norm.

In 1996, Imai and Iseki [1] introduced the notion of BCK-algebras. After the introduced of the concept of fuzzy sets by Zadeh [2], several researches were conducted on the generalization of the notion of fuzzy sets. Many authors considered the fuzzification of ideals and subalgebras in BCK-algebras [3–6]. Triangular norms and co-norms(S-norm) are in operations which generalize the logical conjunction and logical disjunction to fuzzy logic. Recently, Rasuli [7] introduced a study on T-norms over fuzzy implicative ideal, and also positive implicative in the context of BCK-algebras. Satyanarayana et al.’s [8–12] studied a numerus ideals and intuitionistic fuzzy ideals in BCK-algebras.

In this paper, as using (T, S)-norms and , we define under (T, S)-norms fuzzy implicative ideals, under (T, S)-norms fuzzy positive implicative ideals in BCK-algebras. Next we investigate them with under (T, S)- norms intuitionistic implicative ideals, under (T, S)-norms intuitionistic positive implicative ideals in BCK-algebras. Also we investigate them under Intersection, Union, Cartesian product and homomorphisms (image and pre image) and we study related properties.

In this section we cite the fundamental definitions and results that will be used in the sequel. For more details we refer readers to [8–10], [13–16].

Definition 1. By a BCK-algebra we mean a nonempty set with a binary operation ≬ and a constant 0 satisfying the axioms:

1: (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≤ ( 0 ≬ 0), 2: ( 0 ≬ ( 0 ≬ 0) ≤ 0), 3: 0 ≤ 0, 4: 0 ≤ 0 and 0 ≤ 0 imply that 0 = 0, 5: 0 ≤ 0, ∀ 0, 0, 0 ∈ .

A partial ordering ≤ on can be defined by 0 ≤ 0 if and only if 0 ≬ 0 = 0. In any BCK-algebra the following holds:

6: 0 ≬ 0 = 0, 7: 0 ≬ 0 ≤ 0, 8: ( 0 ≬ 0) ≬ 0 = ( 0 ≬ 0) ≬ 0, 9: ( 0 ≬ 0) ≬ ( 0 ≬ 0) ≤ 0 ≬ 0, 10: 0 ≬ ( 0 ≬ ( 0 ≬ 0)) = 0 ≬ 0 11: if 0 ≤ 0, then 0 ≬ 0 ≤ 0 ≬ 0 and 0 ≬ 0 ≤ 0 ≬ 0, ∀ 0, 0, 0 ∈ .

Definition 2. A BCK-algebra is said to be implicative if 0 = 0 ≬ ( 0 ≬ 0), ∀ 0, 0 ∈ .

Definition 3. A BCK-algebra is said to be “positive implicative” if ( 0 ≬ 0) ≬ 0 = ( 0 ≬ 0) ≬ ( 0 ≬ 0) ∀ 0, 0, 0 ∈ .

Definition 4. A non-empty subset ℑ of a BCK- algebra is called an “implicative ideal” of (ℐℐ1) 0 ∈ ℑ, (ℐℐ2) ( 0 ≬ ( 0 ≬ 0)) ≬ 0 ∈ ℑ and 0 ∈ ℑ imply that 0 ∈ ℑ ∀ 0, 0, 0 ∈ .

Definition 5. A non-empty subset ℑ of a BCK-algebra is called a “positive implicative ideal” of ( ℐℐ1) 0 ∈ ℑ, ( ℐℐ2) ( 0 ≬ 0) ≬ 0 ∈ ℑ and 0 ≬ 0 ∈ ℑ imply that 0 ≬ 0 ∈ ℑ ∀ 0, 0, 0 ∈ .

Definition 6. A mapping : → of BCK-algebras is called a homomorphism if ( 0 ≬ 0) = ( 0) ≬ ( 0), ∀ 0, 0 ∈ .

Definition 7. Let be an arbitrary set. A fuzzy subset of , we mean function from into [0, 1]. The set of all fuzzy subsets of is called the [0, 1]-power set of and is denoted [0,1] . For a fixed , ∈ [0,1], the set ξ = { 0 ∈ : ξ( 0) ≥ } and ζ = { 0 ∈ : ζ( 0) ≤ } are called an upper level of ξ and lower level of ζ. Definition 8. Let ς be a function from set into set such that ξ ∈ [0,1] and ζ ∈ [0,1] . ∀ 0 ∈ , 0 ∈ , we define ς(ξ)( 0) = sup{ξ( 0)| 0 ∈ , ς( 0) = 0} and ς (ζ)( 0) = ζ(ς( 0)).

Definition 9. A T-norm is a function : [0,1] × [0,1] → [0,1] having the following four properties:

( 1) ( 0, 1) = 0 (neutral element), ( 2) ( 0, 0) ≤ ( 0, 0) if 0 ≤ 0 (monotonicity), ( 3) ( 0, 0) = ( 0, 0) (commutativity), ( 4) ( 0, ( 0, 0)) = ( ( 0, 0), 0) (associativity), ∀ 0, 0, 0 ∈ [0,1].

It is clear that if 0 ≥ 0 and 0 ≥ 0 , then ( 0 , 0 ) ≥ ( 0 , 0 ).

Example 1.

I. Standard intersection T-norm ( 0, 0) = min{ 0, 0}.

II. Bounded sum T-norm ( 0, 0) = max{0, 0 + 0 − 1}.

III. Algebraic product T-norm ( 0, 0) = 0 0.

0 if 0 = 0, IV. Drastic T-norm ( 0, ) = { if = 0, 0 othewise.

min{ 0, 0} if 0 + 0 > 1, V. Nilpotent minimum T-norm ( 0, ) = { 0 if otherwise.

0, if 0 = 0 = 0, VI. Hamacher product T-norm H ( 0, 0) = { 0 0 0 if otherwise.

0+ 0− 0 0

The drastic T-norm is the pointwise smallest T-norm and the minimum is the pointwise largest T-norm:

( 0, 0) ≤ ( 0, 0) ≤ min( 0, 0) ∀ 0, 0 ∈ [0,1]. We say that be idempotent if ∀ 0 ∈ [0,1] we have

( 0, 0) = 0.

