G be a group. Denote by π(G) the set of prime divisors of |G|. Let GK(G) be the graph with vertex set π(G) such that two primes p and q in π(G) are joined by an edge if G has an element of order p · q. We set s(G) to denote the number of connected components of the prime graph GK(G). Denote by N(G) the set of nonidentity orders of conjugacy classes of elements in G. Alavi and Daneshkhah proved that the groups, An where n = p, p + 1, p + 2 with s(G) ≥ 2, are characterized by N(G). As a development of these topics, we will prove that if G is a finite group with trivial center and N(G) = N(Ap+3) with p + 2 composite, then G is isomorphic to Ap+3.Let  G be a finite group and let Z(G) be its center. For any 1 ≠ x ∈ G, suppose that xG denotes the conjugacy classes in G containing x and CG(x) denotes the centralizer of x in G. We will use N(G) to denote the set {n : G has a conjugacy class of size n}. Thompson in 1987 gave the following conjecture with respect to N(G).In this paper, all groups considered are finite and simple groups are nonabelian simple groups. Let Thompson's Conjecture  = N(L), then G≅L.ure  denote the set of all prime divisors of |G|. Let GK(G) be the graph with vertex set π(G) such that two primes p and q in π(G) are joined by an edge if G has an element of order p · q. We set s(G) to denote the number of connected components of the prime graph GK(G). A classification of all finite simple groups with disconnected prime graph was obtained in  − 1; in particular, exp ≤ p + 2.exp = ∑i=1∞[(p + 3)/ri]  for each r ∈ π(Ap+3)∖{2}. Furthermore, exp < (p − 1)/2, where 3 ≤ r ∈ π(Ap+3). In particular, if r > [(p + 3)/2], then exp = 1.exp(|Let (1) By the definition of Gaussian integer function, we have thatr ∈ π(Ap+3).(2) Similarly, as (1), we have thatr > [(p + 3)/2], exp = 1.If The proof is completed.Sn be the symmetric group of degree n. Assume that the cycle has c1 1-cycles, c2 2-cycles, and so on, up to ck  k-cycles, where 1c1 + 2c2 + ⋯+kck = n. Then the number of conjugacy classes in Sn isAn be the alternating group of degree n.Let x ∈ An. Then, for the size of the conjugacy class xG of x in An, one has the following. i, ci = 0 and, for all odd i, i ∈ {0,1}, then |xG | = z/2.If, for all even xG | = z.In all other cases, |In particular, |xG | ≥z/2.Let See .n ≥ 6 is a natural number, then there are at least s(n) prime numbers pi such that (n + 1)/2 < pi < n. Heres(n) = 6 for n ≥ 48;s(n) = 5 for 42 ≤ n ≤ 47;s(n) = 4 for 38 ≤ n ≤ 41;s(n) = 3 for 18 ≤ n ≤ 37;s(n) = 2 for 14 ≤ n ≤ 17;s(n) = 1 for 6 ≤ n ≤ 13.In particular, for every natural number n > 6, there exists a prime p such that (n + 1)/2 < p < n − 1, and, for every natural number n > 3, there exists an odd prime number p such that n − p < p < n.If See Lemma 1 of .a, b, and n be positive integers such that  = 1. Then there exists a prime p with the following properties:p divides an − bn,p does not divide ak − bk for all k < n,with the following exceptions: a = 2,  b = 1; n = 6 and a + b = 2k; n = 2.Let See .2 − 2(13)4 = −1 and (3)5 − 2(11)2 = 1 every solution of the equationm = n = 2; that is, it comes from a unit p − q · 21/2 of the quadratic field Q(21/2) for which the coefficients p and q are primes.With the exceptions of the relations (239)See , 21.L be a nonabelian simple group and let O denote the order of the outer automorphism group of L.Let L be a nonabelian simple group. Then the orders and their outer automorphism of L are as listed in Tables Let See .In this section, we give the main theorem and its proof.G be a finite group with trivial center and N(G) = N(Ap+3) with p + 2 composite. Then G is isomorphic to Ap+3.Let Ap+3 are characterized by N(G) if p = 7,13,19 . The. TheAp+We divide the proof into the following lemmas.L = Ap+3. Then the following hold.(1)r ≤ [(p + 3)/2], then we can write p + 3 = kr + m with 0 ≤ m < r and conjugacy class sizes of r-elements of L areIf 2 ≠  i with 1 ≤ i ≤ k = [(p + 3)/r].for possible  r is an odd prime divisor of |G|, then conjugacy class sizes of r-element of L areIn particular, if  p + 3 = 2r + k and 0 ≤ k < r.where (2)r = 2, then one can write p + 3 = 2k + m with 0 ≤ m ≤ 1 and conjugacy class sizes of 2-elements of L areIf  i with 1 ≤ i ≤ k = [(p + 3)/2].for possible (3)r > [(p + 3)/2], then one can write p + 3 = r + m with 0 ≤ m < r and conjugacy class sizes of r-elements of L areIf  r = p, then the conjugacy class size of p-elements of L isIn particular, if (4)N(G) are maximality with respect to divisibility.The following numbers from (a)One of the following holds:  p + 3)!/2r2 if 2 · r = p + 3;(  p + 3)!/4r2 if 2 · r + 2 = p + 3;(  p + 3)!/2(k − 1)r2 if 2 · r + k = p + 3 and k = 2n with n ≥ 2;((b)p + 3)!/6p. = N(L). Then |L|∣|G| and π(G) = π(L).Let L | = ∏n∈N(L)n. Since |xG||CG(x)| = |G|, every member from N(G) divides the order of G and |L|∣|G|. So by π(G) = π(L).Note that |G is a finite group with trivial center and N(G) = N(L). Then the following hold.r1, r2, p from π(L) such that r1, r2, p > [(p + 3)/2]. In particular, the Sylow r-subgroup S of G is a cyclic group of order r where r ∈ {r1, r2, p}. There does not exist an element of order r1 · r2, r1 · p, or r2 · p.There exist different primes n ∈ N(G), if n is divisible at most by ra, then the Sylow r-subgroup S of G is of order ra.For all Suppose that r1, r2, p from π(G) such that r1, r2, p > [(p + 3)/2].(1) By r1, r2, p are prime divisors of |G| and r12, r22, p2 do not divide |xG| for all x ∈ G. Then by S is elementary abelian. Therefore if |x | = r, then |xG| is an r′-number.From Lemmas S | ≥p2. Consider an element y of G withLet |p∤|y|. Let x be an element of CG(y) having order p. Then CG(xy) = CG(x)∩CG(y), |xG|∣|(xy)G|, and |yG|∣|(xy)G| by S is abelian, S ≤ CG(x). Hence, p∤|xG|. It follows that |xG| equals (p + 3)!/6p or (p + 1)(p + 2)(p + 3)/3 by Assume that xG| equals (p + 3)!/6p, then (p + 3)!/6∣|(xy)G|. On the other hand, |yG|∣|(xy)G|; then we have thatN(G) such that |xG | ∣ | (xy)G| and |yG | ∣ | (xy)G|.If |xG| equals (p + 1)(p + 2)(p + 3)/3. In the following, we will consider the following three cases.Therefore |Case 1. |yG | = (p + 3)!/2 · r2 if 2r = p + 3.r∣(p + 1)(p + 2)(p + 3)/3. Therefore (p + 3)!/r∣|(xy)G|, a contradiction, since |xG|∣|(xy)G|, |yG|∣|(xy)G|, and the maximality of |yG| = (p + 3)!/2 · r2.Obviously,  Case 2. |yG | = (p + 3)!/4 · r2 if 2r + 2 = p + 3.r∣(p + 1)(p + 2)(p + 3)/3. Therefore (p + 3)!/r∣|(xy)G|. Also we get a contradiction as in Case 1.Obviously, Case 3. |yG | = (p + 3)!/2(k − 1)r2 if 2r + k = p + 3 and k = 2n with n ≥ 2.r∤(p + 1)(p + 2)(p + 3)/3. It follows that |yG | = |(xy)G|. By CG(y) ≤ CG(x) and so |xG | ∣ | yG|, a contradiction.In this case, p∣|y|. Let |y | = p · t. Since S is elementary abelian, the numbers p and t are coprime. Lety = uv and CG(uv) = CG(u)∩CG(v). Therefore,Assume that v of G is of order p. Since the Sylow p-subgroup of G is elementary abelian, then p∤|vG|. It follows thatOn the other hand, the element vG | = (p + 3)!/6p, then |vG|∣|yG|, a contradiction. Hence |vG | = (p + 1)(p + 2)(p + 3)/3. We consider the following three cases.If |Case 1. |yG | = (p + 3)!/2 · r2 if 2r = p + 3.r∣(p + 1)(p + 2)(p + 3)/3. But r∤(p + 3)!/2r2, a contradiction, since |xG|∣|(xy)G|, |yG|∣|(xy)G|, and the maximality of |yG | = (p + 3)!/2 · r2.Obviously,  Case 2. |yG | = (p + 3)!/4 · r2 if 2r + 2 = p + 3.r∣(p + 1)(p + 2)(p + 3)/3. But r∤(p + 3)!/4r2, a contradiction, since |xG|∣|(xy)G|, |yG|∣|(xy)G|, and the maximality of |yG| = (p + 3)!/4 · r2.Obviously,  Case 3. |yG | = (p + 3)!/2(k − 1)r2 if 2r + k = p + 3 and k = 2n with n ≥ 2.r∤(p + 1)(p + 2)(p + 3)/3. It follows that |yG | = |(xy)G| since the maximality of |vG|. By CG(y) ≤ CG(x) and so |xG|∣|yG|, a contradiction.In this case,  p-subgroup of G is of order p.Therefore the Sylow Similarly we can prove the other two cases.r1 · r2, r1 · p, or r2 · p.There does not exist an element of order n is divisible at most by r2.(2) Without loss of generality, we assume that S | ≥r3. Consider an element x of G such thatAssume that |r∤|x|. Then there is an element y of G of order r. By CG(xy) = CG(x)∩CG(y), |xG|∣|(xy)G|, and |yG|∣|(xy)G|.Let y is an r-central element, then S ≤ CG(y) and |yG| is an r′-number. We consider the following three cases.If Case 1. 2r = p + 3.yG | = (p + 3)!/2 · r2. Then we have that |xG | = |yG | = |(xy)G| and so r2∣∣ | CG(x)| = |CG(y)| since y is an r-central element |, then there is a conjugacy class size which is a multiple of the number (p + 3)!/r3 · 3! contradicting 2r = p + 3). Thus r∣|xG| = |yG| since r3∣|G|, a contradiction. In this case |Case 2. 2r + 2 = p + 3.yG | = (p + 3)!/4 · r2. We similarly can rule out this case as “Case 1: 2r = p + 3”. We have that |Case 3. 2r + k = p + 3 and k = 2n with n ≥ 2.yG | = (p + 3)!/2(k − 1)r2 or (2r + 1)(2r + 2) ⋯ (2r + k)/2(k − 2) ). If the former, then |xG | = |yG | = |(xy)G| and so r2∣∣ | CG(x)| = |CG(y)| since y is an r-central element |, then there is a conjugacy class size which is a multiple of the number (p + 3)!/r3 · 3! contradicting 2r + k = p + 3). Thus r∣|xG| = |yG| since r3∣|G|, a contradiction. Then |yG | = (2r + 1)(2r + 2)⋯(2r + k)/2(k − 2). But there is no member from N(G) such that |xG | ∣ | (xy)G| and |yG | ∣ | (xy)G|.In this case, |y is a noncentral r-element, then we choose an element z of order p such that r2∣∣ | zG|. By hypothesis, r∣ | CG(z)| and, obviously, r∤|z|. Then by Lemma 1.2 of  and r ≥ [(p + 3)/2]”.r > [(p + 3)/2] and r ≠ r1, r2, p. In this case, by r-subgroup of G is of order r. Hence there is a Hall {r, p}-subgroup H. Since H must be cyclic, then there is an element of order r · p, a contradiction by the proof of r ≤ [(p + 3)/2] and r ≠ 2,3. If 2r = p + 3, 2r + 2 = p + 3, or 2r + k = p + 3 with k = 2n and n ≥ 2, then the index of A is at most r2. By m of ϕ(p) such that p divides rm − 1, and the subgroup Z(P)∩A, then the order of p. By y lies in the center of a Sylow p-subgroup of G. This contradicts It follows that r ∈ {5,7,…, p}.Let Oπ,π′(G) = Oπ(G). In particular, G is insoluble.Therefore K ≤ H ≤ G such that H/K≅Ap+3.There is a normal series 1 ≤ G | = (p + 3)!/2.By Lemmas M : H/K = S1 × S2 × ⋯×Sk is a direct product of nonabelian simple groups S1, S2,…, Sk. Since G cannot contain a Hall {r1, r2, p}-subgroup, numbers r1, r2, and p divide the order of exactly one of these groups that is listed as in S1. Since G* and M* denote the factor groups M/S1, respectively. If k > 1, then a Sylow 2-subgroup of G* is nontrivial and its center Z has a nontrivial intersection with M*. Consider a nontrivial element y of T = S2 × ⋯×Sk such that its image in Z. Since y centralizes S1, it lies in the center of a Sylow 2-subgroup of p, a contradiction. Thus M = S1, and r1, r2, p∣ | H/K| |, a contradiction, from r1, r2, p∣ | K|, then there is an element of order r · p with r ∈ {r1, r2} contradicting r ∈ π(G) = π(L). In the following, we consider S1 which is listed as in Tables By erefore2HK≤G¯≤Aut(i) Case 1. H/K≅An with n ≥ 6.n = p, p + 1,…, p + k with p + 2, p + 4,… composite and p + k + 1 prime. If k ≥ 4, then (p + k)!/2∣(p + 3)!, a contradiction. Therefore H/K is isomorphic to Ap,  Ap+1, Ap+2, or Ap+3.Then x be an element of order p in H. Then |xH| is p′-number since |H|p = pH/K≅Ap.Let Ap | ∣(p + 3)!, then 3∣ | K|. We have |xH | = (p − 1)!/2. On the other hand, |xG | = (p + 3)!/6p. It follows that |xK | ∣(p + 1)(p + 2)(p + 3)/3 and so there is an element of r · p or of order r′ · p with 3 < r′ < r < p and r and r′ divide one of the prime divisors of the numbers p + 1, p + 2, or p + 3, which contradicts H/K≅Ap+1 and H/K≅Ap+2”.Since |H/K≅Ap+3.Therefore (ii) Case 2. H/K is not isomorphic to a sporadic simple group according to (iii) Case 3. H/K is isomorphic to a simple group of Lie type.q be a prime power.