Notes on (T, S)-Intuitionistic Fuzzy Subhemirings of a Hemiring

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NOTES ON (T, S)-INTUITIONISTIC FUZZY SUBHEMIRINGS OF A HEMIRING K.Umadevi1, V.Gopalakrishnan2 1Assistant

Professor,Department of Mathematics,Noorul Islam University,Kumaracoil,Tamilnadu,India

2Research

Scolar,Department of Civil,Noorul Islam University,Kumaracoil,Tamilnadu,India

---------------------------------------------------------------------***--------------------------------------------------------------------fuzzy sets. The concept of intuitionistic fuzzy subset was Abstract - In this paper, we made an attempt to study the algebraic nature of a (T, S)-intuitionistic fuzzy subhemiring of a hemiring.2000 AMS Subject classification: 03F55, 06D72, 08A72.

Key

Words:

T-fuzzy

subhemiring,

anti

introduced by K.T.Atanassov[4,5], as a generalization of the notion of fuzzy set. The notion of anti fuzzy left h-ideals in hemiring was introduced by Akram.M and K.H.Dar [1]. The

S-fuzzy

notion of homomorphism and anti-homomorphism of fuzzy

subhemiring, (T, S)-intuitionistic fuzzy subhemiring,

and anti-fuzzy ideal of a ring was introduced by

product.

N.Palaniappan & K.Arjunan [16], [17], [18]. In this paper, we introduce the some Theorems in (T, S)-intuitionistic fuzzy

1. INTRODUCTION

subhemiring of a hemiring.

There are many concepts of universal algebras generalizing

2. PRELIMINARIES

an associative ring ( R ; + ; . ). Some of them in particular, nearrings and several kinds of semirings have been proven

2.1 Definition

very useful. Semirings (called also halfrings) are algebras ( R ; + ; . ) share the same properties as a ring except that ( R

A (T, S)-norm is a binary operations T: [0, 1][0, 1]  [0, 1]

; + ) is assumed to be a semigroup rather than a

and

commutative group. Semirings appear in a natural manner

following requirements;

in some applications to the theory of automata and formal

S: [0, 1][0, 1]  [0, 1] satisfying the

(i) T(0, x )= 0, T(1, x) = x (boundary condition)

languages. An algebra (R ; +, .) is said to be a semiring if (R ; +) and (R ; .) are semigroups satisfying a. ( b+c ) = a. b+a. c

(ii) T(x, y) = T(y, x) (commutativity)

and (b+c) .a = b. a+c. a for all a, b and c in R. A semiring R is (iii) T(x, T(y, z) )= T ( T(x,y), z )(associativity)

said to be additively commutative if a+b = b+a for all a, b and c in R. A semiring R may have an identity 1, defined by 1.

(iv)

a = a = a. 1 and a zero 0, defined by 0+a = a = a+0 and a.0

monotonicity).

= 0 = 0.a for all a in R. A semiring R is said to be a hemiring

(v) S(0, x) = x, S (1, x) = 1 (boundary condition)

if it is an additively commutative with zero. After the introduction of fuzzy sets by L.A.Zadeh[23], several

(vi) S(x, y ) = S (y, x )(commutativity)

researchers explored on the generalization of the concept of

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(vii) S (x, S(y, z) )= S ( S(x, y), z ) (associativity)

(iv) A(xy) ≤ S (A(x), A(y) ), for all x and y in R.

(viii) if x  y and w  z, then S (x, w )  S (y, z )(

2.5 Definition

monotonicity).

Let A and B be intuitionistic fuzzy subsets of sets G and

2.2 Definition

H, respectively. The product of A and B, denoted by AB, is defined as AB = {(x, y), AB(x, y), AB(x, y)  /

Let ( R, +, . ) be a hemiring. A fuzzy subset A of R is said

for all x in G and y in H }, where AB(x, y) = min {

to be a T-fuzzy subhemiring (fuzzy subhemiring with

A(x), B(y) } and AB(x, y) = max{ A(x), B(y) }.

respect to T-norm) of R if it satisfies the following

2.6 Definition

conditions: (i) A(x+y) ≥ T(A(x), A(y) ),

Let A be an intuitionistic fuzzy subset in a set S, the

(ii) A(xy) ≥ T(A(x), A(y) ), for all x and y in R.

strongest intuitionistic fuzzy relation on S, that is an intuitionistic fuzzy relation on A is V given by V(x, y)

2.3 Definition

= min{ A(x), A(y) } and V(x, y) = max{ A(x), A(y) }, for all x and y in S.

Let ( R, +, . ) be a hemiring. A fuzzy subset A of R is

2.7 Definition

said to be an anti S-fuzzy subhemiring (anti fuzzy subhemiring with respect to S-norm) of R if it satisfies

Let ( R, +, . ) and ( R , +, . ) be any two hemirings. Let f

the following conditions:

: R → R

(i) A(x+y) ≤ S(A(x), A(y) ),

intuitionistic fuzzy subhemiring in R, V be an

‫׀‬

(ii) A(xy) ≤ S(A(x), A(y) ), for all x and y in R.

