Solitons are of the important significant in many fields of nonlinear science such as nonlinear optics, Bose-Einstein condensates, plamas physics, biology, fluid mechanics, and etc. The stable solitons have been captured not only theoretically and experimentally in both linear and nonlinear Schrödinger (NLS) equations in the presence of non-Hermitian potentials since the concept of the parity-time  In Section of Results, we introduce the NLS equation with third-order dispersion in the presence of complex We focus on the generalized form of the third-order NLS equation5253ψ ≡ ψ is a complex wave function of x, z, z denotes the propagation distance, the real parameter β stands for the coefficient of TOD, the V(x) = V(−x) and W(x) = −W(−x) describing the real-valued external potential and gain-and-loss distribution, respectively, and g > 0 (or <0) is real-valued inhomogeneous self-focusing (or defocusing) nonlinearity. The power of β = 0 becomes the β ≠ 0) and gain-and-loss distribution. Here our following results are also suitable for the case x → t in where Raman effect, nonlinear dispersion terms , and higher-order dispersion terms are neglected495055We start to study the physically interesting potential in V0 and TOD parameter β can be used to modulate the amplitudes of the reflectionless potential V(x)W(x), respectively. Moreover, V(x) and W(x) are both bounded | ≤ |V0|, V(x), W(x) → 0 as |x| → ∞ (see W(x) ~ 0 as |x| > M > 0. The sole difference between the potential (2) and the usual Scraff-II potentialβ sec h x tan h x) in Scarff-II potential for the same amplitudes.where the real constant  → ∞ see . It is eg = 0) in the ψ = Φ(x)eiλz− to yieldWe firstly consider the linear spectrum problem  are the corresponding eigenvalue and eigenfunction, respectively, and limx|→∞|Φ(x) = 0. Since the discrete spectrum of a complex V0, β for which the complex where V0 < 0 such that the shape of potential V(x) seems to be V-shaped with zero boundary conditions (see L (see Methods). V0, β) space. Two almost parallel straight lines (β ≈ ±0.12) separate the limited space {|−0.02 ≤ V0 ≤ −3, |β| ≤ 0.5}. The regions of broken and unbroken β = 0.1 and varying V0, the spontaneous symmetry breaking does not occur from the two lowest states since the maximum absolute value of imaginary parts of λ is less than 6 × 10−14 and they can be regarded as zero /6, and the existent condition g(V0 − κβ + 1) > 0 is required. For the signs of parameter v and nonlinearity g, we find the following four cases for the existent conditions of bright solitons (4) -space); (ii) v = g = −1 and V0 < −α -space); (iii) v = g = 1 and V0 > α -space); (iv) v = −g = 1 and V0 < α -space).where the phase wavenumber is defined by TOD coefficient g = ±1) via the direct propagation of the initially stationary state (4) with a noise perturbation of order about 2% in v = −g = −1 and different parameters V0 and β. For V0 = −0.8, β = 0.1 belonging to the domain of the unbroken linear V0 = −0.8, if we change β = 1.1 corresponding to the domain of the broken linear β increases a little bit , then the nonlinear mode becomes unstable  for this case, ϕ0 can be excited to a stable and weakly oscillatory (breather-like behavior) situation  has no discrete spectra, but we still find the stable nonlinear modes /g > 0 and β > 0 . For v = 1 (or −1), we have S > 0 (or <0), which implies that the pamaeter v change the directions of power flows from gain to loss regions. The power of the solutions (4) is For the above-obtained nonlinear modes (4), we have the corresponding transverse power-flow or Poynting vector given by g = −1, if we choose V0 = 1.1, β = v = 1 and consider the initial condition ϕ given by V0 = −1.5, β = 0.1, v = −1 and consider the initial condition ϕ given by g = 1, if we choose V0 = 1.2, β = v = 1 and consider the initial condition ϕ given by V0 = −0.8, β = 0.1, v = −1 and consider the initial condition ϕ given by We here study the interactions of two bright solitons in the β → β(z) in z, that is, we focus on simultaneous adiabatic switching on the TOD and gain-and-loss distribution, modeled byNowadays we turn to the excitation of nonlinear modes by means of a slow change of the control TOD parameter V(x), W(x) are given by β → β(z), and β(z) is chosen aswhere β1,2 being real constants. It is easy to see that the solutions (4) with β → β(z) do not satisfy z = 0 and z ≥ 1000 solutions (4) indeed satisfy with ψ of β → β(z) given by v = 1 and different parameters g, V0, β1,2, ψ|2 given by z = 0, β = β1 are all of the higher amplitudes and then the amplitudes decrease slowly as z increases such that they reach the alternative stable sates beginning from about z = 1000. For v = −1 and different parameters g, V0, β1,2, z increases such that they reach the stable and weakly oscillatory (breather-like behavior) situations beginning from about z = 1000, but z = 0 to z = 1100 and then the wave slowly increases a little bit to reach another stable and weakly oscillatory (breather-like behavior) feature. In particular, in v = −g = 1, V0 = 0.8, β = 0.7  to the stable states subject to exact solitons (4) of v = −g = 1, V0 = 0.8, β = 1 , that is,We here consider nonlinear modes of the generalized form of β(x) of a Gaussian functionWe here are interested in the TOD coefficient β0 ≠ 0 being a real amplitude of the Gaussian profile and another complex with β = β(x) is given by σ ≠ 0 being a real constant. In fact, we can also consider the general case β(x). We know that the potential V(x) approaches to the harmonic potential x2/2  and W(x) → 0 as |x| → ∞ . For the given amplitude β0 = 0.1 of TOD coefficient, we give the first six lowest eigenvalues  is an error function. In the following we take g = 1 without loss of generality.where g = 1) via the direct propagation of the initially stationary state (10) with a noise perturbation of order about 2% in σ = 0. For σ = β0 = 0.1, in which the potential becomes almost a harmonic potential x2/2  of nonlinear localized modes (10) for self-focusing case  = sgn(σ). Since the gain-and-loss distribution W(x) given by β0 and σ, which generate that there are more one intervals for gain (or loss) distribution, thus though the power flows along the positive (negative) direction for the x axis as σ > 0 (<0), it is of the complicated structures from the gain-and-loss view.For the above-obtained nonlinear modes (10), we have the corresponding transverse power-flow or Poynting vector given by g = 1, if we choose σ = −0.1, β0 = 0.1 and consider the initial condition ϕ given by σ = 0.2, β0 = 0.1 and consider the initial condition ϕ given by Moreover, we also study the interactions of two bright solitons in the β(x) → β in β0 is regarded as a function of propagation distance z, that is, we focus on simultaneous adiabatic switching on the TOD, potential, and gain-and-loss distribution.Nowadays we turn to the excitation of nonlinear modes by means of a slow change of the control TOD parameter σ = 0.1, we consider the varying amplitude β0 → β0(z) in For the given V(x) and gain-or-loss distribution W(x) given by ψ satisfied by which makes the TOD coefficient given by β0 = 0.1, and consider the effect of varying amplitude σ → σ(z) on nonlinear modes:Now we fix the amplitude of TOD, V(x) and gain-or-loss distribution W(x) given by ψ satisfied by β0(z) and σ(z) given by which makes the potential In conclusions, we have introduced some non-Hermitian , we assume that ψ = Φ(x)eiλz−, then we have x| → ∞, reduces to λ = 1/(3β2) and β > 0 , we have its three roots Λ1 = i/β, β depends on the space x, e.g., β(x) = β0 exp(−x2), and V(x), W(x) are given by β(x), W(x) → 0 and V(x) → x2/2 as |x| → ∞. Thus for this case x| → ∞, where the condition For ψ = ϕ(x)eiμz−, where ϕ(x) is a complex field function and μ the corresponding propagation constant. We have ρ(x) and φ(x) being real functions and separate the real and imaginary parts to yieldWe consider the stationary solutions of For the given ψ = ϕ(x)eiμz− of 59δ and F(x) and G(x) are the eigenvalue and eigenfunctions of the linearized eigenvalue problem. We substitute the expression into ϵ to yield the following linear eigenvalue problemTo further study the linear stability of the above-obtained nonlinear stationary solutions δ has no imaginary component, otherwise they are linearly unstable.where How to cite this article: Chen, Y. and Yan, Z. Solitonic dynamics and excitations of the nonlinear SchrÖedinger equation with third-order dispersion in non-Hermitian Sci. Rep. 6, 23478; doi: 10.1038/srep23478 (2016).


As applications, we derive some new fixed point theorems in partially ordered modular metric spaces, Suzuki type fixed point theorems in modular metric spaces and new fixed point theorems for integral contractions. In last section, we develop an important relation between fuzzy metricand modular metric and deduce certain new fixed point results in triangular fuzzy metric spaces. Moreover, some examples are provided here to illustrate the usability of the obtained results.The notion of modular metric spaces being a natural generalization of classical modulars over linear spaces like Lebesgue, Orlicz, Musielak-Orlicz, Lorentz, Orlicz-Lorentz, and Calderon-Lozanovskii spaces was recently introduced. In this paper we investigate theexistence of fixed points of generalized  But the way we approached the concept of modular metric spaces is different. Indeed we look at these spaces as the nonlinear version of the classical modular spaces introduced by Nakano , we will writeλ > 0 and x, y ∈ X.Let ω :  × X × X →  is said to be modular metric on X if it satisfies the following axioms:x = y if and only if ωλ = 0, for all λ > 0;ωλ = ωλ, for all λ > 0, and x, y ∈ X;ωλ+μ ≤ ωλ + ωμ, for all λ, μ > 0 and x, y, z ∈ X.If instead of (i), we have only the condition (i′)ω is said to be a pseudomodular (metric) on X. A modular metric ω on X is said to be regular if the following weaker version of (i) is satisfied:ω is said to be convex if for λ, μ > 0 and x, y, z ∈ X, it satisfies the inequalityω on a set X and any x, y ∈ X, the function λ → ωλ is nonincreasing on . Indeed, if 0 < μ < λ, thenA function Following example presented by Abdou and Khamsi  is an imX be a nonempty set and Σ a nontrivial σ-algebra of subsets of X. Let P be a δ-ring of subsets of X, such that E∩A ∈ P for any E ∈ P and A ∈ Σ. Let us assume that there exists an increasing sequence of sets Kn ∈ P such that X = ⋃Kn. By E we denote the linear space of all simple functions with supports from P. By M∞ we will denote the space of all extended measurable functions; that is, all functions f : X →  such that there exists a sequence {gn} ⊂ E, |gn| ≤ |f|, and gn(x) → f(x) for all x ∈ X. By 1A we denote the characteristic function of the set A. Let ρ : M∞ →  be a nontrivial, convex, and even function. We say that ρ is a regular convex function pseudomodular ifρ(0) = 0;ρ is monotone; that is, |f(x)| ≤ |g(x)| for all x ∈ X implies ρ(f) ≤ ρ(g), where f, g ∈ M∞;ρ is orthogonally subadditive; that is, ρ(f1A∪B) ≤ ρ(f1A) + ρ(f1B) for any A, B ∈ Σ such that A∩B ≠ ∅, f ∈ M;ρ has the Fatou property; that is, |fn(x)|↑|f(x)| for all x ∈ X implies ρ(fn)↑ρ(f), where f ∈ M∞;ρ is order continuous in E; that is, gn ∈ E and |gn(x)| ↓ 0 implies ρ(gn) ↓ 0.Similarly, as in the case of measure spaces, we say that a set A ∈ Σ is ρ-null if ρ(g1A) = 0 for every g ∈ E. We say that a property holds ρ-almost everywhere if the exceptional set is ρ-null. As usual we identify any pair of measurable sets whose symmetric difference is ρ-null as well as any pair of measurable functions differing only on a ρ-null set. With this in mind we definef ∈ M is actually an equivalence class of functions equal to ρ-a.e. rather than an individual function. Where no confusion exists we will write M instead of M. Let ρ be a regular function pseudomodular.