(1)H/K≅Bn(q) with n ≥ 2. π(G) = {2,3, 5,7,…, p} and soIn this situation, by  p∣q or p∣∏i=1n(qi2 − 1). If p∣q, then q is a power of p. Since |Gp | = p by n ≥ 2. Therefore p∣∏i=1n(qi2 − 1). It follows that p∣qt2 − 1 for some 1 ≤ t ≤ n as p is prime. On the other hand, qn2∣r or qn2∣rm. If the former, then q = r and n = 1, a contradiction. It follows that qn2∣rm. Hence r∣q and m = kn2 for some integer k ≥ 1. By kn2 < p/2 < (qn2 − 1)/2, but the equation has no solution in N. Furthermore, since Cn(q) has the same order as Bn(q), we also can rule out.It follows that (2)H/K≅Dn(q) with n ≥ 4. qn − 1))qn(n−1)(qn − 1)∏i=1n−1(qi2 − 1)∣(p + 3)!. Since the Sylow p-subgroup of G is of order p, p∤q as, otherwise, q = p and thus n = 1, a contradiction. It follows that p∣qn − 1 or p∣qt2 − 1 for some integer 1 ≤ t ≤ n − 1.Therefore we have that ∣r < p with r prime, and so qn(n−1) < pn − 1. Thus n = 1, a contradiction.Let  p∣qt2 − 1 for some integer 1 ≤ t ≤ n − 1. Then qn(n−1)∣r < p with r prime, and so qn(n−1) − 1 ≤ qt2 − 1 < qn2 − 1. It follows that n = 1,2, 3, a contradiction.Let (3)H/K≅2An(q) with n ≥ 2. In this situation, p-subgroup of G is of order p and n ≥ 2, we obtain that p∣qn(n+1)(1/2) or p∣qt+1 − (−1)t+1 for some integer 1 ≤ t ≤ n.Since the Sylow  q = p and n = 1, a contradiction.If the former, then we have that  p∣qt+1 − (−1)t+1 for some integer 1 ≤ t ≤ n. Then qn(n+1)(1/2)∣rm for some m. If m = 1, then q = r and n = 1, a contradiction. It follows that qn(n+1)(1/2)∣rm for some m > 1 and r∣q. By r ≤ [p/2] < p/2 and n(n + 1)/2 ≤ m ≤ p/2.Let  t is odd, then p∣rt+1 − 1. By Lemmas q = 2, t = 5, and r = 2. Hence p = 3 or 7. If p = 3, we can rule out this case since p ≥ 5. If p = 7, then n(n + 1) ≤ 3 and so n = 1,2, a contradiction, since t = 5 < n.If  t is even, then p∣rt+1 + 1. So p∣r + 1 or p∣rt − rt−1 + ⋯+1. If the former, then p∣r + 1 < [p/2] + 1, a contradiction. If the latter, then p < rt + 1. It follows that p ≤ rt − 1. Let t = 2k with 1 < k < [p/2]. Then p ≤ rk + 1 or p ≤ rk − 1 and so p∣rk + 1 or p∣rk − 1. If p∣rk − 1, then, by r = 2 and t = 5; we also can rule out this case as above. So p∣rk + 1. It follows that p∣rt − 1. Similarly, as p∣rk − 1, we can rule out this case.If (4)H/K≅E8(q). Therefore we have that It follows that p∣q120, then we can rule out this case since the Sylow p-subgroup of G is of order p. Hence p∣qt − 1, where t ∈ {2,8, 12,14,18,20,24,30}. On the other hand, rm∣q120. If m = 1, then q = r and 1 > 120, a contradiction. If m > 1, then q = r and m ≤ 120. By p and so p ∈ {5,7, 11,13,…, 101,103,107,109,113}. It is easy to rule out this case by considering the orders of G. Similarly, we can exclude that H/K≅E6(q), E7(q), and F4(q).If (5)H/K≅G2(q). q6(q6 − 1)(q2 − 1)∣(p + 3)!. It follows that q6∣p, p∣q6 − 1, or p∣q2 − 1.Then we have  q6∣p, we rule out this case.If  p∣p6 − 1, then there exists a prime r such that q6∣rm for some integer m. Therefore q = r and 6 ≤ m < p by p = 5 and so we have a contradiction by considering the order of G. Similarly, we also can rule out this case “p∣q2 − 1”.If (6)H/K≅2E6(q). q + 1))q36(q12 − 1)(q9 + 1)(q8 − 1)(q6 − 1)(q5 + 1)(q2 − 1)∣(p + 3)!. It follows that p∣qt − 1 with t = 12,8, 6,2, p∣qk + 1 with k = 9,5, or p∣q36. If p∣q36, then we rule out this case since the Sylow p-subgroup of G is of order p. So p∣qt − 1 with t = 12,8, 6,2, p∣qk + 1 with k = 9,5, and so there exists a prime r such that q36∣rm for some integer m. It means that 36 ≤ m ≤ p. Therefore p = 31,29,23,19,17,13,11,7. We also can rule out this case by order consideration.It is easy to see that H/K≅2B2(q) with q = 2m+12. q2(q2 + 1)(q − 1)∣p!. Thus q2∣p, p∣q2 + 1, or p∣q − 1.It follows that  q2∣p, then we rule out this case.If  p∣q2 + 1, then there is a prime r such that q2∣rm and so 2 ≤ m ≤ p by p = 2, a contradiction. Similarly we can rule out this case “p∣q + 1”.If  H/K≇2F4(2m+12).Similarly (8)H/K≅2G2(q), q = 3n+12 with n ≥ 1. q3(q3 + 1)(q − 1)∣p!. Since the Sylow p-subgroup of G is of order p, then p∤q3. It follows that p∣q3 + 1 or p∣q − 1. If p∣q3 + 1, then there exists a prime r such that q3∣rm for some integer m. If m = 1, then 1 > 3, a contradiction. Hence 3 ≤ m ≤ p by p = 3, a contradiction. If p∣q − 1 and r∣q, then there exists a Frobenius group of r · p with a kernel of order r and a complement of order p, respectively, and so there is an element of order r · p, which contradicts We see that (9)H/K≅3D4(q). q12(q8 + q4 + 1)(q6 − 1)(q2 − 1)∣p!. In this case, since G has a Sylow p-subgroup of order p, then p∣q8 + q4 + 1, q∣q6 − 1, or p∣q2 − 1. If p∣q8 + q4 + 1, then there exists a prime r such that rm∣q12 and so m ≤ 12. By p ≤ 12. It follows that p = 5,7, 11. Order consideration rules out these cases “p = 5,7, 11”. Similarly we can rule out this case “p∣q2 − 1”.We have that (10)H/K≅An(q) with n ≥ 1. It is easy to get that p∣qn(n+1)/2 or p∣∏i=1n(qi+1 − 1). If p∣qn(n+1)/2, then n = 1 since the Sylow p-subgroup of G is of order p, a contradiction. Hence p∣∏i=1n(qi+1 − 1) and so p∣qt+1 − 1 for some integer 1 ≤ t ≤ n. It follows that there exists a prime r such that qn(n+1)/2∣rm and so n(n + 1)/2 ≤ m ≤ p/2 by p-subgroup of G is of order p, then n(n + 1)∣p and so n = 1, a contradiction.It follows that This completes the proof of the lemma.Let Consider the following.G≅Ap+3.By K ≤ G into the chief ones. We prove that K = 1. By π(K)⊆{2,3}.So K is a 2-group, in this case, let x centralizes K. It follows that there is an element of 2 · p which contradicts If K is a 3-group. Then similarly as the case “K is a 2-group”, we have that N(G) and CG(x) is abelian. So by Lemma 1.12 of [K ≤ Z(G) = 1.If  1.12 of , K ≤ Z = N(Ap+3) and |G | = |Ap+3|. Then G≅Ap+3.Let An with n = 10,16,22,26 are characterized by N(G). Then by [One knows that the alternating groups  Then by , 15, 21,G be a finite group with trivial center. Assume that N(G) = N(An) with n = p, p + 1, p + 2, p + 3. Then G≅An.Let


H1-Galerkin mixed finite element (H1-GMFE) method to look for the numerical solution of time fractional telegraph equation. We introduce an auxiliary variable to reduce the original equation into lower-order coupled equations and then formulate an H1-GMFE scheme with two important variables. We discretize the Caputo time fractional derivatives using the finite difference methods and approximate the spatial direction by applying the H1-GMFE method. Based on the discussion on the theoretical error analysis in L2-norm for the scalar unknown and its gradient in one dimensional case, we obtain the optimal order of convergence in space-time direction. Further, we also derive the optimal error results for the scalar unknown in H1-norm. Moreover, we derive and analyze the stability of H1-GMFE scheme and give the results of a priori error estimates in two- or three-dimensional cases. In order to verify our theoretical analysis, we give some results of numerical calculation by using the Matlab procedure.We discuss and analyze an  Ω is a bounded domain with boundary ∂Ω and J =  is a given source function, u0(x) and u1(x) are two given initial functions, and the time Caputo fractional-order derivatives ∂t0,αu and ∂t0,α2u are defined, respectively, byα < 1.In this paper, our purpose is to present and discuss a mixed finite element method for the time fractional telegraph equationIn the current literatures, we can see that some numerical methods for solving fractional partial differential equations (PDEs), which include finite element methods –8, mixedH1-GMFE method. This method includes some advantages, such as avoiding the LBB consistency condition, allowing different polynomial degrees of the finite element spaces, and obtaining the optimal a priori estimates in both H1 and L2-norms. In 1/2. When m = 0, we simply write the norm ||·||0 as ||·||.For the need of study, we denote the natural inner product as  in /∂x and split /∂t = ∂u/∂t = 0 and ∂2u/∂t2 = ∂2u/∂t2 = 0, we easily getv/∂x, v ∈ H01, and integrate with respect to space from xL to xR to obtainVh ⊂ H01 and Wh ⊂ H1, which satisfy the following approximation properties: for 1 ≤ p ≤ ∞ and k, r positive integers  with step length Δt = T/M and nodes tn = nΔt, for some positive integer M. For a smooth function ϕ on , define ϕn = ϕ(tn). In the following analysis, for deriving the convenience of theoretical process, we now denoteFor formulating fully discrete scheme, let 0 = t0,ασ at t = tn is approximated by, for 0 < α < 1,The time fractional derivative ∂t0,α2σ at t = tn is estimated by, for 1 < 2α < 2,The time fractional order derivative ∂u and σ.In the next analysis, we will derive and prove some a priori error results for E0n = Eαn + Eα2n.Based on the approximation formulas  and 22)22) of tiuhn, σhn) ∈ Vh × Wh such thatNow, we formulate a fully discrete procedure: find (Making a combination of -25) wit wit25) wIn the following discussion, we will derive the proof for the fully discrete a priori error estimates.σ0 = ux0 and σt(0) = ux1, suppose that un ∈ H01∩Hk+1, σn ∈ Hr+1, σh0 = Rhσ(0), and σht(0) = Lhσt(0), where Lh is the L2 projection defined by  = 0, wh ∈ Wh. Then there exists a positive constant C free of space-time discrete parameters h and Δt such that, for 1/2 < α < 1,α → 1With L2-norm ||ϑn|| and the H1-norm ||ϑn||1. Taking vh = ϑn in ∑k=sily get||ϑn||≤||quality:Γ(2−α)∑k=tion to (−Γ(2−α)∑kNoting (βΓα(ϕn+ϑnrrive atΓ(2−α)∑k=Bkα2 − Bk+1α2 > 0, we arrive atBkα − Bk+1α > 0, we use the similar method to haveB0α = B0α2 = 1, we arrive atμ0||δn||1 ≤ B/||δn||1 ≤ B/||δn|| to getn = 1 in /Bj−1α2)(hk+1 + hr+1 + Δt2 + Δtα1+2) hold for j = 1,2,…, n − 1, we now prove that the inequality  ↦ Vh × Wh such thatIn order to get fully discrete mixed finite element scheme, we now choose the finite element spaces integers ,(59)infIn the following discussion, we will give the stability for the system  and 61)61). Firsq0 = ∇u0 and qt(0) = ∇u1, the following inequality holds:With By a similar discussion as in , we have||qh−1||≤The following stable inequality for the system  and 61)61) holdsvh = ∇uhn and use Cauchy-Schwarz inequality and Poincaré inequality to arrive atwh = qhn and use Cauchy-Schwarz inequality, Young inequality, and proof of ||qhn||+κrrive at||uhn||≤|q.For deriving the a priori error analysis, we define the Ritz projection Let u~h; see , 44; we q0 = ∇u0, qt(0) = ∇u1, un ∈ H01∩Hk+1, and qn ∈ Hr+1, there exists a positive constant C free of space-time meshes h and Δt such that, for 1/2 < α < 1,α → 1With We can use a similar proof as in κ = 1/2 and β = 0, the exact solution u = tα2+sin(2πx), for all ∈×, and the determined source term f by the exact solution in , b = Δtα−/Γ(2 − α), Here, in order to show the numerical performances on the rate of convergence and a priori error estimates, we now choose ution in  and then= 1. Letuhn=∑k=1Nressions :(78)[. From the calculated data, we can find that the rates of convergence, which are higher than the results O(Δtα3−2 + h2) of theory, gradually decrease with the increased α . In u and σ in H1-norm, which are unchanged with the changed α = 0.6,0.7,0.8,0.9, are given.In H1-GMFE method.In view of the above analysis on the numerical results, we now announce that the time fractional telegraph equation can be well solved by the H1-GMFE method for time fractional telegraph equation, especially, has not been made and discussed. In this paper, we give the detailed proof's process of the error analysis on H1-GMFE method for time fractional telegraph equation. Further, we provide a numerical procedure to verify the theoretical results of the studied method.As far as we know, more and more people have proposed and analyzed a lot of numerical methods for fractional partial differential equations. However, the discussions on mixed finite element methods for solving fractional partial differential equations are fairly limited. The a priori error analysis of H1-Galerkin moving mixed finite element method, which is based on a combination of H1-GMFE methods and moving finite element methods.In the near future, our aim is to study an


