Generalizations of 𝒩-Subalgebras in BCK/BCI-Algebras Based on Point 𝒩-Structures

Definition 2.1 (see [1].)By a subalgebra of X based on 𝒩-function f (briefly, 𝒩-subalgebra of X), we mean an 𝒩-structure (X, f) in which f satisfies the following assertion:

For any 𝒩-structure (X, f) and t ∈ [−1,0), the set

is called a closed  t-support of (X, f), and the set is called an open   t-support of (X, f).

Using the similar method to the transfer principle in fuzzy theory (see [7, 8]), Jun et al. [2] considered transfer principle in 𝒩-structures as follows.

Theorem 2.2 (see [2]; 𝒩-transfer principle.)An 𝒩-structure (X, f) satisfies the property $$ if and only if for all α ∈ [−1,0],

Lemma 2.3 (see [1].)An 𝒩-structure (X, f) is an 𝒩-subalgebra of X if and only if every open t-support of (X, f) is a subalgebra of X for all t ∈ [−1,0).

Definition 3.1. An 𝒩-structure (X, f) is called an 𝒩-subalgebra of type

• (i)

  (∈, ∈) (resp., (∈, q) and (∈, ∈∨q)) if whenever two point 𝒩-structures $$ and $$ are 𝒩_∈-subsets of (X, f) then the point 𝒩-structure $$ is an 𝒩_∈-subset (resp., 𝒩_q-subset and 𝒩_∈∨q-subset) of (X, f).

• (ii)

  (q, ∈) (resp., (q, q) and (q, ∈∨q)) if whenever two point 𝒩-structures $$ and $$ are 𝒩_q-subsets of (X, f) then the point 𝒩-structure $$ is an 𝒩_∈-subset (resp., 𝒩_q-subset and 𝒩_∈∨q-subset) of (X, f).

Definition 3.2. An 𝒩-structure (X, f) is called an 𝒩-subalgebra of type (∈, ∈∨q_k) (resp., ( q, ∈∨q_k)) if whenever two point 𝒩-structures $$ and $$ are 𝒩_∈-subsets (resp., 𝒩_q-subsets) of (X, f) then the point 𝒩-structure $$ is an $$-subset of (X, f).

Example 3.3. Consider a BCI-algebra X = {0, a, b, c} with the following Cayley table:

Let (X, f) be an 𝒩-structure in which f is defined by It is routine to verify that (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_−0.2).

Note that if k, r ∈ (−1,0] with k < r, then every 𝒩-subalgebra of type (∈, ∈∨q_k) is an 𝒩-subalgebra of type (∈, ∈∨q_r), but the converse is not true as seen in the following example.

Example 3.4. The 𝒩-subalgebra (X, f) of type (∈, ∈∨q_−0.2) in Example 3.3 is not of type (∈, ∈∨q_−0.4) since (X, a_−0.65) and (X, a_−0.68) are 𝒩_∈-subsets of (X, f), but

is not an $$-subset of (X, f).

Theorem 3.5. Every 𝒩-subalgebra of type (∈, ∈) is of type (∈, ∈∨q_k).

Proof. Straightforward.

Corollary 3.6. Every 𝒩-subalgebra of type (∈, ∈) is of type (∈, ∈∨q).

Example 3.7. Consider the 𝒩-subalgebra (X, f) of type (∈, ∈∨q_−0.2) which is given in Example 3.3. Then (X, f) is not an 𝒩-subalgebra of type (∈, ∈) since (X, a_−0.65) and (X, a_−0.68) are 𝒩_∈-subsets of (X, f), but (X, (a*a) _{⋁{−0.65,−0.68}}) is not an 𝒩_∈-subset of (X, f).

Definition 3.8. An 𝒩-structure (X, f) is called an 𝒩-subalgebra of type (∈, q_k) if whenever two point 𝒩-structure $$ and $$ are 𝒩_∈-subsets of (X, f) then the point 𝒩-structure $$ is an $$-subset of (X, f).

Theorem 3.9. Every 𝒩-subalgebra of type (∈, q_k) is of type (∈, ∈∨q_k).

Proof. Straightforward.

Corollary 3.10. Every 𝒩-subalgebra of type (∈, q) is of type (∈, ∈∨q).

