An Exploration of Ideals and Filters In Triangle Algebras

1. Preliminaries

In this section, we review the basic definitions and results on residuated lattices [2,5,7,18,19], and triangle algebras [16,17,18] that will be used in the sequel.

1.1. Residuated Lattices

A bounded commutative residuated lattice is an algebraic structure $(L,\wedge ,\vee ,\odot ,\to ,0,1)$ of type $(2,2,2,2,0,0)$ satisfying the following conditions:

(RL1) 

$(L,\wedge ,\vee ,0,1)$ is a bounded lattice;

(RL2) 

$(L,\odot ,1)$ is a commutative monoid;

(RL3) 

for every $x,y,z\in L$, $x⩽y\to z\iff x\odot y⩽z$, where ⩽ is the lattice order induced by the lattice operations ∧ and ∨.

Unless otherwise specified, we denote a residuated lattice $(L,\wedge ,\vee ,\odot ,\to ,0,1)$ by L. For all $x\in L$, we define $x^{*}:=x\to 0$, $x^{**}:={\left(x^{*}\right)}^{*}$, $x\oslash y:=x^{*}\to y$, $x\oplus y:={(x^{*}\odot y^{*})}^{*}$, $x^0:=1$ and $x^n=x^{n-1}\odot x$, for $n\ge 1$.

In any bounded commutative residuated lattice L, we have the following rules of calculus (for detailed proofs, see [2,7,10,12,19]):

(r₁)

$x⩽y\iff x\to y=1$;

(r₂)

$x\odot y⩽x,y$;

(r₃)

$x\odot (x\to y)⩽y$;

(r₄)

$x⩽y\Rightarrow x\odot z⩽y\odot z$;

(r₅)

$x⩽y\Rightarrow \left\{\begin{array}{c}y\to z⩽x\to z\hfill \\ z\to x⩽z\to y\hfill \end{array}\right.$;

(r₆)

$x⩽y\Rightarrow y^{*}⩽x^{*}$ ;

(r₇)

$x⩽x^{**}$ and $x⩽y\Rightarrow x^{**}⩽y^{**}$;

(r₈)

$x^{*}=x^{***}$;

(r₉)

$x^{**}\odot y^{**}⩽{(x\odot y)}^{**}$;

(r₁₀)

$(x\to y)\odot (y\to z)⩽x\to z$;

(r₁₁)

$x\to y⩽(x\wedge z)\to (y\wedge z)$;

(r₁₂)

$x\to y⩽(x\odot z)\to (y\odot z)$;

(r₁₃)

$x\to y⩽(x\vee z)\to (y\vee z)$;

(r₁₄)

$x\to y⩽y^{*}\to x^{*}$;

(r₁₅)

${(x\vee y)}^{*}=x^{*}\wedge y^{*}$;

(r₁₆)

$x^{*}\vee y^{*}⩽{(x\wedge y)}^{*}$;

(r₁₇)

$x^{**}\wedge y^{**}={(x^{**}\wedge y^{**})}^{**}$;

(r₁₈)

$(x\vee y)\odot (x\vee z)⩽x\odot (y\vee z)$;

(r₁₉)

${(x\odot y)}^{*}=x\to y^{*}=y\to x^{*}=x^{**}\to y^{*}=y^{**}\to x^{*}$.

Filters and ideals play an important role in the connection between algebras and formal logic, and they have been intensively studied from an algebraic point of view.

A filter of L is any non-empty subset F of L satisfying:

(F1) 

$(\forall x,y\in L)\left(\right(x⩽y$ and $x\in F)\Rightarrow y\in F)$;

(F2) 

$(\forall x,y\in L)\left(\right(x\in F$ and $y\in F)\Rightarrow x\odot y\in F$).

Let us now consider the following formulas: for I a subset of L,

(I1) 

$(\forall x,y\in L)\left(\right(x⩽y$ and $y\in I)\Rightarrow x\in I)$;

(I’1) 

$0\in I$;

(I2) 

$(\forall x,y\in L)\left(\right(x\in I$ and $y\in I)\Rightarrow x\oslash y\in I)$;

(I3) 

$(\forall x,y\in L)\left(\right(x\in I$ and $y\in I)\Rightarrow x\oplus y\in I)$;

(I4) 

$(\forall x,y\in L)\left(\right(x\in I$ and $x^{*}\odot y\in I)\Rightarrow y\in I$);

(I5) 

$(\forall x,y\in L)\left(\right(x\in I$ and ${(x^{*}\to y^{*})}^{*}\in I)\Rightarrow y\in I$);

(I6) 

$(\forall x\in L)(x\in I\iff x^{**}\in I)$.

An ideal of L is a non-empty subset I of L that satisfies (I1) and (I2).

Some alternative characterizations of the notion of ideal in residuated lattices are given in the literature. Particularly, in [5], it is proved that a subset I of L is an ideal of L if and only if (I1) and (I3) are satisfied. In [7], it is proved that a subset I of L is an ideal of L if and only if (I’1) and (I4) are satisfied, if and only if (I’1) and (I5) are satisfied. For any ideal I of L, (I6) is always satisfied.

1.2. Triangle Algebras

Van Gasse et al. [15,16,17,18] introduced triangle algebras as a variety of residuated lattices equipped with approximation operators and with an angular point u different from 0 and 1.

A triangle algebra is an algebraic structure $\mathcal{A}=(A,\wedge ,\vee ,\odot ,\to ,\nu ,\mu ,0,u,1)$ of type (2,2,2,2,1,1,0,0,0) such that $(A,\wedge ,\vee ,\odot ,\to ,0,1)$ is a bounded commutative residuated lattice, and in which the following conditions hold: for all $x,y\in A$:

(T1) $\nu x⩽x$	(T’1) $x⩽\mu x$;
(T2) $\nu x⩽\nu \nu x$;	(T’2) $\mu \mu x⩽\mu x$;
(T3) $\nu (x\wedge y)=\nu x\wedge \nu y$;	(T’3) $\mu (x\wedge y)=\mu x\wedge \mu y$;
(T4) $\nu (x\vee y)=\nu x\vee \nu y$;	(T’4) $\mu (x\vee y)=\mu x\vee \mu y$;
(T5) $\nu u=0$;	(T’5) $\mu u=1$;
(T6) $\nu \mu x=\mu x$;	(T’6) $\mu \nu x=\nu x$;
(T7) $\nu (x\to y)⩽\nu x\to \nu y$;
(T8) $\nu (x\leftrightarrow y)\odot \mu (x\leftrightarrow y)⩽x\leftrightarrow y$;
(T9) $\nu x\to \nu y⩽\nu (\nu x\to \nu y)$.

Where, $x\leftrightarrow y:=(x\to y)\wedge (y\to x)$. Hereafter, we denote by $\mathcal{A}$ a triangle algebra $(A,\wedge ,\vee ,\odot ,\to ,\nu ,\mu ,0,u,1)$, unless otherwise specified.

The following is a simple example of a triangle algebra, which we will use in the sequel.

Example 1. 

 

Let $A_0=\{0,u,1\}$, such that $0<u<1$ and the operators $\nu ,\mu ,\odot$, and → are defined by Table 1; then ${\mathcal{A}}_0=(A_0,\wedge ,\vee ,\odot ,\to ,\nu ,\mu ,0,u,1)$ is a triangle algebra (see [22]).

In general, there is a way to construct a triangle algebra from a residuated lattice by means of the triangularization process; for details, we refer to [16]. In any triangle algebra $\mathcal{A}$, we have the following rules of calculus (see [15,16,17,18]): for all $x,y\in A$,

(p₁)

$\nu (x\to y)=(\nu x\to \nu y)\wedge (\mu x\to \mu y)$;

(p₂)

$\mu (x\to y)=(\mu x\to (\mu (u\odot u)\to \mu y\left)\right)\wedge (\nu x\to \mu y)$;

(p₃)

$\nu (x\odot y)=\nu x\odot \nu y$;

(p₄)

$\mu (x\odot y)=(\nu x\odot \mu y)\vee (\mu x\odot \nu y)\vee (\mu x\odot \mu y\odot \mu (u\odot u\left)\right)$;

(p₅)

$\nu 1=1$, $\mu 0=0$, $\nu 0=0$ and $\mu 1=1$;

(p₆)

$\nu \nu x=\nu x$, $\mu \mu x=\mu x$, and $\nu (\nu x\to \nu y)=\nu x\to \nu y$;

(p₇)

$\nu \left({\left(\nu x\right)}^{*}\right)={\left(\nu x\right)}^{*}$, $\mu \left({\left(\mu x\right)}^{*}\right)={\left(\mu x\right)}^{*}$;

(p₈)

$\mu \left({\left(\nu x\right)}^{*}\right)={\left(\nu x\right)}^{*}$, $\nu \left({\left(\mu x\right)}^{*}\right)={\left(\mu x\right)}^{*}$;

(p₉)

${\left(\mu x\right)}^{*}=\nu \left(x^{*}\right)$.