Ω ⊂ ℝN be a bounded domain, let σ be a nonnegative locally integrable function or, more generally, a locally finite measure on Ω, and let ω be a nonnegative Borel measure. In this paper, we consider the following nonlinear partial differential equations with measure data:P(u) = Pℓ,α,β(u) ∈ Lσ,loc1(Ω) is defined, following (x), which includes fractional Laplacian (−Δ)γ. Therefore, we also can obtain similar results of these integral equations.The four previous theorems are particular case of the more general class of nonlinear Wolff integral equations: p-superharmonic. Theorems σ and ω provided that there exist solutions of . We also denote by B = {x ∈ ℝn : |x − x0| < r} the open ball with center x0 and radius r > 0; when it is not important or clear from the context, we shall omit denoting the center as Br = B.In this section, we first recall some notations and definitions. In the following, we denote by p > 0 and σ be a nonnegative Borel measures in Ω which are finite on compact subsets of Ω. The σ-measure of a measurable set E ⊂ Ω is denoted by σ(E)≔∫Edσ. We denote by Lp  , resp.) the space of measurable functions f such that |f|p is integrable  with respect to σ. When dσ = dx, we write Lp(Ω) (Llocp(Ω), resp.).Let α > 0,  p > 1, such that αp < n, the R-truncated Wolff's potential Wp1,Rμ(x) of a nonnegative Borel measure μ on ℝN is defined by Wp1,[μ](x) the ∞-truncated Wolff's potential.For u : Ω → , p-harmonic in D, the implication holds: h ≤ u on ∂D implies h ≤ u in D. Note that p-superharmonic function u does not necessarily belong to Wlocp1,(Ω), but its truncation Tk(u) = min{u, k} does for every integer k; therefore, we will need the generalized gradient of a p-superharmonic function u defined by Du = limk→∞∇(Tk(u)). For more properties of p-superharmonic, see (x) is the Wolff potential of μ.Let r > 0, we consider a ball B ⊂ Ω and shrinking balls Bj≔Brj(x0), where rj = r2j− with j ≥ 0 is an integer.Given μ be locally finite nonnegative measures on Ω. Then, there exists a constant C = C > 0 such that for any s > 0 we haveϕ = ∑j=0∞cjχBj with cj = C[σ(Bj)/(r2j−)N−p]p−1)1/ > 0 such that if μ = Fk[−u] thenBR3 ⊂ Ω.Let p-superharmonic supersolution, then ω is absolutely continuous with respect to σ. The fact can be used to obtain a characterization of removable singularities for the homogeneous quasilinear equation: In this section, we will give the proof of our main theorem. Firstly, we prove the following integral estimate for solutions of quasilinear equations , which s 2.18 in  and Theo 2.18 in .σ and ω be locally finite nonnegative measures on Ω and p > 1. There exists a constant C = C > 0 such that if u(x) is a solution to  ⊂ Ω. Suppose that u is a positive solution of (x ∈ BR(x0),C depends on N, p.For fixed ution of . In viewution of , we findσ on BR(x0) and let dσ′ = χBR(x0)dσ; thus, taking into account (x ∈ BR(x0),M is defined as𝔐 is a nonlinear integral operator defined by 𝔐f = Wp1,R(fℓβdσ′). The iterates of 𝔐 are denoted by 𝔐if = 𝔐(𝔐i−1f). It is then easy to see from s > 0, ϕ(y) appears in Restrict the integration  account , we obtaaccount (ux≥Cαll!1account (ux≥Cαll!1BR(x0),uxC1+lβ/pIn the following, we will divide the proof into two cases.Case  1 (u(x0) < ∞). In this case,  = ∞). According to  is the sum of the k × k principal minors of D2u, which coincides with the Laplacian F1[u] = Δpu if k = 1, and the Monge-Ampère operator Fn[u] = det(D2u) if k = n.We now move to and Wang –17:(54)FWk/(k+1),k+12R in place of Wp1,R. Therefore, the proof is omitted.The proof of the following theorems is completely analogous to that of . One onlu(x) be a solution of (Ω with p > 1 and ℓβ > k. Suppose that BR4(x0) ⊂ Ω. Then, there exists a constant M = M such that r ≤ R/32.Let ution of  in Ω witu(x) be a solution of (Ω with p > 1 and P(u) = σup−1. Suppose that BR4(x0) ⊂ Ω. Then, there exists a constant M = M such that r ≤ R/32.Let ution of  in Ω wit


EQ-algebras. First,we present several characterizations of fuzzy positiveimplicative prefilters (filters), fuzzy implicative prefilters(filters), and fuzzy fantastic prefilters (filters). Next,using their characterizations, we mainly consider therelationships among these special fuzzy filters. Particularly,we find some conditions under which a fuzzyimplicative prefilter (filter) is equivalent to a fuzzy positiveimplicative prefilter (filter). As applications, weobtain some new results about classical filters in EQ-algebras and some related results about fuzzy filters inresiduated lattices.We introduce and study sometypes of fuzzy prefilters (filters) in  BL-algebras, MTL-algebras, MV-algebras, and so forth are the best known classes of residuated lattices .Aμ be a fuzzy set in L. For all t ∈ , the set μt = {x ∈ L | μ(x) ≥ t} is called a level subset of μ.Let F be a nonempty subset, we denote the characteristic function of F by χF.Let a, b ∈ , we denote max{a, b} and min{a, b} by a∨b and a∧b, respectively.For convenience, for any μ, ν in L, we define μ⊆ν if and only if μ(x) ≤ ν(x) for all x ∈ L.For any fuzzy sets EQ-algebras and give some of their properties that will be used in the sequel.In this section, we introduce the notion of fuzzy prefilters (filters) in L denote an EQ-algebra unless otherwise is specified.In what follows, let μ be a fuzzy set in L. μ is called a fuzzy prefilter of L if it satisfies(FF1)μ(1) ≥ μ(x) for all x ∈ L,(FF2)μ(y) ≥ μ(x)∧μ(x → y) for all x, y ∈ L. μ is called a fuzzy filter if it satisfiesA fuzzy prefilter (FF3)μ((x⊙z)→(y⊙z)) ≥ μ(x → y) for all x, y, z ∈ L.Let EQ-algebras exist.The following examples show that fuzzy prefilters (filters) in L = {0, a, b, 1} be a chain with Cayley tables as Let L, ∧, ⊙, ~, 1) is an EQ-algebra in , μt is either empty or a prefilter (filter) of L.Let The  proof  is straightforward.F be a nonempty subset of L. F is a prefilter (filter) if and only if χF is a fuzzy prefilter (filter) of L.Let EQ-algebra, which will be used in the sequel.Next, we discuss some properties of fuzzy filters in an L be an EQ-algebra and let μ be a fuzzy filter of L; for all x, y, z ∈ L, the following hold: μ(x⊙y) = μ(x)∧μ(y),μ(x → z) ≥ μ(x → y)∧μ(y → z).Let x⊙y ≤ x∧y ≤ x, y, then μ(x⊙y) ≤ μ(x)∧μ(y) by y ≤ 1 → y, it follows that μ(y) ≤ μ(1 → y). Since μ is a fuzzy filter of L, then μ((x⊙1)→(x⊙y)) ≥ μ(1 → y) by (FF3); that is, μ(x → (x⊙y)) ≥ μ(1 → y). By (FF2), we get μ(x⊙y) ≥ μ(x)∧μ(x → (x⊙y)). Hence, we have μ(x⊙y) ≥ μ(x)∧μ(x → (x⊙y)) ≥ μ(x)∧μ(1 → y) ≥ μ(x)∧μ(y). Therefore, μ(x⊙y) = μ(x)∧μ(y).(1) Since x → y)⊙(y → z) ≤ x → z, we have μ(x → z) ≥ μ((x → y)⊙(y → z)) by μ((x → y)⊙(y → z)) = μ(x → y)∧μ(y → z). Therefore, μ(x → z) ≥ μ(x → y)∧μ(y → z).(2) Since  in μ be a fuzzy prefilter in L. μ is called a fuzzy positive implicative prefilter of L if it satisfies(FF4)μ(x → z) ≥ μ(x → (y → z))∧μ(x → y) for all x, y, z ∈ L.Let μ of L is called a fuzzy positive implicative filter if it satisfies (FF4).A fuzzy filter For better understanding of the above definition, we illustrate it by the following example.L be the EQ-algebra and let μ be the fuzzy set of L defined in μ is a fuzzy positive implicative filter of L.Let The following result gives a characterization of fuzzy positive implicative prefilters by fuzzy prefilters.μ be a fuzzy prefilter in L. μ is a fuzzy positive implicative prefilter of L if and only if μa : L →  is a fuzzy prefilter in L, where μa(x) = μ(a → x) for all a, x ∈ L.Let μ is a fuzzy positive implicative prefilter of L. Since a → 1 = 1, then μ(a → 1) = μ(1). It follows that μa(1) = μ(a → 1) = μ(1). From a → x ≤ 1, we have μ(a → x) ≤ μ(1); that is, μa(x) ≤ μ(1). Then, μa(x) ≤ μa(1). On the other hand, since μ is a fuzzy positive implicative prefilter of L, then μ(a → y) ≥ μ(a → (x → y))∧μ(a → x); that is, μa(y) ≥ μa(x → y)∧μa(x). Therefore, μa is a fuzzy prefilter in L.Suppose  that  a ∈ L, since μa is a fuzzy prefilter in L, then μx is a fuzzy prefilter in L. It follows that μx(z) ≥ μx(y → z)∧μx(y). By the definition of μx, we have μ(x → z) ≥ μ(x → (y → z))∧μ(x → y). Therefore, μ is a fuzzy positive implicative prefilter of L.Conversely, for any μa.As a consequence of μ be a fuzzy positive implicative prefilter of L. Then, for any a ∈ L, μa is the fuzzy prefilter containing μ.Let μ is a fuzzy  positive implicative prefilter of L. By μa is a fuzzy prefilter. For any x ∈ L, x ≤ a → x. It follows from μ(x) ≤ μ(a → x); that is, μ(x) ≤ μa(x). And so μ⊆μa. Therefore, μa is the fuzzy prefilter containing μ.Assume that μ, ν be two fuzzy prefilters of L. Then, for any a, b ∈ L, the following statements hold: μa = μ if and only if μ(a) = μ(1),a ≤ b implies that μb⊆μa,μ⊆ν implies that μa⊆νa,μ∩ν)a = μa∩νa, (μ ∪ ν)a = μa ∪ νa. = μa(a) = μ(a → a) = μ(1); hence, μ(a) = μ(1).(1) Suppose that μ(a) = μ(1). For any x ∈ I, since x ≤ a → x and μ is a fuzzy prefilter, we have μ(x) ≤ μ(a → x) = μa(x). Hence, μ⊆μa. On the other hand, since μ is a fuzzy prefilter, then μ(x) ≥ μ(a → x)∧μ(a) = μ(a → x)∧μ(1) = μ(a → x) for all x ∈ L. It follows that μ(x) ≥ μa(x); that is, μ⊇μa. Consequently, we obtain μa = μ.Conversely, assume that a ≤ b. For any x ∈ L, then b → x ≤ a → x. Since μ is a fuzzy prefilter, we have μ(b → x) ≤ μ(a → x). It follows that μb(x) ≤ μa(x). So μb⊆μa.(2) Suppose that (3) and (4) It is easy to prove them, and we hence omit the details.In the following, we give some equivalent conditions of fuzzy positive implicative prefilters (filters) for further discussions.μ be a fuzzy prefilter (filter) in L. The following are equivalent:μ is a fuzzy positive implicative prefilter (filter) of L,μ(x → y) ≥ μ(x → (x → y)) for all x, y ∈ L,μ(x → y) = μ(x → (x → y)) for all x, y ∈ L.Let μ is a fuzzy positive implicative prefilter of L. Then, μ(x → y) ≥ μ(x → (x → y))∧μ(x → x) follows from μ(x → y) ≥ μ(x → (x → y)).  (1)⇒(2)  Suppose  that x → y ≤ x → (x → y), it follows that μ(x → y) ≤ μ(x → (x → y)), which together with (2) leads to μ(x → y) = μ(x → (x → y)).(2)⇒(3) From x → (y → z)≤((y → z)→(x → z))→(x → (x → z)) and x → y ≤ (y → z)→(x → z). Since μ is a fuzzy prefilter in L, then μ(x → (y → z)) ≤ μ(((y → z)→(x → z))→(x → (x → z))) and μ(x → y) ≤ μ((y → z)→(x → z)). From μ(x → (x → z)) ≥ μ((y → z)→(x → z))∧μ((y → z)→(x → z))→(x → (x → z))) ≥ μ(x → y)∧μ(x → (y → z)). Combining (3), we get μ(x → z) ≥ μ(x → y)∧μ(x → (y → z)). Therefore, μ is a fuzzy positive implicative prefilter of L.(3)⇒(1) By μ be a fuzzy set in L. μ is a fuzzy positive implicative prefilter (filter) of L if and only if, for all t ∈ , μt is either empty or a positive implicative prefilter (filter) of L.Let The proof  is easy, and we hence omit the details.F be a nonempty subset of L. F is a positive implicative prefilter (filter) if and only if χF is a fuzzy positive implicative prefilter (filter).Let Now, we continue to characterize fuzzy positive implicative filters.L be an EQ-algebra and let μ be a fuzzy filter. Then, μ is a fuzzy positive implicative filter of L if and only if μ(x∧(x → y) → y) = μ(1) for all x, y ∈ L.Let μ is a fuzzy positive implicative filter of L, then μ(x∧  (x → y) → y) ≥ μ(x∧(x → y)→(x → y))  ∧  μ(x∧(x → y) → x). For any x, y ∈ L, since x∧(x → y) ≤ x → y and x∧(x → y) ≤ x, then x∧(x → y)→(x → y) = 1 and x∧(x → y) → x = 1. Hence, μ(x∧(x → y)→(x → y)) = μ(1) and μ(x∧(x → y) → x) = μ(1). Consequently, we have μ(x∧(x → y) → y) = μ(1).Since μ(x → z) ≥ μ(x → (y∧(y → z)))∧μ((y∧(y → z)) → z) ≥ μ(x → (x∧y))∧  μ((x∧y)→(y∧  (y → z)))  ∧  μ(y∧  (y → z) → z). It follows from x → (y → z)≤(x∧y)→(y∧(y → z)) and x → y = x → (x∧  y). Then, μ((x∧y)→(y∧(y → z))) ≥ μ(x → (y → z)) and μ(x → (x∧y)) = μ(x → y). Combining μ(y∧(y → z) → z) = μ(1), we obtain that μ(x → z) ≥ μ(x → (y → z))∧μ(x → y). Therefore, μ is a fuzzy positive implicative filter of L.Conversely, by EQ-algebra.In order to get some related results about fuzzy filters in residuated lattices see , we charL be a residuated EQ-algebra and μ be a fuzzy filter. The following are equivalent:μ is a fuzzy positive implicative filter of L,μ(x → (x⊙x)) = μ(1) for all x ∈ L,μ(x∧y → (x⊙y)) = μ(1) for all x, y ∈ L,μ(x∧(x → y)→(x⊙y)) = μ(1) for all x, y ∈ L.Let L is a residuated EQ-algebra, we have 1 = (x⊙x)→(x⊙x) ≤ x → (x → (x⊙x)) by μ(1) ≤ μ(x → (x → (x⊙x))), which implies that μ(x → (x → (x⊙x))) = μ(1). From μ(x → (x⊙x)) ≥ μ(x → (x → (x⊙x)))∧μ(x → x) = μ(x → (x → (x⊙x))) = μ(1). Consequently, we have μ(x → (x⊙x)) = μ(1).  (1)⇔(2) Since μ(x → z) ≥ μ(x → (x⊙x))∧μ((x⊙x) → z) = μ(1)∧μ((x⊙x) → z) = μ((x⊙  x) → z) ≥ μ((x⊙  x)→(y⊙  (y → z)))∧μ((y⊙  (y → z)) → z). By y⊙(y → z) ≤ z. Then, μ((y⊙(y → z)) → z) = μ(1). It follows that μ(x → z) ≥ μ((x⊙x)→(y⊙(y → z))). Hence, we obtain that μ(x → z) ≥ μ((x⊙x)→(y⊙(y → z))) ≥ μ((x⊙  x)→(x⊙  y))  ∧  μ((x⊙  y)→(y⊙  (y → z))) by μ is a fuzzy filter, we have μ((x⊙x)→(x⊙y)) ≥ μ(x → y) and μ((x⊙y)→(y⊙(y → z))) ≥ μ(x → (y → z)) by (FF3). Combining them, we get μ(x → z) ≥ μ(x → (y → z))∧μ(x → y). Therefore, μ is a fuzzy positive implicative filter of L.Conversely, by μ is a fuzzy positive implicative filter of L. Then, μ(x∧(x → (x⊙y))→(x⊙y)) = μ(1) by L is a residuated EQ-algebra, we have y ≤ x → (x⊙y) by x∧y ≤ x∧(x → (x⊙y)), which implies that x∧(x → (x⊙y))→(x⊙y) ≤ x∧y → (x⊙y). From μ(x∧y → (x⊙y)) ≥ μ(x∧(x → (x⊙y))→(x⊙y)) = μ(1). Therefore, we have μ(x∧y → (x⊙y)) = μ(1).