summary:We define an ultra $LI$-ideal of a lattice implication algebra and give equivalent conditions for an $LI$-ideal to be ultra. We show that every subset of a lattice implication algebra which has the finite additive property can be extended to an ultra $LI$-ideal

Jun, Young Bae

Qin, Keyun

Xu, Yang

English

Institute of Mathematics AS CR, v. v. i.

Czechoslovak Mathematical JournalKe Yun Qin; Yang Xu; Young Bae JunUltra LI-ideals in lattice implication algebrasCzechoslovak Mathematical Journal, Vol. 52 (2002), No. 3, 463–468Persistent URL: http://dml.cz/dmlcz/127735Terms of use:© Institute of Mathematics AS CR, 2002Institute of Mathematics of the Czech Academy of Sciences provides access to digitized documentsstrictly for personal use. Each copy of any part of this document must contain these Terms of use.This document has been digitized, optimized for electronic delivery andstamped with digital signature within the project DML-CZ: The Czech DigitalMathematics Library http://dml.czCzechoslovak Mathematical Journal, 52 (127) (2002), 463–468ULTRA LI-IDEALS IN LATTICE IMPLICATION ALGEBRASKeyun Qin, Yang Xu, Sichuan, and Young Bae Jun, Chinju(Received April 24, 1999)Abstract. We define an ultra LI-ideal of a lattice implication algebra and give equivalentconditions for an LI-ideal to be ultra. We show that every subset of a lattice implicationalgebra which has the finite additive property can be extended to an ultra LI-ideal.Keywords: lattice implication algebra, (ultra) LI-ideal, finite additive propertyMSC 2000 : 03G10, 06B10, 54E15IntroductionIn order to research a logical system whose propositional value is given in a lattice,Y. Xu [5] proposed the concept of lattice implication algebras, and discussed someof their properties. Also, in [4], Y. Xu discussed the homomorphisms between latticeimplication algebras. Y. Xu and K.Y. Qin [6] introduced the notion of filters in alattice implication algebra, and investigated their properties. In [1], Y. B. Jun et al.proposed the concept of an LI-ideal of a lattice implication algebra and discussedthe relationship between filters and LI-ideals, and studied how to generate an LI-ideal by a set. This paper is devoted to the discussion of ultra LI-ideals of latticeimplication algebras. We give equivalent conditions for an LI-ideal to be ultra. Weshow that every subset of a lattice implication algebra which has the finite additiveproperty can be extened to an ultra LI-ideal.463PreliminariesBy a lattice implication algebra we mean a bounded lattice (L, ∨, ∧, 0, 1) withorder-reversing involution “ ′ ” and a binary operation “→” satisfying the followingaxioms:(I1) x → (y → z) = y → (x → z),(I2) x → x = 1,(I3) x → y = y′ → x′,(I4) x → y = y → x = 1⇒ x = y,(I5) (x → y)→ y = (y → x)→ x,(L1) (x ∨ y)→ z = (x → z) ∧ (y → z),(L2) (x ∧ y)→ z = (x → z) ∨ (y → z)for all x, y, z ∈ L.In the sequel the binary operation “→” will be denoted by juxtaposition. We candefine a partial ordering “” on a lattice implication algebra L by x  y if and onlyif xy = 1.In a lattice implication algebra L, the following relations hold (see [5]):(1) 0x = 1, 1x = x and x1 = 1,(2) x′ = x0,(3) xy  (yz)(xz),(4) x ∨ y = (xy)y,(5) x  y implies yz  xz and zx  zy.In a lattice implication algebra L, if we denote (xy′)′ by x × y and x′y by x+ y,then the following relations are easily proved:(6) x+ y = y + x,(7) (x + y) + z = x+ (y + z),(8) x+ y  x ∨ y,(9) x× y = y × x,(10) (x × y)× z = x× (y × z),(11) x× y  x ∧ y.A subset A of a lattice implication algebra L is called an LI-ideal of L (see [1]) ifit satisfies(LI1) 0 ∈ A,(LI2) (xy)′ ∈ A and y ∈ A imply x ∈ A for all x, y ∈ L.An LI-ideal A of a lattice implication algebra L is said to be proper if A = L.Theorem 2.1. ([1, Theorem 2.2]) Let A be an LI-ideal of a lattice implicationalgebra L and let x ∈ A. If y  x, then y ∈ A for all y ∈ L.Let A be a subset of a lattice implication algebra L. Then the least