The notion of filter in triangle algebras has been introduced in [18]: a filter of $\mathcal{A}$ is any non-empty subset F of A such that for all $x,y\in A$:

(F1) 

$(x⩽y$ and $x\in F)\Rightarrow y\in F$;

(F2) 

$(x\in F$ and $y\in F)\Rightarrow x\odot y\in F$;

(FT) 

$x\in F\Rightarrow \nu x\in F$.

Observe that a filter of a triangle algebra is also a filter of the underlying residuated lattice. We denote by $\mathcal{F}\left(\mathcal{A}\right)$ the set of all filters of $\mathcal{A}$.

2. Ideals in Triangle Algebras

In MV-algebras, filters and ideals are dual notions, but this is not necessarily the case in general residuated lattices (see [6,7,13,20]). The authors in [7,13] have studied ideals and established a link with filters in residuated lattices by using different operators. In this section, we start by introducing the notion of ideal in the framework of triangle algebras. To establish the link between filters and ideals, we first prove that the operators used by [7] and [13] in residuated lattices are equal. Then, we prove that applying this operator to an ideal does not yield a filter in triangle algebras. Hence, we set up a new operator to link ideals to filters in the case of triangle algebras. We end by investigating the link between ideals and congruences, and some characterizations of the ideal generated by a given subset.

2.1. Definition and Characterizations of the Notion of Ideal

Since we want an ideal of a triangle algebra to be an ideal of the underlying residuated lattice, and taking into account the approximation operators $\nu$ and $\mu$, we propose the following definition.

Definition 1. 

An ideal of $\mathcal{A}$ is any non-empty subset I of A satisfying, for all $x,y\in A$:

(I1) 

$(x⩽y$ and $y\in I)\Rightarrow x\in I$;

(I2) 

$(x\in I$ and $y\in I)\Rightarrow x\oslash y\in I$;

(IT) 

$x\in I\Rightarrow \mu x\in I$.

We denote by $\mathcal{I}\left(\mathcal{A}\right)$ the set of all ideals of $\mathcal{A}$.

The following lemma will be helpful in the sequel.

Lemma 1. 

 

The following properties hold in any triangle algebra $\mathcal{A}$, for all $x,y\in A$.

1. 

$\mu \left(x^{*}\right)⩽{\left(\nu x\right)}^{*}$ and ${\left(\mu x\right)}^{*}⩽\mu \left(x^{*}\right)$;

2. 

$\mu (\nu x\oplus \nu y)=\nu x\oplus \nu y$ and $\mu (\mu x\oplus \mu y)=\mu x\oplus \mu y$;

3. 

$\mu (x\oplus y)⩽\mu x\oplus \mu y$ and $\mu (x\oslash y)⩽\mu x\oslash \mu y$ (triangle inequalities).

Proof. 

1.

Since $\nu x⩽x$ (T1), we have $x^{*}⩽{\left(\nu x\right)}^{*}$$\left(r_6\right)$, thus $\mu \left(x^{*}\right)⩽\mu \left({\left(\nu x\right)}^{*}\right)={\left(\nu x\right)}^{*}$ ((T’3) and $\left(p_8\right)$).

Since $x⩽\mu x$, we have ${\left(\mu x\right)}^{*}⩽x^{*}⩽\mu \left(x^{*}\right)$ by ($T^{\prime }1$ and $\left(r_6\right)$).

2.

$$\begin{array}{cc}\hfill \mu \left(\right(\nu x)\oplus (\nu y\left)\right)& =\mu \left({({\left(\nu x\right)}^{*}\odot {\left(\nu y\right)}^{*})}^{*}\right)\hfill \\ \hfill & ⩽{\left(\nu ({\left(\nu x\right)}^{*}\odot {\left(\nu y\right)}^{*})\right)}^{*}({\displaystyle according} {\displaystyle to} 1.{\displaystyle above})\hfill \\ \hfill & ={(\nu \left({\left(\nu x\right)}^{*}\right)\odot \nu \left({\left(\nu y\right)}^{*}\right))}^{*},\left(p_3\right)\hfill \\ \hfill & ={({\left(\nu x\right)}^{*}\odot {\left(\nu y\right)}^{*})}^{*},\left(p_7\right)\hfill \\ \hfill & =\left(\nu x\right)\oplus \left(\nu y\right)\hfill \\ \hfill & ⩽\mu (\left(\nu x\right)\oplus \left(\nu y\right)),\left(T^{\prime }1\right)\hfill \end{array}$$

Hence, we have $\mu \left(\right(\nu x)\oplus (\nu y\left)\right)=\left(\nu x\right)\oplus \left(\nu y\right)$.

By replacing x and y in the above by $\mu x$ and $\mu y$, respectively, and using $\nu \mu x=\mu x$ and $\nu \mu y=\mu y$, we have $\mu (\mu x\oplus \mu y)=\mu x\oplus \mu y$.

3.

Since $x⩽\mu x$ and $y⩽\mu y$$\left(T_1^{\prime }\right)$, we have $x\oplus y⩽\mu \left(x\right)\oplus \mu \left(y\right)$. Therefore, $\mu (x\oplus y)⩽\mu \left(\mu \right(x)\oplus \mu (y\left)\right)=\mu \left(x\right)\oplus \mu \left(y\right)$.

Furthermore,

$$\begin{array}{cc}\hfill \mu (x\oslash y)& =\mu (x^{*}\to y)\hfill \\ \hfill & ⩽\nu \left(x^{*}\right)\to \mu y,\left(p_2\right)\hfill \\ \hfill & ⩽{\left(\mu x\right)}^{*}\to \mu y,\left(p_9\right)\hfill \\ \hfill & =\mu x\oslash \mu y\hfill \end{array}$$

□

The following propositions are some characterizations of the notion of ideal in the case of triangle algebras. Some proofs are omitted, since in any residuated lattice we have the following equivalences [2]:

((I1) + (I2)) ⇔ ((I1) + (I3)) ⇔ ((I’1) + (I4)) ⇔ ((I’1) + (I5)).

Proposition 1. 

A non-empty subset I of A is an ideal of $\mathcal{A}$ iff for all $x,y\in A$:

(I1) 

$(x⩽y$ and $y\in I)\Rightarrow x\in I$;

(I3) 

$(x\in I$ and $y\in I)\Rightarrow x\oplus y\in I$;

(IT) 

$x\in I\Rightarrow \mu x\in I$.

Proposition 2. 

A non-empty subset I of A is an ideal of $\mathcal{A}$ iff for all $x,y\in A$:

(I’1) 

$0\in I$;

(I4) 

$(x\in I$ and $x^{*}\odot y\in I)\Rightarrow y\in I$;

(IT) 

$x\in I\Rightarrow \mu x\in I$.

Proposition 3. 

A non-empty subset I of A is an ideal of $\mathcal{A}$ iff for all $x,y\in A$:

(I’1) 

$0\in I$;

(I5) 

$(x\in I$ and ${(x^{*}\to y^{*})}^{*}\in I)\Rightarrow y\in I$;

(IT) 

$x\in I\Rightarrow \mu x\in I$.

The following result will be useful in some proofs.

Proposition 4. 

Let I be an ideal of $\mathcal{A}$ and $x\in A$, then the following conditions are equivalent:

(N₁)

$x\in I$;

(N₂)

$\mu x\in I$;

(N₃)

$x^{**}\in I$;

(N₄)

${\left(\mu x\right)}^{**}\in I$;

(N₅)

${\left(\nu x^{*}\right)}^{*}\in I$.

Proof. 

From (I6), we have $\left(N_1\right)\iff \left(N_3\right)$, i.e, $x\in I\iff x^{**}\in I$.

From (I1), (IT) and (T’1), we have $\left(N_1\right)\iff \left(N_2\right)$, i.e, $x\in I\&\iff \&\mu x\in I$.

From (I6), we have $\left(N_2\right)\iff \left(N_4\right)$, i.e, $\mu x\in I\iff {\left(\mu x\right)}^{**}\in I$.

Since $x⩽\mu x$$\left(T_1^{\prime }\right)$, we have ${\left(\mu x\right)}^{*}⩽x^{*}$$\left(r_6\right)$. Hence, ${\left(\mu x\right)}^{*}=\nu \left({\left(\mu x\right)}^{*}\right)⩽\nu \left(x^{*}\right)$$(p_8,T3)$, which implies ${\left(\nu \left(x^{*}\right)\right)}^{*}⩽{\left(\mu x\right)}^{**}$$\left(a\right)$.