(1)⇒(3) Suppose that μ(x∧(x → y)→(x⊙(x → y))) = μ(1). By x⊙(x → y) ≤ x∧y, which implies that x∧(x → y)→(x⊙(x → y)) ≤ x∧(x → y)→(x∧y). Then, μ(x∧(x → y)→(x⊙(x → y))) ≤ μ(x∧(x → y)→(x∧y)). Hence, μ(x∧(x → y)→(x∧y)) = μ(1). By μ(x∧(x → y)→(x⊙y)) ≥ μ(x∧(x → y)→(x∧y))∧μ(x∧y → (x⊙y)) = μ(1)∧μ(1) = μ(1). Therefore, we have μ(x∧(x → y)→(x⊙y)) = μ(1).(3)⇒(4) From (3), it follows that x⊙(x → y) ≤ x∧(x → y), it follows that x⊙  (x → y)→(x⊙  y) ≥ x∧(x → y)→(x ⊙  y). Then, we have μ(x ⊙ (x → y)→(x ⊙ y)) ≥ μ(x ∧(x → y)→(x⊙y)) = μ(1), which implies that μ(x⊙(x → y)→(x ⊙ y)) = μ(1). Taking x = y, we have μ(x → (x⊙x)) = μ(1). Therefore, μ is a fuzzy positive implicative filter by (2).(4)⇒(1) From L be a residuated EQ-algebra and μ be a fuzzy filter. The following are equivalent:μ is a fuzzy positive implicative filter of L,μ((x⊙y) → z) = μ((x∧y) → z) for all x, y, z ∈ L,μ(x∧(x → y)→(x∧y)) = μ(1) for all x, y ∈ L,μ(x∧(x → y)→(y∧(y → x))) = μ(1) for all x, y ∈ L.Let μ is a fuzzy positive implicative filter. We have μ(y∧(y → z) → z) = μ(1) by L is a residuated EQ-algebra, then (x⊙y) → z ≤ x → (y → z) for all x, y, z ∈ L by μ((x⊙y) → z) ≤ μ(x → (y → z)). By x → (y → z)≤(x∧y)→(y∧(y → z)). Then, μ(x → (y → z)) ≤ μ((x∧y)→(y∧(y → z))). By μ((x∧y) → z) ≥ μ((x∧y)→(y∧(y → z)))∧μ((y∧(y → z)) → z) ≥ μ((x⊙y) → z)∧μ(1) = μ((x⊙y) → z). On the other hand, since x⊙y ≤ x∧y, then (x∧y) → z ≤ (x⊙y) → z. It follows that μ((x∧y) → z) ≤ μ((x⊙y) → z). Consequently, we have μ((x⊙y) → z) = μ((x∧y) → z).  (1)⇔(2) Suppose that L is a residuated EQ-algebra, then it is a good EQ-algebra. It follows from x⊙(x → y) ≤ y. Then, (x⊙(x → y)) → y = 1. So, μ((x⊙(x → y)) → y) = μ(1). Combining (2), we obtain μ((x⊙(x → y)) → y) = μ(x∧(x → y) → y) = μ(1). By μ is a fuzzy positive implicative filter of L.Conversely, since μ is a fuzzy positive implicative filter. Then, we have μ(x∧(x → y)→(x ⊙y)) = μ(1) for all x, y ∈ L by y ≤ x → y, it follows that x ⊙y ≤ x ⊙(x → y). Then, we have x∧(x → y)→(x ⊙y) ≤ x∧(x → y) → x⊙(x → y). Hence, we get μ(1) = μ(x∧(x → y)→(x⊙y)) ≤ μ(x∧(x → y) → x⊙(x → y)), which implies that μ(x∧(x → y) → x⊙(x → y)) = μ(1). Moreover, by x⊙(x → y) ≤ x∧y. Then, x∧(x → y)→(x ⊙(x → y)) ≤ x ∧(x → y)→(x∧y). It follows that μ(x ∧(x → y)→(x ⊙(x → y))) ≤ μ(x ∧(x → y)→(x∧y)). Therefore, μ(x∧(x → y)→(x∧y)) = μ(1).(1)⇒(3) Suppose that x ≤ y → x, it follows that x∧y ≤ (y → x)∧y. Then, x∧(x → y)→(x∧y) ≤ x∧(x → y)→(y∧(y → x)). Hence, we have μ(x∧(x → y)→(x∧y)) ≤ μ(x∧(x → y)→(y∧(y → x))). Combining (3), we get μ(x∧(x → y)→(y∧(y → x))) = μ(1).(3)⇒(4) From μ(x∧(x → (x⊙y))→((x⊙y)∧((x⊙y) → x))) = μ(1). Then, μ(x∧(x → (x⊙y))→(x⊙y)) = μ(1). Since L is a residuated EQ-algebra, we get y ≤ x → (x⊙y) by x∧(x → (x⊙y))→(x⊙y) ≤ x∧y → (x⊙y). Then, μ(1) = μ(x ∧(x → (x ⊙y))→(x ⊙y)) ≤ μ(x ∧y → (x ⊙  y)), which implies that μ(x∧y → (x⊙y)) = μ(1). Therefore, μ is a fuzzy positive implicative filter of L by (4)⇒(1) By (4), we have EQ-algebras.As an application of Theorems L be a residuated EQ-algebra and let F be a filter of L. The following are equivalent:F is a positive implicative filter of L,x → (x⊙x) ∈ F for all x ∈ L,x∧y → (x⊙y) ∈ F for all x, y ∈ L,x∧(x → y)→(x⊙y) ∈ F for all x, y ∈ L,x∧(x → y) → y ∈ F for all x, y ∈ L,x∧(x → y)→(x∧y) ∈ F for all x, y ∈ L,x∧(x → y)→(y∧(y → x)) ∈ F for all x, y ∈ L,x∧(x → y) → y ∈ F for all x, y ∈ L.Let The extension property for fuzzy positive implicative prefilters is obtained from the following proposition.L be a good EQ-algebra and let μ and ν be two fuzzy prefilters which satisfy μ⊆ν, μ(1) = ν(1). If μ is a fuzzy positive implicative prefilter, then so is ν.Let t = x → (x → y); then, x → (x → (t → y)) = t → (x → (x → y)) = 1. Since μ is a fuzzy positive implicative prefilter, we have μ(x → (t → y)) = μ(x → (x → (t → y))) = μ(1) by μ(x → (t → y)) = μ(1) = ν(1). From μ⊆ν, it follows that ν(x → (t → y)) ≥ μ(x → (t → y)) = ν(1). So, ν(x → (t → y)) = ν(1). Since ν is a fuzzy prefilter, then ν(x → y) ≥ ν(t → (x → y))∧ν(t). It follows that ν(x → y) ≥ ν(1)∧ν(t) = ν(t) = ν(x → (x → y)). By ν is a fuzzy positive implicative prefilter.We set L be an EQ-algebra and let μ and ν be two fuzzy filters which satisfy μ⊆ν, μ(1) = ν(1). If μ is a fuzzy positive implicative filter, then so is ν.Let It follows from EQ-algebras and discuss some of their properties.In this section, we introduce the notion of fuzzy implicative prefilters (filters) in μ be a fuzzy set in L. μ is called a fuzzy implicative prefilter of L if it satisfies(FF1)μ(1) ≥ μ(x) for all x ∈ L,(FF5)μ(x) ≥ μ(z → ((x → y) → x))∧μ(z) for all x, y, z ∈ L.Let μ of L is called a fuzzy implicative filter if it satisfies (FF3).A fuzzy implicative prefilter EQ-algebras exist.The following example shows that fuzzy implicative prefilters in L = {0, a, b, c, 1} be a chain with Cayley tables as Let L, ∧, ⊙, ~, 1) is an EQ-algebra in , μt is either empty or an implicative prefilter (filter) of L.Let The proof is easy, and we hence omit the details.F be a nonempty subset of L. F is an implicative prefilter (filter) if and only if χF is a fuzzy implicative prefilter (filter).Let μ be a fuzzy prefilter in L. We call that μ has weak exchange principle if it satisfies for all x, y, z ∈ L  μ(x → (y → z)) = μ(y → (x → z)).Let μ of a good EQ-algebra L has weak exchange principle.From L be the EQ-algebra and μ be the fuzzy set of L defined in L is not a good EQ-algebra and μ is a fuzzy prefilter of L, which has weak exchange principle.Let EQ-algebra with a bottom element 0.We have the following characterizations of fuzzy implicative prefilters in an L be an EQ-algebra with a bottom element 0 and let μ be a fuzzy prefilter with weak exchange principle. The following are equivalent:μ is a fuzzy implicative prefilter of L,μ(x → z) ≥ μ(x → (¬z → y))∧μ(y → z) for all x, y, z ∈ L,μ(x → z) ≥ μ(x → (¬z → z)) for all x, z ∈ L,μ(x → z) = μ(x → (¬z → z)) for all x, z ∈ L.Let y → z ≤ (x → y)→(x → z) and ¬z → (x → y)≤((x → y)→(x → z))→(¬z → (x → z)), then μ(y → z) ≤ μ((x → y)→(x → z)) and μ(¬z → (x → y)) ≤ μ(((x → y)→(x → z))→(¬z → (x → z))). From μ(¬z → (x → z)) ≥ μ((x → y)→(x → z))∧μ(((x → y)→(x → z))→(¬z → (x → z))) ≥ μ(y → z)∧μ(¬z → (x → y)) ≥ μ(y → z)∧μ(x → (¬z → y)). Notice that ¬(x → z)≤¬z, and we have ¬z → (x → z)≤¬(x → z)→(x → z). Then, μ(¬z → (x → z)) ≤ μ(¬(x → z)→(x → z)). Since μ is a fuzzy implicative prefilter of L, we have μ(x → z) ≥ μ(¬(x → z)→(x → z)) by μ(x → z) ≥ μ(x → (¬z → y))∧μ(y → z).  (1)⇔(2) Since μ satisfies μ(x → z) ≥ μ(x → (¬z → y))∧μ(y → z) for all x, y, z ∈ L. Then, μ(1 → x) ≥ μ(1 → (¬x → x))∧μ(x → x) ≥ μ(¬x → x)∧μ(1) = μ(¬x → x). Since μ is a fuzzy prefilter, then μ(x) ≥ μ(1)∧μ(1 → x) = μ(1 → x). It follows that μ(x) ≥ μ(¬x → x). By μ is a fuzzy implicative prefilter of L.Conversely, suppose that μ is a fuzzy implicative prefilter of L. Then, we have μ(x → z) ≥ μ(x → (¬z → z))∧μ(z → z) = μ(x → (¬z → z))∧μ(1) by (2). Therefore, we obtain μ(x → z) ≥ μ(x → (¬z → z).(1)⇒(3) Suppose that z ≤ ¬z → z, it follows that x → z ≤ x → (¬z → z). Then, μ(x → z) ≤ μ(x → (¬z → z)) as μ is a fuzzy prefilter. Combining (3), we get μ(x → z) = μ(x → (¬z → z)).(3)⇒(4) From μ be a fuzzy prefilter which satisfies μ(x → z) = μ(x → (¬z → z)) for all x, z ∈ L. Then, μ(1 → x) = μ(1 → (¬x → x)). Since μ is a fuzzy prefilter, then μ(x) ≥ μ(1 → x)∧μ(1) = μ(1 → x). From ¬x → x ≤ 1 → (¬x → x), it follows that μ(1 → (¬x → x)) ≥ μ(¬x → x). Consequently, we obtain μ(x) ≥ μ(¬x → x). Using μ is a fuzzy implicative prefilter of L.(4)⇒(1) Let EQ-algebras and discuss the relations among various fuzzy prefilters (filters). Particularly, we find some conditions under which a fuzzy implicative prefilter (filter) is equivalent to a fuzzy positive implicative prefilter (filter). Moreover, as applications, we obtain some relations among corresponding classical filters in EQ-algebras and some related results about fuzzy filters in residuated lattices.In this section, we introduce fuzzy fantastic prefilters (filters) in F be a prefilter in L. F is called a fantastic prefilter of L if it satisfies the following:(F6)y → x ∈ F implies that ((x → y) → y) → x ∈ F for all x, y ∈ L.Let F is a filter and it satisfies (F6), then it is called a fantastic filter of L.If It is time to give some examples of fantastic prefilters.L = {0, a, b, 1} be a chain with Cayley tables as Let L, ∧, ⊙, ~, 1) is an EQ-algebra. One can check that F = {1} is a fantastic prefilter.Then, μ(((x → y) → y) → x) ≥ μ(y → x) for all x, y ∈ L.Let μ of L is called a fuzzy fantastic filter if it satisfies (FF6).A fuzzy filter L, ∧, ⊙, ~, 1) be an EQ-algebra in μ in L as follows: μ(1) = 0.7 and μ(0) = μ(a) = μ(b) = 0.4. One can check that μ is a fuzzy fantastic prefilter of L.Let .μ be a fuzzy prefilter (filter) in L. Then, μ is a fuzzy fantastic prefilter (filter) of L if and only if μ(((x → y) → y) → x) ≥ μ(z → (y → x))  ∧  μ(z) for all x, y, z ∈ L.Let μ is  a fuzzy fantastic prefilter of L. From μ(((x → y) → y) → x) ≥ μ(y → x) for all x, y ∈ L. Since μ is a fuzzy prefilter in L, then μ(y → x) ≥ μ(z → (y → x))∧μ(z). Therefore, we obtain μ(((x → y) → y) → x) ≥ μ(z → (y → x))∧μ(z).Suppose that μ satisfies μ(((x → y) → y) → x) ≥ μ(z → (y → x))∧μ(z) for all x, y, z ∈ L. Taking z = 1, we obtain μ(((x → y) → y) → x)) ≥ μ(1 → (y → x))∧μ(1) = μ(1 → (y → x)). From y → x ≤ 1 → (y → x), it follows that μ(y → x) ≤ μ(1 → (y → x)). Consequently, we obtain μ(((x → y) → y) → x) ≥ μ(y → x). Therefore, μ is a fuzzy fantastic prefilter of L.Conversely, suppose that μ be a fuzzy set in L. μ is a fuzzy fantastic prefilter (filter) of L if and only if, for all t ∈ , μt is either empty or a fantastic prefilter (filter) of L.Let The proof is easy, and we hence omit the details.F be a nonempty subset of L. F is a fantastic prefilter (filter) if and only if χF is a fuzzy fantastic prefilter (filter).Let In what follows, we pay attention to the relations among various special fuzzy prefilters (filters).First, the relationship between fuzzy implicative prefilters and fuzzy fantastic prefilters can be described by the following theorem.EQ-algebra.Each fuzzy implicative prefilter (filter) with weak exchange principle is a fuzzy fantastic prefilter (filter) in an μ  is a fuzzy implicative prefilter with weak exchange principle. From x ≤ ((x → y) → y) → x, it follows that (((x → y) → y) → x) → y ≤ x → y. This implies that ((((x → y) → y) → x) → y)→(((x → y) → y) → x)≥(x → y)→(((x → y) → y) → x). Since μ is a fuzzy implicative prefilter, then μ(((((x → y) → y) → x) → y)→(((x → y) → y) → x)) ≥ μ((x → y)→(((x → y) → y) → x)) by μ((x → y)→(((x → y) → y) → x)) = μ(((x → y) → y)→((x → y) → x)) as μ has weak exchange principle. From y → x ≤ ((x → y) → y)→((x → y) → x), it follows that μ(y → x) ≤ μ(((x → y) → y)→((x → y) → x)). Together with them, we obtain μ(((((x → y) → y) → x) → y)→(((x → y) → y) → x)) ≥ μ(y → x). On the other hand, since μ is a fuzzy implicative prefilter, then μ(((x → y) → y) → x) ≥ μ(1 → ((((x → y) → y) → x) → y)→(((x → y) → y) → x))∧μ(1) = μ(1 → ((((x → y) → y) → x) → y)→(((x → y) → y) → x)) ≥ μ(((((x → y) → y) → x) → y)→(((x → y) → y) → x)). Consequently, we obtain μ(((x → y) → y) → x) ≥ μ(y → x). Therefore, we get that μ is a fuzzy fantastic prefilter by Suppose that From the following example, we can see that the converse of L be the EQ-algebra and let μ be the fuzzy set of L defined in μ is a fuzzy fantastic prefilter of L. But it is not a fuzzy implicative prefilter of L, since μ(b) < μ((b → 0) → b).Let Next, the following theorems show the relationship between fuzzy implicative prefilters (filters) and fuzzy positive implicative prefilters (filters).EQ-algebra.Each fuzzy implicative prefilter with weak exchange principle is a fuzzy positive implicative prefilter in an μ is a fuzzy implicative prefilter of L. Then, μ is a fuzzy prefilter of L by μ((x → (x → z)) ≥ μ(x → y)∧μ((x → y)→(x → (x → z))). Since y → (x → z)≤(x → y)→(x → (x → z)), then