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Annals of Fuzzy Mathematics and Informatics
Volume 12, No. 6, (December 2016), pp. 855–876
ISSN: 2093–9310 (print version)
ISSN: 2287–6235 (electronic version)
http://www.afmi.or.kr
@FMIc© Kyung Moon Sa Co.
http://www.kyungmoon.com
Similarity and dissimilarityrelations in intuitionistic fuzzy
matrices using implicationoperators
P. Murugadas , K. Lalitha
Received 7 April 2016; Revised 14 May 2016; Accepted 27 May 2016
Abstract. In this paper, we study similarity and dissim-ilarity relations in Intuitionistic Fuzzy Matrix(IFM) using twoimplication operators and derive a method to get a max-minidempotent, min-max idempotent IFM from an IFM of ordern.
2010 AMS Classification: 03E72, 15B15
Keywords: Intuitionistic Fuzzy Set, Fuzzy Matrix, IntuitionisticFuzzy Matrices, Implication operator.
Corresponding Author: P. Murugadas (bodi [email protected])
1. Introduction
An m × n matrix A = (aij) whose entries are in the unit interval[0, 1] is called a Fuzzy Matrix (FM). After the introduction of fuzzymatrix (FM) theory using max-min algebra by Thomason [21], severalauthors [13, 16] have developed the fuzzy matrix theory and studiedabout convergence [5, 3], decomposition [6], reduction [7, 8] of a fuzzymatrix (FM). In all the above said study implication operator plays avital role. Atanassov [1] developed the concept of Intuitionistic Fuzzy
P. Murugadas et al./Ann. Fuzzy Math. Inform. 12 (2016), No. 6, 855–876
Set (IFS) theory analogous to fuzzy set theory. In [2] Atanassov studiedintuitionistic fuzzy index matrix on the basis of index matrix, and ex-tended intuitionistic fuzzy index matrix (EIFIM). Im et al. [10], Jeongand Lee Hong [11], Khan S. K, Pal M and Amiya K. Shyamal [12],Meenakshi AR and Gandhimathi [15] and several others have studiedIntuitionistic Fuzzy Matrices (IFM). IFM is very useful in the discussionof intuitionistic fuzzy relation. Pal and Mondal[17] studied the similar-ity relation, invertibility and eigenvalues of IFM.
Nora and Atanassov [19] used implication operators in IFSs and theyhave put forth more than hundred and fifty such implication operatorsand derived some axioms for those implication operators. Sriram andMurugadas [20] used ← implication operator for IFM, studied the con-cept of g-inverse and semi-inverse of an IFM which was a generalizationof FM studied by Hiroshi Hashimoto [9].
In fuzzy set theory, for any two fuzzy elements x, y ∈ [0, 1], x← y ≥ xfor all x, y. However this is not true in the case of fuzzy matrices(FMs)A,B. That means A ← B � A always in the case of implica-tion defined in[9]. Also Hashimoto [4] used ↽ implication operator tofuzzy sets as well as to FMs, in which case also x ↽ y ≤ x for allx, y ∈ [0, 1], but it is not true that A ↽ B ≤ A for all fuzzy matrices(FMs) A,B. In this paper we extend the new implication operator ↽for IFM and discuss the relation with the existing implication operator.Many authors [7, 22, 14] have used ↽ operator component wise in FMand in most of the cases the results are evident from fuzzy set theoreticpoint.
In this paper we try to overcome the above problem and we haveshown the condition under which A ← B ≥ A and A ↽ B ≤ A aretrue using A← B, A ↽ B defined in [9] for FM, we extend this studyto IFM and the relation between the above two implication operators.Also we obtain the unique maximum solution for the Intuitionistic Fuzzyinequality A×X×B ≤ C under max-min composition and unique min-imum solution for A�X �B ≥ C under min-max composition. Further,we investigate the similarity relation for IFM using max-min operationand←, dissimilarity relation for IFM using min-max operation and ↽ .Finally we structured a method to find max-min transitive IFM andmin-max c-transitive IFM from any square IFM A.
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2. Preliminaries
Throughout this paper Fmn denotes IFMs of order m×n, Fn denotesIFMs of order n× n, and entries in IFMs are comparable.
Definition 2.1 ([1]). An Intuitionistic Fuzzy Set(IFS) A in E(universalset) is defined as an object of the following form A = {〈x, µA(x), νA(x)〉|x ∈ E}, where µA(x) : E → [0, 1] and νA(x) : E → [0, 1] define themembership and non-membership of the element x ∈ E respectivelyand for every x ∈ E : 0 ≤ µA(x) + νA(x) ≤ 1.
For simplicity we consider the pair 〈x, x′〉 as membership and non-membership function of an IFS with x+ x′ ≤ 1.
Definition 2.2 ([1]). For 〈x, x′〉, 〈y, y′〉 ∈ IFS, define〈x, x′〉 ∨ 〈y, y′〉 = 〈max{x, y},min{x′, y′}〉,〈x, x′〉 ∧ 〈y, y′〉 = 〈min {x, y} ,max {x′, y′}〉,〈x, x′〉c = 〈x′, x〉.
Definition 2.3 ([1]). Let X = {x1, x2, ...xm} be a set of alternativesand Y = {y1, y2, ...yn} be the attribute set of each element of X. Thenan intuitionistic fuzzy matrix (IFM) is defined by
A = (〈(xi, yj), µA(xi, yj), νA(xi, yj)〉),for i = 1, 2...m and j = 1, 2, ...n,, where µA : X × Y → [0,1] and νA :X × Y → [0,1] satisfy the condition 0 ≤ µA(xi, yj) + νA(xi, yj) ≤ 1. Forsimplicity we denote an intuitionistic fuzzy matrix (IFM) as matrix ofpairs A = (〈aij, a′ij〉) of non negative real numbers satisfying aij+a
′ij ≤ 1
for all i, j.For any two elements A = (〈aij, a′ij〉), B = (〈bij, b′ij〉) ∈ Fmn and
C ∈ Fnp, define : for all 1 ≤ i ≤ m and 1 ≤ j ≤ n,(i) (Component wise addition)
A⊕B = (〈max{aij, bij},min{a′ij, b′ij}〉),(ii) (Component wise multiplication)
A�B = (〈min{aij, bij},max{a′ij, b′ij}〉),(iii) (The universal matrix)
J = (〈1, 0〉) is the matrix in which all entries are 〈1, 0〉,(iv) (Identity matrix)
I = (〈δij, δ′ij〉) , where 〈δij, δ′ij〉 =
{〈1, 0〉 if i = j
〈0, 1〉 if i 6= j,
(v) (The zero matrix)O is the matrix in which all the entries are 〈0, 1〉,
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(vi) A ≥ B, if aij ≥ bij and a′ij ≤ b′ij, for all i, j,A > B, if aij > bij and a′ij < b′ij, for all i, j, in which case A and
B are comparable.(vii) (Complement of A)
Ac = (〈a′ij, aij〉),(viii) (Transpose of A)
AT = (〈aji, a′ji〉),(ix) AC = (〈maxnk=1min{aik, ckj},minn
k=1max{a′ik, c′kj}〉)= (〈
∑nk=1 aikckj,
∏nk=1(a
′ik + c′kj)〉),
(x) A← C = (minnk=1(〈aik, a′ik〉 ← 〈ckj, c′kj〉))
= (∧n
k=1(〈aik, a′ik〉 ← 〈ckj, c′kj〉)),(xi) For any two comparable elements 〈x, x′〉, 〈y, y′〉 ∈ IFS, define
〈x, x′〉 ← 〈y, y′〉 =
{〈1, 0〉 if 〈x, x′〉 ≥ 〈y, y′〉〈x, x′〉 if 〈x, x′〉 < 〈y, y′〉
.