Since $\nu \left(x^{*}\right)⩽x^{*}$, we have $x^{**}⩽{\left(\nu \left(x^{*}\right)\right)}^{*}$$\left(b\right)$.

From (a) and (I1), we have $\left(N_4\right)\Rightarrow \left(N_5\right)$, i.e, ${\left(\mu x\right)}^{**}\in I\Rightarrow {\left(\nu \left(x^{*}\right)\right)}^{*}\in I$.

From (b) and (I1), we have $\left(N_5\right)\Rightarrow \left(N_3\right)$, i.e, ${\left(\nu \left(x^{*}\right)\right)}^{*}\in I\Rightarrow x^{**}\in I$.

Thus, $\left(N_1\right)\iff \left(N_2\right)\iff \left(N_3\right)\iff \left(N_4\right)\iff \left(N_5\right)$. □

2.2. Pseudo-Duality

In the case of residuated lattices, in [7] and [13], the authors studied how to obtain a filter from an ideal, and vice versa, an ideal from a filter.

Let us first recall the procedure followed in [7]. Given F a filter and I an ideal of L, we obtain that $F_{*}:=\{x\in L;\exists y\in F,x^{**}⩽y^{*}\}$ is an ideal and $I_{*}:=\{x\in L;\exists y\in I,y^{*}⩽x^{**}\}$ is a filter of L. One can also prove that $I_{**}=I$, $F\subseteq F_{**}$ and $F_{*}=F_{***}$.

In [13], for F and I a filter and an ideal of L respectively, it was shown that $N\left(I\right):=\{x\in L;x^{*}\in I\}$ is a filter and $N\left(F\right):=\{x\in L;x^{*}\in F\}$ is an ideal of L. This induces the following properties:

$N\left(N\right(I\left)\right)=I$, $F\subseteq N\left(N\right(F\left)\right)$ and $N\left(N\right(N\left(F\right)\left)\right)=N\left(F\right)$.

When a filter F satisfies $N\left(N\right(F\left)\right)=F$, it is said to be an N-involutive filter.

In the following, we prove that for the pseudo-duality in residuated lattices the operators proposed by [7] and [13] are equal.

First, to simplify the proofs, we will use the following characterization of $F_{*}$.

Proposition 5. 

Let F be a filter of a residuated lattice L. Then, $F_{*}=\{x\in$L$;\exists y\in F,x⩽y^{*}\}$.

Proof. 

• Let $z\in F_{*}$, then there exists $a\in F$ such that $z^{**}⩽a^{*}$. Since $z⩽z^{**}$, we have $z⩽a^{*}$. Hence $z\in \{x\in$L$;\exists y\in F,x⩽y^{*}\}$.
• Let $x\in L$ such that there exists $y\in F$, $x⩽y^{*}$, then $x^{**}⩽y^{***}=y^{*}$. Hence $x\in F_{*}$.

□

Proposition 6. 

Let F be a filter and I an ideal of a residuated lattice L, then

(i) 

$N\left(I\right)=I_{*}$;

(ii) 

$N\left(F\right)=F_{*}$.

Proof. 

(i)

Let $x\in N\left(I\right)$, then $x^{*}\in I$. Putting $a=x^{*}$, we have $a^{*}⩽x^{**}$, $x\in I_{*}$.

Reciprocally, let $x\in I_{*}$, then there exists $a\in I$ such that $a^{*}⩽x^{**}$, which implies $x^{*}⩽a^{**}$. Since $a\in I$, we have $a^{**}\in I$, by (I6), thus $x^{*}\in I$, by (I1), i.e., $x\in N\left(I\right)$.

(ii)

Let $x\in N\left(F\right)$, then $x^{*}\in F$. Since $x⩽x^{**}$, we have $x\in F_{*}$, with $a=x^{*}$.

Reciprocally, let $x\in F_{*}$, then there exists $a\in F$ such that $x⩽a^{*}$, which implies $a^{**}⩽x^{*}$. Since $a⩽a^{**}⩽x^{*}$, we have $x^{*}\in F$, i.e., $x\in N\left(F\right)$.

□

Let us now, in the case of triangle algebras, consider the pseudo-duality between filters and ideals. We first show that the image of an ideal by the above operator in the case of triangle algebras is not always a filter.

Example 2. 

Let us consider the triangle algebra ${\mathcal{A}}_0$ of Example 1.

$I=\left\{0\right\}$ is an ideal of ${\mathcal{A}}_0$, but $N\left(I\right)=\{u,1\}$ is not a filter of ${\mathcal{A}}_0$. Indeed, $u\in N\left(I\right)$, but $\nu u=0\notin N\left(I\right)$, so $\left(FT\right)$ is not true.

Hence, in the following we introduce a new operator for the pseudo-duality in the case of triangle algebras.

Proposition 7. 

Let I be an ideal of $\mathcal{A}$. Then $T\left(I\right):=\{x\in A;\exists a\in I,a^{*}⩽{\left(\nu x\right)}^{**}\}$ is a filter of $\mathcal{A}$.

Proof. 

We have to show that $T\left(I\right)$ is a filter of $\mathcal{A}$.

For (F1),

let $x,y\in A$ such that $x\in T\left(I\right)$ and $x⩽y$, we have to show that $y\in T\left(I\right)$.

It holds that $x⩽y$ implies $\nu x⩽\nu y$, thus ${\left(\nu x\right)}^{**}⩽{\left(\nu y\right)}^{**}$. Since $x\in T\left(I\right)$, there exists $a\in I$ such that $a^{*}⩽{\left(\nu x\right)}^{**}$, so by transitivity, $a^{*}⩽{\left(\nu y\right)}^{**}$, thus $y\in T\left(I\right)$.

For (F2),

let $x,y\in T\left(I\right)$, we have to show that $x\odot y\in T\left(I\right)$.

Since $x\in T\left(I\right)$, there exists $a\in I$ such that $a^{*}⩽{\left(\nu x\right)}^{**}$.

Since $y\in T\left(I\right)$, there exists $b\in I$ such that $b^{*}⩽{\left(\nu y\right)}^{**}$. Then, by $\left(r_4\right)$, $a^{*}\odot b^{*}⩽{\left(\nu x\right)}^{**}\odot {\left(\nu y\right)}^{**}$, thus ${(a^{*}\odot b^{*})}^{**}⩽{({\left(\nu x\right)}^{**}\odot {\left(\nu y\right)}^{**})}^{**}$, i.e., ${(a\oplus b)}^{*}⩽{({\left(\nu x\right)}^{**}\odot {\left(\nu y\right)}^{**})}^{**}$.

Since $a,b\in I$ and I is an ideal, $a\oplus b\in I,$ and we have by $\left(r_9\right)$ ${\left(\nu x\right)}^{**}\odot {\left(\nu y\right)}^{**}⩽{(\left(\nu x\right)\odot \left(\nu y\right))}^{**}$, we obtain

$$\begin{array}{cc}\hfill {({\left(\nu x\right)}^{**}\odot {\left(\nu y\right)}^{**})}^{**}& ⩽{\left({(\left(\nu x\right)\odot \left(\nu x\right))}^{**}\right)}^{**},\left(r_9\right)\hfill \\ \hfill & ={(\left(\nu x\right)\odot \left(\nu x\right))}^{**},\left(r_8\right)\hfill \\ \hfill & ={\left(\nu (x\odot y)\right)}^{**},\left(p_3\right)\hfill \end{array}$$

It follows that ${(a\oplus b)}^{*}⩽{\left(\nu (x\odot y)\right)}^{**}$, thus $x\odot y\in T\left(I\right)$.

For (FT),

let $x\in T\left(I\right)$. We have to show that $\nu x\in T\left(I\right)$. Since $x\in T\left(I\right)$, there exists $a\in I$ such that $a^{*}⩽{\left(\nu x\right)}^{**}$. Since $\nu \nu x=\nu x$, $\left(p_6\right)$ we have $a^{*}⩽{\left(\nu \nu x\right)}^{**}$, thus $\nu x\in T\left(I\right).$

□

The following is a characterization of the above operator.

Proposition 8. 

Let I be an ideal of $\mathcal{A}$, then $T\left(I\right)=\{x\in A;{\left(\nu x\right)}^{*}\in I\}$.

Proof. 

• Let $z\in A$ such that ${\left(\nu z\right)}^{*}\in I$. Since ${\left({\left(\nu z\right)}^{*}\right)}^{*}⩽{\left(\nu z\right)}^{**}$, we have $z\in \{x\in A;\exists a\in I,a^{*}⩽{\left(\nu x\right)}^{**}\}=T\left(I\right)$, for $a={\left(\nu z\right)}^{*}$.
• Let $x\in T\left(I\right)$. Hence, there exists $a\in I$ such that $a^{*}⩽{\left(\nu x\right)}^{**}$, which implies ${\left(\nu x\right)}^{*}⩽a^{**}$. Since $a\in I$, by (I6), $a^{**}\in I$, so ${\left(\nu x\right)}^{*}\in I$, by (I1).

