $r$. Moreover, $x\leq i/y\leq r(y)$.

Conversely, if $y$ is in the domain of $r$ and $x\leq r(y)$, then $x\leq i/y$
for some $i\in I$. Thus $xy\in I$ and so $xy$ is undefined in $C^{\prime}$.

  4. (4)

We have that $x\backslash y$ is undefined in $C^{\prime}$ iff it is an element
of $I$, or equivalently, iff there is no $z\in C^{\prime}$ with $xz\leq y$.
Similarly, $y/x$ is undefined iff there is no $z$ with $zx\leq y$.

  5. (5)

We need to show that, if $\ell(x)$ and $r(z)$ are defined: $x\backslash r(z)$
is defined iff $\ell(x)/z$ is defined and in such a case $x\backslash
r(z)=\ell(x)/z$. Suppose first that $x\backslash r(z)$ is defined in
$C^{\prime}$. Then $(x\backslash r(z))z\leq x\backslash i\leq\ell(x)$, for
some $i\in I$, which by residuation implies $x\backslash r(z)\leq\ell(x)/z$.
Thus in particular $\ell(x)/z\in C^{\prime}$.

So we also get that $x(\ell(x)/z)z\leq x\ell(x)\leq j$ for some $j\in I$,
which implies that $x(\ell(x)/z)\leq j/z\leq r(z)$, which again by residuation
implies $\ell(x)/z\leq x\backslash r(z)$; thus the equality $x\backslash
r(z)=\ell(x)/z$ is proved.

Likewise, we can show that if $\ell(x)/z$ is defined in $C^{\prime}$ then
$x\backslash r(z)$ is defined in $C^{\prime}$ and $x\backslash
r(z)=\ell(x)/z$.

  6. (6)

Suppose first that $\ell(x)$ is undefined and $r(z)$ is defined, then
$r(z)=i/z$ for some $i\in I$. Thus $x\backslash(i/z)=(x\backslash i)/z\leq
r(z)$, because necessarily $x\backslash i\in I$, and since $r(z)\leq
x\backslash r(z)$ we have the equality. Similarly one can show the other case.

  7. (7)

Now if $\ell(x)=\max\\{x\backslash i:i\in I\\}$ is defined in $C^{\prime}$,
then there exists $i\in I$ with $x\backslash i\in C^{\prime}$. For every $z\in
C-I$ we have $i<z$ and $x\backslash i\leq x\backslash z$, so $x\backslash z\in
C^{\prime}$, and thus it is defined in $C^{\prime}$, and $\ell(x)\leq
x\backslash z$. The analogous fact for $r$ is proven similarly.

  8. (8)

A meet $x\land y$ is undefined in $C^{\prime}$ iff $x\land y\in I$ iff there
is no $z\in C^{\prime}$ with $z\leq x,y$.

  9. (9)

All other operations are defined.

∎

We say that $({\textbf{C}}^{\prime},\ell_{I},r_{I})$ is the upper-compatible
triple of the upper-compatible pair $({\textbf{C}},I)$.

######  Lemma 3.8.

Every upper-compatible triple $({\textbf{L}},\ell,r)$ is the upper-compatible
triple of the upper-compatible pair $({\textbf{C}},J)$, where $J=\\{0\\}$ is a
one-element set and C is an IRL with operations extending
${\textbf{L}}\cup\\{0\\}$, with $0$ as the bottom element.

###### Proof.

Let $C=L\cup\\{0\\}$, with $0$ an idempotent element strictly below all
elements of $L$. The operations of C are defined to extend the existing
operations of $L$, and we further define:

$\displaystyle x\land y$ $\displaystyle=\begin{array}[]{ll}0&\mbox{ if }x\land
y\mbox{ is undefined in }L\end{array}$ $\displaystyle xy$
$\displaystyle=\begin{array}[]{ll}0&\mbox{ if }x=0\mbox{ or }y=0\mbox{ or
}x\leq r(y)\end{array}$ $\displaystyle x\backslash y$
$\displaystyle=\left\\{\begin{array}[]{ll}1&\mbox{ if }x=0\\\ 0&\mbox{ if
}x\not=0,y=0\mbox{ and }\ell(x)\mbox{ is not defined}\\\ \ell(x)&\mbox{ if
}x\not=0,y=0\mbox{ and }\ell(x)\mbox{ is defined}\end{array}\right.$
$\displaystyle y/x$ $\displaystyle=\left\\{\begin{array}[]{ll}1&\mbox{ if
}x=0\\\ 0&\mbox{ if }x\not=0,y=0\mbox{ and }r(x)\mbox{ is not defined}\\\
r(x)&\mbox{ if }x\not=0,y=0\mbox{ and }r(x)\mbox{ is
defined}\end{array}\right.$

We set $J=\\{0\\}$ and show that C is an integral residuated lattice. The
order defined clearly yields a lattice. Let us show that associativity holds,
i.e. for any $x,y,z\in C$,

$x\cdot(y\cdot z)=(x\cdot y)\cdot z.$

We distinguish the following cases.

  * •

If any of $x,y,z$ is $0$, both sides of the equality are $0$ and thus
associativity holds. The same holds if $x\leq r(y)$ and $y\leq r(z)$.

  * •

Assume $x\leq r(y)$, and $r(z)$ is undefined or $y\not\leq r(z)$. Then
$(x\cdot y)\cdot z=0\cdot z=0$. Moreover $yz$ is defined in $L$, thus $yz\leq
y\leq\ell(x)$ (since there is a Galois connection between $\ell$ and $r$)
hence $x\leq r(yz)$, so $x\cdot(y\cdot z)=0$.

Similarly we verify the case where $y\leq r(z)$ and $x\not\leq r(y)$ or $r(y)$
is undefined.

  * •

Finally, assume that $r(y)$ is undefined or $x\not\leq r(y)$, and $r(z)$ is
undefined or $y\not\leq r(z)$. Then the products $xy,yz$ are defined in $L$,
and then we get that $(xy)z=0$ if $xy\leq r(z)$, and $(xy)z\in L$ otherwise.
Similarly, $x(yz)=0$ if $yz\leq\ell(x)$, and $x(yz)\in L$ otherwise.

The claim is proved by showing that $xy\leq r(z)$ if and only if
$yz\leq\ell(x)$. Indeed, suppose $r(z)$ is defined and $xy\leq r(z)$, by Lemma
3.6 we get that $x\backslash r(z)$ is defined and $y\leq x\backslash r(z)$.
Then if $\ell(x)$ is undefined, by Property 6 we get $x\backslash r(z)=r(z)$,
thus $y\leq r(z)$, a contradiction. Then also $\ell(x)$ is defined, thus by
Property 5 $\ell(x)/z$ is defined and $y\leq x\backslash r(z)=\ell(x)/z$.
Since $\ell(x)/z$ is defined and $y\leq\ell(x)/z$, by Lemma 3.6 we obtain
$yz\leq\ell(x)$ since $yz$ is defined. Similarly one can show the right-to-
left direction.

We now show residuation:

$xy\leq z\mbox{ iff }y\leq x\backslash z\mbox{ iff }x\leq z/y$

  * •

If $x=0$ or $y=0$ the claim is easily shown. Suppose now $z=0$; we need to
show $xy\leq 0$ iff $y\leq x\backslash 0$. If $\ell(x)$ is defined then
$x\backslash 0=\ell(x)$, and we know that $xy\leq 0$ iff $y\leq\ell(x)$. If
$\ell(x)$ is not defined then $x\backslash 0=0$ and and $xy$ is defined to be
$0$, so both inequalities hold.

  * •

Let $x,y,z\neq 0$, $r(y)$ defined, and $x\leq r(y)$ (or equivalently $\ell(x)$
defined and $y\leq\ell(x)$). Then $xy$ is not defined in $L$, thus the first
inequality becomes $0\leq z$ and is true. Since $\ell(x)$ is defined, by
Property 7 $x\backslash z$ is also defined and $y\leq\ell(x)\leq x\backslash
z$, so the second inequality is also true. The proof that $x\leq z/y$ holds is
similar.

  * •

Let $x,y,z\neq 0$, $r(y)$ undefined or $x\not\leq r(y)$; then $xy$ is defined
in $L$. Residuation follows from Lemma 3.6.

Notice that if $c$ is a left $J$-divisor, $\ell_{J}(c)=\max\\{c\backslash
i:i\in J\\}=c\backslash 0=\ell(c)$, and if $d$ is a right $J$-divisor
$r_{J}(d)=\max\\{i/d:i\in J\\}=0\backslash d=r(d)$. Thus we have shown that
$({\textbf{L}},\ell,r)$ is the upper-compatible triple of the upper-compatible
pair $({\textbf{C}},J)$. ∎

######  Corollary 3.9.

If $({\textbf{C}},I)$ is an upper-compatible pair, so is
$(({\textbf{C}}-I)\cup\\{0\\},\\{0\\})$.

###  3.4. The gluing over a filter-ideal pair

We are now ready to introduce the gluing construction over a congruence filter
and an ideal, which is depicted in Figure 2. To ease the notation, we write
the pair of the filter $F$ and the ideal $I$ as $P$:

$P:=(F,I)$

We can then define the gluing of B and C over the pair $P=(F,I)$, or
$(F-I)$-gluing of B and C, as the structure
${\textbf{B}}\oplus_{P}{\textbf{C}}$ where the operations extend the ones of B
and C as follows:

$F$$F$$F$$I$$I$$I$$1$B$C^{-}$$B^{-}$$C^{-}$$B^{-}$$\oplus_{P}$$1$C${=}$$1$${\textbf{B}}\oplus_{P}{\textbf{C}}$
Figure 2. The gluing ${\textbf{B}}\oplus_{P}{\textbf{C}}$ of the algebras B
and C over the pair $P=(F,I)$.

$\displaystyle x\cdot y$ $\displaystyle=\left\\{\begin{array}[]{ll}x\cdot
y&\mbox{ if }x,y\in B,\mbox{ or }x,y\in C\\\ \sigma_{F}(x)&\mbox{ if }y\in
C^{-},x\in B^{-}\\\ \sigma_{F}(y)&\mbox{ if }x\in C^{-},y\in
B^{-}\end{array}\right.$ $\displaystyle x\backslash\,y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\backslash y&\mbox{ if }x,y\in
B,\mbox{ or }x,y\in C\\\ \gamma_{F}(y)&\mbox{ if }y\in B^{-}\mbox{ and }x\in
C^{-}\mbox{ is not a left $I$-divisor}\\\ \ell_{I}(x)&\mbox{ if }y\in
B^{-}\mbox{ and }x\in C^{-}\mbox{ is a left $I$-divisor}\\\ 1&\mbox{ if }x\in
B^{-},y\in C^{-}\end{array}\right.$ $\displaystyle x/\,y$
$\displaystyle=\left\\{\begin{array}[]{ll}x/y&\mbox{ if }x,y\in B,\mbox{ or
}x,y\in C\\\ \gamma_{F}(x)&\mbox{ if }x\in B^{-}\mbox{ and }y\in C^{-}\mbox{
is not a right $I$-divisor}\\\ r_{I}(y)&\mbox{ if }x\in B^{-}\mbox{ and }y\in
C^{-}\mbox{ is a right $I$-divisor}\\\ 1&\mbox{ if }x\in C^{-},y\in
B^{-}\end{array}\right.$ $\displaystyle x\land y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\land y&\mbox{ if }x,y\in B,\mbox{
or }x,y\in C\mbox{ with }x\land y\not\in I\\\ \top_{B}&\mbox{ if }x,y\in
C^{-},x\land y\in I\\\ x&\mbox{ if }x\in B^{-},y\in C^{-}\\\ y&\mbox{ if }y\in
B^{-},x\in C^{-}\end{array}\right.$ $\displaystyle x\lor y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\lor y&\mbox{ if }x,y\in C,\mbox{
or }x,y\in B\mbox{ with }x\lor y\not\in F\\\ \bot_{C}&\mbox{ if }x,y\in
B^{-},x\lor y\in F\\\ y&\mbox{ if }x\in B^{-},y\in C^{-}\\\ x&\mbox{ if }y\in
B^{-},x\in C^{-}\end{array}\right.$

######  Theorem 3.10.

If $({\textbf{B}},F,{\textbf{C}},I)$ is a compatible quadruple, then
${\textbf{B}}\oplus_{P}{\textbf{C}}$ is the gluing of B and C over $F$ and
$I$.