Definition 2.4 ([1]). For A = (〈aij, a′ij〉) ∈ Fmn and C = (〈cij, c′ij〉) ∈Fnp, define
A � C = (∧k
(〈aik, a′ik〉 ∨ 〈ckj, c′kj〉)).
Definition 2.5 ([18]). Let R ∈ Fn. Then(i) If R ≥ In, then R is reflexive.(ii) If R2 ≤ R, then R is transitive.(iii) If R �R ≥ R is c-transitive.(iv) If R �R ≤ R is c-compact.(v) If R2 ≥ R, then R is compact.(vi) If R is reflexive and transitive, then R is idempotent.(vii) In R, if all the diagonal entries are 〈0, 1〉, then R is irreflexive.(viii) R is symmetric if and only if 〈rij, r′ij〉 = 〈rji, r′ji〉, for all i, j.
(ix) R is antisymmetric if and only if R�RT ≤ In,where In is the unit matrix with all entries in the main diagonal as〈1, 0〉 and remaining entries with 〈0, 1〉, and R � RT ≤ In means〈rij, r′ij〉〈rji, r′ji〉 = 〈0, 1〉 for all i 6= j and 〈rii, r′ii〉 ≤ 〈1, 0〉 for all i.So if 〈rij, r′ij〉 = 〈1, 0〉, then 〈rji, r′ji〉 = 〈0, 1〉.
(x) If R is irreflexive and transitive, then R is nilpotent.(xi) If 〈rii, r′ii〉 ≥ 〈rij, r′ij〉, for all i, j, then R is weakly reflexive.
Proposition 2.6 ([20]). For A ∈ Fmn, B ∈ Fmp, and C ∈ Fqn, thefollowing inequality hold:
(1) B × (BT → A← CT )× C ≤ A.858
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(2) B × (BT → A) ≤ A.(3) (A← CT )× C ≤ A.
Lemma 2.7 ([20]). For A ∈ Fmn, A← AT is reflexive and transitive.
Proposition 2.8 ([20]). For any A ∈ Fmn, the following inequalityhold:
(1) A→ AT ≥ Im.(2) A← AT ≥ Im.(3) (A→ AT )A ≥ A.
3. Maximal and minimal solution of intuitionistic fuzzyinequalities
In this section we extend implication operator ↽ used by Hashimoto[9] for fuzzy matrices to IFS and IFM and discuss the relation between← and ↽ .
Lemma 3.1. For A,B ∈ Fn, A ← B ≥ A if and only if A is weaklyreflexive.
Proof. Let A = (〈aij, a′ij〉) and B = (〈bij, b′ij〉). Then
(A← B) ≥ A⇔n∧
k=1
(〈aik, a′ik〉 ← 〈bkj, b′kj〉) ≥ 〈aij, a′ij〉
⇔ 〈ail, a′il〉 ← 〈blj, b′lj〉 ≥ 〈aij, a′ij〉 for some l⇔ 〈ail, a′il ≥ 〈aij, a′ij〉⇔ in particular, 〈all, a′ll〉 ≥ 〈alj, a′lj〉. �
Lemma 3.2. For A,B ∈ Fn, A ↽ B ≤ A if and only if 〈aij, a′ij〉 ≥〈aii, a′ii〉.
Proof. Let A = (〈aij, a′ij〉) and B = (〈bij, b′ij〉). Then
(A ↽ B) ≤ A⇔n∨
k=1
(〈aik, a′ik〉↽ 〈bkj, b′kj〉) ≤ 〈aij, a′ij〉
⇔ 〈ail, a′il〉↽ 〈blj, b′lj〉 ≤ 〈aij, a′ij〉 for some l⇔ 〈ail, a′il ≤ 〈aij, a′ij〉⇔ in particular, 〈all, a′ll〉 ≤ 〈alj, a′lj〉. �
Definition 3.3. For any two comparable elements 〈x, x′〉, 〈y, y′〉 ∈ IFS,define
〈x, x′〉↽ 〈y, y′〉 =
{〈0, 1〉 if 〈x, x′〉 ≤ 〈y, y′〉〈x, x′〉 if 〈x, x′〉 > 〈y, y′〉
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and for A ∈ Fmn and C ∈ Fnp, define
A ↽ C =∨k
(〈aik, a′ik〉↽ 〈ckj, c′kj〉)
andA ⇁ C =
∨k
(〈aik, a′ik〉⇁ 〈ckj, c′kj〉).
Lemma 3.4. For 〈a, a′〉, 〈b, b′〉 ∈ IFS,
(3.1) (〈a, a′〉 ← 〈b, b′〉)c = (〈a, a′〉c ↽ 〈b, b′〉c).
Proof. Case (i): If 〈a, a′〉 ≥ 〈b, b′〉, then 〈a, a′〉 ← 〈b, b′〉 = 〈1, 0〉. Byassumption, (〈a, a′〉)c ≤ (〈b, b′〉)c. Thus
(3.2) (〈a, a′〉 ← 〈b, b′〉)c = 〈0, 1〉.So, by Definition 3.3,
(3.3) (〈a, a′〉)c ↽ (〈b, b′〉)c = 〈0, 1〉.Hence, by (3.2) and (3.3), (3.1) holds.