Example 3.11. Consider the 𝒩-subalgebra (X, f) of type (∈, ∈∨q_−0.2) which is given in Example 3.3. Then (X, a_−0.65) and (X, b_−0.25) are 𝒩-subsets of (X, f), but

is not an $$-subset of (X, f) for k = −0.2 since f(c) − 0.25 − 0.2 + 1 > 0.

Theorem 3.12. An 𝒩-structure (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k) if and only if it satisfies

Proof. Let (X, f) be an 𝒩-structure of type (∈, ∈∨q_k). Assume that (3.6) is not valid. Then there exists a, b ∈ X such that

If ⋁{f(a), f(b)} > (k − 1)/2, then f(a*b) > ⋁{f(a), f(b)}. Hence for some t ∈ [−1,0). It follows that point 𝒩-structures (X, a_t) and (X, b_t) are 𝒩_∈-subsets of (X, f), but the point 𝒩-structure (X, (a*b) _t) is not an 𝒩_∈-subset of (X, f). Moreover, and so (X, (a*b) _t) is not an $$-subset of (X, f). Consequently, (X, (a*b) _t) is not an $$-subset of (X, f). This is a contradiction. If ⋁{f(a), f(b)} ≤ (k − 1)/2, then f(a) ≤ (k − 1)/2, f(b) ≤ (k − 1)/2 and f(a*b) > (k − 1)/2. Thus (X, a_{(k − 1)/2}) and (X, b_{(k − 1)/2}) are 𝒩_∈-subsets of (X, f), but (X, (a*b) _{(k − 1)/2}) is not an 𝒩_∈-subset of (X, f). Also, that is, (X, (a*b) _{(k − 1)/2}) is not an $$-subset of (X, f). Hence (X, (a*b) _{(k − 1)/2}) is not an $$-subset of (X, f), a contradiction. Therefore (3.6) is valid.

Conversely, suppose that (3.6) is valid. Let x, y ∈ X and t₁, t₂ ∈ [−1,0) be such that two point 𝒩-structures $$ and $$ are 𝒩_∈-subsets of (X, f). Then

Assume that t₁ ≥ (k − 1)/2 or t₂ ≥ (k − 1)/2. Then f(x*y) ≤ ⋁{t₁, t₂}, and so $$ is an 𝒩_∈-subset of (X, f). Now suppose that t₁ < (k − 1)/2 and t₂ < (k − 1)/2. Then f(x*y) ≤ (k − 1)/2, and thus that is, $$ is an $$-subset of (X, f). Therefore $$ is an $$-subset of (X, f) and consequently (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k).

Corollary 3.13 (see [3].)An 𝒩-structure (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q) if and only if it satisfies

Proof. It follows from taking k = 0 in Theorem 3.12.

Theorem 3.14. Let S be a subalgebra of X and let (X, f) be an 𝒩-structure such that

• (a)

  (∀x ∈ X)(x ∈ S⇒f(x)≤(k − 1)/2),

• (b)

  (∀x ∈ X)(x ∉ S⇒f(x) = 0).

Then (X, f) is an 𝒩-subalgebra of type (q, ∈∨q_k).

Proof. Let x, y ∈ X and t₁, t₂ ∈ [−1,0) be such that two point 𝒩-structures $$ and $$ are 𝒩_q-subsets of (X, f). Then f(x) + t₁ + 1 < 0 and f(y) + t₂ + 1 < 0. Thus x*y ∈ S because if it is impossible, then x ∉ S or y ∉ S. Thus f(x) = 0 or f(y) = 0, and so t₁ < −1 or t₂ < −1. This is a contradiction. Hence f(x*y)≤(k − 1)/2. If ⋁{t₁, t₂} < (k − 1)/2, then f(x*y) + ⋁{t₁, t₂} − k + 1 < ((k − 1)/2) + ((k − 1)/2) − k + 1 = 0 and so the point 𝒩-structure $$ is an $$-subset of (X, f). If ⋁{t₁, t₂} ≥ (k − 1)/2, then f(x*y)≤(k − 1)/2 ≤ ⋁{t₁, t₂} and so the point 𝒩-structure $$ is an 𝒩_∈-subset of (X, f). Therefore the point 𝒩-structure $$ is an $$-subset of (X, f). This shows that (X, f) is an 𝒩-subalgebra of type (q, ∈∨q_k).

