# An alternative definition of tense operators on residuated lattices

Ismael Calomino  CONICET and Instituto de Ciencias Básicas, Universidad
Nacional de San Juan, 5400 San Juan, Argentina<EMAIL_ADDRESS>,
Gustavo Pelaitay  CONICET and Departamento de Matemática, Facultad de Ciencias
Exactas, Universidad Nacional del Centro, Pinto 399, 7000 Tandil, Argentina.
<EMAIL_ADDRESS>and  William Zuluaga Botero

###### Abstract.

In this paper we introduce and study an alternative definition of tense
operators on residuated lattices. We give a categorical equivalence for the
class of tense residuated lattices, which is motivated by an old construction
due to J. Kalman. The paper concludes with some applications regarding the
description of congruences and a 2-contextual translation.

###### Key words and phrases:

Residuated lattice, tense operator, Kalman’s construction

###### 2010 Mathematics Subject Classification:

Primary 03B44; Secondary 06D20, 06D50

This work was supported by Consejo Nacional de Investigaciones Científicas y
Técnicas (PIP 11220170100195CO and PIP 11220200100912CO, CONICET-Argentina)
and Agencia Nacional de Promoción Científica y Tecnológica
(PICT2019-2019-00882, ANPCyT-Argentina). This project has also received
funding from MOSAIC Project 101007627 (European Union’s Horizon 2020 research
and innovation programme under the Marie Sklodowska-Curie).

##  1\. Introduction

Classical tense logic is a logical system obtained from introducing within the
classical propositional logic the notion of time, i.e., an expansion of
propositional logic by new unary operators which are called tense operators.
It is customary to denote these operators by $G$, $H$, $F$ and $P$, and
usually we define $F$ and $P$ via $G$ and $H$ as $F(x)=\neg G(\neg x)$ and
$P(x)=\neg H(\neg x)$, where $\neg x$ denote the Boolean negation of $x$. It
is well known that the class of tense Boolean algebras provides an algebraic
semantics for classical tense logic [2]. A tense Boolean algebra is a
structure $\langle B,G,H\rangle$ such that $B$ is a Boolean algebra and $G$
and $H$ are unary operators on $B$ satisfying the following conditions:

  1. (TB1)

$G(1)=1$ and $H(1)=1$,

  2. (TB2)

$G(x\wedge y)=G(x)\wedge G(y)$ and $H(x\wedge y)=H(x)\wedge H(y)$,

  3. (TB3)

$x\leq GP(x)$ and $x\leq HF(x)$.

Note that from conditions (TB1) and (TB2) it is the case that both operators
$G$ and $H$ are in fact, certain kind of modal operators, so the class of
tense Boolean algebras is, in particular, a class of Boolean algebras with
operators. Such class of alebras was studied in detail by Jónsson and Tarski
in [22].

Tense operators in intuitionistic logic were introduced by Ewald in [15]. In
this paper it was established the corresponding intuitionistic tense logical
system, called IKt. The axiomatization of Ewald is not minimal and in contrast
to classical tense logic, the tense operators $F$ and $P$ cannot be defined in
terms of $G$ and $H$. In [17], Figallo and Pelaitay gave an algebraic
axiomatization of the IKt system and showed that the algebraic axiomatization
given by Chajda in [9] of the tense operators $P$ and $F$ in intuitionistic
logic is not in accordance with the Halmos approach to existential operators
[21]. Thereafter, the study of tense operators has been extended to different
algebraic structures associated with non-classical logics [10, 14, 8]. This
can be evidenced by the approach applied in [17, 6], and also in [29], in
where a Kalman’s construction for the class of tense distributive lattices
with implication was studied.

Following the methodology proposed in [21, 9, 7, 1] and [28], the research on
algebraic properties of tense operators have been taken into the context of
residuated lattices. Motivated by the results presented in [17, 29], the aim
of this paper is to propose an alternative definition of tense operators on
residuated lattices different from the one presented in [1]. We also establish
a categorical equivalence through an adaptation of the Kalman’s construction
developed in [29] and moreover, we present some applications.

The paper is organized as follows. In Section 2 we review some results on
integral commutative residuated distributive lattices, or ICRDL-algebras for
short. In Section 3 we present a definition of tense operators on ICRDL-
algebras, unalike from the one depicted in [1], and we prove a Glivenko style
theorem for ICRDL-algebras as a generalization of similar results for Heyting
algebras. In Section 3.3, and following the results given in [5, 4], we
introduce the class of tense c-differential residuated lattices, or tense DRL-
algebras for short. In Section 4, we extend the Kalman’s construction to
establish a categorical equivalence between the category of tense ICRDL-
algebras and the category of tense DRL-algebras. In Section 5, we prove that
there is an isomorphism between the congruence lattice of a tense ICRDL-
algebra $\mathbf{L}$ and the congruence lattice of the tense ICRDL-algebra
arising from the Kalman functor $K(\mathbf{L})$. Furthermore, as an
application, we see that there is a correspondence between tense filters of
the classes of tense ICRDL-algebras and tense DRL-algebras. Finally, in
Section 6 we apply the results given in Section 5 to obtain a nontrivial
finite 2-contextual translation between the equational consequence relations
relative to tense DRL-algebras and tense ICRDL-algebras, respectively. The
reader is assumed to be familiar with standard results about adjoint functors
as presented in [25].

##  2\. Preliminaries

In this section we recall some results that will be useful through this paper.
For more details on residuated lattices the reader can see [19].

######  Definition 1.

[19] An integral commutative residuated distributive lattice, or ICRDL-algebra
for short, is a structure $\langle L,\vee,\wedge,\cdot,\to,0,1\rangle$ of type
$(2,2,2,2,0,0)$such that:

  * (R1)

$\langle L,\vee,\wedge,0,1\rangle$ is a bounded distributive lattice,

  * (R2)

$\langle L,\cdot,1\rangle$ is a commutative monoid,

  * (R3)

$x\cdot y\leq z$ if and only if $x\leq y\to z$, for all $x,y,z\in L$.

As usual, in what follows, we will denote ICRDL-algebras by their underlying
set $L$, when it is clear from the context.

######  Proposition 2.

[19] Let $L$ be a ICRDL-algebra. Then, the following hold:

  1. (1)

$1\to x=x$,

  2. (2)

$x\cdot y\leq x,y$,

  3. (3)

$y\leq x\to y$,

  4. (4)

$x\leq y$ if and only if $x\to y=1$,

  5. (5)

$x\to y\leq(x\cdot z)\to(y\cdot z)$,

  6. (6)

$x\leq y$ implies $x\cdot z\leq y\cdot z$,

  7. (7)

$x\leq y$ implies $z\to x\leq z\to y$ and $y\to z\leq x\to z$,

  8. (8)

$x\to(y\to z)=(x\cdot y)\to z$,

  9. (9)

$(x\vee y)\rightarrow z=(x\rightarrow z)\wedge(y\rightarrow z)$.

We recall that in every ICRDL-algebra, we can define a unary operation $\neg$
as $\neg x:=x\to 0$. The subsequent proposition summarizes the properties of
this operation.

######  Proposition 3.

Let $L$ be a ICRDL-algebra. Then, the following hold:

  1. (1)

$x\leq y$ implies $\neg y\leq\neg x$,

  2. (2)

$x\leq\neg\neg x$,

  3. (3)

$\neg\neg\neg x=\neg x$,

  4. (4)

$\neg 1=0$ and $\neg 0=1$,

  5. (5)

$x\cdot\neg x=0$,

  6. (6)

$x\to y\leq\neg y\to\neg x$,

  7. (7)

$\neg(x\vee y)=\neg(\neg\neg x\vee\neg\neg y)=\neg x\wedge\neg y$,

  8. (8)

$\neg\neg(x\wedge y)=\neg\neg x\wedge\neg\neg y$.

If $L$ is a ICRDL-algebra, then the set of regular elements of $L$ is
$Reg(L)=\\{\neg\neg a\colon a\in A\\}$. For any
$\star\in\\{\vee,\wedge,\cdot,\to\\}$, we consider the operation
$x\star_{r}y:=\neg\neg(x\star y)$. So, the structure $\langle
Reg(L),\vee_{r},\wedge_{r},\cdot_{r},\to_{r},0,1\rangle$ is a ICRDL-algebra.
Note that $a\to_{r}b=a\to b$ and $a\wedge_{r}b=a\wedge b$, for all $a,b\in
Reg(L)$. However, $\vee_{r}$ and $\cdot_{r}$ are different, in general, from
$\vee$ and $\cdot$, respectively, and so $Reg(L)$ may not be a subalgebra of
$L$. Nevertheless, from the results given in [13], in some cases $Reg(L)$ can
be obtained as a homomorphic image of $L$.

######  Lemma 4.

[13] Let $L$ be a ICRDL-algebra. Then, the following are equivalent:

  1. (1)

The correspondence $x\mapsto\neg\neg x$ is a homomorphism of $L$ onto
$Reg(L)$,

  2. (2)

$L$ satisfies the equation

$\neg\neg(\neg\neg x\to x)=1.$

Recall that a pseudocomplemented ICRDL-algebra is a ICRDL-algebra $L$ that
satisfies the equation

$x\wedge\neg x=0.$

######  Theorem 5.

[12, 3] Let $L$ be a ICRDL-algebra. Then $L$ is pseudocomplemented if and only
if $Reg(L)$ is a Boolean algebra.

##  3\. Tense ICRDL-algebras

In this section, we present an alternative definition of tense operators in
ICRDL-algebras. We stress that such a definition differs from the one proposed
by Barkhshi in [1].

###  3.1. Definition and properties

In [29], tense operators were defined on Heyting algebras and distributive
lattices with implication. Given that every Heyting algebra is, in particular,
an ICRDL-algebra, it is natural to provide the following definition.

######  Definition 6.

Let $L$ be a ICRDL-algebra. Let $G$, $H$, $F$ and $P$ be unary operations on
$L$ satisfying:

  * (T1)

$P(x)\leq y$ if and only if $x\leq G(y)$,

  * (T2)

$F(x)\leq y$ if and only if $x\leq H(y)$,

  * (T3)

$G(0)=0$ and $H(0)=0$,

  * (T4)

$G(x)\cdot F(y)\leq F(x\cdot y)$ and $H(x)\cdot P(y)\leq P(x\cdot y)$,

  * (T5)

$G(x\vee y)\leq G(x)\vee F(y)$ and $H(x\vee y)\leq H(x)\vee P(y)$,

  * (T6)

$G(x\to y)\leq G(x)\to G(y)$ and $H(x\to y)\leq H(x)\to H(y)$.

An algebra ${\mathbf{L}}=\langle L,G,H,F,P\rangle$ will be called tense ICRDL-
algebra and $G$,$H$, $F$ and $P$ will be called tense operators.

######  Remark 7.

Note that any tense Heyting algebra $\langle A,G,H,F,P\rangle$ (refer to
Definition 8.2 in [29]) is also a tense ICRDL-algebra.

In the following remarks, we can observe that Definition 6 and the one
proposed by Bakhshi in [1] are not necessarily related.

