New Operators on Fuzzy Matrices

Yanping Zhang and Mingwen Zheng

Abstract— The fuzzy matrices are successfully used whenfuzzy uncertainty occurs in a problem. In this paper, two newbinary fuzzy operators ⊕ and � and four new cut sets areintroduced. Several properties on these operators are presented.Also, some results on existing operators along with these newoperators are presented.

I. INTRODUCTION

IT is well known, that matrices play major role in var-

ious areas such as mathematics, physics, statistics, en-

gineering, social sciences and many others. Several authors

presented a number of results on fuzzy matrices. Shyamal

[1] introduced two binary operators on fuzzy matrices and

studied some properties. [2][3] introduced new cut sets of

fuzzy sets and studied some properties. In this paper, two

new binary operators on fuzzy matrices which are denoted

by the symbol ⊕ and �(are different form [1]), and four

new cut sets xα, x[λ], x[λ], x

λ are introduced. Also, some

properties of the fuzzy matrices over these operators and

some pre-defined operators are presented. The structure of

this paper is as follows. Section 2 shows the preliminaries,

it includes notations and basic concepts which will be used

in the following section; Section 3 shows results on new

operators of fuzzy matrix; Section 4 introduces the results

on cut sets of fuzzy matrix.

II. PRELIMINARIES

In this section, as a preparation, we define some opera-

tors on fuzzy matrices whose elements are confined in the

closed interval [0, 1]. For all x, y, α, λ ∈ [0, 1] the following

operators are defined.

(i)x ∨ y = max{x, y} (ii)x ∧ y = min{x, y}(iii)x� y =

{x, x > y,0, x ≤ y.

(iv)x(α) =

{1, x ≥ α,0, x < α.

(v)x(α) =

{x, x ≥ α,0, x < α.

(vi)xC = 1− x

Now, we define two new operators ⊕ and � and four new

cut sets as follows:

x⊕ y = (x+ y) ∧ 1,x� y = (x+ y − 1) ∨ 0,

where the operations ′+′ and ′−′ are ordinary addition,

subtraction respectively.

xα =

{1, x ≥ α,x, x < α.

x[λ] =

{1, λ+ x ≥ 1,0, λ+ x < 1.

x[λ] =

{x, λ+ x ≥ 1,0, λ+ x < 1.

xλ =

{1, λ+ x ≥ 1,x, λ+ x < 1.

Yanping Zhang and Mingwen Zheng are with the School of Science,Shandong University of Technology, Zibo, Shandong 255049, P. R. China.(Email: zyp0543@126.com and sdlgzmw−11@163.com).

Next, we define some operations on fuzzy matrices.

Let A = [aij ] and B = [bij ] be two fuzzy matrices of order

m× n. Then

A⊕B = [(aij + bij) ∧ 1]A�B = [(aij + bij − 1) ∨ 0]A ∨B = [aij ∨ bij ]A ∧B = [aij ∧ bij ]A�B = [aij � bij ]A ≤ B if and only if aij ≤ bij for all i, j.

Ac = [1− aij ]A(α) = [(aij)

(α)] A(α) = [(aij)(α)]Aα = [(aij)α]A[λ] = [(aij)

[λ]] A[λ] = [(aij)[λ]]Aλ = [(aij)

λ]

III. RESULTS ON NEW OPERATORS OF FUZZY MATRIX

In this section, we study some properties for the new

operators of fuzzy matrix.

Proposition 3.1: For any fuzzy matrices A and B(1) A⊕B = B ⊕A,(2) A�B = B �A,Proposition 3.2: For any fuzzy matrices A,B and C(1) A⊕B ≥ A�B,(2) A⊕A ≥ A,(3) A�A ≤ A.(4) if A ≤ B, then A⊕ C ≤ B ⊕ C, A� C ≤ B � C.

Proof. (1)The ijth element of A⊕B is (aij + bij) ∧ 1, and

the ijth element of A � B is (aij + bij − 1) ∨ 0. Since

0 ≤ aij ≤ 1, 0 ≤ bij ≤ 1, then aij + bij − 1 ≤ 1, aij + bij −1 < aij + bij , therefore (aij + bij)∧ 1 ≥ (aij + bij − 1)∨ 0,hence A⊕B ≥ A�B.

(2)The ijth element of A⊕A is 2aij ∧1 ≥ aij . Therefore

A⊕A ≥ A.(3)The ijth element of A�A is aij + aij − 1 ∨ 0, aij +

(aij − 1) ≤ aij and 0 ≤ aij Therefore A�A ≤ A.Proposition 3.3: For any fuzzy matrices A,B and C(1)(A⊕B)⊕ C = A⊕ (B ⊕ C),(2)(A�B)� C = A� (B � C).