‫׀‬

be any function and A be an (T, S)(T,

‫׀‬

S)-intuitionistic fuzzy subhemiring in f(R)= R , defined by V(y) =

2.4 Definition

sup

A(x) and V(y) =

inf

A(x), for

x f 1 ( y )

x f 1 ( y ) ‫׀‬

all x in R and y in R . Then A is called a preimage of V Let ( R, +, . ) be a hemiring. An intuitionistic fuzzy

under f and is denoted by f -1(V).

subset A of R is said to be an (T, S)-intuitionistic fuzzy

2.8 Definition

subhemiring(intuitionistic fuzzy subhemiring with respect to (T, S)-norm) of R if it satisfies the following

Let A be an (T, S)-intuitionistic fuzzy subhemiring of a

conditions: (i)

hemiring

A(x + y)  T (A(x), A(y) ),

(R, +, ∙ ) and a in R.

Then the pseudo (T, S)-intuitionistic fuzzy coset (aA)p is defined by ( (aA)p )(x) = p(a)A(x) and ((aA)p)(x) =

(ii) A(xy)  T (A(x), A(y) ),

p(a)A(x), for every x in R and for some p in P. (iii) A(x + y) ≤ S (A(x), A(y) ),

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3. PROPERTIES

Proof: It is trivial.

3.1 Theorem

2.3 Theorem

Intersection of any two (T, S)-intuitionistic fuzzy

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If A and B are any two (T, S)-intuitionistic fuzzy

subhemirings of a hemiring R is a (T, S)-intuitionistic

subhemirings of the hemirings R1 and R2 respectively,

fuzzy subhemiring of a hemiring R.

then AB is an (T, S)-intuitionistic fuzzy subhemiring

Proof: Let A and B be any two (T, S)-intuitionistic

of R1R2.

fuzzy subhemirings of a hemiring R and x and y in R.

Proof: Let A and B be two (T, S)-intuitionistic fuzzy

Let A = { ( x, A(x), A(x) ) / xR } and

B={

subhemirings of the hemirings R1 and R2 respectively.

(x, B(x), B(x) ) / xR } and also let C = AB = { (x,

Let x1 and x2 be in R1, y1 and y2 be in R2. Then ( x1, y1 )

C(x), C(x) ) / xR }, where min { A(x), B(x) } = C(x)

and (x2, y2) are in R1R2. Now, AB [(x1, y1)+(x2, y2) ] =

and max { A(x), B(x) } = C(x). Now, C(x+y) = min

AB (x1+x2, y1+y2) = min

{A(x+y), B(x+y)}≥ min{ T(A(x), A(y) ), T(B(x),



B(y)) }≥ T(min{ A(x), B(x) }, min { A(y), B(y) }) =

T(min{A(x1), B(y1)}, min{A(x2), B(y2)})= T(AB(x1,

T(C(x), C(y)). Therefore, C(x+y) ≥ T(C(x), C(y) ),

y1), AB(x2, y2)). Therefore, AB[(x1, y1) + (x2, y2)] 

for all x and y in R. And, C(xy) = min { A(xy), B(xy)}

T(AB (x1, y1), AB (x2, y2) ). Also, AB[(x1, y1) (x2, y2) ]

≥ min {T (A(x), A(y) ), T(B(x), B(y)) }≥ T( min {

= AB (x1x2, y1y2) = min { A( x1x2 ), B( y1y2) } min {T

A(x), B(x) }, min {A(y), B(y) }) = T (C(x), C(y) ).

(A(x1), A(x2) ), T (B(y1), B(y2) )} T(min{A(x1),

Therefore, C(xy) ≥ T(C(x), C(y)), for all x and y in R.

B(y1)}, min {A(x2), B(y2)}) = T(AB (x1, y1), AB (x2,

min{T(A(x1),

A(x2)),

{A(x1+x2 ), B(y1+y2)} T(B(y1),

B(y2))

}

y2) ). Therefore, AB[(x1, y1)(x2, y2)] T(AB(x1, y1),

Now, C( x+y ) = max { A(x+y), B(x+y) } ≤ max

AB(x2, y2)). Now, AB[(x1, y1) + (x2, y2) ] = AB(x1+x2,

{S(A(x), A(y) ), S(B(x), B(y))} ≤ S(max{ A(x), B(x) },

y1+ y2 ) = max {A(x1+x2 ), B( y1+ y2 )} ≤ max { S (A(x1),

max { A(y), B(y) }) = S (C(x), C(y) ). Therefore,

A(x2) ), S (B(y1), B(y2) ) }≤ S(max{A(x1), B(y1)},

C(x+y) ≤ S(C(x), C(y) ), for all x and y in R. And,

max{A(x2), B(y2) }) = S(AB (x1, y1), AB (x2, y2) ).

C(xy) = max {A(xy), B(xy)} ≤ max { S(A(x), A(y) ), S

Therefore, AB [(x1, y1)+(x2, y2)] ≤ S (AB (x1, y1), AB

(B(x), B(y))} ≤ S( max {A(x), B(x)}, max {A(y), B(y)

(x2, y2) ). Also, AB [ (x1, y1)(x2, y2) ] = AB ( x1x2, y1y2) =

}) = S(C(x), C(y) ). Therefore, C(xy) ≤ S (C(x), C(y) ),

max { A( x1x2 ), B( y1y2 ) }≤ max { S(A(x1), A(x2 ) ),

for all x and y in R. Therefore C is an (T, S)-

S(B(y1), B(y2)) }≤ S(max {A(x1), B(y1) }, max{A(x2),

intuitionistic fuzzy subhemiring of a hemiring R.