μ(y → (x → z)) ≤ μ((x → y)→(x → (x → z))). In a similar way, we get that μ(x → (x → z)) ≤ μ(((x → x) → z)→(x → z)). Hence, we obtain μ(((x → x) → z)→(x → z)) ≥ μ(x → y)∧μ(y → (x → z)) = μ(x → y)∧μ(x → (y → z)) as μ has weak exchange principle. From ((x → z) → z)→(x → z) ≤ 1 → (((x → z) → z)→(x → z)), it follows that μ(((x → z) → z)→(x → z)) ≤ μ(1 → (((x → z) → z)→(x → z))). Since μ is a fuzzy implicative prefilter of L, then μ(x → z) ≥ μ(1 → (((x → z) → z)→(x → z)))∧μ(1) = μ(1 → (((x → z) → z)→(x → z))). Consequently, we obtain μ(x → z) ≥ μ(x → y)∧μ(x → (y → z)). Therefore, μ is a fuzzy positive implicative prefilter.Suppose that  EQ-algebra.Each fuzzy implicative filter is a fuzzy positive implicative filter in an x∧(x → y) ≤ x and x∧(x → y) ≤ x → y, it follows that x∧(x → y) ≤ x → y ≤ (x∧(x → y)) → y. Then, we have ((x∧(x → y)) → y) → y ≤ (x∧(x → y)) → y, which implies that ((x∧(x → y)) → y) → y)→((x∧(x → y)) → y) = 1. Hence, we obtain μ(((x∧(x → y)) → y) → y)→((x∧(x → y)) → y)) = μ(1). Since μ is a fuzzy implicative filter of L, then μ(((x∧(x → y)) → y) → y)→((x∧(x → y)) → y)) = μ((x∧(x → y)) → y) by μ((x∧(x → y)) → y) = μ(1). Therefore, we get that μ is a fuzzy positive implicative filter by From The following example shows that the converse of the above theorem may not be true.L be the EQ-algebra and let μ be the fuzzy set of L defined in μ is a fuzzy positive implicative filter of L. But it is not a fuzzy implicative filter of L, since μ(a) < μ((a → 0) → a).Let The following result displays the relations among fuzzy positive implicative prefilters (filters), fuzzy implicative prefilters (filters), and fuzzy fantastic prefilters (filters). Also, it provides a condition under which the converse of Theorems L be a good EQ-algebra. Then, μ is a fuzzy implicative prefilter (filter) if and only if μ is both a fuzzy positive implicative prefilter (filter) and a fuzzy fantastic prefilter (filter) of L.Let μ is a fuzzy implicative prefilter. Since L is a good EQ-algebra, then μ has weak exchange principle. Hence, we have that μ is a fuzzy fantastic prefilter of L by μ is a fuzzy positive implicative prefilter.Suppose  that μ is both a fuzzy positive implicative prefilter and a fuzzy fantastic prefilter of L. Since μ is a fuzzy positive implicative prefilter, then μ((x → y) → y) ≥ μ((x → y)→((x → y) → y)) by L is a good EQ-algebra; we obtain x ≤ (x → y) → y by x → y) → x ≤ (x → y)→((x → y) → y). It follows that μ((x → y) → x) ≤ μ((x → y)→((x → y) → y)). Consequently, we obtain μ((x → y) → y) ≥ μ((x → y) → x). On the other hand, since μ is a fuzzy fantastic prefilter of L, we get μ(((x → y) → y) → x)) ≥ μ(y → x) by x → y) → x ≤ y → x, it follows that μ(y → x) ≥ μ((x → y) → x), which implies that μ(((x → y) → y) → x) ≥ μ((x → y) → x). Moreover, since μ is a fuzzy prefilter of L, we have μ(x) ≥ μ(((x → y) → y) → x)∧μ((x → y) → y) and μ((x → y) → x) ≥ μ(z → ((x → y) → x))∧μ(z). Consequently, we obtain μ(x) ≥ μ((x → y) → x)∧μ((x → y) → x) = μ((x → y) → x) ≥ μ((z → ((x → y) → x))∧μ(z). Therefore, μ is a fuzzy implicative prefilter of L by Conversely, suppose that Combining Corollaries L be a good EQ-algebra and let F be a nonempty subset of L. Then, F is an implicative prefilter (filter) if and only if F is both a positive implicative prefilter (filter) and a fantastic prefilter (filter) of L.Let EQ-algebras are good EQ-algebras, in view of Since residuated L be a residuated EQ-algebra and let F be a fantastic filter of L. Then, F is an implicative filter if and only if F is a positive implicative filter of L.Let L be a residuated EQ-algebra and let F be a positive implicative filter of L. Then, F is an implicative filter if and only if F is a fantastic filter of L.Let EQ-algebras.The above results sufficiently show that fuzzy filters are a useful tool to obtain results on classical filters. Moreover, the above results further develop the classical filter theory in In what follows, we continue to find the conditions under which a fuzzy positive implicative prefilter (filter) is a fuzzy implicative prefilter (filter).F be a positive implicative prefilter of an EQ-algebra L. Then, for all a ∈ L, Fa = {x ∈ L | a → x ∈ F} is the least prefilter containing F and a.Let F is a positive implicative prefilter of L. Since a → 1 = 1 ∈ F, then 1 ∈ Fa. For all x, y ∈ L, if x, x → y ∈ Fa, then a → x ∈ F and a → (x → y) ∈ F. Since F is a positive implicative prefilter, we have a → y ∈ F; that is, y ∈ Fa. Therefore, Fa is a prefilter in L.Suppose that x ∈ L, if x ∈ F, since x ≤ a → x, and F is a prefilter, we have a → x ∈ F; that is, x ∈ Fa. Hence, F⊆Fa. Moreover, since a → a = 1 ∈ F, then a ∈ Fa. Therefore, F ∪ {a}⊆Fa. Now, let G be a prefilter of L such that F ∪ {a}⊆G; then, for any x ∈ Fa, we have a → x ∈ F⊆G. Since a ∈ G and G is a prefilter in L, we have x ∈ G; that is, Fa⊆G. Therefore, Fa is the least prefilter containing F and a.For any F of an EQ-algebra L is called maximal if and only if it is proper and no proper prefilter of L strictly contains F; that is, for each prefilter, G ≠ F; if F⊆G then G = L.A prefilter μ be a fuzzy prefilter with weak exchange principle in an EQ-algebra L. The following are equivalent:μ is a fuzzy implicative prefilter and μμ(1) is a maximal prefilter in L,μ is a fuzzy positive implicative prefilter and μμ(1) is a maximal prefilter in L,μ satisfies that μ(x) ≠ μ(1) and μ(y) ≠ μ(1) can imply that μ(x → y) = μ(1) and μ(y → x) = μ(1) for all x, y ∈ L.Let   (1)⇒(2) It follows from μ(x) ≠ μ(1) and μ(y) ≠ μ(1); thus, x ∉ μμ(1) and y ∉ μμ(1). Since μ is a fuzzy positive implicative prefilter of L, we can easily prove that μμ(1) is a positive implicative prefilter in L. It follows from Fy = {t ∈ L | y → t ∈ μμ(1)} is the least prefilter containing μμ(1) and y. Notice that μμ(1) is a maximal prefilter in L; we get Fy = L. Hence, x ∈ Fy; that is, y → x ∈ μμ(1), which implies that μ(y → x) = μ(1). In a similar way, we can get μ(x → y) = μ(1).(2)⇒(3) Suppose that μ is not a fuzzy implicative prefilter of L. Then, by x, y ∈ L such that μ(x) < μ((x → y) → x). Hence, μ(x) ≠ μ(1). Now, we consider two cases: either μ(y) = μ(1) or μ(y) ≠ μ(1).(3)⇒(1) Assume on the contrary that μ(y) = μ(1), since μ(y) ≤ μ(x → y), then μ(x → y) = μ(1). Since μ is a fuzzy prefilter in L, then μ(x) ≥ μ((x → y) → x)∧μ(x → y) = μ((x → y) → x).If μ(y) ≠ μ(1), combining μ(x) ≠ μ(1), we have μ(x → y) = μ(1) by assumption. Similarly, we have μ(x) ≥ μ((x → y) → x).If μ is a fuzzy implicative prefilter of L. It follows from μ is a fuzzy positive implicative prefilter of L. Hence, μμ(1) is a positive implicative prefilter in L. By Fa = {x ∈ L | a → x ∈ μμ(1)} is the least prefilter containing μμ(1) and a.In any case, we have a contradiction. Therefore, μμ(1) is a maximal prefilter in L, it is sufficient to show that, for all a ∈ L − μμ(1), Fa = {x ∈ L | a → x ∈ μμ(1)} = L. For all t ∈ L, if t ∈ μμ(1), then t ∈ Fa = {x ∈ L | a → x ∈ μμ(1)}. If t ∉ μμ(1), then μ(t) ≠ μ(1). Since a ∉ μμ(1), then μ(a) ≠ μ(1). It follows from (3) that μ(a → t) = μ(1); that is, a → t ∈ μμ(1), which implies that t ∈ Fa = {x ∈ L | a → x ∈ μμ(1)}. In any case, we have Fa = L. Therefore, μμ(1) is a maximal prefilter in L. This completes the proof.In order to prove that L be a good EQ-algebra. Suppose that μ is a fuzzy prefilter and μμ(1) is a maximal prefilter in L. Then, μ is a fuzzy implicative prefilter of L if and only if it is a fuzzy positive implicative prefilter of L.Let Next, we further find the conditions under which a fuzzy positive implicative filter is equivalent to a fuzzy implicative filter.L be a good EQ-algebra with a bottom element 0 and let μ be a fuzzy positive implicative filter. Then, μ is a fuzzy implicative filter if and only if it satisfies μ(¬¬x) = μ(x) for all x ∈ L.Let μ is a fuzzy implicative filter. From ¬¬x = ¬x → 0 ≤ ¬x → x, it follows that μ(¬x → x) ≥ μ(¬¬x). By μ(x) = μ(¬x → x). Consequently, we obtain μ(x) ≥ μ(¬¬x). Since L is a good EQ-algebra with a bottom element 0, then x ≤ (x → 0) → 0 by x ≤ ¬¬x. It follows that μ(x) ≤ μ(¬¬x). Therefore, we get μ(¬¬x) = μ(x).Suppose that μ is a fuzzy positive implicative filter and satisfies μ(¬¬x) = μ(x) for all x ∈ L. Since ¬x → x ≤ (x → 0)→(¬x → 0) = ¬x → (¬x → 0), then μ(¬x → (¬x → 0)) ≥ μ(¬x → x). It follows from μ(¬x → 0) = μ(¬x → (¬x → 0)). So we have μ(¬¬x) ≥ μ(¬x → x). Combining μ(¬¬x) = μ(x), we obtain μ(x) ≥ μ(¬x → x). Therefore, μ is a fuzzy implicative filter of L by Conversely, suppose that EQ-algebra L satisfies ¬¬x = x for all x ∈ L. We have the following.Notice that an involutive L be a good involutive EQ-algebra. Then, μ is a fuzzy implicative filter of L if and only if it is a fuzzy positive implicative filter of L.Let EQ-algebras, which are good involutive EQ-algebras, we can obtain some related results about fuzzy filters in residuated lattices.Since involutive residuated lattices are involutive residuated L be an involutive residuated lattice and let μ be a fuzzy filter of L. Then, the following are equivalent:μ is a fuzzy implicative filter of L,μ is a fuzzy positive implicative filter of L,μa is a fuzzy filter for all a ∈ L,μ(x → y) ≥ μ(x → (x → y)) for all x, y ∈ L,μ(x → y) = μ(x → (x → y)) for all x, y ∈ L,μ(x∧(x → y) → y) = μ(1) for all x, y ∈ L,μ(x → (x⊙x)) = μ(1) for all x ∈ L,μ(x∧y → (x⊙y)) = μ(1) for all x, y ∈ L,μ(x∧(x → y)→(x⊙y)) = μ(1) for all x, y ∈ L,μ((x⊙y) → z) = μ((x∧y) → z) for all x, y, z ∈ L,μ(x∧(x → y)→(x∧y)) = μ(1) for all x, y ∈ L,μ(x∧(x → y)→(y∧(y → x))) = μ(1) for all x, y ∈ L,μ(x) ≥ μ((x → y) → x) for all x, y ∈ L,μ(x) = μ((x → y) → x) for all x, y ∈ L,μ(x) ≥ μ(¬x → x) for all x ∈ L,μ(x) = μ(¬x → x) for all x ∈ L,μ(x → z) ≥ μ(x → (¬z → y))∧μ(y → z) for all x, y, z ∈ L,μ(x → z) ≥ μ(x → (¬z → z)) for all x, z ∈ L,μ(x → z) = μ(x → (¬z → z)) for all x, z ∈ L.Let EQ-algebras. Using characterizations of these fuzzy prefilters (filters), we mainly consider the relations among special fuzzy prefilters (filters). We find some conditions under which a fuzzy positive implicative prefilter (filter) is equivalent to a fuzzy implicative prefilter (filter). As applications of our obtained results, we give some new characterizations about classical filters in EQ-algebras and some related results about fuzzy filters in residuated lattices. In our future work, we will introduce the notion of states on EQ-algebras and discuss the relations between fantastic filters and states on EQ-algebras.In this paper, we present several characterizations of some fuzzy prefilters (filters) in


