370 Y. B. Jun generate an LI-ideal by both an LI-ideal and an element. We dene a prime LI-ideal, and give an equivalent condition for a proper LI-idea

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1 J. Korean Math. Soc. 36 (1999), No. 2, pp. 369{380 ON LI-IDEALS AND PRIME LI-IDEALS OF LATTICE IMPLICATION ALGEBRAS Young Bae Jun Abstract. As a continuation of the paper [3], in this paper we investigate the further properties on LI-ideals, and show that how to generate an LI-ideal by both an LI-ideal and an element. We dene a prime LI-ideal, and give an equivalent condition for a proper LIideal to be prime. Using this result, we establish the extension property and prime LI-ideal theorem. 1. Introduction In order to research the logical system whose propositional value is given in a lattice, Y. Xu [6] proposed the concept of lattice implication algebras, and discussed their some properties. Also, in [7], Y. Xu and K. Y. Qin discussed the properties of lattice H implication algebras, and gave some equivalent conditions about lattice H implication algebras. Y. Xu and K. Y. Qin [8] introduced the notion of lters in a lattice implication algebra, and investigated their properties. In [3], Y. B. Jun et al. dened an LI-ideal of a lattice implication algebra and showed that every LI-ideal is a lattice ideal. They gave an example that a lattice ideal may not be an LI-ideal, and showed that every lattice ideal is an LI-ideal in a lattice H implication algebra. They discussed the relationship between lters and LI-ideals, and studied how to generate an LI-ideal by a set. Moreover they constructed the quotient structure by using an LI-ideal, and studied the properties of LI-ideals related to implication homomorphisms. This paper is a continuation of the paper [3]. In this paper we investigate the further properties on LI-ideals, and show that how to Received October 1, Mathematics Subject Classication: 03G10, 06B10, 54E15. Key words and phrases: lattice (H) implication algebra, LI-ideal, lattice ideal, prime LI-ideal.

2 370 Y. B. Jun generate an LI-ideal by both an LI-ideal and an element. We dene a prime LI-ideal, and give an equivalent condition for a proper LI-ideal to be prime. Using this result, we establish the extension property and prime LI-ideal theorem. 2. Preliminaries By a lattice implication algebra we mean a bounded lattice (L; _; ^; 0; 1) with order-reversing involution \ 0 " and a binary operation \!" satisfying the following axioms: (I1) x! (y! z) =y!(x!z); (I2) x! x =1; (I3) x! y = y 0! x 0 ; (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 2 L. Note that the conditions (L1) and (L2) are equivalent to the conditions (L3) x! (y ^ z) =(x!y)^(x!z);and (L4) x! (y _ z) =(x!y)_(x!z);respectively. A lattice implication algebra L is called a lattice H implication algebra if it satises x _ y _ ((x ^ y)! z) =1for all x; y; z 2 L. In the sequel the binary operation \! " will be denoted by juxtaposition. We can dene a partial ordering \ " on a lattice implication algebra L by x y if and only if xy =1. In a lattice implication algebra L, the following hold (see [6]): (1) 0x =1,1x=xand x1 =1. (2) x 0 = x0. (3) xy (yz)(xz): (4) x _ y =(xy)y. (5) ((yx)y 0 ) 0 = x ^ y =((xy)x 0 ) 0 : (6) x y implies yz xz and zx zy. (7) x (xy)y. In a lattice H implication algebra L, the following hold (see [7]): (8) x(xy) =xy.

