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Research article Special Issues

Pythagorean fuzzy incidence graphs with application in illegal wildlife trade

  • Chemical engineers can model numerous interactions in a process using incidence graphs. They are used to methodically map out a whole network of interconnected processes and controllers to describe each component's impact on the others. It makes it easier to visualize potential process paths or a series of impacts. A Pythagorean fuzzy set is an effective tool to overcome ambiguity and vagueness. In this paper, we introduce the concept of Pythagorean fuzzy incidence graphs. We discuss the incidence path and characterize the strongest incidence path in Pythagorean fuzzy incidence graphs. Furthermore, we propose the idea of Pythagorean fuzzy incidence cycles and Pythagorean fuzzy incidence trees in Pythagorean fuzzy incidence graphs and give some essential results. We illustrate the notions of Pythagorean fuzzy incidence cut vertices, Pythagorean fuzzy incidence bridges, and Pythagorean fuzzy incidence cut pairs. We also establish some results about Pythagorean fuzzy incidence cut pairs. Moreover, we study the types of incidence pairs and determine some crucial results concerning strong incidence pairs in the Pythagorean fuzzy incidence graph. We also obtain the characterization of Pythagorean fuzzy incidence cut pairs using α-strong incidence pairs and find the relation between Pythagorean fuzzy incidence trees and α-strong incidence pairs. Finally, we provide the application of Pythagorean fuzzy incidence graphs in the illegal wildlife trade.

    Citation: Ayesha Shareef, Uzma Ahmad, Saba Siddique, Mohammed M. Ali Al-Shamiri. Pythagorean fuzzy incidence graphs with application in illegal wildlife trade[J]. AIMS Mathematics, 2023, 8(9): 21793-21827. doi: 10.3934/math.20231112

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  • Chemical engineers can model numerous interactions in a process using incidence graphs. They are used to methodically map out a whole network of interconnected processes and controllers to describe each component's impact on the others. It makes it easier to visualize potential process paths or a series of impacts. A Pythagorean fuzzy set is an effective tool to overcome ambiguity and vagueness. In this paper, we introduce the concept of Pythagorean fuzzy incidence graphs. We discuss the incidence path and characterize the strongest incidence path in Pythagorean fuzzy incidence graphs. Furthermore, we propose the idea of Pythagorean fuzzy incidence cycles and Pythagorean fuzzy incidence trees in Pythagorean fuzzy incidence graphs and give some essential results. We illustrate the notions of Pythagorean fuzzy incidence cut vertices, Pythagorean fuzzy incidence bridges, and Pythagorean fuzzy incidence cut pairs. We also establish some results about Pythagorean fuzzy incidence cut pairs. Moreover, we study the types of incidence pairs and determine some crucial results concerning strong incidence pairs in the Pythagorean fuzzy incidence graph. We also obtain the characterization of Pythagorean fuzzy incidence cut pairs using α-strong incidence pairs and find the relation between Pythagorean fuzzy incidence trees and α-strong incidence pairs. Finally, we provide the application of Pythagorean fuzzy incidence graphs in the illegal wildlife trade.



    In 1980, Haristchain [10], in order to be able to conveniently handle the vast amount of varied data that defines a spatial structure, a sophisticated form of database was evolved, which called a plenix (also see, [13,19]). In 1984, Nooshin [18] defined the notion of a nexus as a mathematical object that represents the constitution of a plenix, and a nexus was defined axiomatically by using the concept of the address set. In 2009, Bolourian [5] introduced the notion of nexus algebras as an abstract algebraic structure and investigated them. Many authors have worked on nexuses and sub nexuses. In 2019, Bolourian et al. [6] constructed a moduloid on a nexus. In 2020, Kamrani et al. gave a structure of moduloid on a nexus and verified the concepts of sub moduloids, finitely generated sub moduloids, and prime sub moduloids on a nexus [14,15]. In 2020, Norouzi et al. worked on sub nexuses on N-structure [20].

    Many familiar concepts in an abstract algebra are studied deeply in the context of nexus algebra [8,9,11,17,26,27]. In 1965, L. A. Zadeh [29] introduced the concept of fuzzy sets. Later in 1971, A. Rosenfeld [24] used this concept to define a fuzzy sub-groupoid and a fuzzy subgroup. In 1982, W. J. Liu [16] studied fuzzy invariant subgroups, fuzzy ideals and proved some fundamental properties. In 1988, U. M. Swamy and K. L. N. Swamy [28] defined the concept of fuzzy prime ideals of rings and got some useful results. In 1991, S. Abou-Zaid [2] introduced the notion of fuzzy sub-near-rings and ideals. In 2020, S. Abdurrahman [1] defined the notion of fuzzy sub-semi-ring and investigated the related properties. Moreover, they introduced the ideal of a sub-semi-ring induced from the level set. They also characterized fuzzy sub-semi-rings.

    In 2011, A. Saeidi Rashkolia and A. Hasankhani [25] introduced fuzzy sub nexuses and investigated these properties. In 2012, D. Afkhami Taba et al. [3], as a generalization of fuzzy nexuses, defined soft nexuses. In 2014, H. Hedayati and A. Asadi [12] discussed normal, maximal, and product fuzzy sub nexuses of nexuses. In 2015, A. A. Estaji et al. [7] defined the fuzzy sub-nexuses over a nexus and the notion of prime fuzzy sub-nexuses. For some recent papers, see references [4,22,23]. Many authors work in fuzzy ideals. In [23], the authors defined q-rung orthopair fuzzy ideals, and also introduced the notion of q-rung orthopair fuzzy cosete. In [21], the authors defined the concept of Pythagorean fuzzy cosets of a Pythagorean fuzzy ideal and proved that the set of all Pythagorean fuzzy cosets of a Pythagorean fuzzy ideal forms a ring under cetain binary operations. In this paper, we construct a semi-ring on a nexus and then define a fuzzy sub-semi-ring on related nexus N and verify some applications of it. We obtain some relationship between sub-semi-ring and fuzzy sub-semi-ring of N. In Proposition 3.12, we show that A is a sub-semi-ring of N if and only if χA is a fuzzy sub-semi-ring of N. In Proposition 3.14, we show that μ is a fuzzy sub-semi-ring of N if and only if, for every α[0,μ(0)], μα is a sub-semi-ring of N. Since the Propositions 3.12 and 3.14 are not true for ideals and since ideals in semi-rings are very important than sub-semi-rings, we put a condition on fuzzy sub-semi-ring such that Propositions 3.12 and 3.14 be true for ideals. We define this fuzzy and named it by a strong fuzzy sub-semi-ring. Also, for a strong fuzzy sub-semi-ring, for every aN, we have μ(0)μ(a). In [28], the authors define fuzzy ideals in rings, but in this paper, the notion of a fuzzy sub-semi-ring is different from it because N is not a ring (see Remark 4.5). The set of all fuzzy sub-semi-rings of N is denoted by FSUBS(N), and the set of all strong fuzzy sub-semi-rings of N is denoted by FSUBT(N). In the following, for a semi-ring homomorphism f:NM, we show that if μFSUBS(N) then f(μ)FSUBS(M) and if μFSUBS(M), then fμFSUBS(N). Finally, we verify some concepts of fuzzy quotient of a nexus semi-rings.