□

According to the relation between $\mu$ and $\nu$, we introduce the following operator for filters of triangle algebras.

Proposition 9. 

Let F be a filter of $\mathcal{A}$, then $T\left(F\right):=\{x\in A;{\left(\mu x\right)}^{*}\in F\}$ is an ideal of $\mathcal{A}$.

Proof. 

For (I1),

let $x,y\in A$ such that $x⩽y$ and $y\in T\left(F\right)$. We have that ${\left(\mu y\right)}^{*}\in F$ and $x⩽y$ imply ${\left(\mu y\right)}^{*}⩽{\left(\mu x\right)}^{*}$. Hence, ${\left(\mu x\right)}^{*}\in F$, because F is a filter. Thus $x\in T\left(F\right)$.

For (I3),

let $x,y\in T\left(F\right)$. Then, ${\left(\mu x\right)}^{*}\in F$ and ${\left(\mu y\right)}^{*}\in F$. Hence, ${\left(\mu x\right)}^{*}\odot {\left(\mu y\right)}^{*}\in F$. Furthermore,

$$\begin{array}{cc}\hfill {\left(\mu x\right)}^{*}\odot {\left(\mu x\right)}^{*}& ⩽{({\left(\mu x\right)}^{*}\odot {\left(\mu y\right)}^{*})}^{**},\left(r_7\right)\hfill \\ \hfill & ={\left({({\left(\mu x\right)}^{*}\odot {\left(\mu y\right)}^{*})}^{*}\right)}^{*}\hfill \\ \hfill & ={(\mu x\oplus \mu y)}^{*}\hfill \end{array}$$

Therefore, ${(\mu x\oplus \mu y)}^{*}\in F$. From the triangle inequality $\mu (x\oplus y)⩽\mu x\oplus \mu y$, we have ${(\mu x\oplus \mu y)}^{*}⩽{\left(\mu (x\oplus y)\right)}^{*}$. Hence, ${\left(\mu (x\oplus y)\right)}^{*}\in F$. Thus, $x\oplus y\in T\left(F\right)$.

For (IT),

it holds because $\mu \mu x=\mu x$.

□

In the following, we prove that the above operator is equal to the one define, in the case of residuated lattices, on filters, and some characterizations are given.

Proposition 10. 

Let F be a filter of $\mathcal{A}$, then $T\left(F\right)=\{x\in A;\exists a\in F,\mu x⩽a^{*}\}=\{x\in A;\exists a\in F,x⩽{\left(\nu a\right)}^{*}\}=N\left(F\right)$.

Proof. 

• Let $z\in T\left(F\right)$, we have to find $a\in F$ such that $\mu z⩽a^{*}$.
  Since $z\in T\left(F\right)$, we have ${\left(\mu z\right)}^{*}\in F$. Because $\mu z⩽{\left(\mu z\right)}^{**}$, we have $z\in \{x\in A;\exists a\in F,\mu x⩽a^{*}\}$, with $a={\left(\mu z\right)}^{*}$.
• Let $z\in \{x\in A;\exists a\in F,\mu x⩽a^{*}\}$. We have to find $b\in F$ such that $z⩽{\left(\nu b\right)}^{*}$. Since $z\in \{x\in A;\exists a\in F,\mu x⩽a^{*}\}$, there exists $a\in F$ such that $\mu z⩽a^{*}$. Hence, $a^{**}⩽{\left(\mu z\right)}^{*}$. Because $a⩽a^{**}$, we have $a^{**}\in F$, which implies ${\left(\mu z\right)}^{*}\in F$, by (F1). Since ${\left(\nu \left({\left(\mu z\right)}^{*}\right)\right)}^{*}={\left(\mu z\right)}^{**}$$\left(p_8\right)$ and $z⩽\mu z⩽{\left(\mu z\right)}^{**}$$\left(T^{\prime }1\right)$, we have $z⩽{\left(\nu \left({\left(\mu z\right)}^{*}\right)\right)}^{*}$. Thus, $z\in \{x\in A;\exists b\in F,x⩽{\left(\nu b\right)}^{*}\}$, with $b={\left(\mu z\right)}^{*}$.
• Let us prove that $\{x\in A;\exists a\in F,x⩽{\left(\nu a\right)}^{*}\}\subseteq T\left(F\right)$. Let $z\in \{x\in A;\exists a\in F,x⩽{\left(\nu a\right)}^{*}\}$. Then, there exists $a\in F$ such that $z⩽{\left(\nu a\right)}^{*}$. Therefore, $\mu z⩽\mu \left({\left(\nu a\right)}^{*}\right)={\left(\nu a\right)}^{*}$$(T^{\prime }3,p_8)$. Hence, ${\left(\nu a\right)}^{**}⩽{\left(\mu z\right)}^{*}$. Because $a\in F$, we have $\nu a\in F$, by (FT). Since $\nu a⩽{\left(\nu a\right)}^{**}$, we have ${\left(\nu a\right)}^{**}\in F$, and then ${\left(\mu z\right)}^{*}\in F$, by (F1). Thus, $z\in T\left(F\right)$.

Next, we have to prove that $T\left(F\right)=N\left(F\right)$.

• Let $x\in N\left(F\right)$, then $x^{*}\in F$, which implies by (FT) $\nu \left(x^{*}\right)\in F$. Since $\nu \left(x^{*}\right)={\left(\mu x\right)}^{*}$$\left(p_9\right)$, we have ${\left(\mu x\right)}^{*}\in F$. Thus $x\in T\left(F\right)$.
• Let $x\in T\left(F\right)$, then ${\left(\mu x\right)}^{*}\in F$, i.e., $\nu \left(x^{*}\right)\in F$, by $\left(p_9\right)$. Since $\nu \left(x^{*}\right)⩽x^{*}$, we have $x^{*}\in F$, by $\left(F1\right)$. Thus $x\in N\left(F\right)$.

□

Proposition 11. 

Let I be an ideal and F a filter of $\mathcal{A}$. Then,

(1) 

$T\left(T\right(I\left)\right)=I$;

(2) 

$x\in T\left(T\left(F\right)\right)\iff {\left(\nu x\right)}^{**}\in F$;

(3) 

$F\subseteq T\left(T\right(F\left)\right)$;

(4) 

$T\left(T\right(T\left(F\right)\left)\right)=T\left(F\right)$.

Proof. 

(1)

$$\begin{array}{cc}\hfill x\in T\left(T\right(I\left)\right)& \iff {\left(\mu x\right)}^{*}\in T\left(I\right)\hfill \\ \hfill & \iff {\left(\nu \left({\left(\mu x\right)}^{*}\right)\right)}^{*}\in I\hfill \\ \hfill & \iff {\left(\mu x\right)}^{**}\in I,\left(p_8\right)\hfill \\ \hfill & \iff x\in I,\left(Proposition4\right)\hfill \end{array}$$

(2)

Let $x\in A$.

$$\begin{array}{cc}\hfill x\in T\left(T\right(F\left)\right)& \iff {\left(\nu x\right)}^{*}\in T\left(F\right)\hfill \\ \hfill & \iff {\left(\mu {\left(\nu x\right)}^{*}\right)}^{*}\in F\hfill \\ \hfill & \iff {\left(\nu x\right)}^{**}\in F,(\mathrm{by} p_9)\hfill \end{array}$$

(3)

Let $x\in F$. Then, $\nu x\in F$, and since $\nu x⩽{\left(\nu x\right)}^{**}$, we have ${\left(\nu x\right)}^{**}\in F$, which implies $x\in T\left(T\right(F\left)\right)$. Thus $F\subseteq T\left(T\right(F\left)\right)$.

(4)

Since $T\left(F\right)$ is an ideal, then $T\left(T\right(T\left(F\right)\left)\right)=T\left(F\right)$, by Proposition 11 (1).

□

We have in the following example an illustration of the previous Propositions; it also highlighted that in Proposition 11 (3), we do not always have an equality.

Example 3. 

 

Let $A=\left\{\right[0,0];[0,v];[0,a];[0,b];[0,1];[v,v];[v,a];[v,b];[v,1];[a,a];[a,1];[b,b];[b,1];[1,1\left]\right\}$. We define $*\mathrm{and}\Rightarrow$ by the Table 2;

$\forall x,y,z,t\in A,[x,y]\odot [z,t]=[x*z,y*t],\mathrm{and}$

$[x,y]\to [z,t]=\left[\right(x\Rightarrow z)\wedge (y\Rightarrow t),(y\Rightarrow t\left)\right]$, $\mu [x,y]=[y,y]$, $\nu [x,y]=[x,x]$. Then, $\mathcal{A}=(A,\wedge ,\vee ,\odot ,\to ,\nu ,\mu ,[0,0],[0,1],[1,1\left]\right)$ is a triangle algebra [24] whose lattice is given by Figure 1 .