Case (ii): If 〈a, a′〉 < 〈b, b′〉, then 〈a, a′〉 ← 〈b, b′〉 = 〈a, a′〉. Thus
(3.4) (〈a, a′〉 ← 〈b, b′〉)c = 〈a, a′〉c = 〈a′, a〉.By assumption, 〈a, a′〉c > 〈b, b′〉c. So, by Definition 3.3,
(3.5) 〈a, a′〉c ↽ 〈b, b′〉c = 〈a′, a〉.Hence, by (3.4) and (3.5), (3.1) holds. �
Lemma 3.5. For 〈a, a′〉, 〈b, b′〉 ∈ IFS,
(3.6) (〈a, a′〉↽ 〈b, b′〉)c = (〈a, a′〉c ← 〈b, b′〉c).
Proof. Case (i): If 〈a, a′〉 ≤ 〈b, b′〉, then
(3.7) (〈a, a′〉↽ 〈b, b′〉)c = 〈1, 0〉.By assumption, (〈a, a′〉)c ≥ (〈b, b′〉)c. Thus
(3.8) (〈a, a′〉c ← 〈b, b′〉c) = 〈1, 0〉.So, from (3.7) and (3.8), (3.6) holds.
Case (ii): If 〈a, a′〉 > 〈b, b′〉, then 〈a, a′〉c < 〈b, b′〉c. Thus
(3.9) (〈a, a′〉↽ 〈b, b′〉)c = 〈a, a′〉c
and
(3.10) 〈a, a′〉c ← 〈b, b′〉c = 〈a, a′〉c.So, by (3.9) and (3.10), (3.6) holds. �
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Lemma 3.6. For 〈a, a′〉, 〈b, b′〉 ∈ IFS,
(3.11) (〈b, b′〉 → 〈a, a′〉)c = (〈b, b′〉c ⇁ 〈a, a′〉c).
Proof. Case (i): If 〈a, a′〉 ≥ 〈b, b′〉, then
(3.12) (〈b, b′〉 → 〈a, a′〉)c = (〈1, 0〉)c.By assumption, (〈a, a′〉)c ≤ (〈b, b′〉)c. Thus
(3.13) (〈b, b′〉c ⇁ 〈a, a′〉)c = 〈0, 1〉 = 〈1, 0〉c.So, by (3.12) and (3.13), (3.11) holds.
Case (ii): If 〈a, a′〉 < 〈b, b′〉, then
(3.14) (〈b, b′〉 → 〈a, a′〉)c = 〈a, a′〉c
and
(3.15) 〈b, b′〉c ⇁ 〈a, a′〉c = 〈a, a′〉c.Thus, by (3.14) and (3.15), (3.11) holds. �
In the same manner, we prove the following Lemma.
Lemma 3.7. For 〈a, a′〉, 〈b, b′〉 ∈ IFS,
(〈b, b′〉⇁ 〈a, a′〉)c = 〈b, b′〉c → 〈a, a′〉c.
Lemma 3.8. For 〈a, a′〉, 〈b, b′〉, 〈c, c′〉 ∈ IFS,
〈a, a′〉 ∧ 〈b, b′〉 ≤ 〈c, c′〉 ⇔ 〈a, a′〉 ≤ 〈c, c′〉 ← 〈b, b′〉.
Proof. Case(i): If〈a, a′〉 ≤ 〈b, b′〉, then
〈a, a′〉 ∧ 〈b, b′〉 = 〈a, a′〉 ≤ 〈c, c′〉⇔ 〈a, a′〉 ≤ 〈c, c′〉 ← 〈b, b′〉.
Case (ii): If 〈a, a′〉 > 〈b, b′〉, then
〈a, a′〉 ∧ 〈b, b′〉 = 〈b, b′〉 ≤ 〈c, c′〉⇔ 〈c, c′〉 ← 〈b, b′〉 = 〈1, 0〉⇔ 〈a, a′〉 ≤ 〈c, c′〉 ← 〈b, b′〉.
�
Lemma 3.9. For 〈a, a′〉, 〈b, b′〉, 〈c, c′〉 ∈ IFS,
〈a, a′〉 ∨ 〈b, b′〉 ≥ 〈c, c′〉 ⇔ 〈a, a′〉 ≥ 〈c, c′〉↽ 〈b, b′〉.
Proof. Case (i): If 〈a, a′〉 ≥ 〈b, b′〉, then
〈a, a′〉 ∨ 〈b, b′〉 = 〈a, a′〉 ≥ 〈c, c′〉 ⇔ 〈a, a′〉 ≥ 〈c, c′〉↽ 〈b, b′〉.Case (ii): If〈a, a′〉 < 〈b, b′〉, then
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〈a, a′〉 ∨ 〈b, b′〉 = 〈b, b′〉⇔ 〈c, c′〉↽ 〈b, b′〉 = 〈0, 1〉⇔ 〈a, a′〉 ≥ 〈c, c′〉↽ 〈b, b′〉.
�
Now we extend this result to IFMs.
Proposition 3.10. For A ∈ Fmn, B ∈ Fnm, the following results hold:(1) (A← B)c = Ac ↽ Bc.(2) (A ↽ B)c = Ac ← Bc.(3) (A→ B)c = Ac ⇁ Bc.(4) (A ⇁ B)c = Ac → Bc.
Proof. The proof is evident from Lemma 3.4, Lemma 3.5, Lemma 3.6and Lemma 3.7 �
Lemma 3.11. For A ∈ Fmn, B ∈ Fnm,(1) (A← B)T = BT → AT .(2) (A ↽ B)T = BT ⇁ AT .
Proof. (1) Let A = (〈aij, a′ij〉), B = (〈bij, b′ij〉). Then
〈aik, a′ik〉 ← 〈bkj, b′kj〉 =
{〈1, 0〉 if 〈aik, a′ik〉 ≥ 〈bkj, b′kj〉,〈aik, a′ik〉 if 〈aik, a′ik〉 < 〈bkj, b′kj〉.
Thus,
(3.16) (〈aik, a′ik〉 ← 〈bkj, b′kj〉)T =
{〈1, 0〉 if 〈aki, a′ki〉 ≥ 〈bjk, b′jk〉.〈aki, a′ki〉 if 〈aki, a′ki〉 < 〈bjk, b′jk〉.
Now,
(3.17) 〈bjk, b′jk〉 → 〈aki, a′ki〉 =
{〈1, 0〉 if 〈aki, a′ki〉 ≥ 〈bjk, b′jk〉,〈aki, a′ki〉 if 〈aki, a′ki〉 < 〈bjk, b′jk〉.
