Stabilizers in EQ-algebras

Definition 2.1

([1]) An EQ-algebra is an algebra (E, ∧, ⊗, ∼, 1) of type (2, 2, 2, 0) such that for all x, y, z, t ∈ E:

1. (E, ∧, 1) is a commutative idempotent monoid (i.e. ∧-semilattice with top element 1);

2. (E, ⊗, 1) is a commutative monoid and ⊗ is isotone w.r.t. ≤ (where x ≤ y is defined as x ∧ y = x);

3. x ∼ x = 1 (reflexivity axiom);

4. ((x ∧ y) ∼ z) ⊗ (t ∼ x) ≤ z ∼ (t ∧ y) (substitution axiom);

5. (x ∼ y) ⊗ (z ∼ t) ≤ (x ∼ z) ∼ (y ∼ t) (congruence axiom);

6. (x ∧ y ∧ z) ∼ x ≤ (x ∧ y) ∼ x (monotonicity axiom);

7. (x ∧ y) ∼ x ≤ (x ∧ y ∧ z) ∼ (x ∧ z) (monotonicity axiom);

8. x ⊗ y ≤ x ∼ y (boundedness axiom).

Proposition 2.2

([1]) Let E be an EQ-algebra. Then for any x, y, z ∈ E:

1. x ≤ y implies x → y = 1, x ∼ y = y → x, x̄ ≤ ȳ;

2. x → x = 1, x → 1 = 1, x ≤ x̄, 1̄ = 1;

3. x ≤ y → x;

4. x ≤ y implies z → x ≤ z → y and y → z ≤ x → z;

5. x → y ≤ (z → x) → (z → y) and x → y ≤ (y → z) → (x → z);

6. (x → y) ⊗ (y → z) ≤ x → z.

Definition 2.3

([1, 3, 4]) An EQ-algebra E is said to be

• bounded if it has a a bottom element 0;

• good if x̄ = x for all x ∈ E;

• separated if for all x, y ∈ E, x ∼ y = 1 implies x = y;

• residuated if for all x, y, z ∈ E, (x ⊗ y) ∧ z = x ⊗ y iff x ∧ ((y ∧ z) ∼ y) = x;

• prelinear if for all x, y ∈ E, 1 is the unique upper bound in E of the set {x → y, y → x};

• idempotent if for all x ∈ E, x ⊗ x = x;

• involutive (IEQ-algebra) if it contains a bottom element 0 and for all x ∈ E, ¬¬x = x;

• lattice-ordered if the underlying ∧-semilattice is a lattice;

• a ℓEQ-algebra if it is lattice-ordered and for all x, y, z, u ∈ E, ((x ∨ y) ∼ z) ⊗ (u ∼ x) ≤ ((u ∨ y) ∼ z).

Proposition 2.4

([1, 3]) Let E be a good EQ-algebra. Then for any x, y, z ∈ E:

1. x → (y → z) = y → (x → z);

2. x ≤ (x → y) → y;

3. For all indexed families {a_i} in E, provided that {a_i} has supremum in E, we have ∨_ia_i → c = ∧_i(a_i → c);

4. x ⊗ (x → y) ≤ y.

Proposition 2.5

([1, 3]) Let E be a residuated EQ-algebra. Then for any x, y, z ∈ E:

1. x ⊗ y ≤ z iff x ≤ y → z;

2. (x ⊗ y) → z = x → (y → z);

3. x ≤ y → (x ⊗ y);

4. x → y ≤ (x ⊗ z) → (y ⊗ z).

Proposition 2.6

([4]) (1) If E is a prelinear and separated lattice-ordered EQ-algebra, then (x ∧ y) → z = (x → z) ∨ (y → z) for any x, y, z ∈ E.

(2) If E is a prelinear and good lattice-ordered EQ-algebra, then E is an ℓEQ-algebra and x ∨ y = ((x → y) → y) ∧ ((y → x) → x) for any x, y, z ∈ E.

Definition 2.7

([4]) A nonempty subset F of an EQ-algebra E is called a prefilter of E if it satisfies:

1. 1 ∈ F,

2. x ∈ F, x → y ∈ F imply y ∈ F for all x, y ∈ E.

   A prefilter F is called a filter if it satisfies:

3. x → y ∈ F implies (x ⊗ z) → (y ⊗ z) ∈ F for all x, y, z ∈ E.