    Now, we review the basic definitions and some elementary aspects that are necessary for this paper.

    An address is a sequence of N such that ak=0 implies that ai=0, for all ik. The sequence of zero is called the empty address and denoted by (). In other words, every non-empty address is of the form (a1,a2,,an,0,0,), where ai and nN, and. it is denoted by (a1,a2,,an).

    Definition 2.1. ([3,6,18]) A set N of addresses is called a nexus if

    (Ⅰ) (a1,a2,,an1,an)N implies (a1,a2,,an1,t)N, 0tan,

    (Ⅱ) {ai}i=1N,aiN implies nN, 0tan,(a1,,ant)N.

    Let aN. The level of a is said to be:

    (Ⅰ) n, if a=(a1,a2,,an), for some akN,

    (Ⅱ) , if a is an infinite sequence of N,

    (Ⅲ) 0, if a=().

    The level of a is denoted by l(a) and stema=a1. We put st(N)=sup{iN:(i)N}.

    Let a={ai} and b={bi}, iN, be two addresses. Then ab, if l(a)=0 or if one of the following cases is satisfied:

    (Ⅰ) if l(a)=1, that is a=(a1), for some a1N and a1b1,

    (Ⅱ) if 1<l(a)<, then l(a)l(b) and al(a)bl(a) and for any 1i<l(a), ai=bi,

    (Ⅲ) if l(a)=, then a=b.

    Definition 2.2. ([29]) Let N be a set. A fuzzy subset of N is a mapping μ:N[0,1]. If μ and ν are fuzzy subsets of N such that ν(x)μ(x) for all xN, we write νμ or νμ and say that ν is contained in μ or ν is a fuzzy subset of μ.

    Definition 2.3. ([25]) Let μ be a fuzzy subset of a nexus N. Then μ is called a fuzzy sub-nexus of N, if ab implies that μ(b)μ(a), for all a,bN.

    In this section, at first we construct a semi-ring on a nexus N.

    For ()a,bN, let a={at}nt=1 and b={bt}mt=1. We define binary operations "+" and "" on N as following:

    (Ⅰ) ()+a=a+()=a,

    (Ⅱ) a+b=(a1b1),

    (Ⅲ) ab=(a1,,ai1,aibi),

    where i{a,b}=min{t:atbt} (briefly, i{a,b}:=i). If there is no such that i, then a=b and i{a,a}=l(a).

    Example 3.1. Consider a nexus:

    N={(),(1),(2),(3),(1,1),(1,2),(1,2,1),(2,1),(2,2),(2,3),(2,3,1),(3,1)}

    with the following diagram:

    Figure 1.  Diagram of N.

    We have (2,3,1)+(3)=(3), (1,1)+(3,1)=(3), (1,1)+(1,1)=(1).

    i{(2,3),(2,3,1)}=3 and (2,3)(2,3,1)=(2,3).

    i{(2,3,1),(1,2,1)}=1 and (2,3,1)(1,2,1)=(1).

    i{(2,3),(2,3)}=2 and (2,3)(2,3)=(2,3).

    Now, we prove that (N,+,,()) is a semi-ring.

    Lemma 3.2. Let a={at}nt=1, b={bt}nt=1, c={ct}nt=1N. Then

    (Ⅰ) if i{a,b}=i{a,c}=r, then i{b,c}r,

    (Ⅱ) if i{a,b}=r, i{a,c}=s and rs, then i{b,c}=min{r,s}.

    Proof. (Ⅰ) Since i{a,b}=i{a,c}=r, for every 1ir1, we get ai=bi=ci. Consequently, i{b,c}r.

    (Ⅱ) Let r<s. Then ai=bi, for every 1ir1 and bi=ci, for every 1is1. Hence bi=ci, for every 1ir1, arbr and ar=cr. It follows that brcr, and so i{b,c}=r. By a similar argument we can see that if s<r, then i{b,c}=s. Therefore, i{b,c}=min{r,s}.

    Theorem 3.3. The algebra (N,+,,()) is a semi-ring.

    Proof. It is obvious that (N,+,()) is a commutative semi-group. Now, assume a={at}nt=1, b={bt}nt=1, c={ct}nt=1N. Then there are two cases:

    Case 1. Let i{a,b}=i{a,c}=i{b,c}=r. For this we consider 4 sub cases as follows:

    Sub case 1-1. If ab=(a1,,ar) and (a1,,ar)c=(a1,,ar), then ar<br and ar<cr. Thus, (ab)c=(a1,,ar)c=(a1,,ar). On the other hand, we have bc=(b1,,br) or bc=(c1,,cr). Since ar<br and ar<cr, we get a(bc)=(a1,,ar). Therefore, (ab)c=a(bc).

    Sub case 1-2. If ab=(a1,,ar) and (a1,,ar)c=(c1,,cr), then ar<br and cr<ar and so cr<br. Thus, (ab)c=(a1,,ar)c=(c1,,cr). On the other hand, since ar<br, cr<ar and cr<br, we get a(bc)=(a1,,ar)(c1,,cr)=(c1,,cr). Therefore, (ab)c=a(bc).

    Sub case 1-3. If ab=(b1,,br) and (b1,,br)c=(b1,,br), then br<ar and br<cr. Thus, (ab)c=(b1,,br)c=(b1,,br). On the other hand, since br<ar and br<cr, we get a(bc)=(a1,,ar).(b1,,br)=(b1,,br). Therefore, (ab)c=a(bc).

    Sub case 1-4. If ab=(b1,,br) and (b1,,br)c=(c1,,cr), then br<ar and cr<br and so cr<ar. Thus, (ab)c=(b1,,br)c=(c1,,cr). On the other hand, since br<ar, cr<br and cr<ar, we get a(bc)=(a1,,ar)(c1,,cr)=(c1,,cr). Therefore, (ab)c=a(bc).

