We consider the emerging problem of comparing the similarity between (unlabeled) pedigrees. More specifically, we focus on the simplest pedigrees, namely, the 2-generation pedigrees. We show that the isomorphism testing for two 2-generation pedigrees is GI-hard. If the 2-generation pedigrees are monogamous  then the isomorphism testing problem can be solved in polynomial time. We then consider the problem by relaxing it into an NP-complete decomposition problem which can be formulated as the Minimum Common Integer Pair Partition (MCIPP) problem, which we show to be FPT by exploiting a property of the optimal solution. While there is still some difficulty to overcome, this lays down a solid foundation for this research. Pedigrees, or commonly known as family trees, are important tools in evolutionary and computational biology. They are important for geneticists, as with a valid pedigree the recombination events can be deduced more accurately , or diseThere have been many practical methods for reconstructing pedigrees ,4,5,17. It is known that a lot of computations on pedigree graphs are NP-hard ,20,16, set al. [On the other hand, methods for comparing pedigrees are rare. The brute-force method will not work when the data set has size in the thousands ,14. Peopet al. , where tet al. [In this paper, we follow the work by Kirkpatrick et al.  to consiP = (I(P), E(P)) with vertices I(P) and edges E(P), together with a gender function s : I(P) → {male, female} such that:An (unlabeled) pedigree is a directed graph P is acyclic.1 v ∈ I(P), the in-degree of v is either two or zero.2 For all nodes a, c),  ∈ E(P), we have s(a) ≠ s(b).3 For two edges  nodes to represent males . See Figure u ∈ I(P) is monogamous if it mates with exactly one partner, i.e., the number of individuals u', u' ≠ u, such that ,  ∈ E(P) for some x ∈ I(P) is exactly one. A pedigree is monogamous if all the individuals are monogamous. In Figure P = (I(P), E(P)) is generational if there is a function Let g(v) = 1 for all v ∈ I(P) with in-degree zero.1 u, v) ∈ E(P), we have g(v) = g(u) + 1.2 For all (g(v) is called the generation of v. For a generational pedigree P, we use g I(P) to represent the individuals of P whose generation is g. The pedigree on Figure The number P = (I(P), E(P)), P' = (I(P'),E(P')) with the associated gender functions s(−), s'(−) respectively, a bijection ϕ: I(P) → I(P') is a pedigree isomorphism between P and P' if:Given two pedigrees u ∈ I(P), s(u) = s'(ϕ(u)), and1 For every u, v) ∈ E(P) if and only if (ϕ(u), ϕ(v)) ∈ E(P').2 (Graph Isomorphism (GI) is one of the most famous problems in computational complexity whose precise complexity has been open since 1972 ,12. It iIn , it was Theorem 1 Testing the isomorphism between two 2-generation pedigrees is GI-hard.Proof. We reduce bipartite graph isomorphism problem to our problem. Let B1 = , B2 =  be two bipartite graphs (with no isolated nodes). For our construction, we perform the following:U1,U2 are marked male.1 All nodes in V1,V2 are marked female.2 All nodes in B1 is converted into a pedigree P1 = (I(P1), E(P1)) as follows: (3.1) I(P1) = U1 ∪ V1 are the generation-1 nodes; (3.2) E(P1) is initially set as empty; (3.3) for  ∈ E1 we create a new generation-2 node uv such that s(uv) = female, E(P1) ← E(P1) ∪ {, }.3 B2 is converted into a pedigree P2 identically as in step (3).4 B1 and B2 are isomorphic iff P1 and P2 , both 2-generational, are isomorphic. We only show the necessary direction here as the other one is easy. If P1 and P2 are isomorphic, the first property we make use of is that all the generation-2 nodes are female. So, in the isomorphism between P1 and P2, if a generation-2 node uv ∈ I(P1) is mapped to a generation-2 node xy ∈ I(P2), we can simultaneously contract uv, xy to their corresponding male parents in P1 and P2. Consequently, we obtain the isomorphism between B1 and B2.   □We claim that Note that in our construction, all generation-2 individuals are female; moreover, a pair of generation-1 individuals mate with exactly one female child. A simple example on this reduction is shown on Figure Although the isomorphism testing problem is GI-hard even for 2-generation pedigrees, in some situations the problem is not hard to solve. In fact, when both of the 2-generation pedigrees are monogamous then the problem can be solved in linear time.i females and j males, we say that these two parents and the i + j children form an 〈i, j〉-family. In Figure When a pair of generation-1 couple mate to have generation-2 children, for instance Theorem 2 Testing the isomorphism between two 2-generation monogamous pedigrees is polynomial time solvable.Proof. It is easily seen that when a 2-generation pedigree Q1 is monogamous then it is composed of a set of disjoint 〈i, j〉-families. So to test the isomorphism between two monogamous 2-generation pedigrees Q1,Q2 it suffices to check whether two sets of integral pairs are identical, which can be done in O(n log n) time using the standard optimal sorting algorithms in two passes similar to the radix sort. In the first pass, we sort all the pairs according to their first components, and in the second, for each contiguous list of pairs with the same first component, we sort them according to the second components.   □The hardness result in the previous section implies that it might be too much if we use the standard isomorphism to measure the similarity of 2-generation pedigrees. In practice, ambiguities exist in pedigree-related datasets. In fact, it is estimated that 2-10% of people do not know their biological father . For 2-gP, it is not difficult to identify all (not necessarily disjoint) 〈i, j〉-families .  = {n1, n2,..., tn} such that n = 9, {1, 2, 2, 4} is a partition of n. It should be noted that while it is simple to partition an integer, the number of such partitions is usually (counter-intuitively) huge. For instance, the integer 10 has 190569292 distinct partitions [Throughout this paper, for MCIP, we focus on integers in X = {x1, x2,...,px} is a multiset union of all the partitions τ(ix), i.e., ∪i≤p1≤τ(ix). A multiset Z is a common partition of two multisets X = {x1, x2,..., px}, Y = {y1, y2,..., qy} if there are partitions τ1, τ2 with ∪i≤p1≤τ1(ix) = ∪j≤q1≤τ2(jy) = Z. The size of the partition Z is denoted as |Z|. For example, given X = {5, 8}, Y = {3,10}, a common partition of X, Y is Z = {1, 2, 2, 4, 4}, and the size of this partition is 5. It is easily seen that the necessary condition for X and Y to admit a common partition is that the sums of the integers in X and Y are equal. Throughout this paper, whenever we talk about a common partition for sets of integers X and Y, we always assume that this condition is met.A partition of a multiset A and B, and an integer k.Instance: Two multiple sets of integers A, B admit a common partition of size k?Question: Does For the ease of presentation, we use MCIP to represent this instance.a, b〉, the projection S be a set of 2-tuples of integers, Given a 2-tuple of integers, 〈S, T, a common partition of S and T is a set of 2-tuples H = {〈g1, h1〉, 〈g2, h2〉, ⋯, 〈kg, kh〉} such that k is the size of the partition H. Again, it is easily seen that the necessary condition for S and T to admit a common partition is that the sums of the integers in S, T, we always assume that this condition is met.Given two sets of 2-tuples S and T, and an integer k.Instance: Two multiple sets of 2-tuples of integers S, T admit a common partition of size k?Question: Does i, j〉 represents the pedigree of a couple which has i female and j male chilren. Again, we use MCIPP to represent this instance. As MCIPP is a generalization for MCIP, all the known negative results regarding MCIP hold for MCIPP; i.e., MCIP and MCIPP are both NP-complete and APX- hard, following [d-MCIP has also been considered, where the input is d multisets with the same sum. Efficient asymptotic approximation algorithms have been obtained for large d [d + O(1) [d = 2 in this paper.) Also, note that the integer 0 in a solution for MCIP is meaningless while it is possible that 0 can appear either in the input or in the solution for MCIPP. So for MCIPP, we focus on integers in Recall that a 2-tuple 〈ollowing . cn) = O*(f(k)), where f (−) is any computable function on k and c is a constant. FPT algorithms are efficient tools for handling some NP-complete problems, especially when k is small in practical datasets [Finally, a Fixed-Parameter Tractable (FPT) algorithm is an algorithm for a decision problem with input size datasets ,11.a, c, we say a dominates c if a >c. Given a pair of 2-tuples of integers 〈a, b〉 and 〈c, d〉, we say 〈a, b〉 dominates 〈c, d〉 if a ≥ c and b ≥ d. To simplify the writing, we say that 〈a, b〉 and 〈c, d〉 form a dominating pair if either 〈a, b〉 dominates 〈c, d〉 or vice versa. Likewise, 〈a, b〉 and 〈c, d〉 form a non-dominating pair if either a >c, b <d or a <c, b >d.Given a pair of integers We first describe some optimality properties for both the optimization versions of MCIP and MCIPP. When the context is clear, we still use MCIP and MCIPP to denote the corresponding optimization versions of the instances.Lemma 1 Let A, B be the input for MCIP. In any feasible solution, if a partition for some × ∈ A, τ (x) = {x1, x2,...,px}, and a partition for some y ∈ B, tau(y) = {y1, y2,...,qy}, satisfies that |τ(x) ∩ τ(y)| > 1 then this solution for MCIP is not optimal.Proof. Suppose to the contrary that |τ(x) ∩ τ(y)| > 1, and the corresponding partition for A, B is optimal. WLOG, suppose τ(x) = {x1, x2,..., px} and τ(y) = {y1, y2,..., qy} contain r common elements {z1, z2,..., rz} then we can update τ(x) ← τ(x) − {z1, z2,..., rz}∪{z1 + z2 + ... + rz} and τ(y) ← τ(y) − {z1, z2,..., rz}∪{z1 + z2 + ... + rz}. Then the solution size for MCIP on A, B is reduced by r − 1, contradicting the optimality of the assumption.   □x ∈ A and some y ∈ B, they share at most one common element. Notice that this lemma also holds for MCIPP, i.e., in an optimal partition of 〈s1, s2〉 ∈ S and 〈t1, t2〉 ∈ T, τ and τ share at most one common 2-tuple. Similarly, we can assume that in the input for MCIP  ) there is no common pair of integers in A and B (resp. no common pair of 2-tuples in S and T), as it must be put in the optimal solution.With the above lemma, we can now assume that for any optimal partition for some The following property is trivial and holds for both MCIP and MCIPP.Lemma 2 Let |MCIPP*| be the optimal solution size for MCIPP. Then |MCIPP*| >max{|S|, |T|}.a, b〉 and 〈c, d〉, we can use subtraction to partition them into two common pairs. For example, if a ≤ c and b ≤ d, then we can obtain the common partition {〈a, b〉, 〈c − a, d − b〉. So, given 〈2, 4〉 and 〈4, 5〉 we can obtain a partition {〈2,4〉, 〈2,1〉} for 〈4, 5〉. We also say that this is a dominating - partition operation. Apparently, for MCIP, this gives a way to partition a pair of integers as well. For instance, given 2 and 6, we can subtract 2 from 6 to obtain a partition {2, 4} for 6.For a pair of dominating 2-tuples 〈a, b has the property that either a dominates b or vice versa. This is not the case for a non-dominating pairs of 2-tuples, e.g. 〈1, 4〉 and 〈2, 3〉. We start with this fundamental lemma.We next describe some properties on non-dominating pairs of 2-tuples which are unique for MCIPP -- for MCIP, a pair of integers Lemma 3 Let A', B' be a set of positive integers with the same total sum, moreover, let us suppose |A'| = m + 1, |B'| = m. Then there must exist elements a ∈ A', b ∈ B' such that a <b.Proof. As |A'| > |B'|, we can arbitrarily select m elements from A' and match up them with those in B in an one-to-one fashion. As the sum of integers in A and B are the same, in this matching at least one of elements a ∈ A must be smaller than its matched counterpart b ∈ B -- otherwise, the sum of integers in A would be larger than that of B'.   □Corollary 1 Let A, B be two sets of n > 1 positive integers with the same total sum. WLOG, let A = {a1, a2, ⋯, na}, B = {b1, b2, ⋯, nb}. Then there must exist an element b ∈ B' = B − {jb} which is greater than some element a ∈ A' = {a1, a2, ⋯, ai−1, i a− jb, ai+1, ⋯, na}, i where a>jb.Proof. Obviously we have |A'| = n and |B'| = n − 1. Then this corollary follows directly from Lemma 3.   □A, B is obvious -- we can successively find pairs of dominating integers. In fact, in the proof of Corollary 1, once we obtain a' = k a∈ A' and b' = ℓ b∈ B' such that a' <b', we can repeatedly use the above argument to A'' = A' − {a'} = {a1, a2, ⋯, ak−1, ak+1, ⋯, na} and B'' = {b1, b2, ⋯, bℓ−1, bℓ − a', bℓ+1, ⋯, bn}, where |A''| = |B''| = n − 1 and the two sets A, B have the same sum.The implication of Corollary 1 for MCIP with input S, T} is each composed of m non-dominating 2-tuples, then we can find a pair s = 〈s1, s2〉 ∈ S and t = 〈t1, t2〉 ∈ T (assuming s1 >t1 and s2 <t2) such that we can put 〈t1, s2〉 in some solution set while the resulting instance S' =  ∪ {〈s1 − t1, 0〉},T' =  ∪ {〈0, t2 − s2〉} is still a valid instance for MCIPP.  Then, following Corollary 1, there exists a pair of dominating 2-tuples s' ∈ S', t' ∈ T'. Moreover, if we apply the dominating-partition process on these two tuples s', t', following Corollary 1, we can repeatedly find dominating tuples until all tuples in S, T are all commonly partitioned. This is because after we apply the dominating-partition on s', t' (say s' <t') to obtain an MCIPP instance S'',T'', we have Now let us see how this can be applied to MCIPP. When we have an instance of MCIPP whose input {Algorithm Heuristic-MCIPPS, TInput: τ for S, T, initially empty.Output: A common partition S| ≥ 2 and |T| ≥ 21   While |2      Repeats ∈ S and t ∈ T,2.1         select a pair of dominating 2-tuples, 2.2         compute two decomposing 2-tuples by subtraction,S ← S − {s}, T ← (T − {t}) ∪{t − s} if s <t,2.3         update S ← (S − {s}) ∪{s − t}, T ← T − {t} if s >t,2.4         update τ ←τ ∪ {min},2.5         update 3      Until no dominating 2-tuples can be found.S and T4      If there are at least two pairs of non-dominating 2-tuples in 5      Thens' ∈ S, t' ∈ T which leads to successive dominating pairs.5.1         use a brute-force method to select two non-dominating 2-tuples S| = 1, |T| = 1, then find the smaller tuple in S and T, x.6   If |τ ← τ ∪ {x}.7   Return s ∈ S, t ∈ T which can make the process of repeatedly processing dominating pairs possible. Let us show an example, S = {〈9, 4〉, 〈1, 11〉, 〈6, 3〉} and T = {〈2, 8〉, 〈12, 1〉, 〈2, 9〉}. In this example, among the 9 non-dominating pairs between S and T, there are 4 solutions enabling us to successively find dominating pairs. One of them is s = 〈6, 3〉 and t = 〈2, 9〉, which gives us a common partition of size 6. The other 5 solutions all lead to a common partition of size 7.Of course, due to the 'existence' constraint in Lemma 3 and Corollary 1, we would have to use a brute-force method to find a pair The above discussion enables us to design an algorithm Heuristic-MCIPP to prove the next lemma.Lemma 4 Let |MCIPP| be the size of the solution returned by Heuristic-MCIPP. Then |MCIPP| ≤ |S| + |T|.Proof. When there is no non-dominating pairs in the input, with the running of the algorithm Heuristic-MCIPP, we have |MCIPP| ≤ |S| + |T| − 1. The reason is that when each of S and T has at least two 2-tuples, we can use the dominating-partition procedure to obtain two 2-tuples in the solution set for each pair of dominating 2-tuples from S, T. When there are a total of three elements in S, T, say, one in S and two in T, we just need to return the two elements in T as their sum matches the one in S already. (This is certainly true for MCIP as pointed out in [d out in .)p ≥ 2 pairs of non-dominating pairs at Step 4-5 of the Heuristic-MCIPP algorithm, following Corollary 1 and the subsequent arguments, there exists a non-dominating pair s = 〈s1, s2〉 ∈ S, t = 〈t1, t2〉 ∈ T which leads to successive dominating pairs. (We can use the brute-force method to find this in O(p2(|S| + |T|)) time.) In this case, the solution obtained by Heuristic-MCIPP has size at most 1 + (|S| + |T| − 1) = |S| + |T|, where the first one corresponds to  (if s1 <t1) or  (if s1 >t1).   □When there are S| + |T| ≤ 2max{|S|, |T|} ≤ 2|MCIPP*|. On the other hand, designing approximation algorithms is not our focus for this paper; in fact, by a simple modification for the Maximum Packing method in [In fact, the above three lemmas imply that Heuristic-MCIPP provides a factor-2 approximation for MCIPP, as we have |ethod in  we can oMCIP| < |A| + |B| − 1. The latter in fact immediately implies that for MCIP there is always an optimal solution which does not partition at least an integer from either A or B. That further implies that there is a simple FPT algorithm for MCIP based on the bounded-degree search. We will show a stronger property in the next section to improve the FPT algorithm for MCIP, and subsequently, an FPT algorithm for MCIPP can be obtained.Note that Lemma 4 is different from its counterpart for MCIP, which, according to Lemma 2.2 in , states We first give the following lemma for MCIP.Lemma 5 Let A, B be the input for MCIP and let a be the smallest element in A or B. Then there is an optimal solution for MCIP which contains a, i.e., there is an optimal solution which does not partition the smallest element in A and B.Proof. We first show the following claim: in an optimal solution τ for MCIP with input A, B, let a1, a2 be a pair of elements in τ(z), z ∈ A ∪ B, with the condition that (1) a1 + a2 <a, and (2) a2 is the minimum among all pairs of elements in τ satisfying the condition (1), then there is an optimal solution τ' which partitions some element z ∈ A ∪ B with τ'(z) = τ(z) − {a1, a2} ∪ {a1 + a2}.z ∈ A and let a1 ∈ τ(y1) and a2 ∈ τ(y2) for two distinct integers y1, y2 ∈ B. Following the definition of 〈a1, a2〉, τ(y1) contains at least one more element other than a1, say a3; and following (2) we have a3 >a2. Suppose that a3 ∈ τ(x) for some x ∈ A. By Lemma 1, x ≠ z. We replace a3 by a1 by τ' with τ'(x) = τ(x) − {a3} ∪ {a2}, τ'(y1) = τ(y1) − {a1, a3} ∪ {τ'(z) = τ(z) − {a1, a2} ∪ {τ' has the same size as τ, so it is also a minimum size common partition for A, B.The proof for the above claim is as follows. WLOG, let a is partitioned in some optimal partition τ with τ' with t − 1 such steps, we obtain an optimal partition which contains the minimum element a in A ∪ B.   □It is obvious that, as long as the smallest element A = {2, 5, 5}, B = {6, 6}, and an optimal partition τ = {1,1, 5, 5} where τ(2) = {a1 = 1, a2 = 1}. By the construction in the proof of Lemma 5, y1 = 6, y2 = 6, a3 = 5, τ' = {1, 2, 4, 5}, where a = 2 is kept.An example of the above proof is given as follows. We have et al. [MCIP| < |A| + |B| − 1, there must be an optimal solution whose corresponding matching graph between the partitioned elements in A, B contains no cycle, which means there is at least one leaf node. Then this leaf node corresponds to an unpartitioned integer in A or B. The above lemma in fact implies a faster FPT algorithm for MCIP. Pick the smallest element a ∈ A ∪ B (say a ∈ A), we try to partition some other integer z ∈ B by subtracting a from it. Then we repeat over the new problem instance involving z − a. This process is repeated k times when either a solution is founded or we have to report that there is no solution of size k. The running time is O*k) = O*(kk).We comment that we can use Lemma 2.2 by Chen et al.  directlyTo obtain an FPT algorithm for MCIPP, we also need a similar lemma.Lemma 6 Let S, T be the input for MCIPP. Then there is an optimal solution for MCIPP which either contains 〈a, b〉 ∈ S ∪ T or 〈c, d〉 ∈ S ∪ T, or contains 〈a, d〉, where a is the minimum element in and d is the minimum element in Proof. Again, we first show the following claim: in an optimal solution τ for MCIPP with input S, T, let 〈a1, a2〉, 〈b1, b2〉 be two 2-tuples in τ(z), z ∈ S ∪ T, such that (1) a1 + b1 ≤ a, and (2) b1 is the minimum among all pairs of 2-tuples in τ satisfying (1), then there is an optimal solution τ' which partitions some 2-tuple z ∈ S ∪ T with τ'(z) = τ(z) − {〈a1, a2〉, 〈b1, b2〉} ∪ { a1 + b1, a2}. z = 〈z1, z2〉 ∈ S and let 〈a1, a2〉 ∈ τ(y1) and 〈b1, b2〉 ∈ τ(y2) for two distinct 2-tuples y1, y2 ∈ T. Following the definition of , tau(y1) contains at least one more pair 〈c1, c2〉, with c1 ≥ b1. Suppose that 〈c1, c2〉 ∈ τ(x) for some x ∈ S. Again, by Lemma 1, x ≠ z. We replace 〈c1, c2〉 by 〈c1 − b1, c2〉, and 〈a1, a2〉 by 〈a1 + b1, a2〉. Subsequently, we obtain another optimal partition τ' with τ'(x) = τ(x) − {〈c1, c2〉}∪{〈c1 − b1, c2〉, 〈a2, b2〉}, τ'(y1) = τ(y1) − {〈a1, a2〉, 〈c1, c2〉 ∪ {〈a1 + b1, a2〉}, 〈c1 − b1, c2〉, and τ'(z) = τ(z) − {〈a1, a2〉, 〈b1, b2〉 ∪ {〈a1 + b1, a2〉}. Again, τ' is also a minimum size common partition for S, T.WLOG, let Similar to Lemma 5, it is obvious that we can repeatedly apply the above steps to obtain an optimal solution with does not partition the smallest element in a, b〉, 〈c, d〉 ∈ S ∪ T or 〈a, d〉 ∈ S ∪ T, where a is the minimum element in d is the minimum element in a, b〉, 〈c, d〉 or 〈a, d〉. For one step, the running time for the former would be O(k1 + k2) for the first two cases and for the latter would also be O(k1 + k2) -- as 〈a, d〉 could be subtracted from O(k1 + k2) pairs, where k1 = |S|, k2 = |T|. As k1, k2 ≤ k, the running time of this step is bounded by O(2k). Running this for k steps, the running time of the whole algorithm is O*(2kkk). Hence, we have the following theorem.With the above lemma, it is again possible to have an FPT algorithm, Exact-MCIPP, for MCIPP using bounded degree search. At each step, we search for 〈Exact-MCIPPAlgorithm S, T, kInput: τ for S, T, initially empty.Output: A common partition k ≥ 11   While 2      Repeata be the minimum element in 2.1         let d be the minimum element in 2.2         let a, d〉 ∈ S ∪ T then τ ← τ ∪ {〈a, d〉}, delete 〈a, d〉 from S ∪ T, and update S, T and k ← k − 1,2.3         if 〈a, b〉 ∈ S ∪ T then τ ← τ ∪ {〈a, b〉}, delete 〈a, b〉 from S ∪ T, and update S, T and k ← k − 1,2.4         if 〈c, d〉 ∈ S ∪ T then τ ← τ ∪ {〈c, d〉}, delete 〈c, d〉 from S ∪ T, and update S, T and k ← k − 1,2.5         if 〈S = ∅ or T = ∅ or k = 0.3      Until S = ∅ and T = ∅4   If both τ,4.1      Then return 4.2      Else return 'no solution'.Theorem 3 Minimum Common Integer Pair Partition is FPT.k is large. In [k is relatively small.The running time of the above FPT algorithm is still too high to be applied alone to the similarity comparison for arbitrary 2-generation pedigrees, i.e., when arge. In , the salk values, we suggest a combination of the FPT algorithm and approximation algorithms [k is not too large, we can run this FPT algorithm; when k is too large for the FPT algorithm to handle, we can then use the approximation algorithms.  problem, which generalizes the NP-complete problem of Minimum Common Integer Partition (MCIP) problem. We show that MCIPP (hence MCIP) is FPT (Fixed-Parameter Tractable). It would be interesting to significantly improve the running time of the FPT algorithms presented in this paper.We consider the problem of testing the isomorphism and similarity of the simplest possible unlabeled pedigrees. We show that the isomorphism testing is GI-hard, excluding any chance for a polynomial time algorithm . We define a new similarity measure based on 〈The authors declare that they have no competing interests.BZ conceived the study. All authors contributed to the algorithm design and analysis, read and approved the final manuscript.