###### Proof.

We show that ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is an IRL. The fact that
${\textbf{B}}\oplus_{P}{\textbf{C}}$ has an underlying lattice structure is
guaranteed by the order properties of the compatible quadruple. In particular,
it follows that $F$ is strictly above all other elements of $B$ and $C$
implies that if $x,y\in B-F$ (or $x,y\in C-F$) and $x\lor y\in F$, then $F$
has a bottom element $\bot_{F}$ and $x\lor y=\bot_{F}$. Also, if there are
$x,y\in B-F$ with $x\lor_{B}y\in F$, then $x\lor y=\bot_{C}$. This is not in
conflict with the definition of the operations, since we can show that
$\sigma_{F}(x)=\bot_{F}\cdot x=x\cdot\bot_{F}$ and
$\gamma_{F}(x)=\bot_{F}\backslash x=x/\bot_{F}$ in B. First, it is easy to see
that $\bot_{F}\cdot x\in[x]_{F}$; indeed, $(\bot_{F}\cdot x)\backslash x=1$
and $\bot_{F}\leq x\backslash(\bot_{F}\cdot x)$, thus both $(\bot_{F}\cdot
x)\backslash x$ and $x\backslash(\bot_{F}\cdot x)$ are in $F$. Moreover,
$(\bot_{F}\cdot
x)\backslash\sigma_{F}(x)=\bot_{F}\backslash(x\backslash\sigma_{F}(x))=1$
since $x\backslash\sigma_{F}(x)\in F$ and thus $\bot_{F}\cdot
x\leq\sigma_{F}(x)$. Therefore, $\sigma_{F}(x)=\bot_{F}\cdot x$ and the proof
for $x\cdot\bot_{F}$ is analogous.

We now show that $\gamma_{F}(x)=\bot_{F}\backslash x$; the proof for
$x/\bot_{F}$ is similar. Since $x\leq\bot_{F}\backslash x$, we get
$x\backslash(\bot_{F}\backslash x)=1\in F$, and we also have
$(\bot_{F}\backslash x)\backslash x\geq\bot_{F}\in F$; hence
$\bot_{F}\backslash x\in[x]_{F}$. Moreover,
$\gamma_{F}(x)=\max[x]_{F}\leq\bot_{F}\backslash x$, or equivalently,
$\bot_{F}\gamma_{F}(x)\leq x$, since
$\bot_{F}\gamma_{F}(x)=\min[\gamma(x)]_{F}=\min[x]_{F}=\sigma_{F}(x)\leq x$.

Similarly, since $I$ is strictly below all other elements of $B$ and $C$, if
$x,y\in B-I$ (or $x,y\in C-I$) and $x\land y\in I$, then $I$ has a top element
$\top_{I}$ and $x\land y=\top_{I}$. Thus given $x,y\in C-I$ with
$x\land_{C}y\in I$, the meet is redefined as $x\land y=\top_{B}$. This is not
in conflict with the definition of the operations, due to Lemma 3.5 (2), and
Lemma 3.3.

Also, it is clear that $1$ is both the monoidal unit and the top element of
the lattice.

To prove associativity, we need to show that for every $x,y,z\in
B\oplus_{P}C$, $(xy)z=x(yz)$. We distinguish the following cases.

  * •

Let $x\in F,y\in C^{-},z\in B^{-}$. Then $(xy)z=\sigma_{F}(z)$, since $xy\in
C$ from Lemma 3.5(2). Now, $x(yz)=x\sigma_{F}(z)=\sigma_{F}(z)$, given Lemma
3.5(1). Similarly we can show the cases where: $x\in B^{-},y\in C^{-},z\in F$;
$x\in F,y\in B^{-},z\in C^{-}$; $x\in C^{-},y\in F,z\in B^{-}$, $x\in
C^{-},y\in B^{-},z\in F$; $x\in B^{-},y\in F,z\in C^{-}$.

  * •

Let $x,y\in C^{-},z\in B^{-}$. We have that $(xy)z=\sigma_{F}(z)$, if either
$xy\in C^{-}$ (by definition) or if $xy\in I$ (by the compatibility condition
2 for the quadruple). On the other hand,
$x(yz)=x\sigma_{F}(z)=\sigma_{F}(\sigma_{F}(z))=\sigma_{F}(z)$. The proof is
analogous for the case: $x\in B^{-},y,z\in C^{-}$.

  * •

If $x\in C^{-},y,z\in B^{-}$, then $(xy)z=\sigma_{F}(y)z=\sigma_{F}(yz)$,
given that $\sigma_{F}$ is a strong conucleus. Also, $x(yz)=\sigma_{F}(yz)$,
if either $yz\in B^{-}$ (by definition) or $yz\in I$ (by Lemma 3.3). We get a
similar proof for the cases: $x,y\in B^{-},z\in C^{-}$; $x\in B^{-},y\in
C^{-},z\in I$; $x\in I,y\in C^{-},z\in B^{-}$; $x\in I,y\in B^{-},z\in C^{-}$;
$x\in B^{-},y\in I,z\in C^{-}$; $x,z\in B^{-},y\in C^{-}$; $x,z\in C^{-},y\in
B^{-}$.

  * •

Since both B and C are subalgebras with respect to multiplication, the
remaining cases hold automatically.

We now prove that for all $x,y,z$,

$x\cdot y\leq z\mbox{ iff }x\leq z/y\mbox{ iff }y\leq x\backslash z$

We have the following cases:

  * •

Let $x\in F,y\in C^{-},z\in B^{-}$. Then it never happens that $x\cdot y\leq
z$, by Lemma 3.5(2). The other inequalities are also false by definition and
order preservation. An analogous case is given by $x\in C^{-},y\in F,x\in
B^{-}$.

  * •

Let $x\in F,y\in B^{-},z\in C^{-}$. Then all three inequalities are always
true, given the definition of the operations and order preservation. Similar
cases are given by: $x\in C^{-},y\in B^{-},z\in F$; $x,z\in C^{-},y\in B^{-}$;
$x\in B^{-},y\in F,z\in C^{-}$; $x\in B^{-},y\in C^{-},z\in F$; $x\in
B^{-},y,z\in C^{-}$; $x,y\in B^{-},z\in C^{-}$; $x\in B^{-},y\in I,z\in
C^{-}$; $x\in I,y\in C^{-},z\in B^{-}$; $x\in I,y\in B^{-},z\in C^{-}$; $x\in
C^{-},y\in I,z\in B^{-}$.

  * •

Let $x,y\in C^{-},z\in B^{-}$. We distinguish two cases, based on whether
$xy\in I$ or not. If $xy=i$ for some $i\in I$, then all inequalities hold.
Indeed $xy=i\leq z$; moreover $xy=i$ implies $y\leq x\backslash
i\leq\ell_{I}(x)=x\backslash z$ and similarly $x\leq i/y\leq r_{I}(y)=z/y$. If
$xy\in C^{-}$, none of the inequalities hold. Indeed, $xy\not\leq z$ by
definition of the order. Moreover, if $y\leq x\backslash z$ the only
possibility by definition is that $x\backslash z=\ell_{I}(x)$, but
$x\ell_{I}(x)\in I$ thus we would have $xy\in I$, a contradiction. Similarly
it cannot be that $x\leq z/y$.

  * •

Let $x\in C^{-},y,z\in B^{-}$. To show that $xy=\sigma_{F}(y)\leq z$ iff
$x\leq z/y$ it suffices to note that, equivalently, $\sigma_{F}(y)\leq z$ iff
$z/y\in F$. Indeed since $\sigma_{F}(y)/y\in F$, we have that
$\sigma_{F}(y)\leq z$ implies $z/y\in F$. Conversely, if $z/y\in F$ then there
is $f\in F$ such that $f\leq z/y$, thus $fy\leq z$, and so
$\sigma_{F}(y)=\sigma_{F}(fy)\leq fy\leq z$ (where in the first equality we
used Lemma 3.5(1)).

We now show that $x\cdot y=\sigma_{F}(y)\leq z$ iff $y\leq x\backslash z$. If
$\sigma_{F}(y)\leq z$, then since $\sigma_{F}$ and $\gamma_{F}$ form a Galois
connection we get $y\leq\gamma_{F}(z)\leq x\backslash z$, where the second
inequality holds because $\gamma_{F}(z)=x\backslash z$ or $x\backslash z\in
C^{-}$. Conversely, assume $y\leq x\backslash z$. If $x\backslash
z=\ell_{I}(x)$, then by definition $x$ is a left $I$-divisor thus there is a
$c\in C^{-}$ such that $xc\in I$, thus by the compatibility condition (2) for
the quadruple $\sigma_{F}(y)\in I$ thus $\sigma_{F}(y)\leq z$. Otherwise, if
$x\backslash z=\gamma_{F}(z)$ then $y\leq\gamma_{F}(z)$, so by the Galois
connection we have that $\sigma_{F}(y)\leq z$.

An analogous case is given by $x,z\in B^{-},y\in C^{-}$.

  * •

Let $x\in C^{-},y\in B^{-},z\in I$. The fact that $xy\leq z$ iff $x\leq z/y$
can be shown, as in the previous case, using the fact that $\sigma_{F}(y)\leq
z$ if and only if $z/y\in F$.

We now show that $xy=\sigma_{F}(y)\leq z$ iff $y\leq x\backslash z$. Since C
is a subalgebra, either $x\backslash z\in C^{-}$ or $x\backslash z\in I$. If
$x\backslash z\in C^{-}$ then clearly $y\leq x\backslash z$ and by the
compatibility condition 2 for the quadruple, we get $\sigma_{F}(y)=xy\leq
x(x\backslash z)\leq z$.

Otherwise, we have $x\backslash z\in I$, so $y\not\leq x\backslash z$, since
$y\in B^{-}$. Moreover, $x\backslash z=\gamma_{F}(z)$ from Lemma 3.3. Thus
since $y\not\leq\gamma_{F}(z)$, we get that $\sigma_{F}(y)\not\leq z$, given
that the two operators are a residuated pair.

Similarly we prove residuation for the case: $x\in B^{-},y\in C^{-},z\in I$.

  * •

Since both B and C are subalgebras for the divisions the other cases do not
need to be checked.

Thus, ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is an integral residuated lattice.∎

Notice that in the case where $\mathbf{I}$ is empty, we get the proof of
Theorem 2.8. Indeed, if $I$ is empty, no elements of $C-F$ are $I$-divisors
and thus we obtain exactly the hypothesis of Theorem 2.8.

###  3.5. A gluing of partial algebras

$I$$I$$1$K$\top_{I}$$\top_{I}$$\oplus_{\tau}$$c_{L}$$1$$\bot_{L}$$\bot_{L}$L${=}$$c_{L}$$1$${\textbf{K}}\oplus_{\tau}{\textbf{L}}$
Figure 3. The partial gluing ${\textbf{K}}\oplus_{\tau}{\textbf{L}}$, in the
case where L has a coatom $c_{L}$.

In this section, we obtain a different construction that glues together two
structures that intersect at the top $1$ and keeps the same order relation,
but where some of the divisions are redefined. The underlying idea is to
forget the filter $F$ from the previous construction. See Figure 3 for a
pictorial intuition.

We start from a lower-compatible triple $({\textbf{K}},\sigma,\gamma)$ and an
upper-compatible triple $({\textbf{L}},\ell,r)$. Recall that in upper-
compatible triples already some divisions are not defined. Here we will allow
other divisions not to be defined. Precisely, we shall say that $x\backslash
y$ is _strongly undefined_ (in order to distinguish this case in the
definition of the operations) if all elements $z$ in the interval
$[y,1)=\\{z\in L:y\leq z<1\\}$ are such that $xz\leq y$ and there is no
coatom. Similarly, $y/x$ is strongly undefined if all elements $z$ the
interval $[y,1)$ are such that $zx\leq y$ and L has no coatom. We assume:

  1. (A1)

K has an ideal $I\subseteq K$ with an idempotent top element $\top_{I}$ such
that $\sigma(\top_{I})=\top_{I}$.