    Case 2. Let i{a,b}=i{a,c}=r and i{b,c}=s>r. Then br=cr and we consider 2 sub cases as follows:

    Sub case 2-1. If ab=(a1,,ar), then ar<br. Since br=cr, (a1,,ar)c=(a1,,ar). Thus, (ab)c=(a1,,ar)c=(a1,,ar). On the other hand, we have bc=(b1,,bs) or bc=(c1,,cs). Since ar<br and br=cr, we get a(bc)=(a1,,ar). Therefore, (ab)c=a(bc).

    Sub case 2-2. If ab=(b1,,br), then br<ar. Since br=cr, (b1,,br)c=(b1,,br). Thus, (ab)c=(b1,,br)c=(b1,,br). On the other hand, we have bc=(b1,,bs) or bc=(c1,,cs). Since br<ar and br=cr, we get a(bc)=(b1,,br). Therefore, (ab)c=a(bc).

    For distributivity, there are two cases:

    Case 1. Let a1b1c1. We have a(b+c)=(a1,,an)(b1c1)=(a1(b1c1)). On the other hand, ab+ac=((a1b1)(a1c1))=(a1(b1c1)). Therefore, a(b+c)=ab+ac.

    Case 2. a1=b1c1. we consider 2 sub cases as follows:

    Sub case 2-1. Let c1b1. Then a1=b1. We have a(b+c)=(a1,,an)(b1c1)=(a1). Since c1b1, ab+ac=(a1). Therefore, a(b+c)=ab+ac.

    Sub case 2-2. Let b1c1. Then a1=c1. We have a(b+c)=(a1,,an)(b1c1)=(a1). Since b1c1, ab+ac=(a1). Therefore, a(b+c)=ab+ac.

    Since the operations + and are commutative, also we have (b+c)a=ba+ca.

    Therefore, (N,+,,()) is a semi-ring.

    Example 3.4. Let N={(),(1),(2),(1,1),(2,1)}. By defined the binary operations "+ " and "" on N, we have: for every aN

    a+()=()+a=a,

    (1)+(1)=(1)+(1,1)=(1,1)+(1,1)=(1),

    (2)+a=(2,1)+a=(2),

    a()=()a=() and a(1)=(1)a=(1),

    (2)(2)=(2)(2,1)=(2),

    (2)(1,1)=(1,1)(2,1)=(1),

    (1,1)(1,1)=(1,1),

    (2,1)(2,1)=(2,1).

    For the associativity + we have:

    [(1)+(2)]+(1,1)=(2)+(1,1)=(2)=(1)+(2)=(1)+[(2)+(1,1)],

    [(1)+(2)]+(2,1)=(2)+(2,1)=(2)=(1)+(2)=(1)+[(2)+(2,1)],

    [(1)+(1,1)]+(2,1)=(1)+(2,1)=(2)=(1)+(2)=(1)+[(1,1)+(2,1)],

    [(2)+(1,1)]+(2,1)=(2)+(2,1)=(2)=(2)+(2)=(2)+[(1,1)+(2,1)].

    For the associativity we have:

    [(1)(2)](1,1)=(1)(1,1)=(1)=(1)(1)=(1)[(2)(1,1)],

    [(1)(2)](2,1)=(1)(2,1)=(1)=(1)(2)=(1)[(2)(2,1)],

    [(1)(1,1)](2,1)=(1)(2,1)=(1)=(1)(1)=(1)[(1,1)(2,1)],

    [(2)(1,1)](2,1)=(1)(2,1)=(1)=(2)+(1,1)=(2)[(1,1)(2,1)].

    For the distributivity over + we have:

    (1)[(2)+(1,1)]=(1)(2)=(1)=(1)+(1)=(1)(2)+(1)(1,1),

    (2)[(1)+(1,1)]=(2)(1)=(1)=(1)+(1)=(2)(1)+(2)(1,1),

    (1,1)[(1)+(2)]=(1.1)(1)=(1)=(1)+(1)=(1,1)(1)+(1,1)(2),

    (1)[(2)+(2,1)]=(1)(2)=(1)=(1)+(1)=(1)(2)+(1)(2,1),

    (2)[(1)+(2,1)]=(2)(2)=(2)=(1)+(2)=(2)(1)+(2)(2,1),

    (2,1)[(1)+(2)]=(2,1)(2)=(2)=(1)+(2)=(2,1)(1)+(2,1)(2),

    (1)[(1,1)+(2,1)]=(1)(2)=(1)=(1)+(1)=(1)(1,1)+(1)(2,1),

    (1,1)[(1)+(2,1)]=(1,1)(2)=(1)=(1)+(1,1)=(1,1)(1)+(1,1)(2,1),

    (2,1)[(1)+(1,1)]=(2,1)(1)=(1)=(1)+(1,1)=(2,1)(1)+(2,1)(1,1),

    (2)[(1,1)+(2,1)]=(2)(2)=(2)=(1)+(2)=(2)(1,1)+(2)(2,1),

    (1,1)[(2)+(2,1)]=(1,1)(2)=(1)=(1)+(1,1)=(1,1)(2)+(1,1)(2,1),

    (2,1)[(2)+(1,1)]=(2,1)(2)=(2)=(2)+(1,1)=(2,1)(2)+(2,1)(1,1).

    Then (N,+,,()) is a semi-ring.

    Remark 3.5. Notice that the semi-ring (N,+,,()) can not be a ring. Since, if for every ()aN, there exists ()bN, such that a+b=b+a=(), then a=b=(), which is a contradiction.

    Definition 3.6. Let ()XN. We say that X is a sub-ring of N, if for every a,bX, a+b,abX.

    In the sequel, for briefly, we denote the semi-ring (N,+,,()) related to a nexus N only by N and put 0:=().

    Example 3.7. Let N=(1,4),(1,2,3) and X={(),(1),(1,2),(1,3),(1,2,3)}. X is a sub-semi-ring of N but since (1,2,2)(1,2,3)=(1,2,2)X, X is not an ideal of N.

    Definition 3.8. Let ()IN. We say that I is an ideal of N, if it satisfies the following conditions:

    (Ⅰ) for every a,bI, a+bI,

    (Ⅱ) for every aI and every bN, abI.

    An ideal I of N is prime if for every a,bN, abI implies aI or bI.

    Example 3.9. Let N=(1,4),(1,2,3) and I={(),(1),(1,2),(1,3),(1,2,1),(1,2,2),(1,2,3)}. I is an ideal of N.