######  Remark 8.

In [1], the author introduces tense operators in the class of residuated
lattices as follows: a structure $\langle L,G,H\rangle$ is a tense residuated
lattice if $L$ is a residuated lattice, and $G$ and $H$ are unary operations
on $L$ satisfying

  1. (TRL0)

$G(1)=1$ and $H(1)=1$,

  2. (TRL1)

$G(x\to y)\leq G(x)\to G(y)$ and $H(x\to y)\leq H(x)\to H(y)$,

  3. (TRL2)

$x\leq GP(x)$ and $x\leq HF(x)$,

where $F(x)=\neg G(\neg x)$ and $P(x)=\neg H(\neg x)$. If we consider the
following bounded distributive lattice

$b$$0$$a$$c$$d$$1$

and the operations $\cdot=\wedge$,

$\to$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$
$a$ | $b$ | $1$ | $b$ | $1$ | $1$ | $1$
$b$ | $d$ | $d$ | $1$ | $1$ | $d$ | $1$
$c$ | $0$ | $d$ | $b$ | $1$ | $d$ | $1$
$d$ | $b$ | $c$ | $b$ | $c$ | $1$ | $1$
$1$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
$x$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$
---|---|---|---|---|---|---
$G(x)$ | $0$ | $a$ | $0$ | $c$ | $d$ | $1$
$H(x)$ | $0$ | $0$ | $b$ | $c$ | $0$ | $1$
$F(x)$ | $0$ | $c$ | $b$ | $c$ | $1$ | $1$
$P(x)$ | $0$ | $a$ | $b$ | $c$ | $d$ | $1$

then it is easy to check that $\langle L,G,H,F,P\rangle$ is a tense ICRDL-
algebra. However, it is not a tense residuated lattice in the sense of [1]
since

$F(a)=c\neq 1=\neg G(\neg a),$

and

$P(c)=c\neq 1=\neg H(\neg c).$

######  Remark 9.

Consider the chain of four elements $L=\\{0,a,b,1\\}$, where $0<a<b<1$, along
with the binary operations $\cdot$ and $\to$ defined by the following tables:

. | $0$ | $a$ | $b$ | $1$
---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$
$a$ | $0$ | $0$ | $0$ | $a$
$b$ | $0$ | $0$ | $a$ | $b$
$1$ | $0$ | $a$ | $b$ | $1$
$\to$ | $0$ | $a$ | $b$ | $1$
---|---|---|---|---
$0$ | $1$ | $1$ | $1$ | $1$
$a$ | $b$ | $1$ | $1$ | $1$
$b$ | $a$ | $b$ | $1$ | $1$
$1$ | $0$ | $a$ | $b$ | $1$

We define the tense operators $G$ and $H$ as $G(x)=H(x)=1$ for all
$x\in\\{0,a,b,1\\}$. Then the structure $\langle L,G,H\rangle$ is a tense
residuated lattice in the sense of [1], but it is not a tense ICRDL-algebra
since $H(0)=1\neq 0$ and $G(0)=1\neq 0$.

The following proposition can be proved in a similar fashion to Proposition
3.1 in [29], so we leave the details to the reader.

######  Proposition 10.

Let ${\bf{L}}$ be a tense ICRDL-algebra. Then:

  * (T7)

$G(x\to y)\leq F(x)\to F(y)$ and $H(x\to y)\leq P(x)\to P(y)$,

  * (T8)

$G(1)=1$ and $H(1)=1$,

  * (T9)

$G(x\wedge y)=G(x)\wedge G(y)$ and $H(x\wedge y)=H(x)\wedge H(y)$,

  * (T10)

$x\leq GP(x)$ and $x\leq HF(x)$,

  * (T11)

$F(0)=0$ and $P(0)=0$,

  * (T12)

$F(x\vee y)=F(x)\vee F(y)$ and $P(x\vee y)=P(x)\vee P(y)$,

  * (T13)

$FH(x)\leq x$ and $PG(x)\leq x$,

  * (T14)

$x\leq y$ implies $G(x)\leq G(y)$ and $H(x)\leq H(y)$,

  * (T15)

$x\leq y$ implies $F(x)\leq F(y)$ and $P(x)\leq P(y)$,

  * (T16)

$x\cdot F(y)\leq F(P(x)\cdot y)$ and $x\cdot P(y)\leq P(F(x)\cdot y)$,

  * (T17)

$F(x)\cdot y=0$ if and only if $x\cdot P(y)=0$,

  * (T18)

$G(x\vee H(y))\leq G(x)\vee y$ and $H(x\vee G(y))\leq H(x)\vee y$,

  * (T19)

$x\vee H(y)=1$ if and only if $G(x)\vee y=1$.

The following result was proved for Heyting algebras in [18]. We now present
its version for ICRDL-algebras.

######  Lemma 11.

Let $G$ and $H$ be two unary operators on a ICRDL-algebra ${\bf{L}}$ such that
$G$ and $H$ satisfy (T14). Then (T6) is equivalent to

  * (T6’)

$G(x)\cdot G(y)\leq G(x\cdot y)$ and $H(x)\cdot H(y)\leq H(x\cdot y)$.

###### Proof.

Let $a,b\in L$ and suppose that (T6’) is true. Then

$G(a)\cdot G(a\to b)\leq G(a\cdot(a\to b))\leq G(b).$

So, $G(a\to b)\leq G(a)\to G(b)$. Reciprocally, if we suppose that (T6) is
true, then we have $G(a)\leq G(b\to(a\cdot b))\leq G(b)\to G(a\cdot b)$ and
$G(a)\cdot G(b)\leq G(a\cdot b)$. Is similar for the operator $H$. ∎

######  Remark 12.

Note that if ${\bf{L}}$ is a tense ICRDL-algebra, then is a fuzzy dynamic
algebra in the sense of [8].

######  Proposition 13.

Let $\langle L,G,H,F,P\rangle$ be a tense ICRDL-algebra. Then

$G(x)\cdot F(x\to y)\leq F(y)\text{ and }H(x)\cdot P(x\to y)\leq P(y).$

###### Proof.

Let $a,b\in L$. Since $a\leq(a\to b)\to b$, then

$G(a)\leq G((a\to b)\to b)\leq F(a\to b)\to F(b)$

and $G(a)\cdot F(a\to b)\leq F(b)$. Similarly, we can prove the inequality
$H(x)\cdot P(x\to y)\leq P(y)$. ∎

###  3.2. Connections between tense ICRDL-algebras and tense Boolean algebras

We know that if $\bf{L}$ is a ICRDL-algebra, then the set of regular elements
$Reg(L)$ of $\bf{L}$ is a ICRDL-algebra. We conclude this section by showing
some necessary and sufficient conditions for the ICRDL-algebra $Reg(L)$ to be
a tense Boolean algebra.

The following result was proven in [17], within the context of Heyting
algebras.

######  Proposition 14.

Let ${\bf{L}}$ be a tense ICRDL-algebra. Then:

  1. (1)

$G(\neg x)\leq\neg G(x)$ and $H(\neg x)\leq\neg H(x)$,

  2. (2)

$G(\neg x)\leq\neg F(x)$ and $H(\neg x)\leq\neg P(x)$,

  3. (3)

$G(x)\leq\neg F(\neg x)$ and $H(x)\leq\neg P(\neg x)$,

  4. (4)

$F(\neg x)\leq\neg G(x)$ and $P(\neg x)\leq\neg H(x)$.

###### Proof.

We only prove (4). By Proposition 13, we have $G(x)\cdot F(x\to y)\leq F(y)$.
Consequently, according to (T11), it follows that $G(x)\cdot F(x\to 0)\leq
F(0)=0$, and $F(\neg x)\leq G(x)\to 0$, i.e., $F(\neg x)\leq\neg G(x)$. ∎

If ${\bf{L}}$ is a tense ICRDL-algebra, then we know that the structure
$Reg(\mathbf{L})=\langle
Reg(L),\vee_{r},\wedge_{r},\cdot_{r},\to_{r},0,1\rangle$ is a ICRDL-algebra.
Now, we define on $Reg(L)$ the following unary operators

$\displaystyle G_{r}(x)$ $\displaystyle:=$ $\displaystyle\neg\neg G(x),$
$\displaystyle H_{r}(x)$ $\displaystyle:=$ $\displaystyle\neg\neg H(x),$
$\displaystyle F_{r}(x)$ $\displaystyle:=$ $\displaystyle\neg\neg F(x),$
$\displaystyle P_{r}(x)$ $\displaystyle:=$ $\displaystyle\neg\neg P(x),$

and consider the inequalities

  * (A1)

$\neg F(\neg x)\leq G(x)$,

  * (A2)

$\neg P(\neg x)\leq H(x)$.

######  Remark 15.

Not every tense ICRDL-algebra satisfies conditions $(A1)$ and $(A2)$ (see
[17], Example 2). On the other hand, by Proposition 14, in every tense ICRDL-
algebra that satisfies conditions (A1) and (A2), we have the equations
$G(x)=\neg F(\neg x)$ and $H(x)=\neg P(\neg x)$.

######  Theorem 16.

Let ${\bf{L}}$ be a tense ICRDL-algebra. Then, there are equivalent:

  1. (1)

$\bf{L}$ is pseudocomplemented and satisfies conditions (A1) and (A2),

  2. (2)

$\langle Reg(\mathbf{L}),G_{r},H_{r}\rangle$ is a tense Boolean algebra.

###### Proof.

$(1)\Rightarrow(2)$ By Theorem 5, the structure $Reg(\mathbf{L})$ is a Boolean
algebra. By (T8) and (T9) it easily follows (TB1) and (TB2). We prove (TB3).
Let $a\in Reg(L)$. Then, by Remark 15 and as $a=\neg\neg a$, we have

$\neg G_{r}(\neg a)=\neg\neg\neg G(\neg a)=\neg G(\neg a)=\neg\neg F(\neg\neg
a)=\neg\neg F(a)=F_{r}(a),$

i.e., $F_{r}(a)=\neg G_{r}(\neg a)$. Similarly, we have $P_{r}(a)=\neg
H_{r}(\neg a)$. Finally, we see that $a\leq G_{r}P_{r}(a)$. By (T10), (T14),
(T15) and (2) of Proposition 3,

$a\leq GP(a)\leq\neg\neg G(\neg\neg P(a))=G_{r}P_{r}(a).$

Analogously, $a\leq H_{r}F_{r}(a)$. So, $\langle
Reg(\mathbf{L}),G_{r},H_{r}\rangle$ is a tense Boolean algebra.

$(2)\Rightarrow(1)$ It is immediate. ∎

###  3.3. Tense DRL-algebras

The aim of this subsection is to introduce the class of tense DRL-algebras,
which will be crucial for the rest of the paper.

We recall that a centered integral involutive residuated lattice is a
structure $\langle A,\vee,\wedge,\ast,\sim,c,0,1\rangle$ such that:

  1. (1)

$\langle A,\vee,\wedge,\ast,\sim,\Rightarrow,0,1\rangle$ is an integral
commutative residuated lattice, where $x\Rightarrow y=\sim(x\ast(\sim y))$,

  2. (2)

$\sim$ is a dual lattice-automorphism which is also an involution,

  3. (3)

$c$ is a fixed point of the involution.

If there is no risk of confusion, we will denote centered integral involutive
residuated lattices by their underlying set $A$.

A centered integral involutive residuated lattice $A$ is said to be a
c-differential residuated lattice, or DRL-algebra for short, if the following
equation, called Leibniz condition in [5, 4], holds:

  * (LC)

$(x\ast y)\wedge c=((x\wedge c)\ast y)\vee(x\ast(y\wedge c))$.

The following lemma will be useful in Section 6 of this paper.

######  Lemma 17.

Let $A$ be a DRL-algebra. Then, the following hold:

  1. (1)

$[(x\vee c)\ast(y\vee c)]\vee c=(x\ast y)\vee c.$

  2. (2)

$\sim((x\vee c)\ast(\sim(\sim y\vee c)))\wedge\sim((y\vee c)\ast(\sim(\sim
x\vee c)))=\sim(x\ast y)\vee c.$

###### Proof.

On the one hand, observe that

$\begin{array}[]{lcl}[(x\vee c)\ast(y\vee c)]\vee c&=&(x\ast y)\vee(c\ast
y)\vee(x\ast c)\vee(c\ast c)\vee c\\\ &=&(x\ast y)\vee c.\end{array}$

On the other hand, let

$t(x,y)=\sim((x\vee c)\ast(\sim(\sim y\vee c)))\wedge\sim((y\vee
c)\ast(\sim(\sim x\vee c))).$

Then, we have

$\begin{array}[]{lcl}t(x,y)&=&\sim[((x\vee c)\ast(y\wedge c))\vee((y\vee
c)\ast(x\wedge c))]\\\ &=&\sim[(x\ast(y\wedge c))\vee(y\ast(x\wedge
c))\vee(c\ast(x\wedge c))\vee(c\ast(y\wedge c))].\end{array}$

Note that $c\ast(x\wedge c)=\sim(c\Rightarrow(\sim x\vee c))=\sim 1=0$.
Similarly $c\ast(y\wedge c)=0$. Finally, by (LC), $(x\ast(y\wedge
c))\vee(y\ast(x\wedge c))=(x\ast y)\wedge c$. Hence,

$t(x,y)=\sim[((x\ast y)\wedge c)\vee 0]=\sim(x\ast y)\vee c.$

This concludes the proof. ∎

Let $A$ be a DRL-algebra and $G,H\colon A\rightarrow A$ be two unary
operators. We define the operators $F$ and $P$ by $F(x):=\sim G(\sim x)$ and
$P(x):=\sim H(\sim x)$, for any $x\in A$.