LI-idealcontaining A is called the LI-ideal generated by A, denoted by 〈A〉.464The next statement gives a description of the elements of 〈A〉.Theorem 2.2. ([1, Theorem 2.9]) If A is a non-empty subset of a lattice impli-cation algebra L, then〈A〉 = {x ∈ L | a′n(. . . (a′1x′) . . .) = 1 for some a1, ..., an ∈ A}.Ultra LI-idealsWe start by providing a characterization of LI-ideals.Proposition 3.1. Let A be a subset of a lattice implication algebra L. Then Ais an LI-ideal of L if and only if the following implications hold:(i) x ∈ A and y  x imply y ∈ A,(ii) x ∈ A and y ∈ A imply x+ y ∈ A. . If A is an LI-ideal of L, then (i) holds by Theorem 2.1. Let x, y ∈ A.Then((x + y)y)′ = ((x′y)y)′ = (x′ ∨ y)′ = x ∧ y′  x.It follows from Theorem 2.1 that ((x + y)y)′ ∈ A and hence x + y ∈ A by (LI2).Conversely, let A be a subset of L satisfying the conditions (i) and (ii). Since 0  xfor all x ∈ L and hence for all x ∈ A, it follows from (i) that 0 ∈ A. Suppose(xy)′ ∈ A and y ∈ A. Then (xy)′ + y ∈ A by (ii), and(xy)′ + y = ((xy)′)′y = (xy)y = x ∨ y  x.Using (i), we have x ∈ A which proves (LI2), completing the proof. Theorem 3.2. If A is a subset of a lattice implication algebra L, then〈A〉 = {x ∈ L | x  a1 + a2 + . . .+ an for some a1, ..., an ∈ A}. . By Theorem 2.2 it is sufficient to show that(3.1) x  a1 + a2 + . . .+ an ⇐⇒ a′n(. . . (a′1x′) . . .) = 1.We will prove (3.1) by induction on n. If n = 1, thenx  a1 ⇐⇒ xa1 = 1⇐⇒ a′1x′ = 1;465hence (3.1) holds for n = 1. Suppose (3.1) is true for n = k, i.e.,x  a1 + a2 + . . .+ ak ⇐⇒ a′k(. . . (a′1x′) . . .) = 1.Thenx  a1 + a2 + . . .+ ak + ak+1 = ak+1 + a1 + a2 + . . .+ ak⇐⇒ x  a′k+1(a1 + a2 + . . .+ ak) = (a1 + a2 + . . .+ ak)′ak+1⇐⇒ (a1 + a2 + . . .+ ak)′  xak+1 = a′k+1x′⇐⇒ (a′k+1x′)′  a1 + a2 + . . .+ ak⇐⇒ a′k(a′k−1(. . . (a′1(a′k+1x′)) . . .)) = 1⇐⇒ a′k+1(a′k(. . . (a′2(a′1x′)) . . .)) = 1,which shows that (3.1) holds for n = k + 1. This completes the proof. Definition 3.3. A subset A of a lattice implication algebra L is said to have thefinite additive property if a1+ a2+ . . .+ an = 1 for any finite members a1, a2, . . . , anof A.The following corollary is an immediate consequence of Theorem 3.2.Corollary 3.4. For a subset A of a lattice implication algebra L, 〈A〉 is a properLI-ideal of L if and only if A has the finite additive property.Definition 3.5. An LI-ideal A of a lattice implication algebra L is said to beultra if for every x ∈ L, the following equivalence holds:(3.2) x ∈ A ⇐⇒ x′ ∈ A.Theorem 3.6. Let A be a subset of a lattice implication algebra L. Then A isan ultra LI-ideal of L if and only if A is a maximal proper LI-ideal of L. . Suppose that A is an ultra LI-ideal of L. Since 0 ∈ A, we have1 = 0′ ∈ A, and hence A is proper. If B is an LI-ideal of L and A   B, then thereexists x ∈ L such that x ∈ B and x ∈ A. By (3.2) we have x′ ∈ A   B, and so1 = x+x′ ∈ B. It follows that B = L and B is not proper. Therefore A is a maximalproper LI-ideal of L.Conversely, assume that A is a maximal proper LI-ideal of L. For each x ∈ L,we claim that (3.2) is true. Assume x′ ∈ A and let B = A ∪ {x}. Then B has thefinite additive property. In fact, suppose y1, . . . , yn ∈ B. If y1, . . . , yn ∈ A, theny1 + . . . + yn = 1 because A is proper. Now if there exists i  n such that yi = x,theny1 + . . .+ yn = x+ y1 + . . .+ yi−1 + yi+1 + . . .+ yn.466If y1 + . . . + yn = 1 then x′(y1 + . . . + yi−1 + yi+1 + . . . + yn) = 1, i.e., x′ y1+ . . .+ yi−1+ yi+1+ . . .