Let $F=\left\{\right[a,a],[a,1],[1,1\left]\right\}$, $G=\left\{\right[v,v],[v,a],[v,b],[v,1],[a,a],[b,b],[a,1],[b,1],[1,1\left]\right\}$, and $I=\left\{\right[0,0\left]\right\}$ then:

i) 

$F\mathrm{and}G$ are filters and I is an ideal;

ii) 

$T\left(F\right)=I$, $T\left(T\right(F\left)\right)=T\left(I\right)=G$, $T\left(T\right(I\left)\right)=T\left(G\right)=I$, $T\left(T\right(T\left(F\right)\left)\right)=T\left(F\right)$, and $I=T\left(T\right(I\left)\right)$. We have $F\subset T\left(T\right(F\left)\right)$, but $[v,v]\in T\left(T\right(F\left)\right)$ and $[v,v]\notin F$, so $F⊊T\left(T\right(F\left)\right)$.

Guided by the above observations, for pseudo-duality in triangle algebras, we will use the operator T, i.e., for any ideal I and filter F of $\mathcal{A}$,

$$\begin{array}{ccc}\hfill T\left(I\right)& =& \{x\in A;{\left(\nu x\right)}^{*}\in I\}\hfill \\ \hfill T\left(F\right)& =& \{x\in A;{\left(\mu x\right)}^{*}\in F\}.\hfill \end{array}$$

2.3. Ideals and Congruences

Let us now investigate the link between ideals and congruences. In particular, we will explain how to build a congruence from an ideal and vice versa.

Definition 2. 

A congruence relation on $\mathcal{A}$ is an equivalence relation which is compatible with all operations on $\mathcal{A}$.

Lemma 2. 

 

Let $\mathcal{A}=(A,\wedge ,\vee ,\odot ,\to ,\nu ,\mu ,0,u,1)$ be a triangle algebra, θ be an equivalence of $\mathcal{A}$ and $\alpha \in \{\wedge ,\vee ,\odot ,\to \}$.

If for all $x,y,z\in A$, $x\theta y\Rightarrow \left(x\alpha z\right)\theta \left(y\alpha z\right)$ and $\left(z\alpha x\right)\theta \left(z\alpha y\right)$, then, for all $x,y,z,t\in A$, ($x\theta y$ and $z\theta t)\Rightarrow (x\alpha z\left)\theta \right(y\alpha t)$.

Proof. 

let assume $x\theta y\Rightarrow \left(x\alpha z\right)\theta \left(y\alpha z\right)$,for all $x,y,z\in A$. Let $\alpha \in \{\wedge ,\vee ,\odot ,\to \}$ , $a,b,c,d\in A$ such that $a\theta b$ and $c\theta d$; we have to show that $\left(a\alpha c\right)\theta \left(b\alpha d\right)$.

Since $a\theta b$, then, $\left(a\alpha c\right)\theta \left(b\alpha c\right)$, and $\left(c\alpha a\right)\theta \left(c\alpha b\right)$. Also, since $c\theta d$, then $\left(c\alpha b\right)\theta \left(d\alpha b\right)$ and $\left(b\alpha c\right)\theta \left(b\alpha d\right)$. Furthermore $\left\{\begin{array}{cc}\left(a\alpha c\right)\theta \left(b\alpha c\right)\hfill & \\ \left(b\alpha c\right)\theta \left(b\alpha d\right)\hfill \end{array}\right.$ which implies $\left(a\alpha c\right)\theta \left(b\alpha d\right)$. Likely, $\left\{\begin{array}{cc}\left(c\alpha a\right)\theta \left(c\alpha b\right)\hfill & \\ \left(c\alpha b\right)\theta \left(d\alpha b\right)\hfill \end{array}\right.$which implies $\left(c\alpha a\right)\theta \left(d\alpha b\right)$ □

The following proposition shows that any ideal of a triangle algebra induces a congruence relation.

Proposition 12. 

Let I be an ideal of $\mathcal{A}$. Then, the binary relation ${\sim }_I$ defined by $x{\sim }_I{y\iff (\left(\nu (x\to y)\right)}^{*}\in I$ and ${\left(\nu (y\to x)\right)}^{*}\in I)$, for all $x,y\in A$, is a congruence on $\mathcal{A}$.

Proof. 

• ${\sim }_I$ is an equivalence relation.

  Reflexivity: 

  let $x\in A$, we have ${\left(\nu (x\to x)\right)}^{*}={\left(\nu 1\right)}^{*}=0\in I.$

  Symmetry: 

  ${\sim }_I$ is symmetric by definition.

  Transitivity: 

  let $x,y,z\in A$ such that $x{\sim }_{I}y$ and $y{\sim }_{I}z$, we have to show that $x{\sim }_{I}z$.

  Since $x{\sim }_{I}y$, we have ${\left(\nu (x\to y)\right)}^{*}\in I$ and ${\left(\nu (y\to x)\right)}^{*}\in I$;

  since $y{\sim }_{I}z$, we have ${\left(\nu (y\to z)\right)}^{*}\in I$ and ${\left(\nu (z\to y)\right)}^{*}\in I$;

  since $(z\to y)\odot (y\to x)⩽(z\to x)$$\left(r_{10}\right)$, then by (T3) and $\left(p_3\right)$, $\nu (z\to y)\odot \nu (y\to x)⩽\nu (z\to x),$ by residuation $\nu (z\to y)⩽\nu (y\to x)\to \nu (z\to x)$. Furthermore, $\nu (y\to x)\to \nu (z\to x)⩽{\left(\nu (z\to x)\right)}^{*}\to {\left(\nu (y\to x)\right)}^{*}$$\left(r_{14}\right)$, then by transitivity $\nu (z\to y)⩽{\left(\nu (z\to x)\right)}^{*}\to {\left(\nu (y\to x)\right)}^{*}$. Hence, ${({\left(\nu (z\to x)\right)}^{*}\to {\left(\nu (y\to x)\right)}^{*})}^{*}⩽{\left(\nu (z\to y)\right)}^{*}$. Since ${\left(\nu (z\to y)\right)}^{*}\in I$, we have ${({\left(\nu (z\to x)\right)}^{*}\to {\left(\nu (y\to x)\right)}^{*})}^{*}\in I.$

  Since ${\left(\nu (z\to x)\right)}^{*}\to {\left(\nu (y\to x)\right)}^{*}⩽{\left(\nu (y\to x)\right)}^{**}\to {\left(\nu (z\to x)\right)}^{**}$$\left(r_{14}\right)$, we have ${({\left(\nu (y\to x)\right)}^{**}\to {\left(\nu (z\to x)\right)}^{**})}^{*}⩽{({\left(\nu (z\to x)\right)}^{*}\to {\left(\nu (y\to x)\right)}^{*})}^{*}$; then, ${({\left(\nu (y\to x)\right)}^{**}\to {\left(\nu (z\to x)\right)}^{**})}^{*}\in I$. Furthermore, ${\left(\nu (y\to x)\right)}^{**}\to {\left(\nu (z\to x)\right)}^{**}{)}^{*}\in I$ and ${\left(\nu (y\to x)\right)}^{*}\in I$, which implies ${\left(\nu (z\to x)\right)}^{*}\in I$, by Proposition 3.

  Likewise, we can show that ${\left(\nu (x\to z)\right)}^{*}\in I$.

• ${\sim }_I$ is compatible with the operations of $\mathcal{A}$
  Let $x,y,z\in A$, such that $x{\sim }_{I}y$, that is ${\left(\nu (x\to y)\right)}^{*}\in I$ and ${\left(\nu (y\to x)\right)}^{*}\in I$.

  1.

  We have to show that $(x\wedge z){\sim }_I(y\wedge z)$.

  We have $x\to y⩽(x\wedge z)\to (y\wedge z)$$\left(r_{11}\right)$, then $\nu (x\to y)⩽\nu \left(\right(x\wedge z)\to (y\wedge z\left)\right)$, since $\nu$ is increasing. Consequently, ${\left(\nu ((x\wedge z)\to (y\wedge z))\right)}^{*}⩽{\left(\nu (x\to y)\right)}^{*}$. Hence, ${\left(\nu ((x\wedge z)\to (y\wedge z))\right)}^{*}\in I$, since ${\left(\nu (x\to y)\right)}^{*}\in I$. Likewise, we show that ${\left(\nu ((y\wedge z)\to (x\wedge z))\right)}^{*}\in I$. Thus $(x\wedge z){\sim }_I(y\wedge z)$.

