Powered by TCPDF (www.tcpdf.org) Bol. Soc. Mat. Mexicana (3) Vol. 12, 2006 ON THE CHARACTERIZATION OF A SUBVARIETY OF SEMI-DE MORGAN ALGEBRAS CANDIDA´ PALMA AND RAQUEL SANTOS Abstract. In this note we characterize by a new set of axioms the largest subvariety of semi-De Morgan algebras with the congruence extension prop- erty. 1. Introduction The equational class of semi-De Morgan algebras was introduced by Sankap- panavar in [8]. It consists of bounded distributive lattices with an additional unary operation and it contains the variety of pseudocomplemented distribu- tive lattices and K1,1, one of the subvarieties of Ockham algebras which in- cludes De Morgan algebras. In [4] Hobby developed a duality for semi-De Morgan algebras which he used to find the largest subvariety of semi-De Morgan algebras with the con- gruence extension property. This variety, which Hobby denoted by C, contains both K1,1 and the equational class of demi-pseudocomplemented lattices, a generalization of pseudocomplemented lattices studied by Sankappanavar in [9] and [10]. The equations defining principal congruences as well as the subdirectly irreducibles of the variety C were determined by us in [6]; however, the two inequalities (α and β) that characterize this subvariety of semi-De Morgan are rather complicated. In fact Problem 2 in [4] is to find “nicer axioms for C”. We solved this problem algebraically determining a new inequality (γ) such that C can be characterized by γ and β. 2. Preliminaries We start by recalling some definitions and essential results from [8]. Definition (2.1). An algebra L = L, ∨, ∧,0 , 0, 1 is a semi-De Morgan algebra if the following five conditions hold a, b ∈ L : (S1) (L, ∨, ∧, 0, 1) is a distributive lattice with 0, 1. (S2) 00 ≈ 1 and 10 ≈ 0. 0 (S3) a ∨ b ≈ a0 ∧ b0. 00 (S4) a ∧ b ≈ a00 ∧ b00. (S5) a000 ≈ a0. 2000 Mathematics Subject Classification: 06D99, 06D15, 06D30. Keywords and phrases: semi-De Morgan algebras, congruence extension property. This work was done within the project POCTI-ISFL-1-143 of Centro de Algebra´ da Universidade de Lisboa and it was supported by FCT and by FEDER. 149 150 CANDIDA´ PALMA AND RAQUEL SANTOS This equational class of algebras will be denoted by SDMA. The following rules hold in SDMA and some of them are proved in [8] : 0 0 0 (S6) a ∧ b ≈ a00 ∧ b00 ≈ a ∧ b00 . 0 00 (S7) a ∧ b ≈ a0 ∨ b0 . 00 0 (S8) a ∧ b ≈ a0 ∨ b0 . (S9) a ≤ b implies b0 ≤ a0. 0 (S10) a ∧ a ∧ b ≥ a ∧ b0. 00 0 00 00 (S11) a ∨ b ≈ a0 ∧ b0 ≈ a00 ∨ b00 ≈ a ∨ b00 . 