Discriminating groups were introduced by G. Baumslag, A. Myasnikov, and V. Remeslennikov as an outgrowth of their theory of algebraic geometry over groups. Algebraic geometry over groups became the main method of attack on the solution of the celebrated Tarski conjectures. In this paper we explore the notion of discrimination in a general universal algebra context. As an application we provide a different proof of a theorem of Malcev on axiomatic classes of Ω-algebras. A complduced in . In 5]   discrimΩ-algebras .Ω is an ordered triple  where FΩ and CΩ are disjoint sets (possibly empty) and dΩ is a function from FΩ into the set of positive integers. FΩ is the set of function symbols of Ω, CΩ is the set of constant symbols of Ω, and dΩ is the degree function or arity function of Ω. Given an operator domain Ω, an Ω-algebra 𝒜 is an ordered triple A is a nonempty set (the domain or carrier or universe of 𝒜);c ∈ CΩ, cA ∈ A;for each f ∈ FΩ with dΩ(f) = n, fA : An → A is an n-ary operation defined on A.for each An operator domain Ω-algebra whose carrier is a singleton is called trivial. All other Ω-algebras are nontrivial. If 𝒜 and ℬ are Ω-algebras, then 𝒜 is a subalgebra of ℬ provided thatA⊆B;c ∈ CΩ, cA = cB;for each f ∈ FΩ with dΩ(f) = n, fA = fBAn|.for each An 𝒜 and ℬ be Ω-algebras. A function ϕ : A → B is a homomorphism provided thatc ∈ CΩ, ϕ(cA) = cB;for each f ∈ FΩ with dΩ(f) = n and all  ∈ An, ϕ) = fB(ϕ(a1),…, ϕ(an)).for each Let 𝒜 be an Ω-algebra. An equivalence relation R on A is a congruence on 𝒜 provided for all f ∈ FΩ with dΩ(f) = n and all xiRyi for i = 1,…, n, one has Epimorphism, monomorphism, isomorphism, endomorphism, and automorphism are defined in the obvious way. Let R is a congruence on 𝒜, then the quotient set A/R = {[a]R; a ∈ A} can be made into a Ω-algebra ℬ = 𝒜/R by setting cB = [cA]R for all c ∈ CΩ and defining f ∈ FΩ with dΩ(f) = n and all  ∈ An. The function A → A/R given by a ↦ [a]R is an epimorphism 𝒜 → 𝒜/R.If ϕ : 𝒜 → ℬ is a homomorphism, then the image of ϕ is a subalgebra of ℬ; moreover, the relation Ker(ϕ) is defined by xKer(ϕ)y if and only if ϕ(x) = ϕ(y) is a congruence on 𝒜 and A/Ker(ϕ) is isomorphic to the image of ϕ. Let (Ai)i ∈ I be an indexed family of Ω-algebras. Let i ∈ I. Let P = ∏i∈IAi. We make P into an Ω-algebra cP(i) = ci for all i ∈ I and definingf ∈ FΩ with dΩ(f) = n, all  ∈ Pn, and all i ∈ I. 𝒫 is the direct product of the family (𝒜i)i∈I. For each fixed i0 ∈ I the projection πio : 𝒫 → 𝒜i0 given by α ↦ α(i0) is an epimorphism. If all the 𝒜i are the same algebra 𝒜, then 𝒫 = 𝒜I is a direct power of 𝒜. In that event the diagonal map δ : 𝒜 → 𝒜I given by a ↦ δ(a) defined by δ(a)(i) = a for all a ∈ A and all i ∈ I is a monomorphism.If Ω-algebras closed under taking subalgebras, homomorphic images, and direct products is a variety of Ω-algebras. Note that since the trivial Ω-algebra is a homomorphic image of any Ω-algebra whatsoever, every variety contains the trivial algebra.A nonempty class of FΩ = {·,−1} with dΩ(·) = 2 and dΩ(−1) = 1 and CΩ = {1}, then the class of all groups is a variety of Ω-algebras.(1) If R be an integral domain. If FΩ = R ∪ {+, −, [xx]} with dΩ(−) = dΩ(α) = 1 for all α ∈ R and dΩ(+) = dΩ([x]) = 2 and CΩ = {0}, then the class of all Lie algebras over R is a variety of Ω-algebras.(2) Let P(I) be the power set of a set I. A subset M0⊆P(I) will be called an ideal in the ring P(I) if ∅ ∈ M0 and M0 is closed under finite unions and also closed under the formation of subsets. M0 will be a proper ideal in P(I) if is an ideal in P(I) and I ∉ M0. The dual of a proper ideal in a Boolean algebra is a filter. Specifically, if I is a nonempty set and 𝒟 is a family of subsets of I, then 𝒟 is a filter on I provided that∅ ∉ 𝒟;I ∈ 𝒟;A, B) ∈ 𝒟2 implies A∩B ∈ 𝒟;(A⊆B⊆I and A ∈ 𝒟 implies B ∈ 𝒟.Let 𝒫 = ∏𝒜i is the direct product of the indexed family (𝒜i)i∈I of Ω-algebras and 𝒟 is a filter on I, then we may define a congruence R(𝒟) on 𝒫 by α ≅R𝒟 β provided {i ∈ I : α(i) = β(i)} ∈ 𝒟. The quotient algebra 𝒫/R𝒟 is the reduced product of the family (𝒜i)i∈I modulo the filter 𝒟 on I. If 𝒫 = 𝒜I is a direct power, then 𝒫/R𝒟, written 𝒜I/𝒟, may be called the reduced power of 𝒜 modulo the filter 𝒟 on I. In that event one can prove that the mapping d : 𝒜 → 𝒜I/𝒟 defined by a ↦ [δ(a)]R(𝒟) where δ is the diagonal embedding δ : 𝒜 ↦ 𝒜I, δ(a)(i) = a for all a ∈ A, i ∈ I, is an algebra monomorphism.If d : 𝒜 → 𝒜I/D is called the canonical embedding of 𝒜 into the reduced power 𝒜I/𝒟. We give as examples two extreme cases.The map 𝒟 = {I} be the trivial filter on I. Then 𝒫/R(𝒟) is isomorphic to 𝒫 so that direct products (powers) may be viewed as special cases of reduced products (powers).(1) Let 𝒟 on I is called an ultrafilter on I. If D is an ultrafilter on I, then 𝒫/R(𝒟)(𝒜I/𝒟) is called the ultraproduct of the family (𝒜i)i∈I  modulo the ultrafilter 𝒟 on I. If 𝒫 = 𝒜I is a direct power then the ultraproduct is called an ultrapower.(2) A maximal filter ω of nonnegative integers with its natural order as the first limit ordinal. Suppose 𝒜0 is an Ω-algebra and, for each n < ω, 𝒟n is an ultrafilter on an index set In. Let 𝒜n+1 = 𝒜nIn/𝒟n if 𝒜n has already been defined and let dn : 𝒜n → 𝒜n+1 be the canonical embedding. Then the direct limit 𝒜ω of the system 𝒜o with respect to the sequence (𝒟n)n<ω of ultrafilters. Note that in this event the limit map dω : 𝒜0 → Aω is a monomorphism.Now view the set Ω there corresponds the first order language with equality LΩ. Besides Ω we need a countably infinite set {xn : n < ω} of distinct variables. The terms or polynomials or words of LΩ are defined recursively as follows.Constant symbols and variables are terms.f ∈ FΩ and dΩ(f) = n and  is a tuple of terms already defined, then f is a term.If To each operator domain t1 = t2 where  is an ordered pair of terms of LΩ is an atomic formula of LΩ. The negation ~(t1 = t2) of an atomic formula of LΩ is a negated atomic formula of LΩ. If α is either an atomic formula or a negated atomic formula of LΩ, then α is a literal of LΩ. We omit the recursive definition of  formula of LΩ but appeal to the classical result that every formula of LΩ is logically equivalent to one in prenex normal form. Such a formula is of the form Q1y1 … Qnynϕ where each yi = xni is a variable, each Qi is a quantifier, ∀, ∃, and ϕ, the matrix of the formula, is a Boolean combination of atomic formulas. We do not exclude the possibility n = 0. Such formulas are quantifier free. Each quantifier free formula of LΩ is equivalent to a quantifier free formula of LΩ in disjunctive normal form  as well as to a quantifier free formula of LΩ in conjunctive normal form . A formula of LΩ containing no unquantified variables is a sentence of LΩ. We omit the recursive definition of a formula Ψ of LΩ holding in an Ω-algebra 𝒜 under an interpretation of the variables and trust the reader to understand intuitively what it means for a sentence a of LΩ to hold in an Ω-algebra 𝒜. If in Q1y1 … Qnyn all the Qi are ∀, then the above formula is a universal formula of LΩ. Similarly, if all the Qi are ∃, then the above formula is an existential formula of LΩ. A universal  formula of LΩ containing no unquantified variables is a universal  sentence of LΩ. Clearly, the negation of a universal sentence is logically equivalent to an existential sentence and vice versa. Vacuous quantifications are permitted. Thus, a quantifier free formula of LΩ is considered a universal formula of LΩ as well as an existential formula of LΩ. For example, the formula 1 · 1−1 = 1 in the language of group theory is considered both a universal sentence and an existential sentence.An expression of the form LΩ containing at most the variables in LΩ. Clearly a universal sentence of LΩ of the form LΩ containing at most the variables in y, is equivalent to the negation of a primitive sentence of LΩ. We will find it convenient to call such sentences negated primitive. Given an Ω-algebra 𝒜, we let Th∀(𝒜) be the set of all universal sentences of LΩ true in 𝒜, Th∃(𝒜) the set of all existential sentences of LΩ true in 𝒜, and Th(𝒜) the set of all sentences of LΩ true in 𝒜. We call Th∀(𝒜) the universal theory of 𝒜, Th∃(𝒜) is the existential theory of 𝒜 and Th(𝒜) the theory of A. Clearly Th∀(𝒜) = Th∀(ℬ) if and only if Th∃(𝒜) = Th∃(ℬ). In that event we say that 𝒜 and ℬ are universally equivalent and write 𝒜 ≡∀ ℬ. If Th(𝒜) = Th(ℬ) we say that 𝒜 and ℬ are elementarily equivalent and write 𝒜 ≡ ℬ. Being elementarily equivalent is a sufficient but not necessarily necessary condition for being universally equivalent.An existential sentence of the form LΩ whose matrix LΩ. Similarly, if LΩ whose matrix LΩ.Suppose Ω-algebra 𝒜 is a subalgebra of the Ω-algebra ℬ. Then every existential sentence of LΩ holding in 𝒜 holds also in ℬ and every universal sentence of LΩ holding in ℬ holds also in 𝒜. The next result is rather obvious. We omit a proof.Suppose that the Ω-algebra 𝒜 be a subalgebra of the Ω-algebra ℬ. Then the following three statements are equivalent in pairs:𝒜 ≡∀ ℬ;LΩ true in ℬ is also true in 𝒜;every primitive sentence of LΩ true in 𝒜 is also true in ℬ.every negated primitive sentence of Let the LΩ whose matrix LΩ.Now let LΩ has the form si = ti) as (si ≠ ti)) i(si ≠ ti)∧(s = t) so that A universal Horn sentence of 𝒜 is an Ω-algebra, then we have the inclusions I(𝒜) is the set of identities of LΩ true in 𝒜, Q(𝒜) is the set of quasi-identities of LΩ true in 𝒜, and H(𝒜) is the set of universal Horn sentences of LΩ true in 𝒜.  An Ω-algebra 𝒜 is a model of a set S of sentences of LΩ provided every sentence s ∈ S holds in 𝒜. Appealing to the classical Godel-Henkin Completeness Theorem, we see that a set S of sentences of LΩ is consistent if and only if it has a model. Let S be a consistent set of sentences of LΩ. Then M(S) will be the class of all models or the model class of S.It follows from this discussion that if Ω-algebras is axiomatic provided that it is the model class M(S) for at least one consistent set S of sentences of LΩ. Every axiomatic class of Ω-algebras is nonempty and closed under isomorphism. Note that every set of quasi-identities  of LΩ holds in the trivial Ω-algebra, hence it is consistent. A celebrated theorem of Garrett Birkhoff asserts that a class of Ω-algebras is a variety if and only if it is the model class of a set of identities of LΩ. If we define a quasivariety of Ω-algebras to be an axiomatic class of Ω-algebras containing the trivial Ω-algebra and closed under taking subalgebras and direct products, then a well-known characterization, due to Mal'cev, along the lines of Birkhoff's Theorem asserts that a class of Ω-algebras is a quasivariety if and only if it is the model class of a set of quasi-identities of LΩ. Note that the model class operator M applied to sets of sentences reverses inclusions.A class of 𝒜 be an Ω-algebra. Recall that I(𝒜) is the set of identities of LΩ true in 𝒜, Q(𝒜) is the set of quasi-identities of LΩ true in 𝒜 and H(𝒜) is the set of universal Horn sentences of LΩ true in 𝒜, and Th∀(𝒜) is the set of universal sentences of LΩ true in 𝒜. (All of these sets are consistent since they have A as a model.) Applying the model class operator, we get M(I(𝒜)) = var(𝒜) is the least variety of Ω-algebras containing 𝒜. The set M(Q(𝒜)) = qvar(𝒜) is the least quasivariety of Ω-algebras containing 𝒜. The set M(H(𝒜)) = uhc(𝒜) is the least universally axiomatizable Horn class containing 𝒜. Finally, M(Th∀(𝒜)) ) is the least universally axiomatizable class containing 𝒜. With the above notation, we have Now let ϵ : 𝒜 → ℬ is an elementary embedding provided for each formula Ψ and each tuple  from 𝒜 it is the case that Ψ(ϵ(a1),…, ϵ(an)) holds in ℬ if and only if Ψ holds in 𝒜. The existence of an elementary embedding is a sufficient, but in general unnecessary, condition for elementary equivalence. If 𝒜 is a subalgebra of ℬ and the inclusion map embeds 𝒜 elementarily into ℬ, then ℬ is said to be an elementary extension of 𝒜. Obviously isomorphic Ω-algebras are elementarily equivalent. Thus, necessary conditions for a nonempty class 𝒳 of Ω-algebras to be axiomatic are that 𝒳 be closed under elementary equivalence and taking ultraproducts. These conditions are also sufficient. That is the content of Theorem 3, Section 42 of [A monomorphism on 42 of  (see 8]ϵ : 𝒜 → ℬS be a consistent set of sentences of LΩ. A consequence of a theorem of Los is that if every member of the family (𝒜i)i∈I of Ω-algebras lies in M(S), then so does every ultraproduct of the family. From that it is easy to deduce that every ultrapower and every ultralimit of an Ω-algebra 𝒜 must be elementarily equivalent to 𝒜.  The canonical embedding is in fact an elementary embedding .σ be a sentence of LΩ.σ is equivalent to a universal sentence of LΩ if and only if σ is preserved under taking subalgebras.σ is equivalent to a Horn sentence of LΩ if and only if σ is preserved under taking reduced products.σ is a universal sentence of LΩ, then σ is equivalent to a universal Horn sentence of LΩ if and only if σ is preserved under taking direct products.If Let For proofs see, for example, .