The amount of functional genomic information has been growing rapidly but remains largely unused in genomic selection. Genomic prediction and estimation using haplotypes in genome regions with functional elements such as all genes of the genome can be an approach to integrate functional and structural genomic information for genomic selection. Towards this goal, this article develops a new haplotype approach for genomic prediction and estimation.A multi-allelic haplotype model treating each haplotype as an ‘allele’ was developed for genomic prediction and estimation based on the partition of a multi-allelic genotypic value into additive and dominance values. Each additive value is expressed as a function of h − 1 additive effects, where h = number of alleles or haplotypes, and each dominance value is expressed as a function of h(h − 1)/2 dominance effects. For a sample of q individuals, the limit number of effects is 2q − 1 for additive effects and is the number of heterozygous genotypes for dominance effects. Additive values are factorized as a product between the additive model matrix and the h − 1 additive effects, and dominance values are factorized as a product between the dominance model matrix and the h(h − 1)/2 dominance effects. Genomic additive relationship matrix is defined as a function of the haplotype model matrix for additive effects, and genomic dominance relationship matrix is defined as a function of the haplotype model matrix for dominance effects. Based on these results, a mixed model implementation for genomic prediction and variance component estimation that jointly use haplotypes and single markers is established, including two computing strategies for genomic prediction and variance component estimation with identical results.The multi-allelic genetic partition fills a theoretical gap in genetic partition by providing general formulations for partitioning multi-allelic genotypic values and provides a haplotype method based on the quantitative genetics model towards the utilization of functional and structural genomic information for genomic prediction and estimation. With the factorization of a = Wαα and d = Wδδ, genomic additive relationship is a function of WαWα ' and genomic dominance relationship is a function of WδWδ '  × 1 column vector of 1’s, ah = Wαhαh = [h(h + 1)/2] × 1 column vector of additive values , dh = Wδhδh = [h(h + 1)/2] × 1 column vector of dominance values (dominance deviations), Wαh = [h(h + 1)/2] × (h − 1) model matrix of αhwith wαij,k defined by Eqs. h = [h(h + 1)/2] × 1 column vector of dominance values (dominance deviations), Wδh = [h(h + 1)/2] × [h(h − 1)/2] matrix of δh with wδij,kf defined by Eqs. αh = (h − 1) × 1 column vector with αlk defined by Eq. δh = [h(h − 1)/2] × 1 column vector with δkf defined by Eq. For convenience of computer programming, Eqs. lele Eq. , or 1 colele Eq.  or 1 comlele Eq. . SimilarA hypothetical numerical example is used to illustrate the genetic partition of multi-allelic genotypic values described by Eqs. Using Eqs. Using Eqs. The genotypic values calculated as the summation of the additive and dominance values are:g = 1μ + ah + dh = 1μ + Wαhαh + Wδhδh described by Eqs. ij = gji, aij = aji and dij = dji, the genotypic variance (σg2), additive variance (σa2) and dominance variance (σd2) are:By comparing with the genotypic values in Table g2 = σa2 + σd2.It is readily seen that σWαh, Wδh, ah, dh, αh and δh in Eq. th haplotype block has hi haplotypes, hi−1 additive effects, and nδi dominance effects or heterozygous genotypes. Let nα = total number of additive effects of all r haplotype blocks, nδ = total number of dominance effects (or heterozygous genotypes) of all r haplotype blocks. Then, nα = ∑i = 1rhi − r, and nδ = ∑i = 1rnδi. For a given sample of q individuals, the limit number of effects is 2q-1 for additive effects and is the number of heterozygous genotypes for dominance effects. For a sample with N observations on q individuals, the mixed model to implement the multi-allelic haplotype model of Eq. Z = N × q incidence matrix allocating phenotypic observations to each individual = identity matrix for one observation per individual (N = q), αh = nα × 1 column vector of haplotype additive effects, Wαh = q × nα model matrix of αh, δh = nδ × 1 column vector for dominance effects of haplotype genotypes, Wδh = q × nδ model matrix of δh, αs = m × 1 column vector of single-SNP additive effects, b = c × 1 column vector of fixed effects such as heard-year-season in dairy cattle (c = number of fixed effects), and X = N × c model matrix of b. To define two equivalent models with complementary computing advantages and identical GBLUP and GREML results, the mixed model of Eq. ah = Tαhαh = multi-allelic genomic breeding values, dh = Tδhδh = multi-allelic genomic dominance values, and each T matrix can be defined by any of the six definitions of genomic relationships we previously discussed and implemented [T matrices are defined as: Tαh = Wαh/kαh1/2, Tδh = Wδh/kδh1/2, where kαh = the average of diagonal elements of WαhWαh ', and kδh = the average of diagonal elements of WδhWδh '. The genomic relationship matrices of Eq. A mixed model to implement the multi-allelic haplotype model of Eq. The multi-allelic genomic relationships of Eqs. It is important to distinguish between single-locus multi-allelic markers such as microsatellite markers from haplotypes where each haplotype is treated as an ‘allele’ and each haplotype block is treated as a ‘locus’, because recombination between loci within a haplotype block generally exists, leading to lowered haplotype similarity than single-locus similarity among relatives. As the number of loci increases in each haplotype block, genomic relationships using haplotypes are expected to decrease from those using single-locus markers. Therefore, the utility of haplotype genomic relationships using Eqs. y) = Xb, E(αh) = E(δh) = E(αs) = E(δs) = 0, Var(αh) = σαh2Inα, Var(ah) = Gαh = σαh2Ah, Var(δh) = σδh2Inδ, Var(dh) = Gδh = σδh2Dh, and Var(e) = R = σe2IN, where σαh2 = variance of multi-allelic additive effects, σδh2 = variance of multi-allelic dominance effects, σe2 = residual variance, and Inα, Inδ, Im and IN are identity matrices of orders nα, nδ, m and N, respectively. All random effects are assumed to be uncorrelated so that the phenotypic variance-covariance matrix is:To establish mixed models using multi-allelic markers or haplotypes, assumptions for the first and second moments of the mixed model of Eq. αh = τ1, δh = τ2; Tαh = T1, Tδh = T2; ui = Tiτi, i = 1,2; Ah = S1, Dh = S2; and σαh2 = σ12, σδh2 = σ22. Then, Eqs. To simply notations for the two equivalent mixed models, terms in Eqs. Zi = ZTi, an equivalent model of Eqs. By defining y) and V, but these two models have different computational advantages that can be complementary to each other. For each model, two methods can be established for genomic prediction and estimation: the method of conditional expectation (CE) and the method of mixed model equations (MME), yielding a total of four methods for the two equivalent models. Model-I using CE is the best method for large numbers of SNP markers and multiple genetic factors, Model-II using MME is the best method for large numbers of individuals, and Model-I using MME and Model-II using CE have no computing advantage. Therefore, Model-I using CE and Model-II using MME will be used for genomic prediction and estimation. Using our previous naming of these two methods, GBLUP and GREML of Model-I using CE will be referred to as the CE set of formulations, and GBLUP and GREML of Model-II using MME as the QM set of formulation, where QM means ‘q > m’. These two methods yield identical results of prediction and estimation and are applicable to singular genomic relationship matrices. Assuming one observation per individual, CE based on Eqs. α + nδ according to the size of the largest matrix to invert for each method −X ' V− 1, and th type of genetic values on the phenotypic values in the training population. Two equivalent methods with identical results can be used to predict genetic values of individuals without phenotypic observations : placing all individuals with or without records in the same mixed model by setting to zero the Z matrix for the validation population, or calculate predictions separately based on the regressed phenotypic values of the training population [th type of genetic values for individuals in the validation population is calculated as:Si01 = q0 × q genomic relationship matrix between the training and validation populations for the ith type of genetic values .Using the CE method of Model-I Eqs.  and 37, pulation , 39. UsiZg = , λi = σe2/σi2, t = nα, nδ, m and N for i = 1,2, respectively, and ⊕ denotes direct sum that defines a block diagonal matrix. With haplotype and SNP effects from Eq. th type of genetic values for individuals in the training and validation populations are obtained as:Ti0 = the Ti matrix calculated using SNPs of the validation population. Equations ĝ = ∑i = 12ûi = predicted genotypic values of all individuals, and ĝ0 = ∑i = 12ûi0 = predicted genotypic values of the validation population. Prediction reliabilities of additive, dominance and genotypic predictions as the squared correlations between the genomic and true values has the same formulations as the Rai2, Rdi2 and Rgi2 formulae in [Using the QM method −X ', and ti = nα for i = 1 and ti = nδ for i = 2.Using the CE method of Model-I Eqs.  and 37, hod Eqs.  and 39, The EM-REML of Eqs. αs = m × 1 column vector of SNP additive effects, Tαs = q × m model matrix of αs, δs = m × 1 column vector of SNP dominance effects, Tδs = q × m model matrix of δs, Var(αs) = σαs2Im, Var(as) = Gαs = σαs2As, Var(δs) = σδs2Im, Var(ds) = Gδs = σδs2Ds, As = genomic additive relationship matrix, and Ds = SNP genomic dominance relationship matrix, and where As = TαsTαs ' and Ds = TδsTδs '. Let αs = τ3, δ = τ4; ui = Tiτi, i = 1,…,4; As = S3, Dh = S4; and σαs2 = σ32, σδs2 = σ42. The GBLUP and GREML formulations to jointly include haplotype and single SNP additive and dominance effects essentially entails to extending the range of the subscript i from 2 to 4 for Eqs. Haplotype analysis and single SNP analysis can be analyzed jointly for genomic prediction in the same mixed model by adding single SNP effects from our previous work  to the mi2 = σi2/σy2, where σy2 = ∑i = 14σi2 + σe2 = phenotypic variance. The total heritability of all types of genetic effects is the summation of all effect heritabilities, i.e., H2 = ∑i = 14hi2. Genomic heritability estimation has flexibility unavailable from heritability estimation using pedigree relationships: the heritability estimation for a single SNP, a chromosome region, or a set of selected SNPs. Using the GREML formulae of Eqs. GREML estimation using the joint mixed model with haplotype and SNP effects offer flexibility to estimate the heritability for various types of functional genomic information in any given autosome regions based on formulations we implemented in GVCBLUP , e.g., tThe multi-allelic haplotype model can be used for the integration of functional genomic information with genomic prediction and estimation. This integration defines haplotype blocks using functional genomic information under the hypothesis that a chromosome region with functional information required more than a single point to affect a phenotype, followed by genomic prediction and estimation using a haplotype analysis such as the methods developed in this article. Each gene could be a ‘natural haplotype block’ and the use of gene blocks improved the prediction accuracy for some human phenotypes in our preliminary results . Other tThe mixed model approach outlined above allows rare haplotypes. In the extreme case of rare haplotypes with one observation per haplotype or haplotype frequency of 1/h when h is large, the multi-allelic model with the mixed model implementation still is applicable for additive effects and values. Missing genotypic values is a problem for dominance effects and values. The dominance effect defined by Eq. A multi-allelic haplotype model for genomic prediction and estimation is established using the quantitative genetics model that partitions a multi-allelic genotypic value into additive and dominance values, factorizes each additive value into a product between a function of allele frequencies and additive effect, and factorizes each dominance value into a product between a function of allele frequencies and dominance effect. Haplotype genomic additive and dominance relationship matrices and formulations are then derived for GBLUP and GREML utilizing haplotypes in haplotype blocks. These results fill a gap in the theory of quantitative genetics for multi-allelic genetic partition and provide a haplotype approach within the theory of quantitative genetics towards the integration of functional and structural genomic information for genomic selection.The only data set used in this article is shown in Tables


