Li-ideals of implicative almost distributive lattices

Alachew Amaneh Mechderso; Tilahun  Mekonnen Munie

doi:10.12688/f1000research.159175.1

Home Browse Li-ideals of implicative almost distributive lattices

ALL Metrics

-

Views

-

Downloads

Get PDF

Get XML

Export

▬

✚

Research Article

Li-ideals of implicative almost distributive lattices

[version 1; peer review: 1 approved]

Alachew Amaneh Mechderso ¹, Tilahun Mekonnen Munie²

PUBLISHED 10 Feb 2025

Author details Author details

¹ Department of Mathematics, University of KabriDahar, Kabri Dahar, Somalie, Ethiopia
² Department of Mathematics, Bahir Dar University College of Science, Bahir Dar, Amhara, 6000, Ethiopia

Alachew Amaneh Mechderso
Roles: Conceptualization, Writing – Original Draft Preparation, Writing – Review & Editing

Tilahun Mekonnen Munie
Roles: Conceptualization, Visualization, Writing – Review & Editing

OPEN PEER REVIEW

REVIEWER STATUS

Abstract

Background

In this paper, we introduce the concept of H-implicative almost distributive lattices, which are a special class of lattices with both implicative and almost distributive properties. This study is aimed at extending the understanding of lattice structures and their ideals, particularly focusing on LI-ideals, a novel concept within H-implicative almost distributive lattices.

Methods

We define an LI-ideal of an implicative almost distributive lattice L and investigate its properties. The paper demonstrates that every LI-ideal in L is an almost distributive lattice ideal of L. Additionally, we explore the relationship between filters and LI-ideals, and we study the process of generating an LI-ideal from a given set. Lastly, we examine the construction of quotient structures via LI-ideals.

Results

We present several examples showing that every almost distributive lattice ideal is also an LI-ideal in an H-implicative almost distributive lattice. The study establishes key relationships between the concepts of filters and LI-ideals. Furthermore, we provide a method for generating an LI-ideal from a set and construct a quotient structure using an LI-ideal.

Conclusions

The paper introduces new concepts and relationships within the study of H-implicative almost distributive lattices. Our findings demonstrate the interconnection between almost distributive lattice ideals and LI-ideals and offer insights into how these ideals can be generated and used to construct quotient structures. This work provides a deeper understanding of lattice theory and opens new avenues for further research in the area of lattice ideals.

Keywords

Filter, H-implicative almost distributive lattice, LI-ideals and implicative almost distributive lattice..

Corresponding author: Alachew Amaneh Mechderso

Competing interests: No competing interests were disclosed.

Grant information: The author(s) declared that no grants were involved in supporting this work.

Copyright: © 2025 Mechderso AA and Munie TM. This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

How to cite: Mechderso AA and Munie TM. Li-ideals of implicative almost distributive lattices [version 1; peer review: 1 approved]. F1000Research 2025, 14:182 (https://doi.org/10.12688/f1000research.159175.1) First published: 10 Feb 2025, 14:182 (https://doi.org/10.12688/f1000research.159175.1) Latest published: 10 Feb 2025, 14:182 (https://doi.org/10.12688/f1000research.159175.1)

Introduction

In order to research the logical system whose propositional value is given in a lattice, Xu, Y.⁸ proposed the concept of lattice implication algebras and discussed some of their properties. Xu, Y. and Qin, K. Y.⁷ discussed the properties of lattice H-implication algebras. Jun, Y. B. and Xu, et al.^3,9 introduced the notions of a filters in a lattice implication algebra and investigate their properties. Jun, Y. B. et al.⁴ proposed the concept of an LI-ideal of lattice implication algebra. They discussed the relationship between filters and LI-ideals and studied how to generate an LI-ideal by a set. They constructed quotient structure by using an LI-ideal. Kolluru,V. and Bekele, B.⁵the concept of implicative algebras was introduced, and several key properties were established. It was also proven that every implicative algebra is lattice implication algebra. In 1981, Swamy, U. M., and Rao, G. C. introduced the concept of an almost distributive lattice (ADL) as a common abstraction for many of the existing lattice structure.⁶

The existing ring-theoretic and lattice-theoretic generalizations of Boolean algebra have been explored in various studies. Berhanu Assaye, Mihret Alamneh, and Tilahun Mekonnen¹ introduced the concept of implicative almost distributive lattices (IADLs) as a generalization of implicative algebras within the class of almost distributive lattices (ADLs). In this paper, we prove several properties and equivalence conditions within the framework of IADLs.

We also introduced filter, implicative filter in IADL.² In this paper we discuss H-implicative almost distributive lattice (in short, H-IADL) and give equivalent conditions. We introduce LI-ideals in IADL and show that every LI-ideal is an ADL ideal. We give an example that an ADL ideal may not be an LI-ideal, and show that every ADL ideal is an LI-ideal in H-IADL. We discuss the relationship between filters and LI-ideal, and study how to generate an LI-ideal by a set. We construct quotient structure by using an LI-ideal.

In the following, we give some important definitions and results that will be useful in this study.

Preliminaries

Definition 2.1

⁶ An algebra $(L, \lor, \land, 0)$ of type $(2, 2, 0)$ is called an almost distributive lattice (ADL) with $0$ if it satisfies the following axioms:

1. $(x \lor y) \land z = (x \land z) \lor (y \land z)$
2. $x \land (y \lor z) = (x \land y) \lor (x \land z)$
3. $(x \lor y) \land y = y$
4. $(x \lor y) \land x = x$
5. $x \lor (x \land y) = x$
6. $0 \land x = 0, for all x, y, z \in L .$

If (L, ∨, ∧, 0) is an ADL, for any $x, y \in L$ , define $x \leq y$ if and only if $x = x \land y$ or equivalently $x \lor y = y$ , then $\leq$ is a partial ordering on L.

Definition 2.2

⁶ Let L be an ADL. An element $m \in L$ is called maximal if for any $x \in L$ , $m \leq x$ implies $m = x .$

Definition 2.3

⁶ A non-empty subset I of an ADL L is called an ideal of L, if it satisfies the following:

i. $x, y \in I$ implies $x \lor y \in I .$
ii. $x \in I$ and $y \in L$ implies that $x \land y \in I$ . we call I as an $ADL$ ideal of $L .$

The following important property of the ideals is very useful to develop the algebra of ideals. If I is an ideal of ADL L and $a, b \in L$ , then $a \land b \in I$ if and only if $b \land a \in I .$

Definition 2.4

⁶ Let L be an ADL. For any $a \in L$ , principal filter of L generated by a is $[a) = {x \lor a : x \in L}$

Definition 2.5

[6] A non-empty subset F of an ADL L is called a filter of L if it satisfies the following:

i. $x, y \in F$ implies $x \land y \in F$ .
ii. $x \in F$ and $y \in L$ implies that $y \lor x \in F$ .

Definition 2.6

⁶ A non-empty subset S of an ADL L is called a sub ADL of L if

1. $0 \in S$
2. $x \land y \in S$ and $x \lor y \in S$ , for any $x, y \in S$ .

Theorem 2.7

⁶ Let F is filter of an ADL L and $x, y \in L$ . Then $x \lor y \in F$ if and only if $y \lor x \in F .$

Lemma 2.8

⁶ Every ideal (filter) of L is a sub ADL of ADL L.

Definition 2.9

⁵ An algebra $(L, \to,', 0, 1)$ of type $(2, 1, 0, 0)$ is called implicative algebra if it satisfies the following conditions:

1. $x \to (y \to z) = y \to (x \to z)$
2. $1 \to x = x$
3. $x \to 1 = 1$
4. $x \to y = y^{'} \to x^{'}$
5. $(x \to y) \to y = (y \to x) \to x$
6. $0^{'} = 1, for x, y, z \in L$

Theorem 2.10

⁵ Let $(L, \to,', 0, 1)$ be an implicative algebra. Then $(L, \lor, \land, \to,', 0, 1)$ is a lattice implication algebra.

Definition 2.11

¹ Let $(L, \lor, \land, 0, m)$ be an ADL with $0$ and maximal element m. Then an algebra $(L, \lor, \land, \to,', 0, m)$ of type $(2, 2, 2, 1, 0, 0)$ is called implicative almost distributive lattice (IADL) if it satisfies the following conditions:

1. $x \lor y = (x \to y) \to y$
2. $x \land y = {[(x \to y) \to x^{'}]}^{'}$
3. $x \to (y \to z) = y \to (x \to z)$
4. $m \to x = x$
5. $x \to m = m$
6. $x \to y = y^{'} \to x^{'}$
7. $0^{'} = m, for all x, y, z \in L .$

Now we define the relation $\leq$ on an IADL L as follows: $x \leq y \Leftrightarrow x \to y = m$ , for all $x, y \in L$ . The relation $\leq$ on L is a partial ordering.Thus $(L, \leq)$ is a poset.