B(y2)}) = S(AB(x1, y1), AB(x2, y2) ).

Therefore, AB[(x1, y1)(x2, y2)] ≤ S(AB(x1, y1), AB(x2,

3.2 Theorem

y2)). Hence AB is an (T, S)-intuitionistic fuzzy

The intersection of a family of (T, S)-intuitionistic

subhemiring of hemiring of R1R2.

fuzzy subhemirings of hemiring R is an (T, S)intuitionistic fuzzy subhemiring of a hemiring R.

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3.4 Theorem

3.6 Theorem

If A is a (T, S)-intuitionistic fuzzy subhemiring of a hemiring

Let A and B be two intuitionistic fuzzy subsets of the

(R, +, ∙), then A(x) A(0) and A(x) ≥ A(0), for x in R, the

hemirings R1 and R2 respectively and AB is an (T, S)-

zero element 0 in R.

intuitionistic fuzzy subhemiring of R1R2. Then the following are true:

Proof: For x in R and 0 is the zero element of R. Now, A(x) = A(x+0)  T(A(x), A(0)), for all x in R. So, A(x)  A(0) is

(i) if A(x)  B(0‫ ) ׀‬and A(x)  B(0‫) ׀‬, then A is an (T,

only possible. And A(x) = A(x +0)  S( A(x), A(0) ) for all x

S)-intuitionistic fuzzy subhemiring of R1.

in R. So, A(x) ≥ A(0) is only possible.

(ii) if B(x)  A(0) and B(x)  A(0), then B is an (T,

3.5 Theorem

S)-intuitionistic fuzzy subhemiring of R2.

Let A and B be (T, S)-intuitionistic fuzzy subhemiring

(iii) either A is an (T, S)-intuitionistic fuzzy

of the hemirings R1 and R2 respectively. Suppose that 0

subhemiring of R1 or B is an

and 0‫ ׀‬are the zero element of R1 and R2 respectively.

intuitionistic fuzzy subhemiring of R2.

If AB is an (T, S)-intuitionistic fuzzy subhemiring of

(T, S)-

Proof: Let AB be an (T, S)-intuitionistic fuzzy

R1R2, then at least one of the following two

subhemiring of R1R2 and x and y in R1and 0‫ ׀‬in R2.

statements must hold. (i) B(0‫ ) ׀‬ A(x) and B(0‫ ) ׀‬

Then (x, 0‫ ) ׀‬and (y, 0‫ ) ׀‬are in R1R2. Now, using the

A(x), for all x in R1, (ii) A(0)  B(y) and A(0 ) 

property that A(x)  B(0‫ ) ׀‬and A(x)  B(0‫) ׀‬, for all x

B(y), for all y in R2.

in R1. We get, A(x+y) = min{ A(x+y), B(0‫ ׀‬+0‫=}) ׀‬

Proof: Let AB be an (T, S)-intuitionistic fuzzy

AB((x+y), (0‫ ׀‬+0‫ = )) ׀‬AB[ (x, 0‫ ) ׀‬+(y, 0‫ ]) ׀‬ T(AB(x,

subhemiring of R1R2. By contraposition, suppose that

0‫) ׀‬, AB(y, 0‫ = )) ׀‬T(min{A(x), B(0‫}) ׀‬, min{A(y),

none of the statements (i) and (ii) holds. Then we can

B(0‫ = )} ) ׀‬T(A(x), A(y)). Therefore, A(x+y) 

find a in R1 and b in R2 such that A(a)  B(0‫) ׀‬, A(a) 

T(A(x), A(y)), for all x and y in R1. Also, A(xy) =

B(0‫ ) ׀‬and

B(b)  A(0), B(b)  A(0). We have,

min{A(xy), B(0‫ ׀‬0‫ = }) ׀‬AB( (xy), (0‫ ׀‬0‫ = ) ) ׀‬AB[(x,

AB(a, b) = min{A(a), B(b)} min {B(0‫) ׀‬, A(0) }=

0‫() ׀‬y, 0‫ ] ) ׀‬ T(AB(x, 0‫) ׀‬, AB(y, 0‫ = )) ׀‬T(min{ A(x),

min { A(0), B(0‫ =} ) ׀‬AB (0, 0‫) ׀‬. And, AB (a, b) =

B(0‫}) ׀‬,

max{A(a), B(b) } max {B(0‫) ׀‬, A(0) }= max{A(0),

Therefore, A(xy)  T(A(x), A(y) ), for all x and y in

B(0‫ =} ) ׀‬AB(0, 0‫) ׀‬. Thus AB is not an (T, S)-

R1. And, A(x+y) = max{A(x+y), B(0‫ ׀‬+0‫ =}) ׀‬AB(

intuitionistic fuzzy subhemiring of R1R2. Hence either

(x+y), (0‫ ׀‬+0‫ = )) ׀‬AB[ (x, 0‫) ׀‬+( y, 0‫ ]) ׀‬ S(AB(x, 0‫) ׀‬,

B(0‫ ) ׀‬ A(x) and B(0‫ ) ׀‬ A(x), for all x in R1 or

AB(y, 0‫ = )) ׀‬S(max{A(x), B(0‫} ) ׀‬, max

A(0) B(y) and A(0) B(y), for all y in R2.