The concept of Γ-semihyperrings was introduced by Dehkordi and Davvaz as a generalization of semirings, semihyperrings, and Γ-semiring. In this paper, by using the notion of triangular norms, we define the concept of triangular fuzzy sub-Γ-semihyperrings as well as triangular fuzzy Γ-hyperideals of a Γ-semihyperring, and we study a few results in this respect. M be an additive group whose elements are denoted by a, b, c,… and Γ another additive group whose elements are γ, β, α,…. Suppose that aγb is defined to be an element of M and that γaβ is defined to be an element of Γ for every a, b, γ, and β. If the products satisfy the following three conditions: (1)(a1 + a2)γb = a1γb + a2γb, a(γ1 + γ2)b = aγ1b + aγ2b, aγ(b1 + b2) = aγb1 + aγb2; (2)(aγb)βc = aγ(bβc) = a(γbβ)c; (3) if aγb = 0 for any a and b in M, then γ = 0; then M is called a Γ-ring in the sense of Nobusawa  which associates with each point x ∈ X its grade or degree of membership μμ(x) ∈ . Fuzzy sets generalize classical sets since the characteristic functions of classical sets are special cases of the membership functions of fuzzy sets, if the latter only take values 0 or 1. After the introduction of fuzzy sets by Zadeh, reconsideration of the concept of classical mathematics began. In 1971, Rosenfeld  ×  →  satisfying the following conditions: (T1) T = T; (T2) T) = T, z); (T3) T ≤ T if y ≤ z; (T4) T = x, for all x, y, z ∈ . For every t-norm T, we set ΔT = {x ∈ , T = x}. A t-norm on  is called a continuous t-norm if T is a continuous function from  ×  to  with respect to the usual topology. Note that the function “Min” is a continuous t-norm. A triangular conorm (t-norm for short) is a binary operation S on the unit interval , that is, a function S : ×→, which for all x, y, z ∈  satisfies (T1)–(T3) and (S4) S = x. From an axiomatic point of view, t-norms and t-conorms differ only with respect to their respective boundary conditions. In fact, the concepts of t-norms and t-conorms are dual in some sense. Anthony and Sherwood . Then, for any x, y ∈ R and γ ∈ Γ, we havex, y ∈ Γ,x ∉ Γ and y ∈ Γ ,x, y ∉ Γ. Regarding the above cases, we have T = 9/17, T = 15/40, and T = 25/81. Thus, in every case, we obtainμ is a T-fuzzy sub-Γ-semihyperring of R.Suppose that R be a Γ-semihyperring, T be a t-norm, and μ be a T-fuzzy sub-Γ-semihyperring of S. Thenx1,…, xn ∈ R and γ ∈ Γ, whereLet The proof is straightforward by mathematical induction.R be a Γ-semihyperring, T be a t-norm, and μ be a T-fuzzy sub-Γ-semihyperring of S. Let A and B be nonempty subsets of R. Thenγ ∈ Γ.Let The proof is straightforward.R be a Γ-semihyperring, T be a t-norm, and μ be a fuzzy subset of S with imaginable property and b the maximum of Im μ. Then, the following two statements are equivalent:μ is a T-fuzzy sub-Γ-semihyperring of S,μ−1 is a sub-Γ-semihyperring of S whenever a ∈ ΔT and 0 < a ≤ b.Let a ∈ ΔT and 0 < a ≤ b. If x, y ∈ μ−1, then infz∈x+y{μ(z)} ≥ T(μ(x), μ(y)) ≥ T = a, which implies that x + y⊆μ−1. Similarly, assume that a ∈ ΔT and 0 < a ≤ b. If x, y ∈ μ−1 and γ ∈ Γ, then infz∈xγy{μ(z)} ≥ T(μ(x), μ(y)) ≥ T = a. Then, we have xγy⊆μ−1, and so μ−1 is a sub-Γ-semihyperring of R.(1)⇒(2): Suppose that x, y ∈ S and γ ∈ Γ. Since Im μ⊆ΔT, both μ(x) and μ(y) are in ΔT. Now, we haveT(μ(x), μ(y)) ∈ ΔT. Assume that a = T(μ(x), μ(y)). If a = 0, thena = T(μ(x), μ(y)) ≤ μ(x)∧μ(y) ≤ μ(x) ≤ b. Hence x, y ∈ μ−1, which implies that x + y⊆μ−1 and xγy⊆μ−1. Therefore T(μ(x), μ(y)) ≤ infz∈x+y{μ(z)} and T(μ(x), μ(y)) ≤ infz∈xγy{μ(z)}.(2)⇒(1): Suppose that R be a Γ-semihyperring, T be a t-norm, and μ be a fuzzy subset of R. Then(1)μ is called a T-fuzzy left Γ-hyperideal of R if(2)μ is called a T-fuzzy right Γ-hyperideal of R if(3)μ is called a T-fuzzy Γ-hyperideal of R if it is both a T-fuzzy left Γ-hyperideal and a T-fuzzy right Γ-hyperideal of R.Let R be a Γ-semihyperring, T be a t-norm, and μ be a fuzzy subset of S with imaginable property and b the maximum of Im μ. Then, the following two statements are equivalent:μ is a T-fuzzy Γ-hyperideal of R,μ−1 is a Γ-hyperideal of R whenever a ∈ ΔT and 0 < a ≤ b.Let The proof is similar to the proof of μ be a fuzzy subset of R and t ∈ . The set U = {x ∈ R | μ(x) ≥ t} is called a level subset of μ. So, we obtain the following corollary.Let R be a Γ-semihyperring and μ be a fuzzy subset of R. Thenμ is a Min-fuzzy sub-Γ-semihyperring of R if and only if every nonempty level subset is a sub-Γ-semihyperring of R;μ is a Min-fuzzy Γ-hyperideal of R if and only if every nonempty level subset is a Γ-hyperideal of R.Let A be a subset of R. ThenχA is a T-fuzzy sub-Γ-semihyperring of R if and only if A is a sub-Γ-semihyperring of R;the characteristic function χA is a T-fuzzy Γ-hyperideal of R if and only if A is a Γ-hyperideal of R.the characteristic function Let R be a Γ-semihyperring and K be a sub-Γ-semihyperring of R. Let T be the t-norm defined by T = max{0, a + b − 1} and μ be a fuzzy subset of R defined bya, b ∈  and x ∈ R, where r, s ∈  such that s < r. Then, μ is a T-fuzzy sub-Γ-semihyperring of R. In particular, if r = 1 and s = 0, then μ is imaginable.Let The proof is similar to the proof of Theorem  2.6 in .R1 and R2 be Γ1 and Γ2-semihyperrings, respectively. If there exists a map φ : R1 → R2 and a bijection f : Γ1 → Γ2 such thatx, y ∈ R1 and γ ∈ Γ, then we say  is a homomorphism from R1 to R2. Also, if φ is a bijection then  is called an isomorphism and R1 and R2 are isomorphic.Let R1 and R2 be Γ1 and Γ2-semihyperrings, respectively. Let  be a homomorphism from R1 to R2. If λ is a T-fuzzy sub-Γ-semihyperring of R2, then φ−1(λ) is a T-fuzzy sub-Γ-semihyperring of R1 too.Let x, y ∈ R1 and γ ∈ Γ. Then, we haveφ−1(λ) is a T-fuzzy sub-Γ-semihyperring of R1.Suppose that R1 and R2 be Γ1 and Γ2-semihyperrings, respectively. Let  be a homomorphism from R1 to R2. If λ is a T-fuzzy Γ-hyperideal of R2, then φ−1(λ) is a T-fuzzy Γ-hyperideal of R1 too.Let The proof is similar to the proof of ai}i∈I and {bj}j∈J be two sets of real numbers in . Then, we say T is infinitely distributive ifT is continuous, then T is infinitely distributive , we have φ(μ)t = ⋂t>ϵ>0φ(μt−ϵ).Let The proof is similar to the proof of Lemma  3.5 in .R1 and R2 be Γ1 and Γ2-semihyperrings, respectively, and let μ be a fuzzy subset of R1. Let  be an onto homomorphism from R1 to R2. If μ is a Min-fuzzy sub-Γ-semihyperring of R1, then φ(μ) is a Min-fuzzy sub-Γ-semihyperring of R2 too.Let μ is a Max-fuzzy sub-Γ-semihyperring of R1. By φ(μ) is a Max-fuzzy sub-Γ-semihyperring of R2 if every nonempty level subset U(φ((μ); t)) is a sub-Γ-semihyperring of R2. Thus, assume that U(φ((μ); t)) is any nonempty level subset. If t = 0, then U(φ((μ); t)) = R2, and if 0 < t ≤ 1, then by U(φ((μ); t)) = ⋂t>ϵ>0φ). By U for each t > ϵ > 0 is a sub-Γ-semihyperring of R1. Hence, φ) is a sub-Γ-semihyperring of R2. By φ) being an intersection of a family of sub-Γ-semihyperrings is also a sub-Γ-semihyperring of R2 and the proof is completed.Suppose that R1 and R2 be two Γ-semihyperrings and let μ and λ be fuzzy subsets of R1 and R2, respectively. The product of μ and λ is defined to be the fuzzy subset μ × λ of R1 × R2 with (μ × λ) = T(μ(x), λ(x)), for all  ∈ R1 × R2.Let R1 and R2 be two Γ-semihyperrings and let μ and λ be fuzzy subsets of R1 and R2, respectively. Thenif μ and λ are T-fuzzy sub-Γ-semihyperrings of R1 and R2, respectively, then μ × λ is a T-fuzzy sub-Γ-semihyperring of R1 × R2; if μ and λ are T-fuzzy Γ-hyperideals of R1 and R2, respectively, then μ × λ is a T-fuzzy Γ-hyperideal of R1 × R2. Let It is straightforward.A of subsets of a Γ-semihyperring R satisfies the ascending chain condition (or Acc) if there does not exist a properly ascending infinite chain A1 ⊂ A2 ⊂ ⋯ of subsets from A. Recall that a subset B ∈ A is a maximal element of A if there does not exist a subset in A that properly contains B. Similar to  for every n ∈ Z, R = ⋃n∈ZAn, and Γ = Z. Then, R is a Noetherian Γ-semihyperring with respect to the following hyperoperations:x ∈ An and y ∈ Am.Let Ak | k ∈ N} be a family of Γ-hyperideals of a Γ-semihyperring R, where A1⊃A2⊃A3⋯. Let μ be a fuzzy subset of R defined byx ∈ R, where A0 stands for R. Let T be a t-norm with Im μ⊆ΔT. Then, μ is a T-fuzzy Γ-hyperideal of R.Let {x, y ∈ R. Suppose that x ∈ Ak∖Ak+1 and y ∈ Ar∖Ar+1 for k = 0,1, 2,… and r = 0,1, 2,…. Without loss of generality we may assume that k ≤ r. Then, obviously y ∈ Ak. Since Ak is a Γ-hyperideal of R, it follows that x + y⊆Ak and xαy⊆Ak which imply that infz∈x+yμ(z) ≥ k/(k + 1) = T(μ(x), μ(y)) and infz∈xαyμ(z) ≥ k/(k + 1) = μ(y) for all α ∈ Γ. If x ∈ ⋂k=0∞Ak and y ∈ ⋂k=0∞Ak, then x + y⊆⋂k=0∞Ak. Hence, infz∈x+yμ(z) = 1 = T(μ(x), μ(y)) and infz∈xαyμ(z) = 1 ≥ μ(y). If x ∉ ⋂k=0∞Ak and y ∈ ⋂k=0∞Ak, then there exists n ∈ N such that x ∈ An∖An+1. It follows that x + y⊆An which implies that infz∈x+yμ(z) ≥ n/(n + 1) = T(μ(x), μ(y)) and infz∈xαyμ(z) ≥ n/(n + 1) = μ(y) for all α ∈ Γ. Finally, assume that x ∈ ⋂k=0∞Ak and y ∉ ⋂k=0∞Ak, then y ∈ An∖An+1 for some n ∈ N. Therefore, x + y⊆An, and thus infz∈x+yμ(z) ≥ n/(n + 1) = T(μ(x), μ(y)) and infz∈xαyμ(z) ≥ n/(n + 1) = μ(y), for all α ∈ Γ. Hence, μ is a T-fuzzy Γ-hyperideal of R.Let R be a Γ-semiring satisfying descending chain condition, let μ be a fuzzy subset of R, and let T be a t-norm with Im μ⊆ΔT. Let μ be a T-fuzzy Γ-hyperideal of R. If a sequence of elements of Im μ is strictly increasing, then μ has finite number of values.Let tk} be a strictly increasing sequence of elements of Imμ. Then 0 ≤ t1 ≤ t2 ≤ ⋯≤1. Then, U is an ideal of M for all r = 2,3,… Let x ∈ U. Then μ(x) ≥ tr ≥ tr−1, and so x ∈ U. Hence U⊆U. Since tr−1 ∈ Imμ, there exists xr−1 ∈ M such that μ(xr−1) = tr−1. It follows that xr−1 ∈ U but xr−1 ∉ U. Thus, U ⊂ U and so we obtain a strictly descending sequence U⊃U⊃U⊃⋯ of Γ-hyperideals of R which is not terminating. This contradicts the assumption that M satisfies the descending chain condition. Consequently, μ has finite number of values.Let {R be a Γ-semiring, μ be a fuzzy subset of R, and T be a t-norm with Im μ⊆ΔT. Then, the following conditions are equivalent: R is a Noetherian Γ-semihyperring,R is a well-ordered subset of .the set of values of any Γ-hyperideal of Let μ be a T-fuzzy Γ-hyperideal of R. Suppose that the set of values of μ is not a well-ordered subset of . Then, there exists a strictly decreasing sequence {tk} such that μ(xk) = tk. It follows that U ⊂ U ⊂ ⋯ is a strictly ascending chain of Γ-hyperideals of M, where U = {x ∈ M | μ(x) ≥ tr}, for every r = 1,2,…. This contradicts the assumption that R is a Noetherian Γ-semihyperring.