Theorem 2.12

¹ In an IADL L, for all $x, y, z \in L$ the following conditions hold:

1. $[(x \to y) \to y] \land m = [(y \to x) \to x] \land m$
2. $[{((x \to y) \to x^{'})}^{'}] \land m = [{((y \to x) \to y^{'})}^{'}] \land m$
3. $x \to x = m$
4. $m^{'} = 0$
5. ${(x^{'})}^{'} = x$
6. $x^{'} = x \to 0$
7. $0 \to x = m$
8. $x \to y = m = y \to x implies x = y .$
9. $If x \to y = m and y \to z = m, then x \to z = m$
10. $x \leq y if and only if z \to x \leq z \to y and y \to z \leq x \to z$
11. $((x \to y) \to y) \to y = x \to y$
12. $(x \to y) \to ((y \to z) \to (x \to z)) = m, (x \to y) \to ((z \to x) \to (z \to y)) = m$
13. $(x \to z) \to (x \to y) = (z \to x) \to (z \to y) = (x \land z) \to y .$
14. $x \to y \leq (y \to z) \to (x \to z)$
15. ${(x \land y)}^{'} = x^{'} \lor y^{'}, {(x \lor y)}^{'} = x^{'} \land y^{'}$
16. $x \leq y implies y^{'} \leq x^{'} .$
17. $(x \lor y) \to z = (x \to z) \land (y \to z)$
18. $(x \land y) \to z = (x \to z) \lor (y \to z)$
19. $x \to (y \land z) = (x \to y) \land (x \to z)$
20. $x \to (y \lor z) = (x \to y) \lor (x \to z) .$

Definition 2.13

² Let L be an IADL.

1. A subset F of L is called a filter of L if it satisfies:
(F₁) $m \in F$
(F₂) $x \in F$ and $x \to y \in F$ implies $y \in F$ , for all $x, y \in L$ .
2. A subset F of L is called implicative filter of L if it satisfies
(F₁) $m \in F$
(I) $x \to y \in F$ and $x \to (y \to z) \in F$ implies $x \to z \in F$ , for all $x, y, z \in L$ .

Lemma 2.14

² Let F be a non-empty subset of an IADL L. Then F is a filter of L if and only if it satisfies for all $x, y \in F$ and $z \in L$ :

$x \leq y \to z$ implies $z \in F$

Lemma 2.15

² Every filter F of an IADL L has the following property: $x \leq y$ and $x \in F$ implies $y \in F$ .

Definition 2.16

⁷ A lattice implication algebra L is called a lattice H-implication algebra, if for any $x, y, z \in L$ , $x \lor y \lor ((x \land y) \to z) = 1$ .

Theorem 2.17

⁷ Let L be a lattice H-implication algebra. Then for any $x, y, z \in L$ , $x \to (y \to z) = (x \to y) \to (x \to z) .$

Theorem 2.18

¹ Let L be an IADL, then for any $x, y, z \in L$ , $(x \to (y \to z)) \to ((x \to y) \to (x \to z)) = x \lor y \lor ((x \land y) \to z)$ .

Definition 2.19

⁴ Let L be a lattice implication algebra. An LI-ideal A of L is a non-empty subset of L such that

1. $0 \in A,$
2. $y \in A$ and ${(x \to y)}^{'} \in A$ implies $x \in A$ , for $x, y \in L$ .

Theorem 2.20

^4,9 Let L be a lattice implication algebra. Every LI-ideal of L is a lattice ideal.

Theorem 2.21

⁴ In lattice H-implication algebra, every lattice ideal is an LI-ideal.

Definition 2.22

⁶ An ADL $(L, \lor, \land)$ is said to be associative if the operation $\lor$ in L is associative.

Definition 2.23

⁶ An equivalence relation θ on an ADL L is called a congruence relation on L if for all $a, b, c, d \in L$ , $(a, b), (c, d) \in θ$ implies $(a \land c, b \land d); (a \lor c, b \lor d) \in θ$ .

LI-ideals in implicative almost distributive lattices

In this section first we introduce H-implicative almost distributive lattice (H-IADL). Then we introduce LI-ideals in IADLs and discuss some of their properties. Finally, we give some characterizations of LI-ideals.

H-implicative almost distributive lattices

In this subsection we introduce H-implicative almost distributive lattice (H-IADL) and study the conditions of IADL being H-IADL.

Definition 3.1

An implicative almost distributive lattice (IADL) L is called H-implicative almost distributive lattice (H-IADL) if it satisfies:

x \lor y \lor ((x \land y) \to z) = m, for all x, y, z \in L

Example 3.2

Let $L = {0, x, y, m}$ be a set. Define the partially ordered relation on L as $\leq = {(0, x), (0, m), (0, y), (x, m), (y, m), (0, 0), (x, x), (y, y), (m, m)}$ and also define $x \land y = min {x, y}, x \lor y = max {x, y}$ for all $x, y, z \in L$ . Define the unary operation ^′ and binary operation $\to$ as shown in the tables below respectively.

Table 1.

a	a
0	m
x	y
y	x
m	0

Table 2.

→	0	x	y	m
0	m	m	m	m
x	y	m	y	m
y	x	x	m	m
m	0	x	y	m

Then clearly $(L, \lor, \land, \to,', 0, m)$ is an IADL and hence it can be easily verified that L is an H-IADL.

In the following some properties of H-IADLs are discussed.

Theorem 3.3

Let L be an H-IADL. Then $x \to (y \to z) = (x \to y) \to (x \to z)$ for any $x, y, z \in L$ .

Proof.

Let L be an H-IADL and $x, y, z \in L$ . Now $(x \to (y \to z)) \to ((x \to y) \to (x \to z)) = ((x \to (y \to z)) \to ((y^{'} \to x^{'}) \to (z^{'} \to x^{'})) = (y^{'} \to x^{'}) \to ((y \to (x \to z)) \to (x \to z)) = (x \to y) \to (y \lor (x \to z)) = (x \to (y \to y)) \lor ((x \to y) \to (x \to z)) = x \lor y \lor ((x \land y) z) = m$ (by Definition 3.1 and Theorem 2.12). Therefore $(x \to (y \to z)) \leq ((x \to y) \to (x \to z)) .$ (1).

On the other hand, $((x \to y) \to (x \to z)) \to (x \to (y \to z)) = y \to ((x \to y) \to (x \to z)) \to (x \to z)) = y \to ((x \to y) \lor (x \to z)) = (y \to (x \to y)) \lor (y \to (x \to z)) = m \lor (y \to (x \to z)) = m$ .

Therefore, $((x \to y) \to (x \to z)) \leq (x \to (y \to z)) \dots . (2) .$ Hence by (1) and (2), we get $x \to (y \to z) = (x \to y) \to (x \to z)$ . □

Corollary 3.4

Let L be an IADL. Then the following are equivalent:

1. L is an H-IADL;
2. For any $x, y \in L$ , $x \to (x \to y) = x \to y$
3. For any $x, y, z \in L, x \to (y \to z) = (x \to y) \to (x \to z)$
4. For any $x, y, z \in L$ , $x \to (y \to z) = (x \land y) \to z$
5. For any $x, y, z \in L, (x \to (y \to z)) \to ((x \to y) \to (x \to z)) = m$ .

Proof.

Let L be an IADL and $x, y, z \in L .$

$(1) \Rightarrow (2) .$ Suppose L is an H-IADL. Then using Theorem 3.3, we have $x \to (x \to y) = (x \to x) \to (x \to y) = m \to (x \to y) = x \to y$ .

$(1) \Rightarrow (4) .$ Suppose $x \to (x \to y) = x \to y$ . By assumption and Theorem 2.12, $(x \land y) \to z = (x \land y) \to [(x \land y) \to z] = [(x \land y) \to (x \to z)] \lor [(x \land y) \to (y \to z)] = [(x \land y) \to (x \to z)] \lor [(x \land y) \to (y \to z) = [(x \to (x \to z)) \lor ((y \to (x \to z))] \lor [(x \to (y \to z)] \lor [(y \to (y \to z) = [(x \to z) \lor (y \to (x \to z))] \lor [(x \to (y \to z) \lor (y \to z)] = [(x \to z) \to (y \to (x \to z)) \to (y \to (x \to z)] \lor [((x \to (y \to z)) \to (y \to z)) \to (y \to z)] = [y \to ((x \to z) \to (x \to z))] \lor [(x \to (y \to z))] = [y \to m) \to (y \to (x \to z)) \lor [x \to (y \to z)] = [m \to (y \to (x \to z)] \lor [x \to (y \to z)] = [y \to (x \to z)] \lor [x \to (y \to z)] = [x \to (y \to z)] \lor [x \to (y \to z)] = x \to (y \to z) .$

(4) ⇒ (3). Suppose $x \to (y \to z) = (x \land y) \to z .$ Now $(x \land y) \to z = {[(x \to y) \to x^{'}]}^{'} \to z = z^{'} \to [(x \to y) \to x^{'}] = (x \to y) \to (z^{'} \to x^{'}) = (x \to y) \to (x \to z) = x \to (y \to z) .$

$(3) \Rightarrow (4) .$ Suppose $x \to (y \to z) = (x \to y) \to (x \to z) . Now (x \land y) \to z = {[(x \to y) \to x^{'}]}^{'} \to z = z^{'} \to ((x \to y) \to x^{'}) = (x \to y) \to (x \to z) = x \to (y \to z)$ (by Theorem 3.3).