######  Definition 18.

Let $A$ be a DRL-algebra. Let $G$ and $H$ be unary operations on $A$
satisfying:

  * (t0)

$G(1)=1$ and $H(1)=1,$

  * (t1)

$G(c)=c$ and $H(c)=c$,

  * (t2)

$G(x\wedge y)=G(x)\wedge G(y)$ and $H(x\wedge y)=H(x)\wedge H(y)$,

  * (t3)

$x\leq GP(x)$ and $x\leq HF(x)$,

  * (t4)

$G(x\vee y)\leq G(x)\vee F(y)$ and $H(x\vee y)\leq H(x)\vee P(y)$,

  * (t5)

$G(x\Rightarrow y)\leq G(x)\Rightarrow G(y)$ and $H(x\Rightarrow y)\leq
H(x)\Rightarrow H(y)$.

An algebra ${\bf{A}}=\langle A,G,H\rangle$ will be called tense DRL-algebra.

######  Remark 19.

Notice that if ${\bf A}$ is a tense DRL-algebra, then the reduct $\langle
A,\vee,\wedge,\sim,G,H,0,1\rangle$ becomes a tense De Morgan algebra, as
defined in [16].

The following result enables us to provide an equivalent definition of tense
DRL-algebras.

######  Lemma 20.

Let ${\bf{A}}$ be a tense DRL-algebra. Then, axiom (t5) is equivalent to each
of the following axioms:

  * (t6)

$G(x\Rightarrow y)\leq F(x)\Rightarrow F(y)$ and $H(x\Rightarrow y)\leq
P(x)\Rightarrow P(y)$.

  * (t7)

$G(x)\ast F(\sim y)\leq F(x\ast(\sim y))$ and $H(x)\ast P(\sim y)\leq
P(x\ast(\sim y))$,

  * (t8)

$F(x)\ast G(y)\leq F(x\ast y)$ and $P(x)\ast H(y)\leq F(x\ast y)$.

###### Proof.

Let $a,b\in A$. We only prove that ${\rm(t6)}$ is equivalent to ${\rm(t7)}$.
We start by noticing that the following follows straight from the definitions:
$a\ast(\sim b)=\sim(a\Rightarrow b)$ and $\sim F(a)=G(\sim a)$. Therefore, if
$G(a\Rightarrow b)\leq G(a)\Rightarrow G(b)$, then

$G(\sim(a\ast(\sim b)))\leq G(a)\Rightarrow G(b).$

Thus, since $\sim$ is antitone, from the identities above we get
$\sim(G(a)\Rightarrow G(b))\leq F(a\ast(\sim b))$. Hence, $G(a)\ast F(\sim
b)\leq F(a\ast(\sim b))$. The converse uses the same arguments. The details of
the remaining proofs are left to the reader. ∎

The following proposition will be useful in the categorical equivalence that
we will present in the next section.

######  Proposition 21.

Let ${\bf{A}}$ be a tense DRL-algebra. Then:

  * (c1)

$F(c)=c$ and $P(c)=c$,

  * (c2)

$G(x\vee c)=G(x)\vee c$ and $H(x\vee c)=H(x)\vee c$,

  * (c3)

$F(x\wedge c)=F(x)\wedge c$ and $P(x\wedge c)=P(x)\wedge c$.

###### Proof.

We will only prove (c1) and (c2).

(c1): By (t1) and (t3), $F(c)=\sim G(\sim c)=\sim G(c)=\sim c=c$. Similarly,
$P(c)=c$.

(c2): Let $a\in A$. Since $a\leq a\vee c$ and $c\leq a\vee c$, we have that
$G(a)\leq G(a\vee c)$ and $c=G(c)\leq G(a\vee c)$. So, $G(a)\vee c\leq G(a\vee
c).$ On the other hand, from (t4) and (c1), we obtain $G(a\vee c)\leq G(a)\vee
F(c)=G(a)\vee c$. Therefore, $G(a\vee c)=G(a)\vee c$. Similarly, we have
$H(a\vee c)=H(a)\vee c$. ∎

##  4\. Kalman’s construction

In this section we prove some results that establish the connection between
the class of tense ICRDL-algebras and the class of tense DRL-algebras.

Let ${\mathbf{L}}$ be a tense ICRDL-algebra and let us consider

$K(L)=\\{(a,b)\in L\times L\colon a\cdot b=0\\}.$

It is well known from [5, 4] that by defining

$\displaystyle(a,b)\vee(x,y).$ $\displaystyle:=$ $\displaystyle(a\vee
x,b\wedge y),$ $\displaystyle(a,b)\wedge(x,y)$ $\displaystyle:=$
$\displaystyle(a\wedge x,b\vee y),$ $\displaystyle(a,b)\ast(x,y)$
$\displaystyle:=$ $\displaystyle(a\cdot x,(a\to y)\wedge(x\to b)),$
$\displaystyle(a,b)\Rightarrow(x,y)$ $\displaystyle:=$ $\displaystyle((a\to
x)\wedge(y\to b),a\cdot y),$ $\displaystyle\sim(a,b)$ $\displaystyle:=$
$\displaystyle(b,a),$ $\displaystyle 0$ $\displaystyle:=$
$\displaystyle(0,1),$ $\displaystyle 1$ $\displaystyle:=$
$\displaystyle(1,0),$ $\displaystyle c$ $\displaystyle:=$
$\displaystyle(0,0),$

we get that the structure $\mathbf{L}_{K}=\langle
K(L),\vee,\wedge,\ast,\sim,c,0,1\rangle$ is a DRL-algebra. Now, we define on
$K(L)$ the following unary operators given by

$\displaystyle G_{K}((a,b))$ $\displaystyle:=$ $\displaystyle(G(a),F(b)),$
$\displaystyle H_{K}((a,b))$ $\displaystyle:=$ $\displaystyle(H(a),P(b)),$
$\displaystyle F_{K}((a,b))$ $\displaystyle:=$ $\displaystyle(F(a),G(b)),$
$\displaystyle P_{K}((a,b))$ $\displaystyle:=$ $\displaystyle(P(a),H(b)).$

Taking into account the unary operators defined earlier, we obtain the
following result.

######  Lemma 22.

Let ${\mathbf{L}}$ be a tense ICRDL-algebra. Then, for every $(a,b)\in K(L)$:

  1. (1)

$G_{K}(a,b),H_{K}(a,b)\in K(L)$,

  2. (2)

$F_{K}(a,b)=\sim G_{K}(\sim(a,b))$ and $P_{K}(a,b)=\sim H_{K}(\sim(a,b))$,

  3. (3)

$F_{K}(a,b),P_{K}(a,b)\in K(L)$.

###### Proof.

We will only prove (1). If $(a,b)\in K(A)$, then $a\cdot b=0$. Therefore, from
(T4) and (T11), $G(a)\cdot F(b)\leq F(a\cdot b)=F(0)=0$. So, $(G(a),F(b))\in
K(A)$. The proof of $H_{K}(a,b)\in K(A)$ is similar. The details of the proof
of the remaining items are left to the reader. ∎

######  Lemma 23.

Let ${\mathbf{L}}$ be a tense ICRDL-algebra. Then

$K(\mathbf{L})=\langle\mathbf{L}_{K},G_{K},H_{K}\rangle$

is a tense DRL-algebra.

###### Proof.

From Lemma 22, the operators $G_{K}$ and $H_{K}$ are well defined. We will
prove the axioms of tense DRL-algebra. Due to the symmetry of tense operators
$G_{K}$ and $H_{K}$, we will prove the axioms only for the operator $G_{K}$.
We take $(a,b),(x,y)\in K(A)$.

(t0): From (T8) and (T11) it follows

$G_{K}(1)=G_{K}(1,0)=(G(1),F(0))=(1,0)=1.$

Similarly, $H_{K}(1)=1$.

(t1): From (T3) and (T10) it follows

$G_{K}(c)=G_{K}(0,0)=(G(0),F(0))=(0,0)=c.$

Similarly, $H_{K}(c)=c$.

(t2): From (T9) and (T12) we have

$\begin{array}[]{lll}G_{K}((a,b)\wedge(x,y))&=&G_{K}(a\wedge x,b\vee y)\\\
&=&(G(a\wedge x),F(b\vee y))\\\ &=&(G(a)\wedge G(x),F(b)\vee F(y))\\\
&=&(G(a),F(b))\wedge(G(x),F(y))\\\ &=&G_{K}(a,b)\wedge G_{K}(x,y).\\\
\end{array}$

(t3): By properties (T10) and (T13),

$\begin{array}[]{lll}(a,b)\wedge G_{K}P_{K}(a,b)&=&(a,b)\wedge
G_{K}((P(a),H(b)))\\\ &=&(a,b)\wedge(GP(a),FH(b))\\\ &=&(a\wedge GP(a),b\vee
FH(b))\\\ &=&(a,b),\\\ \end{array}$

i.e., $(a,b)\leq G_{K}P_{K}(a,b)$.

(t4): From (T5) and (T4), then

$\begin{array}[]{lll}G_{K}((a,b)\vee(x,y))&=&G_{K}(a\vee x,b\wedge y)\\\
&=&(G(a\vee x),F(b\wedge y))\\\ &\leq&(G(a)\vee F(x),G(b)\wedge F(y))\\\
&=&(G(a),G(b))\vee(F(x),F(y))\\\ &=&G_{K}(a,b)\vee F_{K}(x,y).\\\ \end{array}$

(t5): We start by noticing that $(a,b)\Rightarrow(x,y)=\sim((a,b)\ast(y,x))$.
Thus, it is the case that $G_{K}((a,b)\Rightarrow(x,y))=(G(a\rightarrow
x)\wedge G(y\rightarrow b),F(a\cdot y))$. Further, $G_{K}(a,b)\Rightarrow
G_{K}(x,y)=((G(a)\rightarrow G(x))\wedge(F(y)\rightarrow F(b)),G(a)\cdot
F(y))$. So, from (T6) and (T7), we have $G(a\to x)\leq G(a)\to G(x)$ and
$G(y\to b)\leq F(y)\to F(b)$. Besides, from (T4), $G(a)\cdot F(y)\leq F(a\cdot
y)$. Hence,

$(G(a\to x)\wedge G(y\to b),F(a\cdot y))\leq((G(a)\to G(x))\wedge(F(y)\to
F(b)),G(a)\cdot F(y)).$

Therefore, $G_{K}((a,b)\Rightarrow(x,y))\leq G_{K}(a,b)\Rightarrow
G_{K}(x,y)$. ∎

We write $\mathsf{tICRDL}$ for the category whose objects are tense ICRDL-
algebras, $\mathsf{tDRL}$ for the category whose objects are tense DRL-
algebras and in both cases, the morphisms are the corresponding algebra
homomorphisms. If $\mathbf{L}=\langle L,G,H,F,P\rangle$ and
$\mathbf{M}=\langle M,G,H,F,P\rangle$ are tense ICRLD-algebras and $f\colon
L\rightarrow M$ is a morphism in $\mathsf{tICRDL}$, then it is no hard to see
that the map $K(f)\colon K(L)\rightarrow K(M)$ given by
$K(f)(x,y)=(f(x),f(y))$ is a morphism in $\mathsf{tDRL}$ from $\langle
K(\mathbf{L}),G_{K},H_{K}\rangle$ to $\langle
K(\mathbf{M}),G_{K},H_{K}\rangle$. These assignments establish a functor $K$
from $\mathsf{tICRDL}$ to $\mathsf{tDRL}$.