ρ is a regular function semimodular if ρ(αf) = 0 for every α > 0 implies f = 0  ρ-a.e.We say that ρ is a regular function modular if ρ(f) = 0 implies f = 0  ρ-a.e.We say that The class of all nonzero regular convex function modulars defined on X will be denoted by R. Let us denote ρ = ρ(f1E) for f ∈ M, E ∈ Σ. It is easy to prove that ρ is a function pseudomodular in the sense of Definition 2.1.1 in . On the other hand, Tw ∈  for all w ∈ . Then, α ≥ η. That is, T is an α-admissible mapping with respect to η. If {xn} is a sequence in X such that α ≥ η with xn → x as n → ∞, then Txn, T2xn, T3xn ∈  for all n ∈ N. That is,n ∈ N. Clearly, α ≥ η. Let α ≥ η. Now, if x ∉  or y ∉ , then 1/8 ≥ 1/4, which is a contradiction. So, x, y ∈ . Therefore,T has a unique fixed point.Let Xω, ⪯) be a partially ordered modular metric space. Recall that T : Xω → Xω is nondecreasing if for all x, y ∈ X, x⪯y⇒T(x)⪯T(y). Fixed point theorems for monotone operators in ordered metric spaces are widely investigated and have found various applications in differential and integral equations  be a complete partially ordered modular metric space and T : Xω → Xω self-mapping satisfying the following assertions:(i)T is nondecreasing;(ii)x0 ∈ X such that x0⪯Tx0;there exists (iii)T is continuous function;(iv)H ∈ ΔH such that for all x, y ∈ Xω and λ > 0 with x⪯y we haveassume that there exists  c.where 0 < l < Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have x⪯y and H > 0 for all u > 0, then T has a unique fixed point.Let  be a complete partially ordered modular metric space and T : Xω → Xω self-mapping satisfying the following assertions:(i)T is nondecreasing;(ii)x0 ∈ X such that x0⪯Tx0;there exists (iii)xn} is a sequence in X such that xn⪯xn+1 with xn → x as n → ∞, then eitherif { n ∈ N;holds for all (iv)H ∈ ΔH such that for all x, y ∈ Xω and λ > 0 with x⪯y we haveassume that there exists  l < c.where 0 < Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have x⪯y and H > 0 for all u > 0, then T has a unique fixed point.Let  be a complete partially ordered modular metric space and T : Xω → Xω self-mapping satisfying the following assertions:(i)T is nondecreasing;(ii)x0 ∈ X such that x0⪯Tx0;there exists (iii)xn} is a sequence in X such that xn⪯xn+1 with xn → x as n → ∞, then eitherif { n ∈ N;holds for all (iv)x, y ∈ Xω and λ > 0 with x⪯y we haveassume that for all  ψ ∈ Ψ and 0 < l < c.where Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have x⪯y, then T has a unique fixed point.Let  be a complete partially ordered modular metric space and T : Xω → Xω self-mapping satisfying the following assertions:(i)T is nondecreasing;(ii)x0 ∈ X such that x0⪯Tx0;there exists (iii)xn} is a sequence in X such that xn⪯xn+1 with xn → x as n → ∞, then eitherif { n ∈ N;holds for all (iv)x, y ∈ Xω and λ > 0 with x⪯y we haveassume that for all  ψ ∈ Ψ and 0 < l < c.where Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have x⪯y, then T has a unique fixed point.Let  be a complete partially ordered modular metric space and T : Xω → Xω self-mapping satisfying the following assertions:(i)T is nondecreasing;(ii)x0 ∈ X such that x0⪯Tx0;there exists (iii)xn} is a sequence in X such that xn⪯xn+1 with xn → x as n → ∞, then eitherif { n ∈ N;holds for all (iv)x, y ∈ Xω and λ > 0 with x⪯y we haveassume that for all  a + b < 1 and 0 < l < c.where Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have x⪯y, then T has a unique fixed point.Let  > 0 for all u > 0, then T has a unique fixed point.Let α, η :  × Xω × Xω →  is continuous.The function M denotes the degree of nearness between x and y with respect to t.A 3-tuple  be a fuzzy metric space. The fuzzy metric M is called triangular wheneverx, y, z ∈ X and all t > 0.Let  is nondecreasing on .For all X induces a modular metric.As an application of X, M, ∗) be a triangular fuzzy metric space. Definex, y ∈ X and all λ > 0. Then ωλ is a modular metric on X.Let  is triangular, then we getLet As an application of X, M, ∗) be a complete triangular fuzzy metric space and T : X → X self-mapping satisfying the following assertions:(i)T is an α-admissible mapping with respect to η;(ii)x0 ∈ X such that α ≥ η;there exists (iii)T is an α-η-continuous function;(iv)H ∈ ΔH such that for all x, y ∈ X and λ > 0 with η ≤ α we haveassume that there exists  l < c.where 0 < Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have η ≤ α and H > 0 for all u > 0, then T has a unique fixed point.Let  be a complete triangular fuzzy metric space and T : X → X self-mapping satisfying the following assertions:(i)T is an α-admissible mapping with respect to η;(ii)x0 ∈ X such that α ≥ η;there exists (iii)xn} is a sequence in X such that α ≥ η with xn → x as n → ∞, then eitherif { n ∈ N;holds for all (iv)condition (iv) of Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have η ≤ α and H > 0 for all u > 0, then T has a unique fixed point.Let  be a partially ordered complete triangular fuzzy metric space and T : X → X self-mapping satisfying the following assertions:(i)T is nondecreasing;(ii)x0 ∈ X such that x0⪯Tx0;there exists (iii)T is continuous function;(iv)H ∈ ΔH such that for all x, y ∈ X and λ > 0 with x⪯y we haveassume that there exists  l < c.where 0 < Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have x⪯y and H > 0 for all u > 0, then T has a unique fixed point.Let  be a partially ordered complete triangular fuzzy metric space and T : X → X self-mapping satisfying the following assertions:(i)T is nondecreasing;(ii)x0 ∈ X such that x0⪯Tx0;there exists (iii)xn} is a sequence in X such that xn⪯xn+1 with xn → x as n → ∞, then eitherif { n ∈ N;holds for all (iv)H ∈ ΔH such that for all x, y ∈ Xω and 0λ > 0 with x⪯y we haveassume that there exists  l < c.where 0 < Then T has a fixed point. Moreover, if for all x, y ∈ Fix(T) we have x⪯y and H > 0 for all u > 0, then T has a unique fixed point.Let  be a complete triangular fuzzy metric space and T continuous self-mapping on X. Assume thatx, y ∈ X and λ > 0, where 0 < l < c. Then T has a unique fixed point.Let  be a complete triangular fuzzy metric space and T continuous self-mapping on X. Assume thatx, y ∈ Xω and λ > 0, where 0 < l < c and a + b < 1. Then T has a unique fixed point.Let (


A. We find explicit expressions for the generalized Drazin inverse of the sum a + b, under new conditions on a, b ∈ A. As an application we give some new representations for the generalized Drazin inverse of an operator matrix.We investigate additive properties of the generalized Drazin inverse in a Banach algebra  A be a complex Banach algebra with unite 1. We use σ(a) to denote the spectrum of a ∈ A. The sets of all nilpotent and quasinilpotent elements (σ(a) = {0}) of A will be denoted by Anil and Aqnil, respectively.Let a ∈ A  is the erse see . For a cp = p2 ∈ A is an idempotent, we denote a ∈ A asa11 = pap, If a ∈ Ad and aπ = 1 − aad be the spectral idempotent of a corresponding to {0}. It is well known that a ∈ A can be represented in the following matrix form d under the conditions ab = babπ and ab = aπbabπ, respectively. Then we will apply these formulas to provide some representations for the generalized Drazin inverse of the operator matrix In this paper, we first give the formulas of a ∈ (pAp)d and x and y are generalized Drazin invertible, andIf (ii)x ∈ Ad and a ∈ (pAp)d, then xd and yd are given by (If given by  and 7)..x ∈ Ad Let a, b ∈ Aqnil. If ab = ba or ab = 0, then a + b ∈ Aqnil.Let The following result is a generalization of [a ∈ Aqnil, b ∈ Ad, and ab = babπ, then a + b ∈ Ad andIf b ∈ Aqnil. Therefore, bπ = 1 and from ab = babπ we obtain ab = ba. Using a + b ∈ Aqnil and −1, Now we assume ab = babπ anda1b1 = 0 and a3b1 = 0. Since b1 is invertible, we have a1 = 0 and a3 = 0.Let us representHence we haveab = babπ implies that a4b2 = b2a4. Hence, using a4 + b2 ∈ Aqnil. By a + b ∈ Ad andbd, a, and a + b, we easily obtain formula −1, then a1π = 0.From u reduces toa2b4 = b4a2b4π we get = a2b4(b4d)2 = b4a2b4π(b4d)2 = 0.Hence, the expression of a2b4d = 0, we haveHence, from formula  and a2tituting  and 24)a2b4dtituting , we getu by the above expression and considering matrix representations of a and b, after direct computations, we obtain the formula (a + b)d.Now, replacing  formula  for  and D ∈ B(Y) are generalized Drazin invertible.This section is devoted to the generalized Drazin inverse of 2 × 2 operator matrix:Next we will state some auxiliary lemmas.A and D be generalized Drazin invertible and let M be matrix of form d = 0, (A2Ad)d = Ad, (A2Ad)π = Aπ, and Ind(AAπ) = Ind(A) and Ind(A2Ad) = 1.Let X permits us to write A = S(C ⊕ N)S−1, where S and C are nonsingular, and N is nilpotent with index Ind(A). Thus Ad = S(C−1 ⊕ 0)S−1. Now, it is evident that A2Ad = S(C ⊕ 0)S−1 and AAπ = S(0 ⊕ N)S−1, which lead to the affirmations of this lemma.The Jordan canonical form of Md under conditions BC = 0, DC = 0, and BD = 0. Here we replace the last two conditions by the two weaker conditions DC = DπCAAπ and BD = AAπB.In [A and D be generalized Drazin invertible and let M be matrix of form n+2.From fined in . From DAAπB = BD and DC = DπCAAπ, we obtain PQ = PπQPQπ. Applying Since AdBD = 0, we haveAdBD = 0, we haveFrom nce from , we obtaBC = 0 and BD = AAπB  imply that BDnC = 0. From BDnC = 0 and AdBD = 0, we getThe conditions From , 38), a, a38), aThe proof is finished.SinceA and D be generalized Drazin invertible and let M be matrix of form 47) it foThe proof is finished.A and D be generalized Drazin invertible and let M be matrix of form nC = BDnDπC = BDnC = 0 for any nonnegative integer n; we can apply Q with D replaced by DDπ; we haveBDnC = 0 implies thatNote that rve that  and 51)DDπ is qAAπB = BD2Dd and D2DdC = DπCAAπ, we obtain PQ = PπQPQπ. Applying From Applying P+Qd=QπPdo we getQdP=AdΓΔΔBDnC = 0, we haveBDnC = 0 yieldBy direct computation we verify thatrve that  and BDnQπPd=Pd,  0 yieldQπ∑n=0∞P++2.From .The proof is finished.A and D be generalized Drazin invertible and let M be matrix of form d = 0, so we getXn is defined in 72) it foThe proof is finished.Using  and TheoCA = DπDCAπ, BDd = 0, CB = 0, and BCAπ = 0. ThenIf Using the case of A and D be generalized Drazin invertible and let M be matrix of form , a, a80), aUsing  and TheoA and D be generalized Drazin invertible and let M be matrix of form (ABDπ = BD, DCA = 0, BC = ABCAd, and CB = 0. ThenLet  of form . If ABD