$(3) \Rightarrow (5) .$ Suppose $x \to (y \to z) = (x \land y) \to z$ . $Now (x \to (y \to z)) \to ((x \to y) \to (x \to z)) = ((x \land y) \to z) \to ((x \to y) \to (x \to z)) = ((x \to y) \to (x \to z)) \to ((x \to y) \to (x \to z)) = m .$ Therefore $[x \to (y \to z)] \to [(x \to y) \to (x \to z)] = m$ . Hence (5) holds.

$(3) \Rightarrow (1)$ . Suppose $(x \to (y \to z)) \to [(x \to y) \to (x \to z)] .$ Then using Theorem 2.12 and Definition 3.1, $(x \to (y \to z)) \to ((x \to y) \to (x \to z)) = (x \to y) \to (y \to ((x \to z) \to (x \to z)) = (x \to y) \to (y \lor (x \to z)) = ((x \to y) \to y) \lor ((x \to y) \to (x \to z)) = x \lor y \lor ((x \to y) \to (x \to z))$ = $x \lor y \lor ((x \land y) \to z) = m$ . Therefore, L is an H-IADL. □

Definition 3.5

An IADL $(L, \lor, \land, \to,', 0, m)$ is said to be associative IADL if $\lor$ in L is assocative

Theorem 3.6

Let L be an associative IADL. Then the following statements are equivalent.

1. L is an H-IADL;
2. for any $x, y \in L$ and $z \in [x, m], z \to (x \to y) = x \to y$ ;
3. for any $x, y \in L$ , $(x \to y) \to x = x$ .
4. for any $x \in L, x \lor x^{'} = m$

Proof.

Let L be an IADL and $x, y, z \in L$ .

$(1) \Rightarrow (2$ ). Suppose L is an H-IADL. Let z ∈ [x, m]. Then we have $x \leq z$ . From Theorem 2.12, we have $z \to (x \to y) \leq x \to (x \to y) .$ This implies again $x \to y \leq z \to (x \to y) \leq x \to (x \to y) = x \to y$ (by Theorem 2.12 and Corollary 3.4), i.e., $z \to (x \to y) = x \to y$ . Therefore (2) holds.

$(2) \Rightarrow (1) .$ Putting z = x, we have x → (x → y) = x → y. Then it follows from Corollary 3.4 that L is an H-IADL.

$(3) \to (1) .$ Suppose $(x \to y) \to x = x . Now (x \to (x \to y)) \to (x \to y) = (((x \to y) \to x) \to x) = x \to x = m$ . It follows that $x \to (x \to y) \leq x \to y$ but we know from Theorem 2.12 that $x \to y \leq x \to (x \to y)$ . Therefore $x \to y = x \to (x \to y),$ it follows from Corollary 3.4 that L is an H-IADL.

$(1) \Rightarrow (3) .$ Suppose L is an H-IADL. Using Corollary 3.4 and Theorem 2.12, we have ((x → y) → x) → x = (x → (x → y)) → (x → y) = (x → y) → (x → y) = m. Therefore $(x \to y) \to x \leq x$ and clearly $x \to ((x \to y) \to x) = m$ . This implies $x \to (x \to y) \leq x .$ Hence $(x \to y) \to x = x$ .

$(3) \Rightarrow (4) .$ Suppose $(x \to y) \to x = x$ . Now $x^{'} = (x^{'} \to 0) \to x^{'} = x \to x^{'}$ (by supposition), it follows $x \lor x^{'} = (x \to x^{'}) \to x^{'} = x^{'} \to x^{'} = m$ . That is (4) holds.

$(4) \Rightarrow (3) .$ Suppose $x \lor x^{'} = m$ . Then ${(x \lor x^{'})}^{'} = x^{'} \land x = m^{'} = 0$ . Therefore L is complemented IADL. Hence $\lor$ is associative. Now using Theorem 2.12 and our assumption we get $(x \to y) \to (x^{'} \lor y) = ((x \to y) \to x^{'}) \lor ((x \to y) \to y) = y^{'} \lor x^{'} \lor x \lor y = m and (x^{'} \lor y) \to (x \to y) = (((x \to 0) \to y) \to y) \to (x \to y) = x \to ((((x \to 0) \to y) \to y) \to y) = x \to ((x \to 0) \to y) = x \to ((y \to 0) \to x) = y^{'} \to (x \to x) = y^{'} \to m = m$ (by Theorem 3.3). Hence $x \to y = x^{'} \lor y .$ It follows that $(x \to y) \to x = (x^{'} \lor y) \to x = {(x^{'} \lor y)}^{'} \lor x = (x \land y^{'}) \lor x = ((x \land y^{'}) \to x) \to x = [(x \to x) \lor (y^{'} \to x)] \to x = m \to x = x$ , that is (3) holds. □

Corollary 3.7

Let L be an H-IADL. Then for any $x, y \in L, y \lor {(x \to y)}^{'} = x \lor y .$

Proof.

Let L be an H-IADL and $x, y \in L$ . Now $(y \lor {(x \to y)}^{'}) \to (x \lor y) = (y \lor {(x \to y)}^{'}) \to [(x \to y) \to y] = (x \to y) \to [(y \lor {(x \to y)}^{'}) \to y)] = (x \to y) \to [(y \to y) \land ({(x \to y)}^{'} \to y)] = (x \to y) \to [{(x \to y)}^{'} \to y] = y^{'} \to [(x \to y) \to (x \to y)] = m$ . This implies $y \lor {(x \to y)}^{'} \to (x \lor y) = m$ . Therefore, $y \lor {(x \to y)}^{'} \leq x \lor y$ …(1). Conversely $(x \lor y) \to [y \lor {(x \to y)}^{'}] = ((x \lor y) \to y) \lor [(x \lor y) \to {(x \to y)}^{'}] = (x \to y) \lor [(x \to y) \to (x^{'} \land y^{'})] = (x \to y) \lor [((x \to y) \to x^{'}) \land ((x \to y) \to y^{'})] = [(x \to y) \lor ((x \to y) \to x^{'}] \land [(x \to y) \lor ((x \to y) \to y^{'})] = [(x \to y) \to ((x \to y) \to x^{'}) \to ((x \to y) \to x^{'})] \land [(x \to y) \to ((x \to y) \to y^{'}) \to ((x \to y) \to y^{'})] = [(x \to y) \to x^{'}) \to ((x \to y) \to x^{'})] \land [((x \to y) \to y^{'}) \to ((x \to y) \to y^{'})] = m \land m = m$ . This implies $x \lor y \leq y \lor {(x \to y)}^{'}$ .... (2). Hence by (1) and (2), we have $y \lor {(x \to y)}^{'} = x \lor y$ . □

LI-ideals in implicative almost distributive lattices

In this section we introduce LI-ideal in IADL. We discuss some characterization of LI-ideal with that of H-IADL. We observe relation between LI-ideal and filters in IADL, construction of LI-ideal using a set and lastly we discuss quotient structure of LI-ideal in IADL.

Definition 3.8

Let L be an IADL. An LI-ideal A of L is a subset of L such that

(I₁) $0 \in A$ and

(I₂) $y \in A$ $and {(x \to y)}^{'} \in A$ implies that $x \in A$ .

In this case if $A \neq L$ ,then A is proper LI-ideal of L. A proper LI-ideal A of L is said to be maximal if it is not properly contained in any other proper LI-ideal of L. That is, if I is any other LI-ideal of L such that $A \subseteq I \subseteq L$ implies either $A = I$ or $I = L$ , then A is maximal LI-ideal of L.

Remark 3.9

From the above definition we observe that {0} and L are trivial examples of LI-ideals.

The following example shows that there is a proper LI-ideal in IADL.

Example 3.10

Let $L = {0, x, y, z, w, m}$ be the underlining set with partial ordering $\leq = {(0, w), (0, x), (0, m), (0, z), (0, y), (w, x), (w, m), (w, y), (z, y), (z, m), (y, m), (0, 0), (w, w), (x, x), (z, z) (y, y), (m, m)}$ . Define the unary operation ′ and binary operation $\to$ as shown in the tables below respectively.