  2. (A2)

If there are undefined products in $L$, then $\sigma(x)=\top_{I}$ for all
$x\in K-I$ and $\sigma(\top_{I}y)=\sigma(y\top_{I})=\sigma(y)$ if $y\in I$.

  3. (A3)

If there exists $x,y\in K-\\{1\\}$ such that $x\lor y=1$, then $L$ has a
bottom element $\bot_{L}$.

  4. (A4)

If L has undefined meets, K has a splitting coatom $c_{K}$.

Moreover, we assume that $K\cap L=\\{1\\}$, we set
$\tau:=(\sigma,\gamma,\ell,r)$ and we define the _partial gluing_
${\textbf{K}}\oplus_{\tau}{\textbf{L}}$ to be the structure where the
operations extend the ones of K and L in the following way, here $D_{\ell}$
and $D_{r}$ denote the domains of $\ell$ and $r$, respectively:

$\displaystyle xy$ $\displaystyle=\left\\{\begin{array}[]{ll}xy&\mbox{ if
}x,y\in K\mbox{ or }x,y\in L\mbox{ and $xy$ is defined}\\\ \sigma(x)&\mbox{ if
}y\in L-\\{1\\},x\in K-\\{1\\}\\\ \sigma(y)&\mbox{ if }x\in L-\\{1\\},y\in
K-\\{1\\}\\\ \top_{I}&\mbox{ if }x,y\in L\mbox{ and $xy$ is undefined}\\\
\end{array}\right.$ $\displaystyle x\backslash\,y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\backslash y&\mbox{ if }x,y\in
K\mbox{ or }x,y\in L,\mbox{ and $x\backslash y$ is defined }\\\ c_{L}&\mbox{
if }x,y\in K,L\mbox{ has a coatom }c_{L}\mbox{ and $x\backslash y$ is
undefined}\\\ \ell(x)&\mbox{ if }x\in L-\\{1\\},y\in K-\\{I\cup 1\\}\mbox{ and
}x\in D_{\ell},\mbox{ or }x,y\in L\mbox{ and $x\backslash y$ undefined}\\\
\gamma(y)&\mbox{ if }x\in L-\\{1\\},y\in I\mbox{ or }(y\in K-\\{1\\}\mbox{ and
}x\not\in D_{\ell})\\\ 1&\mbox{ if }x\in K-\\{1\\},y\in L\\\ \mbox{ undefined
}&\mbox{ if }x,y\in L,\mbox{ and $x\backslash y$ strongly
undefined}\end{array}\right.$ $\displaystyle y/\,x$
$\displaystyle=\left\\{\begin{array}[]{ll}y/x&\mbox{ if }x,y\in K\mbox{ or
}x,y\in L,\mbox{ and $y/x$ is defined }\\\ c_{L}&\mbox{ if }x,y\in K,L\mbox{
has a coatom }c_{L}\mbox{ and $y/x$ is undefined}\\\ r(x)&\mbox{ if }x\in
L-\\{1\\},y\in K-\\{I\cup 1\\}\mbox{ and }x\in D_{r}\mbox{ or }x,y\in L\mbox{
and $y/x$ undefined}\\\ \gamma(y)&\mbox{ if }x\in L-\\{1\\},y\in I\mbox{ or
}(\in K-\\{1\\}\mbox{ and }x\not\in D_{r})\\\ 1&\mbox{ if }x\in K-\\{1\\},y\in
L\\\ \mbox{ undefined }&\mbox{ if }x,y\in L,\mbox{ and $y/x$ strongly
undefined}\end{array}\right.$ $\displaystyle x\land y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\land y&\mbox{ if }x,y\in K,\mbox{
or }x,y\in L\\\ c_{K}&\mbox{ if }x,y\in L\mbox{ and }x\land y\mbox{
undefined}\\\ y&\mbox{ if }y\in K-\\{1\\},x\in L\\\ x&\mbox{ if }x\in
K-\\{1\\},y\in L\end{array}\right.$ $\displaystyle x\lor y$
$\displaystyle=\left\\{\begin{array}[]{ll}x\lor y&\mbox{ if }x,y\in K,\mbox{
or }x,y\in L\\\ x&\mbox{ if }y\in K-\\{1\\},x\in L\\\ y&\mbox{ if }x\in
K-\\{1\\},y\in L\\\ \bot_{L}&\mbox{ if }x\lor y=1\mbox{ in
}K.\end{array}\right.$

######  Theorem 3.11.

${\textbf{K}}\oplus_{\tau}{\textbf{L}}$ is a partial IRL, that is total if $L$
has a coatom.

###### Proof.

Notice first that all operations are defined except possibly the divisions. It
is easy to see that the operations $\land,\lor$ define a lattice order, with
$1$ being the top. Let us now prove associativity of multiplication.

  * •

Suppose $x,y,z\in L$. Then by definition:

$\displaystyle(xy)z$ $\displaystyle=\left\\{\begin{array}[]{ll}(xy)z&\mbox{ if
}xy\mbox{ and }(xy)z\mbox{ are defined in }L\\\
\sigma(\top_{I})=\top_{I}&\mbox{ otherwise }\end{array}\right.$ $\displaystyle
x(yz)$ $\displaystyle=\left\\{\begin{array}[]{ll}x(yz)&\mbox{ if }yz\mbox{ and
}x(yz)\mbox{ are defined in }L\\\ \sigma(\top_{I})=\top_{I}&\mbox{ otherwise
}\end{array}\right.$

We will show that $xy$ and $(xy)z$ are defined if and only if $yz$ and $x(yz)$
are defined, or equivalently, $yz$ or $x(yz)$ is undefined iff $xy$ or $(xy)z$
is undefined. Notice that when $(xy)z$ and $x(yz)$ are defined they coincide,
since in upper compatible triples the IRL axioms hold whenever the operations
involved are defined.

We show the left-to-right direction first: we assume that $yz$ is undefined or
$x(yz)$ is undefined; we also assume that $xy$ is defined. If $yz$ is
undefined, then $z\in D_{r}$ and $y\leq r(z)$. Since $xy\leq y$, we get
$xy\leq r(z)$, so $(xy)z$ is undefined. If instead $yz$ is defined and $x(yz)$
is undefined, we have that $yz\in D_{r}$ and $x\leq r(yz)$. Then by Property 1
of an upper compatible triple, $\ell(x)$ is defined and $yz\leq\ell(x)$.
Suppose $r(z)$ is defined, thus using Lemma 3.6 and Property 5,
$y\leq\ell(x)/z=x\backslash r(z)$ and then $xy\leq r(z)$; thus $(xy)z$ is
undefined. We also show that $r(z)$ is necessarily defined, indeed if $r(z)$
is undefined, $\ell(x)/z=\ell(x)$, and $y\leq\ell(x)$ implies $x\leq r(y)$,
thus $xy$ undefined.

The other direction is proved in a similar way.

  * •

For $x,y\in L$ and $z\in K-I$, we show that $x(yz)=\sigma(z)$. If $x\not\leq
r(y)$, then $(xy)z=\sigma(z)$ holds by definition. Otherwise we get
$\top_{I}z=\top_{I}$, since $\top_{I}$ is idempotent; so there is at least an
undefined product in $L$, thus by (A2) we get $\sigma(z)=\top_{I}$. Similarly
we can show the case: $y,z\in L$, $x\in K-I$.

  * •

For $x,y\in L$ and $z\in I$, we show that $x(yz)=\sigma(z)$. If $x\not\leq
r(y)$, then $(xy)z=\sigma(z)$ by definition. Otherwise we get
$\top_{i}z=\sigma(z)$ since $\sigma(z)=\sigma(\top_{I}z)$ implies
$\sigma(z)\leq\top_{I}z$ and $\top_{I}z\leq\sigma(z)$. Similarly we prove the
case: $y,z\in L$ and $x\in I$.

  * •

For $x,z\in L,y\in K$, associativity follows directly from the definition of
multiplication and the idempotency of $\sigma$.

  * •

For $x\in L,y,z\in K$, associativity follows from the definition of
multiplication and the strong conuclear property. Similar cases are: $x,z\in
K,y\in L$; $x,y\in K,z\in L$.

  * •

In the remaining cases all elements belong to $K$.

It easily follows that the product is a monoidal operation with unit $1$. We
now show that residuation holds (when the divisions are defined):

$xy\leq z\mbox{ iff }y\leq x\backslash z\mbox{ iff }x\leq z/y$

  * •

For $x,y,z\in L$, we distinguish two cases, depending whether $xy$ is defined
in $L$ or not. If $xy$ is not defined in $L$, then we get that all three
inequalities hold since they respectively become: $\top_{I}\leq
z,y\leq\ell(x),x\leq r(y)$; here we used Property 3 of upper-compatible
triples.

If $xy$ is defined in $L$, then we get $xy\leq z$ if and only if $x\backslash
z$ is defined and $y\leq x\backslash z$, by Lemma 3.6. Similarly, this is
equivalent to $z/y$ being defined and $x\leq z/y$.

  * •

For $x,y\in L,z\in K-\\{I\cup 1\\}$, we distinguish two cases, based on
whether $xy$ is defined in $L$ or not. If $xy$ is defined in $L$, we have
$xy\not\leq z$ by the definition of the order; also $x\not\leq r(y)$ and thus
$y\not\leq\ell(x)$, by Property 1 (Galois connection). Moreover, $x\backslash
z$ is either $\ell(x)$ or $\gamma(z)$, and in either case $y\not\leq
x\backslash z$, since $\gamma(z)\in K-1$ and $y\not\leq\ell(x)$.

If $xy$ undefined in $L$, then $xy$ is equal to the top element of $I$, and
$\top_{I}\leq z$. Moreover, $y\leq\ell(x)=x\backslash z$, and $x\leq
r(y)=z/y$, using again Property 1,3.

  * •

For $x,y\in L,z\in I$, we distinguish two cases, whether $x\leq r(y)$ or not.
If $x\not\leq r(y)$, the proof is the same as in the previous case. If $x\leq
r(y)$ we distinguish whether $z=\top_{I}$ or not. If $z<\top_{I}$, then
$xy=\top_{I}\not\leq z$, $y\not\leq\gamma(z)=x\backslash z$ and
$x\not\leq\gamma(z)=z/y$. If $z=\top_{I}$, then $xy=\top_{I}$ implies
$x\leq\top_{I}/y=r(y)$ iff $y\leq\ell(x)=x\backslash\top_{I}$.

  * •

If $x,z\in L,y\in K-\\{1\\}$, it follows directly from the definition of the
operations that all inequalities hold, whenever the divisions are defined.
Similarly for the cases: $x\in L,y\in I,z\in K-\\{I\\}$; $x\in K,y,z\in L$;
$x,y\in K,z\in I$.

  * •

For $x\in L,y,z,\in K-\\{I\\}$, it follows directly from the definition of the
operations that $xy=\sigma(y)\leq z$ iff $x\leq z/y$ whenever the division is
defined. Indeed $z/y$ is either $z/_{K}y$ iff $\sigma(y)\not\leq z$ and it is
either undefined or $c_{L}$ if $\sigma(y)\leq z$. Now we show that
$xy=\sigma(y)\leq z$ iff $y\leq x\backslash z$. If $x\in D_{\ell}$ then there
are undefined products in $L$, thus $\sigma(y)=\top_{I}\leq z$ and
$x\backslash z=\ell(x)$ thus $y\leq x\backslash z$. Otherwise, if $x\not\in
D_{\ell}$, $\sigma(y)\leq z$ iff $y\leq\gamma(z)=x\backslash z$ since
$\sigma,\gamma$ form a residuated pair. The following cases are proven
similarly: $x\in L,y\in K,z\in I$; $x,z\in K-\\{I\\},y\in L$; $x\in
K-\\{I\\},y\in L,z\in I$.