Definition 2.8

([5]) A prefilter F of an EQ-algebra E is called

• a positive implicative prefilter of E if for any x, y, z ∈ E, x → (y → z) ∈ F and x → y ∈ F imply x → z ∈ F;

• an obstinate prefilter of E if for any x, y ∈ E, x, y ∉ F implies x → y ∈ F and y → x ∈ F.

Definition 2.9

([6]) Let μ be a fuzzy set of an EQ-algebra E. Then μ is called a fuzzy prefilter of E if it satisfies:

1. μ(1) ≥ μ(x) for all x ∈ E,

2. μ(y) ≥ μ(x) ∧ μ(x → y) for all x, y ∈ E.

   A fuzzy prefilter μ is called a fuzzy filter if it satisfies:

3. μ((x ⊗ z) → (y ⊗ z)) ≥ μ(x → y) for all x, y, z ∈ E.

Proposition 2.10

([6]) Let μ be a fuzzy prefilter of an EQ-algebra E. Then for all x, y ∈ E:

1. x ≤ y implies μ(x) ≤ μ(y);

   Furthermore, if μ is a fuzzy filter, we have:

2. μ(x ⊗ y) = μ(x) ∧ μ(y);

3. μ(x ∧ y) ≥ μ(x) ∧ μ(y).

Theorem 2.11

([18]) Let μ be a fuzzy set of a good EQ-algebra E. If μ satisfied (FF1),

1. x ≤ y implies μ(x) ≤ μ(y),

2. μ(x ⊗ y) ≥ μ(x) ∧ μ(y),

   for any x, y ∈ E, then μ is a fuzzy prefilter of E.

Definition 2.12

([6]) Let μ be a fuzzy set of an EQ-algebra E. For all x, y, z ∈ E, μ is called a

• fuzzy positive implicative prefilter of E if μ is a prefilter of E and μ(x → z) ≥ μ(x → (y → z)) ∧ μ(x → y);

• fuzzy implicative prefilter of E if μ(x → z) ≥ μ(x → (¬z → y)) ∧ μ(y → z).

Theorem 2.13

([6]) Let μ be a fuzzy prefilter of E. Then

1. μ is a fuzzy positive implicative filter of E if and only if μ(x ∧ (x → y) → y) = μ(1) for all x, y ∈ E;

2. μ is a fuzzy implicative filter of E if and only if μ(x) = μ(¬x → x) for all x ∈ E.

Definition 2.14

([18]) A fuzzy relation R on E is called a fuzzy congruence relation if R is a fuzzy equivalence relation satisfying R(x △ u, y △ v) ≥ R(x, y) ∧ R(u, v) for any x, y, u, v ∈ E, where △ ∈ {⊗, ∧, ∼}.

Lemma 2.15

([18]) Let μ be a fuzzy filter of E and μ(1) = 1. Define a fuzzy relation R on E by R(x, y) = μ(x ∼ y) for x, y ∈ E. Then R is a fuzzy congruence relation on E, which is called the generated fuzzy congruence relation by μ.

Lemma 2.16

([18]) Assume that R is a fuzzy congruence relation on a good EQ-algebra E. Define a fuzzy set μ(x) = R(1, x) for x ∈ E. Then μ is a fuzzy prefilter of E.

Definition 3.1

Let E be an EQ-algebra and A be a nonempty subset of E. The right stabilizer and left stabilizer of A are defined as follows:

Example 3.2

Let E = {0, a, b, c, d, 1} with 0 < a < b < d < 1, a < c < d. Define operations ⊗ and ∼ on E as follows:

Then (E, ∧, ⊗, ∼, 1) is an EQ-algebra. One can calculate that the right stabilizer and the left stabilizer of A are St_r(A) = {d, 1}, St_l(A) = {a, 1} by taking A = {c, 1}.

Proposition 3.3

Let A, B, A_i, B_i(i ∈ I) be nonempty subsets of E. We have:

1. If A ⊆ B, then St_r(B) ⊆ St_r(A) and St_l(B) ⊆ St_l(A);

2. St_l(E) = {1} and St_r({1}) = E;

3. If E is good, then St_r(E) = {1} and St_l({1}) = E;

4. St_r(A) = ⋂{St_r({x}) : x ∈ A} and St_l(A) = ⋂{St_l({x}) : x ∈ A};

5. ⋂St_r(B_i) ⊆ St_r(⋂ B_i) and ⋂ St_l(B_i) ⊆ St_l(⋂ B_i);

6. If f : E → E is a homomorphism and x ∈ E, then f(St_r({x})) ⊆ St_r({f(x)}) and f(St_l({x})) ⊆ St_l({f(x)});

7. If E is residuated, then a ∈ St_r(A) implies aⁿ ∈ St_r(A) for any n ∈ N.

Proof

(1) and (4) are straightforward.