Table 3.

a	a
0	m
x	z
y	w
z	x
w	y
m	0

Table 4.

–→	0	x	y	z	w	m
0	m	m	m	m	m	m
x	z	m	y	z	y	m
y	w	x	m	y	x	m
z	x	x	m	m	x	m
w	y	m	m	y	m	m
m	0	x	y	z	w	m

Define the binary operation $\lor$ and $\land$ by $x \lor y = (x \to y) \to y$ $\land y = {[(x \to y) \to x^{'}]}^{'}$ for all $x, y \in L$ . Then clearly $(L, \lor, \land, \to,', 0, m)$ is an IADL. We can easily verify that $A = {0, z}$ is an LI-ideal of L.

Theorem 3.11

Let A be an LI-ideal of IADL L. If $x \in A$ and $y \leq x$ for some $\in L$ , then $y \in A$ .

Proof.

Let A be an LI-ideal of IADL L. Suppose $x \in A$ and $y \leq x$ for some $y \in L .$

Claim: $y \in A$ for some $y \in L$ . Since $y \leq x$ implies ${(y \to x)}^{'} = m^{'} = 0 \in A$ then by definition of LI-ideal (I₂) we have $y \in A$ . □

Theorem 3.12

Let A be a non-empty subset of IADL L. Then A is an LI-ideal of L if and only if it satisfies for all $x, y \in A$ and $z \in L, {(z \to x)}^{'} \leq y$ implies $z \in A$ .

Proof.

Suppose that A is an LI-ideal and $x, y \in A, z \in L$ . If ${(z \to x)}^{'} \leq y$ , then ${(z \to x)}^{'} \in A$ by Theorem 3.11. Using definition 3.8 (I₂) we obtain $z \in A$ . Conversely, suppose that for all $x, y \in A$ and $z \in L, {(z \to x)}^{'} \leq y$ implies $z \in A$ . Since A is a non empty subset of L, we assume $x \in A$ . Because ${(0 \to x)}^{'} \leq x$ , we have $0 \in A$ , and so I₁ holds for A. Let ${(x \to y)}^{'} \in A$ and $y \in A$ . Since ${(x \to y)}^{'} \leq (x \to y)$ ′, we have $x \in A,$ and so (I₂) holds for A. Hence, A is an LI-ideal of L. □

Theorem 3.13

Let L be an IADL. Every LI-ideal of L is an ADL ideal of L.

Proof.

Let A be an LI-ideal of IADL L.

i. Let $x, y \in A$ . Claim: $x \lor y \in A$ .
Now ${((x \lor y) \to y)}^{'} = {(((x \to y) \to y) \to y)}^{'} = {(x \to y)}^{'} \leq {(x^{'})}^{'} = x$ (since ${(x \to y)}^{'} \to x = x^{'} \to (x \to y) = x \to (x^{'} \to y) = y^{'} \to (x \to x) = m)$ . This implies ${((x \lor y) \to y)}^{'} \in A$ (by Theorem 3.11). Thus $x \lor y \in A$ (by Definition 3.8 (I₂)).
ii. Suppose $a \in A$ and $x \in L .$
Claim: $a \land x \in A$ .
Now ${((a \land x) \to a)}^{'} = {((a \to a) \lor (x \to a))}^{'} = {(m \lor (x \to a))}^{'} = m^{'} = 0 \in A$ . This implies $a \land x \in A$ . Hence, from (i) and (ii) A is an ADL ideal of L. □

Remark 3.14

The converse of Theorem 3.13 is not true. Consider Example 3.10. The set A = {0, w} is an ADL ideal. But it is not an LI-ideal since ${(x \to w)}^{'} = w \in A$ and $\notin A$ . □

Now we have the following characterization.

Theorem 3.15

In H-IADL L, every ADL ideal is an LI-ideal.

Proof.

Let $x, y \in L$ and A be an ADL ideal of H-IADL L. Assume that $y \in A$ and ${(x \to y)}^{'} \in A$ . We want to show that $x \in A$ . Clearly $0 \in A$ . From Corollary 3.7, we have $x \lor y = y \lor (x \to y)$ ^′. Then it follows from Definition 2.3 (ii) that $x \lor y = y \lor {(x \to y)}^{'} \in A$ . Since $x = (x \lor y) \land x \in A$ , then we have $x \in A$ . Hence A is an LI-ideal of L. □

Lemma 3.16

Let A be a non empty subset of H-IADL L. Then A is an LI-ideal of L if and only if for every $x, y \in L, x, y \in A$ if and only if $x \lor y \in A$ .

Proof.

Suppose A is an LI-ideal of H-IADL L. Let $x, y \in A$ . Since $((x \lor y) \to y)' = {((x \to y) \land (y \to y))}^{'} = 0 \in A$ and A is an LI-ideal, then we have $x \lor y \in A$ . Also ${(x \to (x \lor y))}^{'} = m^{'} = 0 \in A$ implies $x \in A$ as A is an LI-ideal. Conversely suppose $x, y \in A$ if and only if $x \lor y \in A$ and ${(x \to y)}^{'} \in A$ trivially $x \in A$ . Hence, A is an LI-ideal.

For any A and B of IADL L we set A ∧ B = {a ∧ b|a ∈ A and b ∈ B}.

Theorem 3.17

If A and B are LI-ideals of an H-IADL of L, then so is $A \land B$ .

Proof.

Let A and B be two LI-ideals of H-IADL L. Let $x, y \in A \land B$ . Then x = a₁ ∧ b₁ for a₁ ∈ A and b₁ ∈ B and y = a₂ ∧ b₂ for a₂ ∈ A and b₂ ∈ B. Now x ∨ y = (a₁ ∧ b₁) ∨ (a₂ ∧ b₂) = ((a₁ ∧ b₁) ∨ a₂) ∧ ((a₁ ∧ b₁) ∨ b₂) ∈ A ∧ B as ((a₁ ∧ b₁) ∨ a₂ ∈ A $and ((a 1 \land b 1) \lor b 2) \in B . Thus, x \lor y \in A \land B . Conversely assume x \lor y \in A \land B for all x, y \in L$ . Then $x \lor y = a \land b$ for some $a \in A$ and $b \in B$ . We observe that $x = x \land (x \lor y) = x \land (a \land b) \leq a \in A$ implies $x \in A$ . Similarly one can show that $x \in B$ and that $y \in A$ and $y \in B$ . Thus $x, y \in A \land B .$ Hence, $A \land B$ is an LI-ideal of L. □

Theorem 3.18

If A and B are LI-ideals of an H-IADL L, then $A \land B = A \cap B$ .

Proof.

Let $x \in A \land B$ . Then $x = a \land b$ for some $a \in A$ and $b \in B$ . Since $x = b \land a \leq a \in A$ implies $a \land b \in A$ (as A is an ADL ideal by Theorem 3.15 and properties of ideal in ADL) and $x = a \land b \leq b \in B$ , then from Theorem 3.11, $x \in A$ and also $x \in B$ and hence $x \in A \cap B$ . Therefore, $A \land B \subseteq A \cap B$ . Conversely suppose $x \in A \cap B$ then $x = x \land x \in A \land B$ . Therefore, $A \cap B \subseteq A \land B$ . Hence, $A \land B = A \cap B$ . □

Theorem 3.19

If A is an LI-ideal of an H-IADL L and $a \in L$ , then the set $K = {x \in L | x^{'} \to a \in A}$ is an LI-idel of L.

Proof

Let $x, y \in K$ . Then $x^{'} \to a \in A$ and $y^{'} \to a \in A$ . Using Theorem 2.13 that ${(x \lor y)}^{'} \to a = (x^{'} \land y^{'}) \to a = (x^{'} \to a) \lor (y^{'} \to a) \in A$ (by Theorem 3.24) so that $x \lor y \in K$ . Conversely let $x, y \in L$ be such that $x \lor y \in K$ . Then ${(x \lor y)}^{'} \to a \in A$ . Using Theorem 2.12 and Theorem 3.3, we $have {(x \lor y)}^{'} \to a = a^{'} \to (x \lor y)) = a^{'} \to ((x \to y) \to y) = (a^{'} \to (x \to y)) \to (a^{'} \to y) =$ $(a^{'} \to x) \to (a^{'} \to y) \to (a^{'} \to y) = (a^{'} \to x) \lor (a^{'} \to y) = (x^{'} \to a) \lor (y^{'} \to a) .$ Thus $(x^{'} \to a) \lor (y^{'} \to a) \in A$ which implies that $x^{'} \to a \in A$ and $y^{'} \to a \in A$ because A is an LI-ideal of H-IADL and Lemma 3.16. This implies that $x \in K$ and $y \in K$ . Hence by Theorem 2.13, K is an LI-ideal of L.