  2.

  We have to show that $(x\vee z){\sim }_I(y\vee z)$.

  We have $x\to y⩽(x\vee z)\to (y\vee z)$ by $\left(r_{13}\right)$, then $\nu (x\to y)⩽\nu \left(\right(x\vee z)\to (y\vee z\left)\right)$. Consequently ${\left(\nu ((x\vee z)\to (y\vee z))\right)}^{*}⩽{\left(\nu (x\to y)\right)}^{*}$. Hence, ${\left(\nu ((x\vee z)\to (y\vee z))\right)}^{*}\in I$, since ${\left(\nu (x\to y)\right)}^{*}\in I$. Likewise, we show that ${\left(\nu ((y\vee z)\to (x\vee z))\right)}^{*}\in I$. Thus $(x\vee z){\sim }_I(y\vee z)$.

  3.

  We have to show that $(x\odot z){\sim }_I(y\odot z)$.

  We have $x\to y⩽(x\odot z)\to (y\odot z)$ by $\left(r_{12}\right)$, then $\nu (x\to y)⩽\nu \left(\right(x\odot z)\to (y\odot z\left)\right)$. Hence, ${\left(\nu ((x\odot z)\to (y\odot z))\right)}^{*}⩽{\left(\nu (x\to y)\right)}^{*}$, then ${\left(\nu ((x\odot z)\to (y\odot z))\right)}^{*}\in I$, since ${\left(\nu (x\to y)\right)}^{*}\in I$. Likewise, we show that ${\left(\nu ((y\odot z)\to (x\odot z))\right)}^{*}\in I$. Thus $(x\odot z){\sim }_I(y\odot z)$.

  4.

  We have to show that $(x\to z){\sim }_I(y\to z)$.

  We have $x\to y⩽(y\to z)\to (x\to z)$ by $\left(r_{10}\right)$ and residuation. Hence, $\nu (x\to y)⩽\nu \left(\right(y\to z)\to (x\to z\left)\right)$, then ${\left(\nu ((y\to z)\to (x\to z))\right)}^{*}⩽{\left(\nu (x\to y)\right)}^{*}.$ Therefore ${\left(\nu ((y\to z)\to (x\to z))\right)}^{*}\in I$, since ${\left(\nu (x\to y)\right)}^{*}\in I$. Likewise, we show that ${\left(\nu ((x\to z)\to (y\to z))\right)}^{*}\in I$. Thus $(x\to z){\sim }_I(y\to z)$.

  5.

  We have to show that $z\to x{\sim }_{I}z\to y$.

  We have $y\to x⩽(z\to y)\to (z\to x)$ by $\left(r_{10}\right)$ and residuation. Hence, $\nu (y\to x)⩽\nu \left(\right(z\to y)\to (z\to x\left)\right)$, then ${\left(\nu ((z\to y)\to (z\to x))\right)}^{*}⩽{\left(\nu (y\to x)\right)}^{*}.$ It follows ${\left(\nu ((z\to y)\to (z\to x))\right)}^{*}\in I$, since ${\left(\nu (y\to x)\right)}^{*}\in I$. Likewise, we show that ${\left(\nu ((z\to x)\to (z\to y))\right)}^{*}\in I$. Thus $(z\to x){\sim }_I(z\to y)$.

  6.

  We have to show that $\nu x{\sim }_I\nu y$.

  We have $\nu (x\to y)⩽\nu x\to \nu y$ by (T7). Then ${(\nu x\to \nu y)}^{*}⩽{\left(\nu (x\to y)\right)}^{*}$, hence ${(\nu x\to \nu y)}^{*}\in I$. We have ${\left(\nu (\nu x\to \nu y)\right)}^{*}\in I$, since $\nu (\nu x\to \nu y)=\nu x\to \nu y$, $\left(p_6\right)$. Likewise, we show that ${\left(\nu (\nu y\to \nu x)\right)}^{*}\in I$. Thus $\nu x{\sim }_I\nu y$.

  7.

  We have to show that $\mu x{\sim }_I\mu y$.

  We have $\nu (x\to y)⩽\mu x\to \mu y$ by $\left(p_1\right)$. Therefore, $\nu (x\to y)=\nu \nu (x\to y)⩽\nu (\mu x\to \mu y)$ by $(p_6,T3)$. Hence, ${\left(\nu (\mu x\to \mu y)\right)}^{*}⩽{\left(\nu (x\to y)\right)}^{*}$, and ${\left(\nu (\mu x\to \mu y)\right)}^{*}\in I$.Likewise, we show that ${\left(\nu (\mu y\to \mu x)\right)}^{*}\in I$. Thus $\mu x{\sim }_I\mu y$.

□

For any ideal I of $\mathcal{A}$, the ${\sim }_I$-class of any $x\in A$ is denoted by ${\left[x\right]}_{{\sim }_I}$ and the set of all ${\sim }_I$-class is denoted by $A/I$.

The following proposition establishes a one-to-one correspondence between the set $\mathcal{I}\left(A\right)$ of ideals and the set $\mathcal{C}on\left(A\right)$ of congruences of $\mathcal{A}$.

Proposition 13. 

Let I be an ideal of $\mathcal{A}$, then $I={\left[0\right]}_{{\sim }_I}$.

Proof. 

• Let $x\in I$, then ${\left(\nu (x\to 0)\right)}^{*}={\left(\nu x^{*}\right)}^{*}\in I$, by Proposition 4.
  Moreover, ${\left(\nu (0\to x)\right)}^{*}={\left(\nu 1\right)}^{*}=1^{*}=0\in I$. Hence, $x{\sim }_{I}0$, and consequently $x\in {\left[0\right]}_{{\sim }_I}$.
• Since I is an ideal, it holds that $0\in I$.
  Let $x\in {\left[0\right]}_{{\sim }_I}$, then ${\left(\nu x^{*}\right)}^{*}\in I$, which implies $x\in I$, by Proposition 4.

□

The following result proves that any congruence relation on a triangle algebra induces an ideal.

Proposition 14. 

For any congruence ∼ of $\mathcal{A}$, ${\left[0\right]}_{\sim }$ is an ideal.

Proof. 

For (I1),

let $x,y\in A$ such that $x⩽y$ and $y\in {\left[0\right]}_{\sim }$. Since $y\in {\left[0\right]}_{\sim }$, then $y\sim 0$, which implies $(x\wedge y)\sim (x\wedge 0)$, because ∼ is a congruence. Therefore, $x\sim 0$, because $0⩽x⩽y$. Hence, $x\in {\left[0\right]}_{\sim }$.

For (I3),

let $x,y\in {\left[0\right]}_{\sim }$. Since $x\in {\left[0\right]}_{\sim }$, then $x\sim 0$. Consequently, $x^{*}\sim 0^{*}=1$. Similarly, $y^{*}\sim 1$. Hence, $(x^{*}\odot y^{*})\sim (1\odot 1)=1$, therefore ${(x^{*}\odot y^{*})}^{*}\sim 1^{*}=0$, i.e., $x\oplus y\sim 0$. Thus $x\oplus y\in {\left[0\right]}_{\sim }$.

For (IT),

let $x\in {\left[0\right]}_{\sim }$. Since $x\in {\left[0\right]}_{\sim }$, then $x\sim 0$ thus $\mu x\sim \mu 0=0$, which implies $\mu x\in {\left[0\right]}_{\sim }$.

Hence ${\left[0\right]}_{\sim }$ is an ideal. □

Given a congruence relation ∼ on $\mathcal{A}$, it is not clear whether ${\sim }_{{\left[0\right]}_{\sim }}=\sim$ holds. This remains an open question.

As a consequence of Proposition 12, the following result establishes that the quotient of a triangle algebra by a congruence induced by an ideal is a triangle algebra.

Proposition 15. 

Let I be an ideal of $\mathcal{A}$, then $\mathcal{A}/I=(A/I,\wedge ,\vee ,\odot ,\to ,\nu ,\mu ,I,$${\left[u\right]}_{{\sim }_I},{\left[1\right]}_{{\sim }_I})$ is a triangle algebra, where for all $x,y\in A$, $\theta \left({\left[x\right]}_{{\sim }_I}\right)={\left[\theta \left(x\right)\right]}_{{\sim }_I},\forall \theta \in \{\nu ,\mu \}$ and ${\left[x\right]}_{{\sim }_I}\alpha {\left[y\right]}_{{\sim }_I}={\left[x\alpha y\right]}_{{\sim }_I},\forall \alpha \in \{\wedge ,\vee ,\to ,\odot \}$.

Proof. 