3 (9) x(yz) =(xy)(xz). On LI-ideals and prime LI-ideals LI-ideals We begin with the following proposition. Proposition 3.1. In a lattice implication algebra the following identity holds: (10) ((xy)y)y = xy: Proof. The inequality xy ((xy)y)y follows from (7). Now using (3), (I1) and (I2), we have (((xy)y)y)(xy) x((xy)y) =(xy)(xy)=1 and hence (((xy)y)y)(xy) = 1, i.e., ((xy)y)y xy. This completes the proof. Definition 3.2. (Y. B. Jun et al. [3]) Let L be a lattice implication algebra. An LI-ideal A of L is a non-empty subset of L such that (LI1) 0 2 A, (LI2) (xy) 0 2 A and y 2 A imply x 2 A, for all x; y 2 L. Under this denition f0g and L are the trivial examples of LI-ideals. The following example shows that there is a proper LI-ideal in a lattice implication algebra. Example 3.3. (Y. B. Jun et al. [3, Example 2.1]) Let L := f0; a; b; c; d; 1g be a set with Figure 1asapartial ordering. Dene a unary operation \ 0 " and a binary operation denoted by juxtaposition on L as follows (Tables 1 and 2, respectively):

4 372 Y. B. Jun x x 0 0 a b c d a d b c a b c d 1 c d a b 0 a b c d 1 c 1 b c b 1 d a 1 b a 1 a a 1 1 a 1 b 1 1 b a b c d 1 Figure 1 Table 1 Dene _- and ^-operations on L as follows: Table 2 x _ y := (xy)y; x ^ y := ((x 0 y 0 )y 0 ) 0 ; for all x; y 2 L. Then L is a lattice implication algebra. It is easy to check that A := f0;cg is an LI-ideal of L. Definition 3.4. (S. Burris et al. [1, Denition 8.2]) Let L be a lattice. An ideal I of L is a non-empty subset of L such that (LI3) x 2 I, y 2 L and y x imply that y 2 I, (LI4) x; y 2 I implies x _ y 2 I. Throughout this paper we call this a lattice ideal. Note that in a lattice H implication algebra, an LI-ideal and a lattice ideal coincide (see [3, Theorems 2.4 and 2.6]), also observe that a nonempty subset A of a lattice H implication algebra L is an LI-ideal of L if and only if for every x; y 2 L, (LI5) x; y 2 A, x _ y 2 A: Proposition 3.5. Let L be a lattice H implication algebra and a 2 L. Then there is no proper LI-ideal of L containing a and a 0 simultaneously. Proof. Let A be a proper LI-ideal of L containing a and a 0 simultaneously. Then 1=a_a 0 2A,and hence A = L a contradiction.

5 On LI-ideals and prime LI-ideals 373 For any subsets A and B of a lattice H implication algebra L we set A ^ B = fa ^ bja 2 A and b 2 Bg: Proposition 3.6. If A and B are LI-ideals of a lattice H implication algebra L, then so is A ^ B. Proof. Let x; y 2 A ^ B. Then x = a1 ^ b1 and y = a2 ^ b2 for some a1;a2 2 A and b1;b2 2 B. Since A is an LI-ideal, it follows from (LI5) that a1 _ a2 2 A. Now x _ y =(a1^b1)_(a2^b2)=((a1^b1)_a2)^((a1 ^ b1) _ b2): Noticing that (a1 ^ b1) _ a2 =(a1_a2)^(b1_a2)a1_a2;wehave (a1 ^ b1) _ a2 2 A by (LI3). Similarly we know that (a1 ^ b1) _ b2 2 B. Hence x _ y =((a1^b1)_a2)^((a1 ^ b1) _ b2) 2 A ^ B: Conversely assume that x_y 2 A^B for all x; y 2 L. Then x_y = a^b for some a 2 A and b 2 B. Observe that x = x ^ (x _ y) =x^(a^b)a so from (LI3) that x 2 A. Similarly, one can show that x 2 B and that y 2 A and y 2 B. Thus x; y 2 A ^ B, ending the proof. Proposition 3.7. If A and B are LI-ideals of a lattice H implication algebra L, then A ^ B = A \ B. Proof. Let x 2 A ^ B. Then x = a ^ b for some a 2 A and b 2 B. Observe that x = a ^ b a (also b). It follows from (LI3) that x 2 A (also B) and hence x 2 A\B. Conversely if x 2 A\B then x = x^x 2 A ^ B. Therefore A ^ B = A \ B. Proposition 3.8. If A is an LI-ideal of a lattice H implication algebra L and a 2 L, then the set K := fx 2 Ljx 0 a 2 Ag is an LI-ideal of L.