Proposition 3.20

Let L be an H-IADL and $a \in L$ .Then there is no proper LI-ideal of L containing $a$ and $a^{'}$ simultaneously.

Proof.

Let A be a proper LI-ideal of H-IADL L containing a and a′ simultaneously. Then $m = a \lor a^{'} \in A$ , and hence $A = L$ a contradiction. Therefore, A is not a proper LI-ideal of H-IADL L containing a and a′ simultaneously. □

For any non-empty subset A of an IADL L, we define $A^{'} = {x^{'} | x \in A}$ . It is obvious that every non empty subset A of IADL L is not a filter. Similarly the set A^′ for every subset A of L is not an LI-ideal in general. In fact the dual concept of a filter is one of an LI-ideal in IADL. Then we have the following theorem to verify this ideal.

Theorem 3.21

Let A be a non empty subset of IADL L. Then A is a filter of L if and only if A′ is an LI-ideal of L.

Proof.

Assume that A is a filter of IADL L. Then $m \in A$ , and so $m^{'} = 0 \in A^{'} .$ Let ${(x \to y)}^{'} \in A$ ′ and $y \in A$ ′ for all $x, y \in L$ . Then ${(x \to y)}^{'} = u^{'}$ and $y = v$ ′ for some $u, v \in A$ . Thus $v \to x^{'} = x \to v^{'} = x \to y = {({(x \to y)}^{'})}^{'} = {(u^{'})}^{'} = u \in A$ . Since A is a filter, we have $x^{'} \in A$ and so $x = {(x^{'})}^{'} \in A^{'}$ . This proves that A′ is an LI-ideal of L. Conversely let $x, y \in L$ be such that $x \in A$ and $x \to y \in A$ . We want to show $y \in A$ . Suppose that A′ is an LI-ideal of L. Then $x^{'} \in A^{'}$ and ${(y^{'} \to x^{'})}^{'} = {(x \to y)}^{'} \in A$ ′. As A′ is an LI-ideal, it follows from Definition 3.8 (I₂) that $y^{'} \in A^{'}$ or $y \in A$ . Also $0^{'} = m \in A$ . Hence, A is a filter of L. □

Theorem 3.22

Suppose { $B$ _i $: i \in J}$ for index set J is a non-empty family of LI-ideal of IADL L. Then $A = \cap i$ _∈J $B$ _i is also an LI-ideal of L, for any $i \in J$ of index set J.

Proof.

Let $A = {B$ _i $: B$ _i $is an LI - ideal of L} .$

We need to show A = ∩_i∈J B_i is also an LI-ideal of L.

1. Clearly $0 \in B$ _i for each $\in J$ . This implies $0 \in \cap$ _i∈J $B$ _i
2. Since ${(x \to y)}^{'} \in B$ _i and $y \in B$ _i $implies x \in B$ _i $for each i \in J, then {(x \to y)}^{'} \in \cap i \in J Bi and$ y ∈ ∩_i∈J B_i implies $x \in \cap$ _i∈J $B$ _i.
Therefore, A = ∩_i∈J B_i is an LI-ideal of L.□

Let A be a subset of an IADL L. Then the least LI-ideal containing A is called the LI-ideal generated by A, written $⟨ A ⟩ .$ From Remark 3.14, L is clearly an LI- ideal Containing A. If $A = {a}, ⟨ {a} ⟩$ is written $⟨ a ⟩ .$ In short we shall write [a₁, a₂, a₃, · · ·, a_n, x] for a₁ → (a₂ → (· · · → (a_n → x) · · ·), and write [aⁿ, x] if a₁ = a₂ = · · · = a_n = a. So in our case we define $⟨ A ⟩ = {x \in L | a_{n}^{'} \to (\cdot \cdot (a_{1}^{'} \to$ $x^{'}) \dots) =$ m for some a₁, · · ·, a_n ∈ A} and $⟨ a ⟩ = {x \in L [(a^{'})$ ⁿ $, x^{'}] = m for some n \in N} .$

For any natural number n we define n(x) → y recursively as follows: 1(x) → y = x → y and (n + 1)(x) → y = x → (n(x) → y).

Using definition 2.11(3) repeatedly, we know that the following identity in IADL holds:

i. z → (y₁ → (y₂ → (· · · (y_n → x) · · ·))) = y₁ → (y₂ → (· · · (y_n → (z →x) · · ·)). As a special case of (i) we have
ii. $z \to (n (y) \to x) = n (y) \to (z \to x)$ . □

In the following theorem we can describe elements of $⟨ A ⟩ .$

Theorem 3.23

If A is a non-empty subset of an IADL L, then $⟨ A ⟩$ = {x ∈ L|a₁^′→ $\dots$ $(a'_{1}$ → x^′) · · ·) = m for some a₁, · · ·, a_n $\in A}$ is the least LI-idea of L containing.

Proof.

Let us denote U = {x ∈ L | $(a'_{n}$ → (· · · ( a₁′→ x′) · · ·) = m for some a₁, · · ·, $a$ _n $\in A$ }. We claim $U = ⟨ A ⟩ .$ That is, we need to prove 1) $\subseteq U$ , 2) $U$ is an LI-ideal, 3) $U \subseteq V$ for any LI-ideal V containing A. To prove (1). Since A is non-empty there exists $a \in A$ such $a^{'} \to a^{'} = m$ . This is implies a ∈¹U. Therefore, $\subseteq U$ . To prove 2) For $a \in A, a^{'} \to 0^{'} = m$ implies $0 \in U$ . Let ${(x \to y)}^{'} \in U$ and $y \in U$ , then there exists $a$ _i $\in A (i = 1, 2, 3, \dots, n)$ and $b$ _j $\in A$ ( j = 1, 2, 3, · · ·, p) such that a′ → (· · · (a′ → ((x → y)′)^′ · · ·) = m…( $*) .$

$b_{p}^{'}$ → (· · · ( $b_{1}^{'}$ → y′) · · ·) = m…( $* *) .$ From ( $*),$ we have $a_{n}^{'}$ → (· · · (a′ →((y′ → x′) · · ·) = m … ( $* * *) .$ This implies y′ ≤ $a_{n}^{'}$ → (· · ·(a1^′→ x′) · · ·) … ( $* * * *)$

Combining ( $* *),$ ( $* * * *)$ and properties of IADL, we get m = $b_{p}^{'}$ → (· · · ( $b_{1}^{'}$ →y′) · · ·) ≤ $b_{p}^{'}$ → (· · ·( $b_{1}^{'}$ →→ (· · ·( $a_{1}^{'}$ → x′) · · ·))) · · ·) and hence $b_{p}^{'}$ →(· · · ( $b_{1}^{'}$ →( $a_{n}^{'}$ → (·· · (a′→ x′) · · ·))) · · ·) = m. This shows that $\in U$ .

Therefore, U is an LI-ideal of L containing A. To prove 3) Let V be an LI- ideal containing A and let $\in U$ . Then ( $a_{n}^{'}$ → (· · ·( $a_{1}^{'}$ → x′) · · ·) = m for some $a$ ₁ $, \dots, a$ _n $\in A$ . Thus m = ( $a_{n}^{'}$ →(( $a_{n - 1}^{'}$ → (· · · ( $a_{1}^{'}$ → x′) · · ·))= ( $a_{n}^{'}$ →(( $a_{n - 1}^{'}$ → (· · · ( $a_{1}^{'}$ → x′) · · ·))) = ( $a_{n - 1}^{'}$ → $a_{1}^{'}$ → (· · $a_{1}^{'} \to x^{'}$ )· ·))→a_n, which implies that (( $a_{n - 1}^{'}$ → (· · · ( $a_{1}^{'}$ → x′) · · ·))^′ → a_n)′ = m^′ = $0 \in V$ .

Since $a$ _n $\in A \subseteq V$ and V is an LI-ideal of L, we have ( $a_{n - 1}^{'}$ → (· · · ( $a_{1}^{'}$ →x′) · · · $))')' \in V$ . Now ( $a_{n - 1}^{'}$ → (· · · ( $a_{1}^{'}$ → x′) · · ·))′ = ( $a_{n - 2}^{'}$ → (· · · ( $a_{1}^{'}$ → x′) → a_n−1) · · ·))′, since $a_{n} - 1 \in A \subseteq V$ , it follows from Definition 3.16 I₂ that ( $a_{n - 2}^{'}$ → (· · · (a′→ x′) · · · $))' \in V$ . Repeating the process we conclude that ${(x \to a)}^{'} = m^{'} = 0 \in V$ and $\in V$ . As V is an LI-ideal of L we have $\in V$ . This proves that $U \subseteq V$ , hence $U = ⟨ A ⟩ .$ That is, U is the least LI-ideal containing A. □

The following corollary is immediate from Theorem 3.23.