Ω-algebras under a restriction on the operator domain Ω. Specifically, we consider those Ω whose set CΩ of constant symbols is a singleton CΩ = {θ}. We then consider those varieties of Ω-algebras satisfying, at least, the laws f = θ as f varies over the set FΩ of function symbols of Ω. Recall that quantifier free sentences of LΩ are special cases of universal sentences of LΩ. We say that such varieties contain a zero. For example, the varieties of groups, monoids, and Lie algebras contain a zero but the variety of semigroups does not. We conclude this section with several observations.We will have need in the next section to consider special varieties of Ω-algebra 𝒜 is a subalgebra of the Ω-algebra ℬ. Since 𝒜 ≡∀ ℬ if and only every primitive sentence of LΩ true in ℬ is also true in 𝒜 it follows that necessary and sufficient conditions for 𝒜 ≡∀ ℬ are that every finite system (in finitely many variables) of equations and inequations Pi, pi, Qj, and qj are terms of LΩ, which has a solution in ℬ must already have a solution in 𝒜.Suppose the i(si ≠ ti) contains no atomic formula is false in the trivial Ω-algebra. Hence, if such a sentence is true in an Ω-algebra 𝒜, then uhc(𝒜), the universal Horn class of 𝒜, must be a proper subclass of qvar(𝒜), the quasivariety generated by 𝒜. On the other hand, if 𝒜 lies in a variety containing a zero θ, then 𝒜 contains the trivial subalgebra {θA} so such a sentence could not hold in 𝒜.A negated primitive Horn sentence Ω-algebra ℬ is a direct power of the Ω-algebra 𝒜. For some special 𝒜 we will want to show that 𝒜 and ℬ are universally equivalent. Since the diagonal map δ : 𝒜 → ℬ embeds 𝒜 isomorphically into ℬ every universal sentence of LΩ true in ℬ must also be true in 𝒜. Thus, in the above situation it will suffice for our purposes to show that every universal sentence of LΩ true in 𝒜 is also true in ℬ.Suppose the Ω be an operator domain and let 𝒱 be a variety of Ω-algebras. Throughout this section we will assume that all algebras lie in 𝒱. We will sometimes  assume that 𝒱 contains a zero, which in that event implies, among other things, a restriction on Ω.Let 𝒜 and ℬ be elements of 𝒱.𝒜 separates ℬ provided to every pair x ≠ y of unequal elements of ℬ there is a homomorphism ϕ : ℬ → 𝒜 such that ϕ(x) ≠ ϕ(y).𝒜 discriminates ℬ provided given finitely many pairs xi ≠ yi, i = 1,…, n, of unequal elements of ℬ there is a homomorphism ϕ : ℬ → 𝒜 such that ϕ(xi) ≠ ϕ(yi) for all i = 1,…, n.𝒜 is discriminating provided it discriminates every element of 𝒱 which it separates.Let 𝒜 is discriminating if and only if it discriminates its direct square 𝒜2.The proof is identical to that for groups. See, for example, .𝒜 is discriminating, then 𝒜2 ≡∀ 𝒜; that is, 𝒜 has the same universal theory as its direct square 𝒜2.If 𝒜 with its image in 𝒜2 under the diagonal embedding δ : A → 𝒜2 given by δ(a) = . Viewing 𝒜 as a subalgebra of 𝒜2, it will suffice to show that every finite system 𝒜2 already has a solution in 𝒜.We identify x1,…, xn) =  is a solution in 𝒜2 to the above system. Since 𝒜 discriminates 𝒜2 there is a homomorphism ϕ : 𝒜2 → 𝒜 such that Suppose that  = (ϕ(b1),…, ϕ(bn)) is a solution to the system in 𝒜. Hence, 𝒜2 ≡∀ 𝒜.But then  = uhc(𝒜);ℬ in 𝒱 such that 𝒜 ≡∀ ℬ.there is a discriminating algebra Let Momentarily assuming the theorem, we have the following consequence.𝒱 contains a zero and let 𝒜 be an algebra in 𝒱. Then the following three conditions are equivalent in pairs:𝒜 is squarelike;ucl(𝒜) = qvar(𝒜);ℬ in 𝒱 such that 𝒜 ≡∀ ℬ.there is a discriminating algebra Suppose the variety 𝒜) = qvar(𝒜). Since uhc(𝒜) is axiomatizable by universal Horn sentences it is closed under taking subalgebras and direct products. Since 𝒜 ∈ uhc(𝒜) and the trivial algebra {θA} is a subalgebra of 𝒜, uhc(𝒜) contains the trivial algebra. Thus the axiomatic class uhc(𝒜) is closed under taking subalgebras and direct products and contains the trivial algebra. Hence, it is a quasivariety. Therefore, uhc(𝒜) = qvar(𝒜).Assuming the theorem it will suffice to show that uhc is axiomatizable by universal Horn sentences. Assume deducing a contradiction that u is a universal sentence of LΩ true in 𝒜 but not a consequence of any set of universal Horn sentences of LΩ. Of course then u itself cannot be a Horn sentence. We may assume that the matrix of u is written in conjunctive normal form and hence u has the form u is equivalent to the conjunction u would be a Horn sentence. We claim that at least one of the negated primitive sentences LΩ true in 𝒜. Suppose not. For each i such that Hi be a set of universal Horn sentences of LΩ true in 𝒜 such that Hi. For each i for which i we take H = ∪iHi. It follows then that u would be a consequence of the set H of universal Horn sentences of LΩ true in 𝒜, contrary to hypothesis. The contradiction shows that at least one negated primitive sentence LΩ true in 𝒜.(1) ⇒ (2) Assume i notationally). Let n ≥ 2. For each fixed ν0 = 1,…, n let Ψν0 be the quasi-identity Fix such a conjunct ν0 is true in 𝒜. Then for every tuple 𝒜, 𝒜. Equivalently, for every tuple 𝒜, 𝒜. Since a disjunction is true if at least one of its disjuncts is true we would have in the above event that 𝒜 for every tuple 𝒜. This implies that ν0. But quasi-identities are special cases of universal Horn sentences and LΩ true in 𝒜. The contradiction shows that none of the quasi-identities Ψν can hold in 𝒜. Thus, for each ν = 1,…, n, the existential sentence 𝒜.Suppose deducing a contradiction that Ψ𝒜 such that simultaneously μ = 1,…, m and 𝒜n so that 𝒜n. By 𝒜n ≡∀ A. It follows that the existential sentence 𝒜. But that contradicts the fact that its negation  𝒜. The contradiction shows that ucl(𝒜) must have at least one set of universal Horn axioms. Hence, ucl(𝒜) = uhc(𝒜) if 𝒜 is squarelike.Now let 𝒜) = uhc(𝒜). Now A ∈ uhc(𝒜) and uhc(𝒜) is closed under taking direct products. Let I be an infinite index set and let ℬ = 𝒜I. Then ℬ ∈ ucl(𝒜) so ℬ is a model of Th∀(𝒜) and every universal sentence of LΩ true in 𝒜 is also true in ℬ. Thus, 𝒜 ≡∀ ℬ by ℬ2 is isomorphic to ℬ so that ℬ is discriminating by (2) ⇒ (3) Suppose ucl2 which is elementarily equivalent to ℬ2 by ⋆ℬ)2 ≡∀ ℬ2. But ℬ is discriminating so ℬ2 ≡∀ ℬ by 𝒜 ≡∀ ℬ so ultimately (⋆ℬ)2 ≡∀ 𝒜. Thus, every universal sentence of LΩ true in 𝒜 is also true in (⋆ℬ)2. But 𝒜2 embeds in (⋆ℬ)2 so every universal sentence of LΩ true in 𝒜 must also be true in 𝒜2. Then 𝒜2 ≡∀ 𝒜 by 𝒜 is squarelike.(3) ⇒ (1) Suppose that Exactly as for groups  we have𝒜i)i∈I be an indexed family of Ω-algebras and let 𝒟 be a filter on I. Let 𝒜 be the reduced product 𝒫/R(𝒟) where 𝒫 = ∏i∈I𝒜i. Then 𝒜2 is isomorphic to the reduced product of the family (𝒜i2)i∈I modulo the filter 𝒟 on I.Let i∈I be a family of Ω-algebras and let 𝒟 be an ultrafilter on I. Let 𝒜 be the ultraproduct constructed from this data. If ϕ is a sentence of LΩ let the support of ϕ be the set Supp(ϕ) of all i ∈ I such that ϕ holds in 𝒜i. A consequence of Los' Theorem  ∈ 𝒟. Now consider 𝒜2. By 𝒜2 is isomorphic to the ultraproduct of the family (𝒜2)i∈I modulo the filter 𝒟 on I. Suppose that each 𝒜i lies in 𝒳 so that (𝒜2)i ≡∀ 𝒜i for all i ∈ I. Let ϕ be a universal sentence of LΩ holding in 𝒜. Then Supp(ϕ) ∈ 𝒟. Thus, for each i ∈ Supp(ϕ), ϕ holds in 𝒜i. But Ai ≡∀ (𝒜2)i since Ai ∈ 𝒳. But then ϕ holds in 𝒜2 since Supp(ϕ) ∈ 𝒟. It follows that every universal sentence of LΩ true in 𝒜 must also be true in 𝒜2 and hence 𝒜2 ≡∀ 𝒜 by 𝒳 is closed under taking ultraproducts as well as elementary equivalence. Hence, 𝒳 is axiomatic.Let e, e.g., ) is that𝒜 be a squarelike algebra in 𝒱. Then 𝒜 is elementarily equivalent to a discriminating member of 𝒱. Consequently, the class of squarelike algebras in 𝒱 is the least axiomatic class containing the discriminating members of 𝒱.Let A proof which uses the ultralimit construction is exactly the same as that for groups and may be found, for example, in . See als𝒱 has a set of Horn axioms.The class of squarelike algebras in 𝒱 is closed under taking direct unions. Thus, that class, in addition to having a set of Horn axioms, has a set of universal-existential axioms.A theorem of Los and Suszko asserts that a model class has a set of universal-existential axioms if and only if it is closed under direct unions. For definitions of the relevant terms and a proof of the Los-Suszko Theorem see, for example, . It is e𝒜i)i∈I be a family of squarelike algebras in 𝒱 and let 𝒟 be a filter on I. By the theorem there is, for each i ∈ I, a discriminating algebra ℬi in 𝒱 such that 𝒜i ≡ ℬi. By 𝒜i)i∈I modulo the filter 𝒟 on I is elementarily equivalent to the reduced product of the family (ℬi)i∈I modulo the filter 𝒟 on I. In particular, the reduced product of the family (𝒜i)i∈I modulo the filter 𝒟 on I is universally equivalent to the reduced product of the family (ℬi)i∈I modulo the filter 𝒟 on I. Hence, the reduced product of the family (𝒜i)i∈I is squarelike by 𝒱 is preserved under taking reduced products. Now, as mentioned in page 225 of [Exactly as for groups  one proe 225 of , an axio𝒱 contains a zero θ we can explicitly describe a set of axioms for the class of squarelike algebras in 𝒱 by mimicking the situation for groups. For the remainder of this section we will restrict ourselves to those varieties 𝒱 which contain a zero θ. The class of squarelike algebras in the variety 𝒱 containing a zero is the model class of the laws of 𝒱 together with the sentences Pi, pi, Qj, and qj vary over terms of LΩ containing at most the variables in In the special case when example, .𝒱 be a variety containing a zero. Let 𝒬 be a subquasivariety of 𝒱. An algebra 𝒜 ∈ 𝒬q-discriminates 𝒬 provided that given finitely many quasi-identities 𝒜 such that simultaneously i, j. An algebra 𝒜 in 𝒱 is q-discriminating provided it q-discriminates qvar(𝒜).Let 𝒱 be a variety containing a zero. An algebra 𝒜 in 𝒱 is q-algebraically closed if and only if whenever a finite system ℬ∈ qvar(𝒜) it also has a solution in 𝒜.Let Exactly as for groups we have the following.𝒱 be a variety containing a zero. Let 𝒜 be an algebra in 𝒱. The following conditions are equivalent in pairs.𝒜 is q-discriminating.𝒜 is q-algebraically closed.𝒜 is squarelike.Let See, for example, .Ω-algebras which contains the trivial algebra and is closed under taking subalgebras and direct products must be the model class of at least one set of quasi-identities of LΩ. Granting us the result that an axiomatic class of Ω-algebra is closed under taking subalgebras if and only if it is the model class of at least one set of universal sentences of LΩ (Theorem 5.2.4 of [A2 ≡∀ A implies ucl(A) = uhc(A) can be adapted to provide a proof of Malcev's Theorem.A classical theorem of Malcev asserts that an axiomatic class of 5.2.4 of ) we obseΩ-algebras which contains the trivial algebra and is closed under taking subalgebras and direct products must be the model class of at least one set of quasi-identities of LΩ.An axiomatic class of Ω-algebra is closed under taking subalgebras if and only if it is the model class of at least one set of universal sentences of LΩ (Theorem 5.2.4 of [We assume that an axiomatic class of 5.2.4 of ).τ be an axiomatic class of Ω-algebras containing the trivial algebra and closed under taking subalgebras and direct products. Then τ is the model class of a set of universal sentences of LΩ. Assume deducing a contradiction that u is a universal sentence of LΩ holding in every algebra 𝒜 ∈ τ but that u is not a consequence of any set of quasi-identities of LΩ holding in every algebra 𝒜 ∈ τ. We may assume that the matrix of u is written in conjunctive normal form. Hence u has the form u is equivalent to the conjunction u would be equivalent to a conjunction of quasi-identities true in every algebra 𝒜 ∈ τ contrary to hypothesis.Now let τ. Therefore at least one We claim that it is impossible to have a conjunct LΩ true in every algebra 𝒜 ∈ τ. Suppose not. For each i such that Hi be the set of quasi-identities of Lω true in every algebra 𝒜 ∈ τ such that Hi. For each i for which i we take H = ∪iHi. Then u would be a consequence of the set H of quasi-identities  of LΩ true in every algebra 𝒜 ∈ τ, contrary to hypothesis. The contradiction shows that at least one negated primitive sentence LΩ true in every algebra 𝒜 ∈ τ. Fix such a conjunct i notationally). Let n ≥ 2. For each fixed ν0 = 1,…, n let ψν0 be the quasi-identity We now claim that at least one of the negated primitive sentences ψν0 is true in every algebra 𝒜 ∈ τ. Fix an 𝒜 ∈ τ. Then for every tuple 𝒜𝒜.Suppose deducing a contradiction that 𝒜, 𝒜. Since a disjunction is true if at least one of its disjuncts is true we have in that event that 𝒜 for every tuple 𝒜. Since 𝒜 was arbitrary, ψν0 true in every algebra in τ. But ν = 1,…, n, there is an algebra 𝒜ν ∈ τ such that the existential sentence 𝒜ν. Now let 𝒜ν such that simultaneously μ = 1,…, m and Equivalently, for every tuple ν=1n𝒜ν. It follows that the existential sentence ν=1n𝒜ν ∈ τ.Let 𝒜 ∈ τ. The contradiction shows that τ must be axiomatizable by at least one set of quasi-identities of LΩ completing the proof.But that contradicts the fact that its negation