In this paper, we present protocols which use entanglement more efficiently than teleportation to distinguish some classes of orthogonal product states in m ⊗ n, which are not UPB. For the open question, our results offer rather general insight into why entanglement is useful for such tasks, and present a better understanding of the relationship between entanglement and nonlocality.It is known that there are many sets of orthogonal product states which cannot be distinguished perfectly by local operations and classical communication (LOCC). However, these discussions have left the following open question: What entanglement resources are necessary and/or sufficient for this task to be possible with LOCC? In  That is, with enough entanglement, LOCC can be used to teleportHowever, in recent years, entanglement has been shown to be a valuable resource4144m ⊗ n(m ≤ n) can be distinguished perfectly with a m ⊗ n, which are not UPB.In 2008, Cohen presented that certain classes of unextendible product bases (UPB) in m ⊗ n, we present that only needing a 2 ⊗ 2 maximally entangled state, these states are also distinguishable by LOCC. Each of the locally indistinguishable orthogonal product states that we consider can be extended to a complete “nonlocality without entanglement” basis, and the latter can also be distinguished by the same  protocols. Our results show that the locally indistinguishable quantum states may become distinguishable with a small amount of entanglement resources. And these results also present a better understanding of the relationship between entanglement and nonlocality.In this paper, we will consider the locally indistinguishable orthogonal product states in the general bipartite quantum systems and present the LOCC protocols which, using entanglement as a resource, distinguish these states considerably more efficiently than teleportation. Specifically, in 5 ⊗ 5, we show a set of locally indistinguishable orthogonal product states can be distinguished by LOCC with a 2 ⊗ 2 maximally entangled state. Furthermore, for several classes of locally indistinguishable orthogonal product states on a higher-dimensional ϕ〉 = |0〉A|0 ± 1〉B, but in this paper |ϕ〉 = |0 ± 1〉A|0〉B.In this section, we present that a set of locally indistinguishable orthogonal product states in 5 ⊗ 5, can be perfectly distinguished by LOCC with a 2 ⊗ 2 maximally entangled state. In 5 ⊗ 5, the following 21 orthogonal product states cannot be distinguished by LOCCIn the following, applying the proof method which was presented by Cohenheorem 1 In 5 ⊗ 5, a 2 ⊗ 2 maximally entangled state is sufficient to perfectly distinguish the above 21 states by LOCC.TProof. First of all, Alice and Bob share a 2 ⊗ 2 maximally entangled state Then, Bob performs a two-outcome measurement, each outcome corresponding to a rank-5 projector:ab, and then operating with B1 on systems bB, is that each of the initial states is transformed as:To be precise, the result of bringing in the ancillary systems in state |Ψ〉A1 = |1〉a〈1| ⊗ |3 + 4〉A〈3 + 4|. The only remaining possibility is A2 = |1〉a〈1| ⊗ |3 − 4〉A〈3 − 4|, A3 = |1〉a〈1| ⊗ |1 + 2〉A〈1 + 2|, A4 = |1〉a〈1| ⊗ |1 − 2〉A〈1 − 2|, respectively.Let us now describe how the parties can proceed from here to distinguish the states. Alice makes a seven-outcome projective measurement, and we begin by considering the first outcome, A5 = |0〉a〈0| ⊗ |4〉A〈4| onto the Alice’s Hilbert space, it leaves B and |2 ± 3〉B.Using a rank-1 projector A6 = |0〉a〈0| ⊗ |2〉A〈2| + |0〉a〈0| ⊗ |3〉A〈3| onto the Alice’s Hilbert space, it leaves B61 = |0〉b〈0| ⊗ |1〉B〈1|, B62 = |0〉b〈0| ⊗ |3〉B〈3|, respectively. When Bob uses a projector B63 = |0〉b〈0| ⊗ |0〉B〈0|, it can leave A. In the same way, we can easily distinguish Using a rank-2 projector A7 = |0〉a〈0| ⊗ |0〉A〈0| + |0〉a〈0| ⊗ |1〉A〈1| + |1〉a〈1| ⊗ |0〉A〈0|. This leaves B71 = |0〉b〈0| ⊗ |0〉B〈0|, it leaves A. When Bob uses a projector B72 = |0〉b〈0| ⊗ (|1〉B〈1| + |2〉B〈2|), it can leave A, and leave B73 = |0〉b〈0| ⊗ |3〉B〈3| + |1〉b〈1| ⊗ |4〉B〈4|, it leaves the last two states Alice’s last outcome is a rank-3 projector onto the remaining part of Alice’s Hilbert space B2 on systems bB, it creates new states which differ from the states (3) only by ancillary systems |00〉ab → |11〉ab and |11〉ab → |00〉ab. Thus, the latter can be handled using the exact same method as described above for B1.For operating with That is, we have succeeded in designing a protocol that perfectly distinguishes the states (1) using LOCC with an additional resource of a two-qubit maximally entangled state. This completes the proof.ab = m|00〉 + n|11〉(m ≠ n), is shared, the states |ϕ13,14〉 will be not orthogonal after Bob performs a two-outcome measurement. This means that 1 ebit is necessary for the protocol. Then, it is interesting to design a protocol to distinguish these states with less than one ebit of entanglement, or to prove that there is no any protocol to accomplish it.In addition, a 2 ⊗ 2 maximally entangled state is necessary to distinguish the set of product states for the above protocol. When a partially entangled state, |Ψ〉m ⊗ nk + 1) ⊗ (2l), (2k) ⊗ (2l) and (2k + 1) ⊗ (2l + 1).In this section, we consider the set of locally indistinguishable orthogonal product states in k + 1) ⊗ (2l)In this subsection, we first present the following locally indistinguishable orthogonal product states in (2heorem 2 In (2k + 1) ⊗ (2l), a 2 ⊗ 2 maximally entangled state is sufficient to perfectly distinguish the above 6(k + l)− 6 states by LOCC.TProof. Similarly to the proof of Theorem 1, Alice and Bob first share a 2 ⊗ 2 maximally entangled state Then, Bob performs a two-outcome measurement, each outcome corresponding to a rank-2l projector:B1. To be precise, the result of bringing in the ancillary systems in state |Ψ〉ab, and then operating with B1 on systems bB, is that each of the initial states is transformed as:As Theorem 1, we only need to discuss the projector i = 1, …, 2k by 2k projectors Ai = |1〉a〈1| ⊗ |i + (i + 1)〉A〈i + (i + 1)|, Ai+1 = |1〉a〈1| ⊗ |i − (i + 1)〉A〈i − (i + 1)|, i = 1, 3, …, 2k − 1, respectively.Then, Alice makes a (2k + 3)-outcome projective measurement. Similarly to the proof of Theorem 1, Alice can identify Ak+12 = |0〉a〈0| ⊗ |2k〉A〈2k| onto the Alice’s Hilbert space, it leaves i ± (i + 1)〉B, i = 0, 2, …, 2l − 4.Using a rank-1 projector Ak+22 = |0〉a〈0| ⊗ |2〉A〈2| + |0〉a〈0| ⊗ |3〉A〈3| + … + |0〉a〈0| ⊗ |2k − 1〉A〈2k − 1|, onto the Alice’s Hilbert space, it leaves i = 2k + 2l − 1, …, 6k + 4l − 9 and annihilates other states in (6). Then, Bob can identify i = 4k + 2l − 2, …, 4k + 4l − 6 by 2l − 3 projectors Bk+2)i1 projectors Ak+32 = |0〉a〈0| ⊗ |0〉A〈0| + |0〉a〈0| ⊗ |1〉A〈1| + |1〉a〈1| ⊗ |0〉A〈0|. This leaves Bk+3)12(2 = |0〉b〈0| ⊗ (|1〉B〈1| + |2〉B〈2| + … + |(2l − 4)〉B〈(2l − 4)|), it can leave i = 6k + 4l − 6, …, 6k + 6l − 10. Then, Alice can easily identify A, and leave Bk+3)3(2 = |0〉b〈0| ⊗ |(2l − 3)〉B〈(2l − 3)| + |0〉b〈0| ⊗ |(2l − 2)〉B〈(2l − 2)| + |1〉b〈1| ⊗ |(2l − 1)〉B〈(2l − 1)|, it leaves the last four states A31 = |0〉a〈0| ⊗ |1〉A〈1| and A32 = |0〉a〈0| ⊗ |0〉A〈0| + |1〉a〈1| ⊗ |0〉A〈0| which leave two sets of orthogonal states Alice’s last outcome is a rank-3 projector onto the remaining part of Alice’s Hilbert space Therefore, a 2 ⊗ 2 maximally entangled state is sufficient to perfectly distinguish the states (4) by LOCC. This completes the proof.k) ⊗ (2l)In this subsection, we consider the following locally indistinguishable orthogonal product states in  ⊗ (2l), a 2 ⊗ 2 maximally entangled state is sufficient to perfectly distinguish the above 6(k + l)−9 states by LOCC.TProof. Similarly to the proof of Theorem 2, Alice and Bob first share a 2 ⊗ 2 maximally entangled state B1. In the same way, the initial states are transformed as:As Theorem 1, we only need to discuss the projector i = 1, …, 2k − 2 by 2k − 2 projectors Aj = |1〉a〈1| ⊗ |i + (i + 1)〉A〈i + (i + 1)|, Aj+1 = |1〉a〈1| ⊗ |i − (i + 1)〉A〈i − (i + 1)|, i = 1, 3, …, 2k − 5, j = i, i = 2k − 2, j = i − 1, respectively.Then, Alice makes a (2k + 1)-outcome projective measurement. Similarly to the proof of Theorem 2, Alice can identify Ak2 − 1 = |0〉a〈0| ⊗ |2k〉A〈2k| onto the Alice’s Hilbert space, it leaves i ± (i + 1)〉B, i = 0, 2, …, 2l − 4.Using a rank-1 projector Ak2 = |0〉a〈0| ⊗ |2〉A〈2| + |0〉a〈0| ⊗ |3〉A〈3| + … + |0〉a〈0| ⊗ |2k − 2〉A〈2k − 2|, it leaves i = 2k + 2l − 3,…, 6k + 4l − 12. Similarly to the proof of Theorem 2, these states can also be distinguished by LOCC.When Alice uses a rank-(2k-3) projector Ak+12 = |0〉a〈0| ⊗ |0〉A〈0| + |0〉a〈0| ⊗ |1〉A〈1| + |1〉a〈1| ⊗ |0〉A〈0|, it leaves Bk+1)12(2 = |0〉b〈0| ⊗ (|1〉B〈1| + |2〉B〈2| + … + |(2l − 4)〉B〈(2l − 4)|), it can leave i = 6k + 4l − 9, …, 6k + 6l − 13. Then, Alice can identify A, and leave Bk+1)3(2 = |0〉b〈0| ⊗ |(2l − 3)〉B〈(2l − 3)| + |0〉b〈0| ⊗ |(2l − 2)〉B〈(2l − 2)| + |1〉b〈1| ⊗ |(2l − 1)〉B〈(2l − 1)|, it leaves the last four states For the last rank-3 projector Thus, the states (7) can be perfectly distinguished using LOCC with an additional resource of a 2 ⊗ 2 maximally entangled state. This completes the proof.k + 1) ⊗ (2l + 1)In this subsection, we consider a generalization of (1) to higher-dimensional systems (2heorem 4 In (2k + 1) ⊗ (2l + 1), a 2 ⊗ 2 maximally entangled state is sufficient to perfectly distinguish the above 6(k + l)−3 states by LOCC.Tk + 1) ⊗ (2l + 1), the states (10) are a generalization of the states (1). Thus, the proof is similar to the proof of Theorem 1. In the following, we only give a simple proof.In 〉bB〈0(2l − 1)| + |1(2l)〉bB〈1(2l)| and B2 = |10〉bB〈10| + |11〉bB〈11| + … + |1(2l − 1)〉bB〈1(2l − 1)| + |0(2l)〉bB〈0(2l)|.B1, the initial states are transformed as:For i = 1, …, 4k + 2l − 2, 4k + 4l + 2, …, 6k + 6l − 3.In the same way, it is easily to distinguish Ak+32 = |0〉a〈0| ⊗ |0〉A〈0| + |0〉a〈0| ⊗ |1〉A〈1| + |1〉a〈1| ⊗ |0〉A〈0|, which leaves Bk+3)12(2 = |0〉b〈0| ⊗ (|1〉B〈1| + |2〉B〈2| + … + |(2l − 2)〉B〈(2l − 2)|), it can leave i = 4k + 2l + 1, …, 4k + 4l − 2, 4k + 4l + 1. Then, Alice can easily identify A, and leave Bk+3)3(2 = |0〉b〈0| ⊗ |(2l − 1)〉B〈(2l − 1)| + |1〉b〈1| ⊗ |2l〉B〈2l|, which leaves the last two orthogonal states Alice’s last outcome is a rank-3 projector That is, we have succeeded in designing a protocol to distinguish the states (10) by LOCC with a two-qubit maximally entangled state. This completes the proof.In the proof, the important idea is that entanglement provides multiple Hilbert space, and that the parties can, independently of one another, act on these subspaces in ways that differ from one subspace to the next. This allows an apart of Hilbert space such that initially nonorthogonal pairs of local states end up being orthogonal, aiding the process of distinguishing the set of states. It should be noted that our protocols do not rely on details of the individual states, but only on the general way they are distributed through the space.m ⊗ n for the above protocol. Furthermore, it will be good to do the analysis using n2, which is the optimal probability of converting such a state into a maximally entangled state, where m≥n. However, we have not succeeded in doing so. Hence, it remains an open question whether it is possible to distinguish these states with less than one ebit of entanglement.As Theorem 1, a 2 ⊗ 2 maximally entangled state is also necessary to distinguish these product states in In addition, each of the locally indistinguishable orthogonal product states that we consider can be extended to a complete “nonlocality without entanglement” basism ⊗ n, which are constructed by Wang et al.et al.In this paper, we present that the locally indistinguishable orthogonal product states in How to cite this article: Zhang, Z.-C. et al. Entanglement as a resource to distinguish orthogonal product states. Sci. Rep.6, 30493; doi: 10.1038/srep30493 (2016).