Corollary 3.24

For any element a of an IADL L, we have $⟨ a ⟩ = {x \in L :$ $n (a^{'}) \to x^{'} =$ m for some natural number n}.

Theorem 3.25

Let A be an an LI-ideal of IADL L and $a \in L$ . Then $⟨ A \cup {a} ⟩ =$ ${x \in L | {(n (a^{'}) \to x^{'})}^{'} \in A for some n \in N} .$

Proof.

Let A be an LI-ideal of IADL L and $a \in L$ . Consider $A$ _a $= {x \in L | {(n (a^{'}) \to x^{'})}^{'} \in A for some n \in N}$ . Since ${(n (a^{'}) \to 0^{'})}^{'} = 0 \in A$ , then $0 \in A$ _a. Let ${(y \to x)}^{'} \in A$ _a and $x \in A$ _a. Then there exist $m, n \in N$ such that ${(n (a^{'}) \to {({(y \to x)}^{'})}^{'})}^{'} \in A$ and ${(m (a^{'}) \to x^{'})}^{'} \in A$ . It follows that ${(n (a^{'}) \to (y \to x))}^{'} = u$ and ${(m (a^{'}) \to x^{'})}^{'} = v$ for some $u, v \in A$ , so that $u^{'} = n (a^{'}) \to (y \to x)$ and $v^{'} = m (a^{'}) \to x^{'} .$ Then $m = u^{'} \to u^{'} = u^{'} \to (n (a^{'}) \to (y \to x)) = u^{'} \to (n (a^{'}) \to (x^{'} \to y^{'})) = u^{'} \to (x^{'} \to (n (a^{'}) \to y^{'})) = x^{'} \to (u^{'} \to (n (a^{'}) \to y^{'})), i . e ., x^{'} \leq u^{'} \to (n (a^{'}) \to y^{'}) .$ Using Theorem 2.12 $v^{'} = m (a^{'}) \to x^{'} \leq m (a^{'}) \to (u^{'} \to (n (a^{'}) \to y^{'})) = u^{'} \to (m (a^{'}) \to (n (a^{'}) \to y^{'})) = u^{'} \to ((m + n) (a^{'}) \to y^{'}), which implies that v^{'} \to (u^{'} \to {({((m + n) (a^{'}) \to y^{'})}^{'})}^{'}) = v^{'} \to (u^{'} \to ((m + n) (a^{'}) \to y^{'})) = m$ . Since $u, v \in A$ , it follows from Theorem 3.23 that ${((m + n) (a^{'}) \to y^{'})}^{'} \in ⟨ A ⟩ = A$ . Hence $y \in Aa$ and A_a is LI-ideal of L. Since ${(n (a^{'}) \to a^{'})}^{'} = m^{'} = 0 \in A$ , then $a \in A$ _a. Let $x \in A$ . Since $x^{'} \leq a^{'} \to x^{'} = {({(a^{'} \to x^{'})}^{'})}^{'}$ , it follows from Theorem 3.19 that ${(a^{'} \to x^{'})}^{'} \in A$ i.e., $x \in A$ _a and so $A \subseteq A$ _a. Thus $A \cup {a} \subseteq A$ _a. Finally we show that A_a is the least LI-ideal containing A and a. Let B be any LI-ideal containing A and a, and let $x \in A$ _a. Then ${(n (a^{'}) \to x^{'})}^{'} \in A \subseteq B$ for some $n \in N$ , and hence ${(((n - 1) (a^{'}) \to x^{'}))}^{'} \to a)' = {(a^{'} \to ((n - 1) (a^{'}) \to x^{'}))}^{'} = {(n (a^{'}) \to x^{'})}^{'} \in B$ . Since $a \in B$ it follows from definition of LI-ideal we have $(n - 1) (a^{'}) \to x^{'})' \in B$ . Repeating this process we obtain $x = {(x^{'})}^{'} \in B$ . Therefore, A_a is the least LI- ideal containing A_a and a, i.e., $⟨ A \cup {a} ⟩ = A$ _a.

Lemma 3.26

Let L be an IADL and $a, b, x \in L$ .If $n (a) \to x = m$ and $m (b) \to$ $x = m$ for some $n, m \in N$ , then there exists $k \in N$ such that $k (a \lor b) \to x = m$

Theorem 3.27

Let A be an LI-ideal of an IADL L. Then $⟨ A \cup {a} ⟩ \cap ⟨ A \cup {b} ⟩ =$ $⟨ A \cup {a \land b} ⟩$ for all $a, b \in L$ .

Proof.

Let $x \in ⟨ A \cup {a} ⟩ \cap ⟨ A \cup {b} ⟩ .$ By Theorem 3.23, there are $m, n \in N$ such that ${(n (a^{'}) \to x^{'})}^{'} \in A$ and ${(m (b^{'}) \to x^{'})}^{'} \in A$ . Hence (n(a′) → x′)′ = u and (m(b′) → x′)′ = v for some $u, v \in A$ . It follows that $n (a^{'}) \to x^{'} = u^{'}$ and $m (b^{'}) \to x^{'} = v^{'}$ so that $m = v^{'} \to (u^{'} \to (n (a^{'}) \to x^{'})) = m (a^{'}) \to (v^{'} \to (u^{'} \to x^{'}))$ and $m = u^{'} \to (v^{'} \to (m (b^{'}) \to x^{'})) = m (b^{'}) \to (v^{'} \to (u^{'} \to x^{'}))$ . Using Lemma 3.26, there exists $k \in N$ such that $k (a^{'} \lor b^{'}) \to (v^{'} \to (u^{'} \to x^{'})) = m$ . Since $a^{'} \lor b^{'} = {(a \land b)}^{'},$ we have $m = k (a^{'} \lor b^{'}) \to (v^{'} \to (u^{'} \to x^{'})) = k {(a \land b)}^{'} \to (v^{'} \to (u^{'} \to x^{'})) = v^{'} \to (u^{'} \to (k ({(a \land b)}^{'} \to x^{'})) .$ Applying Theorem 3.21 we get ${(k {({(a \land b)}^{'})}^{'})}^{'} \in ⟨ A ⟩ = A and hence x \in ⟨ A \cup {a \land b} ⟩$ by Theorem 3.23. Thus, $⟨ A \cup {a} ⟩ \cap ⟨ A \cup {b} ⟩ \subseteq ⟨ A \cup {a \land b} ⟩$ . Conversely if $x \in ⟨ A \cup {a \land b} ⟩,$ then ${(n {(a \land b)}^{'} \to x^{'})}^{'} \in A$ for some $\in N$ . Since $a \land b \leq a, b$ , then $a^{'} \leq {(a \land b)}^{'} and b^{'} \leq {(a \land b)}^{'} .$ Using Theorem 2.12(10) repeatedly, we get $n ({(a \land b)}^{'}) \to x^{'} \leq n (a^{'}) \to x^{'} and n ({(a \land b)}^{'}) \to x^{'} \leq n (b^{'}) \to x^{'}$ , which imply that ${(n (a^{'}) \to x^{'})}^{'} \leq {(n ({(a \land b)}^{'}) \to x^{'})}^{'}$ and ${(n (b^{'}) \to x^{'})}^{'} \leq (n {({(a \land b)}^{'} \to x^{'})}^{'} .$ Applying Theorem 3.19, we get ${(n (a^{'}) \to x^{'})}^{'} \in A$ and $n (b^{'}) \to x^{'})' \in A$ , i.e., $x \in ⟨ A \cup {a} ⟩$ and $x \in ⟨ A \cup {b} ⟩$ . Hence, $x \in ⟨ A \cup {a} ⟩ \cap ⟨ A \cup {b} ⟩$ and so $⟨ A \cup {a \land b} ⟩ \subseteq ⟨ A \cup {a} ⟩ \cap ⟨ A \cup {b} ⟩$ . □

Let A be an LI-ideal of IADL L. We define a binary relation “ $\sim$ ” on L as follows: $x \sim y$ if and only if ${(x \to y)}^{'} \in A$ and ${(y \to x)}^{'} \in A$ for all $x, y \in L$ .

Lemma 3.28

The binary relation “∼” on IADL L is an equivalent relation on L.

Proof.