Corollary 3.15 (see [3].)Let S be a subalgebra of X and let (X, f) be an 𝒩-structure such that

• (a)

  (∀x ∈ X)(x ∈ S⇒f(x) ≤ −0.5),

• (b)

  (∀x ∈ X)(x ∉ S⇒f(x) = 0).

Then (X, f) is an 𝒩-subalgebra of type (q, ∈∨_q).

Theorem 3.16. Let (X, f) be an 𝒩-subalgebra of type (q_k, ∈∨q_k). If f is not constant on the open 0-support of (X, f), then f(x) ≤ (k − 1)/2 for some x ∈ X. In particular, f(0) ≤ (k − 1)/2.

Proof. Assume that f(x)>(k − 1)/2 for all x ∈ X. Since f is not constant on the open 0-support of (X, f), there exists x ∈ O(f; 0) such that t_x = f(x) ≠ f(0) = t₀. Then either t₀ < t_x or t₀ > t_x. For the case t₀ < t_x, choose r < (k − 1)/2 such that t₀ + r − k + 1 < 0 < t_x + r − k + 1. Then the point 𝒩-structure (X, 0_r) is an $$-subset of (X, f). Since (X, x₋₁) is an $$-subset of (X, f). It follows from (a1) that the point 𝒩-structure (X, (x*0) _⋁{r,−1}) = (X, x_r) is an $$-subset of (X, f). But, f(x) > (k − 1)/2 > r implies that the point 𝒩-structure (X, x_r) is not an 𝒩_∈-subset of (X, f). Also, f(x) + r − k + 1 = t_x + r − k + 1 > 0 implies that the point 𝒩-structure (X, x_r) is not an $$-subset of (X, f). This is a contradiction. Assume that t₀ > t_x and take r < (k − 1)/2 such that t_x + r − k + 1 < 0 < t₀ + r − k + 1. Then (X, x_r) is an $$-subset of (X, f). Since

(X, (x*x) _⋁{r,r}) is not an 𝒩_∈-subset of (X, f). Since (X, (x*x) _⋁{r,r}) is not an $$-subset of (X, f). Hence (X, (x*x) _⋁{r,r}) is not an $$-subset of (X, f), which is a contradiction. Therefore f(x)≤(k − 1)/2 for some x ∈ X. We now prove that f(0) ≤ (k − 1)/2. Assume that f(0) = t₀ > (k − 1)/2. Note that there exists x ∈ X such that f(x) = t_x ≤ (k − 1)/2 and so t_x < t₀. Choose t₁ < t₀ such that t_x + t₁ − k + 1 < 0 < t₀ + t₁ − k + 1. Then f(x) + t₁ − k + 1 = t_x + t₁ − k + 1 < 0, and thus the point 𝒩-structure $$ is an $$-subset of (X, f). Now we have and f(x*x) = f(0) = t₀ > t₁ = ⋁{t₁, t₁}. Hence $$ is not an $$-subset of (X, f). This is a contradiction, and therefore f(0)≤(k − 1)/2.

Corollary 3.17 (see [3].)Let (X, f) be an 𝒩-subalgebra of type (q, ∈∨_q). If f is not constant on the open 0-support of (X, f), then f(x)≤−0.5 for some x ∈ X. In particular, f(0)≤−0.5.

Theorem 3.18. An 𝒩-structure (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k) if and only if for every t ∈ [(k − 1)/2, 0] the nonempty closed t-support of (X, f) is a subalgebra of X.

Proof. Assume that (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k) and let t ∈ [(k − 1)/2, 0] be such that C(f; t) ≠ ∅. Let x, y ∈ C(f; t). Then f(x) ≤ t and f(y) ≤ t. It follows from Theorem 3.12 that

so that x*y ∈ C(f; t). Therefore C(f; t) is a subalgebra of X.

Conversely, let (X, f) be an 𝒩-structure such that the nonempty closed t-support of (X, f) is a subalgebra of X for all t ∈ [(k − 1)/2, 0]. If there exist a, b ∈ X such that f(a*b) > ⋁{f(a), f(b), (k − 1)/2}, then we can take s ∈ [−1,0] such that

Thus a, b ∈ C(f; s) and s ≥ (k − 1)/2. Since C(f, s) is a subalgebra of X, it follows that a*b ∈ C(f; s) so that f(a*b) ≤ s. This is a contradiction, and therefore f(x*y) ≤ ⋁{f(x), f(y), (k − 1)/2} for all x, y ∈ X. Using Theorem 3.12, we conclude that (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k).