(1)⇒(2): Let R is not a Noetherian Γ-hyperring. There exists a strictly ascending chainR. Note that A = ⋂k=0∞Ak is a Γ-hyperideal of R. Define a fuzzy subset in R byμ is a T-fuzzy Γ-hyperideal of R. Let x, y ∈ R. If x ∈ Ak∖Ak−1 and y ∈ Ak∖Ak−1, then x − y ⊂ Ak and xαy ⊂ Ak. It follows that infz∈x+yμ(z) = 1/k = T(μ(x), μ(y)) and infz∈xαyμ(z) ≥ 1/k ≥ μ(y), for all β ∈ Γ. Suppose that x ∈ Ak and y ∈ Ak∖Ar for all r < k. Since Ak is a Γ-hyperideal of R, it follows that x + y ⊂ Ak. Hence, infz∈x+yμ(z) ≥ 1/k ≥ 1/(k + 1) ≥ μ(y) and so infz∈x+yμ(z) ≥ min{μ(x), μ(y)} ≥ T(μ(x), μ(y)). Similarly, for the case x ∈ Ak∖Ar and y ∈ Ak, we have infz∈x+yμ(z) ≥ min{μ(x), μ(y)} ≥ T(μ(x), μ(y)) and infz∈xαyμ(z) ≥ μ(y), for all β ∈ Γ. Thus, μ is a T-fuzzy Γ-hyperideal of R. Since the chain is not terminating, μ has a strictly descending sequence of values. This contradicts the assumption that the value set of any ideal is well-ordered. Therefore, R is a Noetherian Γ-semihyperring.(2)⇒(1): Suppose that the condition (2) is satisfied and μα | α ∈ Λ} of fuzzy subsets in R, we define the join ∨α∈Λμα and the meet ∨α∈Λμα as follows:x ∈ R, where Λ is any index set.For a family {T-fuzzy Γ-hyperideals in R is a completely distributive lattice with respect to meet “∧” and join “∨”.The family of α∈Λμα and ∧α∈Λμα are T-fuzzy Γ-hyperideals of R for family {μα | α ∈ Λ} of T-fuzzy Γ-hyperideals of R. For any x, y ∈ R, we haveSince  is a completely distributive lattice with respect to the usual ordering in , it is sufficient to show that ∨x, y ∈ M and β ∈ Γ. Thenα∈Λμα and ∧α∈Λμα are T-fuzzy Γ-hyperideals of R. This completes the proof.Now, let


The proof of the main result is based upon a fixed point theorem of a sum operator. It is expected in this paper not only to establish existence and uniqueness of positive solution, but also to show a way to construct a series to approximate it by iteration.It is expected in this paper to investigate the existence and uniqueness of positive solution for the following difference equation: −Δ T > 1 be an integer; ℤa,b : = {a, a + 1,…, b}, where a, b are positive integers. Difference equation appears as natural descriptions of observed evolution phenomena because most measurements of time evolving variables are discrete and so it arises in many physical problems, as nonlinear elasticity theory or mechanics, and engineering topics. In recent years, the study of positive solutions for discrete boundary value problems has attracted considerable attention, but most research dealt with two-point boundary value problem; see [Let lem; see  and the lem; see –7. Howevα < 1, β > 0, and η ∈ ℤT−12,.In this paper, we consider the existence uniqueness and positive solutions for difference equation E = C denote the class of real valued functions ω on ℤT+10, with norm ||ω|| = maxk∈ℤT+10, | ω(k)|. Observe that E is a Banach space. Set P = {u ∈ E : u(t) ≥ 0, t ∈ ℤT+10,} to be the normal cone in E with the normality constant 1. For u, v ∈ E, the notation u ~ v means that there exist λ > 0 and μ > 0 such that λv ≤ u ≤ μv. Clearly, ~ is an equivalence relation. Given h > θ, we denote by Ph the set Ph = {x ∈ E∣x ~ h}.Let ℤT+10, with the discrete topology, every ω ∈ C is continuous.As suggested by the notation, by equipping D = P or D = P0 and let γ be a real number with 0 ≤ γ < 1. An operator A : P → P is said to be γ-concave if it satisfies Let A : E → E is said to be homogeneous if it satisfies A : P → P is said to be subhomogeneous if it satisfies An operator  The main tool of this paper is the following fixed point theorem.P be a normal cone in a real Banach space E, A : P → P an increasing γ-concave operator, and B : P → P an increasing subhomogeneous operator. Assume that (i)h > 0 such that Ah ∈ Ph and Bh ∈ Ph;there is  (ii)δ0 > 0 such that, for any x ∈ P, Ax ≥ δ0Bx.there exists a constant Then the operator equation Ax + Bx = x has a unique solution x* ∈ Ph. Moreover, constructing successively the sequence yn = Ayn−1 + Byn−1, n = 1,2,…, for any initial value y0 ∈ Ph, one has y → x* as n → ∞.Let B is a null operator, When In this section, we will apply T ∈ {4, 5,…}, η ∈ ℤT−12,, and α, β ∈ ℝ are real numbers with β ≠ −1 and (T + 1 − αη) + β(1 − α) ≠ 0, for any y defined in ℤT+10,, the nonlocal boundary value problem If t ∈ ℤT+10,, s ∈ ℤT1,, the Green function G in  (i)G > 0, t ∈ ℤT+10,, s ∈ ℤT1,; (ii)G ≤ r(s)h(t), t ∈ ℤT+10,, s ∈ ℤT1,, where  (iii)s ≥ t, G = p(s)h(t), t ∈ ℤT+10,, s ∈ ℤT1,, where for For s < t, s < η, notice that t − η < t − s, η ≤ s < t, t ≤ s < η, t ≤ s, η ≤ s, t ∈ ℤT+10, and s ∈ ℤT1,, G ≤ r(s)h(t).Since (i) and (iii) are obvious, here we just prove (ii). For  (A1)f, g : ℤT+10, × [0, ∞)→[0, ∞) are continuous and increasing with respect to the second variable, g ≠ 0; (A2)g ≥ λg for λ ∈ ,  t ∈ ℤT+10,, x ∈ [0, ∞) and there exists a constant γ ∈  such that f ≥ λγf for λ ∈ , t ∈ ℤT+10,, x ∈ [0, ∞); (A3)δ0 > 0 such that f ≥ δ0g, t ∈ ℤT+10,, x ∈ [0, ∞).there exists a constant The problem  = t + β, t ∈ ℤT+10,. Moreover, for any initial value u0 ∈ Ph, constructing successively the sequence un(t) → u*(t) as n → ∞, where G is given as  ≠ 0; Aggiven as .A : P → E and B : P → E by u is a solution of (u = Au + Bu. From (A2) and A : P → P and B : P → P. In the sequel we check that A, B satisfy all assumptions of Define two operators P → E by 7(Au)(t)=ution of  if and oA, B are two increasing operators. In fact, by (A1) and u, v ∈ P with u ≥ v, we know that u(t) ≥ v(t), t ∈ ℤT+10,, and obtain Bu ≥ Bv.Firstly, we prove that A is a γ-concave operator and B is a subhomogeneous operator. In fact, for any λ ∈  and u ∈ P, from (A2), we know that A is a γ-concave operator. At the same time, for any λ ∈  and u ∈ P, from (A2), we get B is subhomogeneous.Next we show that Ah ∈ Ph and Bh ∈ Ph. From (A3) and l1 = δ0p(T)g > 0 and l2 = ∑s=1Tr(s)f. Hence we have l1h(t)≤(Ah)(t) ≤ l2h(t), t ∈ ℤT+10,; that is, Ah ∈ Ph. We can similarly prove that Bh ∈ Ph. Thus condition (i) of Now we show that Au ≥ δ0Bu, u ∈ P. By applying Au + Bu = u has a unique solution u* ∈ Ph. Moreover, constructing successively the sequence un = Aun−1 + Bun−1, n = 1,2,…, for any initial value u0 ∈ Ph, we have un → u* as n → ∞. That is, problem  → u*(t) as n → ∞.In the following we show that condition (ii) of 2(Au)(t)=The following result can be obtained by  (A1)′f : ℤT+10, × [0, ∞)→[0, ∞) is continuous and increasing with respect to the second variable, f ≠ 0; (A2)′γ ∈  such that f ≥ λγf for λ ∈ , t ∈ ℤT+10,, x ∈ [0, ∞).there exists a constant Then problem u* ∈ Ph, where h(t) = t + β, t ∈ ℤT+10,. Moreover, for any initial value u0 ∈ Ph, constructing successively the sequence un(t) → u*(t) as n → ∞, where G is given as =αu→[0, ∞) are continuous and increasing with respect to the second variable, g ≠ 0, and for λ ∈ , t ∈ ℤ0,10, y ∈ [0, ∞), Pt+2.Consider the following nonlinear discrete problem: −Δ2y(t−1)α ≠ 1, the nonlocal boundary value problem If t ∈ ℤT+10,, s ∈ ℤT1,, the Green function G in  (i)G > 0, t ∈ ℤT+10,, s ∈ ℤT1,; (ii)t ∈ ℤT+10,, s ∈ ℤT1,, where For We omit it since it is obvious.h(t) = 1, t ∈ ℤT+10,. Moreover, for any initial value n → ∞, where Assume that A1), (A2), and A3) are satisfied; then the problem , a are satigiven as .It is similar to the proof of The following corollary can be obtained by h(t) = 1,  t ∈ ℤT+10,. Moreover, for any initial value n → ∞, where Assume that (A1)′ and (A2)′ are satisfied; then the problem given as .In a similar way, we can get the corresponding results for the difference equation  subject 9u(0)=αu→[0, ∞) are continuous and increasing with respect to the second variable, g ≠ 0, and for λ ∈ , t ∈ ℤ0,5, y ∈ [0, ∞), P1.Consider the following nonlinear discrete problem: −Δ2y(t−1)


Rough set theory is a suitable tool for dealing with the imprecision, uncertainty, incompleteness, and vagueness of knowledge. In this paper, new lower and upper approximation operators for generalized fuzzy rough sets are constructed, and their definitions are expanded to the interval-valued environment. Furthermore, the properties of this type of rough sets are analyzed. These operators are shown to be equivalent to the generalized interval fuzzy rough approximation operators introduced by Dubois, which are determined by any interval-valued fuzzy binary relation expressed in a generalized approximation space. Main properties of these operators are discussed under different interval-valued fuzzy binary relations, and the illustrative examples are given to demonstrate the main features of the proposed operators. Rough set theory proposed by Pawlak  is an exA rough set model is composed of two parts: the approximation space and the approximated object. Rough set theory comes with a lot of extensions and generalizations. Yao et al. researched the generalized rough sets by considering sets and relations of the approximation space and the approximated object , 16. In Most researches on the fuzzy rough set theory focus on point-valued fuzzy sets and point-valued fuzzy binary relations. But the fuzzy notion described by using point values may lose some available information in the real-life information systems sometimes. If the description is done by interval values, it may acquire a better effectiveness than that by using point ones, for example, a self-evolving interval type-2 fuzzy neural network with online structure and parameter learning , encodinIn this paper, we further study the generalized fuzzy rough approximation operators defined in . In partThe rest of the paper is organized as follows. In In this section, we introduce some basic notions and properties related to interval-valued fuzzy sets which will be used in this paper. We first review an interval-valued subset originated by . We firsI be a closed unit interval; that is, I = . [I] = { : a− ≤ a+, a−, a+ ∈ I} is the set of all interval-valued subsets of I. a = ∈[I] is an interval value. When a− = a+, the interval-valued a =  becomes a real number in [I]. In particular, real numbers return intervals of zero length, say 1 =  and 0 = .Let a, b ∈ [I]. a ≤ b if and only if a− ≤ b−, a+ ≤ b+; a = b if and only if a− = b−, a+ = b+; a < b if and only if a ≤ b and a ≠ b.Let a, b ∈ [I]. a≰b indicates that a is not less than or equal to b; a ≮ b indicates that a is not less than b; a≱b indicates that a is not greater than or equal to b; a≯b indicates that a is not greater than b.Let I] may not exhibit order relations, so According to the order relation defined in ai ∈ [I], bi ∈ I, i ∈ J, J = {1,2, ⋯, m}; one definesLet I], ≤) is a complete lattice, and the triple  is an algebraic system, which is derived by  with the maximal element  and the minimum element .Obviously, . In particular, when A = U, A(x) = , for all x ∈ U, and when A = ∅, A(x) = , for all x ∈ U.Let A, B ∈ FI(U), A⊆B means A(x) ≤ B(x) and for all x ∈ U, (A∩B)(x) = A(x)∧B(x), (A ∪ B)(x) = A(x)∨B(x), and (~A)(x) = 1 − A(x).Similar to fuzzy sets, the operators ⊆, ∩, ∪, and complement of interval-valued fuzzy sets are defined as follows. For all α ∈ [I], A ∈ FI(U). αA is called numerical product of α and A and is defined as (αA)(x) = α∧A(x), for all x ∈ U.Let α ∈ [I], A ∈ FI(U). Aα = {x ∈ U : A(x) ≥ α} is called α-cut set of A and α-cut set of A.Let A ∈ FI(U); thenLet x ∈ U,For all Then Similarly, one can show that U and W be two finite and nonempty universes of discourse. Then the mapping IR : U × W → [I] is called an interval-valued fuzzy relation from U to W, where U × W = { : x ∈ U, y ∈ W}. When U = W, IR is called an interval-valued fuzzy relation on U.Let IR from U to W is an interval-valued fuzzy set denoted by IR ∈ FI(U × W). So Definitions IR ∈ FI(U × W) is an interval-valued fuzzy relation. If we see it as an interval-valued fuzzy set, then IRα = { ∈ U × W : R ≥ α}.Obviously, an interval-valued fuzzy relation IR be an interval-valued fuzzy relation from U to W; then IR is said to be serial if and only if for all x ∈ U, there exists y ∈ W such that IR = .Let IR be an interval-valued fuzzy relation on U; then IR is reflexive if and only if IR = , for all x ∈ U; IR is symmetric if and only if IR = IR, for all x, y ∈ U; IR is transitive if and only if x, z ∈ U; IR is Euclidean if and only if y, z ∈ U.Let α-cut set or strong α-cut set to an interval-valued fuzzy relation, for all α ∈ [I], still satisfies the corresponding definition of IR is, respectively, reflexive, symmetric, and transitive, then One can prove that the binary relation obtained by calculating U and W be two finite universes of discourse. If R is an arbitrary binary fuzzy relation from U to W, then the triple  is called a generalized fuzzy approximation space.Let U, W, R) be a generalized fuzzy approximation space, for all x ∈ U; one defines R(x) = {) : y ∈ W}.Let (R(x) is the row of the fuzzy relation which includes x, and obviously Rα(x) = (R(x))α.U, W, R) be a generalized fuzzy approximation space, for all α, β ∈ , A ∈ F(W),Let  lower and upper approximations of A with respect to .U, W, R) be a generalized fuzzy approximation space, for all A ∈ F(W). One definesLet , and the operators The pair The dual properties are quite useful in proving the properties of the approximation operators. When one intends to prove two dual properties, it suffices to prove one of them, which simplifies the proof procedure. The properties of the lower and upper approximation operators are characterized as follows.U, W, R) be a generalized fuzzy approximation space. Then for all A ∈ F(W), Let (x ∈ U.(1) Note that x ∈ U, y ∈ W if R ≥ 1 − α, then A(y) > α; that is, for all y ∈ W, R < 1 − α or A(y) > α.For all Hence we have α ∈ I, so that For the second case, Hence,Similarly,Therefore, (2) Now, we prove the validity of the relationship Hence, Similarly, x is a certain value, the variables y1 and y2 are functions of the variable α. Refer to Suppose that y1 is equal to the maximum of function y2, such that y2 equal to zero at the point β, which makes the maximum of function y2 approach β, but it does not exist. In this paper, the lower and upper approximation operators in In the proof of olds. In , the lowU and W be two finite universes of discourse. If IR is an arbitrary binary interval-valued fuzzy relation from U to W, then the triple  is called a generalized interval-valued fuzzy approximation space. In particular, when U = W, the space is denoted by .Let U, W, IR) be a generalized interval-valued fuzzy approximation space, for all x ∈ U,Let  be a generalized interval-valued fuzzy approximation space, A ∈ FI(W), for all x ∈ U; one definesLet . The operators The pair U, W, IR) be a generalized interval-valued fuzzy approximation space, for all α, β ∈ [I], A ∈ FI(W); one definesLet , respectively.U, W, IR) be a generalized interval-valued fuzzy approximation space, A ∈ FI(W); one definesLet . The operators The pair The approximation operators introduced in U, W, IR) be a generalized interval-valued fuzzy approximation space, A ∈ FI(W); then for all Let ∩Aα ≠ ∅. This means that there exist y ∈ W, IR ≥ α, and A(y) ≥ α.We observe that, for all By the interval-valued operations of Therefore, y ∈ W, IR∧A(y) ≥ α; that is, there exists y ∈ W, so that IR ≥ α and A(y) ≥ α cannot be deduced by Now, we prove that the reverse of Next, we give an example illustrating that the relationship U, W, IR) is a generalized interval-valued fuzzy approximation space,Suppose that (IRα(x1) = Aα = ∅, we have IRα(x1)∩Aα = ∅ and we get On the other hand, since U, W, IR) be a generalized interval-valued fuzzy approximation space, A ∈ FI(W); then, for all α ∈ [I], Let  > 1 − α then A(y) ≥ α; that is, for all y ∈ W, IR ≤ 1 − α or A(y) ≥ α. By the interval-valued operations as in Note that, for all IR∨A(y) ≥ α means that )−∨A(y)− ≥ α− and )+∨A(y)+ ≥ α+, which cannot deduce that 1 − IR ≥ α or A(y) ≥ α in y ∈ W, ~IR∨A(y) ≥ α, we cannot deduce that IR ≤ 1 − α or A(y) ≥ α, and note that, for all y ∈ W, ~IR∨A(y) ≥ α if and only if y ∈ W, IR ≤ 1 − α or A(y) ≥ α cannot hold.Now we prove that the reverse of Next, we show that U, W, IR) is a generalized interval-valued fuzzy approximation space,Suppose that  be a generalized interval-valued fuzzy approximation space, A ∈ FI(W); then Let ∧A(y)∈[I], such that y ∈ IRα(x)∩Aα. We observe that y ∈ IRα(x)∩Aα means that IRα(x)∩Aα ≠ ∅, which can deduce that In fact, for all y, So, for arbitrary value of Therefore, U, W, IR) be a generalized interval-valued fuzzy approximation space, A ∈ FI(W); then Let  We verify that y0 ∈ W, IR > 1 − α1, and from (1−α1)+ = 1, we have IR− > ∨y∈W−∧(1 − A(y)+)).Note that, for all y0 ∈ W, we have IR− > IR−∧(1 − A(y0)+)  .Further from IR− > 1 − A(y0)+  and  1 − A(y0)+ = IR−∧(1 − A(y0)+).Therefore we obtain that α1− = 0, we get A(y0) ≥ α1; that is, y0 ∈ Aα1.Because y0, So, for arbitrary value of (2) Similar to the proof shown in (1), we have x, For any Therefore, According to Theorems U, W, IR) be a generalized interval-valued fuzzy approximation space, A ∈ FI(W); then Let  be a generalized interval-valued fuzzy approximation space; then the lower approximation operator Let , a ∈ [I],For all x ∈ U and x ∈ W.Here (1) We prove that x ∈ U, letx is “for all x” and ∃x is “there exists x,” which are the same as follows.For all D3 = D2 − D1, for all β ∈ D3; two cases appear:Obviously, a, b ∈ D1; we have For the first case, suppose β, it is easy to see that For arbitrary Hence, Similarly, (2) We verify x ∈ U, letFor all We observe thatSimilarly, A⊆B then A⊆B, then Aα⊆Bα. According to (3) We prove that if A⊆B, then Similarly, if (4) From (3), one immediately obtains (4).From U, W, IR) be a generalized interval-valued fuzzy approximation space; then the following conditions are equivalent: IR is serial;A⊆FI(W);Let  be a generalized interval-valued fuzzy approximation space; then the following conditions are equivalent: IR is reflexive;A⊆FI(W);A⊆FI(W).Let  be a generalized interval-valued fuzzy approximation space; then the following properties hold: x, y) ∈ U × W;x, y) ∈ U × W.Let  be a generalized interval-valued fuzzy approximation space and A is an interval-valued fuzzy set on U; then ∀α ∈ [I], Let  is a generalized interval-valued fuzzy approximation space, where U = {x1, x2}, IR(x1) = /x1 + /x2, A = /x1 + /x2,   α = , β1 = , and β2 = . Since IRα(x1) = ∅, Aα = {x1, x2}, we have IRα(x1)∩Aα = ∅. Hence Suppose that (IRβ1(x1) = {x1}, Aβ1 = {x1, x2}, IRβ2(x1) = {x2}, Aβ2 = {x1, x2}, we see that On the other hand, since Note that U, IR) is a generalized interval-valued fuzzy approximation space, where U = {x1, x2, x3}, IR(x1) = /x1 + /x2 + /x3, A = /x1 + /x2 + /x3,α = , β1 = , β2 = .Suppose that  be a generalized interval-valued fuzzy approximation space; then the following conditions are equivalent: IR is transitive;A ∈ FI(U);A ∈ FI(U).Let (A ∈ FI(U), from (1)⇒(2) For all Hence x, y, z ∈ U, let(3)⇒(1) For all y ∈ U, suppose that α = D7(y) = IR∧IR; then IR ≥ α, IR ≥ α; hence α ∈ D6 and by the arbitrary y, For all α ∈ D6, there exists y ∈ U, such that α ≤ IR∧IR. For arbitrary α, For all IR is transitive.Hence, by (2)⇔(3) This conclusion follows immediately from the duality.IR is symmetric, then the approximation operators satisfy A⊆U; if IR is Euclidean, then the approximation operators satisfy A⊆U. These properties do not hold in the interval-valued fuzzy rough sets. Next, we give a counterexample to show it.In , if IR iU, IR) is a generalized interval-valued fuzzy approximation space, U = {x1, x2, x3}, andSuppose that  be a generalized interval-valued fuzzy approximation space; then the following conditions are equivalent: Let ⇒(1) For all By Hence, At the same time, we haveTherefore (1) has been proven.(1)⇒(3) First we prove thatx ∈ U, if IRα(x)∩Aα ≠ ∅; that is, there exists z ∈ IRα(x)∩Aα. So we have IR ≥ α and A(z) ≥ α.Note that, for all Sincewe havey ∈ U, IR+ ≤ 1 − α− or IR− ≥ α−and IR− ≤ 1 − α+ or IR+ ≥ α+ imply that if IR+ > 1 − α−, then IR− ≥ α−, and if IR− > 1 − α+, then IR+ ≥ α+. It follows that IR+ > 1 − α− and IR− > 1 − α+ imply that IR− ≥ α− and IR+ ≥ α+; therefore, IR > 1 − α implies that IR ≥ α, because IR > 1 − α and IR ≥ α are equivalent to z ∈ IRα(y), respectively. If z ∈ IRα(y), and since A(z) ≥ α, z ∈ IRα(y)∩Aα; that is, IRα(y)∩Aα ≠ ∅. So Hence, for all y, it follows that x, we obtain For arbitrary By Therefore (2)⇔(3) This conclusion follows immediately from the duality.U, IR) be a generalized interval-valued fuzzy approximation space; then the following conditions are equivalent: IR = 0  or  IR = 1, x, y ∈ U;Let +)∨IR−) = 1 and −)∨IR+) = 1; namely, for all x ∈ U, IR+ = 0  or  IR− = 1 and IR− = 0 or IR+ = 1.(1)⇔(2) We observe that, for all IR+ = 0, then IR− = 0; we have IR = 0. If IR+ ≠ 0, then IR− = 1 and IR+ = 1; we have IR = 1.On the one hand, if y ∈ U, IR = 0 or IR = 1.Hence, (2)⇒(4) We first prove that x ∈ Aα, for all y ∈ U, IR = 0 or IR = 1 if and only if IR ≠ 0 deduces IR = 1. It follows that if IR > 1 − α, then IR ≥ α; that is, x ∈ IRα(y). Further since x ∈ Aα, IR > 1 − α implies that x ∈ IRα(y)∩Aα. So IRα(y)∩Aα ≠ ∅. Note that IR > 1 − α and IRα(y)∩Aα ≠ ∅ are equivalent to IR > 1 − α, then x ∈ IRα(y)∩Aα. It shows that if y, x, Suppose that On the other hand, in view of x, z ∈ U, from the proof of “(2)⇒(1)” in (3)⇒(1) For all Furthermore, since z = x, we have I], we have When (3)⇔(4) This conclusion follows immediately from the duality.U, IR) be a generalized interval-valued fuzzy approximation space.IR is reflexive and transitive, then A ∈ FI(U).If IR is reflexive and x, z ∈ U, then A ∈ FI(U).If Let  is a generalized interval-valued fuzzy approximation space. (1)IR is reflexive and transitive, thenIf (2)IR is reflexive and x, z ∈ U, thenIf Suppose that  is reflexive, symmetric, and transitive, respectively, under the classical binary relation. Thus, if IR can satisfy the above functions, this technology can be applied in reasoning, learning, and decision-making. In Sections In this paper, one can prove that the binary relation obtained by calculating