Proof. (1)Let dij , eij , fij , and gij be the ijth elements of the

fuzzy matrices A⊕B, B⊕C, (A⊕B)⊕C, and A⊕(B⊕C)respectively.

Then dij = (aij + bij) ∧ 1, eij = (bij + cij) ∧ 1,fij = dij ⊕ cij = (dij + cij) ∧ 1,gij = aij ⊕ eij = (aij + eij) ∧ 1.

Case1. aij + bij ≤ 1then fij = (dij + cij) ∧ 1 = (aij + bij + cij) ∧ 1.Subcase 1.1 aij+bij+cij ≤ 1, then fij = aij+bij+cij , in

this case bij+cij ≤ 1, then gij = aij⊕eij = (aij+eij)∧1 =(aij + bij + cij) ∧ 1 = aij + bij + cij = fij .

Fourth International Workshop on Advanced Computational Intelligence Wuhan, Hubei, China; October 19-21, 2011

978-1-61284-375-9/11/$26.00 @2011 IEEE 295

Subcase 1.2 aij + bij + cij > 1, then fij = 1. In this case

bij + cij ≤ 1 or > 1. If bij + cij ≤ 1, gij = aij ⊕ eij =(aij + eij) ∧ 1 = (aij + bij + cij) ∧ 1 = 1, If bij + cij > 1,gij = aij ⊕ eij = (aij + eij) ∧ 1 = (aij + 1) ∧ 1 = 1.Therefore gij = fij for all i, j.

Case2. aij + bij > 1fij = (dij + cij) ∧ 1 = (1 + cij) ∧ 1 = 1. In this case

aij + bij + cij > 1. the same as Subcase 1.2 , gij = 1.Therefore gij = fij for all i, j.Hence(A⊕B)⊕ C = A⊕ (B ⊕ C).

(2)Let dij , eij , fij , and gij be the ijth elements of the

fuzzy matrices A�B, B�C, (A�B)�C, and A�(B�C)respectively.

Then dij = (aij + bij − 1) ∨ 0, eij = (bij + cij − 1) ∨ 0,fij = dij � cij = (dij + cij − 1) ∨ 0,gij = aij � eij = (aij + eij − 1) ∨ 0.Case1. aij + bij − 1 ≤ 0fij = dij � cij = (0 + cij − 1) ∨ 0 = 0.Subcase 1.1 bij + cij − 1 ≤ 0, then gij = aij ⊕ eij =

aij ⊕ 0 = (aij + 0− 1) ∨ 0 = 0.Subcase 1.2 bij + cij − 1 > 0, then gij = aij ⊕ eij =

aij ⊕ (bij + cij − 1) = (aij + bij + cij − 1 − 1) ∨ 0, since

aij + bij − 1 < 0, and cij − 1 ≤ 0, then gij = 0. Therefore

gij = fij for all i, j.Case2. aij + bij − 1 > 0fij = dij � cij = (dij + cij − 1) ∨ 0 = (aij + bij − 1 +

cij − 1) ∨ 0 = (aij + bij + cij − 2) ∨ 0.Subcase 2.1 aij + bij + cij − 2 > 0, thenfij = aij +

bij + cij − 2, in this case bij + cij > 2− aij ≥ 1, therefore

gij = aij ⊕ eij = aij ⊕ (bij + cij − 1) = (aij + bij + cij −1− 1) ∨ 0 = aij + bij + cij − 2 = fij .

Subcase 2.2 aij + bij + cij − 2 ≤ 0, thenfij = 0.Subcase 2.2.1 bij + cij ≥ 1, then gij = aij ⊕ eij =

aij ⊕ (bij + cij − 1) = (aij + bij + cij − 1− 1) ∨ 0 = 0.Subcase 2.2.2 bij + cij < 1, then gij = aij ⊕ eij =

aij ⊕ 0 = (aij + 0− 1) ∨ 0 = 0.Therefore gij = hij for all i, j.

Hence (A�B)� C = A� (B � C).Proposition 3.4: For any two fuzzy matrices A,B

A�B ≤ A ∧B ≤ A ∨B ≤ A⊕BProof. Let dij , eij , fij , and gij be the ijth elements of the

fuzzy matrices A�B, A∧B, A∨B and A⊕B respectively.