6 374 Y. B. Jun Proof. Let x; y 2 K. Then x 0 a 2 A and y 0 a 2 A. It follows from (L2) that (x _ y) 0 a =(x 0^y 0 )a=(x 0 a)_(y 0 a)2aso that x _ y 2 K. Conversely let x; y 2 L be such that x _ y 2 K. Then (x _ y) 0 a 2 A. Using (I3), (4) and (9), we have (x _ y) 0 a = a 0 (x _ y) =a 0 ((xy)y) =(a 0 (xy))(a 0 y) =((a 0 x)(a 0 y))(a 0 y)=(a 0 x)_(a 0 y)=(x 0 a)_(y 0 a): Thus (x 0 a)_(y 0 a) 2 A which implies that x 0 a 2 A and y 0 a 2 A because A is an LI-ideal. This means that x 2 K and y 2 K, completing the proof. For any natural number n we dene n(x)y recursively as follows: 1(x)y = xy and (n + 1)(x)y = x(n(x)y). Using (I1) repeatedly, we know that in a lattice implication algebra the following identity holds: (11) z(y1(y2((y n x)))) = y1(y2((y n (zx)) )): As a special case of (11) we have (12) z(n(y)x) =n(y)(zx): Proposition 3.9. For any elements a; x; y1;y2; ;y n of a lattice implication algebra L we have (13) y1(y2((y n (xa)) )) = ((y1(y2((y n (xa)) )))a)a: Proof. The inequality y1(y2((y n (xa)) )) ((y1(y2((y n (xa)) )))a)a follows from (7). Using (10), (11), (I1) and (I2), we get (((y1(y2((y n (xa)) )))a)a)(y1(y2((y n (xa)) ))) = y1(y2((y n (x((((y1(y2((y n (xa)) )))a)a)a))) )) = y1(y2((y n (x((y1(y2((y n (xa)) )))a))) )) =(y1(y2((y n (xa)) )))(y1(y2((y n (xa)) ))) =1; and hence ((y1(y2((y n (xa)) )))a)a y1(y2((y n (xa)) )): This completes the proof.

7 On LI-ideals and prime LI-ideals 375 Observation (Y. B. Jun et al. [3, Observation 2.8]) Suppose A is a non-empty family of LI-ideals of a lattice implication algebra L. Then A = \A is also an LI-ideal of L. Let A be a subset of a lattice implication algebra L. Then the least LI-ideal containing A is called the LI-ideal generated by A, written hai. Noticing that L is clearly an LI-ideal containing A, in view of Observation 3.10, we know that the above denition hai of A is well-dened. The next statement gives a description of elements of hai. Proposition (Y. B. Jun et al. [3, Theorem 2.9]) If A is a non-empty subset of a lattice implication algebra L, then hai = fx 2 Lja 0 n(:::(a 0 1 x0 ):::) =1for some a1; :::; a n 2 Ag: The following corollary is immediate from Proposition Corollary For any element a of a lattice implication algebra L, we have hai = fx 2 Ljn(a 0 )x 0 =1for some natural number ng: Note that in a lattice H implication algebra the identity x(xy) = xy holds (see [7, Corollary 1]). Hence using Corollary 3.12 we obtain Corollary Let L be a lattice H implication algebra and a 2 L. Then hai = fx 2 Lja 0 x 0 =1g=fx2Ljxa =1g: Lemma (Y. B. Jun et al. [3, Theorem 2.2]) Let A be an LIideal of a lattice implication algebra L and let x 2 A. If y x (or equivalently x 0 y 0 ), then y 2 A for all y 2 L. The following theorem describes how to generate an LI-ideal by both an LI-ideal and an element. Theorem Let A be an LI-ideal of a lattice implication algebra L and let a 2 L. Then ha [fagi = fx 2 Lj(n(a 0 )x 0 ) 0 2 A for some n 2 Ng:

8 376 Y. B. Jun Proof. Denote A a = fx 2 Lj(n(a 0 )x 0 ) 0 2 A for some n 2 Ng: Clearly 0 2 A a. Let (yx) 0 2 A a and x 2 A a. Then there exist m; n 2 N such that (n(a 0 )((yx) 0 ) 0 ) 0 2 A and (m(a 0 )x 0 ) 0 2 A. It follows that (n(a 0 )(yx)) 0 = u and (m(a 0 )x 0 ) 0 = v for some u; v 2 A, so that u 0 = n(a 0 )(yx) and v 0 = m(a 0 )x 0. Then 1=u 0 (n(a 0 )(yx)) = u 0 (n(a 0 )(x 0 y 0 )) = u 0 (x 0 (n(a 0 )y 0 )) = x 0 (u 0 (n(a 0 )y 0 )); i.e., x 0 u 0 (n(a 0 )y 0 ): Using (6) we have v 0 = m(a 0 )x 0 m(a 0 )(u 0 (n(a 0 )y 0 )) = u 0 (m(a 0 )(n(a 0 )y 0 )) = u 0 ((m + n)(a 0 )y 0 ); which implies that v 0 (u 0 (((m+n)(a 0 )y 0 ) 0 ) 0 )=v 0 (u 0 ((m+n)(a 0 )y 0 )) = 1: Since u; v 2 A, it follows from Proposition 3.11 that ((m + n)(a 0 )y 0 ) 0 2 hai = A. Hence y 2 A a and A a is an LI-ideal of L. Note that (n(a 0 )a 0 ) 0 =1 0 =02Aso that a 2 A a. Let x 2 A. Since x 0 a 0 x 0 = ((a 0 x 0 ) 0 ) 0 ; it follows from Lemma 3.14 that (a 0 x 0 ) 0 2 A, i.e., x 2 A a. Thus A [faga a. Finally we should verify that A a is the least LIideal containing A and a. Let B be any LI-ideal containing A and a, and let x 2 A a. Then (n(a 0 )x 0 ) 0 2 A B for some n 2 N, and hence (((n, 1)(a 0 )x 0 ) 0 a) 0 =(a 0 ((n, 1)(a 0 )x 0 )) 0 =(n(a 0 )x 0 ) 0 2B: Since a 2 B, it follows from (LI2) that ((n, 1)(a 0 )x 0 ) 0 2 B. Repeating this process, we obtain x = (x 0 ) 0 2 B. Therefore A a is the least LIideal containing A and a, i.e., ha [fagi = A a. This completes the proof. Lemma (J. Liu et al. [4, Corollary 1]) Let L be a lattice implication algebra and a; b; x 2 L. If n(a)x = 1 and m(b)x = 1 for some m; n 2 N, then there exists k 2 N such that k(a _ b)x =1. Theorem Let A be an LI-ideal of a lattice implication algebra L. Then for all a; b 2 L. ha [fagi \ ha [fbgi = ha [fa^bgi