    Definition 3.10. A fuzzy subset μ of N is called a fuzzy sub-semi-ring of N if it satisfies the following conditions: for all a,bN,

    (Ⅰ) μ(a+b)min{μ(a),μ(b)},

    (Ⅲ) μ(ab)min{μ(a),μ(b)}.

    The set of all fuzzy sub-semi-rings of N, is denoted by FSUBS(N).

    Example 3.11. We define the fuzzy subset μ of N as follows:

    (ⅰ) For a=(a1,a2,,an)N of level n, μ(a):=1a1a2an, otherwise

    μ(v)={1v=0;0l(v)=.

    Let a=(a1,,an), b=(b1,,bm) and a1b1. Then μ(a+b)=μ((a1))=1a11a1an. Hence μ(a+b)μ(a)min{μ(a),μ(b)}. Let i{a,b}=i and biai. Hence ab=(b1,,bi) and so μ(ab)1b1bm=μ(b)min{μ(a),μ(b)}. Therefore, μFSUBS(N).

    (ⅱ) For a=(a1,a2,,an)N, μ(a):=1l(a), otherwise

    μ(v)={1v=0;0l(v)=.

    Since l(a+b)=1, μ(a+b)=1min{μ(a),μ(b)}. Clearly l(ab)l(a), now we have μ(ab)μ(a)min{μ(a),μ(b)}. Therefore, μFSUBS(N).

    (ⅲ) For a=(a1,a2,,an)N, μ(a):=1stema, otherwise

    μ(v)={1v=0;0l(v)=.

    Let a,bN with stema=a1, stemb=b1 and a1b1. Hence stem(a+b)=a1 and stemab=b1. So μ(a+b)=μ(a)min{μ(a),μ(b)} and μ(ab)=μ(b)min{μ(a),μ(b)}. Therefore, μFSUBS(N).

    Proposition 3.12. Let A be a subset of N with 0A. Then A is a sub-semi-ring of N if and only if χAFSUBS(N), where

    χA(x)={1ifxA;0ifxA.

    Proof. Let A be a sub-semi-ring of N and a,bN. If a,bA, then ab,a+bA and so χA(ab)=χA(a+b)=1. Now, we have 1=χA(a+b)min{χA(a),χA(b)}=1 and so 1=χA(ab)min{χA(a),χA(b)}=1. If aA, we have min{χA(a),χA(b)}=0, hence χA(a+b)min{χA(a),χA(b)}=0 and χA(ab)min{χA(a),χA(b)}=0. Thus, χAFSUBS(N).

    Conversely, let χAFSUBS(N). Suppose that a,bA, then χA(a)=χA(b)=1. Since χA(a+b)min{χA(a),χA(b)}=1, and χA(ab)min{χA(a),χA(b)}=1, χA(a+b)=χA(ab)=1 and so a+b,abA. Therefore, A is a sub-semi-ring of N.

    Definition 3.13. Let μFSUBS(N) and α[0,1]. We define μα={xN:μ(x)α} and Supp(μ)={aN:μ(a)0}.

    Proposition 3.14. Let μ be a fuzzy subset of N such that for every aN, μ(0)μ(a). Then μFSUBS(N) if and only if, for every α[0,μ(0)], μα be a sub-semi-ring of N.

    Proof. Let μ be a fuzzy sub-semi-ring of N, α[0,μ(0)] and a,bμα. Hence 0μα, μ(a)α and μ(b)α. This shows that μ(a+b)min{μ(a),μ(b)} α, and so a+bμα. Similarly, abμα and so μα is a sub-ring of N.

    Conversely, let for every α[0,μ(0)], μα be a sub-semi-ring of N. Let a,bN, μ(a)=α, μ(b)=β and αβ. Then aμα and a,bμβ. So a+bμβ and abμβ. Hence μ(a+b)β=min{μ(a),μ(b)} and μ(ab)β=min{μ(a),μ(b)}. Therefore, μ is a fuzzy sub-semi-ring of N.

    The following examples shows that the Propositions 3.12 and 3.14 are not true for ideals.

    Example 3.15. Let N=(1,4),(1,2,3) and X={(),(1),(1,2),(1,3),(1,2,3)}. X is a sub-ring of N and so χX is a fuzzy sub-semi-ring of N, but (1,2,2)(1,2,3)=(1,2,2)X. This shows that X is not an ideal of N. Hence Proposition 3.12 is not true for ideals.

    Example 3.16. Let N={(),(1),(2),(3),(1,2)}. We define the fuzzy μ on N as following:

    μ(())=1, μ((1))=0.95, μ((2))=0.5, μ((3))=0.75, μ((1,2))=0.85.

    For every aN, we have

    ()+a=a, (1)+(1)=(1)+(1,2)=(1), (1)+(2)=(2)+(1,2)=(2)+(2)=(2), (1)+(3)=(2)+(3)=(3)+(3)=(3)+(1,2)=(3), (1,2)+(1,2)=(1,2). Hence for every a,bN, μ(a+b)min{μ(a),μ(b)}.

    ()a=(), (1)a=(1), aa=a, (2)(3)=(2), (2)(1,2)=(3)(1,2)=(1). Hence for every a,bN, μ(ab)min{μ(a),μ(b)}. Then μFSUBS(N) and so by Proposition 3.14, for every α[0,1], μα is a sub-semi-ring of N. We have μ0.75={(),(1),(3),(1,2)}. Since (2)(3)=(2)μ0.75, μ0.75 is not an ideal of N. Therefore, Proposition 3.14 is not true for ideals.

    In Examples 3.15 and 3.16, we show that Propositions 3.12 and 3.14 are not true for ideals. Since ideals in semi-rings are more important than sub-semi-rings, in this section, we want to put a condition on fuzzy sub-semi-ring such that Propositions 3.12 and 3.14 be true for ideals. See the following definition:

    Definition 4.1. A fuzzy subset μ of N is called a strong fuzzy sub-semi-ring of N if it satisfies the following conditions: for all a,bN,

    (Ⅰ) μ(a+b)min{μ(a),μ(b)},

    (Ⅲ) μ(ab)max{μ(a),μ(b)}.

    The set of all strong fuzzy sub-semi-rings of N is denoted by FSUBT(N).

    Remark 4.2. Let N be a semi-ring and μFSUBT(N). Then, for every aN, μ(0)μ(a), because for every aN, μ(0)=μ(0a)max{μ(a),μ(0)}μ(a).