The objective is to determine the optimal resource allocation and the optimal schedule to minimize a total cost function that dependents on the total completion (waiting) time, the total machine load, the total absolute differences in completion (waiting) times on all machines, and total resource cost. If the number of machines is a given constant number, we propose a polynomial time algorithm to solve the problem. Yang and Kuo  denote the jth job on machine Mi and Ci[j](Wi[j]) denote the completion (waiting) time of job Ji[j]; pi[j], ai[j], bi[j], θi[j], ui[j], and Wi[j] = Ci[j] − pi[j]. Let Li = max{Cij | j = 1,2,…, ni}, TCi = ∑j=1niCij (TWi = ∑j=1niWij), and TADCi = ∑j=1ni∑l=jni | Cij − Cil| (TADWi = ∑j=1ni∑l=jni | Wij − Wil|) be the load, the total completion (waiting) times, and the total absolute differences in completion (waiting) times of machine Mi. Then, the total machine load, the total completion (waiting) time, and the total absolute deviation of job completion (waiting) time on all machines are ∑i=1mLi, ∑i=1mTCi (∑i=1mTWi), and ∑i=1mTADCi (∑i=1mTADWi), respectively. The objective is to determine the optimal resource allocations and the optimal schedule on the machines so that the corresponding value of the following cost functions is optimal:α1 ≥ 0, α2 ≥ 0, α3 ≥ 0, and α4 ≥ 0 are given constants and Gij is the per time unit cost associated with the resource allocation. Using the three-field notation introduced by Graham et al. ; in addition, Ci[j] = ∑l=1jpi[l], Li = ∑j=1nipi[j], and TCi = ∑j=1niCi[j] = ∑j=1ni(ni − j + 1)pi[j]; hence, we haveωij = α1 + α2(ni + 1 − j) + α3(j − 1)(ni − j + 1) andFrom Kanet , we haveFrom , for anyRm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2∑i=1mTCi + α3∑i=1mTADCi + α4∑i=1m∑j=1niGijuij can be determined byui[j]*, i = 1,2,…, m; j = 1,2,…, ni, represents the optimal resource allocation of the job in position j on machine Mi.For a given sequence, the optimal resource allocation of the problem Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2∑i=1mTCi + α3∑i=1mTADCi + α4∑i=1m∑j=1niGijuij, taking the derivative by ui[j] to /ui[j] = α4Gi[j] − βi[j]Ωij for i = 1,2,…, m; j = 1,2,…, ni. Hence, if α4Gi[j] − βi[j]Ωij > 0, we should not allocate any resource to job Ji[j]; if α4Gi[j] − βi[j]Ωij < 0, we will allocate the maximal feasible amount of resource to job Ji[j]; and if α4Gi[j] − βi[j]Ωij = 0, any feasible resource allocation can be optimal.For the problem i[j] to , we havexijr = 1 if job Jj is scheduled in position r on machine Mi, and xijr = 0 otherwise. If the number of jobs on machine Mi is known in advance, then we formulate the Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2∑i=1mTCi + α3∑i=1mTADCi + α4∑i=1m∑j=1niGijuij problem as the following assignment problem:We define bject to∑i=1m ∑r=1,2,…,n,∑j=1nxijrP =  vectors exist. Note that ni may be 0,1, 2,…, n for i = 1,2,…, m. So, if we get the numbers of jobs on the first m − 1 machines, the number of jobs processed on the last machine is then determined uniquely due to n1 + n2 + ⋯+nm = n. Therefore, the upper bound of the number of P vectors is (n + 1)m−1. Based on the above analysis, we have the following result.Next, the question is how many Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2∑i=1mTCi + α3∑i=1mTADCi + α4∑i=1m∑j=1niGijuij can be solved in O(nm+2) time; that is, the problem is polynomially solvable because m is a constant.The problem Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2∑i=1mTCi + α3∑i=1mTADCi + α4∑i=1m∑j=1niGijuij via the following algorithm.Base on the above analysis, we can determine the optimal solution for the problem Consider the following. Step  1. For all the possible vectors , solve the assignment problems .roblems (–16)). T). TStep Step  2. The optimal solution for the problem is the one with the minimum value of the total cost α1∑i=1mLi + α2∑i=1mTCi + α3∑i=1mTADCi + α4∑i=1m∑j=1niGijuij.Step  3. Calculate the optimal resources allocation by using pi[j] (Bagchi [νij = α1 + α2(ni − j) + α3j(ni − j),Similar to the analysis of the problem  (Bagchi ) into  =  is given, the optimal sequence can be determined by solving the following assignment problem:For the problem Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2TWi + α3TADWi + α4∑i=1m∑j=1niGijuij can be solved in O(nm+2) time; that is, the problem is polynomially solvable because m is a constant.The problem Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2TWi + α3TADWi + α4∑i=1m∑j=1niGijuij can be obtained by the following algorithm.The optimal solution for the problem Consider the following. Step  1. For all the possible vectors , solve the assignment problems . roblems –). ). Step  Step  2. The optimal solution for the problem is the one with the minimum value of the total cost α1∑i=1mLi + α2∑i=1mTWi + α3∑i=1mTADWi + α4∑i=1m∑j=1niGijuij. Step  3. Calculate the optimal resources allocation by using  time algorithm for the problems Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2∑i=1mTCi + α3∑i=1mTADCi + α4∑i=1m∑j=1niGijuij and Rm | pij = aijrbij + αt − θijuij | α1∑i=1mLi + α2∑i=1mTWi + α3∑i=1mTADWi + α4∑i=1m∑j=1niGijuij, respectively. The algorithms can also be easily applied to the cases bij > 0 (aging effect) and α < 0. Future research may focus on similar problems with more general processing time model and extend the problems to flow shop machine settings.In this paper, we have studied the problem of scheduling


A scheme is presented and software is documented for representing as integers input decimal numbers that have been stored in a computer as double precision floating point numbers and for carrying out multiplications, additions and subtractions based on these numbers in an exact manner. The input decimal numbers must not have more than nine digits to the left of the decimal point. The decimal fractions of their floating point representations are all first rounded off at a prespecified location, a location no more than nine digits away from the decimal point. The number of digits to the left of the decimal point for each input number besides not being allowed to exceed nine must then be such that the total number of digits from the leftmost digit of the number to the location where round-off is to occur does not exceed fourteen. Mainstream computers base integer and floating point arithmetic on fixed word lengths. As a consequence, only values with a limited number of significant digits can be represented directly, so that the results of arithmetic operations may have to be rounded off or truncated. Such errors can be avoided or, at least, mitigated, by implementing special algorithms for the execution of arithmetic operations. A fully “exact arithmetic”, however, would have to be based on quotients of integers for representing numerical values. In any case, a final limitation is due to finite memory.The need for exact arithmetic became apparent during the development of software for generating triangular and tetrahedral nets from very large point sets. Typically, this need is not due to high accuracy requirements for results—the input data are often noisy or given up to only a few significant digits—but is rather due to the need to maintain the consistency of a combinatorial structure. Building or manipulating such geometry-based combinatorial structures requires the calculation of indicators such as determinants in order to evaluate their sign and to check for zero values: round-off may lead to a false sign or zero value. An example considered later in this paper is to decide whether four given spatial points are coplanar. The approach of using exact computations for implementing computational geometry algorithms in a robust manner has been addressed in , 3], , , 6], , , [3], 4) reduce In this paper, we document software for exact integer arithmetic, accommodating an indeterminate number of digits, for multiplication, addition, subtraction, but excluding division. We found that, in many computational geometry applications, decision variables such as determinants can be calculated without division. Also the sign of a decision variable stated as a quotient but not evaluated is readily derived from the signs of the numerator and denominator.We also describe a preprocessing step, called “two-integer decomposition”, which leads from floating point input to one composed of integers only. At the root of this step lies the concept of space as an integer grid of points, all of which have integer coordinates in some shared unit. After completing the transition from floating point numbers to intermediate representations as pairs of integers—prompted by the fact that Fortran 77 does not provide a double precision integer format—a polynomial decomposition creates the number representations to be used in the exact arithmetic calculations. Software for this preprocessing step together with software for exact integer arithmetic has been successfully incorporated into several computational geometry related programs such as REGTET .In what follows, a “standard computer” is a computer that uses 64 bits of storage for a double precision number and 32 bits for an integer. Given a standard computer, even though it may not store exactly an input decimal number as a double precision floating point number, it is safe to assume that the number will be represented as accurately as possible by a double precision floating point number up to its fourteenth significant digit.x(i), i = 1, …, n, be a double precision array into which input numbers ix, i = 1, …, n, have been read. The two-integer decomposition process is a preprocessing step that takes place before any computations based on the input data are carried out. It rounds off the numbers in the array at a prespecified location of their decimal fractions and decomposes each rounded off number into two integers that are saved in integer arrays, say ix(i), ix2(i), i = 1, …, n. The rounded off numbers are then saved in array x(i), i = 1, …, n.Let k, l, 1 ≤ k ≤ 9, 0 ≤ l ≤ 9, k + l ≤ 14, and assuming each input number ix, i = 1, …, n, has no more that k digits to the left of the decimal point, each number x(i), i = 1, …, n, is rounded off at the lth digit of its decimal fraction and decomposed into two integers in one of two ways according to its size. If the absolute value of x(i) times (10.0d0)l is less than 230(=1073741824), x(i) is multiplied by (10.0d0)l and rounded off at the decimal point. The resulting integer is then placed in ix(i) while ix2(i) is set to zero. Finally, x(i) is redefined to be the double precision value of integer ix(i) divided by (10.0d0)l. On the other hand, if the absolute value of