9 On LI-ideals and prime LI-ideals 377 Proof. Let x 2 ha [fagi \ ha [fbgi. By Theorem 3.15, there are m; n 2 N such that (n(a 0 )x 0 ) 0 2 A and (m(b 0 )x 0 ) 0 2 A. Hence (n(a 0 )x 0 ) 0 = u and (m(b 0 )x 0 ) 0 = v for some u; v 2 A. It follows that n(a 0 )x 0 = u 0 and m(b 0 )x 0 = v 0 so that 1 = v 0 (u 0 (n(a 0 )x 0 )) = n(a 0 )(v 0 (u 0 x 0 )) and 1 = u 0 (v 0 (m(b 0 )x 0 )) = m(b 0 )(v 0 (u 0 x 0 )): Using Lemma 3.16, there exists k 2 N such that k(a 0 _ b 0 )(v 0 (u 0 x 0 )) = 1. Since a 0 _ b 0 =(a^b) 0 ;wehave 1=k(a 0 _b 0 )(v 0 (u 0 x 0 )) = k((a ^ b) 0 )(v 0 (u 0 x 0 )) = v 0 (u 0 (k((a ^ b) 0 )x 0 )): Applying Proposition 3.11 we get (k((a ^ b) 0 )x 0 ) 0 2hAi=Aand hence x 2hA[fa^bgi by Theorem Thus ha [fagi \ ha [fbgi ha [fa^bgi: Conversely if x 2 ha [fa^bgi; then (n((a ^ b) 0 )x 0 ) 0 2 A for some n 2 N. Since a ^ b a; b, therefore a 0 (a ^ b) 0 and b 0 (a ^ b) 0. Using (6) repeatedly, we get n((a ^ b) 0 )x 0 n(a 0 )x 0 and n((a ^ b) 0 )x 0 n(b 0 )x 0 ; which imply that (n(a 0 )x 0 ) 0 (n((a ^ b) 0 )x 0 ) 0 and (n(b 0 )x 0 ) 0 (n((a ^ b) 0 )x 0 ) 0 : Applying Lemma 3.14, we obtain (n(a 0 )x 0 ) 0 2 A and (n(b 0 )x 0 ) 0 2 A, i.e., x 2 ha [fagi and x 2 ha [fbgi. Hence x 2 ha [fagi\ha[fbgi and so ha [fa^bgi ha [fagi\ha[fagi: This completes the proof. Corollary Let A be an LI-ideal of a lattice implication algebra L. For any a1;a2; ;a n 2 L we have n\ ha [fa i gi = ha [fa1^a2^^a n gi: i=1 Proof. Straightforward by induction on n. Corollary Let A be an LI-ideal of a lattice implication algebra L. If a1 ^ a2 ^^a n 2 A for all a1;a2; ;a n 2 L, then n\ n\ ha [fa i gi = A: i=1 Proof. Using Corollary 3.18, we have ha [fa i gi = ha [fa1^a2^^a n gi = hai = A: i=1

10 378 Y. B. Jun 4. Prime LI-ideals Definition 4.1. A proper LI-ideal P of a lattice implication algebra L is said to be prime if whenever x ^ y 2 P then x 2 P or y 2 P for all x; y 2 L. We provide an equivalent condition for a proper LI-ideal to be prime. Theorem 4.2. Let P be a proper LI-ideal of a lattice implication algebra L. Then P is prime if and only if (xy) 0 2 P or (yx) 0 2 P for all x; y 2 L. Proof. Assume that P is a prime LI-ideal of L and let x; y 2 L. Then 1=(x_y)(x _ y) =((x_y)x)_((x _ y)y) =(1^(yx)) _ ((xy) ^ 1) = (yx) _ (xy); and so (yx) 0 ^ (xy) 0 = ((yx) _ (xy)) 0 = 1 0 = 0 2 P: Since P is prime, it follows that (xy) 0 2 P or (yx) 0 2 P. Conversely let P be a proper LI-ideal of L and suppose (xy) 0 2 P or (yx) 0 2 P for all x; y 2 L. Let x; y 2 L be such that x ^ y 2 P. If (yx) 0 2 P, since (y(yx) 0 ) 0 =((yx)y 0 ) 0 =((x 0 y 0 )y 0 ) 0 =(x 0 _y 0 ) 0 =x^y wehave y 2 P by (LI2). Similarly x 2 P whenever (xy) 0 the proof. 2 P; ending Using Theorem 4.2 we have the extension property for prime LIideal. The proof is straightforward and is omitted. Theorem 4.3. (Extension property for prime LI-ideal) Let P be a prime LI-ideal of a lattice implication algebra L. Then every proper LI-ideal containing P is also prime. Theorem 4.4. (Prime LI-ideal theorem) Let A be an LI-ideal of a lattice implication algebra L and S a ^-closed subset of L (i.e., x^y 2 S whenever x; y 2 S) such that A \ S = ;. Then there exists a prime LI-ideal P of L such that A P and P \ S = ;.

