Orthocomplemented lattices with a symmetric difference Milan Matouˇsek 1 ODLs 2 Z2-valued states 3 The variety OML4 1 ODLs Definition 1.1 Let L = (X; ^; _;? ; 0; 1; 4), where (X; ^; _;? ; 0; 1) is an orthocomplemented lattice and 4 : X2 ! X is a binary operation. Then L is said to be an orthocomplemented difference lattice (abbr., an ODL) if the following formulas hold in L: (D1) x 4 (y 4 z) = (x 4 y) 4 z, ? ? (D2) x 4 1 = x , 1 4 x = x , (D3) x 4 y ≤ x _ y. ? (X; ^; _; ; 0; 1) will be denoted by Lsupp and called the support of L. Proposition 1.2 Let L = (X; ^; _;? ; 0; 1; 4) be an ODL. Then the following statements hold true: (1) x 4 0 = x, 0 4 x = x, (2) x 4 x = 0, (3) x 4 y = y 4 x, (4) x 4 y? = x? 4 y = (x 4 y)? , (5) x? 4 y? = x 4 y , (6) x 4 y = 0 , x = y , (7) (x ^ y?) _ (y ^ x?) ≤ x 4 y ≤ (x _ y) ^ (x ^ y)?. Theorem 1.3 Let L be an ODL. Then its support Lsupp is an OML. Proof. We will prove x; y 2 L; x ≤ y; y ^ x? = 0 ) x = y : x ≤ y ) (x^y?)_(y^x?) = y^x? = 0, (x_y)^(x^y)? = y ^ x? = 0. Prop. 1.2: x 4 y = 0 (by (7)) and therefore x = y (by (6)). Let L be an ODL. Then any OML notion can be referred to L as well by applying this notion to the corresponding OML Lsupp. Proposition 1.4 Let L be an ODL, a; b 2 L. (i) a C b ) a 4 b = (a ^ b?) _ (b ^ a?) = (a _ b) ^ (a ^ b)?; (A corollary: For each block B of L, the operation 4 on L acts on B as the standard symmetric difference.) (ii) a ? b (i.e. a ≤ b?) , a 4 b = a _ b. A NATURAL QUESTION: Which OMLs are embed- dable into ODLs ? A MORE GENERAL QUESTION: Let us consider the variety OML4 of OMLs generated by the class fLsupp; L is an ODLg, i.e. OML4 = HSP (fLsupp; L is an ODLg). Which OMLs belong to the variety OML4 ? The main aim of this note is to show that the class OML4 does not contain all OMLs. ? Proof of Prop. 1.2. (D2): 1 4 1 = 1 = 0. (1) x40 = x4(141) = (x41)41 = x?41 = (x?)? = x; the identity 0 4 x = x has an analogous proof. (2) Let us first show that x? 4 x? = x 4 x. We con- secutively obtain x? 4 x? = (x 4 1) 4 (1 4 x) = (x 4 (1 4 1)) 4 x = (x 4 0) 4 x = x 4 x. Moreover, we have x 4 x ≤ x as well as x 4 x = x? 4 x? ≤ x?. This implies that x 4 x ≤ x ^ x? = 0. (3) (x 4 y) 4 (x 4 y) = 0 ) (x 4 y 4 x 4 y) 4 y = 0 4 y ) x 4 y 4 x = y ) (x 4 y 4 x) 4 x = y 4 x ) x 4 y = y 4 x. (4) x 4 y? = x 4 (y 4 1) = (x 4 y) 4 1 = (x 4 y)?. (5) Applying (4), we obtain x? 4 y? = (x? 4 y)? = (x 4 y)?? = x 4 y. (6) If x = y, then x 4 y = 0 by the condition (2). Con- versely, suppose that x 4 y = 0. Then x = x 4 0 = x 4 (y 4 y) = (x 4 y) 4 y = 0 4 y = y. (7) The property (D3) together with the properties (4), (5) imply that x 4 y ≤ x _ y, x 4 y ≤ x? _ y? = (x ^ y)?, x ^ y? ≤ x 4 y, x? ^ y ≤ x 4 y. Proof of Prop. 1.4. (i) It follows from Prop. 1.2. (ii) ()) We use Prop. 1.2. (() x 4 y = x _ y ) y ≤ ? ? ? ? x 4 y ) x 4 y ≤ y . But also y ≤ y . (D3): (x4y?)4y? ≤ y?. Finally, (x4y?)4y? = x4(y?4y?) = x 4 0 = x. 2 Z2-valued states Z2 ≡ the group f0; 1g with the modulo 2 addition ⊕ Definition 2.1 Let L be an OML. (i) Let s : L ! Z2 be a mapping. Then s is said to be a Z2-valued state (abbr., a Z2-state) provided s(1L) = 1 and s(x _ y) = s(x) ⊕ s(y) whenever x; y 2 L, x ≤ y?. (ii) L is called Z2-full if 8x; y 2 L; x 6= y; x 6= 0L; y 6= 1L 9 