  * •

For $x,y,z\in K$, we show that $xy\leq z$ iff $y\leq x\backslash z$, as the
proof the equivalence $xy\leq z\Leftrightarrow x\leq z/y$ is analogous. If
$x\backslash_{K}z$ is defined then residuation holds since $K$ is a partial
IRL. If $x\backslash_{K}z$ is undefined, then $\sigma(x)\leq z$, and
$xy\leq\sigma(x)\leq z$ holds. Moreover $x\backslash z$ is either still
undefined or if there is a coatom $x\backslash z=c_{L}$, thus $y\leq c_{L}$.

We now show that multiplication is order preserving: if $x\leq y$, then
$xz\leq yz$ and $zx\leq zy$ (notice again that all products are defined in the
gluing).

  * •

The cases where $x,y\in L$ and $z\in K-\\{1\\}$ follow directly from the
definition of the operations.

  * •

The cases where $x,y\in K-\\{1\\}$ follow from the order preservation of
$\sigma$.

  * •

If $x,z\in K-\\{1\\},y\in L$, then we get $xz,zx\leq\sigma(z)$, which is a
property of $\sigma$ in a lower compatible triple.

  * •

Let now $x\in K-\\{1\\}$, $y,z\in L$. If $yz$ (equiv. $zy$) is defined in $L$,
clearly $xz\leq yz$ (equiv. $zx\leq zy$). If $yz$ (or $zy$) is undefined in
$L$, the inequalities becomes $\sigma(x)\leq\top_{I}$, which holds by
condition (A2).

  * •

Finally, let $x,y,z\in L$. We show order preservation of right multiplication,
the proof for left multiplication being analogous. If $xz,yz$ are defined in
$L$, then order preservation holds since L is a partial IRL. If $yz$ is
undefined, by the definition of an upper compatible triple, $z\leq\ell(y)$,
and since $x\leq y$, and the domain of $\ell$ is closed downwards, by property
(2) of the definition we also get that $z\leq\ell(x)$ and thus $xz$ is
undefined in $L$. Therefore, $xz=\top_{I}=yz$. Suppose now that $yz$ is
defined and $xz$ is undefined, then the inequality becomes $\top_{I}\leq yz\in
L$, which holds by definition of the order in the gluing.

The fact that (when defined) divisions are order preserving in the numerator
and order reversing in the denominator follows from residuation and the order
preservation of multiplication (which is always defined in
${\textbf{K}}\oplus_{\tau}{\textbf{L}}$). Thus,
${\textbf{K}}\oplus_{\tau}{\textbf{L}}$ is a partial IRL. We note that in the
case where $L$ has a coatom no operation is undefined. ∎

If we take a compatible quadruple $({\textbf{B}},F,{\textbf{C}},I)$ where $I$
has a top element satisfying assumptions $(A1),(A2)$, we can then consider the
lower compatible triple $({\textbf{B}}^{\prime},\sigma_{F},\gamma_{F})$ (where
$B^{\prime}=B-F$) and the upper compatible triple
$({\textbf{C}}^{\prime},\ell_{I},r_{I})$ (where $C^{\prime}=C-(F\cup I)$) and
construct the partial gluing
${\textbf{B}}^{\prime}\oplus_{\tau}{\textbf{C}}^{\prime}$, where
$\tau=(\sigma_{F},\gamma_{F},\ell_{I},r_{I})$.

More generally, starting from any compatible quadruple
$({\textbf{B}},F,{\textbf{C}},I)$, we can always construct a partial gluing.
Indeed either there are no $I$-divisors in C, in which case the assumptions
$(A1),(A2)$ are vacuously true, or otherwise we replace $I$ with
$I^{\prime}:=\\{i\in I:i\neq cd,\mbox{ for }c,d\in C-I\\}\cup\top_{I}$, where
$\top_{I}$ a new element satisfying $(A1),(A2)$. Conditions $(A3),(A4)$ are
implied by the last two compatibility conditions in the definition of a
compatible quadruple. We can then construct the partial gluing
${\textbf{B}}^{\prime}\oplus_{\tau}{\textbf{C}}^{\prime}$, where
$B^{\prime}=B-F$, $C^{\prime}=C-(F\cup I^{\prime})$,
$\tau=(\sigma_{F},\gamma_{F},\ell_{I},r_{I})$.

##  4\. Variations of the constructions

First, notice that the gluing constructions presented that involve a non-empty
ideal $I$ work for both bounded and unbounded integral residuated lattices. In
the case where the ideal is empty (and thus the construction is gluing over a
congruence filter) one still obtains a new structure starting from
$\mathsf{FL}_{w}$-algebras, but one of the two algebras is not a subalgebra
with respect to $0$ anymore.

Importantly, note again that as a special case of the gluing construction,
where the filter is trivially the top element $\\{1\\}$ and the ideal is
empty, we get the $1$-sum construction. This also means that given any pair of
integral residuated lattices B and C (with C having a lower bound or $1$ being
join irreducible in B) we can always glue them. We will call a gluing trivial
if it is a $1$-sum, and non-trivial otherwise.

###  4.1. The congruence filter has a bottom element

We first observe that if $F$ has a bottom element, then $\sigma_{F}$ and
$\gamma_{F}$ have a very transparent definition: also the bottom element of
$F$ multiplies and divides as the elements in $C-F$ in the gluing.

######  Lemma 4.1.

Let $({\textbf{B}},F)$ be a lower-compatible triple where $F$ has a bottom
element $\bot_{F}$. Then given any $x\in B-F$, $\sigma_{F}(x)=\bot_{F}\cdot
x=x\cdot\bot_{F}$ and $\gamma_{F}(x)=\bot_{F}\backslash x=x/\bot_{F}$.

###### Proof.

The proof can be directly extracted from the first paragraphs of the proof of
Theorem 3.10. ∎

It turns out that assuming that $F$ has a bottom element is not a substantial
restriction. Given a lower compatible triple $({\textbf{B}},F)$, where $F$ may
or may not have a bottom element, and a new element $\bot_{F}$, we define the
residuated lattice ${\textbf{B}}_{\bot}$, where for all $b\in B-F,f\in F$

  * •

$b<\bot_{F}<f$,

  * •

$\bot_{F}\cdot\bot_{F}=\bot_{F}\cdot f=f\cdot\bot_{F}=\bot_{F}$ and
$\bot_{F}\cdot b=b\cdot\bot_{F}=\sigma_{F}(b)$,

  * •

$\bot_{F}\backslash b=b/\bot_{F}=\gamma_{F}(b)$, $\bot_{F}\backslash
f=f/\bot_{F}=1$,

  * •

$b\backslash\bot_{F}=\bot_{F}/b=1$, and
$f\backslash\bot_{F}=\bot_{F}/f=\bot_{F}$.

The following result is easy to prove and shows that every lower compatible
triple can be embedded into one where the congruence filter has a bottom
element.

######  Proposition 4.2.

If $({\textbf{B}},F)$ is a lower compatible pair, then
$({\textbf{B}}_{\bot},F\cup\\{\bot_{F}\\})$ is also a lower compatible pair
and B is a subalgebra of ${\textbf{B}}_{\bot}$, except possibly for join if
$F$ has a bottom element.

###  4.2. Non-linear order

We are now going to study whether the order conditions for the congruence
filter $F$ and the ideal $I$ can be weakened. We previously required $F$ to be
strictly above all elements of B and $I$ to be strictly below all elements of
C. This ensures that the product is well-defined and respect the join
operation. Notice that, in particular, asking $F$ to be strictly above all
other elements implies that if there are elements $x,y\in B-F$ such that
$x\lor y\in F$, then $F$ has a bottom element $\bot_{F}$ and $x\lor
y=\bot_{F}$. Lemma 4.1 shows that the product defined in Theorem 3.10
preserves this particular kind of join.

We define a _non-strict_ lower-compatible pair a pair $({\textbf{B}},F)$ that
is lower-compatible except that $F$ is not strictly above all elements in $F$,
but whenever $x,y\in B-F$ are such that $x\lor y=z\in F$, the element $z$ is
such that $zb=bz=\sigma_{F}(b)$ and $z\backslash b=b/z=\gamma_{F}(b)$ for all
$b\in B-F$.

Similarly, if $z,w\in C-I$ are such that $z\land w\in I$, then we required $I$
to have a top element $\top_{I}$ and $z\land w=\top_{I}$; this does not create
issues with respect to the operations, given Lemma 3.5. We can then extend the
construction to also include lattice ideals that are not strictly below all
other elements. In particular, we call a _non-strict_ upper-compatible pair a
pair $({\textbf{C}},I)$ that is upper-compatible except that $I$ is not
strictly below all other elements in $C$.

Let us call a _non-strict compatible quadruple_ a compatible quadruple
$({\textbf{B}},F,{\textbf{C}},I)$ where the upper and lower-compatible pairs
may be non-strict, by redefining all joins of elements $x,y\in B-F$ such that
$x\lor_{B}y\in F$ to be the bottom element of C, $x\lor y=\bot_{C}$, and all
meets of elements $z,w\in C-I$ such that $x\land_{C}w\in I$ to be the top
element of $B-F$, $x\land w=\top_{B}$.

######  Proposition 4.3.

If $({\textbf{B}},F,{\textbf{C}},I)$ is a non-strict compatible quadruple,
then ${\textbf{B}}\oplus_{\mathbf{P}}C$ is the gluing of B and C over $F$ and
$I$.

###### Proof.

The proof of Theorem 3.10 can be adapted to this case. ∎

##  5\. Preservation

In this section we will investigate the interaction of the $(F-I)$-gluing
construction with class operators and equations that are preserved.

###  5.1. Preservation of identities

We identify equations that are preserved by the $(F-I)$-gluing. It is worth
noticing that the gluing construction preserves commutativity. Moreover, we
identify the cases when divisibility and semilinearity are preserved;
linearity is obviously always preserved.

_Semilinear_ integral residuated lattices (i.e., subdirect products of totally
ordered integral residuated lattices) constitute a variety, axiomatized by the
equation:

$[u\backslash(y\backslash x)u]\lor[v(x\backslash y)/v]=1.\qquad{\mbox{(sl)}}$

This equation characterizes semilinearity also in $\mathsf{FL_{w}}$-algebras.
In commutative subvarieties of $\mathsf{IRL}$ and $\mathsf{FL_{w}}$
semilinearity is characterized by the simpler prelinearity identity, obtained
from (sl) by taking $u=v=1$:

$(y\backslash x)\lor(x\backslash y)=1.\qquad{\mbox{(prel)}}$

Commutative prelinear $\mathsf{FL_{w}}$-algebras are called MTL-algebras since
are the equivalent algebraic semantics of Esteva and Godo’s Monoidal t-norm
based logic, the logic of left-continuous t-norms [10]. A residuated lattice A
is called _divisible_ if the lattice order coincides with the inverse
divisibility order:

$a\leq b\qquad\text{if and only if}\qquad\text{there are $c,d\in A$ with
$a=bc$ and $a=db$}.$

Divisibility is characterized equationally by:

$x\land y=x(x\backslash(x\land y))=((x\land y)/x)x\qquad{\mbox{(div)}}$

The latter in integral structures reduces to: $x\land y=x(x\backslash
y)=(y/x)x$. Semilinear, commutative and divisible $\mathsf{FL_{ew}}$-algebras
are called BL-algebras and we denote their variety by $\mathsf{BL}$;
semilinear and divisible CIRLs are called basic hoops, and we refer to their
variety by $\mathsf{BH}$. BL-algebras are the equivalent algebraic semantics
of Hájek’s Basic Logic [18].

######  Proposition 5.1.

If ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is the $P$-gluing of the IRLs B and C,
where $P=(F,I)$, then:

  1. (1)

${\textbf{B}}\oplus_{P}{\textbf{C}}$ is commutative iff both B and C are
commutative.