Then dij = (aij + bij − 1) ∨ 0, gij = (aij + bij) ∨ 1.Since aij + bij − 1 ≤ aij , aij + bij − 1 ≤ bij , then

dij ≤ eij , for all i, j. Thus A�B ≤ A ∧B.Again aij + bij ≥ aij , aij + bij ≥ bij , then gij ≥ fij , for

all i, j. Thus A ∨B ≤ A⊕B.Proposition 3.5: For any fuzzy matrices A,B andC(1)A⊕ (B ∨ C) = (A⊕B) ∨ (A⊕ C),(2)A⊕ (B ∧ C) = (A⊕B) ∧ (A⊕ C),(3)A� (B ∨ C) = (A�B) ∨ (A� C),(4)A� (B ∧ C) = (A�B) ∧ (A� C).

Proof. (1)Let dij , eij , fij , gij and hij be the ijth elements

of the fuzzy matrices B ∨C, A⊕B, A⊕C, A⊕ (B ∨C)and (A⊕B) ∨ (A⊕ C) respectively.

Then dij = max{bij , cij},

eij = (aij + bij) ∧ 1, fij = (aij + cij) ∧ 1,gij = aij ⊕ dij = (aij + dij) ∧ 1,

hij = eij ∨ fij

=

{eij , eij ≥ fij ,fij , eij < fij .

=

{(aij + bij) ∧ 1, eij ≥ fij ,(aij + cij) ∧ 1, eij < fij .

If bij ≥ cij , then eij ≥ fij , dij = bij , therefore gij =(aij + bij) ∧ 1 = hij , for for all i, j

Again, if bij < cij , then eij < fij , dij = cij , therefore

gij = (aij + cij) ∧ 1 = hij , for all i, j.Hence A⊕ (B ∨ C) = (A⊕B) ∨ (A⊕ C).

(2)(3)(4)the same to(1).

Proposition 3.6: For any two fuzzy matrices A and B(1)(A⊕B)c = Ac �Bc,(2)(A�B)c = Ac ⊕Bc.

Proof. Let dij , and eij be the ijth elements of the fuzzy

matrices (A⊕B)c and Ac �Bc.dij = ((aij + bij) ∧ 1)c,eij = (1− aij)� (1− bij) = (1− aij +1− bij − 1)∨ 0 =

(1− aij − bij) ∨ 0.Case1. aij + bij ≤ 1then dij = (aij+bij)

c = 1−(aij+bij), eij = 1−aij−bij ,then dij = eij for all i, j.

Case2. aij + bij > 1then dij = (1)c = 1−1 = 0, eij = (1−aij)�(1−bij) =

(1− aij + 1− bij − 1) ∨ 0 = (1− aij − bij) ∨ 0 = 0, then

dij = eij for all i, j.Hence(A⊕B)c = Ac �Bc,

(2)Let fij and gij be the ijth elements of the fuzzy

matrices (A�B)c and Ac ⊕Bc.fij = (aij + bij − 1 ∨ 0)c

gij = (1 − aij) ⊕ (1 − bij) = (1 − aij + 1 − bij) ∧ 1 =(2− aij − bij) ∧ 1.

Case1. aij + bij ≥ 1then fij = (aij + bij − 1)c = 1 − (aij + bij − 1) =

2− aij − bij , gij = 2− aij − bij , then fij = gij for all i, j.Case2. aij + bij < 1then fij = (0)c = 1−0 = 1, gij = (2−aij−bij)∧1 = 1,

then fij = gij for all i, j.Hence (A�B)c = Ac ⊕Bc.

IV. RESULTS ON CUT SETS OF FUZZY MATRIX

In this section, some results on cut sets of fuzzy matrices

are presented.

Proposition 4.1: For any two fuzzy matrices A and B(1)(A⊕B)(α) ≥ A(α) ⊕B(α),(2)(A⊕B)(α) ≥ A(α) ⊕B(α),(3)(A⊕B)α ≥ Aα ⊕Bα.

Proof. (1)

(aij ⊕ bij)(α) = ((aij + bij) ∧ 1)(α)

=

{1, aij + bij ≥ α,0, aij + bij < α.

296

(aij)(α) ⊕ (bij)

(α) = ((aij)(α) + (bij)

(α)) ∧ 1Case1. aij + bij ≥ α

Subcase 1.1. aij ≥ α, bij ≥ α, therefore (aij)(α) ⊕

(bij)(α) = 1⊕ 1 = (1 + 1) ∧ 1 = 1.

Subcase 1.2. aij ≥ α, bij < α, therefore (aij)(α) ⊕

(bij)(α) = 1⊕ 0 = (1 + 0) ∧ 1 = 1.