B(0‫ = )}) ׀‬S(A(x), A(y) ). Therefore, A(x+y )  S

min{A(y),

B(0‫)}) ׀‬

=

T(A(x),

A(y)).

{A(y),

(A(x), A(y) ), for all x and y in R1. Also, A(xy)= max{A(xy), B(0‫ ׀‬0‫ =} ) ׀‬AB ( (xy), (0‫ ׀‬0‫ = )) ׀‬AB[(x,

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0‫( ) ׀‬y, 0‫ ] ) ׀‬ S(AB(x, 0‫) ׀‬, AB(y, 0‫ = )) ׀‬S( max{A(x),

of R if and only if V is an (T, S)-intuitionistic fuzzy

B(0‫} ) ׀‬, max{A(y), B(0‫ = )}) ׀‬S(A(x), A(y)).

subhemiring

Therefore, A( xy )  S (A(x), A(y) ), for all x and y in R 1.

Hence

A

is

an

(T,

S)-intuitionistic

Proof: Suppose that A is an (T, S)-intuitionistic fuzzy

fuzzy

subhemiring of a hemiring R. Then for any x = (x1, x2)

subhemiring of R1. Thus (i) is proved. Now, using the

and y = (y1, y2) are in RR. We have, V(x+y) = V[(x1,

property that B(x)  A(0) and B(x)  A(0), for all x

x2) + (y1, y2)] = V(x1+y1, x2+y2)

in R2, let x and y in R2 and 0 in R1. Then (0, x) and (0, y)

=

min{A(x1+y1),

A(x2+y2)} min {T(A(x1), A(y1)), T(A(x2), A(y2) ) }

are in R1R2.We get, B(x+y) = min{B(x+y), A(0+0) }=

T(min {A(x1), A(x2)}, min {A(y1), A(y2)}) = T(V (x1,

min{A(0+0), B(x+y) }= AB( (0+0), (x+y)) = AB[(0,

x2), V (y1, y2)) = T(V (x), V (y) ). Therefore, V(x+y) 

x)+(0, y)]  T(AB(0, x), AB(0, y)) = T(min{A(0),

T(V(x), V(y)), for all x and y in RR. And,

B(x) }, min{A(0), B(y)) = T(B(x), B(y) ). Therefore,

V(xy)=V[(x1,

B(x+ y )  S (B(x), B(y) ), for all x and y in R2. Also,

x2)(y1,

min{A(x1y1), A(x2y2)}

B(xy) = min{B(xy), A(00 ) }= min{A(00), B(xy)}=

y2)]

=

V(x1y1,

x2 y 2 )

=

 min { T(A(x1),

A(y1)), T(A(x2), A(y2)) } T(min{ A(x1), A(x2)},

AB((00), (xy))= AB[(0, x)(0, y )]  T(AB(0, x ),

min{A(y1), A(y2)}) = T(V(x1, x2), V (y1, y2) ) =

AB(0, y) ) = T(min{ A(0), B(x) }, min{ A(0 ), B(y) })

T(V(x), V(y) ). Therefore, V(xy)  T(V(x), V(y)), for

= T(B(x), B(y) ). Therefore, B(xy)  T(B(x), B(y) ),

all x and y in RR. We have, V(x+y) = V [(x1, x2) + (y1,

for all x and y in R2. And, B(x+y) = max{B(x+y),

y2)] = V(x1+y1, x2+ y2 ) = max { A(x1+y1), A(x2+y2) }≤

A(0+0 )}= max{A(0+0), B(x+y)}= AB((0+0), (x+y))

max { S(A(x1), A(y1) ), S(A(x2), A(y2))} ≤ S(max

= AB[(0, x)+(0, y)]  S(AB(0, x), AB(0, y ) ) = S (

{A(x1), A(x2)}, max {A(y1), A(y2)})= S(V(x1, x2),

max{A(0), B(x) }, max{A(0), B(y) } ) = S (B(x), B(y)

V(y1, y2)) = S(V(x), V (y) ). Therefore, V(x+y) ≤ S(V

). Therefore, B( x+y )  S (B(x), B(y) ), for all x and y

(x), V (y) ), for all x and y in RR. And, V(xy) = V[(x1,

in R2. Also, B(xy ) = max{B(xy), A(00)}= max{A(00),

x2) (y1, y2)] = V( x1y1 , x2y2 ) = max {A(x1y1), A(x2y2)

B(xy)}= AB((00), (xy) ) = AB[(0, x )(0, y)]  S(AB(0,

}≤ max {S (A(x1), A(y1) ), S (A(x2), A(y2)) }≤ S( max {

x), AB( 0, y)) = S(max{ A(0), B(x)}, max{ A(0),

A(x1), A(x2) }, max{ A(y1 ), A(y2) } ) = S(V(x1, x2),

B(y)}) = S(B(x), B(y)). Therefore, B(xy)  S(B(x),

V(y1, y2) ) = S(V (x), V (y)). Therefore, V(xy) ≤

B(y)), for all x and y in R2. Hence B is an (T, S)-

S(V(x), V(y)), for all x and y in RR. This proves that

intuitionistic fuzzy subhemiring of a hemiring R2. Thus

V is an (T, S)-intuitionistic fuzzy subhemiring of RR.