Let ${\mathbf{A}}$ be a DRL-algebra and define

$C(A)=\\{x\in A\colon x\geq c\\}.$

If we consider $x\cdot y=(x\ast y)\vee c$ and $x\to y=\sim(x\ast(\sim y))$,
then the structure

$\mathbf{A}_{C}=\langle C(A),\vee,\wedge,\cdot,\to,c,1\rangle$

is a ICRDL-algebra (see [5], Theorem 7.6). Moreover, if $f\colon A\rightarrow
B$ is a homomorphism of DRL-algebras, then it is not hard to see that
$C(f)\colon C(A)\rightarrow C(B)$, defined by $C(f)(x)=f(x)$, is a
homomorphism of DRL-algebras. The following lemma shows that the latter
construction can be lifted to the class of tense DRL-algebras.

######  Lemma 24.

Let ${\mathbf{A}}$ be a tense DRL-algebra. Then

$C(\mathbf{A})=\langle\mathbf{A}_{C},G,H,F,P\rangle$

is a tense ICRDL-algebra. Moreover, if $f\colon A\rightarrow B$ is a morphism
in $\mathsf{tDRL}$, then $C(f)\colon C(A)\rightarrow C(B)$ is a morphism in
$\mathsf{tICRDL}$.

###### Proof.

Let $a,b\in A$. We only proof (T4). Notice that from (t3), it is the case that
$F$ preserves all the existing joins. Therefore, from by Lemma 20 we have

$G(x)\cdot F(y)=(G(x)\ast F(y))\vee c\leq F(x\ast y)\vee c=F(x\cdot y),$

as desired. ∎

Let ${\mathbf{A}}$ and ${\mathbf{B}}$ be two tense DRL-algebras. Let $f\colon
A\rightarrow B$ be a homomorphisms of tense DRL-algebras. It is clear now,
from Lemma 24, that the assingments $A\mapsto C(\mathbf{A})$ and $f\mapsto
C(f)$ determine a functor $C$ from $\mathsf{tDRL}$ to $\mathsf{tICRDL}$.

######  Lemma 25.

Let ${\mathbf{L}}$ be a tense ICRDL-algebra. Then the map $\alpha_{L}\colon
L\rightarrow C(K(L))$ given by $\alpha_{L}(x)=(x,0)$ is an isomorphism in
$\mathsf{tICRDL}$.

###### Proof.

Taking into account [5, Theorem 7.6], we only have to prove that $\alpha_{L}$
preserves the tense operators. Let $a\in A$. Then

$\alpha_{L}(G(a))=(G(a),0)=(G(a),F(0))=G_{K}((a,0))=G_{K}(\alpha_{L}(a))$

and

$\alpha_{L}(F(a))=(F(a),0)=(F(a),G(0))=F_{K}((a,0))=F_{K}(\alpha_{L}(a)).$

Similarly, we can prove $\alpha_{L}(H(a))=H_{K}(\alpha_{L}(a))$ and
$\alpha_{L}(P(a))=P_{K}(\alpha_{L}(a))$. ∎

If $\mathbf{A}$ is a DRL-algebra, then

$\beta_{A}\colon A\rightarrow K(C(A))$

given by $\beta_{A}(x)=(x\vee c,\sim x\vee c)$ is an injective homomorphism of
DRL-algebras (see [5], Lemma 7.5). We consider the folllowing condition on the
class of DRL-algebras:

  * (CK⋅)

for every $x,y\geq c$ such that $x\cdot y\leq c$, there exists $z\in A$ such
that $z\vee c=x$ and $\sim z\vee c=y$.

The next lemma follows from the definition of $\beta_{A}$.

######  Lemma 26.

[5] Let $\mathbf{A}$ be a DRL-algebra. Then $\mathbf{A}$ satisfies the
condition (CK⋅) if and only if $\beta_{A}$ is a surjective map.

Next, we will present a crucial lemma for the proof of Theorem 28.

######  Lemma 27.

Let ${\mathbf{A}}$ be a tense DRL-algebra. Then the map $\beta_{A}$ is a
monomorphism in $\mathsf{tDRL}$.

###### Proof.

Let $a\in A$. From (t3), (c2) and (c1) we have that

$\begin{array}[]{lll}\beta_{A}(G(a))&=&(G(a)\vee c,\sim G(a)\vee c)\\\
&=&(G(a)\vee c,F(\sim a)\vee c)\\\ &=&(G(a\vee c),F(\sim a\vee c))\\\
&=&G_{K}((a\vee c,\sim a\vee c)),\\\ \end{array}$

i.e., $\beta_{A}(G(a))=H_{K}(\beta_{A}(a))$. Similarly,
$\beta_{A}(H(a))=H_{K}(\beta_{A}(a))$. ∎

We will denote by $\mathsf{tDRL}_{c}$ for the full subcategory of
$\mathsf{tDRL}$ whose objects satisfy the condition (CK⋅). Straightforward
computations based on previous results of this section prove the following
result.

######  Theorem 28.

The functors K and C establish a categorical equivalence between
$\mathsf{tICRDL}$ and $\mathsf{tDRL_{c}}$ with natural isomorphisms $\alpha$
and $\beta$.

##  5\. A round trip between tense filters

Let $\mathbf{L}$ be a ICRDL-algebra. In Remark 7.12 of [5] it was proved that
Kalman’s construction can be used in order to establish an isomorphism between
the congruence lattice of $\mathbf{L}$ and the congruence lattice of the DRL-
algebra $K(\mathbf{L})$. Moreover, if $\mathbf{A}$ is a DRL-algebra, such a
result can also be extended for proving that there is an isomorphism between
the congruence lattice of $\mathbf{A}$ and the congruence lattice of the
ICRDL-algebra $C(\mathbf{L})$. In this section we are intended to extend these
results to the context of tense DRL-algebras and tense ICRDL-algebras.
Nevertheless, our approach borrows some of the ideas employed along Section
7.1 of [29]. As an application, we provide a detailed description between the
tense filters of the members of the varieties of tense DRL-algebras and tense
ICRDL-algebras.

We start by recalling some notions. Let $\mathbf{L}$ be a bounded distributive
lattice. If $\theta\in\mathsf{Con}(\mathbf{L})$, we can define a congruence
$\gamma_{\theta}$ of $K(\mathbf{L})$ by

$(a,b)\gamma_{\theta}(x,y)\ \text{ if and only if }\ (a,x)\in\theta\ \text{
and }(b,y)\in\theta.$

Reciprocally, if $\gamma\in\mathsf{Con}(K(\mathbf{L}))$, we can also define a
congruence $\theta^{\gamma}$ of $\mathbf{L}$ as

$(a,b)\in\theta^{\gamma}\text{ if and only if }(a,0)\gamma(b,0).$

######  Lemma 29.

Let $\mathbf{L}$ be a tense ICRDL-algebra and let
$\theta\in\mathsf{Con}(\mathbf{L})$. Then
$\gamma_{\theta}\in\mathsf{Con}(K(\mathbf{L}))$.

###### Proof.

We only proof that $\gamma_{\theta}$ is compatible with the product because
the proof of the compatibility with respect the residual is analogue. The
proof of the compatibility of $\theta^{\gamma}$ with the rest of the
operations uses the same ideas of Lemma 7.1 of [29]. Let
$(a_{i},b_{i})\gamma_{\theta}(x_{i},y_{i})$ with $i=1,2$. By definition of
$\gamma_{\theta}$ we have that $(a_{i},x_{i}),(b_{i},y_{i})\in\theta$ for
$i=1,2$. Then, since $\theta$ is a congruence of $\mathbf{L}$ we get that
$(a_{1}a_{2},x_{1}x_{2}),((a_{1}\rightarrow b_{2})\wedge(a_{2}\rightarrow
b_{1}),(x_{1}\rightarrow y_{2})\wedge(x_{2}\rightarrow y_{1}))$ belong to
$\theta$. Therefore,
$(a_{1},b_{1})\ast(a_{2},b_{2})\gamma_{\theta}(x_{1},y_{1})\ast(x_{2},y_{2})$.
∎

######  Lemma 30.

Let $\mathbf{L}$ be a tense ICRDL-algebra and let
$\gamma\in\mathsf{Con}(K(\mathbf{L}))$. Then
$\theta^{\gamma}\in\mathsf{Con}(\mathbf{L})$.

###### Proof.

We will prove the compatibility of $\theta^{\gamma}$ with respect to the
implication and the product. For the compatibility of $\theta_{\gamma}$ with
the rest of the operations, the reader may consult Lemma 7.1 of [29]. To do
so, let $(a_{1},b_{1}),(a_{2},b_{2})\in\theta^{\gamma}$. By definition of
$\theta^{\gamma}$ it is the case that $(a_{i},0)\gamma(b_{i},0)$ with $i=1,2$.
Thus, $(a_{1}\rightarrow a_{2},0)\gamma(b_{1}\rightarrow b_{2},0)$ from the
assumption on $\gamma$. Hence,
$(a_{1},0)\Rightarrow(a_{2},0)\gamma(b_{1},0)\Rightarrow(b_{2},0),$ so
$\theta^{\gamma}$ is compatible with the implication. On the other hand, since
$\gamma\in\mathsf{Con}(K(\mathbf{L}))$, we obtain that
$(a_{1},0)\ast(a_{2},0)\gamma(b_{1},0)\ast(b_{2},0)$ and $(a_{1}a_{2},\neg
a_{1}\wedge\neg a_{2})\gamma(b_{1}b_{2},\neg b_{1}\wedge\neg b_{2})$.
Moreover, we also get that
$(0,a_{2})=\sim(a_{2},0)\gamma\sim(b_{2},0)=(0,b_{2}),$ so
$(a_{1},0)\Rightarrow(0,a_{2})\gamma(b_{1},0)\Rightarrow(0,b_{2})$ and
consequently $(\neg a_{1}\wedge\neg a_{2},0)\gamma(\neg b_{1}\wedge\neg
b_{2},0)$. Finally, by transitivity on $\gamma$ we get
$(a_{1}a_{2},0)\gamma(b_{1}b_{2},0)$, we can conclude
$(a_{1}a_{2},b_{1}b_{2})\in\theta^{\gamma}$, as desired. ∎

The proof of the following result is a straight consequence of Lemmas 29 and
30 and it uses the same ideas deployed in Corollary 5.4 of [6]. So, we leave
the details to the reader.

######  Theorem 31.

Let $\mathbf{L}$ be a tense ICRDL-algebra. Then the assignment
$f:\mathsf{Con}(\mathbf{L})\rightarrow\mathsf{Con}(K(\mathbf{L}))$ defined by
$f(\theta)=\gamma_{\theta}$ is an order isomorphism.

######  Definition 32.

Let $\mathbf{L}$ be a tense ICRDL-algebra. A non-empty subset $S$ of $L$ is a
tense filter provided that $S$ is an up-set, $S$ is closed under
multiplication, $G$ and $H$.

######  Remark 33.

We claim that tense filters of tense ICRDL-algebras are closed under $F$ and
$P$. Indeed, if $S$ is a tense filter of a tense ICRDL-algebra $\mathbf{L}$
and $x\in S$, then since $G(0)=0$ and $H(0)=0$, by (T5) we get that $G(x)\leq
F(x)$ and $H(x)\leq P(x)$. Since $S$ is increasing and closed under $G$ and
$H$, then $F(x),P(x)\in S$, as claimed.

Let $\mathbf{L}$ be a tense ICRDL-algebra. We write $\mathrm{tFi}(\mathbf{L})$
for the poset of tense filters of $\mathbf{L}$ orderded by inclusion and
$\mathsf{Con}(\mathbf{L})$ for the congruence lattice of $\mathbf{L}$. If we
consider the operation $x\circ y=(x\rightarrow y)\cdot(y\rightarrow x)$, then
we have the following result.