Subcase 1.3. aij < α, bij ≥ α, therefore (aij)(α) ⊕

(bij)(α) = 0⊕ 1 = (0 + 1) ∧ 1 = 1.

Subcase 1.4. aij < α, bij < α, therefore (aij)(α) ⊕

(bij)(α) = 0⊕ 0 = (0 + 0) ∧ 1 = 0.

In this case (aij)(α) ⊕ (bij)

(α) ≤ 1, But (aij ⊕ bij)(α) = 1,

therefore (aij ⊕ bij)(α) ≥ (aij)

(α) ⊕ (bij)(α). for all i, j.

Hence (A⊕B)(α) ≥ A(α) ⊕B(α).Case 2. aij + bij < α

Since aij ≤ aij + bij ≤ α, and bij ≤ aij + bij ≤ α, then

(aij)(α) ⊕ (bij)

(α) = 0 ⊕ 0 = (0 + 0) ∧ 1 = 0. for all i, j.Hence (A⊕B)(α) = A(α) ⊕B(α).(2)

(aij ⊕ bij)(α) = ((aij + bij) ∧ 1)(α)

=

⎧⎨⎩

1, aij + bij ≥ 1,aij + bij , α < aij + bij < 1,

0, aij + bij ≤ α.

(aij)(α) ⊕ (bij)(α) = ((aij)(α) + (bij)(α)) ∧ 1Case1. aij + bij ≥ 1

Subcase 1.1. aij ≥ α, bij ≥ α, therefore (aij)(α) ⊕(bij)(α) = aij ⊕ bij = (aij + bij) ∧ 1 = 1.

Subcase 1.2. aij ≥ α, bij < α, therefore (aij)(α) ⊕(bij)(α) = aij ⊕ 0 = (aij + 0) ∧ 1 = aij .

Subcase 1.3. aij < α, bij ≥ α, therefore (aij)(α) ⊕(bij)(α) = 0⊕ bij = (0 + bij) ∧ 1 = bij .

Subcase 1.4. aij < α, bij < α, therefore (aij)(α) ⊕(bij)(α) = 0⊕ 0 = (0 + 0) ∧ 1 = 0.In this case (aij)(α) ⊕ (bij)(α) = 1 or 0 or aij or bij , But

(aij ⊕ bij)(α) = 1, therefore (aij ⊕ bij)(α) ≥ (aij)(α) ⊕(bij)(α). for all i, j. Hence (A⊕B)(α) ≥ A(α) ⊕B(α).Case 2. α < aij + bij < 1,

Subcase 2.1. aij ≥ α, bij ≥ α, therefore (aij)(α) ⊕(bij)(α) = aij ⊕ bij = (aij + bij) ∧ 1 = aij + bij .

Subcase 2.2. aij ≥ α, bij < α, therefore (aij)(α) ⊕(bij)(α) = aij ⊕ 0 = (aij + 0) ∧ 1 = aij .

Subcase 2.3. aij < α, bij ≥ α, therefore (aij)(α) ⊕(bij)(α) = 0⊕ bij = (0 + bij) ∧ 1 = bij .

Subcase 2.4. aij < α, bij < α, therefore (aij)(α) ⊕(bij)(α) = 0⊕ 0 = (0 + 0) ∧ 1 = 0.In this case (aij)(α) ⊕ (bij)(α) = aij + bij or 0 or aij or

bij , But (aij ⊕ bij)(α) = aij + bij , therefore (aij ⊕ bij)(α) ≥(aij)(α) ⊕ (bij)(α). for all i, j. Hence (A⊕B)(α) ≥ A(α) ⊕B(α).Case 3. aij + bij ≤ αIn this case aij ≤ aij+bij ≤ α, bij ≤ aij+bij ≤ α therefore

(aij ⊕ bij)(α) = (aij)(α) ⊕ (bij)(α) = 0. for all i, j. Hence

(A⊕B)(α) = A(α) ⊕B(α).

(3)

(aij ⊕ bij)α = ((aij + bij) ∧ 1)α

=

{aij + bij , aij + bij ≤ α,

1, aij + bij > α.

(aij)α ⊕ (bij)α = ((aij)α + (bij)α) ∧ 1Case1. aij + bij ≤ α

Since aij ≤ aij + bij ≤ α, and bij ≤ aij + bij ≤ α, then

(aij)α⊕ (bij)α = aij ⊕ bij = (aij + bij)∧ 1 = aij + bij . for

all i, j. Hence (A⊕B)α = Aα ⊕Bα.Case 2. aij + bij > α

Subcase 2.1. aij > α, bij > α, therefore (aij)α⊕(bij)α =1⊕ 1 = (1 + 1) ∧ 1 = 1.