R be a prime ring of characteristic different from 2, with extended centroid C, U its two-sided Utumi quotient ring, F a nonzero generalized derivation of R, f a noncentral multilinear polynomial over C in n noncommuting variables, and a, b ∈ R such that ab = 0 for any r1,…, rn ∈ R. Then one of the following holds: (1) a = 0; (2) b = 0; (3) there exists λ ∈ C such that F(x) = λx, for all x ∈ R; (4) there exist q ∈ U and λ ∈ C such that F(x) = (q + λ)x + xq, for all x ∈ R, and f2 is central valued on R; (5) there exist q ∈ U and λ, μ ∈ C such that F(x) = (q + λ)x + xq, for all x ∈ R, and aq = μa, qb = μb.Let  R be a prime ring with center Z(R). We denote by  = ab − ba the simple commutator of the elements a, b ∈ R and by k = k−1, b], for k > 1, the kth commutator of a,  b. Throught this paper we will use the following notation: U will be the (two-sided) Utumi quotient ring of a ring R . The definition, the axiomatic formulation, and the properties of this quotient ring U can be found in  ∈ Z(R), for all x ∈ R, then R is commutative. In k = 0 for all x ∈ L and k ≥ 1 a fixed integer; then char(R) = 2 and R satisfies s4, the standard identity of degree 4.In any case, when f Posner  says thative. In  Lanski gf be a multilinear polynomial over C in n noncommuting variables and denote by f(X) the set of all evaluations of f in X⊆R. In case f is not central valued on R, it is well known that the additive subgroup generated by f(R) contains a noncentral Lie ideal of R. Moreover any noncentral Lie ideal of R contains all the commutators  for x, y in some nonzero ideal of R, unless char(R) = 2 and dimCRC = 4.Let d(x), x]k = 0, in case x ∈ f(I), where I is a two-sided ideal of R. They show that either f is central valued in R or char(R) = 2 and R satisfies s4.In light of this and following the line of investigation of the previous cited papers, in  P. H. LeR is often tightly connected to the behaviour of additive mappings defined on R, which act on suitable subsets of the whole ring. In k = 0, where d is a derivation of R and u ∈ f(I), where I is a one-sided ideal of R. In particular, for I = R, if a ≠ 0 and f is not central valued on R, then char(R) = 2 and R satisfies s4.More recently, Liu  and Wangd is replaced by a generalized derivation F. An additive map F : R → R is said to be a generalized derivation if there is a derivation d of R such that, for all x, y ∈ R, F(xy) = F(x)y + xd(y). A significative example is a map of the form F(x) = ax + xb, for some a, b ∈ R; such generalized derivations are called inner. Generalized derivations have been primarily studied on operator algebras. Therefore any investigation from the algebraic point of view might be interesting .The main result in  is the fR be a prime ring of characteristic different from 2, with extended centroid C, U its two-sided Utumi quotient ring, F ≠ 0 a nonzero generalized derivation of R, f a noncentral multilinear polynomial over C in n noncommuting variables, and a ∈ R such that r1,…, rn ∈ R. Then either a = 0 or one of the following holds: (1)λ ∈ C such that F(x) = λx, for all x ∈ R;there exists (2)q ∈ U and λ ∈ C such that F(x) = (q + λ)x + xq, for all x ∈ R, and f2 is central valued on R.there exist Let We would like to remark that the same conclusions hold in case we consider the right annihilator, more precisely.R be a prime ring of characteristic different from 2, with extended centroid C, U its two-sided Utumi quotient ring, F ≠ 0 a nonzero generalized derivation of R, f a noncentral multilinear polynomial over C in n noncommuting variables, and a ∈ R such that r1,…, rn ∈ R. Then either a = 0 or one of the following holds: (1)λ ∈ C such that F(x) = λx, for all x ∈ R;there exists (2)q ∈ U and λ ∈ C such that F(x) = (q + λ)x + xq, for all x ∈ R, and f2 is central valued on R.there exist Let F(x), x] : x ∈ f(R)} and prove the following.Here we will consider a more general situation, involving a two-sided annihilating condition. More specifically, we study simultaneously left and right annihilators of the set {. Let a, b, c ∈ R, such that c ∉ Z(R) and ab = 0 for all x ∈ S. Then there exists λ ∈ Z(R) such that ac = λa and cb = λb.Let In order to prove K be an infinite field and t ≥ 2. If A1,…, Ak are not scalar matrices in Mt(K), then there exists some invertible matrix B ∈ Mt(K) such that each matrix BA1B−1,…, BAkB−1 has all nonzero entries.Let See Lemma  1.5 in .R be a prime ring with extended centroid C. Suppose ∑i=1naixbi + ∑j=1mcjxdj = 0, for all x ∈ R, where ai, bi, cj, dj ∈ R, for i = 1,…, n and j = 1,…, m. If a1,…, an are C-independent then each bi is C-dependent on d1,…, dm. Analogously if b1,…, bn are C-independent then each ai is C-dependent on c1,…, cm.Let It is Martindale's result contained in .R be a prime ring with extended centroid C. Suppose a+b = 0, for all x, y ∈ R, where a, b ∈ R. Then a = −b ∈ C.Let It is an easy consequence of K be an infinite field, R = Mm(K) the algebra of m × m matrices over K, Z(R) the center of R, and S = . Assume that there exist a,  b,  c,  q nonzero elements of R such that axq + cxb = 0 for all x ∈ S. If q ∈ Z(R) then one of the following holds: (1)a,  b,  c are central matrices and aq + bc = 0;(2)b is a central matrix and aq + bc = 0.Let q ∈ Z(R), by the assumption, we have that aqx + cxb = 0 for all x ∈ S. Clearly if c ∈ Z(R) then aqx + xbc = 0 for all x ∈ S, and by aq = −bc ∈ Z(R); that is, a, b, c ∈ Z(R). On the other hand, if b ∈ Z(R), then (aq + bc)x = 0 for all x ∈ S and it follows easily that aq + bc = 0.Since c and b both nonscalar matrices. We will prove that in this case we get a contradiction.In light of this, we consider eij the usual matrix unit with 1 in the -entry and zero elsewhere.Here we denote by c and b have all nonzero entries, say c = ∑klcklekl and b = ∑klbklekl, for 0 ≠ ckl, 0 ≠ bkl ∈ K.By eji ∈ S for all i ≠ j, then, for any i ≠ j, i, j)-entry of X is cijbij = 0, a contradiction.Since For sake of clearness, we may write the previous lemma as follows.K be an infinite field, R = Mm(K) the algebra of m × m matrices over K, Z(R) the center of R, and S = . Let a1,  a2,  a3,  a4 be nonzero elements of R such that a1xa2 + a3xa4 = 0 for all x ∈ S. Assume there exists i ∈ {1,2, 3,4} such that ai ∈ Z(R). Then a1 = αa3 and a2 = −αa4, for a suitable α ∈ Z(R).Let K be an infinite field, R = Mm(K) the algebra of m × m matrices over K, and Z(R) the center of R. Assume that there exist a,  b,  c,  q nonzero elements of R such that axq + cxb = 0 for all x ∈ S = . If q ∉ Z(R) and b − αq ∈ Z(R), for a suitable α ∈ K, then b − αq = a + αc = 0.Let a + αc is not a scalar matrix. By a + αc and q have all nonzero entries, say a + αc = ∑kltklekl and q = ∑klqklekl, for 0 ≠ tkl, 0 ≠ qkl ∈ K.Assume that b = βI + αq, for a suitable β ∈ K, by our assumption we have that x ∈ S. In particular for x =  = eij, with i ≠ j, j, i)-entry of X is 0 = tjiqji, a contradiction.Since a + αc must be a central matrix. In light of this, there exist β, γ ∈ K such that b = αq + β and a = −αc + γ, so that 0 = (−αc + γ)xq + cx(αq + β) = (βc)x + x(γq), for all x ∈ S = . Once again by q ∉ Z(R), it follows that β = γ = 0; that is, b = αq and a = −αc.Therefore K be an infinite field, R = Mm(K) the algebra of m × m matrices over K, and S = . Suppose there exist a,  b,  c,  q ∈ R such that axq + cxb = 0 for all x ∈ S. Denote akl,  bkl,  ckl, and qkl elements of K. If there are i ≠ j such that qji ≠ 0, cji ≠ 0, and bji = 0, then ari = 0 and brk = 0 for all r ≠ i and k ≠ r .Let x = eij, we have r ≠ i, the -entry of the matrix X is 0 = ariqji + cribji = ariqji. Since qji ≠ 0, one has ari = 0 for all r ≠ i, in particular aji = 0. Thus, in case m = 2 we are done (since bji = aji = 0).Consider the assumption m ≥ 3, and choose x = eit, with t ≠ i, j. Hence we also have (1)s ≠ i, the -entry of the matrix X is ajiqjs + cjibjs = 0;for all (2)s ≠ i, j, the -entry of the matrix Y is ajiqts + cjibts = 0;for all (3)j, i)-entry of the matrix Y is ajiqti + cjibti = 0;the ((4)k ≠ i, t, the -entry of the matrix Y is ajiqtk + cjibtk = 0 .for all By (1) and (2) and since aji = 0 and cji ≠ 0, we have both bjs = 0, for all s ≠ i, and bts = 0 for all t ≠ i, j and s ≠ i, j. So by (3)  bti = 0 for all t ≠ i. Finally by (4), btk = 0 for all t ≠ i, j and k ≠ t.Assume in what follows that K be an infinite field, R = Mm(K) the algebra of m × m matrices over K, and S = . Suppose there exist a, b, c, q ∈ R such that axq + cxb = 0 for all x ∈ S. Denote bkl, ckl, and qkl elements of K. Assume there are i ≠ j such that bji = 0. If qrs ≠ 0, crs ≠ 0 for all r ≠ s, then one of the following holds: (1)a = b = 0;(2)m = 2, cq = 0, and there exists 0 ≠ λ ∈ K such that Let m ≥ 3. The first step is to apply twice b to be a diagonal matrix. In fact bji = 0, qji ≠ 0, and cji ≠ 0 imply that brk = 0 for all r ≠ i and k ≠ r; in particular, since m ≥ 3, there exists t ≠ i such that blt = 0, for all l ≠ t, i. Since qlt ≠ 0, clt ≠ 0, we have brk = 0 for all r ≠ t and k ≠ r, so bik = 0 for all k ≠ i, as required. Say b = ∑kbkkekk.Firstly we consider the case R induced by the invertible matrix P = I + erj, for r ≠ i, j: φ(x) = P−1xP. Of course φ(a)xφ(q) + φ(c)xφ(b) = 0, for all x ∈ S. Moreover the -entries of φ(q), φ(c), and φ(b) are, respectively, qji ≠ 0, cji ≠ 0, and bji = 0. Therefore, again by r, j)-entry of φ(b) is zero, for all r ≠ i. By calculations 0 = (φ(b))rj = bjj − brr; that is, bjj = brr.Consider now the inner automorphism of χ is the inner automorphisms induced by the invertible matrix Q = I + eri, as above χ(a)xχ(q) + χ(c)xχ(b) = 0, for all x ∈ R. Since the -entries of χ(q), χ(c), and χ(b) are, respectively, qij ≠ 0, cij ≠ 0, and bij = 0, and again any -entry of χ(b) is zero, for all r ≠ j; that is, 0 = (φ(b))ri = bii − brr and bii = brr = bjj = β, for all r ≠ i, j. Thus b = βI is a central matrix in R. By b = αq for some α ∈ K or b = 0. Since the first case cannot occur, we get b = 0 and also a = 0 which follows from aq = 0 and q ≠ 0.On the other hand, if m = 2; that is, R = M2(K). In this case it is well known that for any element x ∈  there exist α, β, γ ∈ K such that b21 = 0. In case b12 = 0, then by the same above argument we show that b ∈ Z(R) and we are done again. Thus we consider the case b12 ≠ 0. Moreover, by applying a21 = 0. Hence we may write x = e12 ∈  we have X is 0 = c21b22; that is, b22 = 0 and the -entry of the matrix X is 0 = a11q21; that is, a11 = 0. On the other hand, for x = e21 ∈ , we have Y is 0 = a12q12 + c12b12; that is, a12 ≠ 0 and b12/q12 = −a12/c12. Moreover the -entry of the matrix Y is 0 = a22q12 + c22b12. Therefore, if denoted λ = −b12/q12, one has a22 = λc22 and a12 = λc12.Let now Y is 0 = a12q11 + c12b11. Thus b11 = −λq11 and b12 = −λq12.Analogously, the -entry of the matrix α ≠ 0, we also have cq = 0.Finally, by our assumption and for K be an infinite field, R = M2(K) the algebra of m × m matrices over K, and S = . Let a, b, c ∈ R and denote akl, bkl, ckl, and pkl elements of K. Suppose c ∉ Z(R) and ab = 0 for all x ∈ S. Assume there are i ≠ j such that pji = 0. If ars ≠ 0, brs ≠ 0, for all r ≠ s, then ac = cb = 0.Let acxb − axcb = 0 for all x ∈ S. By ac = cb = 0 or ab = 0 and there exists 0 ≠ λ ∈ K such that ab = 0 implies that the following holds: ac we get cb we also have a11 = 0, by  = 0 and using (b21(c22a11 + c11a11) = 0. Since b21 ≠ 0 and a11 ≠ 0, then c11 = −c22 = μ.By our hypothesis, we have lso have c21b11+c2μ ≠ 0, denoted by I the identity matrix in R, and let Assume c and c′ induce the same inner derivation, then by our assumptions we have that ab = 0 for all x ∈ S. By applying again ac′ = c′b = 0 or ab = 0 and there exists 0 ≠ ν ∈ K such that c′ satisfies the equalities (30)ac'=c0 and by  also c2S: Z is T is c ≠ 0, then c12 ≠ 0. Therefore the sum of  the algebra of t × t matrices over K, and S = . Let a, b, c ∈ R and