Definition 10. By S-norm is a function : [0,1] × [0,1] → [0,1] having the following four properties: ( 1) ( 0, 0) = 0 (neutral element), ( 2) ( 0, 0) ≤ ( 0, 0) if 0 ≤ 0 (monotonicity), ( 3) ( 0, 0) = ( 0, 0) (commutativity), ( 4) ( 0, ( 0, 0)) = ( ( 0, 0), 0) (associativity) ∀ 0, 0, 0 ∈ [0,1], It is clear that if 0 ≤ 0 and 0 ≤ 0 , then ( 0 , 0 ) ≤ ( 0 , 0 ).

Example 2.

I. Standard union S-norm ( 0, 0) = max{ 0, 0}.

II. Bounded sum S-norm ( 0, 0) = min{1, 0 + 0}.

III. Algebraic sum S-norm ( 0, 0) = 0 + 0 − 0 0.

0 if 0 = 0, IV. Drastic S-norm ( 0, 0) = { if = 0, 1 othewise.

The drastic S-norm is the pointwise smallest S-norm and the maximum is the pointwise largest S-norm:

( 0, 0) ≥ ( 0, 0) ≥ max( 0, 0) ∀ 0, 0 ∈ [0,1]. We say that be idempotent if ∀ 0 ∈ [0,1] we have ( 0, 0) = 0.

Definition 11. Let ξ, ζ ∈ [0,1] and define the intersection of ξ and ζ is denoted by ξ ∩ ζ ∈ [0,1] as (ξ ∩ ζ)( 0) = (ξ( 0), ζ( 0)) ∀ 0 ∈ .

Definition 12. Let ξ, ζ ∈ [0,1] and define the union of ξ and ζ is denoted by ξ ∪ ζ ∈ [0,1] as (ξ ∪ ζ)( 0) = (ξ( 0), ζ( 0)) ∀ 0 ∈ .

Definition 13. Let ξ ∈ [0,1] and ζ ∈ [0,1] . Define the Cartesian product of ξ and ζ is denoted by ξ × ζ ∈ [0,1] as (ξ × ζ)( 0, 0) = (ξ( 0), ζ( 0)) and (ξ × ζ)( 0, 0) = (ξ( 0), ζ( 0)) ∀ ( 0, 0) ∈ × .

Lemma 1. Let and be a (T, S)-norms. Then ( ( 0, 0), ( 0, 0)) = ( ( 0, 0), ( 0, 0)) and ( ( 0, 0), ( 0, 0)) = ( ( 0, 0), ( 0, 0)) ∀ 0, 0, 0, 0 ∈ [0,1].

Definition 14. We say that ξ ∈ [0,1] is a “fuzzy implicative ideal” of under T-norm if it satisfies the following inequalities: (FII 1) ξ(0) ≥ ξ( 0), (FII 2) ξ( 0) ≥ (ξ( 0 ≬ ( 0 ≬ 0)), ξ( 0)) ∀ 0, 0, 0 ∈ . Denote by FII ( ), the set of all “fuzzy implicative ideals” of under T-norm .

Definition 15. We say that ξ ∈ [0,1] is a “fuzzy implicative ideal” of under S-norm if it satisfies the following inequalities: (FII 1) ξ(0) ≤ ξ( 0), (FII 2) ξ( 0) ≤ (ξ( 0 ≬ ( 0 ≬ 0)), ξ( 0)) ∀ 0, 0, 0 ∈ . Denote by FII ( ), the set of all “fuzzy implicative ideals” of under S-norm .

Definition 16. Define ξ ∈ [0,1] is a “fuzzy positive implicative ideal” of under T-norm if it satisfies the following inequalities: (FPII 1) ξ(0) ≥ ξ( 0), (FPII 2) ξ( 0 ≬ 0) ≥ (ξ(( 0 ≬ 0) ≬ 0), ξ( 0 ≬ 0)) ∀ 0, 0, 0 ∈ . Denote by FPII ( ), the set of all “fuzzy positive implicative ideals” of under T-norm . Definition 17. Define ξ ∈ [0,1] is a “fuzzy positive implicative ideal (FPII )” of under S-norm if it satisfies the following inequalities: (FPII 1) ξ(0) ≤ ξ( 0), (FPII 2) ξ( 0 ≬ 0) ≤ (ξ(( 0 ≬ 0) ≬ 0), ξ( 0 ≬ 0)), ∀ 0, 0, 0 ∈ . Denote by FPII ( ), the set of all “fuzzy positive implicative ideals” of under S-norm

Throughout this paper, , always mean two BCK-algebras unless otherwise specified.

Definition 18. We say that ξ, ζ ∈ [0,1] is an IFII of under (T, S)-norm and if it satisfies the following inequalities:

(IFII 1) ξ(0) ≥ ξ( 0), and ζ(0) ≤ ζ( 0), (IFII 2) ξ( 0) ≥ (ξ( 0 ≬ ( 0 ≬ 0)), ξ( 0)), (IFII 3) ζ( 0) ≤ (ζ( 0 ≬ ( 0 ≬ 0)), ζ( 0)) ∀ 0, 0, 0 ∈ .

Denote by IFII ( ), the set of all IFII of under (T, S)-norm and .

Example 3. Let = {0,1,2} be a set given by the following cayley table:

Then ( , ≬ ,0) is a BCK-algebra. Define the “intuitionistic fuzzy implicative ideal” : ( , ≬ ,0) → [0,1] and

0.8 if 0 = 0,2 0.7 if 0 = 1,3

ζ: ( , ≬ ,0) → [0,1] as: ξ( 0) = { , and ζ( 0) = {

0.4 if 0 = 1,3 0.3 if 0 = 0,2.

Let ( , ) = ( , ) = and ( , ) = ( , ) = + − . ∀ , ∈ [0,1], then ξ ∈ IFII ( ).

Proposition 1. Let ξ, ζ ∈ [0,1] and and be idempotent. Then ξ, ζ ∈ IFII ( ) if and only if the sets ξ = { 0 ∈ : ξ( 0) ≥ } and ζ = { 0 ∈ : ζ( 0) ≤ } are either empty or an “implicative ideals” of , for every , ∈ [0,1].