+ yn. Thus x′ ∈ A by Theorem 2.1, a contradiction. Thisproves that B has the finite additive property. Using Corollary 3.4, 〈B〉 is a properLI-ideal of L. Since A ⊆ 〈B〉 and A is a maximal proper LI-ideal, we have 〈B〉 = Aand hence x ∈ 〈B〉 = A. Suppose x ∈ A. If x′ ∈ A, then 1 = x + x′ ∈ A; hence Ais not a proper LI-ideal. This is a contradiction. Therefore x′ ∈ A and the proof iscomplete. Theorem 3.7. Let A be a subset of a lattice implication algebra L. If A hasthe finite additive property, then there exists an ultra LI-ideal B of L containing A. . LetΩ = {B | B is a proper LI-ideal of L containing A}.Then 〈A〉 ∈ Ω and hence Ω = ∅. Suppose B1 ⊆ B2 ⊆ . . . is a chain of elements of Ωand let C =⋃iBi. Then (i) A ⊆ C, (ii) 1 ∈ C (because 1 /∈ Bi for all i), (iii) 0 ∈ C,and (iv) if (xy)′, y ∈ C then there exists i such that (xy)′, y ∈ Bi and so x ∈ Bi ⊆ C.This shows that C is a proper LI-ideal of L containing A so that C ∈ Ω. By Zorn’slemma, Ω has a maximal element, say D, which is the desired ultra LI-ideal of L. Since every proper LI-ideal has the finite additive property, we have the followingcorollary.Corollary 3.8. Every proper LI-ideal of a lattice implication algebra can beextended to an ultra LI-ideal.Theorem 3.9. Let A be a proper LI-ideal of a lattice implication algebra L.Then A is ultra if and only if for every a, b ∈ L, whenever a× b ∈ A then a ∈ A orb ∈ A. . Suppose A is ultra and let a, b ∈ L. If a × b ∈ A, then (a × b)′ /∈ A.Since (a × b)′ = ((ab′)′)′ = ab′ = a′ + b′, it follows that a′ /∈ A or b′ /∈ A so thata ∈ A or b ∈ A. Conversely, assume that for every a, b ∈ L, a ∈ A or b ∈ A whenevera× b ∈ A. Then for each x ∈ L, if x′ /∈ A then x′ × x = (x′x′)′ = 1′ = 0 ∈ A, whichimplies that x ∈ A. Clearly if x ∈ A, then x′ /∈ A. This completes the proof. Theorem 3.10. Let f : L → M be an implication homomorphism of latticeimplication algebras satisfying f(0) = 0.(i) If B is an ultra LI-ideal of M , then f−1(B) is an ultra LI-ideal of L.(ii) If f is an isomorphism and if A is an ultra LI-ideal of L, then f(A) is an ultraLI-ideal of M .467 . (i) Clearly 0 ∈ f−1(B). Let x, y ∈ L be such that (xy)′ ∈ f−1(B) andy ∈ f−1(B). Then f(y) ∈ B and (f(x)f(y))′ = (f(xy))′ = f((xy)′) ∈ B. Since B isan LI-ideal of M , it follows from (LI2) that f(x) ∈ B so that x ∈ f−1(B). Hencef−1(B) is an LI-ideal of L. For each x ∈ L, we havex ∈ f−1(B)⇐⇒ f(x) ∈ B ⇐⇒ f(x′) = (f(x))′ /∈ B ⇐⇒ x′ /∈ f−1(B).Hence f−1(B) is an ultra LI-ideal of L.(ii) Note that 0 = f(0) ∈ f(A). Let x, y ∈ M be such that (xy)′ ∈ f(A) andy ∈ f(A). Then there exist u ∈ L and v ∈ A such that f(u) = x and f(v) = y. Itfollows thatf((uv)′) = (f(uv))′ = (f(u)f(v))′ = (xy)′ ∈ f(A)so that (uv)′ ∈ A. Using v ∈ A, we know that u ∈ A and so x = f(u) ∈ f(A). Thusf(A) is an LI-ideal of M . For each y ∈ M , let x ∈ L be such that f(x) = y. Theny ∈ f(A)⇔ x = f−1(y) ∈ A ⇔ x′ /∈ A ⇔ y′ = (f(x))′ = f(x′) /∈ f(A).Therefore f(A) is an ultra LI-ideal of M . This completes the proof. AcknowledgementsThe authors would like to thank the referees for their valuable suggestions.References[1] Y.B. Jun, E.H. Roh and Y. Xu: LI-ideals in lattice implication algebras. Bull. KoreanMath. Soc. 35 (1998), 13–24.[2] Y.B. Jun and Y. Xu: Fuzzy LI-ideals in lattice implication algebras. J. Fuzzy Math. 7(1999), 997–1003.[3] Y.B. Jun, Y. Xu and K.Y. Qin: Positive implicative and associative filters of latticeimplication algebras. Bull. Korean Math. Soc. 35(1) (1998), 53–61.[4] Y. Xu: Homomorphisms in lattice implication algebras. Proc. of 5th Many-Valued Log-ical Congress of China (1992), 206–211.[5] Y. Xu: Lattice implication algebras. J. Southwest Jiaotong University 1 (1993), 20–27.[6] Y. Xu and K.Y. Qin: On filters of lattice implication algebras. J. Fuzzy Math. 1 (1993),251–260.Authors’ addresses:    and , Department of Applied Mathematics, South-west Jiaotong University, Chengdu, Sichuan 610031, P.R. China, e-mail: yxu@center2.swjtu.edu.cn;   	, Department of Mathematics Education, Gyeongsang NationalUniversity, Chinju 660-701, Korea, e-mail: ybjun@nongae.gsnu.ac.kr.468

Ultra $LI$-ideals in lattice implication algebras

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Ultra $LI$ -ideals in lattice implication algebras

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Ultra LILILI-ideals in lattice implication algebras

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Ultra $LI$ -ideals in lattice implication algebras