So, by (3.16) and (3.17), (1) holds.(2) Clearly,
〈aik, a′ik〉↽ 〈bkj, b′kj〉 =
{〈0, 1〉 if 〈aik, a′ik〉 ≤ 〈bkj, b′kj〉,〈aik, a′ik〉 if 〈aik, a′ik〉 > 〈bkj, b′kj〉.
Then
(3.18) (〈aik, a′ik〉↽ 〈bkj, b′kj〉)T =
{〈0, 1〉 if 〈aki, a′ki〉 ≤ 〈bjk, b′jk〉,〈aki, a′ki〉 if 〈aki, a′ki〉 > 〈bjk, b′jk〉.
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and
(3.19) 〈bjk, b′jk〉⇁ 〈aki, a′ki〉 =
{〈0, 1〉 if 〈aki, a′ki〉 ≤ 〈bjk, b′jk〉,〈aki, a′ki〉 if 〈aki, a′ki〉 > 〈bjk, b′jk〉.
Thus, by (3.18) and (3.19), (2) holds. �
Lemma 3.12. For A ∈ Fmn, B ∈ Fnp and C ∈ Fmp,(1) A×B ≤ C ⇔ A ≤ C ← BT ⇔ B ≤ AT → C.(2) A �B ≥ C ⇔ A ≥ C ↽ BT ⇔ B ≥ AT ⇁ C.
Proof. (1) A×B ≤ C⇔ (
∨k(〈aik, a′ik〉) ∧ 〈bkj, b′kj〉) ≤ 〈cij, c′ij〉) for all i, j
⇔ 〈aik, a′ik〉 ≤ 〈cij, c′ij〉 ← 〈bkj, b′kj〉 for all i, j, k, by Lemma 3.8⇔ 〈aik, a′ik〉 ≤
∧j(〈cij, c′ij〉 ← 〈bkj, b′kj〉) for all i, k
⇔ (〈aij, a′ij〉) ≤∧
k〈cik, c′ik〉)← 〈bjk, b′jk〉) for all i, j, k
⇔ A ≤ C ← BT .On one hand,A×B ≤ C ⇔ BT ×AT ≤ CT ⇔ BT ≤ CT ← A⇔ B ≤ (CT ← A)T .
By Lemma 3.11 (1), (CT ← A)T = AT → C. Then
A×B ≤ C ⇔ B ≤ AT → C.
(2) A �B ≥ C⇔ (
∧k(〈aik, a′ik〉) ∨ 〈bkj, b′kj〉) ≥ 〈cij, c′ij〉) for all i, j
⇔ 〈aik, a′ik〉) ∨ 〈bkj, b′kj〉 ≥ 〈cij, c′ij〉 for all i, j, k.⇔ 〈aik, a′ik〉 ≥ 〈cij, c′ij〉↽ 〈bkj, b′kj〉 for all i, j, k⇔ 〈aik, a′ik〉 ≥
∨j(〈cij, c′ij〉↽ 〈bkj, b′kj〉) for all i, k
⇔ 〈aij, a′ij〉 ≥∨
k(〈cik, c′ik〉↽ 〈bjk, b′jk〉) for all i, j.
Then A �B ≥ C ⇔ A ≥ C ↽ BT ⇔ B ≥ AT ⇁ C is obvious. �
Proposition 3.13. For A ∈ Fmn, B ∈ Fpn,(1) (A ↽ BT ) �B ≥ A.(2) B � (BT ⇁ A) ≥ A, for B ∈ Fmp.
Proof. (1) Since A ↽ BT ≥ A ↽ BT , by Lemma 3.12(2), (1) holds.(2) Put D = BT ⇁ A. Then, by Lemma 3.12(2),
B �D ≥ A⇔ D ≥ BT ⇁ A⇔ BT ⇁ A ≥ BT ⇁ A.
Thus B � (BT ⇁ A) ≥ A. �
Remark 3.14. Consider the inequality A × X × B ≤ C where A ∈Fmn, B ∈ Fpq and C ∈ Fmq with unknown X ∈ Fnp.
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By Lemma 3.12(1),
(3.20) X ≤ (AT → C)← BT .
By Proposition 2.6,
(3.21) A× (AT → C ← BT )×B ≤ C.
From (3.20) and (3.21), (AT → C ← BT ) is the maximum solution ofA×X ×B ≤ C.
Remark 3.15. Consider the inequality A � X � B ≥ C with suitablesize and unknown X. Then, by Lemma 3.12(2),
(3.22) X ≥ AT ⇁ C ↽ BT .
By Proposition 3.13,
(3.23) A �(AT ⇁ C ↽ BT
)�B ≥ A �
(AT ⇁ C
)≥ C.
From (3.22) and (3.23), AT ⇁ C ↽ BT is the minimum solution ofA �X �B ≥ C.
4. Intuitionistic fuzzy similarity relation (IFSR) andintuitionistic fuzzy dissimilarity relation(IFDR) using
← and ↽ .
This section provides a method to obtain IFSR and IFDR from anyIFM. Using this IFSR and IFDR we shall obtain max-min idempotent(R×R = R) and min-max idempotent (R �R = R) for any IFM.
Proposition 4.1. For A ∈ Fmn, B ∈ Ffnp,(1) A← (AT ← B) ≥ B.(2) A ↽ (AT ↽ B) ≤ B.(3) (B → AT )→ A ≥ B, for B ∈ Fpn.(4) (B ⇁ AT ) ⇁ A ≤ B, for B ∈ Fpn.
Proof. (1) Since (BT → A) ≤ (BT → A),
(AT ← B)T = BT → (AT )T = BT → A ≤ BT → A.
Since AB ≤ C ⇔ A ≤ C ← BT ⇔ B ≤ AT → C), using Lemma3.12(1),
B(AT ← B)T ≤ Aand
B ≤ A← ((AT ← B)T )T = A← (AT ← B).864
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(2) Since BT ⇁ A ≥ BT ⇁ A or (AT ↽ B)T ≥ BT ⇁ A, usingLemma 3.12(2), B � (AT ↽ B)T ≥ A and B ≥ A ↽ (AT ↽ B).
(3) Since A← BT ≤ A← BT or (B → AT )T ≤ A← BT ,(B → AT )TB ≤ A and B ≤ (B → AT )→ A.