Proof: Let ξ, ζ ∈ IFII ( ) and 0, 0, 0 ∈ . Thus ξ(0) ≥ ξ( 0) ≥ so 0 ∈ ξ and ζ(0) ≤ ζ( 0) ≤ so 0 ∈ ζ . Also let ( 0 ≬ ( 0 ≬ 0)) ≬ 0 ∈ ξ , ∈ ξ and ( 0 ≬ ( 0 ≬ 0)) ≬ 0 ∈ ζ , ∈ ζ . Then ξ( 0) ≥ (ξ (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ξ( 0)) ≥ ( , ) = . Thus 0 ∈ ξ . And ζ( 0) ≤ (ζ (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ζ( 0)) ≤ ( , ) = . Thus 0 ∈ ζ . Then ξ and ζ will be an “implicative ideal” of for every , ∈ [0,1]. Conversely, let ξ and ζ are either empty or an “implicative ideal” of for every , ∈ [0,1]. Let = (ξ (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ξ( 0)) and = (ζ (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ζ( 0)), with ( 0 ≬ ( 0 ≬ 0)) ≬ 0 ∈ ξ , 0 ∈ ξ and ( 0 ≬ ( 0 ≬ 0)) ≬ 0 ∈ ζ , 0 ∈ ζ . Then 0 ∈ ξ and 0 ∈ ζ thus ξ( 0) ≥ = (ξ (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ξ( 0)). And thus ζ( 0) ≤ = (ζ (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ζ( 0)). So ξ, ζ ∈ IFII ( ).

Definition 19. Define ξ, ζ ∈ [0,1] are an IFPII of under (T, S)-norm and if it satisfies the following inequalities:

(IFPII 1) ξ(0) ≥ ξ( 0) and ζ(0) ≤ ζ( 0), (IFPII 2) ξ( 0 ≬ 0) ≥ (ξ(( 0 ≬ 0) ≬ 0), ξ( 0 ≬ 0)), (IFPII 3) ζ( 0 ≬ 0) ≤ (ζ(( 0 ≬ 0) ≬ 0), ζ( 0 ≬ 0)) ∀ 0, 0, 0 ∈ .

Denote by IFPII ( ), the set of all IFPII of under (T, S)-norm and .

Then ( , ≬ ,0) is a BCK-algebra. Define the “intuitionistic fuzzy positive implicative ideal” ξ ∶ ( , ≬ ,0) → [0,1]

0.9 if 0 = 0, 0.5 if 0 = , , ,

and ζ: ( , ≬ ,0) → [0,1] as: ξ( 0) = { and ζ( 0) = {

0.7 if 0 = , , , 0.3 if 0 = 0.

Let ( , ) = ( , ) = , and ( , ) = ( , ) = + − ∀ , ∈ [0,1], then ξ, ζ ∈ IFPII ( ). Proposition 2. Let ξ, ζ ∈ [0,1] and , be idempotent. Then ξ, ζ ∈ IFPII ( ) if and only if the set ξ = { 0 ∈ : ξ( 0) ≥ } and ζ = { 0 ∈ : ζ( 0) ≤ } are either empty or a “positive implicative ideal” of , for every , ∈ [0,1].

Proof: let , ∈ IFPII ( ) and 0, 0, 0 ∈ . Then (0) ≥ ( ) ≥ , so 0 ∈ and (0) ≤ ( 0) ≤ so 0 ∈ . Also let ( 0 ≬ 0) ≬ 0 ∈ , 0 ≬ 0 ∈ and ( 0 ≬ 0) ≬ 0 ∈ , 0 ≬ 0 ∈ . Then ( 0 ≬ 0) ≥ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) ≥ ( , ) = . Thus 0 ≬ 0 ∈ . And ( 0 ≬ 0) ≤ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) ≤ ( , ) = . Thus 0 ≬ 0 ∈ . Then will be a “positive implicative ideal” of for every , ∈ [0,1]. Conversely, let , are either empty or an “positive implicative ideal” of for every , ∈ [0,1]. Let = ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) and = ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)), with ( 0 ≬ 0) ≬ 0 ∈ , 0 ≬ 0 ∈ and ( 0 ≬ 0) ≬ 0 ∈ , 0 ≬ 0 ∈ . Then 0 ≬ 0 ∈ and 0 ≬ 0 ∈ . Thus ( 0 ≬ 0) ≥ ≥ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) and ( 0 ≬ 0) ≤ ≤ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)). So , ∈ IFPII ( ). Proposition 3. Let , ∈ IFI ( ) such that ( 0 ≬ 0) ≥ ( ((( 0 ≬ 0) ≬ 0) ≬ 0) , ( 0)) and ( 0 ≬ 0) ≤ ( ((( 0 ≬ 0) ≬ 0) ≬ 0) , ( 0)) for any 0, 0, 0 ∈ . Then , ∈ IFPII ( ).

Proof: let , , ∈ . As properties 8 and 9 of Definition 1 we get that (( 0 ≬ 0) ≬ 0) ≬ ( 0 ≬ 0) ≤ ( 0 ≬ 0) ≬ 0 = ( 0 ≬ 0) ≬ 0 and from Proposition 2 we get that ((( 0 ≬ 0) ≬ 0) ≬ ( 0 ≬ 0)) ≥ (( 0 ≬ 0) ≬ 0), and ((( 0 ≬ 0) ≬ 0) ≬ ( 0 ≬ 0)) ≤ (( 0 ≬ 0) ≬ 0). Now by hypothesis if we get 0 = 0, and 0 = 0 ≬ 0 we obtain that ( 0 ≬ 0) ≥ ( ((( 0 ≬ 0 ) ≬ 0 ) ≬ ( 0 ≬ 0)) , ( 0 ≬ 0)), and ( 0 ≬ 0) ≤ ( ((( 0 ≬ 0 ) ≬ 0 ) ≬ ( 0 ≬ 0)) , ( 0 ≬ 0)). Then ( 0 ≬ 0) ≥ ( ((( 0 ≬ 0 ) ≬ 0 ) ≬ ( 0 ≬ 0)) , ( 0 ≬ 0)) ≥ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)). And ( 0 ≬ 0) ≤ ( ((( 0 ≬ 0 ) ≬ 0 ) ≬ ( 0 ≬ 0)) , ( 0 ≬ 0)) ≤ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)). Then , ∈ IFPII ( ).