Subcase 2.2. aij > α, bij ≤ α, therefore (aij)α⊕(bij)α =1⊕ bij = (1 + bij) ∧ 1 = 1.

Subcase 2.3. aij ≤ α, bij > α, therefore (aij)α⊕(bij)α =aij ⊕ 1 = (aij + 1) ∧ 1 = 1.

Subcase 2.4. aij < α, bij < α, therefore (aij)α⊕(bij)α =aij ⊕ bij = (aij + bij) ∧ 1 ≤ 1.In this case (aij)α ⊕ (bij)α ≤ 1, But (aij ⊕ bij)α = 1,therefore (aij ⊕ bij)α ≥ (aij)α ⊕ (bij)α for all i, j.Hence (A⊕B)α ≥ Aα ⊕Bα.

Proposition 4.2: For any two fuzzy matrices A and B(1)(A�B)α ≥ Aα �Bα,

(2)(A ∨B)α = Aα ∨Bα

Proof. (1) Case 1. aij > bij , then

(aij � bij)α = (aij)α

=

{aij , aij ≤ α,1, aij > α.

Subcase 1.1. aij > bij ≥ α, therefore (aij)α � (bij)α =1� 1 = 0.

Subcase 1.2. aij > α ≥ bij , therefore (aij)α � (bij)α =1� bij = 1.

Subcase 1.3. α ≥ aij > bij , therefore (aij)α � (bij)α =aij � bij = aij .Case 2. aij ≤ bij , then (aij � bij)α = (0)α = 0.

Subcase 2.1. aij ≤ bij < α, therefore (aij)α � (bij)α =aij � bij = 0.

Subcase 2.2. aij ≤ α < bij , therefore (aij)α � (bij)α =aij � 1 = 0.

Subcase 2.3. α < aij ≤ bij , therefore (aij)α � (bij)α =1� 1 = 0.Hence (aij � bij)α ≥ (aij)α� (bij)α, for all i, j. then (A�B)α ≥ Aα �Bα.(2) Case 1. aij ≥ bij , then

(aij ∨ bij)α = (aij)α

=

{aij , aij ≤ α,1, aij > α.

Subcase 1.1. aij ≥ bij > α, therefore (aij)α ∨ (bij)α =1 ∨ 1 = 1.

Subcase 1.2. aij > α ≥ bij , therefore (aij)α ∨ (bij)α =1 ∨ bij = 1.

Subcase 1.3. α ≥ aij ≥ bij , therefore (aij)α ∨ (bij)α =aij ∨ bij = aij .

297

Case 2. aij < bij ,the same to case 1.

Hence (aij ∨ bij)α = (aij)α ∨ (bij)α, for all i, j,then (A ∨B)α = Aα ∨Bα.

Proposition 4.3: For any two fuzzy matrices A and B

(1)(A⊕B)[λ] ≥ A[λ] ⊕B[λ],

(2)(A�B)[λ] ≥ A[λ] �B[λ],

(3)(A ∨B)[λ] = A[λ] ∨B[λ],

Proof. (1)

(aij ⊕ bij)[λ] = ((aij + bij) ∧ 1)[λ]

=

{1, λ+ aij + bij ≥ 1,0, λ+ aij + bij < 1.

a[λ]ij ⊕ b

[λ]ij = (a

[λ]ij + b

[λ]ij ) ∧ 1

Case1. λ+ aij + bij ≥ 1

Subcase 1.1. λ + aij ≥ 1, λ + bij ≥ 1, therefore a[λ]ij ⊕

b[λ]ij = 1⊕ 1 = (1 + 1) ∧ 1 = 1.

Subcase 1.2. λ + aij ≥ 1, λ + bij < 1, therefore a[λ]ij ⊕

b[λ]ij = 1⊕ 0 = (1 + 0) ∧ 1 = 1.

Subcase 1.3. λ + aij < 1, λ + bij ≥ 1, therefore a[λ]ij ⊕

b[λ]ij = 0⊕ 1 = (0 + 1) ∧ 1 = 1.

Subcase 1.4. λ + aij < 1, λ + bij < 1, therefore a[λ]ij ⊕

b[λ]ij = 0⊕ 0 = (0 + 0) ∧ 1 = 0.

In this case a[λ]ij ⊕ b

[λ]ij = 0 or 1, But (aij ⊕ bij)

[λ] = 1,

therefore (aij ⊕ bij)[λ] ≥ a

[λ]ij ⊕ b

[λ]ij . for all i, j.