G2 . All these four kinds of mappings form a group G5. And all the groups Gi, i = 2,3, 4 are normal subgroups of G5. Moreover, for G5, a generator set is given, which consists of all the involutive negations of the second kind and the standard negation of the first kind. As a subset of the unit square, the interval-valued set is also studied. Two groups are found: one group consists of all the isomorphisms on LI, and the other group contains all the isomorphisms and all the strict negations on LI, which keep the diagonal. Moreover, the former is a normal subgroup of the latter. And all the involutive negations on the interval-valued set form a generator set of the latter group.The main results are about the groups of the negations on the unit square, which is considered as a bilattice. It is proven that all the automorphisms on it form a group; the set, containing the monotonic isomorphisms and the strict negations of the first (or the second or the third) kind, with the operator “composition,” is a group  In , the gro2 is a very special bilattice, which is a subset of the real plane. Thus it is of particular interest. In 2∣x1 ≤ x2}, with the natural order ≤k if and only if x1 ≤ y1, x2 ≤ y2. Their properties could be found in 2∣x1 + x2 ≤ 1}, with the order  ≤t  if and only if x1 ≤ y1, x2 ≥ y2. Both LI and L* are sublattices of the bilattice 2, with orders ≤k and ≤t. Therefore, similar to 2, ≤t and ≤k, defined as, for any x = , y = ∈2,k is the natural order on the unit square.In , 7, threN¬, called reflection, is a unary operator on the square, satisfying the following properties: ∀x = , y = ∈2,x≤ky, then N¬(x) ≥k N¬(y);if x≤ty, then N¬(x) ≤t N¬(y);if N¬ =  and N¬ = .The first kind negation N−, named conflation, is a unary operator satisfying(1′)x≤ky, then N−(x)≥kN−(y);if (2′)x ≤t y, then N−(x) ≥t N−(y);if (3′)N− =  and N− = .The second kind negation N~ is a unary operator satisfying(1′′)x≤ky, then N~(x)≤kN~(y);if (2′′)x ≤t y, then N~(x) ≥t N~(y);if (3′′)N~ =  and N~ = .The last kind negation N on the unit square, as a bilattice .For convenience, in this paper, these three kinds of negations are collectedly called negations N, N′ is defined as (N∘N′) = N). Then, it is not hard to check that N¬∘N− and N−∘N¬ are negations of the third kind; N¬∘N~ and N~∘N¬ are the second kind negations, and N~∘N− and N−∘N~ are of the first kind.The composition of two negations Similar to the definition of strict interval-valued negations in , the strk and ≤t are strict, that is, in k and ≤t are replaced by <k and <t in the premise, then the conclusion will be changed to <k and <t correspondingly.A negation on the unit square is called strict, if it is continuous and both ≤N1 = (n1(x1), n2(x2)) and N2 = (n1(x2), n2(x1)), withci, i = 1,2 are constants in . Then both N1 and N2 are negations, but neither of them is strict.Let N satisfies N(N(x)) = x, then it is called an involutive negation.If a negation Ns,1 defined as Ns,1 = , for all ∈2, is the standard negation of the first kind and named as the first standard negation. The mapping Ns,2 defined as Ns,2 = , for all ∈2, is called the second standard negation. And Ns,3 =  is called the third standard kind negation. It is obvious that Ns,1, Ns,2, and Ns,3 are involutive.The mapping Obviously, each involutive negation is strict, but the converse is not valid.The following negation is strict but not involutive:N¬ = (n1(x2), n2(x1)), with n1, n2 being strict negations on , and every second kind strict negation could be characterized as N− = (n1(x1), n2(x2)), with n1 and n2 being strict negations on .Each of the first kind strict negations could be characterized as N¬ = (n1(x2), n2(x1)), with n1 = n2−1, negations on , and every second kind involutive negation could be characterized as N− = (n1(x1), n2(x2)), with n1 and n2 being involutive negations on .Each of the first kind involutive negations could be characterized as Similarly, each of the third kind negations could be characterized as follows.N~ is a strict negation of the third kind, if and only if there are two isomorphisms ϕ1 and ϕ2 on , s.t. N~ could be represented as N~ = N¬∘N−, in which N¬ is the first kind and N− is the second kind. Then from Since each ni and ϕi, i = 1,2 are bijections on , we can know each strict negation N is a bijection on 2.From Lemmas Note that, the composition of two involutive negations of different kinds is still a negation, but it may not be involutive. The following example shows it.Ns2,,2 → 2 is a monotonic isomorphism on the unit square, if it is bijective and preserves both the orders ≤k and ≤t.A continuous mapping Φ : N∘Φ and Φ′∘Φ as (N∘Φ) = N) and (Φ′∘Φ) = Φ′).Obviously, each monotonic isomorphism is continuous. Define ϕ1 and ϕ2 on the unit interval  such thatIf Φ is a monotonic isomorphism on the unit square, then there are isomorphisms x1, x2) ∈ 2, ≤k. Since Φ preserves ≤k, we can know Φ≤kΦ. Thus, the greatest element of , ≤k) is Φ. Because Φ is a bijection, we have Φ = . Similarly, Φ = , Φ = , and Φ =  can be proven.Firstly, let us show Φ = , Φ = , Φ = , and Φ = . For any ≤k ∈ 2, then Ns,1≥kNs,1. Because Φ preserves ≤k, we haveNs,1 reverses the order ≤k. Similarly, since both Φ and Ns,1 keep the order ≤t, we can get that Φ∘Ns,1 preserves the order ≤t. Since Φ is a bijection, Φ strictly preserves the orders ≤k and ≤t. Similarly, it can be proven that Φ∘Ns,1 strictly preserves the order ≤t and strictly reverses the order ≤k.Suppose  = , Ns,1 = , the following hold:Ns,1 is continuous, strictly reverses the order ≤k, and strictly preserves the order ≤t, show that Φ∘Ns,1 is a strict negation of the first kind, denoted by N¬ = Φ∘Ns,1.Since Φ = , Φ = , and x1, x2) ∈ 2,  =  = Φ)) = Φ, that is, Φ = N¬∘Ns,1. From N¬ = (n1(x2), n2(x1)), with n1, n2 being negations on the unit interval . Since Ns,1 = , we could get Φ = (n1(1 − x1), n2(1 − x2)). Because n1(1 − x1) and n2(1 − x2) are isomorphisms on the unit interval, Φ can be characterized as Φ = (ϕ1(x1),  ϕ2(x2)), in which ϕ1(x1) = n1(1 − x1), ϕ2(x2) = n2(1 − x2) are isomorphisms on the unit interval.Then for any  = (1 − ϕ(x2), 1 − ϕ(x1)). It shows that Ns,1∘Φ is a continuous bijection on 2, which strictly preserves the order ≤k and strictly reverses ≤t and satisfies  =  and  = . Therefore, Ns,1∘Φ is a strict negation of the first kind.If Ns = Ns,2 or Ns = Ns,3. Therefore, Φ∘Ns and Ns∘Φ are strict negations and of the same kind as Ns, for Ns = Ns,i,  i = 1,  2,  3.Similar for N is a negation and Ns is the standard negation of the same kind as N. Then, N∘Ns is a monotonic isomorphism Φ. As a result, for any  ∈ 2,N = Φ∘Ns.Next, let us show the second part. Suppose Based on the notions in the above section, the groups of the negations on the unit square will be discussed.In , the fol1) For every strict negation n on the unit interval, there exist three involutive negations n1, n2 and n3 s.t. n = n1∘n2∘n3.(2) For every isomophism ϕ on the unit interval, there exist four involutive negations n1, n2, n3 and n4 s.t. ϕ = n1∘n2∘n3∘n4. For every strict negation N of the second kind on the unit square, there exist three involutive negations N1, N2, and N3 of the second kind, s.t. N = N1∘N2∘N3.(2) For every monotonic isomorphism Φ on the unit square, there exist four involutive negations N1, N2, N3, and N4 of the second, s.t. Φ = N1∘N2∘N3∘N4., i = 1,2, 3,4, 5 are groups, with the same unit element id. Moreover, G1◃G2,3,4,5◃G5, that is, G1 is a normal subgroup of Gi, i = 2,3, 4,5, and G2,3,4,5 are normal subgroups of G5.f in Gi, i = 1,…, 5, id∘f = f∘id = f, that is, id is the unit element of Gi, i = 1,…, 5.Obviously, for any element 1, Φ2, the composition of them is still a monotonic isomorphism. Thus G1 is closed under the operator “∘”. For two strict negations N1, N2 of the same kind, the composition of them is also a monotonic isomorphism. The composition of a strict negation N and a monotonic isomorphism Φ is still a strict negation of the same kind with N. Therefore, G2, G3, and G4 are closed under the operator “∘”. For any two strict negations N1, N2 of different kind, the composition is a strict negation of the other kind. So G5 is closed.For any two monotonic isomorphisms ΦG5. This proof also shows the associativity of Gi, i = 1,2, 3,4.Now, let us prove the associativity of F = (f1(x1), f2(x2)) or F = (f1(x2), f2(x1)), in which both of f1 and f2 are isomorphisms on  or both are strict negations on . Let F1, F2, F3 be negations or monotonic isomorphisms on 2. Then,F1∘[F2∘F3] = [F1∘F2]∘F3, which shows the associativity of G5.From Lemmas N, obviously Φ−1 and N−1 exist. Moreover, Φ−1 is also a monotonic isomorphism and N−1 is a negation of the same kind as N, that is, the inverse about the composition operator “∘” exists. Therefore, Gi, i = 1,…, 5 are groups.For any monotonic isomorphism Φ and any strict negation S1 ⊂ S2,3,4 ⊂ S5, G1 < G2,3,4 < G5, we only need to prove that they are normal subgroups.Since G1◃G2. For any Φ ∈ S1, obviously, Φ∘G1∘Φ−1 ⊂ G1, because G1 is a group. For N ∈ S2∖S1, from N = (n1(x2), n2(x1)) and N−1 = (n1−1(x2), n2−1(x1)). For any Φ = (ϕ1(x1), ϕ2(x2)) ∈ S1,n1∘ϕ2∘n2−1 and n2∘ϕ1∘n1−1 are isomorphisms on the unit interval , N∘Φ∘N−1 is in S1. Thus, N∘G1∘N−1 ⊂ G1, that is, G1 is a normal subgroup of G2.Firstly, let us show G1◃G3,4. Since, S5 = S2 ∪ S3 ∪ S4, we could know G1◃G5.Similarly, we could prove G2◃G5. Because G2 is a group, for all f ∈ G2 , we have f∘G2∘f−1 ⊂ G2. For any N ∈ G5∖G2, N is a negation of the second kind or the third kind. If N is of the second kind, then N−1 is also of the second kind. Thus for all Φ ∈ G2, N∘Φ∘N−1 ∈ G2, and for all N′ ∈ G2, N∘N′∘N−1 is of the first kind, since N′∘N−1 is of the third kind. If N is of the third kind, then N−1 is also of the third kind. Thus for all Φ ∈ G2, N∘Φ∘N−1 ∈ G2, and for all N′ ∈ G2, N∘N′∘N−1 is of the first kind, since N′∘N−1 is of the second kind. Therefore, for all f ∈ G5, f∘G2∘f−1 ⊂ G2, that is, G2 is a normal subgroup of G5. Similarly, G3◃G5, G4◃G5 could be proven.Now, let us show From SI,2 of all the involutive negations of the second kind could generate the group G3, that is, for any f ∈ G3, there is some involutive negations Ni, i = 1,…, m of the second kind, such that f = N1∘N2∘⋯∘Nm. And if f is a negation, m could be 3; if f is an isomorphism, m could be 4.The set G3. However, the following problem is still unproven.This theorem shows the relation between the involutive negations of the second kind and the group G2 (or G4) be generated by the set of all the involutive negations of the first (or the third) kind?Could G5 is given in the following theorem.A generator set of SI,2 ∪ {Ns,1} is a generator set of the group G5. And SI,2 ∪ {Ns,3} is also a the generator set of G5.The set N of the first or the third kind, there exists a monotonic isomorphism Φ, s.t. N = Ns,1∘Φ or N = Ns,2∘Φ. Since Ns,1∘Ns,2 = Ns,3, from SI,2 ∪ {Ns,1} generates the group G5.By Ns,2∘Ns,3 = Ns,1, SI,2 ∪ {Ns,3} is also a the generator set of G5.Because LI is a sublattice of the unit square 2, so the negations of the interval-valued set could be defined as the restriction of the negations on the unit square.The interval-valued set NI on the interval-valued set is the restriction of some negation N of the first kind on the unit square, and satisfies thatA negation An interval-valued negation is strict, if it is the restriction of some strict negations of the first kind on the unit square and satisfies .NI is involutive, if it satisfiesAn interval-valued negation NI, it keeps the diagonal Δ = { ∈ LI : x1 = x2}, that is,From this definition, each strict interval-valued negation is an injection and each involutive negation is strict . Also, iI on the interval-valued set is an isomorphism, if it is a bijection and keeps the natural order.A mapping ΦI is an isomorphism on the interval-valued set, then ΦI keeps both the orders ≤k and ≤t , because ΦI keeps the diagonal. Define the mapping (ΦI)−1 as (ΦI)−1 = (ϕ−1(x1), ϕ−1(x2)). Then (ΦI)−1 is also an isomorphism on LI, that is, (ΦI)−1 ∈ SI, andI)−1 is the inverse of ΦI. Thus, G1I is a group.Suppose ΦNI be a strict negation on LI, which keeps the diagonal Δ. From n1, n2 on , s.t. NI = (n1(x2), n2(x1)). From (n1 = n2. Then (NI)−1 = (n1−1(x2), n1−1(x1)) is also a strict negation on LI and keeps the diagonal. Also we can check thatNI)−1 is the inverse of NI. Thus, G2I is a group.Let )). From , n1 = nG1I◃G2I could be proven.Similar to the proof of This theorem could be extended to the unit square.1) All the monotonic isomorphisms on the unit square, which keep the diagonal, form a group, called G6.(2) All the strict negations of the first kind and the monotonic isomorphisms on the unit square, which keep the diagonal, form a group, called G7.(The proof is similar to From Theorems 1) The set of all the involutive negations on LI generates the group G2I.(2) The set of all the involutive negations of the first kind on the unit square, which keep the diagonal, generates the group G7.(LI, which keep the diagonal, form a group. Moreover, some generator sets of the groups are given.In this paper, we firstly study the negations on the unit square. The main results are Theorems