The aim of this paper is to introduce power idealization filter topologies with respect to filter topologies and power ideals of lattice implication algebras, and to investigate some properties of power idealization filter topological spaces and their quotient spaces. By generalizing Boolean algebras and Lukasiewicz implication algebras , Xu 2]  2] definNow we recall some definitions and notions of lattice implication algebras and topological spaces.L, ∧, ∨, 0,1) be a bounded lattice with the greatest 1 and the smallest 0. A system  is called a quasi-lattice implication algebra if  ′ : L → L is an order-reserving involution and →:L × L → L is a map  satisfying the following conditions for any x, y, z ∈ L:x → (y → z) = y → (x → z),x → x = 1,x → y = y′ → x′,x → y = y → x = 1 implies x = y,x → y) → y = (y → x) → x. is called a lattice implication algebra if the implication operator → further fulfils the following conditions:(6)x∨y) → z = (x → z)∧(y → z),((7)x∧y) → z = (x → z)∨(y → z). will be simply denoted by L.A lattice implication algebra  is called a topological space. A subset B of τ is called a base of τ, if for each A ∈ τ and each x ∈ A, there exists B ∈ B such that x ∈ B⊆A.Elements of L be an implication algebra. A subset F of L is called a filer, if F satisfies the following: (1) 1 ∈ F; (2) x, x → y ∈ F implies y ∈ F. The collection of all filters in L is denoted by F(L), or F briefly. Clearly, F consists a base of some topology TF(L), briefly TF. Usually, TF is called the filter topology generated by F. And the pair  is called the filter topological space. A subset U⊆L is called TF-neighborhood of x ∈ L, or neighborhood of x in TF if x ∈ U ∈ TF. The set of all TF-neighborhoods of x is denoted by NTF(x). Since F⊆TF and [x) = ∩{F : x ∈ F ∈ F} ∈ F, [x) is the smallest element of NTF(x).Let TF are denoted by c and i. Clearly, for every A⊆L, c(A) = ∩{L∖[x) : x ∈ L, [x)∩A = ∅} and i(A) = ∪{[x) : x ∈ L, [x)⊆A}. The following proposition describes c(A).The closure operator and interior operator of L, TF) be the filer topology generated by F(L). Then for A⊆L, c(A) = {x ∈ L : [x)∩A ≠ ∅}.Let (x) is the smallest TF-neighborhood of x.The proof is trivial since [L be a lattice implication algebra and let 2L be the power set of L. A nonempty subset I of 2L is called a power ideal of L if I satisfies the following: (1) A, B ∈ 2L and A⊆B ∈ I imply A ∈ I; (2) A, B ∈ I implies A ∪ B ∈ I. The collection of all power ideals in 2L is denoted by I(L), or briefly I. Note that I∅ = {∅} is the smallest power ideal and IL = 2L is the greatest power ideal. Moreover, if I, J ∈ I, then (1) I∩J ∈ I; (2) I∨J = {I ∪ J : I ∈ I, J ∈ J} ∈ I.Let L be a lattice implication algebra, let TF be the filter topology, and let I be a power ideal. An operator *on  2L is defined as follows:A⊆L.Let TF and I. A* is called local function of A. We usually write A*(I) or A* instead of A*.The operator *is called the local function with respect to x ∈ A* if and only if [x)∩A ∉ I. Thus A* = {x ∈ L : [x)∩A ∉ I}. The following proposition gives some further details of A*.Clearly, L, TF) be the filter topological space and I, J ∈ I. ThenA*(I∅) = c(A) and A*(IL) = ∅;A⊆B, then A*(I)⊆B*(I);if I⊆J, then A*(J)⊆A*(I);if A*(I) = c(A*(I))⊆c(A);A*)*(I)⊆A*(I); = ∅;if A ∈ TFc, then A*(I)⊆A;if B ∈ I, then (A ∪ B)*(I) = A*(I) = (A∖B)*(I);if A ∪ A*(I))*(I) = A*(I);(A ∪ B)*(I) = A*(I) ∪ B*(I);(A*(I)∖B*(I) = (A∖B)*(I)∖B*(I)⊆(A∖B)*(I);I, then [x)*(I) = L for each x ∈ L;if {1} ∉ I and 1 ∈ A⊆L, then [A)*(I) = [A*(I)) = L.if {1} ∉ Let (x ∈ A*(I∅) if and only if [x)∩A ≠ ∅ if and only if x ∈ c(A). Thus A*(I∅) = c(A). Since [x)∩A ∈ 2L = IL for each x ∈ L, A*(L) = ∅.(1) By A⊆B and x ∈ A*(I). Then [x)∩A ∉ I. Since I is a power ideal and [x)∩A⊆[x)∩B, [x)∩B ∉ I and so x ∈ B*(I). Thus A*(I)⊆B*(I).(2) Let I⊆J and x ∈ A*(I). Then [x)∩A ∉ J. It follows that [x)∩A ∉ I and so x ∈ A*(I). Thus A*(J)⊆A*(I).(3) Let x ∉ c(A), then x ∈ L∖c(A) ∈ TF and so [x)⊆L∖c(A). Thus [x)∩A⊆(L∖c(A))∩A = ∅ ∈ I. This implies x ∉ A*(I) and so A*(I)⊆c(A). Then c(A*(I))⊆c(c(A)) = c(A).(4) If A*(I)⊆c(A*(I)). Next, we prove c(A*(I))⊆A*(I).It is clear that x ∈ c(A*(I)). By x)∩A*(I) ≠ ∅. Then there exists y ∈ [x)∩A*(I). By y ∈ A*(I), [y)∩A ∉ I. By y ∈ [x), [y)⊆[x). Thus [x)∩A ∉ I and so x ∈ A*(I). Therefore c(A*(I))⊆A*(I).Let A*(I))*(I)⊆c(A*(I)) = A*(I).(5) By (4), (x)∩A⊆A ∈ I for each x ∈ L, A*(I) = ∅.(6) Since [x ∈ A*(I)∖A. Then x ∈ L∖A ∈ TF. Thus [x)⊆L∖A and so [x)∩A⊆(L∖A) = ∅ ∈ I. Hence x ∉ A*(I) which is a contradiction. Therefore A*(I)⊆A.(7) Suppose that A∖B)*(I)⊆A*(I)⊆(A ∪ B)*(I). Next, we prove the inverse inclusions.(8) By (2), (x ∉ (A∖B)*(I), then ([x)∩A)∖B = [x)∩(A∖B) ∈ I. Thus [x)∩A⊆I ∪ B ∈ I which follows from I is a power ideal. This implies x ∉ A*(I). Thus A*(I)⊆(A∖B)*(I) and so A*(I) = (A∖B)*(I).If x ∉ A*(I), then [x)∩A ∈ I. Since B ∈ I,x ∉ (A ∪ B)*(I). This implies (A ∪ B)*(I)⊆A*(I) and so (A ∪ B)*(I) = A*(I).If A*(I)⊆(A ∪ A*(I))*(I). Conversely, if x ∉ A*(I), then [x)∩A ∈ I. Let [x)∩A = I. Then A⊆I ∪ (L∖[x)). By (2), (7), (8), and [x) ∈ TF,A ∪ A*(I)⊆(L∖[x)) ∪ A and sox ∉ (A ∪ A*(I))*(I) and so (A ∪ A*(I))*(I)⊆A*(I).(9) Clearly, A*(I) ∪ B*(I)⊆(A ∪ B)*(I) is clear. Conversely, if x ∉ A*(I) ∪ B*(I), then [x)∩A, [x)∩B ∈ I. Thus [x)∩(A ∪ B) = ([x)∩A)∪([x) ∪ B) ∈ I. This implies x ∉ (A ∪ B)*(I). Therefore (A ∪ B)*(I)⊆A*(I) ∪ B*(I).(10) A*(I)∖B*(I)⊆(A∖B)*(I). Assume that x ∈ (A*(I)∖B*(I))∖(A∖B)*(I). Then [x)∩(A∖B) ∈ I and [x)∩B ∈ I. Thusx ∉ A*(I) which is a contradiction. Thus A*(I)∖B*(I)⊆(A∖B)*(I) and so A*(I)∖B*(I)⊆(A∖B)*(I)∖B*(I). Finally, (A∖B)*(I)∖B*(I)⊆A*(I)∖B*(I) follows from (2). Therefore A*(I)∖B*(I) = (A∖B)*(I)∖B*(I).(11) We firstly prove I, 1 ∉ I for each I ∈ I. Assume that there exists x ∈ L such that [x) ≠ L. Let y ∉ L∖[x)*(I). Thus [y)∩[x) ∈ I. Since [x∨y) = [y)∩[x), 1 ∉ [x∨y) which is a contradiction. Therefore [x)*(I) = L for each x ∈ L.(12) Since {1} ∉ y ∈ L∖A*(I). Then 1 ∈ [y)∩A ∈ I which is a contradiction. Thus A*(I) = L and [A*(I)) = L. Since 1 ∈ A, L = [1)*(I)⊆[A)*(I) follows from (12). Therefore [A)*(I) = L.(13) Assume that there exists L, TF) be the filter topological space and I ∈ I. The operator cI*  on 2L, defined by c*(A) = A ∪ A* for A⊆L, satisfies the following statements:c*(∅) = ∅; c*(L) = L,c*(c*(A)) = c*(A)⊆c(A),c*(A))* = c*(A*) = A*,(c*(A ∪ B) = c*(A) ∪ c*(B),A ∈ TFc or A ∈ I implies c*(A) = A.Let (c*(∅) = ∅ follows from ∅* = ∅. c*(L) = L is clear.(1) (2) By (4) and (9) of c*(A*) = A* ∪ (A*)* = A*.(3) By (5) and (9) of (4) By (10) of (5) The result follows from (6) and (7) of L, TF) be the filter topological space and I ∈ I. The operator c* stated in TF and the topology generated by Ic . Such a topology is called a power idealization filter topology and often denoted by TF*, TF*(I), or TF*.Let (TF* = {A⊆L : c*(L∖A) = L∖A}. We prove that TF* is a topology on L.Let ∅, L ∈ TF* follows from (1) of (1) A, B ∈ TF*, then c*(L∖A) = L∖A and c*(L∖B) = L∖B. ThusA∩B ∈ TF*.(2) If At ∈ TF* for t ∈ T, where T is an index set. Then c*(L∖At) = L∖At andt∈TAt ∈ TF*.(3) Let TF, Ic⊆TF*.Finally, by (6) and (7) of L = {0, a, b, c, d, 1}, 0′ = 1, a′ = c, b′ = d, c′ = a, d′ = b, 1′ = 0, and the implication operator → be defined by a → b = a′∨b for a, b ∈ L. Then  is the Hasse lattice implication algebra  be the filter topological space and I, J ∈ I. ThenTF*(I∅) = TF and TF*(IL) = 2L,I⊆J, then TF*(I)⊆TF*(J).if Let (A*(I∅) = c(A) and A*(IL) = ∅. Thus cI∅*(A) = A if and only if c(A) = A. Similarly, cIL*(A) = A for each A⊆L. Therefore (1) holds.(1) By (1) of I⊆J. By (3) of cJ*(A)⊆cI*(A) for A⊆L. Thus if A ∈ TF*(I), then A ∈ TF*(J). Therefore TF*(I)⊆TF*(J).(2) Let I ∈ I satisfies TFc⊆I, then L ∈ I and so I = 2L = IL. Thus by (1) of TF* = Ic = 2L. If TFc∖{L}⊆I, we have the following proposition.Clearly, if L, TF) be the filter topological space. If I ∈ I satisfies TFc∖{L}⊆I, then TF* = Ic ∪ {∅}.Let (Ic ∪ {∅}⊆TF* follows from TF*⊈Ic ∪ {∅}. Then there exists B ∈ TF*∖(Ic ∪ {∅}) such that (L∖B)*⊆L∖B ≠ L. Let y ∈ (L∖B)∖(L∖B)*. Then y ∉ (L∖B)*. Thus [y)∩(L∖B) ∈ I. Put [y)∩(L∖B) = I. We have L∖B⊆I ∪ (L∖[y)). Since TFc∖{L}⊆I, L∖[y) ∈ I. Thus L∖B⊆I ∪ (L∖B) ∈ I and so L∖B ∈ I. Hence B ∈ Ic. It is a contradiction. Therefore TF*⊆Ic ∪ {∅}.L, TF) be the filter topological space and I ∈ I. If A⊆L satisfies A∩I = ∅ for each I ∈ I, then c*(A) = c(A).Let (c*(A)⊆c(A) is clear. Conversely, if x ∉ c*(A), then x ∉ A and x ∉ A*. Thus I = [x)∩A ∈ I and A⊆I ∪ (L∖[x)). Since A∩I = ∅, A⊆L∖[x). Observe that x ∉ L∖[x) and L∖[x) is TF-closed. We have x ∉ c(A). Thus c(A)⊆c*(A). Therefore c*(A) = c(A).I ∈ TFc, then TF* = TF.If TF⊆TF*. Conversely, let IM be the greatest element of I and A⊆L. Thus (A∖IM)∩I = ∅ for each I ∈ I. By c*(A∖IM) = c(A∖IM). By (8) of A∖IM)* = A*. Now, notice that A∩IM ∈ I⊆TFc and thus c(A∩IM) = A∩IM. We havec* = c. Therefore TF* = TF.It is clear that L, TF) be the filter topological space and I ∈ I. If I ∈ I and U ∈ TF, then U∖I ∈ TF*.Let (P = L∖U. Then P ∈ TFc. By (7) and (8) of L∖(P ∪ I) = U∩(L∖I) = U∖I ∈ TF*.Let L, TF) be the filter topological space and I ∈ I. ThenTF*. Moreover,NTF*(x) for each x ∈ L, where NTF*(x) is the set of all TF*-neighborhoods of x in . Clearly, [x)∖Ix is the smallest TF*-neighborhoods of x, where Ix is the greatest element of I satisfying x ∉ Ix.Let (BTF*⊆TF*. Let B⊆L. Then B ∈ TF*⇔L∖B is TF*-closed ⇔(L∖B)*⊆(L∖B)⇔B⊆L∖(L∖B)*. Thus x ∈ B⇒x ∉ (L∖B)*⇒ there exists U ∈ NTF(x) such that (L∖B)∩U ∈ I. Let I = (L∖B)∩U. Then L∖B⊆I ∪ (L∖U) andBTF* is a base of TF*.By BTF*(x)⊆NTF*(x). Let A ∈ NTF*(x) and y ∈ A. Since BTF* is a base of TF*, there are U, V ∈ TF and I, J ∈ I such that x ∈ U∖I⊆A and y ∈ V∖J⊆A. We can assume y ∉ I and x ∉ J . Theny ∈ (U ∪ V)∖(I ∪ J) ∈ BTF*(x) andBTF*(x) is a base of NTF*(x).Clearly, Ix is the greatest element of I satisfying x ∉ Ix, then [x)∖Ix ∈ NTF*(x) is the smallest TF*-neighborhoods of x.Clearly, if L, τ) be a topological space and I ∈ I. The topology that was generated by B = {U∖I : U ∈ τ, I ∈ I} is denoted by T* [T* = TF*(I).Let  . Clearlyψ = {∅, L} be the indiscrete topology on L and I ∈ I. Then T* = {∅} ∪ Ic.Let Tc∖{L} = {∅} ∈ I and By L, τ) be a topological space and I ∈ I. Then T* = τ∨T*, where τ∨T* is the topology generated by the base {U∩V : U ∈ τ, V ∈ T*}.Let . Since BT* = {U∩V : U ∈ τ, V ∈ T*}, BT* is also a base of τ∨T*. Therefore T* = τ∨T*.Clearly, L, TF) be the filter topological space and I ∈ I. Then TF* = TF∨T*.Let  be the filter topological space and I, J ∈ I. ThenT* = T*∨T*,TF*(I∨J) = T*) = T*),TF*(I∨J) = TF*(I)∨TF*(J),T*) = TF*(I).Let (T*(I∨J)⊇T*(I)∨T*(J). Conversely, let ∅ ≠ A ∈ T*(I∨J). By I ∈ I and J ∈ J such thatT*(I∨J)⊆T*(I)∨T*(J).(1) By (2) of (2) By (1), TF*(I∨J) = T*(TF*(J), I).Similarly, TF*(I∨J) = TF*(I)∨TF*(J).(3) By (1) and I = J. Then the proof follows from (2).(4) Let L, TF) be the filter topological space, I, J ∈ I and A⊆L. ThenA* = A* ∪ A*,A* = A*)∩A*).Let  ∪ A*⊆A*. Conversely, x ∉ A* ∪ A*. Then [x)∩A ∈ I and [x)∩A ∈ J. Let [x)∩A = I and [x)∩A = J. Then A⊆I ∪ (L∖[x)) and A⊆J ∪ (L∖[x)). Thusx)∩A⊆I∩J which implies x ∉ A*. Therefore A*⊆A* ∪ A*.(1) By (3) of x ∉ A*. Then [x)∩A ∈ I∨J. Then there exist I ∈ I and J ∈ J such that [x)∩A = I ∪ J. We can assume I∩J = ∅, ). Thus x ∉ I or x ∉ J, . Now, we take x ∉ I for example. Thenx) ∈ TF and x ∈ [x)∖I, [x)∖I ∈ BTF*(I). Thus x ∉ A*) and so x ∉ A*)∩A*). Hence(3) Let x ∉ A*(TF*(I), J). Then there exists I ∈ I such that ([x)∖I)∩A ∈ J. Let ([x)∖I)∩A = J. Then [x)∩A = I ∪ J which implies x ∉ A*. Similarly, if x ∉ A*(TF*(J), I), then x ∉ A*. Therefore A*⊆A*)∩A*).Conversely, let A* = A*).Consider I = J. The proof follows from (2) of Let TF*(I∩J) = TF*(I)∩TF*(J).Consider TF*(I∩J)⊆TF*(I)∩TF*(J). Conversely, if A ∉ TF*(I∩J), thenL∖A)*⊈(L∖A) or (L∖A)*⊈(L∖A). Thus A ∉ TF*(I) or A ∉ TF*(J). Therefore TF*(I)∩TF*(J)⊆TF*(I∩J).By (2) of L1, ∧, ∨,  ′, →1, 01, 11) and  be two lattice implication algebras. A mapping f from L1 to L2 is called lattice implication homomorphism, if f(x→1y) = f(x)→2f(y) for any x, y ∈ L1. The set of all lattice implication homomorphisms from L1 to L2 is denoted by hom.Let . Then, clearly,Let pologies .L1 and L2 be two implication algebras and let f ∈ hom and I ∈ 2L1, J ∈ 2L2 be power ideals.f is injective, then f−1(J) = {f−1(J) : J ∈ J} ∈ I(L1).If f is surjective, then f(I) = {f(I) : I ∈ I} ∈ I(L2).If Let ∅ ∈ J, then ∅ = f−1(∅) ∈ f−1(J).