Let L be an IADL L such that $x, y, z \in L$ and A be an LI-ideal of L. Then

1. $x \sim x \Leftrightarrow {(x \to x)}^{'} = m^{'} = 0 \in A$ . This implies $\sim$ is reflexive on L.
2. $x \sim y \Leftrightarrow {(x \to y)}^{'} \in A$ and ${(y \to x)}^{'} \in A$ if and only if ${(y \to x)}^{'} \in A$ and ${(x \to y)}^{'} \in A$ if and only if $y \sim x$ . Therefore $\sim$ is symmetric on L.
3. Assume $x \sim y$ and $y \sim z$ . We need to show that $x \sim z$ . Now $x \sim y$ and $y \sim z$ implies ${(x \to y)}^{'} \in A$ and ${(y \to x)}^{'} \in A, {(y \to z)}^{'} \in A$ and ${(z \to y)}^{'} \in A$ . Since ${({(x \to z)}^{'} \to {(x \to y)}^{'})}^{'} = {((x \to y) \to (x \to z))}^{'} \leq {(y \to z)}^{'}$ it follows from Theorem 3.19 that ${((x \to y) \to (x \to z))}^{'} \in A$ . Since ${(x \to y)}^{'} \in A$ and A is LI-ideal of L, we have ${(x \to z)}^{'} \in A$ .

Similarly, ${({(z \to x)}^{'} \to {(z \to y)}^{'})}^{'} = {((z \to y) \to (z \to x))}^{'} \leq {(y \to x)}^{'}$ it also follows from Theorem 3.19 that ${(z \to x)}^{'} \in A$ . Therefore $x \sim z$ . Hence $\sim$ is transitive on L. □

Proposition 3.29

If $x \sim u$ and $y \sim v$ in an LI-ideal A of IADL L, then $(x \to y) \sim (u \to v)$ for any $x, y, u, v \in L$ .

Proof.

Let L be an IADL and $x, y, u, v \in L$ . Assume that $x \sim u$ and $y \sim v$ of an LI-ideal A. Then ${(x \to u)}^{'} \in A$ and ${(u \to x)}^{'} \in A, {(y \to v)}^{'} \in A$ and ${(v \to y)}^{'} \in A$ . Since ${((x \to y) \to (x \to v))}^{'} \leq {(y \to v)}^{'}$ and ${((x \to v) \to (x \to y))}^{'} \leq {(v \to y)}^{'}$ , it follows from Theorem 3.19 that ${((x \to y) \to (x \to v))}^{'} \in A$ and ${((x \to v) \to (x \to y))}^{'} \in A$ . This implies that $(x \to y) \sim (x \to v) \dots (1) .$ Similarly Since ${((x \to v) \to (u \to v))}^{'} \leq {(u \to x)}^{'} \in A$ and ${((u \to v) \to (x \to v))}^{'} \leq {(x \to u)}^{'} \in A$ , it follows from Theorem 3.19 that ${((x \to v) \to (u \to v))}^{'} \in A$ and ${((u \to v) \to (x \to v))}^{'} \in A$ . This implies $(x \to v) \sim ((u \to v)$ …(2). From (1), (2) and transitivity of $\sim$ , We conclude that $(x \to y) \sim (u \to v)$ . □

Corollary 3.30

Let $a \sim b$ and $c \sim d$ . Then

1. $a^{'} \sim b^{'}$
2. $(a \lor c) \sim (b \lor d)$
3. $(a \land c) \sim (b \land d)$

Proof.

Let $a, b, c, d \in L$ for an LI-ideal A of IADL L. Assume $a \sim b$ and $c \sim d$ . Then

1. $a \sim b$ iff ${(a \to b)}^{'} \in A$ and ${(b \to a)}^{'} \in A$ iff ${(b^{'} \to a^{'})}^{'} \in A$ and ${(a^{'} \to b^{'})}^{'} \in A$ iff $a^{'} \sim b^{'} .$
2. $a \sim b$ and $c \sim d$ implies $(a \to c) \sim (b \to d)$ from proposition 3.28. we also implies $((a \to c) \to c) \sim ((b \to d) \to d)$ . Therefore $a \lor c) \sim (b \to d) .$
3. $a \sim b$ and $c \sim d$ implies $(a \to c) \sim (b \to d) .$ We also imply $((a \to c) \to a^{'}) \sim ((b \to d) \to b^{'})$ . Again we imply $({[(a \to c) \to a^{'}]}^{'}) \sim ({[b \to d) \to b^{'}]}^{'})$ . Therefore $(a \land c) \sim (b \land d) .$ □

Theorem 3.31

The binary relation $\sim$ is a congruence relation on an IADL L.

Proof.

Using Lemma 3.28, Proposition 3.29 and Corollary 3.30, it is clear that ~ is a congruence relation on L. □

Now we have the following definition. Let A be an LI-ideal of IADL L. Then the equivalent class containing x is denoted by A_x such that $Ax = {y \in L | x \sim y}$ and the set of all equivalent classes of L is denoted by $L | A$ such that $L | A = {Ax | x \in L} .$

Lemma 3.32

A₀ = A and $A$ _m $= {y \in L | y^{'} \in A}$ .

Proof.

${(y \to 0)}^{'} = y \in A$ and ${(0 \to y)}^{'} \in A$ for all $y \in L$ implies $0 \sim y$ . This implies $y \in A$ ₀. Therefore $A \subseteq A$ ₀. Conversely, $x \in A$ ₀ $\Rightarrow {(0 \to x)}^{'} \in A$ and ${(x \to 0)}^{'} \in A \Rightarrow 0 \in A$ and $x \in A \Rightarrow x \in A$ . Therefore, $A 0 \subseteq A$ . Hence, A₀ = A.

$A$ _m $= {y \in L | m \sim y} = {y \in L : {(m \to y)}^{'} \in A$ and ${(y \to m)}^{'} \in A} =$ ${y \in L : y^{'} \in A and 0 \in A} = {y \in L : y^{'} \in A} .$ □

Define binary operations “∪”, “∩”, “⇒” and unary operation “N” on $L / A$ as follows:

A_{x} \cup A_{y} = A_{x \lor y} A_{x} \cap A_{y} = A_{x \land y},

A_{x} \Rightarrow A_{y} = A_{x \to y}, A_{x}^{N} = A_{x}^{'} for all A_{x}, A_{y} \in L | A .

Then we have the following lemma.

Lemma 3.33

Let L be an IADL and A be an LI-ideal of L. Then

1. $(L | A, \cup, \cap, A$ ₀ $, A$ _m $)$ is a bounded ADL.
2. $(L | A, \cup, \cap, \Rightarrow, N, A$ ₀ $, A$ _m $)$ is an IADL which is called quotient implicative ADL.

Proof.

Let L be an IADL and A be an LI-ideal of L. Let $A$ _x $, A$ _y $, A$ _z $\in L | A$ for all $x, y, z \in L$ . Then

1. Since $A$ ₀ $\cup A$ _x $= A$ _0∨x $= A$ _x and $A$ ₀ $\cap A$ _x $= A$ _0∧x $= A$ ₀, for all $x \in L$ . This implies A₀ is the least element of L|A. Also $A$ _m $\cup A$ _x $= A$ _m∨x $= A$ _m and $A$ _m $\cap A$ _x $= A$ _m∧x $= A$ _x, for all $x \in L$ . This implies A_m is the greatest element of L|A. Therefore, $(L | A, \cup, \cap, A$ ₀ $, A$ _m $)$ is bounded. we can easily verify that $(L | A, \cup, \cap, A$ ₀ $, A$ _m $)$ is an ADL.
2. To show $(L | A, \cup, \cap, \Rightarrow, N, A$ ₀ $, A$ _m $)$ is an IADL,
- (a) $A$ _x $\cup A$ _y $= A$ _x∨y $= A$ _(x→y)→y $= (A$ _x $\Rightarrow A$ _y $) \Rightarrow A$ _y
- (b) $A_{x} \cap A_{y} = A_{x \land y} = [A (x \to y) \to x^{'}] = A$ _(x→y)→x′ $= (Ax \Rightarrow Ay) \Rightarrow {Ax}^{'}$ = [(A_x ⇒ A_y) ⇒ A^N]_x^N
- (c) $A$ _x $\Rightarrow (A$ _y $\Rightarrow A$ _z $) =$ A_x→(y→z) $= A$ _y→(x→z) $= A$ _y $\Rightarrow (A$ _x $\Rightarrow A$ _z $)$
- (d) $A$ _m $\Rightarrow A$ _x $= A$ _m→x $= A$ _x
- (e) $A$ _x $\Rightarrow A$ _m $= A$ _x→m $= A$ _m
- (f ) $A$ _x $\Rightarrow A$ _y $= A$ _x→y $= A$ _y′_→x′ $= A$ _y′ $\Rightarrow A$ _x′ $= A$ y^N $\Rightarrow A$ x^N
- (g) $A_{0}^{N}$ = A₀′ = A_m
Therefore, $(L | A, \cup, \cap, \Rightarrow, N, A$ ₀ $, A$ _m $)$ is quotient IADL. □

Conclusion

Many scholars investigated ideals in lattice and ADL. Furthermore, LI-ideals in Lattice implication algebras are discussed by different researchers. For the advancement of many valued logical algebra we extend this notion to ADL concept. That is, first we investigated H-implicative almost distributive lattice. Then we discussed LI-ideals in IADL. Lastly, we found an interesting result that set theoretic LI-ideals is an LI-ideal and some characterization of LI-ideals in IADL. In the future we will have fuzzy version of this notion.