denote akl, bkl, pkl, and qkl elements of K. Suppose c ∉ Z(R) and ab = 0 for all x ∈ S. Then there exists λ ∈ Z(R) such that ac = λa and cb = λb.Let a, b, ac, or cb is a scalar matrix we are done by a, b, ac, and cb are noncentral matrices.Clearly if one of Q ∈ Mt(K) such that QaQ−1 = a′, QbQ−1 = b′, Q(ac)Q−1 = (ac)′, and Q(cb)Q−1 = (cb)′ have all nonzero entries.By ac, a} are linearly Z(R)-dependent if and only if {(ac)′, a′} are linearly Z(R)-dependent; analogously {cb, b} are linearly Z(R)-dependent if and only if {(cb)′, b′} are linearly Z(R)-dependent. Moreover ac = cb = 0 if and only if (ac)′ = (cb)′ = 0. Therefore, in order to prove our result, we may replace a,  b,  ac,  cb, respectively, by a′,  b′,  (ac)′,  (cb)′, so that a,  b,  ac, and cb have all nonzero entries.Notice that {x = eij ∈ S we have j, i)-entry of X is qjibji − ajipji = 0. Denote 0 ≠ η = qji/aji, so that pji = ηbji. Let I be the identity matrix in R and u = c − ηI. Since u and c induce the same inner derivation in R, then ab = 0; that is, a(c − ηI)xb − ax(c − ηI)b = 0, for all x ∈ S. Moreover a and b have all nonzero entries, and the -entry of (c − ηI)b is zero. Thus we may apply Lemmas ac = ηa and cb = ηb, as required.For C is infinite, the conclusion follows from If one assumes that K be an infinite field which is an extension of the field C and let R. Since P is a multilinear generalized polynomial in the indeterminates x1,  x2, then it is a generalized polynomial identity for Now let f(R), the set of all evaluations of the noncentral multilinear polynomial f over C, and assume that F is an inner generalized derivation, so that there exist c, q ∈ U such that F(x) = cx + xq, for all x ∈ R, and f(R) satisfies a,  b are nonzero elements of R.In this section we consider In order to prove the first result we premit the following.R = Mt(C) be the algebra of t × t matrices over C of characteristic different from 2. Notice that the set f(R) = {f : r1,…, rn ∈ R} is invariant under the action of all inner automorphisms of R. Hence if denoted by r =  ∈ R × R × R × ⋯×R = Rn, then for any inner automorphism φ of Mt(C), we have that Let f is not central then, by b = 0, for all r1,…, rn ∈ R.This means that R, generated by the set S = {x2 : x ∈ f(R)}. By  of R is contained in S. In the first case we conclude that f2 is central valued in R and we are done. In either case we have a]b = 0, for all r1, r2 ∈ R, and by Consider the additive subgroup of (R)}. By , either K be an infinite field which is an extension of the field C and let f is central-valued on R if and only if it is central-valued on R. Moreover it is multihomogeneous of multidegree  in the indeterminates x1,…, xn.Now let P is a multilinear generalized polynomial Θ in 2n indeterminates; moreover x1,…, xn, y1,…, yn) is a generalized polynomial identity for R and C) ≠ 2 we obtain P = 0, for all Hence the complete linearization of a ∈ R, 0 ≠ b ∈ R, c, q ∈ U such that ab = 0, for all x ∈ f(R), then R satisfies a nontrivial generalized polynomial identity, unless when one of the following holds: (1)c, q ∈ C;(2)c − q ∈ C and there exists λ ∈ C such that ac = λa,  qc = λb.If there exist 0 ≠ R does not satisfy any nontrivial generalized polynomial identity with coefficients in U. Therefore, R. By calculations x1,…, xn ∈ R. If c ∈ C and q ∈ C, the proof is completed; hence we suppose that c and q are not simultaneously central. By b, qb} are linearly C-independent then R satisfies the trivial generalized polynomial identity af2qb = 0. It means, since a ≠ 0, qb = 0, a contradiction. Analogously, if we suppose {a, ac} linearly C-independent, we get ac = 0, a contradiction.Assume that ulations Φx1,…,xn=α, β ∈ C such that qb = βb and ac = αa; now c, q ∈ Z(R);(2)λ ∈ C such that c − q = λ, and f2 is central valued on R;there exists (3)λ, μ ∈ C such that c − q = λ, ac = μa, and cb = μb.there exist Let 0 ≠ R is not a domain.By R satisfies the nontrivial generalized polynomial identity: U. In case C is infinite, we have P = 0 for all C. Since both U and R by U or C being finite or infinite. Thus we may assume that R is centrally closed over C which is either finite or algebraically closed. By Martindale's theorem b. We remark that H satisfies P(x) = a(cx2 − x2q + x(q − c)x)b = 0 , for any nontrivial idempotent element e = e2 ∈ H, and obtain R, it follows that either ae = 0 or (1 − e)b = 0 or (1 − e)(q − c)e = 0. Here our aim is to prove that in any case (1 − e)(q − c)e = 0. To do this, we firstly assume that ae = 0. In 2b = 0, which implies ace = 0.Assume next that ma  2 in , the set  = 0, for any idempotent element e ∈ H. Since H is not a domain, then H is generated by its minimal idempotent elements; therefore  = (0); that is, q − c ∈ C. Let λ ∈ C such that q = c + λ. By our assumption it follows that H satisfies ab that is H satisfies ab. In this last replace x by x + 1 and obtain that H satisfies ab. Since char(H) ≠ 2, then acrb − arcb = 0, for all r ∈ H. By 2b = 0, for all r1,…, rn ∈ R, then either a = 0 or b = 0.Let U and thus all generalized derivations of R will be implicitly assumed to be defined on the whole U and obtained the following result.In  Lee provF on a dense right ideal of R can be uniquely extended to U and assumes the form F(x) = cx + d(x), for some c ∈ U and a derivation d on U.Every generalized derivation fd the polynomial obtained from f by replacing each coefficient ασ with d(ασ). Thus we write d) = fd + ∑if,…, rn), for all r1,…, rn in R.In this section we denote by In light of this, we finally prove our main result.a ≠ 0 and b ≠ 0. Since R satisfies the generalized differential identity R satisfies d is an inner derivation induced by an element q ∈ U, then R satisfies the generalized polynomial identity: Suppose both d be an outer derivation of R. In this case R satisfies the differential identity:R satisfies the generalized polynomial identity: i = 1,…, n, R satisfies the blended component q ∈ R − Z(R) and replace any yi by . Thus R satisfies q ∈ Z(R).Hence let rem see , 22), R , R d be


An objective measure of the uncertainty, regarding the interval constraint, accounted for by using the HSGM is proposed for the Bayesian inference. For this purpose, we drive a maximum entropy prior of the normal mean, eliciting the uncertainty regarding the interval constraint, and then obtain the uncertainty measure by considering the relationship between the maximum entropy prior and the marginal prior of the normal mean in HSGM. Bayesian estimation procedure of HSGM is developed and two numerical illustrations pertaining to the properties of the uncertainty measure are provided.This paper considers a  Specifically, we assign a normal prior distribution for θ and an inverse gamma (IG) prior for τ2, that is, θ ~ N and τ2 ~ IG, which are commonly used in a normal model as conjugate priors, where all the hyperparameters μ and σ2,  c, and d are assumed to be known in the first place.Consider the following model for normally distributed data:yi=θ+ϵi, θ a priori; a suitable restriction on the parameter space Θ = R, such as using a truncated normal distribution, is expected. However, it is often the case that the actual observations of  and thun  the degree of uncertainty regarding the interval constraint of  Leonard . Thus, sθ in . In θ by using Boltzmann's maximum entropy theorem  = tj, j = 1,…, k, takes the k-parameter exponential family formλ1, λ2,…, λk can be determined, via the k-constraints, in terms of t1,…, tk. See Leonard and Hsu [Assume now that we can specify the partial information concerning a location parameter the formE[tj(θ)]= and Hsu  for the θ of our interest be the normal mean in , has different formula according to the degree of uncertainty regarding the interval constraint.Let the location parameter  mean in . Then thR, and θ ∈ Θ with certainty.Θ = θ that we can specify values for both mean μ and variance σ2. Then the N prior specification is the maximum entropy prior for θ . Thus ta, b], and θ ∈ Θ with certainty.Θ =  = μ and E[(θ − μ)2] = σ2 on the space θ ∈ R by a priori information. Further suppose that we have certain prior information that the parameter θ is concentrated on the region , that is,π(θ). The last condition is equivalent to E[I(a ≤ θ ≤ b)] = 1. Therefore, by Bolzmann's maximum entropy theorem, the prior density of θ for the t1(θ) = θ, t1 = μ, t2(θ) = (θ − μ)2, t2 = σ2, and t3(θ) = I  (a ≤ θ ≤ b). Since t3 = 1, the maximum entropy prior, subject to these three restrictions, is thusλ1 = 0, λ2 = −1/2σ2, λ3 = −ln{Φ(v(b)) − Φ(v(a))}, where v(a) = (a − μ)/σ and v(b) = (b − μ)/σ.On the other hand, when we are completely sure about the priori interval constraint of θ≤b)}by , becauseNa,b) distribution in Johnson et al. , and θ ∈ Θ with (1 − α) × 100% uncertain.Θ = . Then along with the priori moment conditions E[θ] = μ and E[(θ − μ)2] = σ2, the additional uncertain prior information about the interval constraint can be expressed byα (or (1 − α) × 100%) denotes the degree of uncertainty. Thus the priori uncertain interval constraint is equivalent toNow suppose that we have partial priori information that we can specify values for both mean θ, reflecting the uncertain interval constraint in  of X1 and X2 whose joint distribution is a bivariate normal N2,If raint in  is(17)θt1(θ) = θ, t1 = μ, t2(θ) = (θ − μ)2, t2 = σ2, t3(θ) = I  (a ≤ θ ≤ b), and t3 = α, and setting λ1 = 0 and λ2 = −1/2σ2, the maximum entropy prior π(θ) in (Taking  π(θ) in  reduces π(θ) in (π3(θ)∝exp side of  as folloπ(θ) in (π3(θ)∝expondition , becauseξ* = (1 − δ)−1/2 and Thus the maximum entropy prior density for the π1(θ) = π2(θ) = π3(θ) for a = −∞ and b = ∞. This is consistent with our partial priori information that E[θ] = μ and Var(θ) = σ2 for θ ∈ R. Some algebra using the moments of the WNa,b) distribution given in Kim . Also note, from the conditional property of N2, that δ → 1 for finite vales of μ,σ,a, and b with b > a. Applying these two results to π3(θ), we see that it approximates to π2(θ) as δ → 1. It is straightforward to see from π3(θ) = π1(θ)  for δ = 0, because the WNa,b) distribution is equivalent to N for δ = 0.Note from  that ξv*π3(θ)) and Ent(π2(θ)) as a function of δ ∈  for three values of σ2, two cases of Θ = , and μ = 0. In Ent. Since each graph coincides with the results of π1(θ)) > Ent(π3(θ)) > Ent(π2(θ)) for δ ∈ . (ii) The entropy of π3(θ) is a monotone decreasing function of δ. (iii) Each entropy of the three priors increases as σ2 becomes large. (iv) DiffEnt is closely related with degree of uncertainty  × 100%) for it is a monotone decreasing function of δ ∈  and α is a function of δ. (v) DiffEnt is a monotone increasing function of σ2 for the case where a value of δ is given.Each graph of From a relationship between Lemmas σ02 = (1 − δ)σ2, σ12 = δσ2, and δ ∈ . Then the two-stage prior of θ defined by HSGM in (π3(θ) and the degree of uncertainty regarding the interval constraint, θ ∈ , accounted for by the HSGM isLet σ02 = (1 − δ)σ2, σ12 = δσ2, and δ ∈ , the marginal prior distribution of θ in a,b). Thus the prior density p(θ) in (π3(θ) in (π3(θ) reflects uncertainty about the interval constraint θ ∈  by the degree of (1 − α) × 100% with If  p(θ) in  is equal3(θ) in . Now Lemθ accounted for by HSGM, and it shows that the uncertainty measure is different from that of O'Hagan and Leonald }, with (1 − α) × 100% degree of uncertainty, where μ and σ2 are true prior mean and variance of the parameter θ when the uncertain priori interval condition does not exist.Let us revisit the HSGM with the following two stages of a prior hierarchy of θ in θ and τ2 proportional to the product of likelihood and the prior distribution,Dn = {y1,…, yn}, IG is the inverse-gamma density with parameters c and d, and π3(θ) is the density . Note that the joint posterior is not simplified in an analytic form of the known density and thus intractable for posterior inference. Instead, we derive each of the conditional posterior distributions of θ and τ2 in an explicit form, which will be useful for posterior inference such as Gibbs sampling .