3. The variety C D. Hobby determined in [4] the largest subvariety of SDMA with the con- gruence extension property. He characterized this variety, which he denoted by C, by the following inequalities: 0 0 0 (α) a0 ∨ b0 ≥ a ∧ b ∧ (a ∧ c)0 ∧ b ∧ c ∧ b ∧ c0 0 0 (β) a0 ∨ a0 ∧ b ∧ b0 ≥ a ∧ b . It is possible to obtain simpler inequalities characterizing C. The search for these inequalities requires some rather nasty calculations so we consider several lemmas before we can reach our goal. With this aim we will consider first the following identities: 0 0 0 0 0 0 0 00 (α1) a ∨ b = a ∨ b ∨ a ∧ b ∧ (a ∧ c) ∧ b ∧ c ∧ b ∧ c 0 0 00 0 0 00 (β1) a ∨ a ∧ b ∧ b = a ∧ b ∨ a ∧ b ∧ b . These identities are equivalent to α and β, respectively (note that a ≥ a ∧ b 0 implies a0 ≤ a ∧ b ). Now we can prove the following. Lemma (3.1). Let L ∈ C and a, d ∈ L. Then the identity α1 implies 0 0 00 0 00 (α2)(a ∨ d ) ∧ d = (a ∧ d) ∧ d . 0 Proof. By (S3), a0 ∧ d00 ∨ d0 = a ∨ d0 ∨ d0. Replacing b by a ∨ d0, a by d 0 and c by d in the identity α1 and using commutativity, we obtain (a ∨ d0)0 ∨ d0 = (a∨d0)0 ∨ d0 ∨ ((a ∨ d0) ∧ d)0 ∧ (d ∧ d0)0 ∧ ((a ∨ d0) ∧ d0)0 ∧ ((a ∨ d0) ∧ d00)0 0 0 0 = a ∨ d0 ∨ d0 ∨ a ∨ d0 ∧ d ∧ d ∧ d0 ∧ d00 0 0 because a ∨ d0 ∧ d00 = a ∨ d0 ∧ d by (S6). CHARACTERIZATION OF A SUBVARIETY OF SEMI-DE MORGAN ALGEBRAS 151 0 0 But d ∧ d0 ≥ a ∨ d0 ∧ d since d ∧ d0 ≤ a ∨ d0 ∧ d; hence it follows from the previous equation that 0 0 0 a ∨ d0 ∨ d0 = a ∨ d0 ∨ d0 ∨ a ∨ d0 ∧ d ∧ d00 0 0 = a ∨ d0 ∨ d0 ∨ a ∨ d0 ∧ d ∨ d0 (by S3) 0 0 = a ∨ d0 ∨ d0 ∨ a ∨ d0 ∧ d ∨ d0 (by distributivity) 0 0 = d0 ∨ (a ∨ d0) ∧ d ∨ d0 (because (a ∨ d0)0 ≤a ∨ d0 ∧ d ∨ d0 0 = d0 ∨ a ∧ d ∨ d0 (by distributivity) 0 = d0 ∨ a ∧ d ∧ d00 . Therefore we have (a0 ∧ d00) ∨ d0 = ((a ∧ d)0 ∧ d00) ∨ d0. Now, by distributivity, we obtain (a0 ∨ d0) ∧ (d00 ∨ d0) = ((a ∧ d)0 ∨ d0) ∧ (d00 ∨ d0). Since (a ∧ d)0 ≥ d0, it follows that (a0 ∨ d0) ∧ (d00 ∨ d0) = (a ∧ d)0 ∧ (d00 ∨ d0) 00 and meeting the two members with d , we have α2. Lemma (3.2). Let L ∈ SDMA and let a, b, c ∈ L. Then 0 0 00 0 00 (α2)(a ∨ b ) ∧ b = (a ∧ b) ∧ b and 0 0 0 0 0 0 0 0 (β1) a ∨ (a ∧ b ∧ b ) = (a ∧ b) ∨ (a ∧ b ∧ b ) imply 0 (δ) a0 ∧ (b ∧ (c ∨ c0))0 ∨ b0 ∧ (a ∧ c)0 = (a ∧ b)0 ∧ (a ∧ c)0 ∧ (b ∧ (c ∨ c0) . Proof. Let us denote by A and B, respectively, the left and right sides of the identity δ. We are going to prove that β1 and α2 imply A = B using the distributivity of L. First we will verify that the joins of A and B with (a0 ∧ c0 ∧ c00)0 are equal: A ∨ (a0 ∧ c0 ∧ c00)0 = a0 ∧ (b ∧ (c ∨ c0))0 ∨ b0 ∨ (a0 ∧ c0 ∧ c00)0 ∧ (a ∧ c)0 ∨ (a0 ∧ c0 ∧ c00)0 (by distributivity) = a0 ∧ (b ∧ (c ∨ c0))0 ∨ b0 ∨ (a0 ∧ c0 ∧ c00)0 ∧ a0 ∨ (a0 ∧ c0 ∧ c00)0 (by β1 and S6) = a0 ∧ (b ∧ (c ∨ c0))0 ∨ b0 ∧ a0 ∨ (a0 ∧ c0 ∧ c00)0 (by distributivity) = a0 ∧ (b ∧ (c ∨ c0))0 ∨ (a0 ∧ c0 ∧ c00)0 because, by S9, (b ∧ (c ∨ c0))0 ≥ b0 and thus a0 ∧ (b ∧ (c ∨ c0))0 ≥ a0 ∧ b0). B ∨ (a0 ∧ c0 ∧ c00)0 = (a ∧ b)0 ∧ (b ∧ (c ∨ c0))0 ∨ (a0 ∧ c0 ∧ c00)0 ∧ (a ∧ c)0 ∨ (a0 ∧ c0 ∧ c00)0 (by distributivity) 152 CANDIDA´ PALMA AND RAQUEL SANTOS = (a ∧ b)0 ∧ (b ∧ (c ∨ c0))0 ∨ (a0 ∧ c0 ∧ c00)0 ∧ a0 ∨ (a0 ∧ c0 ∧ c00)0 (by β1 and S6) = (a ∧ b)0 ∧ (b ∧ (c ∨ c0))0 ∧ a0 ∨ (a0 ∧ c0 ∧ c00)0 (by distributivity) = a0 ∧ (b ∧ (c ∨ c0))0 ∨ (a0 ∧ c0 ∧ c00)0 because, by S9, (a ∧ b)0 ≥ a0. Thus we have proved that A ∨ (a0 ∧ c0 ∧ c00)0 = B ∨ (a0 ∧ c0 ∧ c00)0. Now we are going to see that the same is true with the meets. We will have to use identity α2, so we must note that by S3, S11, S9 and S5, (a0 ∧ c0 ∧ c00)0 = (a ∨ c ∨ c0)00 = (a ∨ c00 ∨ c0)00 ≥ (a0 ∧ c0 ∧ c00)00 = (a ∨ c ∨ c0)0. Therefore, denoting by d the expression a ∨ c ∨ c0, we will have (a0 ∧ c0 ∧ c00)0 = d00 ≥ d0 and thus A ∧ (a0 ∧ c0 ∧ c00)0 = A ∧ d00 = a0 ∧ (b ∧ (c ∨ c0))0 ∨ (b0 ∧ (a ∧ c)0) ∧ d00 0 = a ∨ (b ∧ (c ∨ c0)) ∨ (b0 ∧ (a ∧ c)0) ∧ d00 (by S3) 0 = (a ∨ b) ∧ (a ∨ c ∨ c0) ∧ d00 ∨ (b0 ∧ (a ∧ c)0) ∧ d00 (by distributivity) 0 = (a ∨ b) ∧ d ∧ d00 ∨ (b0 ∧ (a ∧ c)0) ∧ d00 (by the definition of d ,) 0 0 00 0 0 00 = (a ∨ b) ∨ d ∧ d ∨ (b ∧ (a ∧ c) ) ∧ d (by α2) = (a0 ∧ b0) ∨ d0 ∧ d00 ∨ (b0 ∧ (a ∧ c)0) ∧ d00 (by S3) = (a0 ∧ b0) ∨ d0 ∨ (b0 ∧ (a ∧ c)0 ∧ d00 (by distributivity) = d0 ∨ (b0 ∧ (a ∧ c)0 ∧ d00 (because (a ∧ c)0 ≥ a0) = b0 ∧ (a ∧ c)0 ∧ d00 ∨ (d0 ∧ d00) (by distributivity) = b0 ∧ (a ∧ c)0 ∧ d00 ∨ d0 (because d00 ≥ d0). By a similar process, B ∧ (a0 ∧ c0 ∧ c00)0 = B ∧ d00 = (a ∧ b)0 ∧ (b ∧ (c ∨ c0))0 ∧ d00 ∧ (a ∧ c)0 (by commutativity) = ((a ∧ b) ∨ (b ∧ (c ∨ c0)))0 ∧ d00 ∧ (a ∧ c)0 (by S3) = (b ∧ (a ∨ c ∨ c0))0 ∧ d00 ∧ (a ∧ c)0 (by distributivity) = (b ∧ d)0 ∧ d00 ∧ (a ∧ c)0 (by the definition of d ) 0 0 00 0 = (b ∨ d ) ∧ d ∧ (a ∧ c) (applying α2) CHARACTERIZATION OF A SUBVARIETY OF SEMI-DE MORGAN ALGEBRAS 153 = ((b0 ∧ d00) ∨ (d0 ∧ d00)) ∧ (a ∧ c)0) (by distributivity) = ((b0 ∧ d00) ∨ d0) ∧ (a ∧ c)0) (because d00 ≥ d0 ) = (b0 ∧ d00 ∧ (a ∧ c)0) ∨ (d0 ∧ (a ∧ c)0) (by distributivity ) = (b0 ∧ d00 ∧ (a ∧ c)0) ∨ d0 (because, by S9, d0 = (a ∨ c ∨ c0)0 ≤ (a ∧ c)0). Therefore we have proved that A ∧ (a0 ∧ c0 ∧ c00)0 = B ∧ (a0 ∧ c0 ∧ c00)0. 0 0 00 0 0 0 00 0 By the characterization of θlatL((a ∧ c ∧ c ) , (a ∧ c ∧ c ) ) we conclude that A = B. Lemma (3.3). Let L ∈ SDMA and let a, b, c ∈ L. Then the identity 0 (δ) a0 ∧ (b ∧ (c ∨ c0))0 ∨ b0 ∧ (a ∧ c)0 = (a ∧ b)0 ∧ (a ∧ c)0 ∧ (b ∧ (c ∨ c0) . is equivalent to α. Proof. First note that α is equivalent to (a0 ∨b0) ∧(a ∧b)0 ∧(a ∧c)0 ∧(b ∧c)0 ∧(b ∧c0)0 = (a ∧b)0 ∧(a ∧c)0 ∧(b ∧c)0 ∧(b ∧c0)0 and, by distributivity, this identity is equivalent to a0 ∧ (a ∧ b)0 ∧ (a ∧ c)0 ∧ (b ∧ c)0 ∧ (b ∧ c0)0 ∨ b0 ∧ (a ∧ b)0 ∧ (a ∧ c)0 ∧ (b ∧ c)0 ∧ (b ∧ c0)0 = = (a ∧ b)0 ∧ (a ∧ c)0 ∧ (b ∧ c)0 ∧ (b ∧ c0)0.