Proposition 4. Let , ∈ IFI ( ). Then , ∈ IFPII ( ) if and only if (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≥ (( 0 ≬ 0 ) ≬ 0), and (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≤ (( 0 ≬ 0 ) ≬ 0) ∀ 0, 0, 0 ∈ .

Proof: let (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≥ (( 0 ≬ 0) ≬ 0) and (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≤ (( 0 ≬ 0) ≬ 0) ∀ 0, 0, 0 ∈ . As Property 9 of Definition 1 we get that ( 0 ≬ 0) ≬ ( 0 ≬ 0) ≤ 0 ≬ 0 and from Proposition 3, we get that ( 0 ≬ 0) ≥ ( ( 0 ≬ 0), ( 0 ≬ 0)) ≥ ( ( 0 ≬ 0), (( 0 ≬ 0) ≬ ( 0 ≬ 0))) = ( (( 0 ≬ 0) ≬ ( 0 ≬ 0)), ( 0 ≬ 0)) ≥ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)). Therefore, ( 0 ≬ 0) ≥ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)), and ( 0 ≬ 0) ≤ ( ( 0 ≬ 0), ( 0 ≬ 0)) ≤ ( ( 0 ≬ 0), (( 0 ≬ 0) ≬ ( 0 ≬ 0))) = ( (( 0 ≬ 0) ≬ ( 0 ≬ 0)), ( 0 ≬ 0)) ≤ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)). Therefore, ( 0 ≬ 0) ≤ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)). Thus , ∈ IFPII ( ). Conversely, let , ∈ IFPII ( ) and 0, 0, 0 ∈ with = 0 ≬ ( 0 ≬ 0) and = 0 ≬ 0. By Property 1 of Definition 1 we will have that (( 0 ≬ ( 0 ≬ 0)) ≬ ( 0 ≬ 0)) ≤ 0 ≬ ( 0 ≬ 0) and thus ((( 0 ≬ ( 0 ≬ 0)) ≬ ( 0 ≬ 0)) ≬ 0) ≤ 0 ≬ ( 0 ≬ 0) ≬ 0 = 0, (Definition 1 Property 1) and Proposition 2 gives us ((( 0 ≬ ( 0 ≬ 0)) ≬ ( 0 ≬ 0)) ≬ 0) ≥ (0) and ((( 0 ≬ ( 0 ≬ 0)) ≬ ( 0 ≬ 0)) ≬ 0) ≤ (0). Then (( ≬ ) ≬ 0) = ((( 0 ≬ ( 0 ≬ 0)) ≬ ( 0 ≬ 0)) ≬ 0) ≥ (0) and (( ≬ ) ≬ 0) = ((( 0 ≬ ( 0 ≬ 0)) ≬ ( 0 ≬ 0)) ≬ 0) ≤ (0). Now (( 0 ≬ 0) ≬ ( 0 ≬ 0)) =

( 0 ≬ ( 0 ≬ 0) ≬ 0) = ( ≬ 0) ≥ ( (( ≬ ) ≬ 0), ( ≬ 0)) ≥ ( (0), ( ≬ 0)) = ( ≬ 0) = (( 0 ≬

0) ≬ 0) and (( 0 ≬ 0) ≬ ( 0 ≬ 0)) = ( 0 ≬ ( 0 ≬ 0) ≬ 0) = ( ≬ 0) ≤ ( (( ≬ ) ≬ 0), ( ≬ 0)) ≤ ( (0), ( ≬ 0)) = ( ≬ 0) = (( 0 ≬ 0) ≬ 0). Thus (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≥ (( 0 ≬ 0) ≬ 0) and (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≤ (( 0 ≬ 0) ≬ 0) ∀ 0, 0, 0 ∈ .

Proposition 5. Let , ∈ IFPII ( ) and 0, 0, 0, , ∈ .

I. If (( 0 ≬ 0) ≬ 0) ≬ ≤ , then ( 0 ≬ 0) ≥ ( ( ), ( )) and ( 0 ≬ 0) ≤ ( ( ), ( )).

II. If (( 0 ≬ 0) ≬ 0) ≬ ≤ , then (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≥ ( ( ), ( )) and (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≤ ( ( ), ( )).

Proof: let , ∈ IFPII ( ) and 0, 0, 0, , ∈ .

I. Let (( 0 ≬ 0) ≬ 0) ≬ ≤ then Proposition 4 [12] we get that (( 0 ≬ 0) ≬ 0) ≥ ( ( ), ( )). Then ( 0 ≬ 0) ≥ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) = ( (( 0 ≬ 0) ≬ 0), (0)) = (( 0 ≬ 0) ≬ 0) ≥ ( ( ), ( )). Then ( 0 ≬ 0) ≥ ( ( ), ( )) and (( 0 ≬ 0) ≬ 0) ≤ ( ( ), ( )). Then ( 0 ≬ 0) ≤ ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) = ( (( 0 ≬ 0) ≬ 0), (0)) = (( 0 ≬ 0) ≬ 0) ≤ ( ( ), ( )). Then ( 0 ≬ 0) ≤ ( ( ), ( )).

II. Let (( 0 ≬ 0) ≬ 0) ≬ ≤ so from Proposition 4 [12] we get that (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≥ (( 0 ≬ 0) ≬ 0) ≥ ( ( ), ( )), and (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≤ (( 0 ≬ 0) ≬ 0) ≤ ( ( ), ( )).

Proposition 6. Let , ∈ [0,1] and (( 0 ≬ 0) ≬ 0) ≬ ≤ ∀ 0, 0, 0, , ∈ . If ( 0 ≬ 0) ≥ ( ( ), ( )) and ( 0 ≬ 0) ≤ ( ( ), ( )), then , ∈ IFPII ( ).