(4) Since A ↽ BT ≥ A ↽ BT or (B ⇁ AT )T ≥ A ↽ BT , by Lemma3.12,
(B ⇁ AT )T �B ≥ A and B ≥ (B ⇁ AT ) ⇁ A. �
Proposition 4.2. For A ∈ Fmn, B ∈ Fpn,(1) ((A← BT )B)← BT = A← BT .(2) ((A ↽ BT ) �B) ↽ BT = A ↽ BT .(3) BT → (B(BT → A)) = BT → A, for B ∈ Fmp.(4) BT ⇁ (B � (BT ⇁ A)) = BT ⇁ A, for B ∈ Fmp.
Proof. (1) Clearly, (A ← BT )B ≤ (A ← BT )B. Then, by Lemma3.12(1), (A← BT ) ≤ ((A← BT )B)← BT . Thus, by Proposition 2.6,
(A← BT ) ≤ ((A← BT )B)← BT ≤ A← BT .
So ((A← BT )B)← BT = A← BT .(2) Clearly, ((A ↽ BT ) �B) ≥ (A ↽ BT ) �B. Then
(A ↽ BT ) ≥ ((A ↽ BT ) �B) ↽ BT ≥ A ↽ BT .
Then, by Lemma 3.12(i), and Proposition 3.13(1),
((A ↽ B) �B) ↽ BT = A ↽ BT .
(3) Clearly, B(BT → A) ≤ B(BT → A). Then
(BT → A)) ≤ BT → B(BT → A) ≤ BT → A.
By Proposition 2.6, (B(BT → A) ≤ A. Thus
BT → (B(BT → A)) = BT → A.
(4) Clearly, B � (BT ⇁ A) ≥ B � (BT ⇁ A). Then, by Proposition3.13, BT ⇁ A ≥ BT ⇁ (B � (BT ⇁ A)) ≥ BT ⇁ A. Thus
BT ⇁ (B � (BT ⇁ A)) = BT ⇁ A.
�
Proposition 4.3. (1) For A ∈ Fmn, B ∈ Fpn and C ∈ Fqn,(a) (A← BT )× (B ← CT )× C ≤ A.(b) (A ↽ BT ) � (B ↽ CT ) � C ≥ A.
(2) For A ∈ Fmn, B ∈ Fmp and C ∈ Fmq,(a) C × (CT → B)× (BT → A) ≤ A.(b) C � (CT ⇁ B) � (BT ⇁ A) ≥ A.
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Proof. (1) (a) By proposition 2.6, (A← BT )B ≤ A, (B ← CT )C ≤ B.Then (A← BT )(B ← CT )C ≤ (A← BT )B ≤ A. Thus (a) holds.
(b) By Proposition 2.6, (A ↽ BT ) �B ≥ A ⇒ (B ↽ CT ) � C ≥ B.Then (A ↽ BT ) � (B ↽ CT ) � C ≥ (A ↽ BT )B ≥ A. Thus (b) holds.
(2) (a) By proposition 2.6, B× (BT → A) ≤ A, C × (CT → B) ≤ B.Then C × (CT → B)× (BT → A) ≤ B(BT → A) ≤ A. Thus (a) holds.
(b) By proposition 3.13, B � (BT ⇁ A) ≥ A, C � (CT ⇁ B) ≥ B.Then C � (CT ⇁ B) � (BT ⇁ A) ≥ B � (BT ⇁ A) ≥ A. Thus (b)holds. �
Proposition 4.4. (1) For A ∈ Fmn, B ∈ Fpn and F ∈ Fnq,(a) (A← BT )(B ← F ) ≤ A← F .(b) (A ↽ BT ) � (B ↽ F ) ≥ A ↽ F .
(2) For A ∈ Fmn, B ∈ Fmp and F ∈ Fqm,(a) (F → B)(BT → A) ≤ F → A.(b) (F ⇁ B) � (BT ⇁ A) ≥ F ⇁ A.
Proof. (1) (a) By Proposition 4.3, (A ← BT )(B ← F T )F ≤ A. Then,by Lemma 3.12, (A← BT )(B ← F T ) ≤ A← F T .
(b) Clearly, (A ↽ BT ) � (B ↽ F T ) � F ≥ A. Then (b) holds.(2) (a) By Proposition 4.3, F (F T → B)(BT → A) ≤ A. Then (a)
holds.(b) Clearly, F � (F T ⇁ B) � (BT ⇁ A) ≥ A. Then (b) holds. �
Proposition 4.5. (1) For A ∈ Fmn, B ∈ Fpn and F ∈ Fn,(a) (A← BT )(B ∧ (B ← F )) ≤ A ∧ (A← F ).(b) (A ↽ BT ) � (B ∨ (B ↽ F )) ≥ A ∨ (A ↽ F ).
(2) For A ∈ Fmn, B ∈ Fmp and F ∈ Fm,(a) (B ∧ (F → B))× (BT → A) ≤ A ∧ (F → A).(b) (B ∨ (F ⇁ B)) � (BT ⇁ A) ≥ A ∨ (F ⇁ A).
Proof. (1)(a) It is clear that B∧(B ← F ) ≤ B, B∧(B ← F ) ≤ B ← F .Then
(4.1) (A← BT )(B ∧ (B ← F )) ≤ (A← BT )B ≤ A.
By Proposition 4.4,
(4.2) A← BT )(B ∧ (B ← F )) ≤ (A← BT )(B ← F ) ≤ A← F.
From (4.1) and (4.2),
(A← BT )(B ∧ (B ← F )) ≤ A ∧ (A← F ).866
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(b) Clearly, B ∨ (B ↽ F ) ≥ B and B ∨ (B ↽ F ) ≥ B ↽ F . ByProposition 3.13,
(A ↽ BT ) � (B ∨ (B ↽ F )) ≥ (A ↽ BT ) �B ≥ A.
By Proposition 4.4(1b),
(A ↽ BT ) � (B ∨ (B ↽ F )) ≥ (A ↽ BT ) � (B ↽ F ) ≥ A ↽ F.
Thus (1b) holds.Similarly, we can prove (2a) and (2b), using this similar argument
and proposition 4.4(2a) and 4.4(2b). �
Proposition 4.6. (1) For A ∈ Fmn, B ∈ Fpn, C ∈ Fpn, D ∈ Fmn,E ∈ Fpn, and F ∈ Fnn,
(a) If A ≤ D,C ≤ B,E ≤ B, then
(A← BT )× (C ∧ (E ← F )) ≤ D ∧ (A← F ).
(b) If A ≥ D, C ≥ B, E ≥ B, then
(A ↽ BT ) � (C ∨ (E ↽ F )) ≥ D ∨ (A ↽ F ).