  2. (2)

${\textbf{B}}\oplus_{P}{\textbf{C}}$ is divisible iff both B and C are
divisible, C has no $I$-divisors, and $B=((B-F)\cup\\{1\\})\oplus_{1}F$.

  3. (3)

If $F\not=\\{1\\}$, then ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is semilinear
iff both B and C are semilinear. If $F=\\{1\\}$ and $C^{-}\not=\emptyset$,
then ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is semilinear iff B is linear and C
is semilinear. (If $F=\\{1\\}$ and $C^{-}=\emptyset$, then
${\textbf{B}}\oplus_{P}{\textbf{C}}={\textbf{B}}$.)

###### Proof.

Recall that $B^{-}=B-(F\cup I)$ and $C^{-}=C-(F\cup I)$. For readability we
write $\sigma$ for $\sigma_{F}$, $\gamma$ for $\gamma_{F}$, $\ell$ for
$\ell_{I}$ and $r$ for $r_{I}$.

  1. (1)

$B$ and $C$ are closed under multiplication. Also, $bc=\sigma(b)=cb$ for $b\in
B^{-},c\in C^{-}$; thus the construction preserves commutativity.

  2. (2)

Recall that in integral structures divisibility states that for all $x,y$,

$x\land y=x(x\backslash y)=(y/x)x.$

Since $B$ and $C$ are closed under $\wedge,\cdot,\backslash,/$, the
divisibility of B and C is a necessary condition for the gluing to be
divisible. Note that if $x\in B^{-},$ and $y\in C^{-}$, then $x(x\backslash
y)=x1=x=x\land y$, and similarly $(y/x)x=x=x\land y$, so divisibility holds in
this case.

If $x\in C^{-}$ and $y\in B^{-}$, then $x\land y=y$, while $x(x\backslash y)$
and $(y/x)x$ depend on whether $x$ is an $I$-divisor or not. Notice that if
$x$ is a left $I$-divisor, then $x\ell(x)\neq y$ since C is a subalgebra,
therefore

$x(x\backslash y)=x\ell(x)\neq y=x\land y.$

Similarly, if $x$ is a right $I$-divisor

$(y/x)x=r(x)x\neq y=x\land y$

Thus for the gluing to be divisible, C must have no $I$-divisors. Now, if $x$
is not an $I$-divisor, we get:

$x(x\backslash y)=x\gamma(y)=\sigma(\gamma(y))=\sigma(y)$

and similarly,

$(y/x)x=\gamma(y)x=\sigma(\gamma(y))=\sigma(y).$

Thus for the gluing to be divisible, we need that for all $y\in B^{-}$,
$\sigma(y)=y$.

Notice that the same holds for $x\in C^{-},y\in I$, by Lemma 3.3.
Consequently, for all $f\in F$, $y=\sigma(y)\leq fy\leq y$ and also
$y=\sigma(y)\leq yf\leq y$, thus have that $fy=yf=y$. This implies that B is
the $1$-sum $B=(B-F)\oplus_{1}F$ (including the trivial case where
$F=\\{1\\}$).

Now we show that if B and C are divisible, C has no $I$-divisors and
$B=(B-F)\oplus_{1}F$, then divisibility holds in the gluing. We only need to
check the case where $x\in C^{-},y\in B^{-}$, where we get $x\land y=y$and

$x(x\backslash y)=x\gamma(y)=\sigma(y)=(y/x)x$

since $x$ is not an $I$-divisor. Now, if $B=(B-F)\oplus_{1}F$, all products
between elements $f\in F$ and $x\in B-F$ are such that $fx=xf=x$. Thus, for
$x\in B^{-}$ with $x\mathrel{\theta_{F}}y$, we have $x\backslash y,y\backslash
x\in F$. So $x=x(x\backslash y)\leq y$ and $y=y(y\backslash x)\leq x$, hence
$x=y$. Therefore, $\sigma(y)=\min[y]_{F}=y$ and divisibility holds in the
gluing.

  3. (3)

If ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is semilinear, then C is semilinear,
since it is a subalgebra except possibly for the meet. Also, in verifying
semilinearity in B, if $x,y,u,v\in B$, then $[u\backslash(y\backslash x)u]$,
$[v(x\backslash y)/v]\in B$ and $[u\backslash(y\backslash
x)u]\lor[v(x\backslash y)/v]=1$ in ${\textbf{B}}\oplus_{P}{\textbf{C}}$. Since
$[u\backslash(y\backslash x)u]\lor[v(x\backslash y)/v]\in B$, it follows that
$[u\backslash(y\backslash x)u]\lor[v(x\backslash y)/v]=1$ in B; hence
semilinearity holds in B.

Moreover, in the particular case where ${\textbf{F}}=\\{1\\}$, let $a,b\in B$
be incomparable. Therefore, $a\not\leq b$ and $b\not\leq a$, so
$1\not=a\backslash b$ and $1\not=b\backslash a$; hence $a\backslash
b,b\backslash a\in B-F$, so $(a\backslash b)\lor(b\backslash a)\leq c$ for
every $c\in C^{-}$. Since $C^{-}\not=\emptyset$ we have $c<1$ for some $c\in
C^{-}$, so $(a\backslash b)\lor(b\backslash a)\leq c<1$. By the semilinearity
of ${\textbf{B}}\oplus_{P}{\textbf{C}}$, we get $(a\backslash
b)\lor(b\backslash a)=1$, a contradiction.

For the converse direction, suppose both B and C are semilinear. We check
whether in ${\textbf{B}}\oplus_{P}{\textbf{C}}$ we have (sl):

$[u\backslash(y\backslash x)u]\lor[v(x\backslash y)/v]=1.$

If all elements belong to C, the equation follows from the semilinearity of C.
If $x$ and $y$ are comparable, then $y\leq x$ or $x\leq y$, so $y\backslash
x=1$ or $x\backslash y=1$; hence for all $u,v$, $u\backslash(y\backslash
x)u=1$ or $v(x\backslash y)/v=1$ and so (sl) holds.

If $x\in B^{-},y\in C^{-}$ or vice versa, then $x,y$ are comparable, so (sl)
holds. It remains to verify (sl) for $x,y\in B^{-}$.

If $F=\\{1\\}$ and B is linear, then $x,y$ are comparable, so (sl) holds.

We now assume that $F\neq\\{1\\}$. For $x,y\in B^{-}$, if they are comparable,
(sl) holds. If they are incomparable, then $y\backslash x\not=1$ and
$x\backslash y\not=1$ and $y\backslash x,x\backslash y\in B$. The
semilinearity of B yields

$(y\backslash x)\lor(x\backslash y)=1.$

Note that if $b_{1},b_{2}\in B$, $b_{1}\not=1$, $b_{2}\not=1$ and $b_{1}\lor
b_{2}=1$, then $b_{1},b_{2}\in F$. (If, say, $b_{1}\not\in F$ and $b_{2}\in
F$, then since $B-F$ is strictly below $F$, we get $b_{1}\lor
b_{2}=b_{2}\not=1.$ If $b_{1},b_{2}\not\in F$, then since $B-F$ is strictly
below $F\neq\\{1\\}$, there is $f\in F-\\{1\\}$ such that $b_{1}\lor b_{2}\leq
f<1$.) The same holds for elements $c_{1},c_{2}\in C$.

Therefore, $y\backslash x,x\backslash y\in F$. If $u\in B$ (or $u\in C$),
since B (respectively, C) is semilinear, we can apply Lemma 6.5 in [4] and
obtain that

$[u\backslash(y\backslash x)u]\lor(x\backslash y)=1.$

If $[u\backslash(y\backslash x)u]=1$, then (sl) holds. If not, then
$[u\backslash(y\backslash x)u]$ and $(x\backslash y)$ are non-identity
elements of $B$ ($C$, respectively) that join to $1$, so by above fact they
are both in $F$. Given the semilinearity of B (respectively, C), we can apply
Lemma 6.5 in [4], which states that whenever semilinearity holds, if $a\lor
b=1$, also $\gamma_{1}(a)\lor\gamma_{2}(b)=1$, for any iterated conjugates
$\gamma_{1},\gamma_{2}$. With $v\in B$ (or $v\in C$), this yields precisely
(sl).

∎

Since the components of the gluing are subalgebras (except in the mentioned
cases for the lattice operations), most one-variable equations are preserved.

######  Proposition 5.2.

The $P$-gluing ${\textbf{B}}\oplus_{P}{\textbf{C}}$, where $P=(F,I)$, of two
IRLs preserves all one-variable equations not involving the lattice
operations. Whenever $B-F$ is closed under joins, and $C-I$ is closed under
meets, all one-variable equations satisfied by both B and C are preserved.

###### Proof.

Follows directly from the definition of the operations. ∎

Thus, for example, the gluing construction preserves $n$-potency,
$x^{n}=x^{n+1}$, for every $n\geq 1$. In fact, the gluing also preserves all
monoidal equations, given the idempotency and the absorbing properties of the
conucleus $\sigma$.

######  Proposition 5.3.

The $P$-gluing ${\textbf{B}}\oplus_{P}{\textbf{C}}$, with $P=(F,I)$, preserves
all monoid equations valid in both B and C.

###### Proof.

If the equation has a variable that appears in only one side, then setting all
the other variables equal to $1$, we obtain a consequence of the form
$x^{n}=1$, for some $n\not=1$, and the only model of that equation is the
trivial algebra. Therefore, we consider equations where all variables appear
on both sides.

Since $B$ and $C$ are closed under multiplication, if an equation holds in the
gluing then it also holds in B and in C. We now assume that some equation
holds in B and in C. If under some evaluation all variables are chosen from
$B$ or all variables are chosen from $C$, then the equation holds true. Now
suppose that at least one variable is assigned to an element of $C^{-}$ and
one variable is assigned to an element $B^{-}$. Assume that $X$, is the set of
variables in the equation, $v$ is the evaluation, and that $X_{C}=\\{x\in
X:v(x)\in C^{-}\\}$ corresponds to the variables that are mapped to elements
of $C^{-}$; $X^{c}_{C}$ denotes the complement of $X_{C}$. We focus on the
position of values $v(x)$, where $x\in X_{C}$, inside the equation; we group
together the elements in between these $v(x)$ as follows. Given that $B$ is
closed under multiplication, the evaluation of each side of the equation takes
the form

$b^{\prime}_{1}c_{1}b^{\prime}_{2}c_{2}\cdots b^{\prime}_{n},$

where each $c_{i}$ is of the form $v(x)$, for some $x\in X_{C}$ and each
$b^{\prime}_{i}$ is a product of elements of the form $v(x)$, for certain
$x\in X^{c}_{C}$, so $v(x)\in B$; hence $c_{i}\in C^{-}$ and
$b^{\prime}_{i}\in B$.

By focusing on the elements adjacent to the $c_{i}$’s and using that
$cb=bc=\sigma(b)\in B-F$, for $c\in C^{-}$ and $b\in B-F$, and that $fc,cf\in
C^{-}$ for $c\in C^{-}$ and $f\in F$, the evaluation of the equation is
reduced to a form that does not contain any elements of $C^{-}$ (recall that
at least one variable is assigned to an element of $C^{-}$ and at least one
variable is assigned to an element of $B^{-}$). Then, we use that
$b_{1}\sigma(b_{2})=\sigma(b_{1}b_{2})=\sigma(b_{1})b_{2}$, for
$b_{1},b_{2}\in B-F$, and that $f\sigma(b)=\sigma(b)f=\sigma(b)$, for $b\in
B-F$ and $f\in F$, and the idempotency of $\sigma$. In the end, the evaluation
of the equation takes the form $\sigma(b_{1}b_{2}\cdots
b_{n})=\sigma(b_{n+1}b_{n+2}\cdots b_{m})$, where the $b_{i}$’s are exactly
the elements of the form $v(x)$, for $x\in X^{c}_{C}$, in the exact order they
appear in the equation. Therefore, by substituting $1$ for all $x$ with $x\in
X_{C}$, and the appropriate value $b_{i}$ for the other variables, the
original equation (which is valid in B), yields $b_{1}b_{2}\cdots
b_{n}=b_{n+1}b_{n+2}\cdots b_{m}$, so $\sigma(b_{1}b_{2}\cdots
b_{n})=\sigma(b_{n+1}b_{n+2}\cdots b_{m})$ is also valid. ∎

###  5.2. HSPU

Constructions as the one presented in this paper are particularly interesting
when they help us better understand and describe the structure theory of the
algebras. In what follows, we will just call “gluing” a gluing over a filter
and an ideal. In this section we characterize when a gluing is subdirectly
irreducible. We will also study the subalgebras, homomorphic images and
ultrapowers of a gluing. We recall that ${\textbf{Fil}}({\textbf{A}})$ denotes
the lattice of congruence filters of A.