    It is easy to see that all fuzzy μ that are defined in Examples 3.2(ⅰ)–(ⅲ) are strong fuzzy sub-semi-rings.

    Proposition 4.3. Let A be a subset of N with 0A. Then A is an ideal of N if and only if, χAFSUBT(N).

    Proof. Let A be an ideal of N and a,bN. If a,bA, then ab,a+bA and it is show that χA(ab)=χA(a+b)=1. Hence, 1=χA(a+b)min{χA(a),χA(b)}=1 and also we have 1=χA(ab)max{χA(a),χA(b)}=1. If aA and bA, then abA and so χA(ab)=1. Hence, χA(ab)max{χA(a),χA(b)}. Also, we have χA(a+b)min{χA(a),χA(b)}=0. If aA and bA, then min{χA(a),χA(b)}=max{χA(a),χA(b)}=0. Now, we have χA(ab)max{χA(a),χA(b)} and χA(a+b)min{χA(a),χA(b)}. Therefore, χAFSUBT(N).

    Conversely, let χAFSUBT(N). If a,bA, then χA(a)=χA(b)=1. Since χA(a+b)min{χA(a),χA(b)}=1, then χA(a+b)=1 and so a+bA. Let aA and bN, then max{χA(a),χA(b)}=1. Then χA(ab)=1, since χA(ab)max{χA(a),χA(b)}=1, and so abA. Therefore, A is an ideal of N.

    Proposition 4.4. Let μ be a fuzzy subset on N. Then μFSUBT(N) if and only if, for every α[0,μ(0)], μα be an ideal of N.

    Proof. Let μFSUBT(N), α[0,μ(0)] and a,bμα. Hence, 0μα, μ(a)α and μ(b)α. So μ(a+b)min{μ(a),μ(b)} α. Hence, a+bμα. Now, let aμα and bN. So μ(ab)max{μ(a),μ(b)}α. Hence, abμα. Therefore, μα is an ideal of N.

    Conversely, let for every α[0,μ(0)], μα be an ideal of N. Let a,bN, μ(a)=α, μ(b)=β and αβ. Then aμα and a,bμβ. So, a+bμβ and abμα. Hence, μ(a+b)β=min{μ(a),μ(b)} and μ(ab)α=max{μ(a),μ(b)}. Therefore, μFSUBT(N).

    Proposition 4.5. Let 0μFSUBT(N). Then Supp(μ) is an ideal of N.

    Proof. Let μ(0)=0. Since for every aN, μ(0)μ(a), μ(a)=0 and so μ=0, which is a contradiction. Hence, μ(0)0 and 0Supp(μ). Let a,bSupp(μ). This shows that μ(a+b)min{μ(a),μ(b)}0 and so a+bSupp(μ). Now, let aSupp(μ) and bN. Since μ(a)0, μ(ab)max{μ(a),μ(b)}0. Then abSupp(μ). Therefore, Supp(μ) is an ideal of N.

    The following example shows that the converse of Proposition 4.5, is not true.

    Example 4.6. Consider N and μ of Example 3.16. We have Supp(μ)=N is an ideal of N. By Example 3.16, μ0.75 is not an ideal of N. Hence, by Proposition 4.3, μ is not a strong fuzzy sub-semi-ring of N.

    The following example shows that Supp(μ) is an ideal of N, but μ is not a fuzzy sub-semi-ring of N.

    Example 4.7. Suppose that N is a semi-ring and for every a=(a1,a2,,an)N. We put [a]=[1a1+1a2++1an], where [] is the bracket function. We define the fuzzy subset μ on N such that for every aN,

    μ(a)={1if[a]N;(1a1+1a2++1an)[a]if[a]N.

    Suppose that N=(2,4), hence Supp(μ)=N is an ideal of N. Since [12+14]=[34]=0 and [12+13]=[56]=0, μ((2,4))=34 and μ((2,3))=56. On the other hand, we have (2)=(2,3)+(2,4) and since [12]=0, μ((2))=12.

    Hence, μ((2,3)+(2,4))=μ((2))=12<34=min{34,56}=min{μ((2,3)),μ((2,4))}. Therefore, μ is not a fuzzy sub-semi-ring of N.

    Proposition 4.8. Let μ be a fuzzy subset of N. Then μ is a strong fuzzy sub-semi-ring if and only if it is a fuzzy sub-semi-ring and a fuzzy sub-nexus of N.

    Proof. Let μ be a strong fuzzy sub-semi-ring of N. Clearly it is a fuzzy sub-semi-ring. If ab, then ab=a. Hence μ(a)=μ(ab)max{μ(a),μ(b)}. So μ(a)μ(b) and so μ is a fuzzy sub-nexus of N.

    Conversely, let μ be a fuzzy sub-semi-ring and fuzzy sub-nexus of N. For every a,bN, since aba,b, and μ is a fuzzy sub-nexus of N, we have μ(ab)μ(a) and μ(ab)μ(b). Hence, μ(ab)max{μ(a),μ(b)}. Therefore, μFSUBT(N).

    The following example give a fuzzy sub-semi-ring μ such that it is not a fuzzy sub-nexus.

    Example 4.9. Consider N and μ of Example 3.16. We have (2)(3), 0.5=μ((2)) and 0.75=μ((3)), but 0.50.75. Hence, μ is not a fuzzy sub-nexus of N.

    Definition 4.10. Suppose that N is a semi-ring and μ is a fuzzy subset of N. We say that μ is an integer fuzzy if for every a,bN, μ(a)=0 and μ(b)=0, imply μ(ab)=0. Let α[0,μ(0)), we say that μ is a prime fuzzy with respect to α, if for every a,bN, μ(a)<α and μ(b)<α imply μ(ab)<α.

    Proposition 4.11. Suppose that NI is an ideal of N, μ is a strong fuzzy sub-semi-ring of N and α(0,μ(0)]. Then

    (Ⅰ) I is a prime ideal of N if and only if χI is an integer fuzzy of N,

    (Ⅱ) NSupp(μ) is a prime ideal of N if and only if μ is an integer fuzzy of N,

    (Ⅲ) μα is a prime ideal of N if and only if μ is a prime fuzzy with respect to α.

    Proof. By Definition 4.10, it is clear.

    Proposition 4.12. If N is a cyclic nexus and μ is a strong fuzzy sub-semi-ring, then for every α[0,μ(0)), μαN is a prime ideal of N.

    Proof. Let a,bN and abμα, hence μ(ab)α. Since N is cyclic nexus, let ab. Hence, μ(ab)=μ(a)α and so aμα. Then, μα is a prime ideal of N.