######  Theorem 34.

Let $\mathbf{L}$ be a tense ICRDL-algebra, $S\in\mathrm{tFi}(\mathbf{L})$ and
$\theta\in\mathsf{Con}(\mathbf{L})$. Then:

  1. (1)

$S_{\theta}:=1/\theta$ is a tense filter of $\mathbf{L}$,

  2. (2)

$\theta_{S}:=\\{(x,y)\in L^{2}\colon x\circ y\in S\\}$ is a congruence of
$\mathbf{L}$,

  3. (3)

The assignments $\theta\mapsto S_{\theta}$ and $S\mapsto\theta_{S}$ establish
a mutually-inverse poset isomorphism between $\mathrm{tFi}(\mathbf{L})$ and
$\mathsf{Con}(\mathbf{L})$.

###### Proof.

It is proved in a similar way to Lemma 11 of [17]. ∎

Let $\mathbf{A}$ be a tense DRL-algebra. Note that since by definition,
$\mathbf{A}$ is a tense ICRDL-algebra, and by Theorem 34, we can obtain a
description of the lattice of congruences of $\mathbf{A}$ by means of tense
filters.

######  Corollary 35.

Let $\mathbf{A}$ be a tense DRL-algebra. Then the assignments $\theta\mapsto
D_{\theta}$ and $D\mapsto\theta_{D}$ establish a mutually-inverse poset
isomorphism between $\mathrm{tFi}(\mathbf{A})$ and $\mathsf{Con}(\mathbf{A})$.

Let $\mathbf{A}$ be a tense DRL-algebra. Our next goal is to describe
explicitly how from a congruence of $\mathbf{A}$ we can get a unique tense
filter of $C(\mathbf{A})$. We start by recalling from Theorem 28 that the map
$\beta_{\mathbf{A}}\colon\mathbf{A}\rightarrow K(C(\mathbf{A}))$ defined by
$\beta_{\mathbf{A}}(x)=(x\vee c,\sim x\vee c)$ is an isomorphism of tense DRL-
algebras. Further, from general reasons the map
$h_{\mathbf{A}}=\beta_{\mathbf{A}}\times\beta_{\mathbf{A}}$ which is defined
by $h_{\mathbf{A}}(x,y)=(\beta_{\mathbf{A}}(x),\beta_{\mathbf{A}}(y))$,
induces an isomorphism between $\mathsf{Con}_{\mathrm{tDRL}}(\mathbf{A})$ and
$\mathsf{Con}_{\mathrm{tDRL}}(KC(\mathbf{A}))$. Now, from Theorem 31, the map

$g\colon\mathsf{Con}_{\mathrm{tDRL}}(K(C(\mathbf{A})))\rightarrow\mathsf{Con}_{\mathrm{tICDRL}}(C(\mathbf{A}))$

defined by $g(\gamma)=\theta^{\gamma}$ is an isomorphism. Consider the
composite $w=gh_{\mathbf{A}}$, i.e.,

$w\colon\mathsf{Con}_{\mathrm{tDRL}}(\mathbf{A})\rightarrow\mathsf{Con}_{\mathrm{tICDRL}}(C(\mathbf{A})).$

######  Lemma 36.

Let $\mathbf{A}$ be a tense DRL-algebra and let
$D\in\mathrm{tFi}(\mathbf{A})$. Then

$S_{D}=D\cap C(A)$

is a tense filter of $C(\mathbf{A})$.

###### Proof.

Notice that from Corollary 35 and the discussion of above it is no hard to see
that $w(\theta_{D})=\theta_{D}\cap C(A)^{2}$. So, from Lemma 34 and the fact
that

$1/(\theta_{D}\cap C(A)^{2})=1/\theta_{D}\cap C(A)=D\cap C(A)$

the result follows. ∎

Let $\mathbf{A}$ be a tense DRL-algebra. Now, we proceed to describe
explicitly how from a tense filter of $C(\mathbf{A})$ we can obtain a unique
congruence of $\mathbf{A}$. To do so, let $S$ be a tense filter of
$C(\mathbf{A})$. Recall that from Lemma 34 the map

$\Theta\colon\mathrm{tFi}(C(\mathbf{A}))\rightarrow\mathsf{Con}(C(\mathbf{A}))$

is an isomorphism. By Proposition 31, the map

$k\colon\mathsf{Con}(C(\mathbf{A}))\rightarrow\mathsf{Con}(K(C(\mathbf{A})))$

defined by $k(\theta)=\gamma_{\theta}$ is a poset isomorphism. Let $(a,b)\in
K(C(A))$. Then, since $\beta_{\mathbf{A}}$ is an isomorphism, from condition
(CK⋅) there exists a unique $z\in A$ such that $z\vee c=a$ and $\sim z\vee
c=b$. Thus, it is the case that the map $r_{\mathbf{A}}\colon
K(C(\mathbf{A}))\rightarrow\mathbf{A}$ defined by $r_{\mathbf{A}}(a,b)=z$ is
in fact $\beta_{\mathbf{A}}^{-1}$ and therefore an isomorphism. By general
reasons, the map $w_{\mathbf{A}}=r_{\mathbf{A}}\times r_{\mathbf{A}}$ induces
an isomorphism between $\mathsf{Con}(K(C(\mathbf{A})))$ and
$\mathsf{Con}(\mathbf{A})$. Consider then the composite $m=w_{A}k\Theta$,
i.e.,

$m\colon\mathrm{tFi}(C(\mathbf{A}))\rightarrow\mathsf{Con}(\mathbf{A}).$

######  Lemma 37.

Let $\mathbf{A}$ be a tense DRL-algebra and let
$S\in\mathrm{tFi}(C(\mathbf{A}))$. Then

$D_{S}=\\{u\in A\colon u\vee c,c\Rightarrow u\in S\\}$

is a tense filter of $\mathbf{A}$.

###### Proof.

From the discussion of above, notice that $m(S)$ is a congruence of
$\mathbf{A}$. We shall prove that $m(S)=\theta_{S}$. To do so, observe first
that from Theorem 31 and Lemma 34 we have

(1) $k\Theta(S)=\\{((a,b),(x,y))\in K(C(T))^{2}\colon a\circ x,b\circ y\in
S\\}.$

Thus, if we write $r_{\mathbf{A}}(a,b)=u$ and $r_{\mathbf{A}}(x,y)=v$, it is
the case that $u\vee c=a$, $v\vee c=x$, $\sim u\vee c=b$ and $\sim v\vee c=y$.
Since $c\leq v\vee c$, then $c\Rightarrow(v\vee c)=1$. So, we get

$\begin{array}[]{rcl}a\Rightarrow x&=&(u\vee c)\Rightarrow(v\vee c)\\\
&=&(u\Rightarrow(v\vee c))\wedge(c\Rightarrow(v\vee c))\\\
&=&(u\Rightarrow(v\vee c))\wedge 1\\\ &=&u\Rightarrow(v\vee c).\end{array}$

By the same arguments of above, we can prove also that $x\Rightarrow
a=v\Rightarrow(u\vee c)$, $b\Rightarrow y=\sim u\Rightarrow(\sim v\vee c)$ and
$y\Rightarrow b=\sim v\Rightarrow(\sim u\vee c)$. Then, since $S$ is an up-
set, we obtain

$m(S)=\\{(u,v)\in T^{2}\colon u\Rightarrow(v\vee c),v\Rightarrow(u\vee
c),u\Rightarrow(\sim v\vee c),\sim v\Rightarrow(\sim u\vee c)\in S\\}.$

Hence $m(S)=\theta_{S}$. We conclude by noticing that from Corollary 35 and
the fact that $\sim u\Rightarrow c=c\Rightarrow u$ we get
$D_{S}=1/\theta_{S}$, as claimed. ∎

######  Theorem 38.

Let $\mathbf{A}$ be a tense DRL-algebra, $D\in\mathrm{tFi}(\mathbf{A})$ and
$S\in\mathrm{tFi}(C(\mathbf{A}))$. Then the assignments $D\rightarrow S_{D}$
and $S\mapsto D_{S}$ establish a mutually-inverse poset isomorphism between
$\mathrm{tFi}(\mathbf{A})$ and $\mathrm{tFi}(C(\mathbf{A}))$.

###### Proof.

We shall prove that $D_{S_{D}}=D$. On the one hand, let $u\in D$. Then from
the fact that $D$ is an up-set and $u\leq c\Rightarrow u$, we get that $u\vee
c$ and $c\Rightarrow u$ belong to $D\cap C(A)$. Thus, $u\in D_{S_{D}}$. On the
other hand, if $u\vee c,c\Rightarrow u\in D\cap C(A)$, since $(u\vee
c)\ast(c\Rightarrow u)\leq u$ and $D$ is a tense filter, we conclude $u\in D$.
The proof of $S_{D_{S}}=S$ is analogue. Straightforward calculations prove
that both assignments are monotone. ∎

We conclude this section by noticing that Theorems 31 and 34, and Corollary 35
may also be applied in order to make explicit the relation between the tense
filters of a tense ICRDL-algebra $\mathbf{L}$ and the tense filters of
$K(\mathbf{L})$. To this end, let $\Theta$ be the lattice isomorphism between
$\mathrm{tFi}(\mathbf{L})$ and $\mathsf{Con}(\mathbf{L})$ of Theorem 34, $w$
be the isomorphism between $\mathsf{Con}(\mathbf{L})$ and
$\mathsf{Con}(K(\mathbf{L}))$ of Theorem 31 and $\mu$ be the isomorphism
between $\mathsf{Con}(K(\mathbf{L}))$ and $\mathrm{tFi}(K(\mathbf{L}))$ of
Corollary 35. Then, if $g$ denotes the composition of these, it is clear that
$g$ establishes an isomorphism between $\mathrm{tFi}(\mathbf{L})$ and
$\mathrm{tFi}(K(\mathbf{L}))$. Moreover, if $S\in\mathrm{tFi}(\mathbf{L})$ and
$J\in\mathrm{tFi}(K(\mathbf{L}))$, straightforward calculations show that

$g(S)=\\{(x,y)\in K(A)\colon x\in S,\neg y\in S\\}=J_{S}$

and

$g^{-1}(J)=\\{a\in L\colon(a,0)\in J\\}=S_{J}.$

The latter discussion may be summarized in the following result.

######  Theorem 39.

Let $\mathbf{L}$ be a tense ICRDL-algebra, $S\in\mathrm{tFi}(\mathbf{L})$ and
$J\in\mathrm{tFi}(K(\mathbf{A}))$. Then the assignments $S\rightarrow J_{S}$
and $J\mapsto S_{J}$ establish a mutually-inverse poset isomorphism between
$\mathrm{tFi}(\mathbf{L})$ and $\mathrm{tFi}(K(\mathbf{L}))$.

##  6\. A contextual translation

Let $\mathcal{V}$ and $\mathcal{W}$ be varieties. Let us write $\mathsf{V}$
and $\mathsf{W}$ for the categories of algebras and homomorphisms of
$\mathcal{V}$ and $\mathcal{W}$, respectively. In [26], Moraschini proved that
there is a correspondence between the nontrivial adjunctions between
$\mathsf{V}$ and $\mathsf{W}$ (i.e. pairs of functors
$F\colon\mathsf{V}\rightarrow\mathsf{W}$ and
$G\colon\mathsf{W}\rightarrow\mathsf{V}$ such that $F$ is left adjoint to $G$
(denoted by $F\dashv G$)) and _nontrivial $\kappa$-contextual translations_
from the equational consequence relation relative to $\mathcal{V}$ into the
equational consequence relation relative to $\mathcal{W}$, denoted by
$\models_{\mathcal{V}}$ and $\models_{\mathcal{W}}$, respectively. In this
setting, $\kappa$ is assumed to be a non necessarily finite cardinal.
Classical examples of $\kappa$-contextual translations are the Kolmogorov
translation induced by the well known adjunction coming from the Kalman
construction for distributive lattices [11], the Glivenko translation induced
by Glivenko’s functor that relates Boolean algebras and Heyting algebras [23]
and the Gödel translation which relates the equational consequence relative to
Heyting algebras and interior algebras [20].