Hence (A⊕B)[λ] ≥ A[λ] ⊕B[λ].

Case 2. λ+ aij + bij < 1,

Since λ + aij ≤ λ + aij + bij < 1, and λ + bij ≤ λ +

aij + bij < 1, then a[λ]ij ⊕ b

[λ]ij = 0 ⊕ 0 = (0 + 0) ∧ 1 = 0.

Hence (A⊕B)[λ] = A[λ] ⊕B[λ].(2) Case 1. aij > bij , then

(aij � bij)[λ] = (aij)

[λ]

=

{1, λ+ aij ≥ 1,0, λ+ aij < 1.

Since λ+ aij > λ+ bij , then

Subcase 1.1. λ+aij > λ+bij ≥ 1, therefore a[λ]ij �b

[λ]ij =

1� 1 = 0.

Subcase 1.2. λ+aij ≥ 1 > λ+bij , therefore a[λ]ij �b

[λ]ij =

1� 0 = 1.

Subcase 1.3. 1 > λ+aij > λ+bij , therefore a[λ]ij �b

[λ]ij =

0� 0 = 0.Case 2. aij ≤ bij , then

(aij � bij)[λ] = (0)[λ] =

{1, λ = 1,0, 0 ≤ λ < 1.

Since λ+ aij ≤ λ+ bij , then

Subcase 2.1. 1 ≤ λ+aij ≤ λ+bij , therefore a[λ]ij �b

[λ]ij =

1� 1 = 0.

Subcase 2.2. λ+aij < 1 ≤ λ+bij , therefore a[λ]ij �b

[λ]ij =

0� 1 = 0.

Subcase 2.3. λ+aij ≤ λ+bij < 1, therefore a[λ]ij �b

[λ]ij =

0� 0 = 0.Hence (aij � bij)

[λ] ≥ a[λ]ij � b

[λ]ij , for all i, j.

Then (A�B)[λ] ≥ A[λ] �B[λ].(3) Case 1. aij ≥ bij , then

(aij ∨ bij)[λ] = (aij)

[λ]

=

{1, λ+ aij ≥ 1,0, λ+ aij < 1.

Since λ+ aij ≥ λ+ bij , then

Subcase 1.1. λ + aij ≥ 1, λ + bij ≥ 1, therefore a[λ]ij ∨

b[λ]ij = 1 ∨ 1 = 1.

Subcase 1.2. λ + aij ≥ 1, λ + bij < 1, therefore a[λ]ij ∨

b[λ]ij = 1 ∨ 0 = 1.

Subcase 1.3. λ + aij < 1, λ + bij < 1, therefore a[λ]ij ∨

b[λ]ij = 0 ∨ 0 = 0.

Case 2. aij < bij , the same to case 1.

Hence (aij ∨ bij)[λ] = a

[λ]ij ∨ b

[λ]ij , for all i, j.

then (A ∨B)[λ] = A[λ] ∨B[λ].Proposition 4.4: For any two fuzzy matrices A and B(1)(A⊕B)[λ] ≥ A[λ] ⊕B[λ],

(2)(A�B)[λ] = A[λ] �B[λ],

(3)(A ∨B)[λ] = A[λ] ∨B[λ]

Proof. (1)

(aij ⊕ bij)[λ]

= ((aij + bij) ∧ 1)[λ]

=

⎧⎨⎩

1, aij + bij ≥ 1,0, aij + bij < λ+ aij + bij < 1,

aij + bij , aij + bij < 1 and λ+ aij + bij ≥ 1.

(aij)[λ] ⊕ (bij)[λ] = ((aij)[λ] + (bij)[λ]) ∧ 1Case1. aij + bij ≥ 1

Subcase 1.1. λ+aij ≥ 1, λ+bij ≥ 1, therefore (aij)[λ]⊕(bij)[λ] = aij ⊕ bij = (aij + bij) ∧ 1 = 1.

Subcase 1.2. λ+aij ≥ 1, λ+bij < 1, therefore (aij)[λ]⊕(bij)[λ] = aij ⊕ 0 = (aij + 0) ∧ 1 = aij .

Subcase 1.3. λ+aij < 1, λ+bij ≥ 1, therefore (aij)[λ]⊕(bij)[λ] = 0⊕ bij = (0 + bij) ∧ 1 = bij .