######  Proposition 5.4.

Consider the $P$-gluing ${\textbf{B}}\oplus_{P}{\textbf{C}}$, with $P=(F,I)$,
of two integral residuated lattices B and C. We distinguish two cases:

  1. (1)

If $C-I$ is a congruence filter of C, then
${\textbf{Fil}}({\textbf{B}}\oplus_{\mathbf{P}}{\textbf{C}})$ is isomorphic to
${\textbf{Fil}}({\textbf{B}})\oplus{\textbf{Fil}}({\textbf{C}}-I)$, the poset
ordinal sum of the two lattices.

  2. (2)

Otherwise,
${\textbf{Fil}}({\textbf{B}}\oplus_{\mathbf{P}}{\textbf{C}})\cong{\textbf{Fil}}({\textbf{C}})$.

###### Proof.

The first claim follows from the definition of the order and operations in the
gluing construction, see Figure 2. Now, if $C-I$ is not a congruence filter,
then the congruence filter generated by $C-I$ has non-empty intersection with
$I$, i.e., either some conjugate or some product of elements in $C^{-}$ are in
$I$. Since $C$ is closed under multiplication and divisions, this is also true
in the gluing ${\textbf{B}}\oplus_{P}{\textbf{C}}$. Thus the congruence filter
$\langle C-I\rangle$ generated by $C-I$ in the gluing also has nonempty-
intersection with the ideal $I$. Since filters are closed upwards, $B^{-}$ is
contained in $\langle C-I\rangle$ and the second claim follows. ∎

######  Corollary 5.5.

For $P=(F,I)$, if $F\neq\\{1\\}$, then ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is
subdirectly irreducible iff F is subdirectly irreducible iff B is subdirectly
irreducible iff C is subdirectly irreducible. If $F=\\{1\\}$, then
${\textbf{B}}\oplus_{P}{\textbf{C}}$ is subdirectly irreducible iff C is
subdirectly irreducible.

###### Proof.

Assume first that $F$ is not trivial. By Proposition 5.4 and standard
universal algebraic results (see Theorem 8.4 in [6]),
${\textbf{B}}\oplus_{\mathbf{P}}{\textbf{C}}$ is subdirectly irreducible iff F
is subdirectly irreducible as an IRL, thus the claim follows since $F$ is a
congruence filter of both B and C, (strictly) above all their other elements.
For $F=\\{1\\}$, it follows from the definition of the operations and order of
the gluing (see also Figure 2) that ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is
subdirectly irreducible iff C is subdirectly irreducible. ∎

We can now describe the homomorphic images of a gluing.

######  Proposition 5.6.

Let $h$ be a homomorphism having as domain a gluing
${\textbf{B}}\oplus_{P}{\textbf{C}}$ and $H$ the associated congruence filter
(the preimage of $1$). If $B/H,C/H,F/H,I/H$ denote the images under $h$, we
have that the homomorphic image of ${\textbf{B}}\oplus_{P}{\textbf{C}}$ via
$h$ is isomorphic to

  1. (1)

${\textbf{C}}/H$, if $H\cap I\not=\emptyset$.

  2. (2)

${\textbf{B}}/H$, if $H\cap I=\emptyset$ and $H\cap B^{-}\not=\emptyset$.

  3. (3)

${\textbf{B}}/H\oplus_{P/H}{\textbf{C}}/H$, where $P/H=(F/H,I/H)$, if $H\cap
B^{-}=\emptyset$.

###### Proof.

The fact that the homomorphic image through $h$ is given by the gluing
${\textbf{B}}/H\oplus_{P/H}{\textbf{C}}/H$ follows from Proposition 5.4.
Moreover, notice that (1) and (2) are particular cases of (3). ∎

We call a subalgebra S of C _divisor-special_ if whenever it contains an
element $c$ that is a left $I$-divisor, it also contains $\ell_{I}(c)$, and
similarly if $d\in S$ where $d$ is a right $I$-divisor, then also $r(d)\in S$.
We call a subalgebra T of B _$\sigma$ -special_ if for all $b\in T-F$, also
$\sigma(b)\in T$, and _$(\sigma,\gamma)$ -special_ if also $\gamma(b)\in T$.

######  Proposition 5.7.

Let ${\textbf{B}}\oplus_{P}{\textbf{C}}$, with $P=(F,I)$, be the $P$-gluing of
IRLs B and C. Then a subalgebra S of ${\textbf{B}}\oplus_{P}{\textbf{C}}$ is
one of the following:

  1. (1)

S is a subalgebra of C that does not include elements whose meet is the top
element of $B-F$.

  2. (2)

S is a subalgebra of B that does not include elements whose join is the bottom
element of $C-I$.

  3. (3)

A gluing ${\textbf{B}}_{1}\oplus_{P_{1}}{\textbf{C}}_{1}$, with
$P_{1}=(F_{1},I_{1})$, where:

     * •

$F_{1}\subseteq F,I_{1}\subseteq I$ and $F_{1}\cup I_{1}$ is a subalgebra of
$F\cup I$;

     * •

${\textbf{C}}_{1}$ is a divisor-special subalgebra of C containing at least an
element that is not an $I$-divisor;

     * •

${\textbf{B}}_{1}$ is a (nonempty) special $(\sigma,\gamma)$-subalgebra of B.

  4. (4)

A gluing ${\textbf{B}}_{2}\oplus_{P_{2}}{\textbf{C}}_{2}$, with
$P_{2}=(F_{2},I_{2})$, where:

     * •

$F_{2}\subseteq F,I_{2}\subseteq I$ and $F_{2}\cup I_{2}$ is a subalgebra of
$F\cup I$;

     * •

${\textbf{C}}_{2}$ is a divisor-special subalgebra of C containing only
$I$-divisors;

     * •

${\textbf{B}}_{2}$ is a (nonempty) $\sigma$-special subalgebra of B.

###### Proof.

The first two claims follow from the fact that B and C are subalgebras except
possibly for those joins and meets. The other two claims follow from the
definition of the operations. For example, whenever a subalgebra S of
${\textbf{B}}\oplus_{P}{\textbf{C}}$ contains both an element $b\in B^{-}$ and
an element $c\in C^{-}$ that is not an $I$-divisor, then both the minimal,
$\sigma(b_{F})$, and the maximal, $\gamma_{F}(b)$, element of the equivalence
class $[b]_{F}$ also belong to S, since $bc=cb=\sigma(b)$ and $c\backslash
b=b/c=\gamma(b)$. Moreover, if in the subalgebra S there is at least an
element $b\in B^{-}$, and $c\in C^{-}$ is a left $I$-divisor, then
$c\backslash b=\ell(b)$ and similarly if $d\in C^{-}$ is a right $I$-divisor
then $b/d=r(d)$, thus such elements need to be in $S$. ∎

We are now going to show that an ultrapower of a gluing
${\textbf{B}}\oplus_{P}{\textbf{C}}$ is a gluing of ultrapowers of B and C.

######  Proposition 5.8.

$P_{U}({\textbf{B}}\oplus_{P}{\textbf{C}})\subseteq
P_{U}({\textbf{B}})\oplus_{(P_{U}(F),P_{U}(I))}P_{U}({\textbf{C}})$

###### Proof.

We sketch the proof. Let ${\textbf{A}}=\prod_{j\in
J}{\textbf{B}}\oplus_{P}{\textbf{C}}$, and let $U$ be an ultrafilter on $J$.
For $x=(x_{j})_{j\in J}\in A$ and $j\in J$, we distinguish cases according to
whether $x_{j}$ is in $B^{-},C^{-}$, $F$ or $I$, and partition $J$ in the
sets:

$J_{B}(x)=\\{j\in J:x_{j}\in B^{-}\\},\,J_{C}(x)=\\{j\in J:x_{j}\in C^{-}\\},$
$J_{F}(x)=\\{j\in J:x_{j}\in F\\},J_{I}(x)=\\{j\in J:x_{j}\in I\\}.$

Since $U$ is an ultrafilter, for each $x\in A$ only one of these sets belongs
to $U$; also if $[x]_{U}=[y]_{U}$, then the corresponding sets are both in $U$
or neither in $U$. This allows us to define the sets

$B_{U}=\\{[x]_{U}:J_{B}(x)\in U\\},C_{U}=\\{[x]_{U}:J_{C}(x)\in U\\},$
$F_{U}=\\{[x]_{U}:J_{F}(x)\in U\\},I_{U}=\\{[x]_{U}:J_{I}(x)\in U\\}.$

It is easy to see that $B_{U}\cup F_{U}\cup I_{U}$ and $C_{U}\cup F_{U}\cup
I_{U}$ are IRLs with the inherited operations, that $(B_{U}\cup F_{U}\cup
I_{U},F_{U},C_{U}\cup F_{U}\cup I_{U},I_{U})$ is a compatible quadruple and
that $B_{U}\cup F_{U}\cup I_{U}\in P_{U}({\textbf{B}})$ and $C_{U}\cup
F_{U}\cup I_{U}\in P_{U}({\textbf{C}})$. Finally, it can also be shown that
${\textbf{A}}/U$ is isomorphic to the gluing $(B_{U}\cup F_{U}\cup
I_{U})/U\oplus_{(F_{U},I_{U})}(C_{U}\cup F_{U}\cup I_{U})/U$ (see Proposition
3.3 in [1] for a similar instance). ∎

##  6\. Amalgamation property

The gluing construction can be seen as a way of finding a (strong) amalgam of
two algebras B and C in the particular case where the common subalgebra A
corresponds to the union of a congruence filter and an ideal of both B and C.

More precisely, let $({\textbf{B}},F,{\textbf{C}},I)$ be a compatible
quadruple, and let us call P the subalgebra of B (equivalently, C) with domain
$F\cup I$. Then P embeds into both B and C; let us name the embeddings with
$i,j$ respectively. Moreover, by construction both B and C embed in the gluing
${\textbf{B}}\oplus_{P}{\textbf{C}}$. Let us denote these embeddings by $h,k$,
respectively. With this notation in mind:

######  Proposition 6.1.

Let $({\textbf{B}},F,{\textbf{C}},I)$ be a compatible quadruple as above, and
let P be the subalgebra of B (equivalently, C) with domain $F\cup I$.
$({\textbf{B}}\oplus_{P}{\textbf{C}},h,k)$ is a (strong) amalgam of
$({\textbf{P}},{\textbf{B}},{\textbf{C}},i,j)$.

In this section we present two applications of the gluing and partial gluing
constructions, respectively, where the gluing constructions shed some light on
when amalgamation holds in classes of (bounded) IRLs.

###  6.1. Generalized rotations

We observe that the generalized $n$-rotation construction introduced in [7] is
actually an example of gluing and we generalize this construction to the non-
commutative case, using the gluing perspective.

The _generalized $n$-rotation_, for $n\geq 3$, defined in [7] is itself
inspired by ideas in Wroński’s reflection construction for BCK-algebras [28]
and generalizes in this context the (dis)connected rotation construction
developed by Jenei [20, 21] for ordered semigroups. In these constructions
given a CIRL, and also more generally in [17] given a topped residuated
lattice (not necessarily commutative or integral), a bounded involutive
structure is produced, obtained by attaching below the original CIRL a rotated
copy of it. On the other hand, the generalized rotation takes an CIRL and
generates a bounded CIRL, which is not necessarily involutive, by attaching
below it a rotated (possibly proper) _nuclear image_ of the original. The
generalized $n$-rotation, for $n\geq 3$, further adds a Łukasiewicz chain of
$n$ elements, $n-2$ of which are between the original structure and its
rotated nuclear image (see Figure 4 for a sketch).