    Proposition 4.13. If N is a semi-ring and consider μ of Example 3.2(ⅲ). Then, for every α[0,μ(0)), μαN is a prime ideal of N.

    Proof. Let a,bN and α[0,μ(0)). If μ(a)<α and μ(b)<α, hence stema>α and stemb>α. Thus, stemab>α. Then, μ(ab)<α and μ is a prime fuzzy with respect to α. Hence, by Proposition 4.11(Ⅲ), μα is a prime ideal of N.

    Proposition 4.14. If N is a semi-ring and consider μ of Example 3.2(ⅱ). Then, for every α[0,μ(0)), μαN is a prime ideal of N if and only if there exists a unique aN with l(a)=n such that {bN:l(a)>n}qa.

    Proof. At first we assume that there exist unique aN such that l(a)=n and for every bN with l(b)>n, bqa. We show that μ1n is a prime ideal of N. By Proposition 4.11(Ⅲ), it is sufficient to show that μ is a prime fuzzy with respect to 1n. Let b,bN with μ(b)<1n and μ(b)<1n. Hence, l(b)>n and l(b)>n. Thus, b,bqa and i{b,b}>n, and so l(bb)>n. Then, μ(bb)<1n. Therefore, μ is a prime fuzzy with respect to 1n and μ1n is a prime ideal.

    Conversely, let there exist a,aN such that aa, l(a)=l(a)=n, bqa and bqa with l(b)>n and l(b)>n. Since aa, l(a)=l(a)=n, i{a,a}n and since bqa and bqa with l(b)>n and l(b)>n, i{b,b}n. Hence, l(bb)n, and so μ(bb)1n. Then, bbμ1n, but b,bμ1n. Therefore, μ1n is not prime ideal of N and the proof is complete.

    Example 4.15. Consider μ of Example 3.11(ⅱ) and N=(1,2,3),(1,1,4). Let a=(1,2,3) and b=(1,1,4), then μ(ab)=μ((1,1))=12, but μ(a)=μ(b)=13. So, abμ12, but a,bμ12. Hence μ12 is not a prime ideal of N. In the other hand, since for every aN, stema=1, μ1={(),(1)} and it is easy to see that ab=(1) implies a=(1) or b=(1). So, μ1 is a prime ideal of N.

    Proposition 4.16. Suppose that st(N)=n1. Consider μ of Example 3.2(ⅱ), δi=max{l(a):stema=i}. Suppose that j is such that 1δj=max{δi:1in}. Also, suppose that k is such that 1δk=max{δi:1in,ij}. Then for every 1mxyδk, μ1m is not a prime ideal of N.

    Proof. Let a,bN such that l(a)=δj and l(b)=δk. Hence, stema=j, stemb=k, μ(a)=1δj and μ(b)=1δk. Then ab=(jk), and so μ(ab)=11m, but μ(a)=1δj<1m, μ(b)=1δk<1m. Therefore, abμ1m, but a,bμ1m. Thus, μ1m is not a prime ideal of N.

    Example 4.17. Suppose that N=(5,4,3,2,1),(6,2,1,1,1,1). By notation of Proposition 4.16, we have st(N)=6, j=6, k=5, δ6=6 and δ5=5. Let μ as Example 3.11(ⅱ). Now by Proposition 4.16, for every 1m4, ν1m, is not a prime ideal of N.

    Definition 4.18. Suppose that N and M are two semi-rings, the function f:NM is a semi-ring homomorphism if f satisfies the following conditions:

    (Ⅰ) f(0)=0,

    (Ⅱ) f(a+b)=f(a)+f(b),a,bN,

    (Ⅲ) f(ab)=f(a)f(b),a,bN.

    Proposition 4.19. Suppose that N and M are two semi-rings and f:NM is a semi-ring homomorphism:

    (Ⅰ) if μFSUBS(M), then μfFSUBS(N) such that for every aN, (μf)(a)=μ(f(a)),

    (Ⅱ) if μFSUBS(N), then f(μ)FSUBS(M) such that for every bM,

    f(μ)(b)={0ifbImf;sup{μ(a):f(a)=b}ifbImf.

    Proof. (Ⅰ) Let a,bN, then

    (μf)(a+b)=μ(f(a+b))=μ(f(a)+f(b))min{μ(f(a)),μ(f(b))}=min{(μf)(a),(μf)(b)}.

    (μf)(a.b)=μ(f(ab))=μ(f(a)f(b))min{μ(f(a)),μ(f(b))}=min{(μof)(a),(μof)(b)}.

    Therefore, μfFSUBS(N).

    (Ⅱ) Let a,bM. If aImf or bImf, then min{f(μ)(a),f(μ)(b)}=0 and hence f(μ)(a+b)min{f(μ)(a),f(μ)(b)} and f(μ)(ab)min{f(μ)(a),f(μ)(b)}. Now, let a,bImf and f(μ)(a)f(μ)(b). We show that f(μ)(a+b)f(μ)(a) and f(μ)(ab)f(μ)(a).

    Let ϵ>0. Since f(μ)(a)=sup{μ(x):f(x)=a} and f(μ)(b)=sup{μ(y):f(y)=b}, there exist x,yN such that f(μ)(a)ϵμ(x)<f(μ)(a) and f(μ)(b)ϵμ(y)<f(μ)(b). Since f(x+y)=a+b, f(μ)(a+b)=sup{μ(z):f(z)=a+b}μ(x+y)min{μ(x),μ(y)}.

    Case 1: If μ(x)μ(y), then f(μ)(a+b)μ(x)f(μ)(a)ϵ.

    Case 2: If μ(x)μ(y), then f(μ)(a+b)μ(y)f(μ)(b)ϵf(μ)(a)ϵ.

    Since for every ϵ>0, f(μ)(a+b)f(μ)(a)ϵ, we get

    f(μ)(a+b)f(μ)(a)=min{f(μ)(a),f(μ)(b)}.

    Similarly, since f(xy)=ab, we have f(μ(ab))=sup{μ(z):f(z)=ab}μ(xy)min{μ(x),μ(y)}f(μ)(a)ϵ.

    Hence, f(μ)(ab)f(μ)(a)=min{f(μ)(a),f(μ)(b)}.

    Therefore, f(μ)FSUBS(N).