(2) For A ∈ Fmn, B ∈ Fnp, C ∈ Fnp, D ∈ Fmn, E ∈ Fmp, andF ∈ Fnm,
(a) If A ≤ D, C ≤ B, E ≤ B, then
(C ∧ (F → E))× (BT → A) ≤ D ∧ (F → A).
(b) If A ≥ D, C ≥ B, E ≥ B, then
(C ∨ (F ⇁ E)) � (BT ⇁ A) ≥ D ∨ (F ⇁ A).
Proof. (1) (a) Since (C ∧ (E ← F )) ≤ C ≤ B and E ≤ B,
(C ∧ (E ← F )) ≤ (E ← F ) ≤ B ← F.
Then
(A← BT )× (C ∧ (E ← F )) ≤ (A← BT )C
≤ (A← BT )B
≤ A ≤ D.
Thus
(4.3) (A← BT )× (C ∧ (E ← F )) ≤ D.867
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Similarly, by Proposition 4.4),
(A← BT )× (C ∧ (E ← F )) ≤ (A← BT )× (E ← F )
≤ (A← BT )× (B ← F )
≤ A← F.
So
(4.4) (A← BT )× (C ∧ (E ← F )) ≤ A← F.
Hence, from (4,3) and (4.4),
(A← BT )(C ∧ (E ← F )) ≤ D ∧ (A← F ).
(b) Clearly, C ∨ (E ↽ F ) ≥ C and C ∨ (E ↽ F ) ≥ E ↽ F .Also, C ∨ (E ↽ F ) ≥ B and C ∨ (E ↽ F ) ≥ B ↽ F.Then
(A ↽ BT ) � (C ∨ (E ↽ F )) ≥ (A ↽ BT ) �B ≥ A ≥ D.
Thus
(4.5) (A ↽ BT ) � (C ∨ (E ↽ F )) ≥ D.
Using Proposition 4.4,
(A ↽ BT ) � (C ∨ (E ↽ F )) ≥ (A ↽ BT ) � (E ↽ F )
≥ (A ↽ BT ) � (B ↽ F )
≥ A ↽ F.
So
(4.6) (A ↽ BT ) � (C ∨ (E ↽ F )) ≥ A ↽ F.
Hence, from (4.5) and (4.6), we get
(A ↽ BT ) � (C ∨ (E ↽ F )) ≥ D ∨ (A ↽ F ).
By similar argument, we can prove (2a) and (2b). �
Theorem 4.7. (1) For A ∈ Fmn, B ∈ Fpn, C ∈ Fpn,D ∈ Fmn, E ∈ Fpn, F ∈ Fnn and C ∈ Fmp and G ∈ Fmp,
(a) If A ≤ D, C ≤ B, E ≤ B, then
(G ∧ (A← BT ))× (C ∧ (E ← F )) ≤ D ∧ (A← F ).
(b) If A ≥ D, C ≥ B, E ≥ B,
(G ∨ (A ↽ BT )) � (C ∨ (E ↽ F )) ≥ D ∨ (A ↽ F ).
(2) For A ∈ Fmn, B ∈ Fmp, C ∈ Fmp, D ∈ Fmn, E ∈ Fnp,F ∈ Fmm and G ∈ Fpn,
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(a) If A ≤ D, C ≤ B, E ≤ B, then
(C ∧ (F → E))× (G ∧ (BT → A)) ≤ D ∧ (F → A).
(b) If A ≥ D, C ≥ B, E ≥ B, then
(C ∨ (F ⇁ E)) � (G ∨ (BT ⇁ A)) ≥ D ∨ (F ⇁ A).
Proof. (1) (a) By Proposition 4.6(1a),
G ∧ (A← BT ) ≤ G and G ∧ (A← BT ) ≤ A← BT .
Then
(G ∧ (A← BT ))(C ∧ (E ← F T )) ≤ (A← BT )(C ∧ (E ← F ))
≤ D ∧ (A← F ).
(b) By Proposition 4.6 (1b),
(G ∨ (A ↽ BT )) � (C ∨ (E ↽ F T )) ≥ (A ↽ BT ) � (C ∨ (E ↽ F )
≥ D ∧ (A ↽ F )).
(2) (a)
(C ∧ (F → E))× (G ∧ (BT → A)) ≤ (C ∧ (F → E))× (BV → A)
≤ D ∧ (F → A).
(b) By Proposition 4.6 (2b),
(C ∨ (F ⇁ E)) � (G ∨ (BT ⇁ A)) ≥ (C ∨ (F ↽ ET )) � (BT ⇁ A))
≥ D ∨ (F ⇁ A).
�
Using the above Theorem 4.7, we obtain the following proposition.
Proposition 4.8. For A,B,C ∈ Fn,(1) (a) If A ≤ C ≤ B, then
(C ∧ (A← BT ))× (C ∧ (A← BT ) ≤ C ∧ (A← BT ).
(b) If A ≥ C ≥ B, then
(C ∨ (A ↽ BT )) � (C ∨ (A ↽ BT ) ≥ C ∨ (A ↽ BT ).
(2) (a) If A ≤ C ≤ B, then
(C ∧ (BT → A))× (C ∧ (BT → A)) ≤ C ∧ (BT → A).
(b) If A ≥ C ≥ B, then
(C ∨ (BT ⇁ A)) � (C ∨ (BT ⇁ A)) ≥ C ∨ (BT ⇁ A).869
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This proposition shows that C ∧ (A ← BT ) and C ∧ (BT → A) aretransitive and C ∨ (BT ⇁ A) and C ∨ (A ↽ BT ) are c-transitive.
Remark 4.9. In Proposition 4.8, if we replace CandB by A, we getA ∧ (A ← AT ) and A ∧ (AT → A) are transitive, and A ∨ (AT ⇁ A)and A ∨ (A ↽ AT ) are c-transitive.
Theorem 4.10. (1) For A ∈ Fmn, B ∈ Fpn, C ∈ Fpn and F ∈ Fnq,(a) If C ≤ B, then (A← BT )(C ← F ) ≤ A← F.(b) If C ≥ B, then (A ↽ BT ) � (C ↽ F ) ≥ A ↽ F.
(2) For any IFM A ∈ Fmn, B ∈ Fmp, C ∈ Fmp and F ∈ Fqm,(a) If C ≤ B, then (F → C)(BT → A) ≤ F → A.(b) If C ≥ B, then (F ↽ C) � (BT ↽ A) ≥ A ↽ F.
Proof. (1) (a) By Proposition 4.4(1a),
(A← BT )(C ← F ) ≤ (A← BT )(B ← F ) ≤ A← F.