Proof: first, we prove that , ∈ IFI ( ). Let 0, 0, 0 ∈ such that 0 ≬ 0 ≤ 0. Definition 1 and Property 1 gives us that (( 0 ≬ 0) ≬ 0) ≬ 0 ≬ 0 = ( 0 ≬ 0) ≬ 0 = 0. Thus (( 0 ≬ 0) ≬ 0) ≬ 0 ≤ 0. Put 0 = 0, = 0, = 0 in hypothesis then ( 0) = ( 0 ≬ 0) ≥ ( ( ), ( 0)) and ( 0) = ( 0 ≬ 0) ≤ ( ( ), ( 0)). Thus from Proposition 5 we get that , ∈ IFI ( ). As ((( 0 ≬ 0) ≬ 0) ≬ (( 0 ≬ 0) ≬ 0)) ≬ 0 = 0, so ((( 0 ≬ 0) ≬ 0) ≬ (( 0 ≬ 0) ≬ 0)) ≤ 0 ∀ 0, 0 ∈ . Using hypothesis will gives us ( 0 ≬ 0) ≥ ( (( 0 ≬ 0) ≬ 0), (0)) = (( 0 ≬ 0) ≬ 0), and ( 0 ≬ 0) ≤ ( (( 0 ≬ 0) ≬ 0), (0)) = (( 0 ≬ 0) ≬ 0). Therefore , ∈ IFPII ( ).

Proposition 6. Let , ∈ [0,1] and (( 0 ≬ 0) ≬ 0) ≬ ≤ ∀ 0, 0, 0, , ∈ . If (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≥ ( ( ), ( )) and (( 0 ≬ 0) ≬ ( 0 ≬ 0)) ≤ ( ( ), ( )), then , ∈ IFPII ( ).

Proof: let (( 0 ≬ 0) ≬ 0) ≬ ≤ ∀ 0, 0, 0, , ∈ . Then ((( 0 ≬ 0) ≬ 0) ≬ ) ≬ = 0. Now ( 0 ≬ 0) = (( 0 ≬ 0) ≬ 0) = ( 0 ≬ 0) ≥ ( ( ), ( )) and ( 0 ≬ 0) = (( 0 ≬ 0) ≬ 0) = ( 0 ≬ 0) ≤ ( ( ), ( )). And as Proposition 5 we will have that , ∈ IFPII ( ).

Proposition 7. Let , ∈ IFII ( ). Then ∩ , ∪ ∈ IFII ( ).

Proof: let 0, 0 ∈ . Then I. ( ∩ )(0) = ( (0), (0)) ≥ ( ( 0), ( 0)) = ( ∩ )( 0), thus ( ∩ )(0) ≥ ( ∩ )( 0), and ( ∪ )(0) = ( (0), (0)) ≤ ( ( 0), ( 0)) = ( ∪ )( 0). Thus ( ∪ )(0) ≤ ( ∪ )( 0).

II. ( ∩ )( 0) = ( ( 0), ( 0)) ≥ ( ( (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ( 0)) , ( (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ( 0))) = ( ( (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , (( 0 ≬ ( 0 ≬ 0)) ≬ 0)) , ( ( 0), ( 0))) (Lemma 1) = (( ∩ ) (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ( ∩ )( 0)), so ( ∩ )( 0) ≥ (( ∩ )( 0 ≬ 0), ( ∩ )( 0)).

III. ( ∪ )( 0) = ( ( 0), ( 0)) ≤ ( ( (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ( 0)) , ( (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ( 0))) = ( ( (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , (( 0 ≬ ( 0 ≬ 0)) ≬ 0)) , ( ( 0), ( 0))) (Lemma 1) = (( ∪ ) (( 0 ≬ ( 0 ≬ 0)) ≬ 0) , ( ∪ )( 0)), so ( ∪ )( 0) ≤ (( ∪ )( 0 ≬ 0), ( ∪ )( 0)). Then ∩ , ∪ ∈ IFII ( ).

Proposition 8. Let , ∈ IFPII ( ). Then ∩ , ∪ ∈ IFPII ( ).

Proof: let 0, 0 ∈ . Then I. ( ∩ )(0) = ( (0), (0)) ≥ ( ( 0), ( 0)) = ( ∩ )( 0). Thus ( ∩ )(0) ≥ ( ∩ )( 0), and ( ∪ )(0) = ( (0), (0)) ≤ ( ( 0), ( 0)) = ( ∪ )( 0), thus ( ∪ )(0) ≤ ( ∪ )( 0).

II. ( ∩ )( 0 ≬ 0) = ( ( 0 ≬ 0), ( 0 ≬ 0)) ≥ ( ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) , ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0))) = ( ( (( 0 ≬ 0) ≬ 0), (( 0 ≬ 0) ≬ 0)) , ( ( 0 ≬ 0), ( 0 ≬ 0))) (Lemma 1) = (( ∩ )(( 0 ≬ 0) ≬ 0), ( ∩ )( 0 ≬ 0)) so ( ∩ )( 0 ≬ 0) ≥ (( ∩ )(( 0 ≬ 0) ≬ 0), ( ∩ )( 0 ≬ 0)).

III. ( ∪ )( 0 ≬ 0) = ( ( 0 ≬ 0), ( 0 ≬ 0)) ≤ ( ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0)) , ( (( 0 ≬ 0) ≬ 0), ( 0 ≬ 0))) = ( ( (( 0 ≬ 0) ≬ 0), (( 0 ≬ 0) ≬ 0)) , ( ( 0 ≬ 0), ( 0 ≬ 0))) (Lemma 1) = (( ∪ )(( 0 ≬ 0) ≬ 0), ( ∪ )( 0 ≬ 0)), so ( ∪ )( 0 ≬ 0) ≤ (( ∪ )(( 0 ≬ 0) ≬ 0), ( ∪ )( 0 ≬ 0)). Then ∩ , ∪ ∈ IFPII ( ). Proposition 9. Let , ∈ IFII ( ) and , ∈ IFII ( ). Then × , × ∈ IFII ( × ).