x(i) times (10.0d0)l exceeds or equals 230, x(i) is truncated at the decimal point. The resulting integer  is placed in ix(i). In addition, the signed decimal fraction obtained by subtracting the double precision value of this integer from the initial value of x(i) is multiplied by (10.0d0)l and rounded off at the decimal point. The resulting integer  is placed in ix2(i). Next, x(i) is redefined to be the double precision value of integer ix(i) plus the value obtained by dividing the double precision value of integer ix2(i) by (10.0d0)l. Finally, if the integer ix2(i) is zero then ix2(i) is set to 230 so that ix2(i) is zero if and only if the initial absolute value of x(i) (before the two-integer decomposition process) times (10.0d0)l is less than 230.Given integers d0, when the signed decimal fraction of a number is obtained by truncating the number at its decimal point and subtracting the result from the initial value of the number, and when a number is rounded off with the two-integer decomposition process. However once the two-integer decomposition process is completed all computations that follow, exact and otherwise, are carried out in terms of the arrays x(i), ix(i), ix2(i), i = 1, …, n, under the assumption that for the purposes of the user for each i, i = 1, …, n, x(i) represents closely enough the input number ix rounded off at the lth digit of its decimal fraction, and an integer (not necessarily stored by the computer) in terms of ix(i) and ix2(i) represents closely enough the input number ix times 10l rounded off at the decimal point.The following is Fortran code for carrying out the two-integer decomposition process. Variables are either integer or double precision following convention. It is noted that while some loss in precision may occur at the time the input numbers are read and transformed into double precision floating point numbers, some additional loss in precision may occur here as well when the decimal point in a number is shifted by dividing or multiplying it by a multiple of 10.0  mfull=1073741824  if(l.lt.0 .or. l.gt.9) stop 10  isclu = 1  dscle = 1.0d0  if(l.eq.0) go to 200  do 100 i = 1, l  isclu = 10*isclu  dscle = 10.0d0*dscle100 continue200 continue  dfull = dble(mfull)  dfill=dfull/dscle  do 300 i = 1, n  ix2(i) = 0  if(dabs(x(i)).lt.dfill) then    ix(i) = idnint(dscle*x(i))   if(iabs(ix(i)).lt.mfull) then     x(i) = dble(ix(i))/dscle    go to 300   endif  endif  if(dabs(x(i)).ge.dfull) stop 20  ix(i) = idint(x(i))  if(iabs(ix(i)).ge.mfull) stop 30  decml = (x(i) - dint(x(i)))*dscle  ix2(i) = idnint(decml)  if(iabs(ix2(i)).eq.0) then   x(i) = dble(ix(i))   ix2(i) = mfull  else   x(i) = dble(ix(i)) + (dble(ix2(i))/dscle)  endif300 continuel, 0 ≤ l ≤ 9, let ix, i = 1, …, n, be input numbers whose double precision floating point representations have been rounded off at the lth digit of their decimal fractions through the two-integer decomposition process. Let x(i), ix(i), ix2(i), i = 1, …, n be the arrays produced by the two-integer decomposition process that contain the rounded off numbers and the two-integer decompositions. For each i, i = 1, …, n, an integer J is symbolically defined as follows . If ix2(i) equals zero then J is set equal to ix(i). If ix2(i) equals 230 then J is set equal to ix(i) · 10l. Finally, if ix2(i) is neither zero nor 230 then J is set to ix(i) · 10l + ix2(i). In all cases for each i, i = 1, …, n, J is considered to approximate closely enough (for the purposes of the user) the input number ix times 10l rounded off at the decimal point.Given an integer M to 215. Given J, 1 ≤ i ≤ n, the polynomial decomposition process is a procedure (presented below in the form of Fortran subroutine decmp2) that decomposes the integer J into a unique collection of integers isga, isga in {−1, 0, 1}, ika, ika > 0, ka, 0 ≤ ka < M, k = 1, …, ika, such that J equals isgaisga the sign of J. Integers ka, k = 1, …, ika, are saved in an integer array, say ia(k), k = 1, …, ika, and the collection of integers isga, ika, ia(k), k = 1, …, ika, and the symbolic expression isgaJ, with isga as the sign of the representation. For each i, i = 1, …, n, the polynomial decomposition of the integer J is identified each time an exact computation involving additions, subtractions, or multiplications is required that references the input number ix. During one such computation, for each i, 1 ≤ i ≤ n, if the number ix is referenced in the computation, once the polynomial decomposition of the corresponding integer J is identified, each reference of ix in the computation is replaced by the symbolic polynomial representation of J. The computation then takes effect as a sequence of additions, subtractions, or multiplications of symbolic polynomial representations with the final result being itself the symbolic polynomial representation of some integer. This final result can usually be used in only one of two ways. If it is known that for some positive integer m the integer that is equal to the final symbolic polynomial representation is approximately equal to the product of (10l)m and the true value of the computation, then this integer is computed approximately as a double precision floating point number from its symbolic polynomial representation and the true value of the computation is then approximately obtained by dividing it by ((10.0d0)l)m. On the other hand, if the purpose of the computation is simply that of obtaining the sign of the true result then the sign of the final symbolic polynomial representation is a satisfactory answer.Set J, 1 ≤ i ≤ n, can also be defined for any integer K (not necessarily stored by the computer) in the same manner. Accordingly, the following is a Fortran subroutine called decomp for finding the polynomial decomposition isga, ia(k), k = 1, 2,  of an integer iwi (stored by the computer) with absolute value less than 230. Here mhalf equals 215(=32768).The concepts of polynomial decomposition and symbolic polynomial representation defined above for subroutine decompinteger ia(*), isga, iwi, mhalf, iviif(iwi.gt.0) then isga = 1 ivi = iwielseif(iwi.lt.0) then isga =-1 ivi = -iwielse isga = 0 ia(1) = 0 ia(2) = 0 returnendifia(2) = ivi/mhalfia(1) = ivi - ia(2)*mhalfreturnendisclu is set to 10l then isclu is less than 230 (since l ≤ 9) so that the polynomial decomposition isgu , iu(i), i = 1, 2,  of isclu can be obtained by calling subroutine decomp with a Fortran instruction as follows.In particular if call decompisga, ika, ia(k), k = 1, …, ika, of the integer J, 1 ≤ i ≤ n. Here iwi equals ix(i), iwi2 equals ix2(i), mhalf equals 215, mfull equals 230, and iu(k), k = 1, 2, is an array such that the polynomial decomposition of 10l is iu(1), iu(2) . In addition, it is assumed that subroutines mulmul and muldif (presented below) exist for multiplying and subtracting, respectively, two symbolic polynomial representations.Finally, the following is a Fortran subroutine called decmp2 for finding the polynomial decomposition   subroutine decmp2  integer nkmax  parameter (nkmax=5)  integer ia(*), isga, ika, iwi, iwi2, mhalf, mfull, iu(*)  integer ie(nkmax), io(nkmax), isge, isgo, ike, iko, isgu, iku  call decomp  ika = 2  if(iwi2.ne.0) then   isgu = 1   iku = 2   call mulmul   if(iwi2.eq.mfull) iwi2 = 0   call decomp   isgo = -isgo   iko = 2   call muldif  endif  return  endisga, ika, ia(k), k = 1, …, ika, and isgb, ikb, ib(k), k = 1, …, ikb, of two integers K1 and K2, respectively, the following is a Fortran subroutine called mulmul that produces the polynomial decomposition isgo, iko, io(k), k = 1, …, iko, of the integer K1 · K2 by multiplying the symbolic polynomial representation of K1 by that of K2  to produce a symbolic polynomial representation of K1 · K2 from which the polynomial decomposition of K1 · K2 can be obtained. Here nkmax is the dimension of the arrays ia, ib, io in the calling routine and mhalf equals 215. It is noted that the value of mhalf is of importance here since given integers i, j, 1 ≤ i ≤ ika, 1 ≤ j ≤ ikb, then 0 ≤ ia(i) < 215, 0 ≤ ib(j) < 215, so that the product ia(i) · ib(j) is less than 230 and therefore can be stored in a 32 bit integer word.Given the polynomial decompositions   subroutine mulmul  integer ia(*), ib(*), io(*)  integer isga, isgb, isgo, ika, ikb, iko, nkmax, mhalf  integer i, ipt, ipr, iko1, k, j  if(isga.eq.0.or.isgb.eq.0)then   isgo=0   iko = 2   io(1) = 0   io(2) = 0   return  endif  iko = ika + ikb  if(iko.gt.nkmax) stop 110  if(isga.gt.0)then   if(isgb.gt.0)then     isgo = 1   else     isgo =-1   endif  else   if(isgb.gt.0)then    isgo =-1   else    isgo = 1   endif  endif   iko1 = iko - 1   ipr = 0   do 200 i = 1, iko1   ipt = ipr   k = i   do 180 j = 1, ikb   if(k .lt. 1) go to 190   if(k .gt. ika) go to 150   ipt = ipt + ia(k)*ib(j)150   continue   k = k - 1180   continue190   continue   ipr = ipt/mhalf   io(i) = ipt - ipr*mhalf200   continue   io(iko) = ipr   if stop 120   iko1 = iko   do 300 i = iko1, ika+1, -1   if(io(i) .ne. 0) go to 400   iko = iko - 1300   continue400   continue   return   endisga, ika, ia(k), k = 1, …, ika, and isgb, ikb, ib(k), k = 1, …, ikb, of two integers K1 and K2, respectively, the following is a Fortran subroutine called muldif that produces the polynomial decomposition isgo, iko, io(k), k = 1, …, iko, of the integer K1 − K2 by subtracting the symbolic polynomial representation of K2 from that of K1  to produce a symbolic polynomial representation of K1 − K2 from which the polynomial decomposition of K1 − K2 can be obtained. Here nkmax is the dimension of the arrays ia, ib, io in the calling routine and mhalf equals 215. It is noted that by setting isgb equal to −isgb the polynomial decomposition of K1 + K2 can also be obtained with this subroutine.Given the polynomial decompositions   subroutine muldif  integer ia(*), ib(*), io(*)  integer isga, isgb, isgo, ika, ikb, iko, nkmax, mhalf  integer i, iko1, irel  if(isgb.eq.0)then   if(isga.eq.0)then    isgo=0    iko = 2    io(1) = 0    io(2) = 0    return   endif   isgo = isga   iko = ika   do 100 i=1,iko    io(i) = ia(i)100   continue  elseif(isga.eq.0)then   isgo =-isgb   iko = ikb   do 200 i=1,iko    io(i) = ib(i)200   continue  else   iko = ika   if(ikb.lt.ika) then    do 300 i=ikb+1,ika     ib(i) = 0300  continue   elseif(ika.lt.ikb) then    iko = ikb    do 400 i=ika+1,ikb     ia(i) = 0400  continue   endif   if(isga*isgb.gt.0)then    irel = 0    do 500 i = iko, 1, -1     if(ia(i).gt.ib(i))then      irel = 1      go to 600    elseif(ia(i).lt.ib(i))then      irel = -1      go to 600    endif500  continue600  continue    if(irel.eq.0)then     isgo = 0     do 700 i=1,iko      io(i) = 0700     continue    