(i) θ is given byThe full conditional posterior distribution of whereω∗2 = σ2τ2/(nσ2 + τ2).(ii)τ2 is the inverse-Gamma distributionThe full conditional posterior distribution of Given the joint posterior distribution , we haveθ given that τ2 and Dn is proportional to(i) The unnormalized conditional density of τ2 isc + 2/n, d + ∑i=1n(yi − θ)2) distribution.(ii) It is straightforward to see from  that theθ, τ2) because their full conditional distributions are given in θ, τ2). The only difficulty would lie in generating random samples from WN distribution, that is, θ  ∣   ~ WNa,b) as given in ,γ*(b)) with γ*(a) = (a − θ1*)/ω11*, γ*(b) = (b − θ1*)/ω11*, θ1* = (1 − δ)μ + δθ*, and ω11∗2 = δ(1 − δ)σ2 + δ2ω∗2. Johnson et al. [ϕ(·) denotes the density of N. It is seen that each of the first terms in  is put on θ (e.g. [π3(θ). That is each of the second terms in the posterior mean and covariance vanishes when θ is assumed to be N a priori without any interval constraint. In this sense, the second terms in , in, inθ, τiven in . The HSGM  is based on the two stages of a prior hierarch advocated by O'Hagan and Leonard [This paper considered the normal models which include the two-stage prior of the normal mean, referred to as Leonard  and elic Leonard . By expl


X), where Bin(X) is the collection of all groupoids. Finally, we define a fuzzy-d-subset of a groupoid and investigate some of its properties.The notion of a fuzzy upper bound over a groupoid is introduced and some properties of it are investigated. We also define the notions of an either-or subset of a groupoid and a strong either-or subset of a groupoid and study some of their related properties. In particular, we consider fuzzy upper bounds in Bin( Iséki and Tanaka introduced two classes of abstract algebras: BCK-algebras and BCI-algebras , 2.d-algebras which is a useful generalization of BCK-algebras and then investigated several relations between d-algebras and BCK-algebras as well as several other relations between d-algebras and oriented digraphs. Han et al. ) is a BCK-algebra.The notion of the semigroup x∗x = 0,(D2)x = 0,0∗(D3)x∗y = 0 and y∗x = 0 imply x = y for all x, y ∈ X.A algebra   is a nond-algebra X satisfying the following additional axioms:(D4)x∗y)∗(x∗z))∗(z∗y) = 0, (((D5)x∗(x∗y))∗y = 0 for all x, y, z ∈ X.  has the multiplicationx∗x = e; e∗x = e; x∗y = y∗x = e yield (x − y)(x − e) = 0, (y − x)(y − e) = 0 and x = y, or x = e = y; that is, x = y; that is,  is a d-algebra.Consider the real numbers X, we let Bin(X) the collection of all groupoids , where ∗ : X × X → X is a map and where ∗ is written in the usual product form. Given elements  and  of Bin(X), define a product “□” on these groupoids as follows:x, y ∈ X. Using the notion, Kim and Neggers . Then μ(0) ≥ μ(x) for all x ∈ X; that is, μ is a fub over .Let , we consider the following conditions: for any x, y ∈ X, A)  (μ(x∗y) = μ(y∗x) = μ(0) implies μ(x) = μ(y),B)  (μ(x∗y) = μ(y∗x) = μ(0) implies x = y,C)  (μ(x∗y) = μ(y∗x) implies μ(x) = μ(y),D)  (μ(x∗y) = μ(y∗x) implies x = y. We have a diagram of implications as follows:A)⇒(B), (B)⇒(D), (A)⇒(C), and (C)⇒(D).Given a fub X, ∗, 0) be a groupoid with |X| ≥ 2 and let μ : X →  be a constant map; that is, μ(x) = α for some α ∈  for all x ∈ X. Then (A) holds as does (C). On the other hand, x ≠ y implies (B) fails as does (D). Thus we have countered examples to (A)⇒(B) and (C)⇒(D).(a) Let  having the condition (B). But μ does not have the condition (D), since μ(1∗2) = μ(2) = μ(2∗1) and 1 ≠ 2. Hence (B)⇒(D) is false.(b) Let X = 2A, where A = {1,2, 3,4}. Define a binary operation “∗” on X by x∗y : = {t | t ∈ x, t ∉ y},  ∀x, y ∈ X. Thus, if x⊆y, then x∗y = ∅. Let μ(x): = 1/(1 + |x|)  , where |x| is the cardinality of x. Assume μ(x∗y) = μ(y∗x) = μ(∅). Since μ(∅) = 1, we have |x∗y| = 0 = |y∗x| and so x∗y = ∅ = y∗x. Therefore x⊆y, y⊆x. Hence x = y. On the other hand, if x = {1,2, 3}, y = {3,4, 5}, then x∗y = {1,2}, y∗x = {4,5}, and μ(x∗y) = 1/3 = μ(y∗x), but x ≠ y. This shows that (B) is true, but (D) is false.(c) Let X, ∗, 0) be a left-zero-semigroup; that is, for any x, y ∈ X, x∗y = x, and let μ : X →  be a fub over . Assume that μ(x∗y) = μ(y∗x) for any x, y ∈ X. Since  is a left-zero-semigroup, we have μ(x) = μ(x∗y) = μ(y∗x) = μ(y). Hence condition (C) holds for any fub μ over .(d) Let  = , where R is the set of all real numbers and “−” is the usual subtraction on R. Also, if μ(x) = ex2−, then μ(0) = 1 ≥ μ(x) for all x ∈ R; that is, μ is a fub over . Note that μ(x − y) = μ(y − x) = μ(0) implies x − y = y − x = 0 and x = y, so that μ(x) = μ(y) as well; that is, the conditions (B) and (A) hold. Now if we let y : = −1 and x : = 2 + y, then μ(x − y) = μ(2 + y − y) = μ(2) = e−4 and μ(y − x) = μ(y − 2 − y) = μ(−2) = e−4, so that μ(x − y) = μ(y − x). Since ey+1)−4 = μ(2 + y) = ey)2−(2+ = ey2−ey+1)−4( < ey2− = μ(y). Thus (C) fails to hold.(e) Let (A), we give two conditions (A)* and (A)α*, 0 ≤ α ≤ 1 as follows:A)*  (μ(x∗y) = μ(y∗x) = μ(0) implies μ(x) = μ(y) = μ(0),A)α*  (μ(x∗y) = μ(y∗x) = μ(0) implies μ(x) = μ(y) ≥ α.As a refinement of the condition  = α for all x ∈ X, then for any groupoid   μ satisfies the condition (A)*. Since (A)*⇒(A), μ satisfies the condition (A).(a) If X, ∗, 0): = , where R is the set of all real numbers and “−” is the usual subtraction on R, and let μ(x): = ex2−. Then μ(0) = 1 ≥ μ(x) > 0 for any x ∈ R, and μ(x) = 1 if and only if x = 0. Assume that μ(x − y) = μ(y − x) = μ(0). Then ex−y)2−( = ey−x)2−( = 1 and so x − y = 0. Hence x = y. Therefore μ(x) = μ(y); that is, (A) holds. Let x = y ≠ 0 imply μ(y) = μ(x) = ex2− < 1 = μ(0). Thus (A)* does not hold. Since limx→∞μ(x) = 0, given α > 0, there is an x such that μ(x) < α; that is, (A)α* does not hold for α > 0.(b) Let , let μβ(x): = (ex2− + β)/(1 + β), where β ≥ 0. Then μβ(0) = 1 ≥ μβ(x) for all x ∈ R and limx→∞μβ(x) = β/(1 + β) ≤ μβ(x) for all x ∈ X. If we take α : = β/(1 + β), then μβ(x) ≥ α for all x ∈ X. If we assume μβ(x − y) = μβ(y − x) = μβ(0), then (x − y)2 = (y − x)2 = 0 and hence μβ(x) = μβ(y); that is, the condition (A)α* holds. If α1 < α2, then (A)α2*⇒(A)α1* holds clearly. If we set β1 : = α1/(1 − α1) and β2 : = α2/(1 − α2), then β1 < β2 and limx→∞μβ1(x) = limx→∞((ex2− + β1)/(1 + β1)) = (β1)/(1 + β1) = α < α2. Take an x ∈ R so that α1 < μβ1(x) < α2. Then ex2− + β1 < α2(1 + β1). It follows that ex2− < (β2/(1 + β2))(1 + β1) − β1 < (β2(1 + β1) − β1(1 + β2))/(1 + β2) = (β2 − β1)(1 + β1)/(1 + β2); that is, x2 > ln[(1 + β2)/(β2 − β1)(1 + β1)]. Hence, if A)α1*⇒(A)α2* does not hold.(c) For the groupoid  and that they both satisfy the condition (A). If μ : = λμ1 + (1 − λ)μ2, where 0 ≤ λ ≤ 1, then μ is also a fub over  having the condition (A).Suppose that λ < 1. For any x ∈ X, we have μ(0) = λμ1(0)+(1 − λ)μ2(0) ≥ λμ1(x)+(1 − λ)μ2(x) = μ(x). Hence μ is a fub over .Let 0 < μ(x∗y) = μ(y∗x) = μ(0), but μ1(x∗y) < μ1(0) or μ2(x∗y) < μ2(0). If μ1(x∗y) < μ1(0), then λμ1(x∗y) < λμ1(0) and thus (1 − λ)μ2(x∗y) = μ(x∗y) − λμ1(x∗y) > μ(x∗y) − λμ1(0) =  μ(0) − λμ1(0) = (1 − λ)μ2(0). This shows that μ2(x∗y) > μ2(0), which is a contradiction. Similarly, μ2(x∗y) > μ2(0) implies μ1(x∗y) > μ1(0), which is also a contradiction. Hence we obtain μ1(x∗y) = μ1(0), μ2(x∗y) = μ2(0). Similarly, we obtain μ1(y∗x) = μ1(0), μ2(y∗x) = μ2(0). Since μi has the condition (A), we have μi(x) = μi(y)  . Hence μ(x) = λμ1(x)+(1 − λ)μ2(x) = λμ1(y)+(1 − λ)μ2(y) = μ(y). Thus μ has the condition (A).Assume that μ1 and μ2 are fubs over a groupoid  and that μ2 has the condition (A)* with μ1(0) < μ2(0). If μ : = μ1∨μ2, where (μ1∨μ2)(x): = max{μ1(x), μ2(x)}, then μ is a fub over  having the condition (A)*.Suppose that μ = μ1∨μ2, then μ(0) = μ1(0)∨μ2(0), μ(x) = μ1(x)∨μ2(x). If μ1(x) ≥ μ2(x), then μ(x) = μ1(x) and μ(0) ≥ μ1(0) ≥ μ1(x) = μ(x) implies μ(0) ≥ μ(x) for any x ∈ X. Similarly, if μ2(x) ≥ μ1(x), then μ(0) ≥ μ(x) for any x ∈ X. Hence μ is a fub over If μ(x∗y) = μ(y∗x) = μ(0). Then μ1(x∗y)∨μ2(x∗y) = μ(x∗y) = μ(0) =  μ1(0)∨μ2(0) = μ2(0). If μ2(x∗y) < μ2(0), then μ1(x∗y) = μ2(0) and μ1(0) ≥ μ1(x∗y) = μ2(0), which is a contradiction. Hence μ2(x∗y) ≥ μ2(0); that is, μ2(x∗y) = μ2(0). Similarly, μ(y∗x) = μ(0) implies μ2(0) = μ2(y∗x). Hence μ2(x∗y) = μ2(y∗x) = μ2(0). Since μ2 has the condition (A)*, we have μ2(x) = μ2(y) = μ2(0) and so μ(x) = μ(0). Similarly, we obtain μ(y) = μ(0). Thus μ has the condition (A)*.Assume that A and B are subsets of X such that A∖B, A∩B, and B∖A are nonempty. Let 0 ∈ A∩B and let μ1 : = χA, μ2 : = χB, where x ∈ A∖B, y ∈ B∖A, and  is the left-zero-semigroup. It follows that μ(x∗y) = μ1(x∗y)∨μ2(x∗y) = μ(x) = μ1(x)∨μ2(x) = 1∨0 = 1 and that μ(y∗x) = 1 = μ(y), μ(0) = 1 = μ1(1) = μ2(1). Hence μ(x) = μ(y) = μ(0); that is, the condition (A)* holds.(a) Suppose R, −, 0) be the set of all real numbers equipped with subtraction and the zero element 0. Let μ1(x): = 1 + x − ⌈x⌉ and μ2(x): = 1 + ⌊x⌋ − x. If x ∈ Z, then x = ⌈x⌉ = ⌊x⌋ and hence μ1(x) = μ2(x) = 1 = μ(0). If x = n + α where n ∈ Z, 0 < α < 1/2, then μ1(x) = α and μ2(x) = 1 − α. Hence μ(x) = μ1(x)∨μ2(x) = 1 − α. If x = n + β where n ∈ Z, 1/2 < β < 1, then μ1(x) = β and μ2(x) = 1 − β. Hence μ(x) = μ1(x)∨μ2(x) = β. Assume μ(x − y) = μ(y − x) = μ(0). Since μ(0) = 1, we have x − y, y − x ∈ Z, say y − x = n ∈ Z. If we let x : = m + γ where m ∈ Z, 0 < γ < 1, then y = x + n = (n + m) + γ. It follows that μ(x) = μ(y) has the value either γ or 1 − γ. Hence μ satisfies the condition (A), but not the condition (A)*.