Proof: let ( 0, 0) ∈ × . Then ( × )(0,0) = ( (0), (0)) ≥ ( ( 0), ( 0)) = ( × )( 0, 0). Thus ( × )(0,0) ≥ ( × )( 0, 0), and ( × )(0,0) = ( (0), (0)) ≤ ( ( 0), ( 0)) = ( × )( 0, 0). Thus ( × )(0,0) ≤ ( × )( 0, 0). Also let 0 ∈ , and 0 ∈ for = 1,2,3. Now ( × )( 0 , 0 ) = ( ( 0 ), ( 0 ))

1 1 1 1

≥ ( ( ( 0 ≬ ( 0 ≬ 0 )) , ( 0 )) , ( ( 0 ≬ ( 0 ≬ 0 )) , ( 0 )))

= ( ( ( 0 ≬ ( 0 ≬ 0 )) , ( 0 ≬ ( 0 ≬ 0 ))) , ( ( 0 ), ( 0 ))) (Lemma 1)

= (( × ) (( 0 ≬ ( 0 ≬ 0 )) , ( 0 ≬ ( 0 ≬ 0 ))) , ( × )( 0 , 0 ))

= (( × ) (( 0 , 0 ) ≬ (( 0 , 0 ) ≬ ( 0 , 0 ))) , ( × )( 0 , 0 )). Thus ( × )( 0 , 0 )

≥ (( × ) (( 0 , 0 ) ≬ (( 0 , 0 ) ≬ ( 0 , 0 ))) , ( × )( 0 , 0 )), and

( × )( 0 , 0 ) = ( ( 0 ), ( 0 ))

1 1 1 1

≤ ( ( ( 0 ≬ ( 0 ≬ 0 )) , ( 0 )) , ( ( 0 ≬ ( 0 ≬ 0 )) , ( 0 )))

= ( ( ( 0 ≬ ( 0 ≬ 0 )) , ( 0 ≬ ( 0 ≬ 0 ))) , ( ( 0 ), ( 0 ))) (Lemma 1)

= (( × ) (( 0 ≬ ( 0 ≬ 0 )) , ( 0 ≬ ( 0 ≬ 0 ))) , ( × )( 0 , 0 ))

= (( × ) (( 0 , 0 ) ≬ (( 0 , 0 ) ≬ ( 0 , 0 ))) , ( × )( 0 , 0 ))

Thus ( × )( 0 , 0 ) ≤ (( × ) (( 0 , 0 ) ≬ (( 0 , 0 ) ≬ ( 0 , 0 ))) , ( × )( 0 , 0 )).

Then × , × ∈ IFII ( × ).

Proposition 10. Let , ∈ IFPII ( ) and , ∈ IFPII ( ). Then × , × ∈ IFPII ( × ). Proof: let ( 0, 0) ∈ × . Then (ξ × ϖ)(0, 0) = (ξ(0), ϖ(0)) ≥ ( ( 0), ( 0)) = (ξ × ϖ)( 0, 0). Thus (ξ × ϖ)(0, 0) ≥ (ξ × ϖ)( 0, 0), and (ζ × ϑ)(0,0) = (ζ(0), ϑ(0)) ≤ ( ( 0), ( 0)) = ( × )( 0, 0). Thus ( × )(0, 0) ≤ ( × )( 0, 0). Also let 0 ∈ , and 0 ∈ for = 1,2,3. Then ( × ) (( 0 , 0 ) ≬ ( 0 , 0 )) = ( × )( 0 ≬ 0 , 0 ≬ 0 ) = ( ( 0 ≬ 0 ), ( 0 ≬ 0 ))

1 3 1 3

≥ ( ( (( 0 ≬ 0 ) ≬ 0 ) , ( 0 ≬ 0 )) , ( (( 0 ≬ 0 ) ≬ 0 ) , ( 0 ≬ 0 )))

= ( ( (( 0 ≬ 0 ) ≬ 0 ) , (( 0 ≬ 0 ) ≬ 0 )) , ( ( 0 ≬ 0 ), ( 0 ≬ 0 ))) (Lemma 1)

= (( × ) ((( 0 ≬ 0 ) ≬ 0 ) , (( 0 ≬ 0 ) ≬ 0 )) , ( × ) (( 0 ≬ 0 ), ( 0 ≬ 0 )))

= (( × ) ((( 0 , 0 ) ≬ ( 0 , 0 )) ≬ ( 0 , 0 )) , ( × ) (( 0 , 0 ) ≬ ( 0 , 0 ))), so

( × ) (( 0 , 0 ) ≬ ( 0 , 0 )) ≥

1 1 3 3

(( × ) ((( 0 , 0 ) ≬ ( 0 , 0 )) ≬ ( 0 , 0 )) , ( × ) (( 0 , 0 ) ≬ ( 0 , 0 ))),

and ( × ) (( 0 , 0 ) ≬ ( 0 , 0 )) = ( × )( 0 ≬ 0 , 0 ≬ 0 ) = ( ( 0 ≬ 0 ), ( 0 ≬ 0 ))

1 1 3 3 1 3 1 3 1 3 1 3

≤ ( ( (( 0 ≬ 0 ) ≬ 0 ) , ( 0 ≬ 0 )) , ( (( 0 ≬ 0 ) ≬ 0 ) , ( 0 ≬ 0 )))

= ( ( (( 0 ≬ 0 ) ≬ 0 ) , (( 0 ≬ 0 ) ≬ 0 )) , ( ( 0 ≬ 0 ), ( 0 ≬ 0 ))) (Lemma 1)

= (( × ) ((( 0 ≬ 0 ) ≬ 0 ) , (( 0 ≬ 0 ) ≬ 0 )) , ( × ) (( 0 ≬ 0 ), ( 0 ≬ 0 )))

= (( × ) ((( 0 , 0 ) ≬ ( 0 , 0 )) ≬ ( 0 , 0 )) , ( × ) (( 0 , 0 ) ≬ ( 0 , 0 )))

( × ) (( 0 , 0 ) ≬ ( 0 , 0 )) ≤

1 1 3 3

(( × ) ((( 0 , 0 ) ≬ ( 0 , 0 )) ≬ ( 0 , 0 )) , ( × ) (( 0 , 0 ) ≬ ( 0 , 0 ))).

Then × , × ∈ IFPII ( × ).

Proposition 11. If , ∈ IFII ( ) and : → be a homomorphism of BCK-algebras. Then ( ), ( ) ∈ IFII ( ).