Author contributions

All the authors are contributed equally in this manuscript and also both authors read and approved the final manuscript.

Ethics and consent

Ethics and consent were not required.

Data availability

No data are associated with this article.

References

1. Assaye B, Alamneh M, Mekonnen T: Implicative Almost Distributive Lattice. Int. J. Comput. Sci. Appl. Math. 2018; 4(1): 19–22.
2. Assaye B, Alamneh M, Mekonnen T: Positive implicative and associative filters of implicative almost distributive lattice. Bull. Inter. Math. Virtual Inst. 2019; 9(1): 111–119.
3. Jun YB: Implicative filters of lattice implication algebras. Bull. Korean Math. Soc. 1997; 34(2): 193–198.
4. Jun YB, Roh EH, Xu Y: LI-ideals in lattice implication algebras. Bull. Korean Math. Soc. 1998; 35(1): 13–24.
5. Kolluru V, Bekele B: Implicative algebras. Momona Ethiop. J. Sci. 2012; 4(1): 90–101. Publisher Full Text
6. Swamy UM, Rao GC: Almost Distributive Lattices. J. Aust. Math. Soc. (Series A). 1981; 31(1): 77–91. Publisher Full Text
7. Xu Y, Qin KY: Lattice H-implication algebras and lattice implication algebra classes. J. Hebin Mining Civil Eng. Institute. 1992; 2: 139–143.
8. Xu Y: Lattice implication algebras. J. South West Jiaotong University. 1993; 28(1): 20–27.
9. Xu Y, Ruan D, Qin KY, et al.: Lattice-Valued Logic: An Alternative Approach to Treat Fuzziness and Incomparability. Berlin: Springer; 2003.

Comments on this article Comments (0)

Version 1

VERSION 1 PUBLISHED 10 Feb 2025

Author details Author details

¹ Department of Mathematics, University of KabriDahar, Kabri Dahar, Somalie, Ethiopia
² Department of Mathematics, Bahir Dar University College of Science, Bahir Dar, Amhara, 6000, Ethiopia

Alachew Amaneh Mechderso
Roles: Conceptualization, Writing – Original Draft Preparation, Writing – Review & Editing

Tilahun Mekonnen Munie
Roles: Conceptualization, Visualization, Writing – Review & Editing

Competing interests

No competing interests were disclosed.

Grant information

The author(s) declared that no grants were involved in supporting this work.

Article Versions (1)

version 1

Published: 10 Feb 2025, 14:182

https://doi.org/10.12688/f1000research.159175.1

Copyright

© 2025 Mechderso AA and Munie TM. This is an open access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Download

Export To

metrics

	Views	Downloads
F1000Research	-	-
PubMed Central Data from PMC are received and updated monthly.	-	-

Citations

0

SEE MORE DETAILS

CITE

how to cite this article

Mechderso AA and Munie TM. Li-ideals of implicative almost distributive lattices [version 1; peer review: 1 approved]. F1000Research 2025, 14:182 (https://doi.org/10.12688/f1000research.159175.1)

NOTE: If applicable, it is important to ensure the information in square brackets after the title is included in all citations of this article.

track

receive updates on this article

Track an article to receive email alerts on any updates to this article.

Open Peer Review

Version 1

VERSION 1

PUBLISHED 10 Feb 2025

Views

3

Reviewer Report 21 Mar 2025

Yang Xu, Southwest Jiaotong University, Chengdu, China

Liu Yi, College of Mathematics and Information Sciences, Neijiang Normal University, Neijiang, Sichuan, China

Approved

https://doi.org/10.5256/f1000research.174866.r368840

Ideals and Filters of two basic research direction in logical algebra. In current paper, authors investigated the LI-ideals of ADL concepts. For the advancement of many valued logical algebra we extend this notion to ADL concept. Firstly, authors investigated H-implicative almost distributive ... Continue reading

Ideals and Filters of two basic research direction in logical algebra. In current paper, authors investigated the LI-ideals of ADL concepts. For the advancement of many valued logical algebra we extend this notion to ADL concept. Firstly, authors investigated H-implicative almost distributive lattice; LI-ideals in IADL are discussed. Authors also found
that set theoretic LI-ideals is an LI-ideal and some characterizations of LI-ideals in IADL.

All results are soundness. The proofs are correct. So I think this paper can be accepted subjected to a minor revised. Some references regarding the ideals and filters in residued lattices should be added.

Is the work clearly and accurately presented and does it cite the current literature?

Partly
Is the study design appropriate and is the work technically sound?

Yes
Are sufficient details of methods and analysis provided to allow replication by others?

No
If applicable, is the statistical analysis and its interpretation appropriate?

I cannot comment. A qualified statistician is required.
Are all the source data underlying the results available to ensure full reproducibility?

No source data required
Are the conclusions drawn adequately supported by the results?

Yes

Competing Interests: No competing interests were disclosed.

Reviewer Expertise: logical algebras

We confirm that we have read this submission and believe that we have an appropriate level of expertise to confirm that it is of an acceptable scientific standard.

CITE

Report a concern

Respond or Comment

Comments on this article Comments (0)

Version 1

VERSION 1 PUBLISHED 10 Feb 2025

Open Peer Review

Reviewer Status

Reviewer Reports

	Invited Reviewers
	1
Version 1 10 Feb 25	read

Yang Xu, Southwest Jiaotong University, Chengdu, China

Liu Yi, Neijiang Normal University, Neijiang, China

Comments on this article

All Comments(0)

Add a comment

Sign up for content alerts

Browse by related subjects

Back to all reports

Reviewer Report

3 Views

21 Mar 2025 | for Version 1

Yang Xu, Southwest Jiaotong University, Chengdu, China

Liu Yi, College of Mathematics and Information Sciences, Neijiang Normal University, Neijiang, Sichuan, China

3 Views Cite this report Responses(0)

Approved

Ideals and Filters of two basic research direction in logical algebra. In current paper, authors investigated the LI-ideals of ADL concepts. For the advancement of many valued logical algebra we extend this notion to ADL concept. Firstly, authors investigated H-implicative almost distributive lattice; LI-ideals in IADL are discussed. Authors also found
that set theoretic LI-ideals is an LI-ideal and some characterizations of LI-ideals in IADL.

All results are soundness. The proofs are correct. So I think this paper can be accepted subjected to a minor revised. Some references regarding the ideals and filters in residued lattices should be added.

Is the work clearly and accurately presented and does it cite the current literature?

Partly
Is the study design appropriate and is the work technically sound?

Yes
Are sufficient details of methods and analysis provided to allow replication by others?

No
If applicable, is the statistical analysis and its interpretation appropriate?

I cannot comment. A qualified statistician is required.
Are all the source data underlying the results available to ensure full reproducibility?

No source data required
Are the conclusions drawn adequately supported by the results?

Yes

Competing Interests

No competing interests were disclosed.

Reviewer Expertise

logical algebras

We confirm that we have read this submission and believe that we have an appropriate level of expertise to confirm that it is of an acceptable scientific standard.

Respond to this report

Responses (0)

[1] 1. Assaye B, Alamneh M, Mekonnen T: Implicative Almost Distributive Lattice. Int. J. Comput. Sci. Appl. Math. 2018; 4(1): 19–22.

[2] 2. Assaye B, Alamneh M, Mekonnen T: Positive implicative and associative filters of implicative almost distributive lattice. Bull. Inter. Math. Virtual Inst. 2019; 9(1): 111–119.

[3] 3. Jun YB: Implicative filters of lattice implication algebras. Bull. Korean Math. Soc. 1997; 34(2): 193–198.

[4] 4. Jun YB, Roh EH, Xu Y: LI-ideals in lattice implication algebras. Bull. Korean Math. Soc. 1998; 35(1): 13–24.

[5] 5. Kolluru V, Bekele B: Implicative algebras. Momona Ethiop. J. Sci. 2012; 4(1): 90–101. Publisher Full Text

[6] 6. Swamy UM, Rao GC: Almost Distributive Lattices. J. Aust. Math. Soc. (Series A). 1981; 31(1): 77–91. Publisher Full Text

[7] 7. Xu Y, Qin KY: Lattice H-implication algebras and lattice implication algebra classes. J. Hebin Mining Civil Eng. Institute. 1992; 2: 139–143.

[8] 8. Xu Y: Lattice implication algebras. J. South West Jiaotong University. 1993; 28(1): 20–27.

[9] 9. Xu Y, Ruan D, Qin KY, et al.: Lattice-Valued Logic: An Alternative Approach to Treat Fuzziness and Incomparability. Berlin: Springer; 2003.