    In Proposition 4.19(Ⅰ), if μFSUBT(M), then by a similar argument we can show that μfFSUBT(N). However, Proposition 4.19(Ⅱ), is not true for μFSUBT(N). See the following example:

    Example 4.20. Let N=(1,2), M=(2,3) and μ be as Example 3.11(ⅱ). Let f:NM be such that f(0)=0 and for every aN, f(a)=(2). Let b=(2) and b=(1). Since bb=bImf, we have f(μ)(bb)=0<max{f(μ)(b),f(μ)(b)}=1. Then μ is a strong fuzzy sub-semi-ring of N, but f(μ) is not a strong fuzzy sub-semi-ring of M.

    Let I be an ideal of N and aN. We put a+I={a+b:bI} and NI={a+I:aN}.

    Lemma 4.21. Let I be an ideal of N and a,aN. Then

    (I) a+I=I if and only if a=().

    (II) a+I=a+I if and only if stema=stema.

    Proof. (Ⅰ) Clearly, ()+I=I. Let a+I=I. Then there exist bI such that a+b=(). Hence a=b=().

    (Ⅱ) Assume a1=stema=stema=a1. Hence for every bI with stemb=b1, we have a+b=(a1b1)=(a1b1)=a+b. Therefore, a+I=a+I.

    Conversely, let a+I=a+I. Then Since ()I, we have (a1)=a+()a+I=a+I. Hence there exist bI, such that (a1)=a+b=(a1b1), with b1=stemb. Then a1a1. Similarly, a1a1. Thus, a1=a1. Therefore, stema=stema.

    Example 4.22. Consider a nexus:

    N={(),(1),(2),(3),(4),(1,1),(1,2),(1,1,1),(1,1,2),(1,1,3),(2),(2,1),(2,1,1),(2,1,2),(3,1),(3,2),(3,3),(4,1)} with the following diagram.

    Figure 2.  Diagram of N.

    If I:={(),(1),(2),(1,1),(1,2),(2),(2,1)}, then I is an ideal of N. Using Lemma 4.21, we can see that

    ()+I=(1,1)+I=(1,2)+I=(1,1,1)+I=(1,1,2)+I=(1,1,3)+I,

    (2)+I=(2,1)+I=(2,1,1)+I=(2,1,2)+I,

    (3)+I=(3,1)+I=(3,2)+I=(3,3)+I,

    (4)+I=(4,1)+I.

    Then NI={I,(1)+I,(2)+I,(3)+I,(4)+I}.

    Definition 4.23. For every a+I, b+INI, we define the binary operations "" and "" on NI with the following:

    (I) (a+I)(b+I)=a+b+I,

    (II) (a+I)(b+I)=ab+I.

    Proposition 4.24. Let I be an ideal of N. Then (NI,,,I) is a semi-ring.

    Proof. Let a+I=a+I and b+I=b+I. By Lemma 4.21, we have a1=stema=stema=a1 and b1=stemb=stemb=b1. Hence, a1b1=a1b1 and a1b1=a1b1. This shows that a+b+I=a+b+I and ab+I=ab+I. Therefore, the binary operations and are well defined. By routine calculation we can see that (NI,,I) is a commutative semi-group, and we have (()+I)(a+I)=()+a+I=a+I and (a+I)(()+I)=a+()+I=a+I. Thus, it is a commutative monoid. Furthermore, (NI,) is a semi-group, in while the operation is distributive with respect to . Therefore, (NI,,,I) is a semi-ring.

    Example 4.25. Let I is an ideal of N and NI is the quotient semi-ring. Let f:NNI be such that for every aN, f(a)=a+I. Clearly f is a semi-ring epimorphism. Let μ be a fuzzy sub-semi-ring of N and ν:NI[0,1] be such that for every aN, ν(a+I)=sup{μ(b):b+I=a+I}. Since ν=f(μ), by Proposition 4.19(Ⅱ), ν is a fuzzy sub-semi-ring of NI.

    Example 4.26. Consider a nexus N={(),(1),(2),(3),(4),(1,1),(1,2),(1,1,1),(1,1,2),(1,1,3), (2),(2,1),(2,1,1),(2,1,2),(3,1),(3,2),(3,3),(4,1)} with the following diagram:

    Figure 3.  Diagram of N.

    If I={(),(1),(2),(1,1),(1,2),(2),(2,1)}, by Lemma 4.21, we have NI={I,(1)+I,(2)+I,(3)+I,(4)+I}.

    Now we define ν:NI[0,1] such that ν(I)=1, ν((1)+I)=1, ν((2)+I)=12, ν((3)+I)=13, ν((4)+I)=14.

    Consider μ in Example 3.11(ⅲ), and f:NNI in Example 4.26. It is easy to see that ν=f(μ) and by Example 4.26, νFSUBS(NI).

    In this section, we verify some concepts of fuzzy quotient of a nexus semi-rings.

    Definition 5.1. Let aN and μFSUBS(N). We define the fuzzy a+μ:N[0,1] such that for every xN, (a+μ)(x)=μ(a+x).

    Proposition 5.2. Suppose that a+μ is the fuzzy of Definition 5.1. Then a+μFSUBS(N).

    Proof. Let x,xN, stema=a1, stemx=x1 and stemx=x1. We have

    (a+μ)(x+x)=μ(a+(x1x1))=μ((a1x1)(a1x1))=μ((a+x)+(a+x))min{μ(a+x),μ(a+x)}=min{(a+μ)(x),(a+μ)(x)}.

    (a+μ)(xx)=μ(a+(xx))=μ((a1(x1x1)))=μ((a1x1)(a1x1))=μ((a+x)(a+x))min{μ(a+x),μ(a+x)}=min{(a+μ)(x),(a+μ)(x)}

    Therefore, a+μFSUBS(N).

    Definition 5.3. Let aN and μFSUBS(N). We define Nμ={a+μ:aN}. Also we define operations and on Nμ such that for every a,bN, (a+μ)(b+μ)=(a+b)+μ and (a+μ)(b+μ)=(ab)+μ.

    Proposition 5.4. Let μFSUBS(N). Then, (Nμ,,,0+μ) is a semi-ring.

    Proof. Suppose that a,a,b,bN and a+μ=a+μ and b+μ=b+μ. Hence, for every xN, we have μ(a+x)=μ(a+x) and μ(b+x)=μ(b+x). Let tN. If x=b+t, then μ(a+b+t)=μ(a+b+t) and if x+a+t, then μ(b+a+t)=μ(b+a+t). Now, for every tN, we have μ(a+b+t)=μ(a+b+t) and so ((a+b)+μ)(t)=((a+b)+μ)(t). Hence (a+b)+μ=(a+b)+μ.