Proof: let 0 ∈ , and 0 ∈ with ( 0) = 0. Now ( )(0) = sup{ (0)|0 ∈ , (0) = 0} ≥ sup{ ( 0)| 0 ∈ , ( 0) = 0} = ( )( 0). Thus ( )(0) ≥ ( )( 0) and ( )(0) = inf{ (0) | 0 ∈ , (0) = 0} ≤ inf{ ( 0) | 0 ∈ , ( 0) = 0} = ( )( 0). Thus ( )(0) ≤ ( )( 0). Also let 0, 0 , 0 ∈ such that ( 0) = 0, ς( 0 ) = 0 , ς( 0 ) = 0 . Then

1 2 1 1 2 2

( )( 0) = sup{ ( 0)| 0 ∈ , ( 0) = 0} ≥ sup { ( ( 0 ≬ ( 0 ≬ 0)) , ( 0 )) | 0, 0 , 0 ∈ , ( 0) = 0, ( 0 ) = 0 , ( 0 ) = 0 }

= (sup { ( 0 ≬ ( 0 ≬ 0)) | 0, 0 ∈ , ( 0) = 0, ( 0 ) = 0 }, sup{ ( 0 )| 0 ∈ , ( 0 ) = 0 } = (sup { ( 0 ≬ ( 0 ≬ 0)) | 0, 0 ∈ , ( 0 ≬ ( 0 ≬ 0)) = 0 ≬ ( 0 ≬ 0)},

1 1 1 1

sup{ ( 0)| 0 ∈ , ( 0 ) = 0 }) = ( ( ) ( 0 ≬ ( 0 ≬ 0)) , ( )( 0 )), therefore ( )( 0) ≥ ( ( ) ( 0 ≬ ( 0 ≬ 0)) , ( )( 0 )) and ( )( 0) = inf{ ( 0)| 0 ∈ , ( 0) = 0}

≤ inf { ( ( 0 ≬ ( 0 ≬ 0)) , ( 0 )) | 0, 0 , 0 ∈ , ( 0) = 0, ( 0 ) = 0 , ( 0 ) = 0 }

= (inf { ( 0 ≬ ( 0 ≬ 0)) | 0, 0 ∈ , ( 0) = 0, ( 0 ) = 0 }, inf{ ( 0 )| 0 ∈ , ( 0 ) = 0 }) = (inf { ( 0 ≬ ( 0 ≬ 0)) | 0, 0 ∈ , ( 0 ≬ ( 0 ≬ 0)) = 0 ≬ ( 0 ≬ 0)},

1 1 1 1

inf{ ( 0 ) | 0 ∈ , ( 0 ) = 0 }) = ( ( ) ( 0 ≬ ( 0 ≬ 0)) , ( )( 0 )). Therefore

2 2 2 2 1 2

( )( 0) ≤ ( ( ) ( 0 ≬ ( 0 ≬ 0)) , ( )( 0 )). Therefore ( ), ( ) ∈ IFII ( ).

Proposition 12. If ϖ, ϑ ∈ IFII ( ) and : → be a homomorphism of BCK-algebras. Then

−1 −1

( ), ( ) ∈ IFII ( ).

−1 −1

Proof: let 0 ∈ . Then ς (ϖ)( 0) = ϖ(ς(0)) ≥ ϖ(ς( 0)) = ς (ϖ)( 0) and

−1 −1 −1

ς (ϑ)( 0) = ϑ(ς(0)) ≤ ϑ(ς( 0)) = ς (ϑ)( 0). Let 0, 0 , 0 ∈ . As ς (ϖ)( 0) = ϖ(ς( 0))

≥ (ϖ (ς( 0) ≬ (ς( 0 ) ≬ ς( 0))) , ϖ (ς( 0 ))) = (ϖ (ς ( 0 ≬ ( 0 ≬ 0))) , ϖ (ς( 0 )))

1 2 1 2 −1 −1 −1 −1 −1

= (ς (ϖ) ( 0 ≬ ( 0 ≬ 0)) , ς (ϖ)( 0 )), so ς (ϖ)( ) ≥ (ς (ϖ) ( 0 ≬ ( 0 ≬ 0)) , ς (ϖ)( 0 )), and

1 2 1 2

ς (ϑ)( 0) = ϑ(ς( 0)) ≤ (ϑ (ς( 0) ≬ (ς( 0 ) ≬ ς( 0))) , ϑ (ς( 0 )))

−1 −1

= (ϑ (ς ( 0 ≬ ( 0 ≬ 0))) , ϑ (ς( 0 ))) = (ς (ϑ) ( 0 ≬ ( 0 ≬ 0)) , ς (ϑ)( 0 )), so

1 2 1 2 −1 −1 −1 −1 −1

ς (ϑ)( 0) ≤ (ς (ϑ) ( 0 ≬ ( 0 ≬ 0)) , ς (ϑ)( 0 )). Therefore ς (ϖ), ς (ϑ) ∈ IFII ( ).

Proposition 13. If , ∈ IFPII ( ) and : → be a homomorphism of BCK-algebras. Then ( ), ( ) ∈ IFPII ( ).

Proof: let 0 ∈ , and ∈ with ( 0) = 0. Now ς(ξ)(0) = sup{ξ(0)|0 ∈ , ς(0) = 0} ≥ sup{ξ( 0)| 0 ∈ , ς( 0) = 0} = ς(ξ)( 0), thus ς(ξ)(0) ≥ ς(ξ)( 0) and ς(ζ)(0) = inf{ζ(0)|0 ∈ , ς(0) = 0} ≤ inf{ζ( 0)| 0 ∈ , ς( 0) = 0} = ς(ζ)( 0), thus ς(ζ)(0) ≤ ς(ζ)( 0). Also let 0 , 0 , 0 ∈ such that

1 2 3

ς( 0 ) = 0 , ς( 0 ) = 0 , ς( 0 ) = 0 . Then

1 1 2 2 3 3

ς(ξ)( 0 ≬ 0 ) = sup{ξ( 0 ≬ 0 )| 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 } ≥ sup { (ξ (( 0 ≬ 0 ) ≬ 0 ) , ξ( 0 ≬ 0 )) | 0 , 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 , ς( 0 ) = 0 }

1 2 3 2 3 1 2 3 1 1 2 2 3 3

= (sup {ξ (( 0 ≬ 0 ) ≬ 0 ) | 0 , 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 , ς( 0 ) = 0 },

1 2 3 1 2 3 1 1 2 2 3 3

sup{ξ( 0 ≬ 0 )| 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 }) = (sup {ξ (( 0 ≬ 0 ) ≬ 0 ) | 0 , 0 , 0 ∈ , ς (( 0 ≬ 0 ) ≬ 0 ) = ( 0 ≬ 0 ) ≬ 0 },

1 2 3 1 2 3 1 2 3 1 2 3

{ ( 0 ≬ 0 )| 0 , 0 ∈ , ( 0 ≬ 0 ) = 0 ≬ 0 , }) = (ς(ξ) (( 0 ≬ 0 ) ≬ 0 ) , ς(ξ)( 0 ≬ 0 )).