(ii) is proved. (iii) is clear.

Conversely assume that V is an (T, S)-intuitionistic

3.7 Theorem

fuzzy subhemiring of RR, then for any x = (x1, x2) and y = (y1, y2) are in RR, we have min{ A(x1 + y1), A(x2+

Let A be an intuitionistic fuzzy subset of a hemiring R

y2) } = V( x1+ y1 , x2+ y2 ) = V [(x1, x2) + (y1, y2)] = V

and V be the strongest intuitionistic fuzzy relation of

(x+y)  T(V (x), V(y)) = T(V (x1, x2), V(y1, y2)) = T(

R. Then A is an (T, S)-intuitionistic fuzzy subhemiring

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we get, A(x1+ y1)  T (A(x1), A(y1) ), for all x1 and y1

(iii) if A(x+y)=1, then either A (x)= 1or A (y) = 1,for

in R. And, min { A(x1y1), A(x2y2)} = V(x1y1, x2y2) =

all x and y in R.

V[(x1, x2)(y1, y2)] = V(xy)  T(V(x), V(y)) = T(V(x1,

(iv) if A(xy) = 1, then either A (x)= 1or A (y) = 1,for

x2), V(y1, y2)) = T(min{A(x1), A(x2)}, min {A(y1),

all x and y in R.

A(y2) }). If x2 =0, y2=0, we get, A(x1y1)  T (A(x1), A(y1) ), for all x1 and y1 in R.

Proof: It is trivial.

We have, max { A(x1+ y1), A(x2+ y2)}= V( x1+ y1, x2+

3.10 Theorem

y2) = V [(x1, x2)+ (y1, y2)] = V (x+y) ≤ S (V (x), V (y)) =

If A is an (T, S)-intuitionistic fuzzy subhemiring of a

S(V(x1, x2), V(y1, y2)) = S(max{ A(x1), A(x2)}, max

hemiring

{A(y1), A(y2) }). If x2 =0, y2=0, we get, A(x1+y1) ≤

A(x)  1and A(x) = 0} is either empty or a

S(A(x1), A(y1) ), for all x1 and y1 in R.

Proof: It is trivial.

x2)(y1, y2)] = V(xy) ≤ S(V(x), V(y) ) = S(V(x1, x2),

3.11 Theorem

V(y1, y2)) = S(max {A(x1), A(x2)}, max {A(y1), A(y2) }). If x2 = 0, y2 = 0, we get A(x1y1) ≤ S(A(x1), A(y1)),

If A is an (T, S)-intuitionistic fuzzy subhemiring of a

for all x1 and y1 in R. A

is

hemiring an

(T,

S)-intuitionistic

fuzzy

R.

3.8 Theorem

Proof: It is trivial.

If A is an (T, S)-intuitionistic fuzzy subhemiring of a

3.12 Theorem

(R, +, . ), then H = { x / xR: A(x) =

If A is an (T, S)-intuitionistic fuzzy subhemiring of a

1, A(x) = 0} is either empty or is a subhemiring of R.

hemiring

Proof: It is trivial.

( R, +, . ), then H = {  x, A(x)  : 0 <

A(x)  1} is either empty or an anti S-fuzzy subhemiring of R.

3.9 Theorem

Proof: It is trivial.

If A be an (T, S)-intuitionistic fuzzy subhemiring of a hemiring

(R, +, .) then H = {  x, A(x)  : 0 <

A(x)  1} is either empty or a T-fuzzy subhemiring of

subhemiring of R.

hemiring

3.13 Theorem

(R, +, . ), then (i) if A(x+y)=0, then

either A(x)= 0 or A(y) = 0,for all x and y in R.

If A is an (T, S)-intuitionistic fuzzy subhemiring of a (R,+, . ), then A is an (T, S)-

(ii) if A(xy) = 0, then either A(x) = 0 or A(y) = 0,for

hemiring

all x and y in R.

intuitionistic fuzzy subhemiring of R.

© 2016, IRJET

T-

fuzzy subhemiring of R.

And, max {A(x1y1), A(x2y2)} = V(x1y1, x2y2) = V[(x1,

Therefore

(R, +, . ), then H = {  x, A(x)  : 0 <

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Proof: Let A be an (T, S)-intuitionistic fuzzy

T(B(x), B(y)). Therefore, B(x+y) ≥ T(B(x), B(y)),

subhemiring of a hemiring R. Consider A = {  x, A(x),

for all x and y in R. And A(xy) ≤ S(A(x), A(y)), for all x

A(x)  }, for all x in R, we take A=B ={ x, B(x), B(x) 

and y in R, which implies that 1–B(xy) ≤ S((1–B(x)),

}, where B(x) = A(x), B(x) = 1 A(x). Clearly,

(1–B(y) ) ), which implies that B(xy) ≥ 1– S( (1–

B(x+y) ≥ T(B(x), B(y) ), for all x and y in R and

B(x)), (1–B(y))) ≥ T(B(x), B(y)). Therefore, B(xy)≥

B(xy) ≥ T(B(x), B(y) ), for all x and y in R. Since A is

T(B(x), B(y) ), for all x and y in R. Hence B = A is an

an

(T, S)-intuitionistic fuzzy subhemiring of a hemiring R.