Z2-state s on L: s(x) = 1 & s(y) = 0. Definition 2.2 Let L be an ODL and let I ⊆ L. I is a 4-ideal if 0L 2 I & 8a; b 2 I : a 4 b 2 I; I is a proper 4-ideal if I is a 4-ideal and 1L 62 I; I is a maximal 4-ideal if I is proper 4-ideal and for any proper 4-ideal J with I ⊆ J we have I = J. Lemma 2.3 Suppose that L is an ODL and I is a proper 4-ideal in L. Suppose that x 2 L and neither x nor x? belongs to I. Let us write J = I [ fa 4 x; a 2 Ig. Then J is also a proper 4-ideal in L and, moreover, x 2 J and x? 62 J. Proposition 2.4 Let L be an ODL and let I be a proper 4-ideal in L. Then I is a maximal 4-ideal in L iff card(fx; x?g \ I) = 1 for any x 2 L. Proposition 2.5 (Stone's lemma) Let L be an ODL, let I0 be a proper 4-ideal in L and let b 2 L, b 62 I0. Then there is a maximal 4-ideal I such that I0 ⊆ I and b 62 I. Proposition 2.6 Let L be an ODL and I be a maximal 4-ideal in L. Let us define a mapping s : L ! Z2 as follows: s(a) = 0 (resp., s(a) = 1) if a 2 I (resp., a 62 I). Then s(x 4 y) = s(x) ⊕ s(y) for any x; y 2 L. A consequence: The mapping s is a Z2-state on Lsupp. Theorem 2.7 Let L be an ODL. Then its support Lsupp is Z2-full. Proof of Lemma 2.3. The set J is obviously a 4-ideal. Let us see that 1L 62 J. Suppose on the contrary that 1L 2 J. Then 1L = a 4 x for some element a 2 I. The ? ? equality 1L = a 4 x implies that a = x . But x does not belong to I which is a contradiction. Thus, 1L 62 J. ? ? Further, x = 0L 4 x 2 J. If x 2 J, then 1L = x 4 x 2 J { a contradiction again. Proof of Prop. 2.4. ()) Suppose that I is maximal and x 2 L. Suppose further that x 62 I and, also, x? 62 I. Then (Lemma 2.3) there is a 4-ideal J such that I ⊆ J and I 6= J. As a result, at least one element of the set fx; x?g belongs to I. Looking for a contradiction, sup- pose that fx; x?g ⊆ I. Then x 4 x? = 1 which means that 1 2 I { a contradiction (I is supposed proper). (() Let us suppose that card(fx; x?g \ I) = 1 for any x 2 L. Suppose further that I ⊂ J for a proper 4-ideal J. J is strictly larger than I ) 9b 2 L : b 2 J; b 62 I. card(fb; b?g \ I) = 1 and b 62 I ) b? 2 I. Now, both the elements b and b? belong to J and there- ? fore 1L = b 4 b 2 J. This means that J is not proper. Proof of Prop. 2.5. Write I = fJ ⊆ L; J is a proper 4-ideal, I0 ⊆ J and b 62 Jg. Then I0 2 I and therefore I 6= ;. Zorn's lemma: the set I ordered by inclusion contains a maximal element, I. (1) I is a proper 4-ideal, I0 ⊆ I and b 62 I. (2) b? 2 I (otherwise the 4-ideal I0 = I [ fc 4 b?; c 2 Ig extends I, Lemma 2.3, and I0 belongs to the system I). (3) I is a maximal 4-ideal (Suppose therefore that I ⊂ J for a proper 4-ideal J. Thus, J is strictly larger than I and therefore J 62 I. Therefore b 2 J and since b? 2 I ⊆ ? J, we see that 1L = b 4 b 2 J. This means that J is not proper.) Proof of Prop. 2.6. Let us consider x; y 2 L. We are to prove the equality s(x4y) = s(x)⊕s(y). We will argue by cases. For example, let us suppose that x 2 I and y 62 I. Then x 4 y 62 I (indeed, should x 4 y be an element of I, then y = x 4 (x 4 y) 2 I which is a contradiction). Hence s(x 4 y) = 1 = 0 ⊕ 1 = s(x) ⊕ s(y). It remains to show that s is a Z2-state on Lsupp. Of course, s(1) = 1. Let us take x; y 2 L with x ? y. Prop.