Corollary 3.19 (see [4].)An 𝒩-structure (X, f) is an 𝒩-subalgebra of type (∈, ∈∨_q) if and only if for every t ∈ [−0.5,0] the nonempty closed t-support of (X, f) is a subalgebra of X.

Theorem 3.20. Let S be a subalgebra of X. For any t ∈ [(k − 1)/2, 0), there exists an 𝒩-subalgebra (X, f) of type (∈, ∈∨q_k) for which S is represented by the closed t-support of (X, f).

Proof. Let (X, f) be an 𝒩-structure in which f is given by

for all x ∈ X where t ∈ [(k − 1)/2, 0). Assume that f(a*b) > ⋁{f(a), f(b), (k − 1)/2} for some a, b ∈ X. Since the cardinality of the image of f is 2, we have f(a*b) = 0 and ⋁{f(a), f(b), (k − 1)/2} = t. Since t ≥ (k − 1)/2, it follows that f(a) = t = f(b) so that a, b ∈ S. Since S is a subalgebra of X, we obtain a*b ∈ S and so f(a*b) = t < 0. This is a contradiction. Therefore f(x*y) ≤ ⋁{f(x), f(y), (k − 1)/2} for all x, y ∈ X. Using Theorem 3.12, we conclude that (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k). Obviously, S is represented by the closed t-support of (X, f).

Corollary 3.21 (see [4].)Let S be a subalgebra of X. For any t ∈ [−0.5,0), there exists an 𝒩-subalgebra (X, f) of type (∈, ∈∨_q) for which S is represented by the closed t-support of (X, f).

Proof. It follows from taking k = 0 in Theorem 3.20.

Theorem 3.22. Let (X, f) be an 𝒩-subalgebra of type (∈, ∈∨q_k) such that f(x)>(k − 1)/2  for all x ∈ X. Then (X, f) is an 𝒩-subalgebra of type (∈, ∈).

Proof. Let x, y ∈ X and t ∈ [−1,0) be such that $$ and $$ are 𝒩_∈-subsets of (X, f). Then f(x) ≤ t₁ and f(y) ≤ t₂. It follows from Theorem 3.12 and the hypothesis that

so that $$ is an 𝒩_∈-subset of (X, f). Therefore (X, f) is an 𝒩-subalgebra of type (∈, ∈).

Corollary 3.23 (see [4].)Let (X, f) be an 𝒩-structure of type (∈, ∈∨_q) such that f(x)>−0.5 for all x ∈ X. Then (X, f) is an 𝒩-subalgebra of type (∈, ∈).

Proof. It follows from taking k = 0 in Theorem 3.22.

Theorem 3.24. Let {(X, f_i)∣i ∈ Λ} be a family of 𝒩-subalgebras of type (∈, ∈∨q_k). Then (X, ⋃_i∈Λf_i) is an 𝒩-subalgebra of type (∈, ∈∨q_k), where ⋃_i∈Λf_i is an 𝒩-function on X given by (⋃_i∈Λf_i)(x) = ⋁_i∈Λf_i(x) for all x ∈ X.

Proof. Let x, y ∈ X and t₁, t₂ ∈ [−1,0) be such that $$ and $$ are 𝒩_∈-subsets of (X, ⋃_i∈Λf_i). Assume that $$ is not an $$-subset of (X, ⋃_i∈Λf_i). Then $$ is neither an 𝒩_∈-subset nor an $$-subset of (X, ⋃_i∈Λf_i). Hence (⋃_i∈Λf_i)(x*y) > ⋁{t₁, t₂} and