(T, S)-intuitionistic fuzzy subhemiring

of R, we have A(x+y ) ≥ T(A(x), A(y) ), for all x and y

3.15 Theorem

in R, which implies that 1– B(x+y) ≥ T ( ( 1– B(x) ), ( 1– B(y) ) ), which implies that B( x+y ) ≤ 1– T ( ( 1–

Let ( R, +, . ) be a hemiring and A be a non empty

B(x) ), ( 1– B(y) ) ) ≤ S (B(x), B(y) ). Therefore,

subset of R. Then A is a subhemiring of R if and only if

B(x+y) ≤ S (B(x), B(y) ), for all x and y in R. And

B =   A ,  A  is an (T, S)-intuitionistic fuzzy

A(xy) ≥ T (A(x), A(y) ), for all x and y in R, which

subhemiring of R, where

implies that 1– B(xy) ≥ T((1– B(x)), (1– B(y)) )

A

is the characteristic

function. which implies that B( xy ) ≤ 1– T ( ( 1– B(x) ), ( 1–

Proof: It is trivial.

B(y) ) ) ≤ S( B(x), B(y) ). Therefore, B(xy) ≤ S(B(x), B(y) ), for all x and y in R. Hence B = A is an

3.16 Theorem

(T, S)-intuitionistic fuzzy subhemiring of a hemiring R.

Let A be an (T, S)-intuitionistic fuzzy subhemiring of a

3.14 Theorem

hemiring H and f is an isomorphism from a hemiring R onto H. Then A◦f is an

If A is an (T, S)-intuitionistic fuzzy subhemiring of a (R, +, .), then A is an (T, S)-

hemiring

(T, S)-

intuitionistic fuzzy subhemiring of R.

intuitionistic fuzzy subhemiring of R.

Proof: Let x and y in R and A be an (T, S)-intuitionistic

Proof: Let A be an (T, S)-intuitionistic fuzzy

fuzzy subhemiring of a hemiring H. Then we have,

subhemiring of a hemiring R.That is A = {  x, A(x),

(A◦f )(x+y) = A(f(x+y)) = A(f(x)+f(y)) ≥ T(A(f(x) ),

A(x)  }, for all x in R. Let A =B= {  x, B(x), B(x)  },

A( f(y) ) ) = T((A◦f)(x), (A◦f)(y)), which implies that

where B(x) =1A(x), B(x) = A(x). Clearly, B(x+y) ≤

(A◦f)(x+y) ≥ T((A◦f)(x), (A◦f)(y) ). And, (A◦f)(xy) =

S (B(x), B(y) ), for all x and y in R and B(xy)≤ S(B(x),

A(f(xy)) = A(f(x)f(y)) ≥ T(A(f(x)), A(f(y))) =

B(y)), for all x and y in R. Since A is an (T, S)-

T((A◦f)(x), (A◦f)(y) ), which implies that (A◦f)(xy) ≥

intuitionistic fuzzy subhemiring of R, we have A(x+y)

T( (A◦f )(x), (A◦f)(y) ). Then we have, (A◦f )(x+y) =

≤ S ( A(x), A(y) ), for all x and y in R, which implies

A( f(x+y)) = A(f(x)+f(y)) ≤ S(A(f(x) ), A(f(y)))

that 1–B(x+y) ≤ S((1– B(x) ), (1– B(y))) which

= S((A◦f)(x), (A◦f)(y) ), which implies that (A◦f )(x+y)

implies that B(x+y) ≥ 1– S((1– B(x)), (1– B(y))) ≥

≤ S((A◦f)(x), (A◦f )(y) ). And (A◦f)(xy) = A(f(xy)) =

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A(f(x)f(y)) ≤ S( A(f(x) ), A( f(y))) = S ( (A◦f )(x),

(T, S)-intuitionistic fuzzy subhemiring of a hemiring R,

(A◦f)(y)), which implies that (A◦f )(xy) ≤ S( (A◦f )(x),

for every a in R.

(A◦f )(y) ). Therefore (A◦f) is an (T, S)-intuitionistic

Proof: Let A be an (T, S)-intuitionistic fuzzy

fuzzy subhemiring of a hemiring R.

subhemiring of a hemiring R.

3.17 Theorem

For every x and y in R, we have, ( (aA)p )(x + y) =

Let A be an (T, S)-intuitionistic fuzzy subhemiring of a

p(a)A( x+ y ) ≥ p(a) T ( ( A(x), A(y) ) = T ( p(a)A(x),

hemiring H and f is an anti-isomorphism from a

p(a)A(y) ) = T ( ( (aA)p )(x), ( (aA)p )(y) ). Therefore,

hemiring R onto H. Then A◦f is an

( (aA)p )(x+y) ≥ T ( ( (aA)p )(x), ( (aA)p )(y) ). Now, (

(T,

(aA)p )(xy) = p(a)A( xy ) ≥ p(a) T (A(x), A(y) ) = T (

S)-intuitionistic fuzzy subhemiring of R.

p(a)A(x), p(a)A(y) ) = T ( ( (aA)p )(x), ( (aA)p )(y) ).