    Then the operation is well-defined. Now we show that the operation is well defined. Let stema=a1, stemb=b1, stema=a1, stemb=b1, stemx=x1. We consider two cases: Case 1: Let a1b1 and a1b1. Then

    ((ab)+μ)(x)=μ((a1b1)x1)=μ(b1x1)=μ(b+x)=μ(b+x)=μ(b1x1)=μ((ab)+x)=(ab+μ)(x).

    Hence (ab)+μ=(ab)+μ.

    Case 2: Let a1b1 and a1b1. Then ((ab)+μ)(x)=μ((ab)+x)=μ((a1b1)x1)=μ(b1x1)=μ(b+x)=μ(b+x)=μ((ab)+x)=μ(a+b+x)=μ(a+b+x)=μ(a1b1x1)=μ(a+x)=μ(a+x)=μ(a1x1)=μ((a1b1)x1)=μ((ab)+x)=((ab)+μ)(x).

    Hence (ab)+μ=(ab)+μ.

    Then the operation is well-defined.

    Also, for every a,b,cN, we have

    (c+μ)((a+μ)(b+μ))=(c+μ)((a+b)+μ))=(c(a+b))+μ=(ca+cb)+μ=((ca)+μ)((cb)+μ)=((c+μ)(a+μ))((c+μ)(b+μ)).

    Therefore, (Nμ,,,0+μ) is a semi-ring.

    Example 5.5. Consider the semi-ring N of Example 4.26 and μ of Example 3.11(ⅰ). For every a,bN, a+μ=b+μ if and only if, for every xN, μ(a+x)=μ(b+x). If x=0, then μ(a)=μ(b). Let a=(a1,...,an) and b=(b1,...,bm). So a1...an=b1...bm. If x=a, we have 1a1=μ((a1))=μ(a+a)=μ(a+b)=μ((a1b1))=1a1b1. Hence, a1b1. If x=b, similarly we have b1a1. Hence a1=b1. Then (1)+μ=(1,1)+μ=(1,1,1)+μ, (1,2)+μ=(1,1,2)+μ, (2)+μ=(2,1)+μ=(2,1,1)+μ, (3)+μ=(3,1)+μ, (4)+μ=(4,1)+μ. Therefore, Nμ={μ,(1)+μ,(2)+μ,(3)+μ,(4)+μ,(1,2)+μ,(1,1,3)+μ,(2,1,2)+μ,(3,2)+μ,(3,3)+μ}.

    Proposition 5.6. Let μ,νFSUBS(N) and νμ. Let for every a,bN, ν(a)=ν(b) implies μ(a)=μ(b). Then μν:Nν[0,1] with μν(a+ν)=μ(a)FSUBS(Nν).

    Proof. Let a,bN and a+ν=b+ν. Then, for every xN, (a+ν)(x)=(b+ν)(x) and so ν(a+x)=ν(b+x). If x=0, then ν(a)=ν(b), and so μ(a)=μ(b). Thus, μν is well-defined. Now, since μFSUBS(N), we have

    μν((a+ν)(b+ν))=μν((ab)+ν)=μ(ab)min{μ(a),μ(b)}=min{μν(a+ν),μν(b+ν)}. Similarly, μν((a+ν)(b+ν))min{μν(a+ν),μν(b+ν)}. Therefore, μνFSUBS(Nν).

    Proposition 5.7. Let μ,νFSUBS(N) and νμ. Let for every a,bN, ν(a)=ν(b) implies μ(a)=μ(b). Then, the semi-rings Nμ and (Nν)μν are isomorphic.

    Proof. Let f:(Nν)μνNμ be such that for every aN, f((a+ν)+μν)=a+μ. Let a,bN and (a+ν)+μν=(b+ν)+μν. If xN, ((a+ν)+μν)(x+ν)=((b+ν)+μν)(x+ν), , and so μν(a+x+ν)=μν(b+x+ν). Thus, μ(a+x)=μ(b+x). Then a+μ=b+μ, and so f is well-defined. Now, we have

    f(((a+ν)+μν)((b+ν)+μν)))=f((a+b+ν)+μν))=(a+b)+μ=(a+μ)(b+μ),

    f(((a+ν)+μν)((b+ν)+μν)))=f((((ab)+ν)+μν))=(ab)+μ=(a+μ)(b+μ). Hence, f is semi-ring homomorphism. Clearly, f is onto. Now, let f((a+ν)+μν)=f((b+ν)+μν). This shows that a+μ=b+μ. Then, for every xN, μ(a+x)=μ(b+x). Thus, ((a+ν)+μν)(x+ν)=((b+ν)+μν)(x+ν). Consequently, (a+ν)+μν=(b+ν)+μν. Thus, f is one-to-one. Therefore, f is a semi-ring isomorphism and the semi-rings Nμ and (Nν)μν are isomorphic.

    We define a semi-ring on nexus N and fuzzy sub-semi-ring of this semi-ring. The set of all fuzzy sub-semi-rings of N is denoted by FSUBS(N). In Proposition 3.12, we show that A is a sub-semi-ring of N if and only if χA is a fuzzy sub-semi-ring of N. In Proposition 3.14, we show that μ is a fuzzy sub-semi-ring of N if and only if, for every α[0,μ(0)], μα is a sub-semi-ring of N. Since the Propositions 3.12 and 3.14 are not true for ideals and since ideals in semi-rings are more important than sub-semi-rings, we put a condition on a fuzzy sub-semi-ring such that Propositions 3.12 and 3.14 are true for ideals. We define this fuzzy and named it by a strong fuzzy sub-semi-ring. The set of all strong fuzzy sub-semi-rings of N, is denoted by FSUBT(N). In the following, for a semi-ring homomorphism f:NM, we show that if μFSUBS(N) then f(μ)FSUBS(M) and if μFSUBS(M) then fμFSUBS(N). Finally, we verify some concepts of fuzzy quotient of a nexus semi-ring. In future, for semi-ring related to nexus N, we define polynomial semi-ring N[x]={a0+a1x++anxn:nN{0},aiN(1in)}. Then, we characterize all prime ideals and prime elements of N.

    Vajiheh Nazemi Niya: Conceptualization, Methodology, Investigation, Writing original draft preparation; Hojat Babaei: Conceptualization, Methodology, Investigation; Akbar Rezaei: Conceptualization, Methodology, Investigation, Writing review and editing. All authors have read and agreed to the published version of the manuscript.

    The authors declares they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors declare that they have no conflicts of interest.



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