The aim of this section is to show that Theorem 28 can be employed to obtain a
finite nontrivial 2-contextual translation between the equational consequence
relations relative to tense DRL-algebras (or tDRL-algebras, for short) and
tense ICDRL-algebras (or tIRL-algebras, for short). In order to make this
section self-contained, we proceed to recall the definitions that we will
require on the following.

Let $X$ be a set of variables. If $\mathcal{L}$ is a propositional language we
write $\mathbf{Fm}_{\mathcal{L}}(X)$ for the absolutely free algebra of
$\mathcal{L}$-terms. If $\mathcal{V}$ is a variety, we write
$\mathbf{F}_{\mathcal{V}}(X)$ for the $\mathcal{V}$-free algebra over $X$.
Further, if $\mathcal{L}$ is the language of $\mathcal{V}$ and
$t(x_{1},...,x_{n})$ is an $\mathcal{L}$-formula, we also denote by
$t(x_{1},...,x_{n})$ its image under the natural map
$\mathbf{Fm}_{\mathcal{L}}(X)\rightarrow\mathbf{F}_{\mathcal{V}}(X)$ from the
term algebra $\mathbf{Fm}_{\mathcal{L}}(X)$ over $X$ onto
$\mathbf{F}_{\mathcal{V}}(X)$. In particular, if $X=\\{x_{1},...,x_{k}\\}$,
then we write $\mathbf{F}_{\mathcal{V}}(k)$ instead of
$\mathbf{F}_{\mathcal{V}}(X)$. We also denote by $\mathcal{V}$ the category
associated to the variety. Let $\mathbf{A}\in\mathcal{V}$. If $S\subseteq
A\times A$, we write $\mathsf{Cg}^{\mathbf{A}}(S)$ for the congruence
generated by $S$. We also write $\mathsf{Cg}^{\mathbf{A}}(\vec{a},\vec{b})$,
for the congruence generated by all pairs $(a_{1},b_{1}),\dots,(a_{N},b_{N})$
where $\vec{a},\vec{b}\in A^{N}$. We say that a congruence $\theta$ on
$\mathbf{A}$ is _finitely generated_ if $\theta=\mathsf{Cg}^{\mathbf{A}}(S)$
for some finite set $S\subseteq A\times A$. We recall that an algebra
$\mathbf{A}$ in $\mathcal{V}$ is a _finitely generated free algebra_ if it is
isomorphic to $\mathbf{F}_{\mathcal{V}}(m)$ for some finite $m$, and _finitely
presented_ if it is isomorphic to an algebra of the form
$\mathbf{F}_{\mathcal{V}}(k)/\theta$, for some $k$ finite and $\theta$
finitely generated congruence on $\mathbf{F}_{\mathcal{V}}(k)$.

If $l\colon X\rightarrow Y$ is a function, it is well known that there is a
unique homomorphism
$\mathbf{F}_{\mathcal{V}}(l)\colon\mathbf{F}_{\mathcal{V}}(X)\rightarrow\mathbf{F}_{\mathcal{V}}(Y)$
extending the map $\alpha(x)=l(x)$. If $(x_{1},...,x_{n})\in X^{n}$ and
$t(x_{1},...,x_{n})$ is a $\mathcal{L}$-term with variables in
$\\{x_{1},...,x_{n}\\}$, then it is well known that
$\mathbf{F}_{\mathcal{V}}(l)(t(x_{1},...,x_{n}))=t(l(x_{1}),...,l(x_{n}))$.
Let $\mathcal{F}=\\{\mathbf{A}_{i}\\}_{i\in I}$ be a family of algebras of
$\mathcal{V}$ and let us assume w.l.o.g. that their universes are pair-wise
disjoint. Let
$\pi_{i}\colon\mathbf{F}_{\mathcal{V}}(A_{i})\rightarrow\mathbf{A}_{i}$ be the
unique onto-homomorphism extending the identity of $A_{i}$. Then, if
$X=\bigcup_{i\in I}A_{i}$ and
$\gamma=\mathsf{Cg}^{\mathbf{F}_{\mathcal{V}(X)}}(\bigcup_{i\in
I}\text{Ker}(\pi_{i}))$ it is well known that
$\mathbf{F}_{\mathcal{V}}(X)/\gamma$ toghether with the homomorphisms
$f_{i}\colon\mathbf{A}_{i}\rightarrow\mathbf{F}_{\mathcal{V}}(X)/\gamma$
defined by $a\mapsto a/\gamma$, is isomorphic to the coproduct of
$\mathcal{F}$, denoted by $\sum_{i\in I}\mathbf{A}_{i}$. In particular, if
$I=\\{1,...,n\\}$, then we write $\mathbf{A}_{1}+\ldots+\mathbf{A}_{n}$
instead of $\sum_{i\in I}\mathbf{A}_{i}$.

###  6.1. 2-contextual translations

Let $\mathcal{K}$ be a class of similar algebras and let
$\mathcal{L}_{\mathcal{K}}$ be the language of $\mathcal{K}$. Then
$\mathcal{L}_{\mathcal{K}}^{2}$ is the algebraic language whose $n$-ary
operations (for every $n\in\mathbb{N}$) are all pairs of
$\mathcal{L}_{\mathcal{K}}$-terms built up with the variables
$X_{n}=\\{x_{j}^{1}\colon 1\leq j\leq n\\}\cup\\{x_{j}^{2}\colon 1\leq j\leq
n\\}$. That is to say,

$f((x_{1}^{1},x_{1}^{2}),...,(x_{n}^{1},x_{n}^{2}))=(t_{1}(x_{1}^{1},x_{1}^{2},...,x_{n}^{1},x_{n}^{2}),t_{2}(x_{1}^{1},x_{1}^{2},...,x_{n}^{1},x_{n}^{2})).$

where $t_{1},t_{2}\in Tm_{\mathcal{L}_{\mathcal{K}}}(X_{n})$. Thus, for
instance, if we set the language of binary operations
$\mathcal{L}_{\mathcal{K}}=\\{+,\cdot\\}$, then the operation $\oplus$,
defined as

$\oplus((x_{1}^{1},x_{1}^{2}),(x_{2}^{1},x_{2}^{2})):=(x_{2}^{2}\cdot(x_{1}^{1}+x_{2}^{1}),x_{1}^{2}\cdot
x_{2}^{1}),$

is a binary operation on $\mathcal{L}_{\mathcal{K}}^{2}$.

Let $\mathbf{A}\in\mathcal{K}$. We write $\mathbf{A}^{[2]}$ for the algebra of
type $\mathcal{L}_{\mathcal{K}}^{2}$ with universe $A^{2}$ where every $n$-ary
operation $f\in\mathcal{L}_{\mathcal{K}}^{2}$ is interpreted as

$f^{\mathbf{A}^{[2]}}((a_{1}^{1},a_{1}^{2}),...,(a_{n}^{1},a_{n}^{2})):=(t_{1}^{\mathbf{A}}(a_{1}^{1},a_{1}^{2},...,a_{n}^{1},a_{n}^{2}),t_{2}^{\mathbf{A}}(a_{1}^{1},a_{1}^{2},...,a_{n}^{1},a_{n}^{2})),$

for every $(a_{j}^{1},a_{j}^{2})\in A^{2}$, with $1\leq j\leq n$.

By the _2-matrix power_ of $\mathcal{K}$ me mean the class
$\mathbb{I}\\{\mathbf{A}^{[2]}\colon\mathbf{A}\in\mathcal{K}\\}$. Such a class
will be denoted by $\mathcal{K}^{[2]}$. It is well known that
$\mathcal{K}^{[2]}$ is a variety if and only if $\mathcal{K}$ is a variety
(Theorem 2.3 (iii) of [24]). Moreover, if
$\mathbf{A},\mathbf{B}\in\mathcal{K}$ and
$h\colon\mathbf{A}\rightarrow\mathbf{B}$ is a homomorphism, then the map
$h\times h\colon\mathbf{A}^{2}\rightarrow\mathbf{B}^{2}$ is a homomorphism.
So, if we set $h^{[2]}=h\times h$, then it is easy to see that the assignments
$\mathbf{A}\mapsto\mathbf{A}^{[2]}$ and $h\mapsto h^{[2]}$ establish a functor
${}^{[2]}\colon\mathcal{K}\rightarrow\mathcal{K}^{[2]}$.

Let $\mathcal{L}_{\mathcal{V}}$ and $\mathcal{L}_{\mathcal{W}}$ be the
languages of the varieties $\mathcal{V}$ and $\mathcal{W}$, respectively. A
_2-translation_ of $\mathcal{L}_{\mathcal{V}}$ into
$\mathcal{L}_{\mathcal{W}}$ is a map
$\tau\colon\mathcal{L}_{\mathcal{V}}\rightarrow\mathcal{L}_{\mathcal{W}}^{2}$
that preserves the arities of function symbols. That is to say, if $\psi$ is a
$n$-ary function symbol of $\mathcal{L}_{\mathcal{V}}$, then

$\tau(\psi)((x_{1}^{1},x_{1}^{2}),...,(x_{n}^{1},x_{n}^{2}))=(t_{1}(x_{1}^{1},x_{1}^{2},...,x_{n}^{1},x_{n}^{2}),t_{2}(x_{1}^{1},x_{1}^{2},...,x_{n}^{1},x_{n}^{2}))$

for some $t_{1},t_{2}\in Tm_{\mathcal{L}_{\mathcal{K}}}(X_{n})$. It is worth
recalling that $\tau$ maps constant symbols into pairs of constant symbols.
Therefore for a translation to exists, both languages are required to contain
at least one constant symbol. Observe that 2-translations extend to arbitrary
terms as follows: if $\lambda$ is a cardinal, we write $X_{\lambda}$ for the
set $\\{x_{j}^{1}\colon j<\lambda\\}\cup\\{x_{j}^{2}\colon j<\lambda\\}$.
Thus, we define recursively a map

$\tau_{\ast}\colon Tm(\mathcal{L}_{\mathcal{V}},\lambda)\rightarrow
Tm(\mathcal{L}_{\mathcal{W}},X_{\lambda})^{2}$

as

$\begin{array}[]{rclc}\tau_{\ast}(x_{j})&:=&(x_{j}^{1},x_{j}^{2}),&\text{for
every $j<\lambda$,}\\\ \tau_{\ast}(c)&:=&\tau(c).&\end{array}$

For variables and constant symbols. If $\psi\in\mathcal{L}_{\mathcal{V}}$ is
$n$-ary and $\varphi_{1},...,\varphi_{n}\in
Tm(\mathcal{L}_{\mathcal{V}},X_{\lambda})$, then

$\begin{array}[]{rcl}\tau_{\ast}(\psi(\varphi_{1},...,\varphi_{n}))&:=&\tau(\psi)(\tau_{\ast}(\varphi_{1}),...,\tau_{\ast}(\varphi_{n})).\end{array}$

Furthermore, the map $\tau_{\ast}$ may be raised to sets of equations in order
to produce a new map

$\tau^{\ast}:\mathcal{P}(\mathrm{Eq}(\mathcal{L}_{\mathcal{V}},\lambda))\rightarrow\mathcal{P}(\mathrm{Eq}(\mathcal{L}_{\mathcal{W}},X_{\lambda}))$

through the assignment

$\begin{array}[]{ccc}\Phi&\longmapsto&\\{\tau_{\ast}(\varepsilon)\approx\tau_{\ast}(\delta)\colon\varepsilon\approx\delta\in\Phi\\}.\end{array}$

A _2-contextual translation_ of $\models_{\mathcal{V}}$ into
$\models_{\mathcal{W}}$ is a pair $(\tau,\Theta)$ where $\tau$ is a
2-translation of $\mathcal{L}_{\mathcal{V}}$ into $\mathcal{L}_{\mathcal{W}}$
and $\Theta\subseteq\mathrm{Eq}(\mathcal{L}_{\mathcal{W}},2)$ is a set of
equations written with variables among $\\{x_{j}\colon j<\lambda\\}$ such
that:

  1. (1)