We introduce the non-commutative version of the generalized $n$-rotation,
building on the construction in [17], and we apply it to IRLs.

We first recall from [17] that the disconnected rotation of an IRL A is the
$\sf{FL}_{ew}$-algebra ${\textbf{A}}^{*}$ whose lattice reduct is given by the
union of $A$ and its disjoint copy $A^{\prime}=\\{a^{\prime}:a\in A\\}$ with
dualized order, placed below $A$: for all $a,b\in A$,

$a^{\prime}<b,\mbox{ and }a^{\prime}\leq b^{\prime}\mbox{ iff }b\leq a.$

In particular, the top element of ${\textbf{A}}^{*}$ is the top $1$ of A and
the bottom element of ${\textbf{A}}^{*}$ is the copy $0:=1^{\prime}$ of the
top $1$. A is a subalgebra, the products in $A^{\prime}$ are all defined to be
the bottom element $0=1^{\prime}$, and furthermore, for all $a,b\in A$,

$a\cdot b^{\prime}=(b/a)^{\prime},\quad b^{\prime}\cdot a=(a\backslash
b)^{\prime};$ $a\backslash b^{\prime}=a^{\prime}/b=(b\cdot a)^{\prime},\quad
a^{\prime}\backslash b^{\prime}=a/b,\quad b^{\prime}/a^{\prime}=b\backslash
a.$

A nucleus on a residuated lattice
${\textbf{A}}=(A,\wedge,\vee,\cdot,\,\backslash,1)$ is a closure operator
$\delta$ on A that satisfies $\delta(x)\delta(y)\leq\delta(xy)$, for all
$x,y\in A$. It is known that then
${\textbf{A}}_{\delta}=(\delta[A],\wedge,\vee_{\delta},\cdot_{\delta},\,\backslash,\delta(1))$
is a residuated lattice, where $x\vee_{\delta}y=\delta(x\vee y)$ and
$x\cdot_{\delta}y=\delta(xy)$.

The _generalized disconnected rotation_ ${\textbf{A}}^{\delta}$ of a IRL A
with respect to a nucleus $\delta$ on A serves as a non-commutative version of
the construction given in [2] (which in turn was inspired by [9]). It differs
from the disconnected rotation above in that it replaces $A^{\prime}$ with
$\delta[A]^{\prime}=\\{\delta(a)^{\prime}:a\in A\\}$, where
$\delta(a)^{\prime}$ is short for $(\delta(a))^{\prime}$. It is easy to see
that then with respect to the above order we have
$\delta(a)^{\prime}\land\delta(b)^{\prime}=\delta(\delta(a)\lor\delta(b))^{\prime}$.
Moreover, for all $a\in A$, $b\in\delta[A]$,

$a\backslash b^{\prime}=(\delta(ba))^{\prime}\qquad\mbox{ and }\qquad
b^{\prime}/a=(\delta(ab))^{\prime}.$

The proof that ${\textbf{A}}^{\delta}$ is a residuated lattice is a very small
variation of the analogous proof for ${\textbf{A}}^{*}$, given in Section 6 of
[17]. In particular, the product is well-defined given the fact that
$a\backslash b^{\prime}=a^{\prime}/b=(\delta(b\cdot a))^{\prime}$. For an in-
depth analysis of this and other rotation constructions, see [15].

Clearly, the disconnected rotation is the special case of a generalized
disconnected rotation where the nucleus is the identity map. Now, the
_generalized $n$-rotation_ ${\textbf{A}}^{\delta}_{n}$ of an IRL A with
respect to a nucleus $\delta$ and $n\geq 3$ is defined on the disjoint union
of $A^{\delta}$ and $\\{\ell_{i}:0<i<n-1\\}$. We also set $\ell_{0}=0$ and
$\ell_{n-1}=1$, the bounds of ${\textbf{A}}^{\delta}$. The order extends the
order of ${\textbf{A}}^{\delta}$ by

$b<\ell_{1}<\ldots<\ell_{n-2}<a,$

for all $a\in A$ and $b\in\delta[A]$; see the rightmost structure of Figure 4.
The operations extend those of ${\textbf{A}}^{\delta}$, of the $n$-element
Łukasiewicz chain Łn, where
$0=\ell_{0}<\ell_{1}<\ldots<\ell_{n-2}<\ell_{n-1}=1$, and for $0<i<n-1$:

$a\ell_{i}=\ell_{i}=\ell_{i}a,\quad b^{\prime}\ell_{i}=0=\ell_{i}b^{\prime}.$

The proof that the resulting structure is an $\mathsf{FL_{ew}}$-algebra is an
easy combination of the proofs of [17] and [7], but it also follows from
Proposition 6.2 below. We mention that in [7] the generalized $n$-rotation is
defined with respect to nuclei that preserve the lattice operations, thus the
construction we propose here is more general also in the commutative case. The
subvariety $\mathsf{MVR_{n}}$ of $\mathsf{FL_{ew}}$ generated by the
generalized $n$-rotations of CIRLs where the nuclei preserve the lattice
operations is axiomatized in [7]. This class of algebras contains as
subvarieties, among others, the varieties of: Gödel algebras, product
algebras, the variety generated by perfect MV-algebras, nilpotent minimum
algebras, $n$-contractive BL-algebras, and Stonean residuated lattices.

We now show that the generalized $n$-rotation is a special case of a gluing.
We refer to $1$-sums of the kind Ł${}_{n}\oplus_{1}{\textbf{A}}$ as _$n$
-liftings _of an IRL A. Then generalized $n$-rotations are gluings of
disconnected rotations and $n$-liftings.

######  Proposition 6.2.

The generalized $n$-rotation ${\textbf{A}}^{\delta}_{n}$ of an IRL A with
respect to a nucleus $\delta$ for $n\geq 3$ is isomorphic to the gluing
${\textbf{A}}^{\delta}\oplus_{(A,\\{0\\})}(\textbf{\L}_{n}\oplus_{1}{\textbf{A}})$
of the generalized disconnected rotation ${\textbf{A}}^{\delta}$ and the
$1$-sum Ł${}_{n}\oplus_{1}{\textbf{A}}$ over A, $\\{0\\}$.

###### Proof.

First we show that the conditions of the gluing are satisfied. Note that $A$
is a congruence filter strictly above all other elements of
${\textbf{A}}^{\delta}$ and Ł${}_{n}\oplus_{1}{\textbf{A}}$, and $\\{0\\}$ is
a shared lattice ideal. For all $x\in A^{\delta}-A=\delta[A]^{\prime}$, there
is $y\in A$ with $x=\delta(y)^{\prime}$, so $x\backslash
0=(\delta(y))^{\prime}\backslash 1^{\prime}=\delta(y)/1=\delta(y)\in A$; also
$0\backslash x=1\in A$. Therefore, $x\mathrel{\theta_{A}}0$ and
$\sigma_{A}(x)=\min[x]_{A}=0$. Furthermore, since all elements in
(Ł${}_{n}\oplus_{1}{\textbf{A}})-(A\cup\\{0\\})$ are $0$-divisors, $\gamma$
does not need to be defined. Moreover, since $\sigma_{A}(x)=0$, for all $x\in
A^{\delta}-A$, $\sigma$ is clearly absorbing. Thus $({\textbf{A}}^{\delta},A)$
is a weak lower-compatible pair.

To show that (Ł${}_{n}\oplus_{1}{\textbf{A}},\\{0\\})$ is an upper-compatible
pair, first note that $\\{0\\}$ is a lattice ideal strictly below all other
elements. Moreover, since being an $I$-divisor here means being a $0$-divisor,
we have $\ell(x)=x\backslash 0$ and $r(x)=0/x$, for all $x\in$
(Ł${}_{n}\oplus_{1}A)-(A\cup\\{0\\})=$ Ł${}_{n}-\\{0,1\\}$.

Now we prove that $({\textbf{A}}^{\delta},A,$
Ł${}_{n}\oplus_{1}{\textbf{A}},\\{0\\})$ is a compatible quadruple:

  1. (1)

All elements of Ł${}_{n}-\\{1\\}$ are 0-divisors, thus the first condition is
satisfied.

  2. (2)

If $c,d\in$ Ł${}_{n}-\\{0\\}$, with $cd=0$, then $0x=x0=0=\sigma_{A}(x)$ for
all $x\in\delta[A]^{\prime}-\\{0\\}$.

  3. (3)

$\delta[A]^{\prime}$ is closed under join, so this condition is vacuously
true.

  4. (4)

(Ł${}_{n}\oplus_{1}A)-\\{0\\}$ has a least element $\ell_{1}$, so this
condition is also vacuously true.

It is clear that ${\textbf{A}}^{\delta}$ and Ł${}_{n}\oplus_{1}{\textbf{A}}$
are subalgebras of the gluing and also of the generalized $n$-rotation and
they are ordered the same way in both of these structures. Finally, for
$\ell_{i}\in$ Ł${}_{n}-\\{1,0\\}$, $1<i<n-1$ and
$\delta(a)^{\prime}\in\delta[A]^{\prime}$, we have
$\ell_{i}\delta(a)^{\prime}=0$ in both the generalized $n$-rotation and in the
gluing. ∎

$1$$1^{\prime}=0$$\oplus_{\mathbf{P}}$$0$$1$${=}$$1$$0$ Figure 4. The gluing
of the disconnected rotation of an IRL A with its $8$-lifting:
${\textbf{A}}^{*}\oplus_{\mathbf{P}}($Ł${}_{8}\oplus_{1}{\textbf{A}})$, where
$\mathbf{P}=({\textbf{A}},0)$.

In [3, Theorem 3.11], it is shown that a variety ${\mathsf{V}}$ of semilinear
CIRLs has the AP if and only if the variety generated by generalized
$n$-rotations of chains in ${\mathsf{V}}$ with respect to a nucleus definable
by a term in the language of residuated lattices, has the AP. This allows to
transfer known results about the AP in relatively tame varieties of CIRLs, to
varieties of $\mathsf{FL_{ew}}$-algebras that are more complicated to study.
For instance, since basic hoops, Wajsberg hoops, cancellative hoops, Gödel
hoops, all have the AP, so do the varieties generated by their generalized
$n$-rotations ([3], Corollary 3.12).

We show that one can go in the same direction also in the non-commutative
case, with the following bridge result.

######  Proposition 6.3.

Let $\delta$ be a term-defined nucleus for a class $\mathsf{K}$ of IRLs, and
let $n\geq 3$. The following are equivalent:

  1. (1)

${\mathsf{K}}$ has the amalgamation property;

  2. (2)

the class of generalized $n$-rotations
$\\{{\textbf{A}}^{\delta}_{n}:{\textbf{A}}\in{\mathsf{K}}\\}$ has the
amalgamation property.

  3. (3)

the class $\\{{\textbf{A}}^{\delta}_{m}:{\textbf{A}}\in{\mathsf{K}},m-1\mbox{
divides }n-1\\}$ has the amalgamation property.

###### Proof.

In this proof, let us denote with ${\mathsf{K}}^{\delta}_{n}$ the class of the
generalized $n$-rotations via $\delta$ of algebras in ${\mathsf{K}}$. We first
show (1) $\Leftrightarrow$ (2). The key idea here is that homomorphisms of
generalized $n$-rotations are uniquely determined by their restriction to the
upper-compatible triple in the gluing: i.e., the IRL and the Łn chain. This is
due to the fact that the lower-compatible triple is a rotation whose domain is
$A\cup\delta[A]^{\prime}$ for some IRL A, and the elements
$\delta(a)^{\prime}\in A$ are such that $\delta(a)^{\prime}=a\backslash 0$,
thus their homomorphic images are determined by those on $A$.