Proof: Let x and y in R and A be an (T, S)-intuitionistic

Therefore, ( (aA)p )(xy) ≥ T ( ( (aA)p )(x), ( (aA)p )(y)

fuzzy subhemiring of a hemiring H. Then we have,

). For every x and y in R, we have, ( (aA)p )(x+y) =

(A◦f )( x+ y) = A(f(x+y)) = A(f(y)+f(x)) ≥ T(A(f(x) ),

p(a)A(x+y ) ≤ p(a) S ( (A(x), A(y) ) = S ( p(a)A(x),

A( f(y)) ) = T( (A◦f )(x), (A◦f )(y)), which implies that

p(a)A( y ) ) = S ( ( (aA)p )( x ), ( (aA)p )(y) ).

(A◦f )(x+y) ≥ T( A◦f )(x), (A◦f )(y) ). And, (A◦f)(xy) =

Therefore,( (aA)p )( x + y) ≤

A(f(xy)) = A( f(y)f(x)) ≥ T(A(f(x)), A( f(y))) =

S ( ( (aA)p)(x),

( (aA)p )(y) ). Now, ( (aA)p )( xy ) = p(a) A( xy ) ≤ p(a)

T((A◦f )(x), (A◦f )(y)), which implies that (A◦f)(xy) ≥

S (A(x), A(y) ) = S ( p(a) A(x), p(a) A(y) ) = S ( (

T( ( A◦f ) (x), ( A◦f ) (y) ). Then we have, (A◦f)(x+y) =

(aA)p )(x), ( (aA)p )(y) ). Therefore,

A(f(x+y))= A( f(y)+f(x) ) ≤ S ( A( f(x) ), A( f(y) ) ) =

( (aA)p )( xy )

≤ S ( ( (aA)p )(x), ( (aA)p )(y) ). Hence (aA)p is an (T,

S((A◦f )(x), (A◦f )(y)), which implies that (A◦f)(x+y) ≤

S)-intuitionistic fuzzy subhemiring of a hemiring R.

S((A◦f )(x), (A◦f )(y)).

And,(A◦f)(xy) = A(f(xy)) = A(f(y)f(x)) ≤ S( A(f(x)),

3.19 Theorem

A(f(y)) ) = S ( (A◦f )(x), (A◦f )(y) ) ,which implies that (A◦f )(xy) ≤ S ( (A◦f ) (x), (A◦f ) (y) ). Therefore A◦f is

Let ( R, +, . ) and ( R , +, .) be any two hemirings. The

an (T, S)-intuitionistic fuzzy subhemiring of the

homomorphic image of an (T, S)-intuitionistic fuzzy

hemiring R.

subhemiring of R is an (T, S)-intuitionistic fuzzy

‫׀‬

‫׀‬

subhemiring of R .

3.18 Theorem ‫׀‬

Proof: Let ( R, +, . ) and ( R , +, . ) be any two Let A be an (T, S)-intuitionistic fuzzy subhemiring of a

hemirings. Let f : R  R be a homomorphism. Then, f

hemiring

(R, +, . ), then the pseudo (T, S)-

(x+y) = f(x) + f(y) and f(xy) = f(x) f(y), for all x and y in

intuitionistic

fuzzy

R. Let V = f(A), where A is an (T, S)-intuitionistic fuzzy

coset

(aA)p

is

‫׀‬

an

subhemiring of R. We have to prove that V is an (T, S)-

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‫׀‬

intuitionistic fuzzy subhemiring of R . Now, for f(x),

p-ISSN: 2395-0072

A(y) ), since v(f(x)) = A(x) which implies that A(xy) ≤

‫׀‬

f(y) in R , v( f(x) + f(y)) = v( f(x+y) ) ≥ A(x + y) ≥ T

S (A(x), A(y) ). Hence A is an

(A(x), A(y) ) which implies that v(f(x) + f(y)) ≥ T

intuitionistic fuzzy subhemiring of R.

(v( f(x) ), v( f(y) ) ). Again, v( f(x)f(y) ) = v( f(xy) ) ≥ A(xy)

(T, S)-

3.21 Theorem

≥ T ( A(x), A(y) ),which implies that

‫׀‬

v(f(x)f(y)) ≥ T ( v(f(x) ), v( f(y) ) ).

Let ( R, +, . ) and ( R , +, . ) be any two hemirings. The anti-homomorphic image of an (T, S)-intuitionistic

‫׀‬

Now,for f(x),f(y) in R , v( f(x)+f(y) ) = v( f(x+y) ) ≤

fuzzy subhemiring of R is an

A(x+ y) ≤ S ( A(x), A(y) ), v(f(x) + f(y)) ≤ S ( v( f(x)

(T, S)‫׀‬

intuitionistic fuzzy subhemiring of R .

), v( f(y) ) ).

‫׀‬

Proof: Let ( R, +, . ) and ( R , +, . ) be any two

Again, v( f(x)f(y) ) = v( f(xy) ) ≤ A(xy) ≤ S (A(x),

‫׀‬

hemirings. Let f : R  R be an anti-homomorphism.