Subcase 1.4. λ+aij < 1, λ+bij < 1, therefore (aij)[λ]⊕(bij)[λ] = 0⊕ 0 = (0 + 0) ∧ 1 = 0.In this case (aij)[λ] ⊕ (bij)[λ] = 1 or 0 or aij or bij , But

(aij⊕bij)[λ] = 1, therefore (aij⊕bij)[λ] ≥ (aij)[λ]⊕(bij)[λ].for all i, j. Hence (A⊕B)[λ] ≥ A[λ] ⊕B[λ].

Case 2. aij + bij < λ+ aij + bij < 1,Since λ+aij ≤ λ+aij+bij < 1, and λ+bij ≤ λ+aij+

bij < 1, then (aij)[λ] ⊕ (bij)[λ] = 0⊕ 0 = (0 + 0) ∧ 1 = 0.Hence (A⊕B)[λ] = A[λ] ⊕B[λ].

Case 3. aij + bij < 1 and λ+ aij + bij ≥ 1Subcase 3.1. λ+aij ≥ 1, λ+bij ≥ 1, therefore (aij)[λ]⊕

(bij)[λ] = aij ⊕ bij = (aij + bij) ∧ 1 = aij + bij .Subcase 3.2. λ+aij ≥ 1, λ+bij < 1, therefore (aij)[λ]⊕

(bij)[λ] = aij ⊕ 0 = (aij + 0) ∧ 1 = aij .Subcase 3.3. λ+aij < 1, λ+bij ≥ 1, therefore (aij)[λ]⊕

(bij)[λ] = 0⊕ bij = (0 + bij) ∧ 1 = bij .Subcase 3.4. λ+aij < 1, λ+bij < 1, therefore (aij)[λ]⊕

(bij)[λ] = 0⊕ 0 = (0 + 0) ∧ 1 = 0.In this case (aij)[λ]⊕(bij)[λ] = aij+bij or 0 or aij or bij , But

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(aij⊕bij)[λ] = 1, Therefore (aij⊕bij)[λ] ≥ (aij)[λ]⊕(bij)[λ].Hence (A⊕B)[λ] ≥ A[λ] ⊕B[λ].(2) Case 1. aij > bij , then

(aij � bij)[λ] = (aij)[λ]

=

{aij , λ+ aij ≥ 1,0, λ+ aij < 1.

Since λ+ aij > λ+ bij , then

Subcase 1.1. λ+ aij > λ+ bij ≥ 1, therefore (aij)[λ] �(bij)[λ] = aij � bij = aij .

Subcase 1.2. λ+ aij ≥ 1 > λ+ bij , therefore (aij)[λ] �(bij)[λ] = aij � 0 = aij .

Subcase 1.3. 1 > λ+ aij > λ+ bij , therefore (aij)[λ] �(bij)[λ] = 0� 0 = 0.Case 2. aij ≤ bij , then

(aij � bij)[λ] = (0)[λ] = 0.Since λ+ aij ≤ λ+ bij , then

Subcase 2.1. 1 ≤ λ+ aij ≤ λ+ bij , therefore (aij)[λ] �(bij)[λ] = aij � bij = 0.

Subcase 2.2. λ+ aij < 1 ≤ λ+ bij , therefore (aij)[λ] �(bij)[λ] = 0� bij = 0.

Subcase 2.3. λ+ aij ≤ λ+ bij < 1, therefore (aij)[λ] �(bij)[λ] = 0� 0 = 0.Hence (aij � bij)[λ] = (aij)[λ] � (bij)[λ], for all i, j. then

(A�B)[λ] = A[λ] �B[λ].(3) Case 1. aij ≥ bij , then

(aij ∨ bij)[λ] = (aij)[λ]

=

{aij , λ+ aij ≥ 1,0, λ+ aij < 1.

Since λ+ aij ≥ λ+ bij , then

Subcase 1.1. λ+aij ≥ 1, λ+bij ≥ 1, therefore (aij)[λ]∨(bij)[λ] = aij ∨ bij = aij .

Subcase 1.2. λ+aij ≥ 1, λ+bij < 1, therefore (aij)[λ]∨(bij)[λ] = aij ∨ 0 = aij .

Subcase 1.3. λ+aij < 1, λ+bij < 1, therefore (aij)[λ]∨(bij)[λ] = 0 ∨ 0 = 0.Case 2. aij < bij ,the same to case 1.

Hence (aij ∨ bij)[λ] = (aij)[λ] ∨ (bij)[λ], for all i, j.Then (A ∨B)[λ] = A[λ] ∨B[λ].