(b) By Proposition 4.4(1b),
(A ↽ BT ) � (C ↽ F ) ≥ (A ↽ BT ) � (B ↽ F ) ≥ A← F .
Similarly, we can prove (2a) and (2b). �
The following Proposition is evident from the Theorem 4.10.
Proposition 4.11. (1) For A ∈ Fmn, B ∈ Fmn and F ∈ Fnq,(a) If A ≤ B, then (A← BT )(A← F ) ≤ A← F .(b) If A ≥ B, then (A ↽ BT ) � (A ↽ F ) ≥ A ↽ F .
(2) For any IFMs A ∈ Fmn, B ∈ Fmn and F ∈ Fqm,(a) If A ≤ B, then (F → A)(BT → A) ≤ F → A.(b) If A ≥ B, then (F ⇁ A) � (BT ⇁ A) ≥ F ⇁ A.
Remark 4.12. If we replace F by BT in Proposition 4.11, we get(1) (a) If A ≤ B, then (A← BT )(A← BT ) ≤ A← BT .
(b) If A ≥ B, then (A ↽ BT ) � (A ↽ BT ) ≥ A ↽ BT .(2) (a) If A ≤ B, then (BT → A)(BT → A) ≤ BT → A.
(b) If A ≥ B, then (BT ⇁ A) � (BT ⇁ A) ≥ BT ⇁ A.Similarly,
(a) If A ≥ B, then (A← BT )(A← BT ) ≥ A← BT .(b) If A ≤ B, then (A ↽ BT ) � (A ↽ BT ) ≤ A ↽ BT .(c) If A ≥ B, then (BT → A)(BT → A) ≥ BT → A.(d) If A ≤ B, then (BT ⇁ A) � (BT ⇁ A) ≤ BT ⇁ A.
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Remark 4.13. In the above Remark 4.12, if A = B, we get(1) (a) (A← AT )(A← AT ) = A← AT .
(b) (A ↽ AT ) � (A ↽ AT ) = A ↽ AT .(2) (a) (AT → A)(AT → A) = AT → A.
(b) (AT ⇁ A) � (AT ⇁ A) = AT ⇁ A.
Remark 4.14. (1) In Proposition 4.8 (1a), if we replace C by (A ←AT )T , then
((A← AT )T ∧ (A← AT )× ((A← AT )T ∧ (A← AT ))
≤ (A← AT )T ∧ (A← AT )).
(2) In Proposition 4.8 (1b), if we replace C by (A ↽ AT )T and B byA, then
((A ↽ AT )T ∨ (A ↽ AT ) � ((A ↽ AT )T ∨ (A ↽ AT ))
≥ (A ↽ AT )T ∨ (A ↽ AT ).
Let R be an n×n IFM. If R×R ≤ R(transitive), RT = R (symmetric)and I ≤ R( reflexive), then R represents a similarity relation or anintuitionistic fuzzy equivalence relation. Clearly
(A← AT ) ∧ (A← AT )T ≥ I ,((A← AT ) ∧ (A← AT )T )T = (A← AT ) ∧ (A← AT )T
and((A← AT ) ∧ (A← AT )T )((A← AT ) ∧ (A← AT )T )
= ((A← AT ) ∧ (A← AT )T ).Thus ((A← AT ) ∧ (A← AT )T ) represents a similarity relation.Also (AT → A) ∧ (AT → A)T represents a similarity relation.
If R � R ≥ R (c-transitive), RT = R and Ic ≥ R, then R representsa dissimilarity relation. That is, if R = (〈rij, r′ij〉) is a dissimilaritymatrix, then〈rik ∨ rkj, r′ik ∧ r′kj〉 ≥ 〈rij, r′ij〉, 〈rij, r′ij〉 = 〈rji, r′ji〉
and〈rii, r′ii〉 = 〈0, 1〉 for i, j, k = 1, 2, ..., n.
Thus (A ↽ AT ) ∨ (A ↽ AT )T and (AT ⇁ A) ∨ (AT ⇁ A)T representsdissimilarity relations.
Proposition 4.15. (1) For A ∈ Fmn, F ∈ Fnq,(a) (A← AT )(A← F ) = A← F.(b) (A ↽ AT ) � (A ↽ F ) = A ↽ F.
(2) For A ∈ Fmn, F ∈ Fqm,(a) (F → A)(AT → A) = F → A.
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(b) (F ↽ A) � (AT ↽ A) = F ↽ A.
Proof. In Proposition 4.11(1a), if we set B = A, then
(A← AT )(A← F ) ≤ A← F.
Since A← AT is reflexive, A← AT ≥ I (A← AT )(A← F ) ≥ A← F.Thus (A← AT )(A← F ) = A← F.
Similarly, setting in Proposition 4.11(1b), B = A. Then we have
(A ↽ AT ) � (A ↽ F ) ≥ A ↽ F.
Since Ic ≥ A ↽ AT , (A ↽ F ) ≥ (A ↽ AT ) � (A ↽ F ). So (1b) holdsBy similar argument, we can prove (2a) and (2b). �
Proposition 4.16. For A ∈ Fmn,(1) (a) (A← AT )A = A.
(b) (A ↽ AT ) � A = A.(2) (a) A(AT → A) = A.
(b) A � (AT ⇁ A) = A.
Proof. (1) (a) Clearly, A← AT ≥ I. Then (A← AT )A ≥ A. Thus, ByProposition 2.6, (A← AT )A ≤ A. So (A← AT )A = A.
(b) Clearly, A ↽ AT ≤ Ic and (A ↽ AT ) � A ≤ A. By Proposition3.13, (A ↽ AT ) � A ≥ A. Thus (A ↽ AT ) � A = A.
Similarly, we can prove (2a) and (2b). �
The following Proposition is obvious from Proposition 4.16.
Proposition 4.17. For A ∈ Fmn,(1) (A← AT )A(AT → A) = A.(2) (A ↽ AT ) � A � (AT ⇁ A) = A.
Proposition 4.18. For A ∈ Fmn,(1) (a) A← (AT → A)T = A← (AT ← A) = A.
(b) A ↽ (AT ⇁ A)T = A ↽ (AT ↽ A) = A.(2) (a) (A← AT )→ A = (A→ AT )→ A = A.
(b) (A ↽ AT )T ⇁ A = (A ⇁ AT ) ⇁ A = A.
Proof. By Proposition 4.1(1), A← (AT ← A) ≥ A. Since (AT ← A) ≥I, A← (AT ← A) ≥ A← I = A. Thus A← (AT ← A) = A.