(1) Since I2 ∈ f−1(J) and I1⊆I2, then there exist J2 ∈ J such that I2 = f−1(J2). Thus f(I1)⊆f(I2) = J2 and f(I1) ∈ J. Since f is injective, I1 = f−1(f(I1)) ∈ f−1(J).If I1, I2 ∈ f−1(J), then there exist J1, J2 ∈ J such that I1 = f−1(J1) and I2 = f−1(J2). One has I1 ∪ I2 = f−1(J1) ∪ f−1(J2) = f−1(J1 ∪ J2). Since J1 ∪ J2 ∈ J, I1 ∪ I2 ∈ f−1(J). Therefore f−1(J) is a power ideal.If ∅ ∈ I, ∅ = f(∅) ∈ f(I).(2) By J2 ∈ f(J) and J1⊆J2, then there exists I2 ∈ I such that J2 = f(I2). Let I1 = f−1(J1). Then I1⊆I2 and I1 ∈ I. Since f is surjective, J1 = f(I1) ∈ f(I).If J1, J2 ∈ f(J), then there exist I1, I2 ∈ I such that J1 = f(I1) and J2 = f(I2). Thus I1 ∪ I2 ∈ I. Hence J1 ∪ J2 = f(I1) ∪ f(I2) = f(I1 ∪ I2) ∈ f(I). Therefore f(I) is a power ideal.If L1 and L2 be two implication algebras and f ∈ hom. Thenf is injective, then for each x ∈ L1, f−1([f(x))) ∈ Tl(TF(L2), f) is the smallest Tl-neighborhood of x;if f is bijective, then for each y ∈ L2, f([f−1(y))) ∈ Tr(TF(L1), f) is the smallest Tr-neighborhood of y.if Let f(x)) is the smallest TF(L2)-neighborhood of f(x). Then x ∈ f−1([f(x))) ∈ Tl(TF(L2), f). Let x ∈ V ∈ Tl(TF(L2), f). Then f(x) ∈ f(V) ∈ TF(L2) and [f(x))⊆f(V). Thus f−1([f(x)))⊆f−1(f(V)) = V. Therefore f−1([f(x))) is the smallest one.(1) Clearly, [f−1(y)) is the smallest TF(L1)-neighborhood of f−1(y). Thus y ∈ f([f−1(y))) ∈ Tr(TF(L1), f). Now, let y ∈ U ∈ Tr(TF(L1), f). Then f−1(y) ∈ f−1(U) ∈ TF(L1). Thus [f−1(y))⊆f−1(U) and so f([f−1(y)))⊆f(f−1(U)) = U. Therefore (2) holds.(2) Clearly, [L1 and L2 be two implication algebras and f ∈ hom.f is injective and J ∈ I(L2), then for each x ∈ L1, [f(x))∖JM ∈ TF*) is the smallest neighborhood of f(x), where JM is the greatest element of J satisfying f(x) ∉ JM.If f is bijective and I ∈ I(L1), then for each y ∈ L2, [f−1(y))∖IM ∈ TF*) is the smallest neighborhood of f−1(y), where IM is the greatest element of I satisfying f−1(y) ∉ IM.If Let L1 and L2 be two implication algebras and f ∈ hom.(1)f is injective and J ∈ I(L2), thenIf (2)f is bijective and I ∈ I(L1), thenIf Let c2* and cl be the closure operators of the left side and the right side of the equation. We only need to prove c1* = cl.(1) Let A⊆L1 and x ∉ c1*(A). Then x ∉ A and x ∉ A*(f−1(J), Tl(TF(L2), f)). By (1) of f−1([f(x)))∩A ∈ f−1(J). Thus there exists J ∈ J such that f−1([f(x)))∩A = f−1(J). Since x ∉ A and f is injective, f(x) ∉ J. Let JM ∈ J be the greatest one satisfying f(x) ∉ JM. Then f−1([f(x)))∩A⊆f−1(JM). Thusx ∉ cl(A). Therefore cl(A)⊆c1*(A).Let y ∉ cl(A). By x ∉ A, f−1([f(x)))∩A⊆f−1(JM) and [f(x))∩f(A)⊆JM. Then J1 = [f(x))∩f(A) ∈ J. Since f is injective, f−1([f(x)))∩A = f−1(J1). By (1) of x ∉ A*(f−1(J), Tl(TF(L2), f)). Therefore x ∉ c1*(A) and c1*(A)⊆cl(A).Conversely, let c2* and cr be the closure operators of the left side and the right side of the equation. We only need to prove c2* = cr.(2) Let y ∉ c2*(A). Then y ∉ A and y ∉ A*((f(I), Tr(TF(L1), f))). By (2) of f([f−1(y)))∩A ∈ f(I). Thus there exists I ∈ I such that f([f−1(y)))∩A = f(I). Since f is bijective, [f−1(y))∩f−1(A) = I. By y ∉ A, f−1(y) ∉ f−1(A) and so f−1(y) ∉ I. Since IM ∈ I is the greatest element of I satisfying f−1(y) ∉ IM, [f−1(y))∩f−1(A)⊆IM and [f−1(y))∩(L1∖IM)∩f−1(A) = ∅. By f being bijective again, we havey ∉ cr(A). Hence c2*(A)⊆cr(A).Let z ∉ cr(A). Since IM is the greatest element of I satisfying f−1(z) ∉ IM, by (2) of f−1(z))∖IM ∈ TF*) is the smallest neighborhood of f−1(z). Thus f([f−1(z))∖IM)∩A = ∅. Since f is bijective,f−1(z))∩f−1(A)⊆IM and [f−1(z))∩f−1(A) ∈ I. Thus f([f−1(y)))∩A ∈ f(I) which implies y ∈ A*((f(I), Tr(TF(L1), f))). Since z ∉ cr(A), z ∉ A. Therefore z ∉ c2*(A) and so c2*(A)⊆cr(A).Conversely, let f ∈ hom is surjective but not bijective, then (2) of Generally, if L1 = {01, a, b, c, d, 11} be the Hasse lattice implication algebra of L2 = {02, e, h, 12} and 02′ = 12, e′ = h, h′ = e, and 12′ = 02. The Hasse diagram and the implication operator of L2 are shown by Let f from L1 to L2 is defined asf ∈ hom and f is surjective. Let I = {∅, {c}}. Then I ∈ I(L1) and f(I) = {∅, {h}} ∈ I(L2).A mapping h} ∈ f(I), by b, c}*) = {01, b, c, d}⊈{b, c}. We havef−1 = {02, a, d, 12}, {02, e, 12} ∉ Tr), f).Since {Tr), f) ≠ T*(f(I), Tr(TF(L1), f)).In fact, we haveL1 and L2 be two implication algebras and let f ∈ hom be bijective.(1)J ∈ I(L2), thenIf (2)I ∈ I(L1), thenIf Let The proof follows from L1 and L2 be two implication algebras and f ∈ hom.(1)f is injective and J1, J2 ∈ I(L2), thenIf (2)f is bijective and I1, I2 ∈ I(L1), thenIf Let f−1(J1∨J2) = f−1(J1)∨f−1(J2). By (3) of (1) Clearly, (2) Is similar to (1).L1 and L2 be two implication algebras and let f ∈ hom be bijective. If J ∈ I(L2) and I ∈ I(L1), thenLet The proof follows from (4) of


Unfortunately, the original version of this article  containeThe results subsection of the Abstract also contains a mistake as a result of the mislabel. The first sentence “The 422 participants  had mean age (± standard deviation) of 38.3 ± 20.5 and 42.9 ± 20.7 years, respectively” should read “The 422 participants  had mean age (± standard deviation) of 38.3 ± 20.5 and 42.9 ± 20.7 years, respectively”. This error has been corrected in the original article.The ‘Baseline Characteristics of Participants’ subsection in the results contains an error. Reading from line 4, the sentence “The mean age amongst participant was 42.9 ± 20.7 and 38.3 ± 20.5 for females and males, respectively” should be corrected to “The mean age amongst participant was 42.9 ± 20.7 and 38.3 ± 20.5 for males and females, respectively”. We regret these errors which has now been corrected.


ℐ-regular pre-ℐ-open sets in which the notion of pre-ℐ-open set is involved. We characterize these sets and study some of their fundamental properties. We also present other notions called extremally pre-ℐ-disconnectedness, locally pre-ℐ-indiscreetness, and pre-ℐ-regular sets by utilizing the notion of pre-ℐ-open and pre-ℐ-closed sets by which we obtain some equivalence relation for pre-ℐ-regular pre-ℐ-open sets.We deal with the new class of pre- ℐ on a topological space  is a nonempty collection of subsets of X which satisfies (i) A ∈ ℐ and B ⊂ A implies B ∈ ℐ and (ii) A ∈ ℐ and B ∈ ℐ implies A ∪ B ∈ ℐ. Given a topological space  with an ideal ℐ on X and if ℘(X) is the set of all subsets of X, a set operator (·)⋆ : ℘(X) → ℘(X), called a local function [A with respect to τ and ℐ, is defined as follows: for A ⊂ X, A⋆ = {x ∈ X∣U∩𝒜 ∉ ℐ}, for every {U ∈ τ(x)} where τ(x) = {U ∈ τ∣x ∈ U}. A Kuratowski closure operator cl⋆ for a topology τ⋆, called the ⋆-topology, finer than τ is defined by cl⋆(A) = A ∪ A⋆ [A⋆ for A⋆ and τ⋆ for τ⋆. If ℐ is an ideal on X, then  is called an ideal space. A subset A of an ideal space  is said to be ℐ-open [A ⊂ int(A⋆). A subset A of an ideal space  is said to be pre-ℐ-open [A ⊂ int(cl⋆(A)). The complement of pre-ℐ-open set is called pre-ℐ-closed. The family of all pre-ℐ-open sets in  is denoted by PℐO or simply PℐO(X). Clearly τ ⊂ PℐO(X). The largest pre-ℐ-open set contained in A, denoted by pℐint(A), is called the pre-ℐ-interior of A. The smallest pre-ℐ-closed set containing A, denoted by pℐcl(A), is called the pre-ℐ-closure of A. A subset A of an ideal space  is said to be α-ℐ-open [A ⊂ int(cl⋆(int(𝒜))). The family of all α-ℐ-open sets is a topology finer than τ. We will denote the α-ℐ-interior subset of A of X by αℐint(A). A subset A of an ideal space  is τ⋆-dense [cl⋆(A) = X. A space  is ℐ-submaximal [τ⋆-dense subset of X is open.The subject of ideals in topological space has been studied by Kuratowski  and Vaidfunction  of A wit⋆ ; when the ℐ-open  if A ⊂ ie-ℐ-open  if A ⊂ iα-ℐ-open  if A ⊂ i⋆-dense  if cl⋆ be an ideal space and let A be a subset of X. Then pℐint(A) = A∩int(cl⋆(A)) [Let (cl⋆(A)) .ℐ-open set A of a space  is said to be pre-ℐ-regular pre-ℐ-open if A = pℐint(pℐcl(A)). The complement of a pre-ℐ-regular pre-ℐ-open set is called pre-ℐ-regular pre-ℐ-closed set, equivalently A = pℐcl(pℐint(A)). A subset A of a space  with an ideal ℐ is said to be pre-ℐ-regular if it is pre-ℐ-open and pre-ℐ-closed. Clearly, X and ∅ are pre-ℐ-regular pre-ℐ-open.A pre-ℐ-regular set is pre-ℐ-regular pre-ℐ-open.Also every pre-A is pre-ℐ-regular that implies A is pre-ℐ-open and pre-ℐ-closed. A is pre-ℐ-open which implies A = pℐint(A) = pℐint(pℐcl(A)). Hence A is pre-ℐ-regular pre-ℐ-open.Assume But the converse is not true as shown by X, τ, ℐ) where X = {a, b, c}, τ = {∅, {a}, {b}, {a, b}, X}, and ℐ = {∅, {a}}. Here {a} is pre-ℐ-regular pre-ℐ-open but not pre-ℐ-closed.Consider the ideal space  where X = {a, b, c}, τ = {∅, {b, c}, X}, and ℐ = {∅, {a}}. Here PℐO(X) = {∅, {b}, {c}, {a, b}, {a, c}, {b, c}, X}. Thus, A = {a, b}, B = {a, c} are both pre-ℐ-regular pre-ℐ-open. But A∩B is not, since it is not even pre-ℐ-open.Consider the ideal space  with an ideal ℐ as in b, c} is ℐ-open but not pre-ℐ-regular pre-ℐ-open. Consider the space X = {a, b, c} and τ = {∅, {a}, {b}, {a, b}, X}, ℐ = {∅, {a}}. Here {a} is pre-ℐ-regular pre-ℐ-open but not ℐ-open. Observe that every pre-ℐ-regular pre-ℐ-open set is pre-ℐ-open but the converse need not be true. Here {b, c} is pre-ℐ-open but not pre-ℐ-regular pre-ℐ-open.The notions of pre-X, τ, ℐ) be an ideal space and let A, B be any subsets of X. Then the following hold.(a)A ⊂ B, then pℐint(pℐcl(A)) ⊂ pℐint(pℐcl(B)).If (b)A ∈ PIO(X), then A ⊂ pℐint(pℐcl(A)).If (c)A ∈ PℐO(X), pℐint(pℐcl(pℐint(pℐcl(A)))) = pℐint(pℐcl(A)).For every (d)A and B are disjoint pre-ℐ-open sets, then pℐint(pℐcl(A)) and pℐint(pℐcl(B)) are disjoint.If (e)A is a pre-ℐ-regular pre-ℐ-open, then pℐcl(X − A) is pre-ℐ-regular pre-ℐ-closed.If (f)A is pre-ℐ-regular pre-ℐ-open, then pℐint(A) is pre-ℐ-regular pre-ℐ-open.If Let ((a)A ⊂ B⇒pℐcl(A) ⊂ pℐcl(B). Therefore pℐint(pℐcl(A)) ⊂ pℐint(pℐcl(B)).Suppose (b)A ∈ PℐO⇒A = pℐint(A) ⊂ pℐint(pℐcl(A)).Suppose that (c)pℐint(pℐcl(A)) ∈ PℐO, so by (b) we have pℐint(pℐcl(A))⊂pℐint(pℐcl(pℐint(pℐcl(A)))). On the other hand pℐint(pℐcl(A)) ⊂ pℐcl(A) which implies pℐcl(pℐint(pℐcl(A)))⊂pℐcl(pℐcl(A)) = pℐcl(A). Therefore, pℐint(pℐcl(pℐint(pℐcl(A))))⊂pℐint(pℐcl(A)). Hence pℐint(pℐcl(pℐint(pℐcl(A)))) = pℐint(pℐcl(A)).It is obvious that (d)A and B are disjoint pre-ℐ-open sets, we have A∩B = ∅ which implies A∩pℐcl(B) = ∅⇒A∩pℐint(pℐcl(B)) = ∅. Since pℐint(pℐcl(B)) is pre-ℐ-open, pℐcl(A)∩pℐint(pℐcl(B)) = ∅. Hencepℐint(pℐcl(A))∩pℐint(pℐcl(B)) = ∅.Since (e)A is pre-ℐ-regular pre-ℐ-open, A = pℐint(pℐcl(A)) implies X − A = X − pℐint(pℐcl(A)) = pℐcl(pℐint(X − A)). Therefore, pℐcl(X − A) = pℐcl(pℐint(pℐcl(X − A))). Hence pℐcl(X − A) is pre-ℐ-regular pre-ℐ-closed.Given that (f)A is pre-ℐ-regular pre-ℐ-open, then pℐcl(X − A) is pre-ℐ-regular pre-ℐ-closed. Hence X − (pℐcl(X − A)) is pre-ℐ-regular pre-ℐ-open that implies pℐint(A) is pre-ℐ-regular pre-ℐ-open.By (e) if X, τ, ℐ) the following are equivalent. (a)ℐ-open set is open.Every pre-(b)τ⋆-dense set is open.Every For an ideal topological space (A be a τ⋆-dense set that implies cl⋆(A) = X which implies int(cl⋆(A)) = X, so that A ⊂ int(cl⋆(A)) = X. By (a) every pre-ℐ-open set is open and hence A is open.(a) ⇒ (b): let B be a pre-ℐ-open subset of X, so that B ⊂ int(cl⋆(B)) = G, say. Then cl⋆(B) = cl⋆(G), so that cl⋆((X − G) ∪ B) = cl⋆(X − G) ∪ cl⋆(B) = (X − G) ∪ cl⋆(B) = X, and thus (X − G) ∪ B is τ⋆-dense in X. Thus (X − G) ∪ B is open. Now B = ((X − G) ∪ B)∩G is the intersection of two open sets, so that B is open.