This is a direct consequence of the fact that ${\sim }_I$ is a congruence. □

Definition 3. 

[25]

A homomorphism between $\mathcal{A}=(A,\vee ,\wedge ,*,\to ,\nu ,\mu ,0,u,1)$ and $\mathcal{B}=(B,\bigsqcup ,\sqcap ,\odot ,\Rightarrow ,\overline{\nu },\overline{\mu },0,\overline{u},1)$ two triangle algebras is any mapping $f:A\to B$, that satisfies for all a, $b\in A$, $f(a\vee b)=f\left(a\right)\bigsqcup f\left(b\right)$, $f(a\wedge b)=f\left(a\right)\sqcap f\left(b\right)$, $f(a*b)=f\left(a\right)\odot f\left(b\right)$, $f(a\to b)=f\left(a\right)\Rightarrow f\left(b\right)$, $f\left(\nu a\right)=\overline{\nu }\left(f\left(a\right)\right)$, $f\left(\mu a\right)=\overline{\mu }\left(f\left(a\right)\right)$; the kernel of f, denoted by $ker\left(f\right)$, is define by $ker\left(f\right):=\{x\in A;f(x)=0\}$.

The following establish that the kernel of any homomorphism is an ideal.

Proposition 16. 

Let $\mathcal{A},B$ be two triangle algebras and $f:A\to B$ a homomorphism, then $Ker\left(f\right)$ is an ideal of $\mathcal{A}$.

Proof. 

1.

Let $x,y\in A$, such that $x⩽y$ and $y\in Ker\left(f\right)$.

Since $x⩽y$, we have $x\wedge y=x$. Hence, $f(x\wedge y)=f\left(x\right)$. Because f is a homomorphism, we have $f\left(x\right)\wedge f\left(y\right)=f\left(x\right)$. Then $f\left(x\right)⩽f\left(y\right)$.

Since $y\in Ker\left(f\right)$, we have $f\left(y\right)=0$. Therefore, $f\left(x\right)=0$. Hence, $x\in Ker\left(f\right)$.

2.

Let $x,y\in Ker\left(f\right)$. Then $f\left(x\right)=0$ and $f\left(y\right)=0$. Furthermore, because f is a homomorphism, we have $f(x\oplus y)=f\left(x\right)\oplus f\left(y\right)=0$. Thus, $x\oplus y\in Ker\left(f\right)$.

3.

Let $x\in Ker\left(f\right)$. We have $f\left(\mu x\right)=\mu \left(f\right(x\left)\right)=\mu \left(0\right)=0$. Therefore, $\mu x\in Ker\left(f\right)$.

Thus, $Ker\left(f\right)$ is an ideal of $\mathcal{A}$. □

The following proposition establishes that any ideal is the kernel of a homomorphism of triangle algebras.

Proposition 17. 

Let I be an ideal of a triangle algebra $\mathcal{A}$, then there exists a homomorphism f of triangle algebras such that $I=Ker\left(f\right)$.

Proof. 

Let us consider

$$\begin{array}{cccc}\pi :& \mathcal{A}& ⟶& \mathcal{A}/I\\ & x& ⟼& {\left[x\right]}_{{\sim }_I}\end{array}$$

$\pi$ is a homomorphism between the triangle algebras $\mathcal{A}$ and $\mathcal{A}/I$. Let $x\in Ker\left(\pi \right)$, then $\pi \left(x\right)=I={\left[0\right]}_{{\sim }_I}$, i.e., $\nu (x\to 0)*={\left(\nu x^{*}\right)}^{*}\in I$. Hence, by Proposition 4, $x\in I$. □

2.4. Ideals Generated by a Subset

Before we dive into the case of ideals, let us first highlight that we have identified an inaccuracy in the characterization of the filter generated by a subset of a triangle algebra as proposed in [22]. Here, we provide an example to counter the arguments in the proof given in ([22], Proposition 4.2). Subsequently, we provide some accurate characterizations of the filter generated by a subset of a triangle algebra. Furthermore, we describe the ideal induced by a subset of a triangle algebra.

Recall that for any subset X of a triangle algebra $\mathcal{A}$, the filter generated by X which we will denote $\left[X\right)$ is the smallest filter of $\mathcal{L}$ containing X.

In ([22], Proposition 4.2) they mentioned that for any subset S of a triangle algebra $\mathcal{A}$, $\left[S\right)=\{x\in L\mid \exists (n,x_i)\in {\mathbb{N}}^{*}\times S,\nu x\ge x_1\odot x_2\odot \cdots \odot x_n,\forall i\in \{1,2,\cdots ,n\}\}$, which we claim is not accurate, as we can see in Example 4 below.

Example 4. 

Let us consider the triangle algebra ${\mathcal{A}}_0$ of Example 1. In the following two cases, we have $\left[S\right)\ne \{x\in A;s_1\odot s_2\odot \cdots \odot s_n⩽\nu x,n\in \mathbb{N},i\in 1;n,s_i\in S\}$.

1. 

For $S=\{u,1\}$, S is a filter of the residuated lattice $(A_0,\wedge ,\vee ,\odot ,\to ,0,1)$, but S is not a filter of the triangle algebra ${\mathcal{A}}_0$ (see Example 2). Using the fact that $\left[S\right)$ is the smallest filter of $\mathcal{A}$ containing S, we have $\left[S\right)=A_0$; however, using the characterization above given in [22], we have $0\notin \{x\in A_0;s_1\odot s_2\odot \cdots \odot s_n⩽\nu x,n\in \mathbb{N},i\in 1;n,s_i\in S\}$. Hence, $\left[S\right)\ne \{x\in A_0;s_1\odot s_2\odot \cdots \odot s_n⩽\nu x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$.

2. 

For $S=\left\{u\right\}$. Since $u\odot u=u$, we have $u^n=u$, $\forall n\in {\mathbb{N}}^{*}$. Therefore, $\forall n\in {\mathbb{N}}^{*}$, ${\underbrace{u\odot \cdots \odot u}}_{n-times}=u>0=\nu u$. Hence, $u\notin \{x\in A;s_1\odot s_2\odot \cdots \odot s_n⩽\nu x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$. Thus $S\neg \subseteq \{x\in A;s_1\odot s_2\odot \cdots \odot s_n⩽\nu x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$.

To mend this problem, we introduce the following characterization of the filter induced by a subset of a triangle algebra.

Proposition 18. 

Let S be a non-empty subset of A. Then,

$$\left[S\right)=\{x\in A;\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}.$$

Proof. 

Let $B=\{x\in A;\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$. We will show that $B=\left[S\right)$.

• First, we show that B is a filter.

  For (F1),

  let $x,y\in A$ such that $x⩽y$ and $x\in B$. Therefore, there exists $n\in \mathbb{N}$, $s_1,\cdots ,s_n\in S$ such that $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽x$. Since $x⩽y$, then $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽y$. Thus $y\in B$.

  For (F2),

  let $x,y\in B$. Then, there exists $n\in \mathbb{N}$, $s_i\in S,i\in 〚1;n〛$ such that $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽x$ and there exists $m\in \mathbb{N}$, $t_i\in S,i\in 〚1;m〛$ such that $\nu t_1\odot \nu t_2\odot \cdots \odot \nu t_m⩽y$. By $\left(r_4\right)$, we have $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n\odot \nu t_1\odot \nu t_2\odot \cdots \odot \nu t_m⩽x\odot y$. Thus $x\odot y\in B$.

  For (FT),

  let $x\in B$. Therefore, there exists $n\in \mathbb{N}$, $s_i\in S,i\in 〚1;n〛$ such that $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽x$. Hence, by (T3), we have $\nu (\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n)⩽\nu x$ which implies, by $\left(p_6\right)$ and $\left(p_3\right)$: $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽\nu x$. Thus $\nu x\in B$.

• Next, we show that $S\subseteq B$.
  Let $s\in S$. Since $\nu s⩽s$$\left(T1\right)$, we have $s\in B$.
• Let F be a filter of $\mathcal{A}$ containing S. Let $x\in B$, then there exists $n\in \mathbb{N}$, $s_i\in S,i\in 〚1;n〛$, such that $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽x$. Let $i\in 〚1;n〛$, then $s_i\in S\subseteq F$, thus $s_i\in F$, so by (FT) $\nu s_i\in F$. Because this is true for all $i\in 〚1;n〛$, we have, by (F2): $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n\in F$. Since $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽x$, we have, by (F1): $x\in F$.

Concluding, $B=\left[S\right)$. □

Below is another characterization of the filter induced by a subset of a triangle algebra.

Proposition 19. 

Let S be a non-empty subset of A.

Then $\left[S\right)=\{x\in A;\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽\nu x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$.

Proof. 