More precisely, any homomorphism $h:A\to B$, for ${\textbf{A}},{\textbf{B}}$
IRLs, extends to a homomorphism
$\bar{h}:{\textbf{A}}^{\delta}_{n}\to{\textbf{B}}^{\delta}_{n}$ in the
following way:

$\bar{h}(a)=h(a),\quad\bar{h}(\delta(a)^{\prime})=\delta(h(a))^{\prime}\quad\bar{h}(l_{i})=l_{i}\mbox{
for all }i:1\ldots n-1$

And vice versa, given any homomorphism
$k:{\textbf{A}}^{\delta}_{n}\to{\textbf{B}}^{\delta}_{n}$ the restriction
$k_{A}$ to A is a homomorphism from $A$ to $B$. Indeed, given $a\in A$,
suppose $k(a)\in{\textbf{B}}^{\delta}_{n}-B$. Then
$k(a^{n})=k(a)^{n}=0_{{\textbf{B}}^{\delta}_{n}}$, but $a^{n}\in A$, since $A$
is a congruence filter of the disconnected rotation. This leads to a
contradiction, since $\neg(a^{n})=a^{n}\backslash 0=\delta(a^{n})^{\prime}$,
thus $(\neg(a^{n}))^{2}=0$, but

$k((\neg(a^{n}))^{2})=(\neg(k(a^{n})))^{2}=(\neg 0)^{2}=1^{2}=1\neq k(0)=0.$

So, $a\in A$ implies $k(a)\in B$. Moreover, $h$ is an embedding iff $\bar{h}$
is an embedding, and if $k$ is an embedding then clearly $k_{A}$ is an
embedding.

Thus, suppose that ${\mathsf{K}}$ has the amalgamation property, and consider
a V-formation in ${\mathsf{K}}^{\delta}_{n}$:
${\textbf{A}}^{\delta}_{n},{\textbf{B}}^{\delta}_{n},{\textbf{C}}^{\delta}_{n}$
with embeddings
$i:{\textbf{A}}^{\delta}_{n}\to{\textbf{B}}^{\delta}_{n},j:{\textbf{A}}^{\delta}_{n}\to{\textbf{C}}^{\delta}_{n}$.
Then one can consider the restrictions of $i$ and $j$ to A, and obtain a
V-formation in ${\mathsf{K}}$, given by
${\textbf{A}},{\textbf{B}},{\textbf{C}}$ and the embeddings $i_{A},j_{A}$.
This has an amalgam, say D with embeddings $f:B\to D,g:C\to D$ such that
$f\circ i=g\circ j$. Thus it follows from what was shown before that
${\textbf{D}}^{\delta}_{n}$ is going to be an amalgam for the V-formation in
${\mathsf{K}}^{\delta}_{n}$, with embeddings $\bar{f},\bar{g}$.

Similarly, supposing that ${\mathsf{K}}^{\delta}_{n}$ has the amalgamation
property, we consider a V-formation in ${\mathsf{K}}$:
${\textbf{A}},{\textbf{B}},{\textbf{C}}\in{\mathsf{K}}$ and embeddings $k:A\to
B,l:A\to C$. We take the corresponding V-formation in
${\mathsf{K}}^{\delta}_{n}$:
${\textbf{A}}^{\delta}_{n},{\textbf{B}}^{\delta}_{n},{\textbf{C}}^{\delta}_{n}$
with embeddings $\bar{k},\bar{l}$. The amalgam in ${\mathsf{K}}^{\delta}_{n}$
is going to be some ${\textbf{D}}^{\delta}_{n}$, with D an IRL, and embeddings
$s:{\textbf{B}}^{\delta}_{n}\to{\textbf{D}}^{\delta}_{n},t:{\textbf{C}}^{\delta}_{n}\to{\textbf{D}}^{\delta}_{n}$.
Thus D with embeddings $s_{B},t_{C}$ are an amalgam for the V-formation in
${\mathsf{K}}$. Therefore, (1) and (2) are equivalent.

While (3) clearly implies (2), (2) $\Rightarrow$ (3) can be shown again via
the fact that homomorphisms of generalized $n$-rotations are uniquely
determined by their restriction to the upper-compatible triple in the gluing.
In particular, consider a V-formation
${\textbf{A}}^{\delta}_{j},{\textbf{B}}^{\delta}_{k},{\textbf{C}}^{\delta}_{l}$,
with ${\textbf{A}}^{\delta}_{j}$ embedding in the other two. If the amalgam in
${\mathsf{K}}$ of ${\textbf{A}},{\textbf{B}},{\textbf{C}}$ is D, then it is
routine to check that the desired amalgam is given by
${\textbf{D}}^{\delta}_{\textrm{lcm\\{k,l\\}}}$. ∎

###  6.2. A $2$-potent variety of $\sf{FL}_{ew}$-algebras

We are now going to study a variety in which the subdirectly irreducible
members can be characterized as partial gluings of a class of algebras. More
precisely, as partial gluings of simple 2-potent (i.e., satisfying
$x^{2}=x^{3}$) CIRL-chains. Let us consider totally ordered
$\sf{FL}_{ew}$-algebras that are 2-potent, and such that for every $x,y$

$x=1\mbox{ or }x\cdot(x\land y)\leq(x\land y)^{2}.$

Equivalently, for $y\leq x<1$, we have $xy=y^{2}$. The above is a positive
universal first-order formula, thus by results in [13] these structures
generate a variety of MTL-algebras that satisfy $x^{2}=x^{3}$ and

$x\lor\;((x\cdot(x\land y))\backslash(x\land y)^{2})=1.$

We will call this variety of $\mathsf{FL}_{ew}$-algebras $\mathsf{GL_{2}}$.
One can easily see that the finite chains in this variety are of the form
shown in Figure 5.

$1$$x$$x^{2}=x^{3}$$y$$y^{2}=y^{3}$$z$$z^{2}=0$ Figure 5. A subdirectly
irreducible algebra in $\mathsf{GL}_{2}$.

More in detail, we will show that any $\mathsf{GL_{2}}$-chain consists of
subintervals made of simple 2-potent CIRL-chains, and it can be characterized
as _partial gluings_ of such chains. Using this representation, we will show
that the amalgamation property fails for the class of
$\mathsf{GL_{2}}$-chains. In particular, using their representation as partial
gluings will allow us to fully characterize their subalgebras and determine
exactly which V-formations have an amalgam.

First, let us show how we can iterate the partial gluing construction in this
case. For example, we consider three simple 2-potent CIRL-chains
${\textbf{S}}_{1},{\textbf{S}}_{2}$ and ${\textbf{S}}_{3}$, as in Figure 6.

$1$$x$$x^{2}=0$${\textbf{S}}_{1}$$1$$y$$y^{2}=0$${\textbf{S}}_{2}$$1$$z$$z^{2}=0$${\textbf{S}}_{3}$$1$$x$$x^{2}=0$${\textbf{S}}_{1}$$\oplus_{\tau_{2}}$$1$$y$$y^{2}$${\textbf{S}}_{2}$${=}$$1$$y$$y^{2}$$x$$x^{2}=0$$1$$z$$z^{2}=0$${\textbf{S}}_{3}$$\oplus_{\tau_{3}}$$1$$y$$y^{2}$$x$$x^{2}=0$${\textbf{S}}_{1}\oplus_{\tau_{2}}{\textbf{S}}_{2}$${=}$$1$$z$$z^{2}$$y$$y^{2}$$x$$x^{2}=0$
Figure 6. The iterated partial gluing of three simple 2-potent chains.

Consider the triple $({\textbf{S}}_{1},\sigma_{1},\gamma_{1})$ where the
implication in ${\textbf{S}}_{1}$ is redefined to be: $x\to y=1$ iff $x\leq
y$, and undefined otherwise, and furthermore for all $a$ with $x^{2}\leq a\leq
x$, we have

$\sigma_{1}(a)=y^{2},\quad\gamma_{1}(a)=y,\quad\sigma_{1}(1)=\gamma_{1}(1)=1.$

Moreover, consider the triple $({\textbf{S}}_{2},\ell_{2},r_{2})$ where the
maps $\ell_{2},r_{2}$ have empty domain.

######  Lemma 6.4.

$({\textbf{S}}_{1},\sigma_{1},\gamma_{1})$ is a lower-compatible triple and
$({\textbf{S}}_{2},\ell_{2},r_{2})$ is an upper-compatible triple.

###### Proof.

For all $a,b\in S_{1}-\\{1\\}$, we have $\sigma(a)\leq b\leq\gamma(a)$, which
implies that the two operators form a residuated pair. In a lower-compatible
triple the implication $x\to y$ is defined iff $\sigma(x)\leq y$ and
$x\not\leq y$, and we can show that this holds in
$({\textbf{S}}_{1},\sigma_{1},\gamma_{1})$. Indeed, notice that for all $x,y$
such that $y^{2}\leq x$, we have $\sigma_{1}(x)\leq y$ and by definition $x\to
y$ is undefined if and only if $x\not\leq y$. Also, it follows from direct
computation that $\sigma_{1}$ is a strong conucleus, $\gamma_{1}$ is a closure
operator, and $cd=dc\leq\sigma_{1}(c)$ for all $c,d\in S_{1},d\neq 1$. Thus
$({\textbf{S}}_{1},\sigma_{1},\gamma_{1})$ is a lower-compatible triple.

$({\textbf{S}}_{2},\ell_{2},r_{2})$ is an upper-compatible triple since all
the products are defined and all other properties are vacuously true. ∎

Moreover, we can consider the ideal $I=\\{0\\}$ with $\top_{I}=0$, where $0$
is the bottom element of the chain, and both assumptions $A1,A2$ are
satisfied. Also, conditions $A3,A4$ are trivially satisfied since the algebras
considered are chains. Thus, letting
$\tau_{2}=(\sigma_{1},\gamma_{1},\ell_{2},r_{2})$, we can define the partial
gluing ${\textbf{S}}_{1}\oplus_{\tau_{2}}{\textbf{S}}_{2}$, that is a total
IRL since ${\textbf{S}}_{2}$ has a coatom (see Theorem 3.11).

Similarly, we consider the upper-compatible triple
$({\textbf{S}}_{3},\ell_{3},r_{3})$ where again $\ell,r$ have empty domain.
Now, we also consider ${\textbf{S}}_{1}\oplus_{\tau_{2}}{\textbf{S}}_{2}$
where $x\to y$ is defined and equal to $1$ iff $x\leq y$, and where for all
$a:x^{2}\leq a\leq x$, and $b:y^{2}\leq b\leq y$ we have

$\sigma(a)=x^{2},\quad\gamma(a)=x,\quad\sigma(b)=y^{2},\quad\gamma(b)=y,\quad\sigma(1)=\gamma(1)=1.$

Building on the same line of reasoning as Lemma 6.4,
$({\textbf{S}}_{1}\oplus_{\tau_{2}}{\textbf{S}}_{2},\sigma,\gamma)$ is a lower
compatible triple, and we can define the partial gluing
$({\textbf{S}}_{1}\oplus_{\tau_{2}}{\textbf{S}}_{2})\oplus_{\tau_{3}}{\textbf{S}}_{3}$
where $\tau_{3}=(\sigma,\gamma,\ell,r)$. This process can be iterated, and
provides a way of constructing a partial gluing of a finite family of simple
chains, indexed by a totally ordered set of indexes. Let us now give a more
general definition, in order to be able to construct a partial gluing of a
family of algebras, indexed by an arbitrary totally ordered chain with a
largest element.

We consider a family of algebras $\\{{\textbf{A}}_{i}\\}_{i\in\mathbf{I}}$,
where: each ${\textbf{A}}_{i}$ is a simple 2-potent CIRL-chain with a coatom
$c_{i}$ and a bottom $0_{i}$; $A_{i}\cap A_{j}=\\{1\\}$ for all $i,j\in I$;
$\mathbf{I}=(I,\leq)$ is a totally ordered index set with largest element
$i_{0}$. We will now define the _iterated partial gluing of
$\\{{\textbf{A}}_{i}\\}_{i\in I}$_, and denote it with