(b) ⇒ (a): let X, τ) with an ideal ℐ is ℐ-submaximal, then any finite intersection of pre-ℐ-open set is pre-ℐ-open.If a space  with an ideal ℐ is ℐ-submaximal, then any finite intersection of pre-ℐ-regular p-ℐ-open set is pre-ℐ-regular pre-ℐ-open.If a space ). Also, for each i = 1,2,…, n, ∩Oi ⊂ Oi which implies pℐint(pℐcl(∩Oi))⊂pℐint(pℐcl(Oi)). Also, each Oi is pre-ℐ-regular pre-ℐ-open that implies Oi = pℐint(pℐcl(Oi)) which implies pℐint(pℐcl(∩Oi))⊂∩Oi and so ∩Oi = pℐint(pℐcl(∩Oi)). Hence ∩Oi is pre-ℐ-regular pre-ℐ-open.Let {ℐ-regular pre-ℐ-open set is pre-ℐ-regular pre-ℐ-open. But the intersection of two pre-ℐ-regular pre-ℐ-closed sets fails to be pre-ℐ-regular pre-ℐ-closed as shown by It should be noted that an arbitrary union of pre-X, τ, ℐ) as in a, c}, {a, b} are pre-ℐ-regular pre-ℐ-closed but their intersection is not pre-ℐ-regular pre-ℐ-closed.Consider the ideal space .(a)A is pre-ℐ-closed, then pℐint(A) is pre-ℐ-regular pre-ℐ-open.If (b)A = pℐint(A), then pℐcl(A) is pre-ℐ-regular pre-ℐ-closed.If (c)A and B are pre-ℐ-regular pre-ℐ-closed sets, then A ⊂ B if and only if pℐint(A) ⊂ pℐint(B).If (d)A and B are pre-ℐ-regular pre-ℐ-open sets, then A ⊂ B if and only if pℐcl(A) ⊂ pℐcl(B).If The following hold for a subset A is pre-ℐ-closed, A = pℐcl(A).(a) Since pℐint(pℐcl(pℐint(A))) = pℐint(pℐint(A)) = pℐint(A). Hence pℐint(A) is pre-ℐ-regular pre-ℐ-open.Now, pℐcl(pℐint(pℐcl(A))) = pℐcl(pℐcl(A)) = pℐcl(A). Hence pℐcl(A) is pre-ℐ-regular pre-ℐ-closed.(b) Now A and B are pre-ℐ-regular pre-ℐ-closed sets, therefore, A = pℐcl(pℐint(A)) and B = pℐcl(pℐint(B)). Clearly, pℐint(A) ⊂ pℐint(B) if A ⊂ B.(c) Given that pℐint(A) ⊂ pℐint(B). Now A = pℐcl(pℐint(A)) ⊂ pℐcl(pℐint(B)) ⊂ B. Hence A ⊂ B.Conversely, A and B are pre-ℐ-regular pre-ℐ-open, therefore, A = pℐint(pℐcl(A)) and B = pℐint(pℐcl(B)). Suppose A ⊂ B, pℐcl(A) = pℐcl(pℐint(pℐcl(A)))⊂pℐcl(pℐint(pℐcl(B)))⊂pℐcl(B). Therefore, pℐcl(A) ⊂ pℐcl(B).(d) Given that pℐcl(A) ⊂ pℐcl(B). Now A = pℐint(pℐcl(A)) ⊂ pℐint(pℐcl(B)) ⊂ B.Conversely, A of an ideal topological space  is said to be ℐ-rare if it has no interior points in τ⋆.A subset X, τ, ℐ) be an ideal space. Then the following hold.(a)ℐ-regular pre-ℐ-open.The empty set is the only subset which is nowhere dense and pre-(b)A is pre-ℐ-regular pre-ℐ-closed, then every ℐ-rare set is pre-ℐ-open.If Let (A is nowhere dense and A is pre-ℐ-regular pre-ℐ-open. Then A = pℐint(pℐcl(A)) = pℐcl(A)∩int(cl⋆(pℐcl(A))), by A ⊂ pℐcl(A)∩int(cl⋆(cl(A)))⊂pℐcl(A)∩int(cl(A)) = pℐcl(A)∩∅ = ∅.(a) Suppose A is pre-ℐ-regular pre-ℐ-closed. Then A = pℐcl(pℐint(A)) = pℐint(A)∪cl(int⋆(pℐint(A)))⊂pℐint(A) ∪ cl(int⋆(A)) = pℐint(A) ∪ ∅ = pℐint(A). Therefore, A = pℐint(A). Hence A is pre-ℐ-open.(b) Suppose X, τ, ℐ) is called extremally pre-ℐ-disconnected if the pre-ℐ-closure of every pre-ℐ-open set is pre-ℐ-open.An ideal space  the following are equivalent.(a)X, τ, ℐ) is extremally pre-ℐ-disconnected.((b)ℐ-regular pre-ℐ-open subset is pre-ℐ-regular.Every pre-For a topological space  is extremally pre-ℐ-disconnected. Suppose A is pre-ℐ-regular pre-ℐ-open. Then A is pre-ℐ-open and so pℐcl(A) is a pre-ℐ-open set. Hence A = pℐint(pℐcl(A)) = pℐcl(A). Hence A is pre-ℐ-closed which implies A is pre-ℐ-regular.(a) ⇒ (b): assume (A is pre-ℐ-open. Then pℐcl(A) is pre-ℐ-regular pre-ℐ-closed which implies X − pℐcl(A) is pre-ℐ-regular pre-ℐ-open. Hence X − pℐcl(A) is pre-ℐ-regular. Therefore, X − pℐcl(A) is pre-ℐ-closed and so pℐcl(A) is pre-ℐ-open. Hence  is extremally pre-ℐ-disconnected.(b) ⇒ (a): suppose X, τ, ℐ) be an extremally pre-ℐ-disconnected space and A ⊂ X. Then the following are equivalent:(a)A is pre-ℐ-regular,(b)A = pℐcl(pℐint(A)),(c)X − A is pre-ℐ-regular pre-ℐ-open,(d)A is pre-ℐ-regular pre-ℐ-open.Let (A is pre-ℐ-regular. Then A is pre-ℐ-open and pre-ℐ-closed and so A = pℐint(A) and A = pℐcl(A). Hence A = pℐcl(pℐint(A)).(a) ⇒ (b): suppose A = pℐcl(pℐint(A)). Then X − A = X − pℐcl(pℐint(A)) = pℐint(pℐcl(A)) so X − A is pre-ℐ-regular pre-ℐ-open.(b) ⇒ (c): let (c) ⇒ (d) is clear.(d) ⇒ (a) follows from X, τ, ℐ) is called locally pre-ℐ-indiscrete if every pre-ℐ-open subset of X is pre-ℐ-closed (or) if every pre-ℐ-closed subset of X is pre-ℐ-open.An ideal space  be an ideal space. Then the following are equivalent.(a)X, τ, ℐ) is locally pre-ℐ-indiscrete.((b)ℐ-open subset is pre-ℐ-regular.Every pre-(c)ℐ-open subset is pre-ℐ-regular pre-ℐ-open.Every pre-(d)pℐint(pℐcl({x})) ≠ ∅, for every x ∈ X.(e)X.The empty set is the only nowhere dense subset of Let  is locally pre-ℐ-indiscrete. Let A be a pre-ℐ-open subset of X. By hypothesis, A is pre-ℐ-closed. Hence A is pre-ℐ-regular.(a) ⇒ (b): assume that . Also by hypothesis, A is pre-ℐ-closed. Therefore, A = pℐint(pℐcl(A)). Hence A is pre-ℐ-regular pre-ℐ-open.(b) ⇒ (c): if x} is preopen, {x} is pre-ℐ-open. By (c), {x} is a pre-ℐ-regular pre-ℐ-open set. Therefore, {x} = pℐint(pℐcl({x})).(c) ⇒ (d): since {ℐ-regular pre-ℐ-open.(d) ⇒ (e): by A is a pre-ℐ-closed set. Now int(cl(A − pℐint(A))) = int(cl(A − (A∩int(cl⋆(A))))) = int(cl(A∩(X − A)∪(int(cl⋆(X − A))))) = int(cl(A∩(X − A)∪(A − int(cl⋆(A))))) = int(cl(A − int(cl⋆(A))))⊂int(cl(A) − int(cl⋆(A))) = int(cl(A) − cl(int(cl⋆(A))))⊂int(cl(A) − (int(cl⋆(A)))) = ∅. Therefore A − pℐint(A) is nowhere dense which implies A = pℐint(A), and so A is pre-ℐ-open. Hence is A locally pre-ℐ-indiscrete.(e) ⇒ (a): suppose that X, τ, ℐ) is said to be pℐR-door if every subset of X is either pre-ℐ-regular pre-ℐ-open or pre-ℐ-regular pre-ℐ-closed.An ideal space  be a pℐR-door space; then every pre-ℐ-open set in the space is pre-ℐ-regular pre-ℐ-open.Let ) which implies that pℐintA = pℐint(pℐint(pℐcl(A))). Since A is pre-ℐ-open, A = pℐint(pℐcl(A)). Hence A is pre-ℐ-regular pre-ℐ-open.Let X, τ, ℐ) be an ideal space. A subset A of X is both α-ℐ-open and α-ℐ-closed; then A is a pre-ℐ-regular pre-ℐ-open set.Let  be an ideal space. A subset A of X is α-ℐ-open and pre-ℐ-regular pre-ℐ-open; then A = int(cl⋆(int(A))).Let (A is α-ℐ-open. Then A ⊂ int(cl⋆(int(A))). And A is pre-ℐ-regular pre-ℐ-open which implies A = pℐint(pℐcl(A))⊃pℐint(cl⋆(int(pℐcl(A))))⊃pℐint(cl⋆(int(A)))⊃int(cl⋆(int(A))). Therefore A = int(cl⋆(A)).Suppose


Based on our decomposition of stochastic processes and our asymptotic representations of Fourier cosine coefficients, we deduce an asymptotic formula of approximation errors of hyperbolic cross truncations for bivariate stochastic Fourier cosine series. Moreover we propose a kind of Fourier cosine expansions with polynomials factors such that the corresponding Fourier cosine coefficients decay very fast. Although our research is in the setting of stochastic processes, our results are also new for deterministic functions. From this, we see that hyperbolic cross approximation with polynomial factors can reconstruct the stochastic process on 2 by using the least Fourier cosine coefficients.Finally, based on our decomposition of stochastic processes, we propose Fourier cosine expansions with polynomial factors (see ) whose hThis paper is organized as follows. In We recall some concepts in calculus of stochastic processes and stochastic Fourier cosine series.ξ, we denote its expectation, second-order moment, and variance by E[ξ], E[ξ2], and Var(ξ), respectively. If ξ(t) is a stochastic variable for each t ∈ d, then we say ξ(t) is a stochastic process on d. In this paper, we always assume that a stochastic process ξ(t) is real-valued and satisfies E[ξ2(t)] < ∞ for each t. This ensures that its expectation, variance, and second-order moment always exist. Calculus of stochastic processes is a generalization of classical calculus. Let {ξn}1∞ be a sequence of stochastic variables and let ξ be a stochastic variable. If limn→∞E[|ξn−ξ|2] = 0, we say that ξ is the limit of {ξn}1∞. Starting from the concept of the limit, one defines continuity, derivatives, partial derivatives, integrals, and double integrals of stochastic processes  and the derivative ξ′ is continuous on , thenξ(t) be a continuously differentiable stochastic process on  and let f(t) be a continuously differentiable deterministic function on . Then [For a stochastic variable rocesses . Moreove1]. Then (6)(ξ(t)tions in . Xiu revtions in .ξ(t) is a univariate stochastic process on  and E[∫01ξ2(t)dt] < ∞, then it can be expanded into the Fourier cosine seriesIf ξ is a bivariate stochastic process on 2 and Eξ2dt1dt2] < ∞, then it can be expanded into the Fourier cosine seriesIf ξ be a stochastic process on 2. Partial sums of its Fourier cosine series areSNr)((ξ) is N2.Let SNh)((ξ) is of order NlogN. The hyperbolic cross approximations have been widely used in multivariate function approximation ξ, denote its mixed derivative ∂l1+l2ξ/∂t1l1∂t2l2 by ξl1,l2) represents the set of continuous stochastic processes on 2. If ξl,l), then we say ξ has the smoothness index l on 2.For a bivariate stochastic process an} and {bn} be two sequences. If K1 | an | ≤|bn | ≤K2 | an| for any n, then we say an ~ bn; if |an | ≤K3 | bn| for any n, then we say an = O(bn); here Ki are constants independent of n. If an → 0, bn → 0, and an/bn → 0 as n → ∞, then we say an = o(bn). If an1,n2 → 0 and bn1,n2 → 0, and an1,n2/bn1,n2 → 0 as an1,n2 = o.Let {In order to study stochastic Fourier cosine series, we give a decomposition of stochastic processes. Although this decomposition is given in the setting of stochastic processes, it is also new for deterministic functions.ξ be a bivariate stochastic process on 2 and ξl,l) for some l ≥ 1. First, based on a fundamental polynomial p(t) = t2/2 − 1/6, we construct a stochastic polynomial Pξ as follows:ξ at vertexes {0,1}2.Let Denoteξ be a bivariate stochastic process on 2 with smoothness index l ≥ 1 and ξ1 be stated in  = ξ1p′(1) = ξ1  (0 ≤ t1 ≤ 1). Therefore, by ξ3 = ξ1 − ξ2, we get ξ3 = 0  . Similarly, ξ3 vanishes on other sides of the square 2.Consider the bottom side 0 ≤ 2 as follows.From  and 21)21), we gξ be a stochastic process on 2 and ξl,l) for some l ≥ 1. Then the decompositionPξ, ξ2, and ξ3 are stated in , a, aξ be atated in , respectξ is a stochastic process on  and the derivative ξl) for some l ≥ 2. Its Fourier cosine coefficients are as follows:E[ξ′′(t)] ∈ C, using the Riemann-Lebesgue lemma [rn) ≤ E[rn2], we have Var(rn) = o(1/n4). By  = ∑k=0N−1ck(ξ)cos(πnt) satisfyE[||SN(ξ)−ξ||22] = o(1/N3) if and only if ξ′(0) = ξ′(1) = 0.Suppose that ue lemma –12, we gn,wheregn=−2(nπ)follows:cn(ξ)=2∫0, we getE[rn2] =4ξ is a bivariate stochastic process on 2 and ξl,l) for some l ≥ 2. We expand ξ into the Fourier cosine seriesλn1,n2 is stated in , we will precisely compute the first two terms and estimate the expectation and variance of the last term on the right-hand side of cn1,n2(Pξ), by the representation  at {0,1}2 and signs for addition and subtraction are determined by odevity of n1 and n2. Since ∫01p(t)dt = 0, we get c0,0(Pξ) = cn1,0(Pξ) = cn20,(Pξ) = 0.(i) For the first term entation  of stochcn1,n2(ξ2), by (ξ2 is the sum of products of separated variables.(ii) For the second term ξ2), by , we knowξ2 are equal tocn(φi) is the Fourier cosine coefficient of the univariate stochastic process φi and each cn(p) and each cn(p(1 − ·)) are both Fourier cosine coefficients of univariate deterministic functions p and p(1 − ·). A direct computation shows that, for n1 ≠ 0 and n2 ≠ 0,Denotecn1,n2(ξ3), using the integration by parts, we deduce thatξ3 = ξ3 = 0  (0 ≤ t2 ≤ 1), the interior integral is equal toξ3 = 0  (0 ≤ t1 ≤ 1), we have ξ3 = 0  (0 ≤ t1 ≤ 1). This implies thatξ3 ∈ Cs that E[ξ3] is a continuous function on 2 and E[ξ3ξ3] is a continuous on 4. By the Riemann-Lebesgue lemma, we havecn1,n2(ξ3)) ≤ E.(iii) For the last term cn1,n2(ξ) satisfies , (ii), and (iii), we getatisfies , where ηtated in  and eachtated in .cn(φi).Now we further estimate these four univariate Fourier cosine coefficients φi be stated as in (cn(φi) satisfyLet each ed as in . Then Foi = 1, since φ1 = ξ1. By ξ1 = ξ − Pξ, we have φ1′(0) = φ1′(1) = 0. From this, we deduce that Fourier cosine coefficients satisfyBy similarity, we only prove the case We will deduce these asymptotic representations of Fourier cosine coefficients.ξ be a stochastic process on 2 and ξl,l) for some l ≥ 2, and let ηn1,n2 and φi be stated in ((i) for n1 → ∞, cn1,n2(ξ) = bn1,n2 + τn1,n2, where τn1,n2 satisfiesand (ii) for n2 → ∞,  and (iii) for n1 → ∞ and n2 → ∞, cn1,n2(ξ) = (4/π2n12n22)ηn1,n2 + τn1,n2* and τn1,n2* satisfiesLet tated in  and 38)ξ be a stn1 → ∞, denotecn1,n2(ξ) = bn1,n2 + τn1,n2. Using E = o(1/n12)(1/n22) andWhen (ξ3).By , cn1,n2From this, we can give asymptotic representations of expectation, second-order moment, and variance of Fourier cosine coefficients.