else     isgo=isga*irel     io(1) = (ia(1)-ib(1))*irel     do 800 i=2,iko      if(io(i-1).lt.0) then       io(i) =-1       io(i-1) = io(i-1) + mhalf      else       io(i) = 0      endif      io(i) = io(i) + (ia(i)-ib(i))*irel800     continue     if(io(iko).lt.0) stop 210    endif  else    isgo=isga    io(1) = ia(1)+ib(1)    do 900 i=2,iko     if(io(i-1).ge.mhalf) then      io(i) = 1      io(i-1) = io(i-1) - mhalf     else     io(i) = 0     endif     io(i) = io(i) + ia(i)+ib(i)900  continue     if(io(iko).ge.mhalf) then     iko = iko+1     if(iko.gt.nkmax) stop 220     io(iko) = 1     io(iko-1) = io(iko-1) - mhalf   endif  endif endif if(iko .eq. 2) go to 990 iko1 = iko do 950 i = iko1, 3, -1  if(io(i) .ne. 0) go to 990  iko = iko - 1950continue990continue   return   endn, n ≥ 4, let S be a set of n points in 3-dimensional space. Given an integer l, 0 ≤ l ≤ 9, let ix, iy, iz, i = 1, …, n, be the input decimal coordinates of the points in S, and assume that their double precision floating point representations have been rounded off at the lth digit of their decimal fractions through applications, one per coordinate, of the two-integer decomposition process. Accordingly, let x(i), ix(i), ix2(i), y(i), iy(i), iy2(i), z(i), iz(i), iz2(i), i = 1, …, n, be the arrays produced by the three applications of the two-integer decomposition process that contain the rounded off x-, y-, z-coordinates and their two-integer decompositions.Given an integer p1, p2, p3 in S that are vertices of a non-degenerate triangle, a fundamental problem in computational geometry is that of finding the location of a point p4 in S relative to the plane H that contains the triangle. Let H+ be the open half-space defined by H for which p1, p2, p3 appear in a counterclockwise direction around the boundary of the triangle when looking at the triangle from H+. Let H− be the other half-space defined by H. Determining in which of H, H+, H−, the point p4 is located may not on occasion be satisfactorily done using floating point arithmetic. Accordingly, the following is a Fortran subroutine called crsinn for doing this using polynomial decompositions. On output the sign isgo  of some polynomial decomposition determines the location of p4 .Given points isgox, ikox, iox(k), k = 1, …, ikox, isgoy, ikoy, ioy(k), k = 1, …, ikoy, isgoz, ikoz, ioz(k), k = 1, …, ikoz, of integers that are the coordinates of a vector v pointing into H+ and perpendicular to H. It also produces the polynomial decomposition isgo, iko, io(k), k = 1, …, iko, of an integer whose sign isgo determines the location of p4 and whose value when divided by both 10l and the length of v is the perpendicular distance from p4 to H. Here mhalf equals 215, mfull equals 230, and if ir, isec, ithi, ifou are locations in the arrays ix, ix2, etc. corresponding to the points p1, p2, p3, p4, respectively. In addition, isclp(k), k = 1, 2, is an array such that the polynomial decomposition of 10l is isclp(1), isclp(2) .This routine actually does more. It produces polynomial decompositions   subroutine crsinn  integer ix(*), iy(*), iz(*), ix2(*), iy2(*), iz2(*)  integer io(*), iox(*),ioy(*), ioz(*)  integer ifir, isec, ithi, ifou  integer isclp(*), mhalf, mfull, nkmax  parameter (nkmax = 30)  integer iu(nkmax), iv(nkmax), iw(nkmax)  integer ixt(nkmax), iyt(nkmax), izt(nkmax)  integer ix3(nkmax), iy3(nkmax), iz3(nkmax)  integer ix4(nkmax), iy4(nkmax), iz4(nkmax)  integer ixf(nkmax), iyf(nkmax), izf(nkmax)  integer ixfiw, iyfiw, izfiw, ixsew, iysew, izsew  integer ixthw, iythw, izthw, ixfow, iyfow, izfow  integer ixfi2, iyfi2, izfi2, ixse2, iyse2, izse2  integer ixth2, iyth2, izth2, ixfo2, iyfo2, izfo2  integer isgxf, isgyf, isgzf, ikxf, ikyf, ikzf  integer isgx2, isgy2, isgz2, ikx2, iky2, ikz2  integer isgx3, isgy3, isgz3, ikx3, iky3, ikz3  integer isgx4, isgy4, isgz4, ikx4, iky4, ikz4  integer isgo, iko, isgox, ikox, isgoy, ikoy, isgoz, ikoz  integer isgu, isgv, isgw, iku, ikv, ikw  ixfiw = ix(ifir)  iyfiw = iy(ifir)  izfiw = iz(ifir)  ixsew = ix(isec)  iysew = iy(isec)  izsew = iz(isec)  ixthw = ix(ithi)  iythw = iy(ithi)  izthw = iz(ithi)  ixfow = ix(ifou)  iyfow = iy(ifou)  izfow = iz(ifou)  ixfi2 = ix2(ifir)  iyfi2 = iy2(ifir)  izfi2 = iz2(ifir)  ixse2 = ix2(isec)  iyse2 = iy2(isec)  izse2 = iz2(isec)  ixth2 = ix2(ithi)  iyth2 = iy2(ithi)  izth2 = iz2(ithi)  ixfo2 = ix2(ifou)  iyfo2 = iy2(ifou)  izfo2 = iz2(ifou)  call decmp2  call decmp2  call decmp2  call decmp2  call muldif  call decmp2  call muldif  call decmp2  call muldif  call decmp2  call muldif  call decmp2  call muldif  call decmp2  call muldif  call decmp2  call muldif  call decmp2  call muldif  call decmp2  call muldif  call mulmul  call mulmul  call muldif  call mulmul  call mulmul  call mulmul  call muldif  call mulmul  isgu =-isgu  call muldif  call mulmul  call mulmul  call muldif  call mulmul  isgu =-isgu  call muldif  return  endp4 relative to the plane H it may be desirable to know the perpendicular distance from p4 to H. The following is Fortran code for this purpose. It uses the polynomial decompositions that are part of the output of subroutine crsinn. Variables here are either integer or double precision following convention. Here r215 equals (2.0d0)15, dscle equals (10.0d0)l, and dist is the resulting signed perpendicular distance. In addition, it is assumed that subroutine doubnm (presented below) exists for transforming the polynomial decomposition of an integer into the double precision floating point value of the integer.Sometimes besides knowing the location of the point   call crsinn  call doubnm  call doubnm  call doubnm  call doubnm  dnux = dmax1(dabs(xnum),dabs(ynum),dabs(znum))  xnum = xnum/dnux  ynum = ynum/dnux  znum = znum/dnux  dnom = dsqrt(xnum**2+ynum**2+znum**2)  dist = ((dnum/dnux)/dnom)/dscleThe following is subroutine doubnm that was called above.  subroutine doubnm  integer io(*)  double precision dnum, r215, rpwr  integer isgo, iko, i  if(isgo.eq.0) then    dnum = 0.0d0    go to 900  else    if(iko .lt. 2) stop 310    if(iko .gt. 68) stop 320    rpwr = 1.0d0    dnum = dble(io(1))    do 100 i = 2, iko       rpwr = rpwr*r215       dnum = dnum + dble(io(i))*rpwr100    continue  endif  if(isgo.lt.0) dnum = -dnum900 continue  return  endx-, y-, z-coordinates of the point, in that order. Given i, 1 ≤ i ≤ 12, point i is the point corresponding to the ith line. It is assumed that the coordinates of the twelve points are read into double precision arrays x(i), y(i), z(i), i = 1, …, 12, so that x(i), y(i), z(i) contain the x-, y-, z-coordinates, respectively, of point i.Twelve lines follow, each line containing three numbers. Each line corresponds to a point in 3-dimensional space, and the three numbers in the line correspond to the l equal to 8, through the two-integer decomposition process, the numbers above are rounded off at the lth = 8th digit of their decimal fractions and saved in x(i), y(i), z(i), i = 1, …, 12, so that then they appear as follows.Given i, i = 1, …, 12, the first two integers in the ith line are the two integers into which the x-coordinate of point i is decomposed. Similarly the next two integers correspond to the y-coordinate, and the last two to the z-coordinate. The two-integer decompositions of the twelve points are then saved into ix(i), ix2(i), iy(i), iy2(i), iz(i), iz2(i), i = 1, …, 12, in the obvious manner. It is noted that when mfull = 230 = 1073741824 appears as the second integer corresponding to a coordinate it is to be interpreted as a zero.Each rounded off coordinate is also decomposed into two integers. Twelve lines follow, each line containing six integers. For each l as above equal to 8 and setting isclu to 10l = 108, by calling subroutine decomp the polynomial decomposition of isclu is found to be isgcl, isclp(1), isclp(2), with isgcl equal to 1, isclp(1) equal to 24832 and isclp(2) equal to 3051 .Given x-coordinate of point 2 and the y-coordinate of point 4 minus the product of the y-coordinate of point 2 and the x-coordinate of point 4 is described. Using rounded-off numbers the result of the computation should equal x(2) · y(4) − y(2) · x(4), i. e.,As an example of how exact computations are carried out that reference coordinates of the input points given above, the computation that is the product of the (38.000000000) · (−49.840807040) − (7.049967880) · (85.557213020).8 rounded off at the decimal point. Since the resulting integer may be too big to be saved into a 32 bit integer word, its polynomial decomposition in the form of two or more 32 bit integer words is obtained instead. Thus, for example, the polynomial decomposition of the integer which is the x-coordinate of point 4 times 108 rounded off at the decimal point can be obtained by calling subroutine decmp2 using the two-integer decomposition of the coordinate, i.e. the integers 85 and 55721302, and the polynomial decomposition of 108 as obtained above. The resulting polynomial decomposition is then found to be isgox4, ikox4, iox4(k), k = 1, …, ikox4, with isgox4 equal to 1, ikox4 equal to 3, iox4(k), k = 1, 2, 3, equal to 29270, 31723, and 7  + 7 · (215)2). In the same manner the polynomial decompositions associated with the x-, y-coordinates of point 2, and the y-coordinate of point 4 are found to be, respectively, isgox2, ikox2, iox2(k), k = 1, …, ikox2, isgoy2, ikoy2, ioy2(k), k = 1, …, ikoy2, isgoy4, ikoy4, ioy4(k), k = 1, …, ikoy4, with isgox2 equal to 1, ikox2 equal to 3, iox2(k), k = 1, 2, 3, equal to 26112, 17662, 3, isgoy2 equal to 1, ikoy2 equal to 2, ioy2(k), k = 1, 2, equal to 26036, 21514, isgoy4 equal to −1, ikoy4 equal to 3, ioy4(k), k = 1, 2, 3, equal to 2368, 21030, 4. Finally, using these polynomial decompositions as input the desired result is obtained by calling subroutines mulmul and muldif as follows. Here nkmax is the dimension of all of the arrays (input and output).On the other hand, if exact computations are required then each of the four numbers involved must first be converted into an integer which is the number times 10call mulmulcall mulmulcall muldif8)2 rounded off at the decimal point is then found to be isgo, iko, io(k), k = 1, …, iko, with isgo equal to −1, iko equal to 5, io(k), k = 1, …, 5, equal to 21112, 15183, 31880, 21597, 21. The desired result is then approximately equal to this integer, i.e. 21112 + 15183 · (215) + 31880 · (215)2 + 21597 · (215)3 + 21 · (215)4, divided by (108)2. By calling subroutine doubnm the integer can be approximated by a double precision number which when divided by (108)2 is approximately −2497.1262712133, an approximation of the desired result.The polynomial decomposition of an integer that approximates