Therefore ς(ξ)( 0 ≬ 0 ) ≥ (ς(ξ) (( 0 ≬ 0 ) ≬ 0 ) , ς(ξ)( 0 ≬ 0 )) and

1 3 1 2 3 2 3 ς(ζ)( 0 ≬ 0 ) = inf{ζ( 0 ≬ 0 )| 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 } ≤ inf { (ζ (( 0 ≬ 0 ) ≬ 0 ) , ζ( 0 ≬ 0 )) | 0 , 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 , ς( 0 ) = 0 }

1 2 3 2 3 1 2 3 1 1 2 2 3 3

= (inf {ζ (( 0 ≬ 0 ) ≬ 0 ) | 0 , 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 , ς( 0 ) = 0 },

1 2 3 1 2 3 1 1 2 2 3 3

inf{ζ( 0 ≬ 0 )| 0 , 0 ∈ , ς( 0 ) = 0 , ς( 0 ) = 0 }) = (inf {ζ (( 0 ≬ 0 ) ≬ 0 ) | 0 , 0 , 0 ∈ , ς (( 0 ≬ 0 ) ≬ 0 ) = ( 0 ≬ 0 ) ≬ 0 },

1 2 3 1 2 3 1 2 3 1 2 3

inf{ζ( 0 ≬ 0 )| 0 , 0 ∈ , ς( 0 ≬ 0 ) = 0 ≬ 0 }) = (ς(ζ) (( 0 ≬ 0 ) ≬ 0 ) , ς(ζ)( 0 ≬ 0 )).

Therefore ς(ζ)( 0 ≬ 0 ) ≤ (ς(ζ) (( 0 ≬ 0 ) ≬ 0 ) , ς(ζ)( 0 ≬ 0 )).

1 3 1 2 3 2 3

Therefore ς(ξ), ς(ζ) ∈ IFPII ( ).

Proposition 14. If ϖ, ϑ ∈ IFPII ( ) and ς: → be a homomorphism of BCK-algebras. Then

−1 −1

ς (ϖ), ς (ϑ) ∈ IFPII ( ).

−1 −1 −1

Proof: let 0 ∈ . Then ς (ϖ)(0) = ϖ(ς(0)) ≥ ϖ(ς( 0)) = ς (ϖ)( 0), and ς (ϑ)(0) = ϑ(ς(0))

−1 −1

≤ ϑ(ς( 0)) = ς (ϑ)( 0). Let 0 , 0 , 0 ∈ . As ς (ϖ)( 0 ≬ 0 ) = ϖ (ς( 0 ≬ 0 ))

1 2 3 1 3 1 3

= ϖ (ς( 0 ) ≬ ς( 0 )) ≥ (ϖ ((ς( 0 ) ≬ ς( 0 )) ≬ ς( 0 )) , ϖ (ς( 0 ) ≬ ς( 0 )))

1 3 1 2 3 2 3 −1 −1

= (ϖ(ς( 0 ≬ 0 ) ≬ 0 ), ϖ (ς( 0 ≬ 0 ))) = (ς (ϖ) (( 0 ≬ 0 ) ≬ 0 ) , ς (ϖ)( 0 ≬ 0 )), so

−1 −1 −1 −1

ς (ϖ)( 0 ≬ 0 ) ≥ (ς (ϖ) (( 0 ≬ 0 ) ≬ 0 ) , ς (ϖ)( 0 ≬ 0 )), and ς (ϑ)( 0 ≬ 0 )

1 3 1 2 3 2 3 1 3

= ϑ (ς( 0 ≬ 0 )) = ϑ (ς( 0 ) ≬ ς( 0 )) ≤ (ϑ ((ς( 0 ) ≬ ς( 0 )) ≬ ς( 0 )) , ϑ (ς( 0 ) ≬ ς( 0 )))

1 3 1 3 1 2 3 2 3 −1 −1

= (ϑ(ς( 0 ≬ 0 ) ≬ 0 ), ϑ (ς( 0 ≬ 0 ))) = (ς (ϑ) (( 0 ≬ 0 ) ≬ 0 ) , ς (ϑ)( 0 ≬ 0 )), so

−1 −1 −1

ς (ϑ)( 0 ≬ 0 ) ≤ (ς (ϑ) (( 0 ≬ 0 ) ≬ 0 ) , ς (ϑ)( 0 ≬ 0 )).

1 3 1 2 3 2 3 −1 −1

Therefore ς (ϖ), ς (ϑ) ∈ IFPII ( ).

This study has introduced and explored “fuzzy implicative ideals and fuzzy positive implicative ideals in BCK-algebras under (T, S)-norms,” establishing connections with “intuitionistic implicative and positive implicative ideals.” The investigation of these ideals under various operations, including “intersection, union, Cartesian product, and homomorphisms,” has provided valuable insights into their properties and behavior, contributing to the advancement of BCK-algebra research and fuzzy logic.

The findings of this research have the potential to enrich the theory of “BCK-algebras and fuzzy logic,” paving the way for applications in decision-making, artificial intelligence, and other areas where uncertainty and imprecision are inherent. The introduced concepts and established relationships can serve as a foundation for further investigations, fostering advancements in algebraic structures and fuzzy set theory.

Future research directions include exploring applications of these ideals in real-world problems, investigating further properties and characterizations in different algebraic structures, and developing new approaches to fuzzy set theory and BCK-algebras using (T, S)-norms and other mathematical tools, potentially leading to novel insights and applications.

All author equally contributated. All authors have read and agreed to the published version of the manuscript.

The authors declare no conflict of interest.