which imply that Let $$ is an 𝒩_∈-subset of  (X, f_i)} and $$ is an $$ is not an 𝒩_∈-subset  of  (X, f_j)}. Then Λ = A₁ ∪ A₂ and A₁∩A₂ = ∅. If A₂ = ∅, then $$ is an 𝒩_∈-subset of (X, f_i) for all i ∈ Λ, that is, f_i(x*y) ≤ ⋁{t₁, t₂} for all i ∈ Λ. Thus (⋃_i∈Λf_i)(x*y) ≤ ⋁{t₁, t₂}. This is a contradiction. Hence A₂ ≠ ∅, and so for every i ∈ A₂, we have f_i(x*y) > ⋁{t₁, t₂} and f_i(x*y) + ⋁{t₁, t₂} − k + 1 < 0. It follows that ⋁{t₁, t₂} < (k − 1)/2. Since $$ is an 𝒩_∈-subset of (X, ⋃_i∈Λf_i), we have for all i ∈ Λ. Similarly, f_i(y)<(k − 1)/2 for all i ∈ Λ. Next suppose that t : = f_i(x*y) > (k − 1)/2. Taking (k − 1)/2 < r < t, we know that (X, x_r) and (X, y_r) are 𝒩_∈-subsets of (X, f_i), but (X, (x*y) _⋁{r,r}) = (X, (x*y) _r) is not an $$-subset of (X, f_i). This contradicts that (X, f_i) is an 𝒩-subalgebra of type (∈, ∈∨q_k). Hence f_i(x*y) ≤ (k − 1)/2 for all i ∈ Λ, and so (⋃_i∈Λf_i)(x*y)≤(k − 1)/2 which contradicts (3.22). Therefore $$ is an $$-subset of (X, ⋃_i∈Λf_i) and consequently (X, ⋃_i∈Λf_i) is an 𝒩-subalgebra of type (∈, ∈∨q_k).

Theorem 3.25. An 𝒩-structure (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k) if and only if the ∈∨q_k-support of (X, f) related to t is a subalgebra of X for all t ∈ [−1,0).

Proof. Suppose that (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k). Let $$ for t ∈ [−1,0). Then (X, x_t) and (X, y_t) are $$-subsets of (X, f). Hence f(x) ≤   t or f(x) +   t  −   k  +   1   <  0, and f(y) ≤ t or f(y) + t − k + 1 < 0. Then we consider the following four cases:

•  

  (c1) f(x) ≤ t and f(y) ≤ t,

•  

  (c2) f(x) ≤ t and f(y) + t − k + 1 < 0,

•  

  (c3) f(x) + t − k + 1 < 0 and f(y) ≤ t,

•  

  (c4) f(x) + t − k + 1 < 0 and f(y) + t − k + 1 < 0.

Combining (3.6) and (c1), we have f(x*y) ≤ ⋁{t, (k − 1)/2}. If t ≥ (k − 1)/2, then f(x*y) ≤ t and so (X, (x*y) _t) is an 𝒩_∈-subset of (X, f). Hence $$. If t < (k − 1)/2, then f(x *  y)≤(k − 1)/2 and so f(x *  y) + t − k + 1 < ((k − 1)/2)+((k − 1)/2) − k + 1 = 0, that is, (X, (x*y) _t) is an $$-subset of (X, f). Therefore $$. For the case (c2), assume that t < (k − 1)/2. Then and so f(x*y) + t − k + 1 < 0. Thus (X, (x*y) _t) is an $$-subset of (X, f). If t ≥ (k − 1)/2, then and thus $$ or x*y ∈ C(f; t). Consequently, $$. For the case (c3), it is similar to the case (c2). Finally, for the case (c4), if t ≥ (k − 1)/2, then k − 1 − t ≤ (k − 1)/2 ≤ t. Hence which implies that x*y ∈ C(f; t). If t < (k − 1)/2, then t < (k − 1)/2 < k − 1 − t. Therefore that is, f(x*y) + t − k + 1 < 0, which means that (X, (x*y) _t) is an $$-subset of (X, f). Consequently, the ∈∨q_k-support of (X, f) related to t is a subalgebra of X for all t ∈ [−1,0).

Conversely, let (X, f) be an 𝒩-structure for which the ∈∨q_k-support of (X, f) related to t is a subalgebra of X for all t ∈ [−1,0). Assume that there exist a, b ∈ X such that f(a*b) > ⋁{f(a), f(b), (k − 1)/2}. Then

for some s ∈ [(k − 1)/2, 0). It follows that $$ but a*b ∉ C(f; s). Also, f(a*b) + s − k + 1 > 2s − k + 1 ≥ 0, that is, $$. Thus $$ which is a contradiction. Therefore for all x, y ∈ X. Using Theorem 3.12, we conclude that (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q_k).

Corollary 3.26 (see [4].)An 𝒩-structure (X, f) is an 𝒩-subalgebra of type (∈, ∈∨q) if and only if the ∈∨q-support of (X, f) related to t is a subalgebra of X for all t ∈ [−1,0).