For every $n$-ary $\psi\in\mathcal{L}_{\mathcal{V}}$:

$\Theta(x_{1}^{1},x_{1}^{2})\cup\ldots\cup\Theta(x_{n}^{1},x_{n}^{2})\models_{\mathcal{W}}\Theta(\tau_{\ast}(\psi)((x_{1}^{1},x_{1}^{2}),...,(x_{n}^{1},x_{n}^{2}))).$

  2. (2)

For every cardinal $\lambda$ and
$\Phi\cup\\{\varepsilon\approx\delta\\}\subseteq\mathrm{Eq}(\mathcal{L}_{\mathcal{V}},\lambda)$:

$\begin{array}[]{ccc}\Phi\models_{\mathcal{V}}\epsilon\approx\delta&\Rightarrow&\tau_{\ast}(\Phi)\cup\bigcup\limits_{j<\lambda}\Theta(x_{j}^{1},x_{j}^{2})\models_{\mathcal{W}}\tau_{\ast}(\varepsilon\approx\delta).\end{array}$

The set $\Theta$ is called the _context_ of the 2-contextual translation
$(\tau,\Theta)$. We say that a 2-contextual translation is _nontrivial_ if
there is a (nonempty) sequence $\vec{\varphi}\in
Tm(\mathcal{L}_{\mathcal{W}},\emptyset)^{2}$ of constant symbols such that if
$\mathcal{W}\models\Theta(\vec{\varphi})$ then there is $i_{0}\in\\{1,2\\}$
and sequences of variables $\vec{x}=(x^{1},x^{2})$ and $\vec{y}=(y^{1},y^{2})$
such that

$\Theta(\vec{x})\cup\Theta(\vec{y})\not\models_{\mathcal{W}}x^{i_{0}}\approx
y^{i_{0}}.$

Now, we proceed to present some technical results that we will require in
order to establish the main result of this section.

######  Lemma 40.

Let $\mathbf{A}$ be a tense ICDRL-algebra. Then, for all $a,b\in A$ such that
$a\cdot b=0$, there exists a unique homomorphism $\psi\colon
C(\mathbf{F}_{t\mathcal{DRL}}(1))\rightarrow\mathbf{A}$ such that $\psi(z\vee
c)=a$ and $\psi(\sim z\vee c)=b$.

###### Proof.

Let $a,b\in A$ such that $a\cdot b=0$ and consider the assignment
$\varepsilon(z)=(a,b)$. If we take
$h\colon\mathbf{F}_{t\mathcal{DRL}}(1)\rightarrow K(\mathbf{A})$ as the unique
homomorphism that extends $\varepsilon$, since $C\dashv K$ (by Theorem 28),
there exists a unique homomorphism $\psi\colon
C(\mathbf{F}_{t\mathcal{DRL}}(1))\rightarrow\mathbf{A}$. For general reasons,
it follows that $\psi=\alpha^{-1}_{\mathbf{A}}C(h)$, where
$\alpha_{\mathbf{A}}$ is the map of Lemma 25. Observe that

$\begin{array}[]{ccl}\psi(\sim z\vee c)&=&\alpha^{-1}_{\mathbf{A}}(h(\sim
z\vee c))\\\ &=&\alpha^{-1}_{\mathbf{A}}(\sim h(z)\vee(0,0)))\\\
&=&\alpha^{-1}_{\mathbf{A}}(\sim(a,b)\vee(0,0))\\\
&=&\alpha^{-1}_{\mathbf{A}}((b,a)\vee(0,0))\\\
&=&\alpha^{-1}_{\mathbf{A}}(b,0)\\\ &=&b.\\\ \end{array}$

In a similar fashion it can be proved that $\psi(z\vee c)=a$. Now, suppose
that there exists another homomorphism $\psi^{\prime}\colon
C(\mathbf{F}_{t\mathcal{DRL}}(1))\rightarrow\mathbf{A}$ such that
$\psi^{\prime}(z\vee c)=a$ and $\psi^{\prime}(\sim z\vee c)=b$. Then, since
$C\dashv K$ (Theorem 28), there exist a unique
$\psi^{\prime}_{\ast}\colon\mathbf{F}_{t\mathcal{DRL}}(1)\rightarrow
K(\mathbf{A})$. We stress that from general reasons,
$\psi^{\prime}_{\ast}=K(\psi^{\prime})\beta_{\mathbf{F}_{t\mathcal{DRL}}(1)}$.
Therefore,

$\begin{array}[]{ccl}\psi^{\prime}_{\ast}(z)&=&K(\psi^{\prime})(z\vee c,\sim
z\vee c)\\\ &=&(\psi^{\prime}(z\vee c),\psi^{\prime}(\sim z\vee c))\\\
&=&(a,b)\\\ &=&\varepsilon(z)\\\ \end{array}$

so $\psi^{\prime}_{\ast}=h$ and consequently, $\psi^{\prime}=\psi$, as
claimed. ∎

######  Corollary 41.

Let $\mathbf{A}$ be a tense ICDRL-algebra. Then, for all
$a_{1},a_{2},b_{1},b_{2}\in A$ such that $a_{1}\cdot b_{1}=0$ and $a_{2}\cdot
b_{2}=0$, there exists a unique homomorphism $\mu\colon
C(\mathbf{F}_{t\mathcal{DRL}}(2))\rightarrow\mathbf{A}$ such that $\mu(x\vee
c)=a_{1}$, $\mu(y\vee c)=a_{2}$, $\mu(\sim x\vee c)=b_{1}$ and $\mu(\sim y\vee
c)=b_{2}$.

###### Proof.

Consider the assignment defined by $\varepsilon(x)=(a_{1},b_{1})$ and
$\varepsilon(y)=(a_{2},b_{2})$, and take
$h\colon\mathbf{F}_{t\mathcal{DRL}}(2)\rightarrow K(\mathbf{A})$ as the unique
homomorphism extending $\varepsilon$. It is no hard to see that the rest of
the proof is analogue to the one of Lemma 40. ∎

Let $\mathcal{V}$ be a variety, $\mathbf{A}\in\mathcal{V}$ and
$\vec{a},\vec{b}\in A^{N}$, with $N\in\mathbb{N}$. It is well known that the
canonical homomorphism
$\nu\colon\mathbf{A}\rightarrow\mathbf{A}/\mathsf{Cg}^{\mathbf{A}}(\vec{a},\vec{b})$
has the universal property of identify $\vec{a}$ with $\vec{b}$, in the sense
that for every homomorphism $h\colon\mathbf{A}\rightarrow\mathbf{B}$ such that
$g(a_{i})=g(b_{i})$ for every $1\leq i\leq N$, there exists a unique
homomorphism
$g\colon\mathbf{A}/\mathsf{Cg}^{\mathbf{A}}(\vec{a},\vec{b})\rightarrow\mathbf{B}$
such that $h\nu=g$. This fact implies that in order to prove that an algebra
$\mathbf{C}$ of $\mathcal{V}$ is isomorphic to
$\mathbf{A}/\mathsf{Cg}^{\mathbf{A}}(\vec{a},\vec{b})$ it is enough to find a
homomorphism with domain $\mathbf{A}$ with the universal property mentioned
above. This observation will be crucial for showing the following result.

######  Theorem 42.

$C(\mathbf{F}_{t\mathcal{DRL}}(1))$ is isomorphic to
$\mathbf{F}_{t\mathcal{ICRL}}(2)/\mathsf{Cg}^{\mathbf{F}_{t\mathcal{IRL}}(2)}(x\cdot
y,0)$.

###### Proof.

In order to prove our claim, regard the assignment $\alpha(x)=z\vee c$ and
$\alpha(y)=\sim z\vee c$ and let $h\colon\mathbf{B}\rightarrow
C(\mathbf{F}_{t\mathcal{DRL}}(1))$ be the homomorphism that extends $\alpha$.
Notice that for every $t(x,y)\in\mathbf{B}$, it is the case that
$h(t(x,y))=t^{C(\mathbf{F}_{t\mathcal{DRL}}(1))}(z\vee c,\sim z\vee c)$. In
particular we have that $h(x\cdot y)=c=0^{C(\mathbf{F}_{t\mathcal{DRL}}(1))}$.
In order to prove our claim, we will show that $h$ is universal with respect
to the homomorphisms $f\colon\mathbf{B}\rightarrow\mathbf{A}$ such that
$f(x\cdot y)=0^{\mathbf{A}}$. To do so, if we consider such an $f$, then
$f(x)\ast^{\mathbf{A}}f(y)=0^{\mathbf{A}}$, thus from Lemma 40, there exists a
unique homomorphism $\psi\colon
C(\mathbf{F}_{t\mathcal{DRL}}(1))\rightarrow\mathbf{A}$ such that $\psi(z\vee
c)=f(x)$ and $\psi(\sim z\vee c)=f(y)$. We will prove that the diagram below

$\textstyle{\mathbf{B}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\scriptstyle{f}$$\textstyle{C(\mathbf{F}_{t\mathcal{DRL}}(1))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\psi}$$\textstyle{\mathbf{A}}$

commutes. Indeed, if $t(x,y)\in\mathbf{B}$, then we have

$\begin{array}[]{ccl}\psi
h(t(x,y))&=&\psi(t^{C(\mathbf{F}_{t\mathcal{DRL}}(1))}(z\vee c,\sim z\vee
c))\\\ &=&t^{\mathbf{A}}(\psi(z\vee c),\psi(\sim z\vee c))\\\
&=&t^{\mathbf{A}}(f(x),f(y))\\\ &=&f(t(x,y)).\\\ \end{array}$

This concludes the proof. ∎

######  Corollary 43.

Let $X_{2}$ be the set $\\{x^{1}_{1},x^{1}_{2},x^{2}_{1},x^{2}_{2}\\}$ and let
$\mathbf{C}=\mathbf{F}_{t\mathcal{IRL}}(X_{2})$. Then, if
$\theta=\mathsf{Cg}^{\mathbf{C}}(x^{1}_{1}\cdot
x^{2}_{1},0)\vee\mathsf{Cg}^{\mathbf{C}}(x^{1}_{2}\cdot x^{2}_{2},0)$, the
algebras $C(\mathbf{F}_{t\mathcal{DRL}}(2))$ and $\mathbf{C}/\theta$ are
isomorphic.

###### Proof.

Let $\mathbf{B}=\mathbf{F}_{t\mathcal{ICRL}}(2)$ and
$\delta=\mathsf{Cg}^{\mathbf{F}_{t\mathcal{IRL}}(2)}(x\cdot y,0)$. Then, from
Theorem 42, the fact that $C$ and $\mathbf{F}_{t\mathcal{DRL}}$ preserve
coproducts, and the well known description on coproducts of algebras, the
following holds:

$\begin{array}[]{rcl}C(\mathbf{F}_{t\mathcal{DRL}}(2))&\cong&C(\mathbf{F}_{t\mathcal{DRL}}(1))+C(\mathbf{F}_{t\mathcal{DRL}}(1))\\\
&\cong&\mathbf{B}/\delta+\mathbf{B}/\delta\\\
&\cong&\mathbf{C}/\mathsf{Cg}^{\mathbf{C}}((x^{1}_{1}\cdot
x^{2}_{1},0),(x^{1}_{2}\cdot x^{2}_{2},0))\\\
&\cong&\mathbf{C}/\theta,\end{array}$

i.e., $C(\mathbf{F}_{t\mathcal{DRL}}(2))$ and $\mathbf{C}/\theta$ are
isomorphic. ∎

Next, we describe the procedure to obtain the contextual 2-translation
associated to the adjunction $C\dashv K$. If $\psi$ is a $n$-ary function
symbol on the language of tense DRL-algebras, by Theorem 4.3 of [27], the
2-translation $\tau$ is determined by a homomorphism $\tau(\psi)$ that makes
the following diagram commutes:

$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces\mathbf{F}_{tICRL}(2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tau(\psi)}$$\scriptstyle{h}$$\textstyle{C(\mathbf{F}_{tDRL}(1))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{C(\psi)}$$\textstyle{\mathbf{F}_{tICRL}(X_{n})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi_{n}}$$\textstyle{C(\mathbf{F}_{tDRL}(n))}$

We stress that the existence of $\tau(\psi)$ is granted because
$\mathbf{F}_{tICRL}(2)$ is onto-projective in $\mathsf{tICRL}$. Furthermore,
the map $\tau\colon\mathcal{L}_{tDRL}\rightarrow\mathcal{L}_{tICRL}^{2}$ can
be identified with its values on the generators. I.e.
$\tau(\psi):=(\tau(\psi)(x_{1}^{1}),\tau(\psi)(x_{1}^{2}))$. Notice that in
order to find such a map, it is enough to find terms
$t(x_{1}^{1},x_{1}^{2},\ldots,x_{n}^{1},x_{n}^{2})$ and
$s(x_{1}^{1},x_{1}^{2},\ldots,x_{n}^{1},x_{n}^{2})$ in
$\mathbf{F}_{tICRL}(X_{n})$ such that
$C(\psi)(h(x))=h(t(x_{1}^{1},x_{1}^{2},\ldots,x_{n}^{1},x_{n}^{2}))$ and
$C(\psi)(h(y))=h(s(x_{1}^{1},x_{1}^{2},\ldots,x_{n}^{1},x_{n}^{2}))$. In the
next result we will see that in our case, such terms can be obtained from the
operations we defined in Section 4 at the moment of establishing the functor
$K$.