(i) for n1 → ∞ and n2 ≠ 0, bn1,n2 is stated in  for n2 → ∞ and n1 ≠ 0, o” is uniform for n1.where tated in  and “o” Under conditions of By similarity, we only prove .n1 → ∞,   n2 ≠ 0,From o we get .l increases.From ξ be a stochastic process on 2 and ξl,l) for some l ≥ 2. We expand ξ into the Fourier cosine series:Let SNr)((ξ) of Fourier cosine series ((i)PN. The first term PN on the right-hand side of the formula  and  and , we haveK1 > 0 such thatn2=1∞E[(bn2(1))2 + (bn2(2))2] converges, and denote its sum by A. SoBy , we deduPk21 = (A/12π4k3) + o(1/k3). Notice thatE = O(1/n14n24) (see (Pk+121 = Pk21 + o(1/k3). This implies thatbn2(1) and bn2(2) are stated in  (see ). We havtated in .o” is uniform for n1 andE = O(1/n14n24), we have PN3 = o(1/N3). Finally, by QN and RN in cn1By (01p(t)dt = 0, we have cn1,0(Pξ) = 0. By (78) PN2=(ξ3).By  and ∫01 = 0. By , we haveSimilar to , we havegain, by , we haveplies by  that for some l ≥ 2. Then the partial sums SN(ξ) of its Fourier cosine series satisfyM = (1/6π4)(A + B + 4C + 4D) and A, B, C, D are stated in ,  for some l ≥ 2. We consider hyperbolic cross truncations of its Fourier cosine series:JN = ∑n2=1N−1∑n1=[(N−1)/n2]∞E. We rewrite JN in the formSuppose that tity and , 86), a, aξ is atity and , we haveJN1. By o” is uniform for n2 andμn2 = −(1/π2n22)  E[ω1(cn2(φ1) − cn2(φ2))] + E[(cn2(φ1) − cn2(φ2))2].We first consider E[(cn2(φ1) − cn2(φ2))2] ≤ 2E[cn22(φ1) + cn22(φ2)] = o(1/n24). Therefore, μn2 = o(1/n24). From this, we haven2 ≤ N,By using the Schwarz inequality in Probability theory,tituting  and 97)(96)|E ~ (logN/N3). This result cannot be improved as the smoothness index l increases.From SNh)((ξ) isNc ~ logN. From this, we getBy , the numSNr)((ξ) is Nc = N2, by Since the number of Fourier cosine coefficients in the partial sum ξ is a stochastic process on 2 and ξl,l) for some l ≥ 3. By using notation  into bivariate Fourier cosine series. Finally, we obtain the Fourier cosine expansion of ξ with polynomial factors:p(t) = (t2/2) − (1/6). Notice that φ1(t2) = ξ1 and ξ1 ∈ Cs, where ξ1 = ξ − Pξ. Using φ1′′′ ∈ Cs and φ1′(0) = φ1′(1) = 0, and soE[cn(φ1)] = o(1/n3) andcn(φ2),cn(φ3), and cn(φ4), we getIn order to use the least Fourier cosine coefficients to reconstruct stochastic processes, we introduce the Fourier cosine expansion with polynomial factors. Suppose that notation , the decnotation  can be rcn1,n2(h3) decay fast. By ξl,l)  (l ≥ 3) and Now we show that bivariate Fourier cosine coefficients SN(φi) is the first N terms partial sums of Fourier cosine series of φi and SNh)((ξ3) is the hyperbolic cross truncation of Fourier cosine series of bivariate stochastic process of ξ3 which is stated in (TNh)((ξ) is a combination of stochastic polynomials and cosine polynomials. By (n=1kan)2 ≤ k2∑n=1kan2, we getξ3 = ξ3 = ξ3 = ξ3 = 0. Based on (E[||TNh)((ξ)−ξ||22] = o((logN)/N5). Noticing that the number of Fourier cosine coefficients in TN : Nc ~ NlogN, we get the following.Take the hyperbolic cross truncations of the expansion ,117)TNtated in . The hypials. By  and  for some l ≥ 3. ThenNc is the number of Fourier cosine coefficients in TNh)((ξ).Suppose that Comparing ξ be a stochastic process on 2 and ξl,l) for some l ≥ 3. ThenSNr)((ξ) and SNh)((ξ) are partial sums and hyperbolic cross truncations of Fourier cosine series of ξ, respectively, TNh)((ξ) is the hyperbolic cross truncations of Fourier expansion of ξ with polynomial factors, and Nc is the number of Fourier cosine coefficients in each sum.Let ξ by TNh)((ξ), we need the least Fourier cosine coefficients.From this, we see that if we reconstruct


E-inversive semigroups. We prove that group fuzzy congruences and normal fuzzy subsemigroups determined each other in E-inversive semigroups. Moreover, we show that the set of group t-fuzzy congruences and the set of normal subsemigroups with tip t in a given E-inversive semigroup form two mutually isomorphic modular lattices for every t∈ .The aim of this paper is to investigate the lattices of group fuzzy congruences and normal fuzzy subsemigroups on  The investigation of fuzzy sets is initiated by Zadeh in . As specSeveral authors investigated fuzzy congruences for some special classes of semigroups. In 2000, Zhang  characteE-inversive semigroups is a very wide class of semigroups which contains groups, inverse semigroups, and regular semigroups as proper subclasses and some kinds of crisp congruences on this class of semigroups have been investigated extensively; see  the set of group t-fuzzy congruences and the set of normal fuzzy subsemigroups with tip t on S form two mutually isomorphic modular lattices. Our results generalize and enrich several results obtained in . A fuzzy set μ in a set S is said to be contained in a fuzzy set η if μ(x) ≤ η(x) for all x in S and this is denoted by μ⊆η. The union μ ∪ η and the intersection μ∩η of two fuzzy sets μ and η in a set S are defined byx in S. Further, if μi is a fuzzy subset in S for i ∈ I where I is an index set, then ⋂i∈Iμi is defined byx ∈ S.Zadeh  defined  semigroup is a nonempty set with an associative binary operation. A semigroup S is called E-inversive if for all a ∈ S there exists a′ ∈ S such that a′aa′ = a. In this case, we denote W(a) = {a′ ∈ S | a′aa′ = a′} and call the elements in W(a) the weak inverses of a for any a ∈ S. It is easy to see that groups, regular semigroups, and semigroups with zeros are all E-inversive semigroups. For more details on E-inversive semigroups, see . A t-fuzzy equivalence on S is a fuzzy subset in S × S which satisfies the following conditions: a ∈ S)  ρ = t,  ρ = ρ ≤ t,  ρ ≥ ρ∧ρ.a, b, c ∈ S)  ρ ≤ ρ∧ρ,a, b, c, d ∈ S)  ρ ≥ ρ∧ρ. = ρ for all x ∈ S.ρa = ρb if and only if ρ = t for all a, b ∈ S.S/ρ = {ρa | a ∈ S} is a semigroup with the multiplication ρaρb = ρab for any a, b ∈ S.Let S. In particular, we show that the set of group t-fuzzy congruences on S forms a modular lattice. Firstly, we give the concept of group t-fuzzy congruences which is parallel to that of usual group fuzzy congruences defined in Kuroki GFCt(S).A t-fuzzy congruences on S.The following result provides a characterization of group t-fuzzy congruence ρ on S is a group t-fuzzy congruence if and only if e, f ∈ E(S))  ρ = t;(∀a ∈ S)(∀a′ ∈ W(a))  ρ = t. . This implies that ρ = t for all e, f ∈ E(S) by a′a ∈ E(S) for any a ∈ S and a′ ∈ W(a), it follows that ρa′a is the identity of S/ρ and so ρa = ρaρa′a = ρaa′a. This yields that ρ = t by If a ∈ S, a′ ∈ W(a), and e ∈ E(S). Then by condition (1) and ρaa′ = ρa′a = ρe. By condition (2) and S/ρ is a group.Conversely, let ρ ∈ GFCt(S), a, b ∈ S, and e ∈ E(S). ThenLet ρ ∈ GFCt(S), by a′ ∈ W(a) and b′ ∈ W(b). This implies thatρ = ρ. By similar arguments, we can showa, b ∈ S and e ∈ E(S).Since σ, τ ∈ GFCt(S), we define σ∘τ as follows:a, b ∈ S. Then we have the following.As usual, for σ, τ ∈ GFCt(S). σ∘τ = τ∘σ.σ∘τ is the least group t-fuzzy congruence of  S containing σ and τ. σ∩τ is the greatest group t-fuzzy congruence of S contained in σ and τ.Let a, b ∈ S and c′ ∈ W(c), by σ∘τ = τ∘σ for all a, b ∈ S.(1) For all a, b, c ∈ S. Sinceσ∘τ = t. Similarly, we haveσ∘τ is the least t-fuzzy congruence on S containing σ and τ. Finally, we can easily show that σ∘τ ∈ GFCt(S) by σ∘τ is the least group t-fuzzy congruence of S containing σ and τ.(2) Let (3) This is clear.By GFCt(S), ⊆) forms a modular lattice for any t in . and a, b ∈ S. ρ = ρ for all a′, a* ∈ W(a). ρ(a′b(a′b)′, a′b) = ρ for all a′ ∈ W(a) and (a′b)′ ∈ W(a′b). Let ρ ∈ GFCt(S), by ρ ≥ ρ = t whence ρ = t. Thus,ρ = ρ.(1) Since a′b(a′b)′, aa′ ∈ E(S), by ρ(a′b(a′b)′, a′b) = ρ.(2) In view of the fact that E-inverse semigroups.In this section, we consider some basic properties of normal fuzzy subsemigroups of t ∈ . A fuzzy subset μ in S is called a normal  fuzzy  subsemigroup  with  tip  t in S if x, y ∈ S)  μ(xy) ≥ μ(x)∧μ(y),  t ≥ μ(a) ≥ μ(xay)∧μ(xy),(∀e ∈ E(S))  μ(e) = t.(∀Let t in S by NFSt(S) and let NFS(S) = ⋃t∈NFSt(S).We denote the set of normal fuzzy subsemigroups with tip In fact, normal fuzzy subsemigroups with tip 1 are introduced in Zhang  where thG be a group with identity e, t ∈  and let μ be a fuzzy set in G. From Ajmal and Thomas  and let μ be a fuzzy subset in G. Then μ is a normal fuzzy subsemigroup with tip t of G if and only if μ is a normal fuzzy subgroup of G with tip t.Let e is the unique idempotent in G and the inverse a−1 of a is certainly the unique weak inverse of a for all a ∈ G. If μ is a normal fuzzy subsemigroup with tip t of G, then by μ is a normal fuzzy subgroup of G with tip t. Conversely, let μ be a normal fuzzy subgroup of G with tip t and a, x, y ∈ S. Thenμ is a normal fuzzy subsemigroup of G with tip t.Observe that ondition . This imi∈Iσi ∈ NFSt(S) for σi ∈ NFSt(S), i ∈ I, and the elementNFSt(S), we have the following theorem by Since ⋂NFSt(S), ⊆) forms a lattice., e, f ∈ E(S), a, b ∈ S, and a′, a* ∈ W(a), b′ ∈ W(b).μ(ef) = t. μ(a′b) = μ(b′a). μ(a′b) = μ(a*b). Let e, f ∈ E(S), we have μ(e) = μ(f) = t. This implies that μ(ef) ≥ μ(e)∧μ(f) = t whence μ(ef) = t.(2)The result follows from the facts that(3)This follows from the proof of Theorem 2.12 in Zhang .(1) Since GFCt(S) is isomorphic to NFSt(S) as lattices whence NFSt(S) is modular for all t in . We first give some useful propositions.In this section, we show that μ ∈ NFSt(S) andρμ ∈ GFCt(S), where a′ ∈ W(a).Let ρμ is well defined. Now, let a, b, c ∈ S and a′ ∈ W(a), b′ ∈ W(b), c′ ∈ W(c). Then we have the following facts:(1)a′a ∈ E(S), ρμ = μ(a′a) = t.Since (2)By In view of b′b ∈ E(S), it follows that μ(b′b) = t and(3) Since ac)′ ∈ W(ac), we have ac(ac)′ ∈ E(S) andμ(a′b) ≤ μ((ac)′bc). Dually, we have μ(a′b) ≤ μ((ca)′cb). This implies thatρμ ≤ ρμ∧ρμ.(5)e ∈ W(e), we have ρμ = μ(ef) = t.By (6)ρμ = μ(a′aa′a) = μ(aa′) = t.From the above six items, we can see that ρμ ∈ GFCt(S) by (4) For any (ρ ∈ GFCt(S) and defineμρ ∈ NFSt(S), where a′ ∈ W(a).Let μρ is well defined. Now, letμρ(xy) ≥ μρ(x)∧μρ(y). On the other hand, also by t ≥ ρ = μρ(a) ≥ μρ(xay)∧μρ(xy). Finally, since e ∈ W(e) for all e ∈ E(S), we have μρ(e) = ρ = ρ = t for all e ∈ E(S). Therefore, μρ ∈ NFSt(S).By At this stage, we can give the main result of this paper.μρ and ρμ are defined as in Propositions The mappingsμ ∈ NFSt(S), a ∈ S, and a′ ∈ W(a). Then aa′ ∈ W(aa′). This implies thata, b ∈ S and a′ ∈ W(a), (a′b)′ ∈ W(a′b), we havet and Φt are mutually inverse bijections. Obviously, Ψt and Φt preserve the inclusion relations.From Propositions GFCt(S), ⊆) and (NFSt(S), ⊆) are isomorphic. As a consequence, the lattice (NFSt(S), ⊆) is also modular.The lattices (GFCt(S), ⊆) is isomorphic to the lattice (NFSt(S), ⊆). Thus, the lattice (NFSt(S), ≤) is also modular by By GFC(S), ⊆) and (NFS(S), ⊆) are two mutually isomorphic lattices.(ρi ∈ GFCti(S), i ∈ I, and σj ∈ NFStj(S), j ∈ J, where I and J are index sets. Moreover,GFC(S), ⊆) and (NFS(S), ⊆), respectively. By GFC(S), ⊆) and (NFS(S), ⊆) are two lattices. Moreover, if we letGFC(S), ⊆) and (NFS(S), ⊆) are isomorphic from It is routine to check thatWe end this section by giving an example to illustrate our previous results.S be a semigroup with the following multiplication table:S is an E-inversive semigroup which is nonregular. Moreover,Let t in the interval . For every s ∈  with s ≤ t, define a fuzzy set μs in S as follows:μs is a normal fuzzy subsemigroup with tip t in S. Furthermore, in view of the fact thatμs ∈ NFSt(S). By ρμs satisfies thatρμs ∈ GFCt(S). By GFCt(S) = {ρμs | s ≤ t, s ∈ }. By virtue of GFCt(S), ⊆) and (NFSt(S), ⊆) are isomorphically modular lattices.Fix an element E-inversive semigroups. Our results generalize the corresponding results of groups and regular semigroups. From the results presented in the paper, the lattices of group fuzzy congruences and normal fuzzy subsemigroups on E-inversive semigroups can be regarded as a source of possibly new modular lattices. On the other hand, this paper also leaves some questions which can be considered as future works. For example, from Ajmal and Thomas [S is a group, then (NFS(S), ⊆) is a modular lattice. Thus, the following question would be interesting: is (NFS(S), ⊆) also modular for an E-inversive semigroup S?In this paper, we have introduced and investigated the lattices of group fuzzy congruences and normal fuzzy subsemigroups on d Thomas , if S is


In addition, there is an error in a should read “µg/cm-2.” Please see the correct There is an error in