The concepts of -fuzzy soft (invertible) n-ary subhypergroups over a commutative n-ary hypergroup are introduced and some related properties and characterizations are obtained. The homomorphism properties of -fuzzy soft (invertible) n-ary subhypergroups are also derived.Maji et al. introduced the concept of fuzzy soft sets as a generalizationof the standard soft sets and presented an application of fuzzy soft sets in a decisionmaking problem. The aim of this paper is to apply the concept of fuzzy soft sets to  To solve complicated problems in economics, engineering, and environment, we cannot successfully use classical methods because of various uncertainties typical for those problems. There are three theories: theory of probability, theory of fuzzy sets, and the interval mathematics, which we can consider as mathematical tools for dealing with uncertainties. However, all of these have their advantages as well as inherent limitations in dealing with uncertainties. One major problem shared by those theories is their incompatibility with the parameterization tools. To overcome these difficulties, Molodtsov  introducd-algebras, respectively. Acar et al. r = {x ∈ H∣xr∈γ∨qδF(ε)}. The next theorem presents the relationships between -fuzzy soft (invertible) n-ary subhypergroups over H and crisp (invertible) n-ary subhypergroups of H.For any fuzzy soft set 〈F, A〉∈FS. Then,F, A〉 is an -fuzzy soft (invertible) n-ary subhypergroup over H if and only if nonempty subset F(ε)r is an (invertible) n-ary subhypergroup of H for all ε ∈ A and r ∈ -fuzzy soft (invertible) n-ary subhypergroup over H if and only if nonempty subset 〈F(ε)〉r is an (invertible) n-ary subhypergroup of H for all ε ∈ A and r ∈ -fuzzy soft (invertible) n-ary subhypergroup over H if and only if nonempty subset [F(ε)]r is an (invertible) n-ary subhypergroup of H for all ε ∈ A and r ∈  and (3). (1) can be easily proved.δ = 1 + γ. Let 〈F, A〉 be an -fuzzy soft n-ary subhypergroup over H and assume that 〈F(ε)〉r ≠ ∅ for some ε ∈ A and r ∈ 〉r. Then rx1qδF(ε),…, rxnqδF(ε); that is, F(ε)(x1) + r > 2δ,…, F(ε)(xn) + r > 2δ. Since 〈F, A〉 is an -fuzzy n-ary subhypergroup over H, we have max{F(ε)(x), γ} ≥ min{F(ε)x1xn, δ} for all x ∈ f(x1n). Hence, by r > δ,r ≤ 1 = 2δ − γ, that is, r + γ ≤ 2δ, we have F(ε)(x) + r > 2δ and so x ∈ 〈F(ε)〉r. Similarly we can show that e ∈ 〈F(ε)〉r and, for all y, y1n−1 ∈ 〈F(ε)〉r, there exists x ∈ 〈F(ε)〉r such that y ∈ f. Therefore, 〈F(ε)〉r is an n-ary hypergroup of H.(2) Assume that 2x1n ∈ H and ε ∈ A. If there exists x ∈ f(x1n) such that max{F(ε)(x), γ} < min{F(ε)x1xn, δ}, take r = 2δ − max{F(ε)(x), γ}. Then r ∈ (x) ≤ 2δ − r, F(ε)(x1) > max{G(ε)(x), γ} = 2δ − r,…, F(ε)(xn) > max{G(ε)(x), γ} = 2δ − r, that is, x1n ∈ 〈F(ε)〉r, but  x ∉ 〈F(ε)〉r, a contradiction. Hence max{F(ε)(x), γ} ≥ min{F(ε)x1xn, δ} and so max{infx∈f(x1n)F(ε)(x), γ} ≥ min{F(ε)x1xn, δ}. Similarly we can show that conditions (F1a) and (F3a) hold. Therefore, 〈F, A〉 is an -fuzzy soft n-ary subhypergroup over H.Conversely, assume that the given condition holds. Let F, A〉 be an -fuzzy soft n-ary subhypergroup over H and assume that [F(ε)]r ≠ ∅ for some ε ∈ A and r ∈ ]r. Then rx1∈γ∨qδF(ε),…, rxn∈γ∨qδF(ε); that is, F(ε)(x1) ≥ r > γ or F(ε)(x1) > 2δ − r ≥ 2δ − (2δ − γ) = γ,…, F(ε)(xn) ≥ r > γ or F(ε)(xn) > 2δ − r ≥ 2δ − (2δ − γ) = γ. Since 〈F, A〉 is an -fuzzy n-ary subhypergroup over H, we have max{F(ε)(x), γ} ≥ min{F(ε)x1xn, δ} for all x ∈ f(x1n) and so F(ε)(x) ≥ min{F(ε)x1xn, δ} since γ < min{F(ε)x1xn, δ} in any case. Now we consider the following cases.(3) Let 〈Case  1 . In this case, 2δ − r ≥ δ ≥ r. F(ε)(x1) ≥ r or ⋯ or F(ε)(xn) ≥ r, then F(ε)(x) ≥ min{F(ε)x1xn, δ} ≥ r. Hence zr∈γF(ε).If F(ε)(xi) + r > 2δ, for all 1 ≤ i ≤ n, then F(ε)(x) ≥ min{F(ε)x1xn, δ} = δ ≥ r. Hence zr∈γF(ε).If Case  2 . In this case, r > δ > 2δ − r. F(ε)(xi) ≥ r, for all 1 ≤ i ≤ n, then F(ε)(x) ≥ min{F(ε)x1xn, δ} = δ > 2δ − r. Hence zrqδF(ε).If F(ε)(x1) + r > 2δ or ⋯ or F(ε)(xn) + r > 2δ, then F(ε)(x) ≥ min{F(ε)x1xn, δ} > 2δ − r. Hence zrqδF(ε).If xr∈γ∨qδF(ε); that is, x ∈ [F(ε)]r. Similarly, we can show that e ∈ [F(ε)]r and, for all y, y1n−1 ∈ [F(ε)]r, there exists x ∈ [F(ε)]r such that y ∈ f. Therefore, [F(ε)]r is an n-ary hypergroup of H.Thus, in any case, x1n ∈ H and ε ∈ A. If there exists x ∈ f(x1n) such that max{F(ε)(x), γ} < r = min{F(ε)x1xn, δ}, then F(ε)(x1) ≥ r > γ,…, F(ε)(xn) ≥ r > γ, F(ε)(x) < r, and F(ε)(x) + r < 2r ≤ 2δ, that is, (x1)r∈γF(ε),…, (xn)r∈γF(ε), but x1n ∈ [F(ε)]r, but x ∉ [F(ε)]r, a contradiction. Hence max{F(ε)(x), γ} ≥ min{F(ε)x1xn, δ} and so max{infx∈f(x1n)F(ε)(x), γ} ≥ min{F(ε)x1xn, δ}. Similarly, we can show that conditions (F1a) and (F3a) hold. Therefore, 〈F, A〉 is an -fuzzy soft n-ary subhypergroup over H.Conversely, assume that the given condition holds. Let As a direct consequence of γ, γ′, δ, δ′ ∈  be such that γ < δ, γ′ < δ′, γ < γ′, and  δ′ < δ. Then every -fuzzy (invertible) n-ary soft subhypergroup over H is an -fuzzy soft (invertible) n-ary subhypergroup over H.Let P ∈ P*(H). Then P is an (invertible) n-ary subhypergroup of H if and only if Σ is an -fuzzy soft (invertible) n-ary subhypergroup over H for any A⊆E and r ≥ δ.Let F, A〉 ∈ FS. Theni ≤ n.Let 〈It is straightforward.F, A〉 be an -fuzzy soft n-ary subhypergroup over H. Theni ≤ n;x ∈ H and 1 ≤ i ≤ n.Let 〈F, A〉 be an -fuzzy soft n-ary subhypergroup over H, ε ∈ A, and 1 ≤ i ≤ n. Since max{F(ε)(e), γ} ≥ min{F(ε)(x), δ}, for all x ∈ H, we have(1) Let 〈x ∈ H and 1 ≤ i ≤ n. By (1), we have(2) Let F, A〉 be an -fuzzy soft n-ary subhypergroup over H. Then 〈F, A〉 is invertible if and only ifx, y ∈ H, ε ∈ A, and 1 ≤ i ≤ n. ProofLet 〈It is straightforward by Lemmas F, A〉,〈F1, A〉,…, 〈Fi, A〉  (2 ≤ i ≤ n) be -fuzzy soft invertible n-ary subhypergroups over H. ThenF, e∗)(y), δ}, γ} = max{min{F, e∗)(x),δ},γ}, for all x, y ∈ H and ε ∈ A;max{min{x, y, y1n−1 ∈ H such that y ∈ f and ε ∈ A, one has max{F(ε)(x), γ} ≥ min{F(ε)y1yn−1, F(ε)(y), δ}.for all  Let 〈x, y ∈ H and ε ∈ A. Then, by F, e∗)(y), δ}, γ} ≥ max{min{F, e∗)(x), δ}, γ}. In a similar way, we have max{min{F, e∗)(x), δ}, γ} ≥ max{min{F, e∗)(y), δ}, γ}. Hence max{min{F, e∗)(y), δ}, γ} = max{min{F1i, e∗)(x), δ}, γ}.(1) Let x, y, y1n−1 ∈ H be such that y ∈ f and ε ∈ A. Then (2) Let Fk, A〉 be -fuzzy soft invertible n-ary subhypergroup over H for all 1 ≤ k ≤ n and 2 ≤ i ≤ n. Then γ, ∈γ∨qδ)-fuzzy soft invertible n-ary subhypergroup over H.Let 〈Fk, A〉 is an -fuzzy soft invertible n-ary subhypergroup over H for all  1 ≤ k ≤ n, let ε ∈ A. It is clear that max{F(F1i(ε),e∗)(e),γ} ≥ min{F(F1i(ε),e∗)(x),δ} for all x ∈ H. From j ≤ n, we havex, y ∈ H; that is, F(F1i(ε), e∗) is invertible. In fact, by Lemmas Since each 〈x, y, y1n−1 ∈ H be such that y ∈ f. Then by F(F1i(ε), e∗)(x), γ} ≥ min{F(F1i(ε),e∗)y1yn−1, F(F1i(ε), e∗)(y), δ}.Next, let γ, ∈γ∨qδ)-fuzzy soft invertible n-ary subhypergroup over H.Summing up the above arguments, F, A〉 and 〈G, B〉 be two -fuzzy soft invertible n-ary subhypergroups over H. Then both F, A〉⋒〈G, B〉 are -fuzzy soft n-ary subhypergroups over H.Let 〈F, A〉 and 〈G, B〉 be -fuzzy soft invertible n-ary subhypergroups over H and ε ∈ A ∪ B, we consider the following cases.Let 〈Case  1 (ε ∈ A − B). In this case, L(ε) = F(ε). It follows that L(ε) satisfies conditions (F1a), (F2a), and (F3a) since 〈F, A〉 is an -fuzzy soft invertible n-ary subhypergroup over S. Case  2 (ε ∈ B − A). In this case, L(ε) = G(ε). It follows that L(ε) satisfies conditions (F1a), (F2a), and (F3a) since 〈G, B〉 is an -fuzzy soft invertible n-ary subhypergroup over S. Case  3 (ε ∈ A∩B). In this case, L(ε) = F(ε)∩G(ε). Now we show that F(ε)∩G(ε) satisfies conditions (F1a), (F2a), and (F3a). (1)x ∈ H, we haveFor any (2)x, x1n ∈ H be such that x ∈ f(x1n). ThenLet It follows that max{infx∈f(x1n)(F(ε)∩G(ε))(x), γ} ≥ min{(F(ε)∩G(ε))x1xn, δ}.(3)x, y, y1n−1 ∈ H be such that y ∈ f. Analogous to (2), we haveLet γ, ∈γ∨qδ)-fuzzy soft n-ary subhypergroup over H. The case for 〈F, A〉⋒〈G, B〉 can be similarly proved.Therefore, γ, ∈γ∨qδ)-fuzzy soft (invertible) subhypergroups. Let us first introduce the following definition.In this section, we consider the homomorphism properties of  and FS be two fuzzy soft classes, and let φ : U → U′ and ϕ : E → E′ be mappings. Define a mapping  : FS → FS by the following: for 〈F, A〉 ∈ FS, the image of 〈F, A〉 under , denoted by 〈F, A〉 = 〈φ(F), ϕ(A)〉, is a fuzzy soft set in FS given byε′ ∈ ϕ(A) and x′ ∈ U′. For 〈F′, A′〉 ∈ FS, the inverse image of 〈F′, A′〉 under , denoted by −1〈F′, A′〉 = 〈φ−1(F′), ϕ−1(A′)〉, is a fuzzy soft set in FS given byε ∈ ϕ−1(B) and x ∈ U.Let The following results can be easily deduced.F, A〉, 〈G, B〉∈FS and φ : U → U′ and ϕ : E → E′ be two mappings. ThenF, A〉⋐−1〈F, A〉) and 〈F, A〉 = −1〈F, A〉) if both φ and ϕ are injective;〈φ, ϕ)⋐〈F, A〉⋒〈G, B〉;〈F, A〉⋐γ,δ)〈G, B〉.if 〈Let 〈F′, A′〉, 〈G′, B′〉 ∈ FS and φ : U → U′ and ϕ : E → E′ be two mappings. Thenφ, ϕ)−1〈F′, A′〉)⋐〈F′, A′〉 and −1〈F′, A′〉) = 〈F′, A′〉 if both φ and ϕ are surjective;−1 = −1〈F′, A′〉⋒−1〈G′, B′〉;−1〈F′, A′〉⋐γ,δ)−1〈G′, B′〉.if 〈Let 〈H, f) and  be two n-ary hypergroups with identities e and e′, respectively, and φ a mapping from H to H′. Then, φ is called a homomorphism if φ(e) = e′ and φ(f(x1n)) = f′(φ(x1),…, φ(xn)) for all x1n ∈ H. If such a homomorphism is surjective, injective, or bijective, it is called an epimorphism, a monomorphism, or an isomorphism.Let  and  be two n-ary hypergroups with identities e and e′, respectively, φ : H → H′ a homomorphism, and ϕ : E → E′ a mapping. Then )⋐F′〈F1, A1〉,…, 〈Fn, An〉) for all 〈F,A〉1n ∈ FS such that ⋂i=1nAi ≠ ∅ and ) = F′〈F1, A1〉,…, 〈Fn, An〉) if ϕ is injective.Let , ε′ ∈ ϕ(⋂i=1nAi)⊆⋂i=1nϕ(Ai), and x′ ∈ H′. If φ−1(x′) = ∅, then F′(φ(F1),…, φ(Fn))(ε′)(x′) = 0 = φ(F(F1n)(ε′))(x′). Otherwise, we haveφ, ϕ))⋐F′〈F1, A1〉,…, 〈Fn, An〉). If ϕ is injective, then the symbol “≥” in the above proof can be replaced with “=”; hence ) = F′〈F1, A1〉,…, 〈Fn, An〉).Let 〈H, f) and  be two commutative n-ary hypergroups with identities e and e′, respectively, φ : H → H′ a homomorphism, and ϕ : E → E′ an injective mapping. Let 〈F, A〉 be an -fuzzy n-ary soft subhypergroup over H which has the sup-property. Then, 〈F, A〉 is an -fuzzy n-ary subhypergroup over H′. If φ is an epimorphism from H onto H′ and 〈F, A〉 is invertible, then 〈F, A〉 is also invertible.Let -fuzzy n-ary soft subhypergroup over H which has the sup-property and ε′ ∈ ϕ(A). Then there exists a unique ε ∈ A such that ϕ(ε) = ε′ and we have the following.(1)φ(F)(ε′)(e′), γ} ≥ min{φ(F)(ε′)(x′), δ} for all x′ ∈ H.It is clear that max{(2)By (3)y′, y1′,…, yn−1′ ∈ H′. If φ−1(yj′) = ∅  (1 ≤ j ≤ n − 1) or φ−1(y′) = ∅, thenLet for y′ ∈ f′. Otherwise, there exist y, y1n−1 ∈ H such that φ(y) = y′, φ(yi) = yi′, φ(F)(ε′)(y′) = F(ε)(y), and φ(F)(ε′)(yi′) = F(ε)(yi) for all 1 ≤ i ≤ n − 1 since 〈F, A〉 has the sup-property. Now, for y, y1n−1 ∈ H, since 〈F, A〉 is an -fuzzy soft n-ary subhypergroup over H, there exists x ∈ H such that y ∈ f and max{F(ε)(x), γ} ≥ min{F(ε)y1yn−1, F(ε)(y), δ}. Hence we haveφ, ϕ) is an -fuzzy soft n-ary subhypergroup over H′.Assume that 〈φ is an epimorphism from H onto H′ and that 〈F, A〉 is an invertible -fuzzy soft n-ary subhypergroup over H. Then, for any x′, y′ ∈ H′, we haveNext assume that φ, ϕ)〈F, A〉 is invertible.Therefore,  and  be two commutative n-ary hypergroups with identities e and e′, respectively, φ : H → H′ a homomorphism, and  ϕ : E → E′ a mapping. Let 〈F′, A′〉 be an -fuzzy soft n-ary subhypergroup over H′. Then −1〈F′, A′〉 is an -fuzzy soft n-ary subhypergroup over H. If both φ and ϕ are injective and 〈F′, A′〉 is an -fuzzy soft (invertible) n-ary subhypergroup over H′, then −1〈F′, A′〉 is an -fuzzy soft (invertible) n-ary subhypergroup over H.Let -fuzzy soft n-ary subhypergroup over H′. Let ε ∈ ϕ−1(A′). Then we have the following.(1)x ∈ H, max{φ−1(F′)(ε)(e), γ} = max{F′(ϕ(ε))(φ(e)), γ} ≥ min{F′(ϕ(ε))(φ(x)), δ} = min{φ−1(F′)(ε)(x), δ}.For any (2)x, x1n ∈ H be such that x ∈ f(x1n). Then φ(x) ∈ φ(f(x1n)) = f′(φ(x)1n) and soLet It follows that max{infx∈f(x1n)φ−1(F′)(ε)(z), γ} ≥ min{φ−1(F′)(ε)x1xn, δ}.(3)x, y, y1n−1 ∈ H be such that y ∈ f. Then φ(y) ∈ f′(φ(x), φ(y1),…, φ(yn−1)). Since 〈F′, A′〉 is invertible, by Let Summing up the above analysis, −1〈F′, A′〉 is an -fuzzy n-ary soft subhypergroup over H. If both φ and ϕ are injective and 〈F′, A′〉 is an -fuzzy soft invertible n-ary subhypergroup over H′, it is easy to check that −1〈F′, A′〉 is an -fuzzy soft invertible n-ary subhypergroup over H.Assume that 〈n-ary hypergroups which generalize some algebra structures: soft groups, fuzzy n-ary hypergroups, fuzzy hypergroups, and so on. The goal is to explain new methodological development in n-ary hypergroups which will also be of growing importance in the future. The results presented in this paper can hopefully provide more insight into and a full understanding of fuzzy soft set theory and algebraic hyperstructures. Our future work on this topic will focus on studying some other classes of fuzzy soft n-ary hypergroups.In this paper, our aim is to promote the research and development of fuzzy soft technology by studying fuzzy soft