Proposition 4.5: For any two fuzzy matrices A and B(1)(A⊕B)λ ≤ Aλ ⊕Bλ,

(2)(A�B)λ ≥ Aλ �Bλ,

(3)(A ∨B)λ = Aλ ∨Bλ,

Proof. (1)

(aij ⊕ bij)λ = ((aij + bij) ∧ 1)λ

=

{1, λ+ aij + bij ≥ 1,

aij + bij , λ+ aij + bij < 1.

aλij ⊕ bλij = (aλij ⊕ bλij) ∧ 1Case1. λ+ aij + bij ≥ 1

Subcase 1.1. λ+aij ≥ 1, λ+bij ≥ 1, therefore aλij⊕bλij =1⊕ 1 = (1 + 1) ∧ 1 = 1.

Subcase 1.2. λ+aij ≥ 1, λ+bij < 1, therefore aλij⊕bλij =1⊕ bij = (1 + bij) ∧ 1 = 1.

Subcase 1.3. λ+aij < 1, λ+bij ≥ 1, therefore aλij⊕bλij =aij ⊕ 1 = (aij + 1) ∧ 1 = 1.

Subcase 1.4. λ+aij < 1, λ+bij < 1, therefore aλij⊕bλij =aij ⊕ bij = (aij + bij) ∧ 1 ≤ 1.In this case aλij ⊕ bλij ≤ 1, but (aij ⊕ bij)

λ = 1, therefore

(aij ⊕ bij)λ ≥ aλij ⊕ bλij . for all i, j. Hence (A ⊕ B)λ ≥

Aλ ⊕Bλ.Case 2. λ+ aij + bij < 1,Since λ+aij ≤ λ+aij + bij < 1 and λ+ bij ≤ λ+aij +

bij < 1, then aλij⊕bλij = aij⊕bij = (aij+bij)∧1 = aij+bij .Hence (A⊕B)λ = Aλ ⊕Bλ.(2) Case 1. aij > bij , then

(aij � bij)λ = (aij)

λ

=

{aij , λ+ aij ≤ 1,1, λ+ aij > 1.

Since λ+ aij > λ+ bij , then

Subcase 1.1. λ+aij > λ+ bij > 1, therefore aλij � bλij =1� 1 = 0.

Subcase 1.2. λ+ aij > 1 ≥ λ+ bij therefore aλij � bλij =1� bij = 1.

Subcase 1.3. 1 ≥ λ+ aij > λ+ bij therefore aλij � bλij =aij � bij = aij .Case 2. aij ≤ bij , then (aij � bij)

λ = (0)λ = 0.Since λ+ aij ≤ λ+ bij , then

Subcase 2.1. 1 ≤ λ+aij ≤ λ+ bij , therefore aλij � bλij =1� 1 = 0.

Subcase 2.2. λ+aij < 1 ≤ λ+ bij , therefore aλij � bλij =aij � 1 = 0.

Subcase 2.3. λ+aij ≤ λ+ bij < 1, therefore aλij � bλij =aij � bij = 0.Hence (aij � bij)

λ ≥ aλij � bλij , for all i, j.Then (A�B)λ ≥ Aλ �Bλ.(3) Case 1. aij ≥ bij , then

(aij ∨ bij)λ = (aij)

λ

=

{aij , λ+ aij ≤ 1,1, λ+ aij > 1.

Since λ+ aij ≥ λ+ bij , then

Subcase 1.1. λ+aij ≥ 1, λ+bij ≥ 1, therefore aλij∨bλij =1 ∨ 1 = 1.

Subcase 1.2. λ+aij ≥ 1, λ+bij < 1, therefore aλij∨bλij =1 ∨ bij = 1.

Subcase 1.3. λ+aij < 1, λ+bij < 1, therefore aλij∨bλij =aij ∨ bij = aij .Case 2. aij < bij ,the same to case 1.

Hence (aij ∨ bij)λ = aλij ∨ bλij , for all i, j.

Then (A ∨B)λ = Aλ ∨Bλ.

REFERENCES

[1] A. K. Shyamal and M. Pal, “ Two new operators on fuzzy matrices,” J.Appl. Math. and Computing, vol. 15, no. 1-2, pp. 91-107, 2004.

[2] Yuan. X, Li. H, and E. Stanley Lee, “ Three new cut sets of fuzzy sets and new theories of fuzzy sets,” Computers and Mathematics with Application, vol. 57, pp. 691-701, 2009.

[3] Li. X, Yuan. X, and E.Stanley Lee, “ The three-dimensional fuzzy sets and their cut sets,” Computers and Mathematics with Application, vol. 58, pp. 1349-1359, 2009.

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