(b) Let (μ1(0) = μ2(0) = 1. This shows that the condition μ1(0) < μ2(0) is a very necessary condition in Note that, in X, ∗) be a groupoid and let ∅ ≠ U⊆X. Then U is said to be either-or if x∗y ∈ U, y∗x ∈ U implies either {x, y}⊆U or {x, y}⊆Uc for any x, y ∈ X.Let  and that μ satisfies the condition (A). Let Kerμ : = {x ∈ X | μ(x) = μ(0)}. Then Kerμ is an either-or subset of X.Suppose that x∗y, y∗x ∈ Kerμ for any x, y ∈ X. Then μ(x∗y) = μ(0) = μ(y∗x). Since μ satisfies the condition (A), we have μ(x) = μ(y). Thus, also, if x ∈ Kerμ, then y ∈ Kerμ, and if x ∉ Kerμ, then y ∉ Kerμ. Therefore Kerμ is an either-or subset of X.Let μ in Note that 0 ∈ KerU is said to be with alternative if there are elements {x, y}⊆U, {u, v}⊆Uc such that {x∗y, y∗x}⊆U, {u∗v, v∗u}⊆U.An either-or subset Z, +) is the groupoid of all integers with respect to the usual addition and if U : = 2Z is the set of all even integers, then U is an either-or subset of X. In fact, let x + y = y + x = 2u ∈ U and x = 2v, where u, v ∈ Z. Then y = 2u − 2v = 2(u − v) ∈ U. Thus, U is an either-or subset of . On the other hand, Uc consists of all odd integers. Now if x + y = y + x is odd for any x, y ∈ Z and if x is odd, then y = (x + y) − x is even. If x is even, then y is odd. Hence Uc fails to be an either-or subset of . The subset U = 2Z of  is an either-or subset with alternative. Both Z and Zc = ∅ are either-or subsets, but without alternative.If . Suppose that U is an either-or subset of the groupoid  and that μ = χU is the characteristic function of U; that is, μ(x) = 1 if x ∈ U and μ(x) = 0 otherwise. Then μ satisfies the condition (A).Let 0 ∈ U be a “selected” element; that is, μ(0) = 1. Then μ(x∗y) = μ(y∗x) = μ(0) implies x∗y, y∗x ∈ U. Since U is an either-or subset of the groupoid , we have either {x, y}⊆U or {x, y}⊆Uc; that is, μ(x) = μ(y) = 1 or μ(x) = μ(y) = 0. In any case μ(x) = μ(y) and the condition (A) holds.Let 0 ∈ X, ∗) be a groupoid and let ∅ ≠ S⊆X. Let EO(S) be the collection of all either-or subsets of the groupoid , sometimes denoted as EO), which contain the subset S of X. Then, since  is an either-or subset of , it follows that EOR(S): = ∩i{Ui | Ui ∈ EO)}⊇S.Let  be a groupoid and let ∅ ≠ S⊆U. Then EOR) is an either-or subset of  containing S.Let ). If {x, y}⊆Uc for some either-or subset U of  which contains S, then EOR)⊆U implies Uc⊆EOR)c and {x, y}⊆EOR)c. If this is not the case, then {x, y} ⊈ Uc for any either-or subset U of  containing S, whence {x, y}⊆U. Thus {x, y}⊆EOR). This shows that EOR) is an either-or subset of .Let We have the following corollaries.X, ∗) be a groupoid. If ∅ ≠ S⊆T, then EOR)⊆EOR).Let  be a groupoid. Then we haveLet  be the left-zero-semigroup; that is, x∗y = x for all x, y ∈ X, and let 0 be a fixed element for which μ(0) ≥ μ(x) for all x ∈ X, where μ : X →  is a fuzzy subset of X. Then μ(x∗y) = μ(x) = μ(y∗x) = μ(y) = μ(0) implies μ(x) = μ(y) = μ(0) and μ satisfies the condition (A)*. If U is any nonempty subset of X, then x∗y, y∗x ∈ U implies {x, y}⊆U, so that U is an either-or subset without alternative. Hence EOR) = U for any ∅ ≠ U⊆X whatever. Of course, if  is the right-zero-semigroup, that is, x∗y = y for any x, y ∈ X, then the same conclusions hold.(a) Let  be the group of all integers with respect to addition. We claim that nZ = {nx | x ∈ Z} is an either-or subset of . Given n ∈ Z, if n = 0, and if x + y, y + x ∈ {0} and x = 0, then y = 0 and hence {x, y} = {0}⊆{0}. On the other hand, if x ≠ 0, then y = −x implies {x, y} = {x, −x}⊆{0}c = Z − {0}. Hence {0} = 0Z is an either-or subset of the groupoid . Let n ≠ 0. If x + y = y + x = nu, and if x = nv, then y = (x + y) − x = nu − nv = n(u − v); that is, y ∈ nZ as well. Thus {x, y}⊆nZ. Assume that x ∉ nZ; that is, x = nu′ + α for some u′ ∈ Z and α ∈ Z such that 0 < α < n. Then y = nu − x = n(u − u′) − α ∉ nZ. Hence {x, y}⊆(nZ)c.(b) Let ) = nZ for all n ∈ Z. Since {0} is the smallest either-or subset of , EOR) = {0} = 0Z. Let n ≠ 0 in Z. Then 0 ∈ EOR). In fact, if Ui is an either-or subset of  containing n, then either {n, 0}⊆Ui or {n, 0}⊆Uic, since n + 0 = 0 + n = n ∈ Ui. It follows that 0 ∈ Ui for all either-or subset Ui of ; that is, 0 ∈ EOR).We claim that EOR). If Ui is an either-or subset of , either {n, −n}⊆Ui or {n, −n}⊆Uic for any either-or subset Ui of , since n + (−n) = (−n) + n = 0 ∈ Ui. Since n ∈ Ui, we have −n ∈ Ui for all either-or subset Ui. This proves that −n ∈ EOR). Hence {−n, 0, n}⊆EOR).We claim that −B is an either-or subset of  containing n. Then it is easy to see that 0, −n ∈ B. If we let x : = n, y : = −2n, then x + y = y + x = −n ∈ B. Since x = n ∈ B and B is an either-or subset of , we obtain −2n = y ∈ B. Similarly, if we let x = −n and y : = 2n, then we obtain 2n = y ∈ B. Similarly, we obtain ±3n, ±4n,…, ∈B. Thus nZ⊆B; that is, B is an either-or subset of  containing nZ. It follows that nZ = ∩{Ui | Ui :  either-or  subset  containing  n} = EOR).Assume that N : = {1,2, 3,…} and let a ∈ N. If x + y = a, then x < a, y < a and thus {a} is an either-or subset of N without alternative, since {x, y}⊆{a}c. If Uak : = {a, 2a,…, ka} for any a ∈ N, let x + y = y + x ∈ Uak. Then there exists h ∈ N such that x + y = ha, h ≤ k. If x = ja, j < h, then y = ha − x = (h − j)a ∈ Uak. Hence {x, y}⊆Uak. If x ∉ Uak, then y ∉ Uak and {x, y}⊆(Uak)c, so that Uak is an either-or subset of N with alternative if k ≥ 2.(c) Let X, ∗) be a groupoid and let U⊆X such that x∗y, y∗x ∈ U implies x = y. Then U is an either-or subset of .Let .We call such a subset X, ∗, f) be a leftoid, and let U = {a} be a singleton. If x∗y = f(x) ∈ U and y∗x = f(y) ∈ U, then f(y) = f(x) = a. Hence, if |f−1(a)| = 1, then x = y, and U is a strong either-or subset of .(a) Let  Let X, ∗, 0) be a groupoid and let U be a strong either-or subset of X with 0 ∈ U. If μ : = χU is the characteristic function of U, then μ is a fub over . Furthermore, μ satisfies the condition (B) and Kerμ = U.Let  = 1 ≥ μ(x) for all x ∈ X and μ is a fub over . Let μ(x∗y) = μ(y∗x) = μ(0) for any x, y ∈ X. Since μ(0) = 1, we have x∗y, y∗x ∈ U. Hence x = y, since U is a strong either-or subset of X. Therefore μ satisfies the condition (B) and in that case Kerμ = μ−1(μ(0)) = U.Since 0 ∈ X, ∗, e) be a group all of whose elements have finite order and let μ : X →  be a fuzzy subgroup. Then μ has the condition (A) and Kerμ is an either-or subset of X.Let (μ be a fuzzy subgroup of X. Then μ(x∗y) ≥ min{μ(x), μ(y)} for any x, y ∈ X. Thus μ(x2) = μ(x∗x) ≥ μ(x) and μ(xn) ≥ min{μ(xn−1), μ(x)}. By induction μ(xn−1) ≥ μ(x) implies μ(xn) ≥ μ(x), so that μ(e) = μ(xn) ≥ μ(x) implies μ(e) ≥ μ(x) for all x ∈ X; that is, μ is a fub over .Let μ = μ−1(μ(e)) = U, then μ(x∗y) = μ(y∗x) = μ(e) implies μ((x∗y)∗y−1) = μ(x) ≥ min{μ(x∗y), μ(y−1)) and hence μ(x) ≥ μ(y−1) ≥ μ(y), so that μ(x) = μ(y); the condition (A) holds for μ, and Kerμ is an either-or subset of .If Kerμ be a fub over  with the condition (A) and a fub over  with the condition (A)*. If : = □, then μ is a fub over  with the condition (A)*.Let μ(x□y) = μ(y□x) = μ(0) for any x, y ∈ X. Then μ((x∗y)•(y∗x)) = μ((y∗x)•(x∗y)) = μ(0) implies μ(x∗y) = μ(y∗x) = μ(0), so that μ(x) = μ(y) = μ(0). Hence μ satisfies the condition (A)* over .Assume that μ be a fub over both  and  with the condition (A)*. If : = □, then μ is an lub over  with the condition (A)*.Let Straightforward.X be a nonempty set. If 〈μ; (A)〉 and 〈μ; (A)*〉 are the collections of groupoids  such that μ satisfies the condition (A) *), then 〈μ; (A)*〉⊆〈μ; (A)〉 andμ; (A)*〉, □) is a subsemigroup of (Bin(X), □).Let μ be a fub over  with the condition (B), and let μ be a fub over  with the condition (A)*. If : = □, then μ is a fub over  with the condition (B).Let μ(x□y) = μ(y□x) = μ(0) for any x, y ∈ X. Then μ((x∗y)•(y∗x)) = μ((y∗x)•(x∗y)) = μ(0). Since μ satisfies the condition (A)* over , we have μ(x∗y) = μ(y∗x) = μ(0). Since μ has the condition (B) over , we obtain x = y, proving the theorem.Assume that X, ∗, 0), the following are interesting properties in fuzzy subgroupoids μ : X → :μ(x∗x) = μ(0) for all x ∈ X,μ(0∗x) = μ(0) for all x ∈ X,μ(x∗y) = μ(y∗x) = μ(0), then μ(x) = μ(y) for all x, y ∈ X.if Given a groupoid  is called a fuzzy-d-subset of the groupoid  if it satisfies conditions (1), (2), and (3).A fub  X : = R be the set of all real numbers and let “+” be the usual addition on R. Then every fuzzy-d-subset μ of  is constant. In fact, for all x ∈ X, we have μ(0) = μ(x + x) = μ(2x). If we let y : = 2x, then μ(y) = μ(0) for all y ∈ X; that is, μ is a constant function on X.(a) Let X, ∗, 0) be a group with identity 0. Let μ be a fuzzy subset of X with (2); then μ(x) = μ(0∗x) = μ(0) for all x ∈ X, whence μ is a constant function. Therefore μ is a fuzzy-d-subset of .(b) Let  be a d-algebra. If μ = χ{0} is the characteristic function of {0}; that is, μ(0) = 1 and μ(x) = 0 otherwise, then μ is a fuzzy-d-subset of X and Kerμ = {0}.Let  = μ(0), μ(0∗x) = μ(0). If μ(x∗y) = μ(y∗x) = μ(0), then x∗y = y∗x = 0. Since  is a d-algebra, we obtain x = y. Hence μ(x) = μ(y). This shows that μ is a fuzzy-d-subset of  and Kerμ = μ−1(μ(0)) = μ−1(1) = {0}.Let S be an either-or subset of a groupoid  such that Δ : = {x∗x | x ∈ X} and Δ0: = {0∗x | x ∈ X} are subsets of S. Let μ : = χS be the characteristic function of S and 0 ∈ S. Then μ is a fuzzy-d-subset of .Let S, we have μ(0) = 1 ≥ μ(x) for all x ∈ X. Since Δ, Δ0⊆S, we obtain μ(x∗x) = μ(0), μ(0∗x) = μ(0). Assume that μ(x∗y) = μ(y∗x) = μ(0). Then x∗y, y∗x ∈ S. Since S is an either-or set over , either {x, y}⊆S or {x, y}⊆Sc. It follows that either μ(x) = μ(y) = 1 or μ(x) = μ(y) = 0. Hence μ is a fuzzy-d-subset of .Since 0 ∈


It has been shown that a normal S-iterative method converges to the solution of a mixed type Volterra-Fredholm functional nonlinear integral equation. Furthermore, a data dependence result for the solution of this integral equation has been proven. The scientists working in almost every field of science are faced with nonlinear problems, because nature itself is intrinsically nonlinear. Such problems can be modelled as nonlinear mathematical equations. Solving nonlinear equations is, of course, considered to be a matter of the uttermost importance in mathematics and its manifold applications. There are numerous systematic approaches which are classified as direct and iterative methods to solve such equations in the existing literature. Indeed, by using direct methods, finding solutions to a complicated nonlinear equation can be an almost insurmountable challenge. In this context, iterative methods have become very important mathematical tools for finding solutions to a nonlinear equation. For a comprehensive review and references to the extensive literature on the iterative methods, the interested reader may refer to some recent works –8.Recently, Sahu  and KhanX be an ambient space and let T be a self-map of X. A normal S-iterative method is defined byξn}n=0∞ is a real sequence in  satisfying certain control condition(s).Let It has been shown both analytically and numerically in , 10 thatThis iterative method, due to its simplicity and fastness, has attracted the attention of many researchers and has been examined in various aspects; see –20.a1; b1] × ⋯× is an interval in Rm, K, H :  × ⋯× ×  × ⋯× × R → R continuous functions, and F :  × ⋯× × R3 → R.In this paper, inspired by the performance and achievements of normal S-iterative method , we willAlso we give a data dependence result for the solution of integral equation  with theWe end this section with some known results which will be useful in proving our main results.A1);A2);A3)(α + (βLK + γLH)(b1 − a1) ⋯ (bm − am) < 1.We suppose that the following conditions are satisfied:x* ∈ C.Then  has a unβn}n=0∞ be a nonnegative sequence for which one assumes there exists n0 ∈ N, such that for all n ≥ n0 one has satisfied the inequalityμn ∈ , for all n ∈ N, ∑n=0∞μn = ∞, and γn ≥ 0, for all n ∈ N. Then the following inequality holds:Let {A1)–(A5) in ξn}n=0∞ ⊂  be a real sequence satisfying ∑n=0∞ξn = ∞. Then  and normal S-iterative method , where ||·||C is Chebyshev's norm. Let {xn}n=0∞ be an iterative sequence generated by normal S-iterative method –(A4), we have thatA5),ξk ∈  for all k ∈ N, assumption (A5) yieldsex ≥ 1 − x for all x ∈ , we can rewrite , an, an2), aave that||xn+1−x∗yn−x∗||,||yn−x∗||, we get||xn+1−x|We now prove the data dependence of the solution for integral equation  with theB be as in the proof of T, K, H, Let F, K, and H be defined as in xn}n=0∞ be an iterative sequence defined by normal S-iterative method  1/2 ≤ ξn, for all n ∈ N, and (ii) ∑n=0∞ξn = ∞. One supposes further that (iii) there exist nonnegative constants ε1 and ε2 such that u ∈ R and for all t, s ∈  × ⋯×.Let e method  associatx* and If quations  and 16)x* and x~A1)–(A4) and (iii), we obtainA5) and 1/2 ≤ ξn in the resulting inequality, we getUsing , 15), , , 17), a, a17), ae obtain||xn+1−x~(bi−ai),||yn−y~n|, we get||xn+1−x~