The concept of soft translations of soft subalgebras and soft ideals over BCI/BCK-algebras is introduced and some related properties are studied. Notions of Soft extensions of soft subalgebras and soft ideals over BCI/BCK-algebras are also initiated. Relationships between soft translations and soft extensions are explored. Recently soft set theory has emerged as a new mathematical tool to deal with uncertainty. Due to its applications in various fields of study researchers and practitioners are showing keen interest in it. As enough number of parameters is available here, so it is free from the difficulties associated with other contemporary theories dealing with uncertainty. Prior to soft set theory, probability theory, fuzzy set theory, rough set theory, and interval mathematics were common mathematical tools for dealing with uncertainties, but all these theories have their own difficulties. These difficulties may be due to lack of parametrization tools , 2. To oAt present, work on the soft set theory is progressing rapidly. Maji et al.  describeThe study of BCI/BCK-algebras was initiated by Imai and Iseki  as the gTranslations play a vital role in reducing the complexity of a problem. In geometry it is a common practice to translate a system to some new position to study its properties. In linear algebra translations help find solution to many practical problems. In this paper idea of translations is being extended to soft BCI/BCK algebras.This paper is arranged as follows: in First of all some basic concepts about BCI/BCK-algebra are given. For a comprehensive study on BCI/BCK-algebras  is a verX, ∗, 0) is called a BCI-algebra if it satisfies the following conditions:x, y, z ∈ X)  (((x∗y)∗(x∗z))∗(z∗y) = 0),  ((x∗(x∗y))∗y = 0),(∀x ∈ X)  (x∗x = 0),  .x ∈ X)  (0∗x = 0),  (x∗0 = x),  (x∗y = 0⇒(x∗z)∗(y∗z) = 0, (z∗y)∗(z∗x) = 0),  ((x∗y)∗z = (x∗z)∗y),  (((x∗z)∗(y∗z))∗(x∗y) = 0).  .(∀A subset U be a universe and E be a set of parameters. Let P(U) denote the power set of U and let A, B be nonempty subsets of E.Now we recall some basic notions in soft set theory. Let F, A) is called a soft set over U, where F is a mapping given by F : A → P(U).A pair  is called a relative null soft set (with respect to the parameters set A), denoted by ∅A, if F(a) = ∅, for all a ∈ A. is called a relative whole soft set (with respect to the parameters set A), denoted by UA, if G(e) = U, for all e ∈ A. is denoted by c and is defined by c = , where Fc : A → P(U) is a mapping given by Fc(a) = U − F(a), ∀a ∈ A. Clearly, c)c = .The complement of a soft set  over U is called a full soft set if ⋃a∈AF(a) = U.A soft set  be set valued map defined asA⊆X. Then FA also denotes a soft set over a BCI/BCK algebra X. From here onward a soft set will be denoted by symbols like FA, unless stated otherwise.Let FA over a BCI/BCK-algebra X is called a soft subalgebra of X if it satisfiesX =  denote a BCI/BCK-algebra, and for any soft set FA over X, we denote T : = X − ∪{FA(x)∣x ∈ X} unless otherwise specified.A soft set T = (⋃x∈XFA(x))c = ⋂x∈XFAc(x).That is T∩FA(x) = ∅ for all x ∈ X. If FA is a full soft set then T is an empty set. Therefore throughout this paper only those soft set are considered which are not full.It is easy to see that FA be a soft set over X and let U1⊆T. A mapping FU1T : X → P(X) is called a soft U1-translation of FA if, for all x ∈ X,Let U1⊆T and FA be a soft set over X, then FA(x) ∪ U1⊇FA(y) ∪ U1 implies FA(x)⊇FA(y), for all x, y ∈ X.Let U1⊆T, FA(x)∩U1 = ∅ and FA(y)∩U1 = ∅. Let a ∈ FA(y) then a ∈ FA(y) ∪ U1⊆FA(x) ∪ U1 this implies a ∈ FA(x) or a ∈ U1 but a ∉ U1 because FA(y)∩U1 = ∅. So a ∈ FA(x) that is FA(x)⊇FA(y), for all x, y ∈ X.Since FA be a soft subalgebra of X and U1⊆T. Then the soft U1-translation FU1T of FA is a soft subalgebra of X.Let x, y ∈ X. ThenFU1T is a soft subalgebra of X.Let FA be a soft set over X such that the U1-translation FU1T of FA is a soft subalgebra of X for some U1⊆T. Then FA is a soft subalgebra of X.Let FU1T is a soft subalgebra of X for some U1⊆T. Let x, y ∈ X, we havex, y ∈ X. Hence FA is a soft subalgebra of X.Assume From Propositions FA of X is a soft subalgebra of X if and only if U1-translation FU1T of FA is a soft subalgebra of X for some U1⊆T.A soft set FA and GB be two soft sets over X. If FA(x)⊆GB(x) for all x ∈ X, then we say that GB is a soft extension of FA.Let X = {0,1, 2,3} presented as follows:FA and GB of X as in Consider a BCI/BCK-algebra FA(0)⊆GB(0), FA(1)⊆GB(1), FA(2)⊆GB(2), and FA(3)⊆GB(3), which implies that GB is a soft extension of FA.Here S-extension is being introduced.Next the concept of soft FA and GB be two soft sets over X. Then GB is called a soft S-extension of FA, if the following conditions hold:GB is a soft extension of FA.FA is a soft subalgebra of X, then GB is a soft subalgebra of X.If Let FU1T(x)⊇FA(x) for all x ∈ X. As a consequence of As we know FA be a soft subalgebra of X and U1⊆T. Then the soft U1-translation FU1T of FA is a soft S-extension of FA.Let The converse of X = {0,1, 2,3} given as follows:FA of X by Consider a BCI/BCK-algebra FA is a soft subalgebra of X. For soft set FA, T = {3}. Let GB be a soft set over X given by Then GB is a soft S-extension of X. But it is not a soft U1-translation of FA for any nonempty U1⊆T.Then FA of X, U1⊆T and U2 ∈ P(X) with U2⊇U1, letFA is a soft subalgebra of X, then it is clear that UU1 is a subalgebra of X for all U2 ∈ P(X) with U2⊇U1. But, if we do not give condition that FA is a soft subalgebra of X, then UU1 may not be a subalgebra of X as seen in the following example.For a soft set X = {0,1, 2,3, 4} be a BCI/BCK-algebra presented as follows:Let FA of X by Define a soft subset FA is not a soft subalgebra of X with T = {1}. Since FA(3∗4) = {0}⊉{0,4} = FA(3)∩FA(4) For U2 = {1,4} and U1 = {1}, we obtain UU1 = {3,4} which is not a subalgebra of X since 3∗3 = 0 ∉ UU1.Then U1-translations and UU1 is studied in case of soft subalgebra.In the following theorem, relationship between FA be a soft set over X and U1⊆T. Then the soft U1-translation FU1T of FA is a soft subalgebra of X if and only if UU1 is a subalgebra of X for all U2 ∈ P(U) with U2⊇U1.Let U1-translation FU1T of FA is a soft subalgebra of X. Then by FA is a soft subalgebra of X if FU1T is a soft subalgebra of X. Further let a, b ∈ UU1, then FA(a)⊇U2 − U1 and FA(b)⊇U2 − U1 are subalgebras of X for all U2 ∈ P(U) with U2⊇U1. Considera∗b ∈ UU1, which shows that UU1 is a subalgebra of X, for all U2⊆P(U), with U2⊇U1.Assume that the soft UU1 is a subalgebra of X for all U2⊆P(U) with U2⊇U1. Now assume that there exist a, b ∈ X such thatFA(a)⊇U2 − U1 and FA(b)⊇U2 − U1 but FA(a∗b) ⊂ U2 − U1. This shows that a, b ∈ UU1 and a∗b ∉ UU1, which is a contradiction and so FU1T(a∗b)⊇FU1T(a)∩FU1T(b) for all a, b ∈ X. Hence FU1T is a soft subalgebra of X.Conversely, suppose that FA be a soft subalgebra of X and let U1, U2⊆T. If U1⊇U2, then the soft U1-translation FU1T of FA is a soft S-extension of the soft U2-translation FU2T of FA.Let U1⊇U2, this implies FU1T(x)⊇FU2T(x), for all x ∈ X. So U1-translation is an extension of U2-translation, and from FU1T and FU2T are soft subalgebras of FA. Hence soft U1-translation FU1T of FA is a soft S-extension of the soft U2-translation FU2T of FA.Since FA of X and U2⊆T, the soft U2-translation FU2T of FA is a soft subalgebra of X. If GB is a soft S-extension of FU2T and then there exists U1⊆T such that U1⊇U2 and GB(x)⊇FU1T(x), for all x ∈ X. Thus, we have the following theorem.For every soft subalgebra FA be a soft subalgebra of X and U2⊆T. For every soft S-extension GB of soft U2-translation FU2T of FA, there exists a U1⊆T such that U1⊇U2 and GB are a soft S-extension of U1-translation of FA.Let FA of X and U2⊆T, the soft U2-translation FU2T of FA is a soft subalgebra of X. If GB is a soft S-extension of FU2T and then there exists U1⊆T such that U1⊇U2 and GB(x)⊇FU1T(x), for all x ∈ X. Then by GB is a soft S-extension of U1-translation of FA.For every soft subalgebra S-extension GB of a soft subalgebra FA of X is said to be normalized if there exists x0 ∈ X such that GB(x0) = X.A soft FA be a soft subalgebra of X. A soft set GB of X is called a maximal soft S-extension of FA if it satisfies the following conditions:GB is a soft S-extension of FA,X which is a soft extension of GB.there does not exist another soft subalgebra of Let Z+ be a set of positive integers and let “∗” be a binary operation on Z+ defined byx, y ∈ Z+, where  is the greatest common divisor of x and y. Then  is a BCK-algebra. Let FA and GB be soft sets of Z+ which are defined by FA(x) = {1,2, 3} and GB(x) = Z+ for all x ∈ Z+. Clearly, FA and GB are soft subalgebras of Z+. By using definition of maximal soft S-extension, then it is easy to see that GB is a maximal soft S-extension of FA.Let GB of X is a normalized soft S-extension of a soft subalgebra FA of X, then GB(0) = X.If a soft set GB is a normalized soft S-extension of a soft subalgebra FA of X then there exists x0 ∈ X such that GB(x0) = X, for some x0 ∈ X. ConsiderGB(0) = X.Assume that FA be a soft subalgebra of X. Then every maximal soft S-extension of FA is normalized.Let S-extensions.This follows from the definitions of the maximal and normalized soft Now concept of translation of a soft ideal of a BCI/BCK-algebra is introduced.FA of a BCI/BCK-algebra is called a soft ideal of X, denoted by FA⊲SX, if it satisfies:x ∈ X) (FA(0)⊇FA(x)), (F(x)⊇(FA(x∗y)∩FA(y))). = FA(0) ∪ U1⊇FA(x) ∪ U1 = FU1T(x) andFU1T is a soft ideal of X for some U1⊆T. Let x, y ∈ X. ThenFA(0)⊇FA(x). NextFA(x)⊇FA(x∗y)∩FA(y)  by . Hence FIn this section concept of soft ideal extension is being introduced and some of its properties are studied.FA and GB be the soft subsets of X. Then GB is called the soft ideal extension of FA, if the following conditions hold:GB is a soft extension of FA.FA⊲SX⇒GB⊲SX.Let FA of X, U1⊆T and U2 ∈ P(X) with U2⊇U1, define EU1: = {x ∈ X∣FA(x) ∪ U1⊇U2}.For a soft subset FA⊲SX, then UU1⊲X for all U2 ∈ P(U) with U2⊇U1.It is clear that if U1⊆T, let FU1T be the soft U1-translation of FA. Then the following are equivalent:FU1T⊲SX.U2 ∈ P(U)) ⊲X).(∀For FU1T⊲SX and let U2 ∈ P(U) be such that U2⊃U1. Since FU1T(0)⊇FU1T(x) for all x ∈ X, we havex ∈ EU1.x, y ∈ X be such that x∗y ∈ EU1 and y ∈ EU1. Then FA(x∗y) ∪ U1⊇U2 and FA(y) ∪ U1⊇U2, that is, FU1T(x∗y) = FA(x∗y) ∪ U1⊇U2 and FU1T(y) = FA(y) ∪ U1⊇U2. Since FU1T⊲SX, it follows thatFA(x) ∪ U1⊇U2 so that x ∈ EU1. Therefore EU1⊲X.(1)⇒(2) Consider EU1⊲X for every U2 ∈ P(U) with U2⊇U1. If there exists x ∈ X with U3⊇U1 such that FU1T(0) ⊂ U3⊆FU1T(x) and then FA(x) ∪ U1⊇U3 but FA(0) ∪ U1 ⊂ U3. This shows that x ∈ EU1 and 0 ∉ EU1. This is a contradiction, and so FU1T(0)⊇FU1T(x), for all x ∈ X.(2)⇒(1) Suppose that a, b ∈ X such that FU1T(a) ⊂ U4⊆FU1T(a∗b)∩FU1T(b). Then FA(a∗b) ∪ U1⊇U4 and FA(b) ∪ U1⊇U4, but FA(a) ∪ U1 ⊂ U4. Hence a∗b ∈ EU1 and b ∈ EU1, but a ∉ EU1. This is impossible and therefore FU1T(x)⊇FU1T(x∗y)∩FU1T(y), for all x, y ∈ X. Consequently FU1T⊲SX.Now assume that there exist FA⊲SX and U1, U2⊆T. If  U1⊇U2, then the soft U1-translation FU1T of FA is a soft ideal extension of the soft U2-translation FU2T of FA.Let U1⊇U2, this implies that (FU1T(x)⊇FU2T(x))  (∀x ∈ X). This shows that FU1T is a soft extension of FU2T.SinceFU2T is a soft ideal of X, then FU1T(0) = FA(0) ∪ U1⊇FA(x) ∪ U1 = FU1T(x) for all x ∈ X, so we have (FU1T(0)⊇FU1T(x)). ConsiderFU1T(x)⊇FU1T(x∗y)∩FU1T(y))   so FU1T is a soft ideal of X. Hence FU1T is a soft ideal extension of FU2T.Now, let Soft set theory is a mathematical tool to deal with uncertainties. Translation and extension are very useful concepts in mathematics to reduce the complexity of a problem. These concepts are frequently employed in geometry and algebra. In this papers, we presented some new notions such as soft translations and soft extensions for BCI/BCK-algebras. We also examined some relationships between soft translations and soft extensions. Moreover, soft ideal extensions and translations have been introduced and investigated as well. It is hoped that these results may be helpful in other soft structures as well.