######  Lemma 44.

The adjunction $C\dashv K$ induces the 2-translation
$\tau\colon\mathcal{L}_{tDRL}\rightarrow\mathcal{L}_{tICRL}^{2}$ defined as
follows:

$\begin{array}[]{rcl}\tau(\wedge)((x_{1}^{1},x_{1}^{2}),(x_{2}^{1},x_{2}^{2}))&:=&(x_{1}^{1}\wedge
x_{2}^{1},x_{1}^{2}\vee x_{2}^{2})\\\
\tau(\vee)((x_{1}^{1},x_{1}^{2}),(x_{2}^{1},x_{2}^{2}))&:=&(x_{1}^{1}\vee
x_{2}^{1},x_{1}^{2}\wedge x_{2}^{2})\\\
\tau(\ast)((x_{1}^{1},x_{1}^{2}),(x_{2}^{1},x_{2}^{2}))&:=&(x_{1}^{1}\cdot
x_{2}^{1},(x_{1}^{1}\rightarrow x_{2}^{2})\wedge(x_{2}^{1}\rightarrow
x_{1}^{2}))\\\
\tau(\Rightarrow)((x_{1}^{1},x_{1}^{2}),(x_{2}^{1},x_{2}^{2}))&:=&((x_{1}^{1}\rightarrow
x_{2}^{1})\wedge(x_{2}^{2}\rightarrow x_{1}^{2}),x_{1}^{1}\cdot x_{2}^{2})\\\
\tau(G)(x_{1}^{1},x_{1}^{2})&:=&(Gx_{1}^{1},Fx_{1}^{2})\\\
\tau(F)(x_{1}^{1},x_{1}^{2})&:=&(Fx_{1}^{1},Gx_{1}^{2})\\\
\tau(H)(x_{1}^{1},x_{1}^{2})&:=&(Hx_{1}^{1},Px_{1}^{2})\\\
\tau(P)(x_{1}^{1},x_{1}^{2})&:=&(Px_{1}^{1},Hx_{1}^{2})\\\
\tau(\sim)(x_{1}^{1},x_{1}^{2})&:=&(x_{1}^{2},x_{1}^{1})\\\
\tau(c)&:=&(0,0)\\\ \tau(0)&:=&(0,1)\\\ \tau(1)&:=&(1,0).\\\ \end{array}$

###### Proof.

We start by recalling that $C(G)\colon C(\mathbf{F}_{tDRL}(1))\rightarrow
C(\mathbf{F}_{tDRL}(1))$ maps a term $u(z)$ into $u(G(z))$ and by Theorem 42,
$h\colon\mathbf{F}_{tICRL}(2)\rightarrow C(\mathbf{F}_{tDRL}(1))$ maps a term
$t(x_{1}^{1},x_{1}^{2})$ into $t^{C(\mathbf{F}_{t\mathcal{DRL}}(1))}(z\vee
c,\sim z\vee c)$. Now, let us consider $t(x_{1}^{1},x_{1}^{2})=G(x_{1}^{1})$
and $s(x_{1}^{1},x_{1}^{2})=F(x_{1}^{2})$. Observe that from Proposition 21,
it is the case that $h(G(x_{1}^{1}))=C(G)(h(x_{1}^{1}))$. On the other hand,
from Definition 18 (t1) and (t2), we have

$\begin{array}[]{rcl}h(F(x_{1}^{2}))&=&F(\sim z\vee c)\\\ &=&\sim G(z\wedge
c)\\\ &=&\sim(G(z)\wedge c)\\\ &=&\sim G(z)\vee c\\\ &=&C(G)(\sim z\vee c)\\\
&=&C(G)(h(x_{1}^{2})).\\\ \end{array}$

Hence, we can set $\tau(G)(x_{1}^{1},x_{1}^{2})=(Gx_{1}^{1},Fx_{1}^{2})$ as
claimed. The proof of the statements about $\tau(H)$, $\tau(F)$, $\tau(P)$ and
$\tau(\sim)$ is similar. Now, in order to show that

$\tau(\ast)((x_{1}^{1},x_{1}^{2}),(x_{2}^{1},x_{2}^{2}))=(x_{1}^{1}\cdot
x_{2}^{1},(x_{1}^{1}\rightarrow x_{2}^{2})\wedge(x_{2}^{1}\rightarrow
x_{1}^{2})),$

we will proceed as next we will describe. Let us consider the following
diagram

$\textstyle{\mathbf{F}_{t\mathcal{IRL}}(2)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{h}$$\scriptstyle{\tau(\ast)}$$\textstyle{C(\mathbf{F}_{t\mathcal{DRL}}(1))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{C(\ast)}$$\textstyle{\mathbf{F}_{t\mathcal{IRL}}(X_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varphi_{\theta}}$$\scriptstyle{\pi_{2}}$$\textstyle{\mathbf{F}_{t\mathcal{IRL}}(X_{2})/\theta\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\lambda}$$\textstyle{C(\mathbf{F}_{t\mathcal{DRL}}(2))}$

in which $\theta$ denotes $\mathsf{Cg}^{\mathbf{F}_{tICRL}(X_{2})}(x_{1}\cdot
y_{1},0)\vee\mathsf{Cg}^{\mathbf{F}_{tICRL}(X_{2})}(x_{2}\cdot y_{2},0)$,
$\varphi_{\theta}$ is its respective quotient homomorphism and $\lambda$ is
the isomorphism of Corollary 43. Having into account that
$\pi_{2}=\lambda\varphi_{\theta}$, it is clear that proving
$\pi_{2}\tau(\ast)=C(\ast)h$ is equivalent to proof that
$\lambda\varphi_{\theta}\tau(\ast)=C(\ast)h$. Thus, since $C(\ast)\colon
C(\mathbf{F}_{tDRL}(1))\rightarrow C(\mathbf{F}_{tDRL}(2))$ maps a term $u(z)$
into $u(x\ast y)$, if $s(x^{1}_{1},x^{2}_{1},x^{1}_{2},x^{2}_{2})$ and
$t(x^{1}_{1},x^{2}_{1},x^{1}_{2},x^{2}_{2})$ are terms of $F_{tICRL}(X_{2})$
such that $\pi_{2}(s(x^{1}_{1},x^{2}_{1},x^{1}_{2},x^{2}_{2}))=C(\ast)(h(x))$
and $\pi_{2}(t(x^{1}_{1},x^{2}_{1},x^{1}_{2},x^{2}_{2}))=C(\ast)(h(y))$ the
latter equation leads us to prove the following

$\begin{array}[]{c}s^{C(\mathbf{F}_{tDRL}(2))}(\lambda(x^{1}_{1}/\theta),\lambda(x^{2}_{1}/\theta),\lambda(x^{1}_{2}/\theta),\lambda(x^{2}_{2}/\theta))=(x\ast
y)\vee c,\\\ \\\
t^{C(\mathbf{F}_{tDRL}(2))}(\lambda(x^{1}_{1}/\theta),\lambda(x^{2}_{1}/\theta),\lambda(x^{1}_{2}/\theta),\lambda(x^{2}_{2}/\theta))=\sim(x\ast
y)\vee c.\end{array}$

Now, since $x^{1}_{1}/\theta\cdot x^{2}_{1}/\theta=0/\theta$ and
$x^{1}_{2}/\theta\cdot x^{2}_{2}/\theta=0/\theta$, from Corollary 41, we can
assume w.l.o.g. that $\lambda(x^{1}_{1}/\theta)=x\vee c$,
$\lambda(x^{1}_{2}/\theta)=y\vee c$, $\lambda(x^{2}_{1}/\theta)=\sim x\vee c$
and $\lambda(x^{2}_{2}/\theta)=\sim y\vee c$. Therefore, by taking
$s(x^{1}_{1},x^{2}_{1},x^{1}_{2},x^{2}_{2})=x^{1}_{1}\cdot x^{1}_{2}$,
$t(x^{1}_{1},x^{2}_{1},x^{1}_{2},x^{2}_{2})=(x_{1}^{1}\rightarrow
x_{2}^{2})\wedge(x_{2}^{1}\rightarrow x_{1}^{2})$ and having in mind that
$\lambda$ is a homomorphism, in order to prove our claim, we shall prove the
following

$\begin{array}[]{rcl}\lambda(x^{1}_{1}/\theta)\cdot\lambda(x^{1}_{2}/\theta)&=&(x\ast
y)\vee c,\\\ \end{array}$
$\begin{array}[]{rcl}(\lambda(x_{1}^{1}/\theta)\Rightarrow\lambda(x_{2}^{2}/\theta))\wedge(\lambda(x_{2}^{1}/\theta)\Rightarrow\lambda(x_{1}^{2}/\theta))&=&\sim(x\ast
y)\vee c.\end{array}$

But notice that these are precisely the equations (1) and (2) of Lemma 17. So

$\tau(\ast)((x_{1}^{1},x_{1}^{2}),(x_{2}^{1},x_{2}^{2}))=(x_{1}^{1}\cdot
x_{2}^{1},(x_{1}^{1}\rightarrow x_{2}^{2})\wedge(x_{2}^{1}\rightarrow
x_{1}^{2})),$

as desired. The proof of the statements about $\tau(\Rightarrow)$,
$\tau(\vee)$ and $\tau(\wedge)$ are analogue. The details on the proof of
$\tau(c)$, $\tau(0)$ and $\tau(1)$ are left to the reader. ∎

In addition to Lemma 44, we stress that from Theorem 42, we may identify the
generating set $\\{(x\cdot y,0)\\}$ with the set equations $\Theta=\\{x\cdot
y\approx 0\\}$. Then, from Theorem 4.3 of [26], it turns out that the pair
$(\tau,\Theta)$ is a 2-contextual translation of $\models_{tDRL}$ into
$\models_{tICRL}$. Further is true. Since $C$ is faithful (Theorem 28), then
by Lemma 6.4 of [27] we are able to conclude:

######  Theorem 45.

For every cardinal $\lambda$ and
$\Phi\cup\\{\varepsilon\approx\delta\\}\subseteq\mathrm{Eq}(\mathcal{L}_{tDRL},\lambda)$,

$\Phi\models_{tDRL}\varepsilon\approx\delta\Longleftrightarrow\tau_{\ast}(\Phi)\cup\\{(x_{j}^{1}\cdot
x_{j}^{2},x_{j}^{1}\cdot x_{j}^{2})\approx(0,0)\colon
j<\lambda\\}\models_{tICRL}\tau_{\ast}(\varepsilon\approx\delta).$

where $(\tau,\Theta)$ is the contextual translation of $\models_{tDRL}$ into
$\models_{tICRL}$ induced by $C$.

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