A(y)), which implies that v( f(x)f(y) ) ≤ S (v( f(x) ),

Then, f (x+y) = f (y) + f (x) and f(xy) = f(y) f(x), for all x

v( f(y) ) ). Hence V is an (T, S)-intuitionistic fuzzy

and y in R. Let V = f(A), where A is an (T, S)-

‫׀‬

subhemiring of R .

intuitionistic fuzzy subhemiring of R. We have to prove that V is an (T, S)-intuitionistic fuzzy

3.20 Theorem

‫׀‬

subhemiring ‫׀‬

‫׀‬

of R . Now, for f(x), f(y) in R ,

Let ( R, +, . ) and ( R , +, . ) be any two hemirings. The

v(f(x)+f(y)) = v(f(y+x) ) ≥ A(y+x) ≥ T(A(y), A(x)) =

homomorphic preimage of an (T, S)-intuitionistic

T(A(x), A(y)), which implies that v( f(x) + f(y)) ≥

‫׀‬

fuzzy subhemiring of R is a

(T, S)-

T(v( f(x) ), v( f(y))). Again, v( f(x)f(y)) = v(f(yx) ) ≥

intuitionistic fuzzy subhemiring of R.

A(yx) ≥ T(A(y), A(x)) = T(A(x), A(y)), which implies that v(f(x)f(y)) ≥ T(v(f(x)), v(f(y))). Now for

Proof: Let V = f(A), where V is an (T, S)-intuitionistic

‫׀‬

fuzzy subhemiring of R . We have to prove that A is an

f(x), f(y) in R , v(f(x)+f(y)) = v(f(y+x)) ≤ A(y+ x ) ≤ S(

(T, S)-intuitionistic fuzzy subhemiring of R. Let x and y in

A(y), A(x) ) = S(A(x), A(y) ), which implies that v(

R. Then, A(x+y) = v(f(x+y)) = v(f(x)+f(y)) ≥ T(v( f(x)

f(x)+f(y)) ≤ S(v( f(x) ), v( f(y) ) ).

‫׀‬

), v(f(y)) ) = T(A(x), A(y) ), since v(f(x)) = A(x),

Again, v( f(x)f(y) ) = v( f(yx)) ≤ A(yx) ≤ S(A(y),

which implies that A(x+y) ≥ T(A(x), A(y)). Again,

A(x) ) = S(A(x), A(y)), which implies that v( f(x)f(y)

A(xy) = v(f(xy) ) = v( f(x)f(y)) ≥ T ( v( f(x) ), v( f(y)

) ≤ S (v( f(x) ), v(f(y)) ). Hence V is an (T, S)-

) ) = T (A(x), A(y)), since v(f(x)) = A(x) which implies

‫׀‬

intuitionistic fuzzy subhemiring of R .

that A(xy) ≥ T(A(x), A(y)). Let x and y in R. Then,

3.22 Theorem

A(x+y) = v( f(x+y) ) = v( f(x)+f(y) ) ≤ S(v( f(x) ), v(f(y))) = S(A(x), A(y) ), since v(f(x)) = A(x) which

‫׀‬

Let ( R, +, . ) and ( R , +, . ) be any two hemirings. The

implies that A(x+ y) ≤ S(A(x), A(y) ). Again, A(xy) =

anti-homomorphic preimage of an (T, S)-intuitionistic

v(f(xy) ) = v( f(x)f(y) ) ≤ S(v( f(x) ), v( f(y))) = S(A(x),

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fuzzy subhemiring of R is an

www.irjet.net (T, S)-

p-ISSN: 2395-0072 6.

intuitionistic fuzzy subhemiring of R. 7.

Proof: Let V = f(A), where V is an (T, S)-intuitionistic ‫׀‬

fuzzy subhemiring of R . We have to prove that A is an

8.

(T, S)-intuitionistic fuzzy subhemiring of R. Let x and y in R. Then, A(x+y) = v(f(x+y)) = v( f(y)+f(x)) ≥ T(v(f(y)), v( f(x))) = T(v(f(x)), v( f(y))) = T(A(x),

9.

A(y)), which implies that A(x+y) ≥ T(A(x), A(y) ).

Again, A(xy) = v(f(xy)) = v(f(y)f(x)) ≥ T(v(f(y)), v(f(x))) = T(v( f(x) ), v( f(y) )) = T(A(x), A(y)), since v(f(x)) = A(x) which implies that A(xy) ≥ A(y)).

T(A(x),

Then,

A(x+y)

=

v(f(x+y))

10.

=

v(f(y)+f(x) ) ≤ S(v(f(y) ), v(f(x))) = S(v(f(x)),

11.

v(f(y))) = S(A(x), A(y)) which implies that A(x+y) ≤ S(A(x), A(y)). Again, A(xy) = v(f(xy)) = v(f(y)f(x)) ≤ 12.

S(v(f(y) ), v(f(x))) = S(v( f(x)), v(f(y))) = S(A(x), A(y)), which implies that A(xy) ≤ S(A(x), A(y)).

Hence A is an

(T, S)-intuitionistic fuzzy

13.

subhemiring of R.

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