11 On LI-ideals and prime LI-ideals 379 Proof. The existence of an LI-ideal P being the maximal element of the family of all LI-ideals that contains A and have empty intersection with S follows from an application of Zorn's lemma. We now show that P is prime. Suppose P is not prime. Then there exist x; y 2 L n P such that x ^ y 2 P. By means of the maximality of P, we know that S \hp [fxgi 6= ; and S \hp [fygi 6= ;. Taking u and v in S \hp [fxgi and S \hp [fygi, respectively, then u ^ v 2 S because S is ^-closed. Since u ^ v u; v, it follows that u ^ v 2hP[fxgi and u ^ v 2hP[fygi. Using Theorem 3.17, one can know that u ^ v 2hP[fxgi\hP [fygi = hp [fx^ygi = P: Hence u ^ v 2 S \ P, which is a contradiction. Therefore P is prime, ending the proof. Corollary 4.5. Let A be an LI-ideal of a lattice implication algebra L. If x 2 L n A, then there is a prime LI-ideal P of L such that A P and x =2 P. Proof. Let S = fy 2 Ljxy = 1g. If y1;y2 2 S, then xy1 = 1 and xy2 = 1. It follows that x(y1 ^ y2) = (xy1) ^ (xy2) = 1 so that y1 ^ y2 2 S; i.e., S is ^-closed. Let y 2 S. Then xy = 1 and hence (xy) 0 = 1 0 = 0 2 A. Since x =2 A, we have y =2 A by (LI2). Hence A \ S = ;. Using Theorem 4.4, there is a prime LI-ideal P of L such that A P and P \ S = ;. Since x 2 S, the identity P \ S = ; implies x =2 P. This completes the proof. Theorem 4.6. For a lattice implication algebra L the following are equivalent: (i) LI-ideal f0g is prime. (ii) every proper LI-ideal of L is prime. (iii) (L; ) is a totally ordered set. Proof. (i) ) (ii) is by Theorem 4.3, and (ii) ) (i) is obvious. Assume that (L; ) isatotally ordered set. Then xy =1or yx =1,and hence (xy) 0 =1 0 =02f0gor (yx) 0 =1 0 =02f0gfor all x; y 2 L. It follows from Theorem 4.2 that f0g is a prime LI-ideal of L. Conversely if f0g is a prime LI-ideal of L, then (xy) 0 2f0gor (yx) 0 2f0gfor all x; y 2 L, that is, (xy) 0 = 0 or (yx) 0 = 0; hence xy = 1 or yx = 1

12 380 Y. B. Jun for all x; y 2 L. This shows that (L; ) is a totally ordered set. This completes the proof. References [1] S. Burris and H. P. Sankappanavar, A course in universal algebra, Springer- Verlag, New York (1981). [2] Y. B. Jun, Implicative lters of lattice implication algebras, Bull. Korean Math. Soc. 34(2) (1997), [3] Y. B. Jun, E. H. Roh and Y. Xu, LI-ideals in lattice implication algebras, Bull. Korean Math. Soc. 35(1) (1998), [4] J. Liu and Y. Xu, Filters and structure of lattice implication algebras, Chinese Science Bulletin 42(18) (1997), [5] Y. Xu, Homomorphisms in lattice implication algebras, Proc. of 5th Many- Valued Logical Congress of China (1992), [6], Lattice implication algebras, J. Southwest Jiaotong University 1 (1993), [7] Y. Xu and K. Y. Qin, Lattice H implication algebras and lattice implication algebra classes, J. Hebei Mining and Civil Engineering Institute 3 (1992), [8], On lters of lattice implication algebras, J. Fuzzy Math. 1 (1993), Department of Mathematics Education Gyeongsang National University Chinju , Korea ybjun@nongae.gsnu.ac.kr