(b) By Proposition 4.1(2), A ↽ (AT ↽ A) ≤ A. Since Ic ≥ AT ↽ A,
A ↽ Ic ≥ A ↽ (AT ↽ A) and A ≥ A ↽ (AT ↽ A).
Thus A ↽ (AT ↽ A) = A. �872
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Remark 4.19. By Proposition 4.16 and 4.18, we have
(A← AT )A = A← (AT ← A) and A(AT → A) = (A→ AT )→ A.
Lemma 4.20. For A,B ∈ Fn, (A ∧B)2 ≤ A2 ∧B2.
Proof. A ∧B = (〈(aij ∧ bij), (a′ij ∨ b′ij)〉). Then
(A ∧B)2
= (〈∨k
((aik ∧ bik) ∧ (akj ∧ bkj)),∧k
((a′ik ∨ b′ik) ∨ (a′kj ∨ b′kj)〉).
Without loss of generality, assume AandB are 2× 2 matrices.Consider∨
k
((aik ∧ bik) ∧ (akj ∧ bkj))
= [(ai1 ∧ a1j) ∧ (bi1 ∧ b1j)] ∨ [(ai2 ∧ a2j) ∧ (bi2 ∧ b2j)]= [(ai1∧a1j)∨[(ai2∧a2j)∧(bi2∧b2j)]]∧[(bi1∧b1j)∨[(ai2∧a2j)∧(bi2∧b2j)]]= [(ai1 ∧ a1j) ∨ (ai2 ∧ a2j)] ∧ [(ai1 ∧ a1j) ∨ (bi2 ∧ b2j)] ∧ [(bi1 ∧ b1j)∨
(ai2 ∧ a2j)] ∧ [(bi1 ∧ b1j) ∨ (bi2 ∧ b2j)]≤ [(ai1 ∧ a1j) ∨ (ai2 ∧ a2j)] ∧ [(bi1 ∧ b1j) ∨ (bi2 ∧ b2j)]= (∨k
(aik ∧ akj)) ∧ (∨k
(bik ∧ bkj)).
Similarly,∧k
[(a′ik ∨ b′ik) ∨ (a′kj ∨ b′kj)] ≥ (∧k
(a′ik ∨ a′kj)) ∨ (∧k
(b′ik ∨ b′kj)).
Thus
(A ∧B)2
≤ ((∨k
(aik ∧ akj)) ∧ (∨k
(bik ∧ bkj)), (∧k
(a′ik ∨ a′kj)) ∨ (∧k
(b′ik ∨ b′kj))
= A2 ∧B2.
�
In dual fashion we can prove the following Lemma.
Lemma 4.21. For A,B ∈ Fn, (A ∨B) � (A ∨B) ≥ (A �A) ∨ (B �B).
Proposition 4.22. (1) For A ∈ Fn, B ∈ Fmn,(a) R = A ∧ (A← AT ) ∧ (BT → B)⇒ R×R ≤ R.(b) R = A ∨ (A ↽ AT ) ∨ (BT ⇁ B)⇒ R �R ≥ R..
(1) For any IFMs A ∈ Fn, B ∈ Fnm,(a) R = A ∧ (A→ AT ) ∧ (BT ← B)⇒ R×R ≤ R.(b) R = A ∨ (A ⇁ AT ) ∨ (BT ↽ B)⇒ R �R ≥ R.
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Proof. (1) (a) Since (A ∧ (A← AT ) and (BT → B) are transitive,
R×R = ((A ∧ (A← AT )) ∧ (BT → B))2
≤ (A ∧ (A← AT ))2 ∧ (BT → B)2
≤ (A ∧ (A← AT )) ∧ (BT → B)
= R.
Similarly, we can prove the other results. �
In the above Proposition, if we replace B by A, we get the following.
Proposition 4.23. For any IFMs A ∈ Fn,(1) R = A ∧ (A← AT ) ∧ (AT → A)⇒ R×R ≤ R.(2) R = A ∨ (A ↽ AT ) ∨ (AT ⇁ A)⇒ R �R ≥ R.
Lemma 4.24. If R ∈ Fn is transitive and symmetric, then it is weaklyreflexive.
Proof. Since R is transitive, R2 ≤ R. Then 〈rik∧rki, r′ik∨r′ki〉 ≤ 〈rii, r′ii〉.Using the symmetricity condition of R, we get 〈rik, r′ik〉 ≤ 〈rii, r′ii〉. Thusthe result holds. �
Lemma 4.25. If R ∈ Fn is transitive and weakly reflexive, then R iscompact.
Proof. Since R is weakly reflexive, 〈rik, r′ik〉 ≤ 〈rii, r′ii〉. Then
〈rii, r′ii〉 ∧ 〈rik, r′ik〉 ≥ 〈rik, r′ik〉.This is true for all i, k. thus R2 ≥ R. �
Lemma 4.26. If R ∈ Fn is c-transitive and symmetric, then R�R = R.
Proof. If R is c-transitive, then 〈rik, r′ik〉 ∨ 〈rkj, r′kj〉 ≥ 〈rij, r′ij〉.In particular, 〈rik, r′ik〉 ∨ 〈rki, r′ki〉 ≥ 〈rii, r′ii〉. By symmetry of R,〈rik, r′ik〉 ≥ 〈rii, r′ii〉. Thus 〈rii, r′ii〉 ∨ 〈rik, r′ik〉 ≤ 〈rik, r′ik〉. So R � R ≤ R.Hence R �R = R. �
From the above Lemmas the following proposition is trivial.
Proposition 4.27. For any symmetric A ∈ Fn,(1) R = A ∧ (A← AT ) ∧ (AT → A)⇒ R×R = R.(2) R = A ∨ (A ↽ AT ) ∨ (AT ⇁ A)⇒ R �R = R.
The Proposition 4.27 shows that one can derive a max-min idempo-tent IFM from any symmetric IFM and min-max idempotent IFM froma symmetric IFM. If we replace replace A by Ac ∧ (AT )c in Proposition
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4.27(1) and A by Ac∨ (AT )c in Proposition 4.27 we get idempotent andmin-max idempotent IFM for any IFM.
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P. Murugadas 1 (bodi [email protected])Assistant Professor, Department of Mathematics, Annamalai Univer-sity, Annamalainagar, Tamil Nadu-608002, India
K. Lalitha 2 (sudhan [email protected])Assistant Professor, Mathematics Section (FEAT), Annamalai Univer-sity, Annamalainagar, Tamil Nadu-608002, India
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