• Let $a\in \left[S\right)$, then (FT), $\nu a\in \left[S\right)$, so $a\in \{x\in A;\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽\nu x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$.
• Let $a\in \{x\in A;\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽\nu x,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$, then there exists $n\in \mathbb{N}$, $s_i\in S,i\in 〚1;n〛$, such that $\nu s_1\odot \nu s_2\odot \cdots \odot \nu s_n⩽\nu a$. Since $\nu a⩽a$ (by (F1)), it holds that $a\in \left[S\right)$.

□

We recall that, for any subset S of a triangle algebra $\mathcal{A}$, the ideal generated by S is the smallest ideal of $\mathcal{A}$ containing S, it is denoted by $\left(S\right]$. Next, we introduce a characterization of the ideal induced by a non-empty subset of a triangle algebra.

Proposition 20. 

Let S be a non-empty subset of a triangle algebra $\mathcal{A}$. We have $\left(S\right]=\{x\in A;x⩽\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$.

Proof. 

Let $I=\{x\in A;x⩽\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n,n\in \mathbb{N},i\in 〚1;n〛,s_i\in S\}$.

• We first show that I is an ideal.

  For (I1),

  this is obvious.

  For (I3),

  let $x,y\in I$. Therefore, $x⩽\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n$, with $n\in \mathbb{N},i\in 〚1;n〛,s_i\in S$, and $y⩽\mu t_1\oplus \mu t_2\oplus \cdots \oplus \mu t_m$, with $m\in \mathbb{N},i\in 〚1;m〛,t_i\in S$. Hence, $x\oplus y⩽\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n\oplus \mu t_1\oplus \mu t_2\oplus \cdots \oplus \mu t_m$. Thus $x\oplus y\in I$.

  For (IT),

  let $x\in I$. Then, we have $x⩽\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n$, with $n\in \mathbb{N},i\in 〚1;n〛,s_i\in S$. Hence,

  $$\begin{array}{cc}\hfill \mu x& ⩽\mu (\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n)\hfill \\ & ⩽\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n,(Lemma\left(3\right),p_6)\hfill \end{array}$$

  Thus $\mu x\in I$.

• Let $s\in S$. Since $s⩽\mu s$, (T’1), we have $s\in I$. Thus $S\subseteq I$.
• Let J be an ideal of $\mathcal{A}$ containing S. Let $x\in I$, we have $x⩽\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n$, with $n\in \mathbb{N},i\in 〚1;n〛,s_i\in S$. Let $i\in 〚1;n〛$, then $s_i\in S\subseteq J$. Because J is an ideal of $\mathcal{A}$, $\mu s_i\in J$ (IT), for all $i\in 〚1;n〛$. Hence, by (I3), $\mu s_1\oplus \mu s_2\oplus \cdots \oplus \mu s_n\in J$ and by (I1), $x\in J$. Thus $I\subseteq J$.

In conclusion, $\left(S\right]=I$. □

Any ideal generated by $\left\{a\right\}$ will be denoted by $\left(a\right]$ and will be called principal ideal.

Let us now explore the algebraic structure of $\mathcal{I}\left(A\right)$, the set of ideals of $\mathcal{A}$. One can observe that, if I and J are two ideals of $\mathcal{A}$, then $I\cap J$ is an ideal of $\mathcal{A}$.

In what follows, we establish that the set of ideals of $\mathcal{A}$ is a Heyting algebra.

Proposition 21. 

$\left(\mathcal{I}\right(A),\wedge ,\vee ,\{0\},A)$ is a complete lattice, with $I\wedge J=I\cap J$, $I\vee J=(I\cup J]$, for all $I,J\in \mathcal{I}\left(A\right)$.

Proof. 

Let $I,J,K\in \mathcal{I}\left(A\right)$;

Idempotence: 

$I\wedge I=I\cap I=I$; $I\vee I=(I\cup I]=\left(I\right]=I$

Symmetry: 

$I\wedge J=I\cap J=J\cap I=J\wedge I$; $I\vee J=(I\cup J]=(J\cup I]=J\vee I$

Associativity: 

$I\wedge (J\wedge K)=I\cap (J\cap K)=(I\cap J)\cap K$. Moreover,

Since $J\subseteq I\cup J\subseteq I\vee J\subseteq (I\vee J)\vee K$, and $K\subseteq (I\vee J)\vee K$, we have $J\cup K\subseteq (I\vee J)\vee K$, thus $(J\cup K]\subseteq (I\vee J)\vee K$, i.e., $J\vee K\subseteq (I\vee J)\vee K$. Moreover, $I\subseteq (I\vee J)\vee K$, thus $I\cup (J\vee K)\subseteq (I\vee J)\vee K$, consequently, $(I\cup (J\vee K\left)\right]\subseteq (I\vee J)\vee K$, i.e., $I\vee (J\vee K)\subseteq (I\vee J)\vee K$.

We show likewise $(I\vee J)\vee K\subseteq I\vee (J\vee K)$, therefore, $I\vee (J\vee K)=(I\vee J)\vee K$.

Absorption laws: 

$I\wedge (I\vee J)=I\cap (I\vee J)=I$; $I\vee (I\wedge J)=(I\cup (I\wedge J\left)\right]=\left(I\right]=I$.

Bounds: 

Let ${\left\{I_i\right\}}_{i\in \Gamma }$ a be non-empty family of elements of $\mathcal{I}\left(A\right)$. For this family the infimum is given by $\underset{i\in \Gamma }{\bigwedge }{I}_i:=\bigcap _{i\in \Gamma }{I}_i$ and the supremum $\underset{i\in \Gamma }{\bigvee }{I}_i:=\left(\bigcup _{i\in \Gamma }{I}_i\right]$.

□

Proposition 22. 

$\left(\mathcal{I}\right(\mathcal{A}),\wedge ,\vee ,\to ,\{0\left\}\right)$ is a Heyting algebra, where $J\to K:=\{a\in A;J\wedge (a]\subseteq K\}$, for all $J,K\in \mathcal{I}\left(\mathcal{A}\right)$ .

Proof. 

• Let us show first that $\left(\mathcal{I}\right(\mathcal{A}),\wedge ,\vee ,\wedge ,\to ,\{0\},A)$ is a residuated lattice.

  For RL1,

  $\left(\mathcal{I}\right(\mathcal{A}),\wedge ,\vee ,0,A)$ is a bounded lattice.

  For RL2,

  $\left(\mathcal{I}\right(\mathcal{A}),\wedge ,A)=\left(\mathcal{I}\right(\mathcal{A}),\cap ,A)$ is a commutative ordered monoid.

  For RL3,

  let $I,J,K\in \mathcal{I}\left(\mathcal{A}\right)$. We will prove that $I\wedge J\subseteq K\iff I\subseteq J\to K$.

  *

  Assume that $I\wedge J\subseteq K$ and let $a\in I$.

  Let $x\in J\wedge \left(a\right]$, then $x\in J$ and $x\in \left(a\right]\subseteq I$. Hence, $x\in I\wedge J\subseteq K$. Therefore, $J\wedge \left(a\right]\subseteq K$, i.e., $a\in J\to K$. Thus $I\subseteq J\to K$.

  *

  Assume that $I\subseteq J\to K$.

  Let $x\in I\wedge J$, then $x\in I$ and $x\in J$. Then $x\in J\to K$, since $x\in I\subseteq J\to K$. Hence, $x\in J\wedge \left(x\right]\subseteq K$. Thus $I\wedge J\subseteq K$.

• Since $\left(\mathcal{I}\right(\mathcal{A}),\wedge ,\vee ,\wedge ,\to ,0,A)$ is a residuated lattice and the product and the conjunction coincide, it holds that $\left(\mathcal{I}\right(\mathcal{A}),\wedge ,\vee ,\to ,\{0\left\}\right)$ is a Heyting algebra.

□

3. Conclusion and Future Work

In this paper, we aimed to introduce and examine the notion of ideals within triangle algebras. We delved into the relations between ideals and filters, as well as the correlations between ideals and congruences. Notably, we have rectified an existing error in the filter construction procedure proposed by Zahiri et al. [22]; as a result, we give an alternative characterization of the filter generated by a subset of a triangle algebra, and this leads to offer a more accurate framework for understanding the relationship between ideals and filters. Particularly, we have proved that the operators given by Liu et al.[7] and Tchoua et al. [13] in the case of residuated lattices are equal, moreover we have proved that in the case of triangle algebras, this operator changes filters to an ideals but the converse is not true. Furthermore, we used the characterization of the ideal generated by a subset to prove that the set of all ideals of a triangle algebra is a Heyting algebra.

In our future work, we plan to study how to use the notion of ideal to develop a logic that formally characterizes antitologies in triangle algebras; moreover, we plan to look about solving equations in algebraic structures as done by [8,11].