the desired result times isgo equals −1 so that point 12 must be in H−. In addition iko equals 7, io(k), k = 1, …, 7, equals 21844, 27853, 3870, 5372, 13630, 9887, 213, isgox equals −1, ikox equals 5, iox(k), k = 1, …, 5, equals 11868, 15341, 2677, 15631, 62, isgoy equals 1, ikoy equals 5, ioy(k), k = 1, …, 5, equals 11577, 8756, 364, 27887, 63, isgoz equals −1, ikoz equals 5, ioz(k), k = 1, …, 5, equals 5921, 23919, 26934, 16812, 19. By calling subroutine doubnm the integer whose polynomial decomposition is isgo, iko, io(k), k = 1, …, iko, can be approximated by a double precision number dnum. Similarly, the three integers, say ix, iy, iz, whose polynomial decompositions are isgox, ikox, iox(k), k = 1, …, ikox, isgoy, ikoy, ioy(k), k = 1, …, ikoy, isgoz, ikoz, ioz(k), k = 1, …, ikoz, can be approximated, respectively, by double precision numbers xnum, ynum, znum. The vector , which is perpendicular to the plane H and points into H+, can then be approximated by the vector . Finally, by dividing dnum by both the length of the vector  and 108 the number −25.047402554921 is approximately obtained, an approximation of the signed perpendicular distance from point 12 to plane H, the negative sign indicating that point 12 is in H−.On output http://math.nist.gov/~JBernal/JBernal_Sft.html. Besides being used for locating a point relative to a plane, in these programs the scheme has also been used for locating a point relative to a sphere, for computing the intersection of a line and a plane, for computing the center of a sphere, etc.A scheme has been presented and software has been documented for transforming into a series of integers input decimal numbers that have been read into a computer as double precision floating point numbers and for carrying out multiplication, addition and subtraction operations based on these numbers using their integer representations. The total number of significant digits of each input number must not be greater than 14, and the number of digits to the left of the decimal point must not exceed 9. Through a preprocessing step the double precision floating point representation of each input decimal number is rounded off at a prespecified location of its decimal fraction, a location no more than 9 digits to the right of the decimal point, and the rounded off number is decomposed into two integers. All operations that follow involving the input number are then carried out in terms of the rounded off double precision floating point number and when this is not satisfactory in terms of the two integers. This scheme has been successfully incorporated into several computational geometry related programs such as REGTET [


Scientific Reports5: Article number: 1061310.1038/srep10613; published online: 06032015; updated: 01122016This Article contains typographical errors in the keys of Figure 4c, 4d, 4e, and 4f.6 ‘ηor’ have been incorrectly given as ‘α’.In Figure 4c, the values of the exponents for power-law decay of GK ‘ηp’ have been incorrectly given as ‘α’.In Figure 4e the values of the exponents for power-law decay of Gp’ and ‘ξor’ which were incorrectly given as ‘[p’ and ‘[or’ respectively.Lastly in Figures 4d and 4f, there are typographical errors in the symbols ‘ξThe correct Figure 4 appears below as


In Volume 93, Issue 10, October 2015, throughout pages 674−83 “Rivaroxavan” should read: “Rivaroxaban”.


T∧-fuzzy (implicative) ideal is proposed in MV-algebras. The relationships between falling fuzzy (implicative) ideals and T-fuzzy (implicative) ideals are discussed, and conditions for a falling fuzzy (implicative) ideal to be a T∧-fuzzy (implicative) ideal are provided. Some characterizations of falling fuzzy (implicative) ideals are presented by studying proprieties of them. The product ⊛ and the up product ⊚ operations on falling shadows and the upset of a falling shadow are established, by which T-fuzzy ideals are investigated based on probability spaces.Based on the falling shadow theory, the concept of falling fuzzy (implicative) ideals as a generalization of that of a  Then there is a unique operation x → y satisfying the condition T ≤ y if and only if z ≤ x → y for all x, y, z ∈ ; namely, x → y = max{z∣T ≤ y}.Let x → y from t-norm.The operation A, ⊕, ¬, 0) of type  satisfying the following equations: for any x, y, z ∈ A, (MV1)x ⊕ (y ⊕ z) = (x ⊕ y) ⊕ z;(MV2)x ⊕ y = y ⊕ x;(MV3)x ⊕ 0 = x;(MV4)x = x;¬¬(MV5)x ⊕ ¬0 = ¬0;(MV6)x ⊕ y) ⊕ y = ¬(¬y ⊕ x) ⊕ x.¬ be an MV-algebra. For any x, y ∈ A, we define 1 = ¬0, x ⊗ y = ¬(¬x ⊕ ¬y), x → y = ¬x ⊕ y, x ⊖ y = x ⊗ ¬y, x∨y = ¬(¬x ⊕ y) ⊕ y = (x ⊖ y) ⊕ y, x∧y = ¬(¬x∨¬y), 0x = 0, (n + 1)x = nx ⊕ x, x0 = 1, and xn+1) is a bounded distributive lattice 2. There are many types of probability distributions, but only three types are the most classic ones   x, y ∈ I implies x ⊕ y ∈ I; (2)  x ≤ y and y ∈ I imply x ∈ I. For purpose of convenience, let ∅ be an ideal of A in the the rest of the sections. If an ideal I satisfies the following condition: (x ⊖ y) ⊖ z ∈ I and y ⊖ z ∈ I imply x ⊖ z ∈ I for any x, y, z ∈ A, then I is called an implicative ideal of A.A nonempty set I be a nonempty set of A. Then the following statements are equivalent: I is an ideal of A;I; ∀x, y ∈ A, x ⊖ y ∈ I, and y ∈ I imply x ∈ I;0 ∈ x, y, z ∈ A, if z ⊖ x ≤ y and x, y ∈ I, then z ∈ I.∀Let T-fuzzy ideals and falling fuzzy ideals of MV-algebras. The relationships between T-fuzzy ideals and falling fuzzy ideals are provided, and some characterizations of falling fuzzy ideals are displayed.In this section, we will introduce the notions of μ of A is called a T-fuzzy ideal of A, if μ satisfies ∀x, y ∈ A, μ(x ⊕ y) ≥ T(μ(x), μ(y));x ≤ y implies μ(x) ≥ μ(y).A fuzzy set T-fuzzy ideal of A is a T∧-fuzzy ideal when T = T∧ and a T∧-fuzzy ideal is also called a fuzzy ideal , L is an ideal of A, where L = {x ∈ A∣μ(x) ≥ t}.∀Let T-fuzzy ideals for the further discussion.In the following, we give some properties of μ1, μ2 be fuzzy sets of A. If μ1, μ2 are T-fuzzy ideals of A, then μ1⊓μ2 is a T-fuzzy ideal of A.Let The proof is straightforward.As a direct consequence of μi  (i ∈ Λ) be fuzzy sets of A, where Λ is an index set. If μi is a T-fuzzy ideal of A, then ⊓i∈Λμi is a T-fuzzy ideal of A.Let A = {0, a, b, 1} where 0 < a < 1, 0 < b < 1. Define the operations ⊕ and ¬ on A as follows:Let A, ⊕, ¬, 0) is an MV-algebra. Define fuzzy sets μ1 and μ2 of A as follows:It is clear that (a ⊕ b) = 0.3 < (μ1⊔μ2)(a)∧(μ1⊔μ2)(b) = 0.6∧0.5 = 0.5.Then T-fuzzy ideals is not a T-fuzzy ideal. In order to investigate the algebraic properties of the set of all T∧-fuzzy ideals in BL-algebras,  and m is the usual Lebesgue measure. Denote A1 = { ∈ R2∣y > 0} and A2 = { ∈ R2∣y < 0}. The mapping ξ : Ω → P(A) is defined byξ(t) is an ideal of A for any t ∈ . Thus A, where Let  be a probability space and F(A): = {f∣f : Ω → A}. Define the operations ⊞ and ~ on F(A) as follows: ∀ω ∈ Ω, f, g ∈ F(A),Let (F(A) be defined by 0(ω) = 0 for any ω ∈ Ω. Then it is easy to check that (F(A), ⊞, ~, 0) is an MV-algebra. Define the operation ⊟ on F(A) by (f⊟g)(ω) = f(ω) ⊖ g(ω), for any ω ∈ Ω, f, g ∈ F(A).Let 0 ∈ S of A and f ∈ F(A), let Sf : = {ω ∈ Ω∣f(ω) ∈ S} andSf ∈ A.For any subset Ω, A, P) be a probability space, S a nonempty subset of A, and ω ∈ Ω. If S is an ideal of A, then ξ(ω) = {f ∈ F(A)∣f(ω) ∈ S} is an ideal of F(A).Let  = 0 ∈ S for any ω ∈ Ω, we have that 0 ∈ ξ(ω). Let f, g ∈ ξ(ω) be such that f⊟g ∈ ξ(ω) and g ∈ ξ(ω). For any ω ∈ Ω, we have f(ω) ⊖ g(ω) = (f⊟g)(ω) ∈ A and g(ω) ∈ S. Hence f(ω) ∈ S; that is, f ∈ ξ(ω). Thus ξ(ω) is an ideal of F(A).Suppose that ξ−1(f) = {ω ∈ Ω∣f ∈ ξ(ω)} = {ω ∈ Ω∣f(ω) ∈ S} = Sf ∈ A and ξ is a random set of F(A), we get that F(A), where Noticing that μ be a T∧-fuzzy ideal of A. Then μ is a falling fuzzy ideal of A.Let Ω, A, P) = , where A is a Borel field on  and m is the usual Lebesgue measure. Since μ is a T∧-fuzzy ideal of A, then L is an ideal of A for any t ∈ , by ξ :  → P(A) by ξ(t) = L for any t ∈ ; then μ is a falling fuzzy ideal of A.Consider the probability space  be the power set of E. Let ⊕, ¬, and 0 denote, respectively, the join, the complement, and the smallest element in A : = P(E). It is clear that  is an MV-algebra.Let Ω, A, P) = , where A is a Borel field on  and m is the usual Lebesgue measure. The mapping ξ : Ω → P(A) is defined byLet (ξ(t) is an ideal of A for any t ∈ . Thus A, where Then T∧-fuzzy ideal of A since But Ω, A, P) be a probability space and ξ : Ω → P(A). For any x ∈ A, let Ω = {ω ∈ Ω∣x ∈ ξ(ω)}; then Ω ∈ A. In what follows, we give a number of equivalent conditions of falling fuzzy ideals for further discussion.Let (ξ : Ω → P(A). Then A if and only if (1)  0 ∈ ξ(ω), (2)  y ∈ ξ(ω) and x ∈ A∖ξ(ω) imply x ⊖ y ∈ A∖ξ(ω) for any ω ∈ Ω, x, y ∈ A.Let A, then ξ(ω) is an ideal of A for any ω ∈ Ω, and it follows that 0 ∈ ξ(ω). Let x, y ∈ A be such that y ∈ ξ(ω) and x ∈ A∖ξ(ω). Supposing that x ⊖ y ∈ ξ(ω), then x ∈ ξ(ω), a contradiction, and thus (2) is valid.Assuming that ω ∈ Ω, let x, y ∈ A be such that x ⊖ y ∈ ξ(ω) and y ∈ ξ(ω). Supposing that x ∈ A∖ξ(ω), it follows that x ⊖ y ∈ A∖ξ(ω), which is a contradiction, and thus x ∈ ξ(ω). Hence A.Conversely, assume that (1) and (2) are valid. For any ξ : Ω → P(A) be a random set and ξ. Then A if and only if z ⊖ x ≤ y implies Ω∩Ω⊆Ω for any x, y, z ∈ A.Let x, y, z ∈ A be such that z ⊖ x ≤ y. For any ω ∈ Ω∩Ω, we get x ∈ ξ(ω), y ∈ ξ(ω). Considering that A, we have that ξ(ω) is an ideal of A. By z ∈ ξ(ω); that is, ω ∈ Ω. Hence Ω∩Ω⊆Ω.Let z ⊖ x ≤ y implies Ω∩Ω⊆Ω for any x, y, z ∈ A. Let ω ∈ Ω be such that ξ(ω) ≠ ∅. Then there exists x ∈ ξ(ω); that is, ω ∈ Ω. Since 0 ⊖ x = 0 ≤ x, then Ω∩Ω = Ω⊆Ω. It follows that ω ∈ Ω; that is, 0 ∈ ξ(ω). Let x, y ∈ A be such that x ⊖ y ∈ ξ(ω) and y ∈ ξ(ω). Then ω ∈ Ω∩Ω. Since x ⊖ (x ⊖ y) ≤ y, it follows that Ω∩Ω⊆Ω. Thus ω ∈ Ω; that is, x ∈ ξ(ω). Hence ξ(ω) is an ideal of A, and so A.Conversely, assume that ξ : Ω → P(A) be a random set and ξ. If A, then the following statements hold: for any x, y ∈ A, Ω⊆Ω;Ω∩Ω⊆Ω.Let A, we have that ξ(ω) is an ideal of A for any ω ∈ Ω. It follows that 0 ∈ ξ(ω); that is, ω ∈ Ω. Hence Ω⊆Ω.Noticing that ω ∈ Ω∩Ω, we get x ⊖ y ∈ ξ(ω) and y ∈ ξ(ω). Since ξ(ω) is an ideal of A, then x ∈ ξ(ω); that is, ω ∈ Ω. Thus Ω∩Ω⊆Ω.For any ξ : Ω → P(A). Then A if and only if, for any x, y ∈ A, Ω∩Ω⊆Ω;x ≤ y implies Ω⊆Ω.Let A, then ξ(ω) is an ideal of A for any ω ∈ Ω. For any x, y ∈ A, if x ∈ ξ(ω), y ∈ ξ(ω), then x ⊕ y ∈ ξ(ω). That is, ω ∈ Ω and ω ∈ Ω imply ω ∈ Ω. Hence Ω∩Ω⊆Ω. If x ≤ y, that is, x ⊖ 0 ≤ y, it follows that Ω⊆Ω by Assuming that ω ∈ Ω, let x, y ∈ A be such that x ∈ ξ(ω) and y ∈ ξ(ω). It follows that ω ∈ Ω∩Ω. Thus ω ∈ Ω; that is, x ⊕ y ∈ ξ(ω). If x ≤ y and y ∈ ξ(ω) for any x, y ∈ A, then ω ∈ Ω⊆Ω. Thus x ∈ ξ(ω). Hence ξ(ω) is an ideal of A, and so A.Conversely, assume that (1) and (2) are valid. For any ξ : Ω → P(A). Then A if and only if, for any x, y ∈ A, Ω∩Ω⊆Ω;Ω⊆Ω.Let x ≤ y if and only if x∧y = x for any x, y ∈ A.The proof is obvious since ξ : Ω → P(A). Then A if and only if, for any x, y ∈ A, Ω∩Ω⊆Ω;Ω⊆Ω.Let x ⊗ y ≤ y. Conversely, assume that conditions (1) and (2) hold. By hypothesis, we have Ω⊆Ω(y ⊗ (¬y ⊕ x); ξ) for any x, y ∈ A. Now let x ≤ y and ω ∈ Ω. Then y ∈ ξ(ω) and y ⊗ (¬y ⊕ x) = x∧y = x ∈ ξ(ω). It means that x ≤ y implies Ω⊆Ω, and so A by One direction is clear since ξ : Ω → P(A). If the following conditions are valid: Ω = Ω;Ω∩Ω⊆Ω, for any x, y ∈ A,then A.Let ω ∈ Ω, we have ω ∈ Ω; that is, 0 ∈ ξ(ω). For any x, y ∈ A, if x ⊖ y ∈ ξ(ω) and y ∈ ξ(ω), then ω ∈ Ω, ω ∈ Ω. It follows that ω ∈ Ω∩Ω⊆Ω. Hence ω ∈ Ω; that is, x ∈ ξ(ω). Thus ξ(ω) is an ideal of A, and so A.Assume that conditions (1) and (2) are valid. For any ξ : Ω → P(A). If A, then the following relationships hold: for any x, y, z ∈ A, Ω∩Ω⊆Ω;Ω = Ω, then Ω⊆Ω;if Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω((x ⊖ z) ⊖ z; ξ);Ω∩Ω = Ω∩Ω;Ω∩Ω = Ω;x ≤ y, then Ω∩Ω = Ω.if Let x, y, z ∈ A, we have Ω((x ⊖ z) ⊖ z; ξ)⊇Ω((x ⊖ z) ⊖ y; ξ)∩Ω = Ω((x ⊖ y) ⊖ z; ξ)∩Ω by (1).Here we only prove (3), and the other cases directly follow from According to ξ : Ω → P(A). If the following conditions are valid: Ω = Ω;Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω((x ⊖ z) ⊖ z; ξ), for any x, y, z ∈ A,then A.Let Ω∩Ω = Ω((x ⊖ y) ⊖ 0; ξ)∩Ω⊆Ω((x ⊖ 0) ⊖ 0; ξ) = Ω, for any x, y ∈ A. It follows from A.By hypothesis, we get T∧-fuzzy ideals. Under what conditions a falling fuzzy ideal becomes a T-fuzzy ideal, we will give some answers to the questions in the following.From ξ : Ω → P(A). If A, then Tm-fuzzy ideal of A.Let A, we have that Ω⊆Ω for any x ∈ A, by Ω∩Ω⊆Ω for any x, y ∈ A by ω ∈ Ω∣x ⊖ y ∈ ξ(ω)}∩{ω ∈ Ω∣y ∈ ξ(ω)}⊆{ω ∈ Ω∣x ∈ ξ(ω)}. Thus P(ω∣x ⊖ y ∈ ξ(ω) or Tm-fuzzy ideal of A.Noticing that ξ : Ω → P(A). If A, then the following statements hold: ∀x, y ∈ A, Ω⊆Ω or Ω⊆Ω, then T∧-fuzzy ideal of A;if Ω and Ω are independent random events, then Tp-fuzzy ideal of A.if Let x ∈ A. According to hypothesis and Ω⊆Ω or Ω⊆Ω for any x, y ∈ A. If Ω⊆Ω, then Ω⊆Ω, then T∧-fuzzy ideal of A.By x ∈ A. Since A, then {ω ∈ Ω∣x ⊖ y ∈ ξ(ω)}∩{ω ∈ Ω∣y ∈ ξ(ω)}⊆{ω ∈ Ω∣x ∈ ξ(ω)}, for any x, y ∈ A. It follows that Tp-fuzzy ideal of A.It is directly obtained that Ω, A, P) be a probability space, and let ξ, ξ1, ξ2 : Ω → P(A), respectively. Then the product of Let  be a probability space and ξ : Ω → P(A). Then x ≤ y implies x, y ∈ A.Let  be a probability space and ξ : Ω → P(A). If the probability distribution P of two-dimensional random variables is diagonal , then T-fuzzy ideal of A if and only if where T ∈ {T∧, Tm, Tp}.Let  be a probability space and ξ : Ω → P(A). If the probability distribution P of two-dimensional random variables is diagonal, then T∧-fuzzy ideal of A if and only if Let , and then T∧-fuzzy ideal of A.Conversely, assume that Ω, A, P) be a probability space and ξ : Ω → P(A). If P of two-dimensional random variables is antidiagonal (or independent), then Tm-fuzzy ideal  of A if and only if Let ; then Tm-fuzzy ideal of A.We only consider that Tm-fuzzy ideal of A. The proof of Conversely, suppose that Ω, A, P) be a probability space, where the probability distribution P of two-dimensional random variables is diagonal. Let ξ1, ξ2 : Ω → P(A), respectively. If T∧-fuzzy ideals of A, then T∧-fuzzy ideal of A.Let  become falling fuzzy implicative ideals.In the section, we introduce the notion of falling fuzzy implicative ideals as a generalization of μ in A is called a T-fuzzy implicative ideal of A if it satisfies ∀x, y, z ∈ A, μ(0) ≥ μ(x);μ(x ⊖ z) ≥ T(μ((x ⊖ y) ⊖ z), μ(y ⊖ z)).A fuzzy set Ω, A, P) be a probability space and ξ : Ω → P(A) be a random set. If ξ(ω) is an implicative ideal of A for any ω ∈ Ω, then the falling shadow ξ, that is, ∀u ∈ A,A.Let  is an MV-algebra.Routine computations prove that  = , where A is a Borel field on  and m is the usual Lebesgue measure. The mapping ξ : Ω → P(A) is defined byLet (ξ(t) is an implicative ideal of A for any t ∈ . Thus A, where Then Ω, A, P) be a probability space, S a nonempty subset of A, and ω ∈ Ω. If S is an implicative ideal of A, then ξ(ω) = {f ∈ F(A)∣f(ω) ∈ S} is an implicative ideal of F(A).Let . Then A if and only if (1)  0 ∈ ξ(ω), (2)  y ⊖ z ∈ ξ(ω) and x ⊖ z ∈ A∖ξ(ω) imply (x ⊖ y) ⊖ z ∈ A∖ξ(ω) for any ω ∈ Ω, x, y, z ∈ A.Let A, it follows that ξ(ω) is an implicative ideal of A for any ω ∈ Ω, and thus 0 ∈ ξ(ω). Let x, y, z ∈ A be such that y ⊖ z ∈ ξ(ω) and x ⊖ z ∈ A∖ξ(ω). If (x ⊖ y) ⊖ z ∈ ξ(ω) hold, then x ⊖ z ∈ ξ(ω), which is a contradiction, and so (2) is valid.Suppose that x, y, z ∈ A be such that (x ⊖ y) ⊖ z ∈ ξ(ω) and y ⊖ z ∈ ξ(ω) for any ω ∈ Ω. If x ⊖ z ∈ A∖ξ(ω), it follows from hypothesis that (x ⊖ y) ⊖ z ∈ A∖ξ(ω), a contradiction, and so x ⊖ z ∈ ξ(ω). Therefore A.Conversely, let ξ : Ω → P(A). If A, then we have, for any x, y ∈ A, Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω;Ω = Ω, for any n ≥ 1;Ω∩Ω = Ω = Ω;Ω = Ω, for any n ≥ 1;Ω∩Ω = Ω;Ω((x ⊖ y) ⊖ y; ξ)⊆Ω.Let ω ∈ Ω((x ⊖ y) ⊖ z; ξ)∩Ω, we have (x ⊖ y) ⊖ z ∈ ξ(ω) and y ⊖ z ∈ ξ(ω). Since ξ(ω) is an ideal of A, it follows that x ⊖ z ∈ ξ(ω); that is, ω ∈ Ω, and so Ω((x ⊖ y) ⊖ z; ξ)  ∩  Ω  ⊆  Ω.For any n = 1. For the case of n = 2, we have Ω = Ω  ⊇  Ω((1 ⊖ ¬x)⊖¬x; ξ)  ∩  Ω = Ω by (1). On the other hand, since x2 ≤ x, then Ω⊆Ω by Ω = Ω. Suppose that n ≥ 3 and Ω = Ω, then we get Ω = Ω  =  Ω  ⊇  Ω((xn−2 ⊖ ¬x)  ⊖  ¬x; ξ)  ∩  Ω  =  Ω. The other direction is that Ω⊆Ω since xn ≤ x, which shows that (2) is valid.It is true for x, y ∈ A, we have x, y ≤ x∨y ≤ x ⊕ y, and so Ω  ⊆  Ω  ⊆  Ω∩Ω. On the other hand, since (x ⊕ y) ⊖ x = ¬x∧y, then Ω  ⊇  Ω((x ⊕ y) ⊖ x; ξ)  ∩  Ω  ⊇    ∪  Ω)  ∩  Ω  =  ∩Ω)  ∪  ∩Ω)  =  Ω  ∪  ∩Ω)  =  Ω∩Ω, and so it proves (3).For any It is straightforward by (3).Ω  =  Ω  ⊇  Ω((1 ⊖ (1 ⊖ y))⊖¬x; ξ)  ∩  Ω((1 ⊖ y)⊖¬x; ξ)  =  Ω∩Ω. On the other hand, since x ⊗ y ≤ x and x ⊖ y ≤ x, we have Ω⊆Ω  ∩  Ω; that means (5) holds.On the one hand, ω ∈ Ω((x ⊖ y) ⊖ y; ξ), we have (x ⊖ y) ⊖ y ∈ ξ(ω). Since y ⊖ y = 0 ∈ ξ(ω) and ξ(ω) is an implicative ideal of A, it follows that x ⊖ y ∈ ξ(ω), that is ω ∈ Ω, and thus Ω((x ⊖ y) ⊖ y; ξ)⊆Ω.For any Some of equivalent conditions of falling fuzzy implicative ideals are given in the next theorem.ξ : Ω → P(A). Then A if and only if for any x, y, z ∈ A, Ω⊆Ω;Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω.Let A, it follows that Ω⊆Ω by Assume that ω ∈ Ω and x ∈ ξ(ω). Then ω ∈ Ω⊆Ω; that is, 0 ∈ ξ(ω). For any x, y, z ∈ A, if (x ⊖ y) ⊖ z ∈ ξ(ω) and y ⊖ z ∈ ξ(ω), then ω ∈ Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω, and so x ⊖ z ∈ ξ(ω). It follows that ξ(ω) is an implicative ideal of A. Thus A.Conversely, assume that (1) and (2) hold. Let As a direct consequence of ξ : Ω → P(A). If Ω = Ω;Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω, for any x, y ∈ A,then A.Let The following result provides a condition for a falling shadow to be a falling fuzzy implicative ideal.ξ : Ω → P(A) be a random set and ξ. Then A if and only if it satisfies Ω⊆Ω for any x ∈ A;Ω((x ⊖ (y ⊖ x)) ⊖ z; ξ)∩Ω⊆Ω for any x, y, z ∈ A.Let A. (1) is directly from ω ∈ Ω((x ⊖ (y ⊖ x)) ⊖ z; ξ)∩Ω, we have (x ⊖ (y ⊖ x)) ⊖ z ∈ ξ(ω) and z ∈ ξ(ω); therefore x ⊖ (y ⊖ x) ∈ ξ(ω). Since (y ⊖ (y ⊖ x))⊖(y ⊖ x)⊖(x ⊖ (y ⊖ x))≤(y ⊖ (y ⊖ x)) ⊖ x = 0 ∈ ξ(ω), we obtain (y ⊖ (y ⊖ x))⊖(y ⊖ x) ∈ ξ(ω), and so y ⊖ (y ⊖ x) = x ⊖ (x ⊖ y) ∈ ξ(ω) by ω ∈ Ω(x ⊖ (x ⊖ y); ξ). Noticing that ((x ⊖ y) ⊖ z)⊖(x ⊖ (y ⊖ x)) = (x ⊖ (x ⊖ (y ⊖ x))) ⊖ y ⊖ z = (x∧(y ⊖ x)) ⊖ y ⊖ z = 0 ∈ ξ(ω), we get (x ⊖ y) ⊖ z ∈ ξ(ω). Since z ∈ ξ(ω), it follows that x ⊖ y ∈ ξ(ω); that is, ω ∈ Ω. Considering that ξ(ω) is an implicative ideal of A, we have ω ∈ Ω∩Ω(x ⊖ (x ⊖ y); ξ)⊆Ω by Suppose that x, y, z ∈ A, we get Ω(x ⊖ (y ⊖ x); ξ)⊆Ω and Ω∩Ω = Ω((x ⊖ (x ⊖ x)) ⊖ z; ξ)∩Ω⊆Ω, and so A. For any ω ∈ Ω((x ⊖ y) ⊖ z; ξ)∩Ω, we obtain (x ⊖ y) ⊖ z ∈ ξ(ω) and y ⊖ z ∈ ξ(ω). Since (x ⊖ z) ⊖ (x ⊖ (x ⊖ z)) ⊖ (y ⊖ z) ≤ (x ⊖ y) ⊖ (x ⊖ (x ⊖ z)) = (x∧(x ⊖ z)) ⊖ y ≤ (x ⊖ z) ⊖ y = (x ⊖ y) ⊖ z, and ξ(ω) is an ideal of A, then (x ⊖ z)⊖(x ⊖ (x ⊖ z)) ∈ ξ(ω); that is, ω ∈ Ω((x ⊖ z)⊖(x ⊖ (x ⊖ z)); ξ)⊆Ω. Thus Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω, it follows that A.Conversely, for any We can also give some conditions for a falling fuzzy ideal to be a falling fuzzy implicative ideal.ξ : Ω → P(A). If A, then the following conditions are equivalent: A,Ω((y ⊖ ¬x) ⊖ z; ξ)∩Ω⊆Ω, for any x, y, z ∈ A;Ω = Ω((x ⊕ x) ⊖ nx; ξ) for any x ∈ A and n ≥ 1;Ω(x ⊖ (y ⊖ x); ξ)⊆Ω, for any x, y ∈ A;Ω((x ⊖ y) ⊖ y; ξ)⊆Ω, for any x, y ∈ A.Let A. For any x, y, z ∈ A, we have Ω((y ⊖ ¬x) ⊖ z; ξ)∩Ω = Ω((¬z ⊖ ¬y)⊖¬x; ξ)∩Ω⊆Ω = Ω, by (1)⇔(2) Suppose that x, y, z ∈ A be such that (x ⊖ y) ⊖ z ∈ ξ(ω) and y ⊖ z ∈ ξ(ω). It follows that ω ∈ Ω((x ⊖ y) ⊖ z; ξ)∩Ω = Ω((¬y ⊖ z)⊖¬x; ξ)∩Ω⊆Ω = Ω, and so x ⊖ z ∈ ξ(ω). Noticing that ξ(ω) is an ideal of A, thus ξ(ω) is an implicative ideal, and therefore A.Conversely, let ω ∈ Ω, we have 0 ∈ ξ(ω). Since x ⊖ x = 0 ∈ ξ(ω) and ((x ⊕ x) ⊖ x) ⊖ x = (x ⊕ x)⊖(x ⊕ x) = 0 ∈ ξ(ω), taking into consideration that ξ(ω) is an implicative ideal, we have (x ⊕ x) ⊖ x ∈ ξ(ω); that is, ω ∈ Ω((x ⊕ x) ⊖ x; ξ), and thus Ω⊆Ω((x ⊕ x) ⊖ x; ξ). Since (x ⊕ x) ⊖ nx ≤ (x ⊕ x) ⊖ x, then Ω⊆Ω((x ⊕ x) ⊖ x; ξ)⊆Ω((x ⊕ x) ⊖ nx; ξ) by Ω((x ⊕ x) ⊖ nx; ξ)⊆Ω; therefore, (2) holds.(1)⇔(3) For any n = 1, we get Ω = Ω((x ⊕ x) ⊖ x; ξ) for any x ∈ A. It follows that (x ⊕ x) ⊖ x ∈ ξ(ω) for any ω ∈ Ω. Let x, y, z ∈ A be such that (x ⊖ y) ⊖ z ∈ ξ(ω) and y ⊖ z ∈ ξ(ω). By x ⊖ (z ⊕ z))⊖((x ⊖ y) ⊖ z)⊖(y ⊖ z) = ((x ⊖ z)⊖(y ⊖ z)) ⊖ z ⊖ ((x ⊖ y) ⊖ z)≤(x ⊖ y) ⊖ z ⊖ ((x ⊖ y) ⊖ z) = 0. Noticing that ξ(ω) is an ideal of A, we obtain x ⊖ (z ⊕ z) ∈ ξ(ω). Since (z ⊕ z) ⊖ z ∈ ξ(ω) and x ⊖ z ≤ ((z ⊕ z) ⊖ z)⊕(x ⊖ (z ⊕ z)), then x ⊖ z ∈ ξ(ω). Hence ξ(ω) is an implicative ideal, and thus A.Conversely, given (1)⇒(4) It is immediate from the proof of x ⊖ y)⊖(x ⊖ (x ⊖ y)) = (x ⊖ y) ⊖ y for any x, y ∈ A, then Ω((x ⊖ y) ⊖ y; ξ)⊆Ω((x ⊖ y)⊖(y ⊖ (x ⊖ y)); ξ)⊆Ω, and thus (5) is valid.(4)⇒(5) Since (x ⊖ (y ⊖ z) ⊖ z) ⊖ z = (x ⊖ z)⊖(y ⊖ z) ⊖ z ≤ (x ⊖ y) ⊖ z, we have Ω((x ⊖ y) ⊖ z; ξ)⊆Ω((x ⊖ (y ⊖ z) ⊖ z) ⊖ z; ξ) = Ω(((x ⊖ z) ⊖ z)⊖(y ⊖ z); ξ). It follows that Ω((x ⊖ y) ⊖ z; ξ)∩Ω⊆Ω(((x ⊖ z) ⊖ z)⊖(y ⊖ z); ξ)∩Ω⊆Ω((x ⊖ z) ⊖ z; ξ)⊆Ω, and thus A.(5)⇒(1) Noticing that (ξ : Ω → P(A). If A, then one has Tm-fuzzy implicative ideal of A;Ω⊆Ω or Ω⊆Ω for any x, y ∈ A, then T∧-fuzzy implicative ideal of A;if x, y ∈ A, if Ω and Ω are independent random events, then Tp-fuzzy implicative ideal of A.for any Let The proof is similar to that of