(2) {1} ⊆ St_l(E) follows from x → 1 = 1 for all x ∈ E. Let a ∈ St_r(E). Then x → a = a for all x ∈ E. Hence a = 1 by taking x = a. Similarly, St_l({1}) = E.

(3) {1} ⊆ St_r(E) follows from 1 → x = x for all x ∈ E. Let a ∈ St_r(E). Then a → x = x for all x ∈ E. Hence a = 1 by taking x = a. Similarly, St_l({1}) = E.

(5) It is clear by (1).

(6) Let b ∈ f(St_r({x})). Then there is a ∈ St_r({x}), namely, a → x = x such that b = f(a). Hence f(a) → f(x) = f(a → x) = f(x), which implies b = f(a) ∈ St_r({f(x)}). The proof of f(St_l({x})) ⊆ St_l({f(x)}) is similar.

(7) Let a ∈ St_r(A). Then for all x ∈ A, a → x = x. Hence a² → x = a ⊗ a → x = a → (a → x) = a → x = x, which implies a² ∈ St_r(A). Now let a^n–1 ∈ St_r(A). Then aⁿ → x = a ⊗ a^n–1 → x = a → (a^n–1 → x) = a → x = x. It follows that aⁿ ∈ St_r(A).□

Theorem 3.4

Let E be a good EQ-algebra and A ⊆ E. Then St_r(A) is a prefilter of E.

Proof

It follows from 1 → x = x for all x ∈ A that 1 ∈ St_r(A). Let a, a → b ∈ St_r(A) for any a, b ∈ E. Then a → x = x and (a → b) → x = x for all x ∈ A. Hence b → x ≤ (a → b) → (a → x) = a → ((a → b) → x) = a → x = x. On the other hand, x ≤ b → x. So we obtain x = b → x for any x ∈ A. That is, b ∈ St_r(A). Therefore St_r(A) is a prefilter of E.□

Theorem 3.5

Let E be an idempotent and residuated EQ-algebra. Then for A ⊆ E, St_r(A) is a positive implicative prefilter of E.

Proof

According to Theorem 3.4, St_r(A) is a prefilter of E. Now let a → (b → c), a → b ∈ St_r(A) for any a, b, c ∈ E. Then (a → (b → c)) → x = x and (a → b) → x = x for all x ∈ A. Hence from Proposition 2.4 and Proposition 2.5, (a → b) ⊗ (b → (a → c)) → x = (b → (a → c)) → ((a → b) → x) = (a → (b → c)) → ((a → b) → x) = (a → (b → c)) → x = x. So by (p6) (a → c) → x = (a ⊗ a → c) → x = (a → (a → c)) → x ≤ (a → b) ⊗ (b → (a → c)) → x = x, and we have (a → c) → x ≤ x. Combining (a → c) → x ≥ x, it follows that (a → c) → x = x. This shows a → c ∈ St_r(A) and therefore St_r(A) is a positive implicative prefilter of E.□

Definition 3.6

([18]) Let E be a lattice-ordered EQ-algebra. A proper prefilter F of E is called prime if x ∨ y ∈ F implies x ∈ F or y ∈ F for any x, y ∈ E.

Theorem 3.7

Let E be a good lattice-ordered EQ-algebra and a be a co-atom in E. Then St_r({a}) is a prime prefilter of E.

Proof

According to Theorem 3.4 St_r({a}) is a proper prefilter of E. Since a is a co-atom, we have a ≠ 1. Now we put x ∨ y ∈ St_r({a}), but x ∉ St_r({a}) and y ∉ St_r({a}). That is, x, y → a ≠ a. Hence by (p3) a < x → a and a < y → a. Thus, x → a = 1 and y → a = 1 as a is a co-atom. Combining (3) of Proposition 2.4, we get x ∨ y → a = (x → a) ∧ (y → a) = 1, which contracts to x ∨ y → a = 1 and a ≠ 1. Therefore St_r({a}) is a prime prefilter of E.□

Theorem 3.8

1. If E is an IEQ-algebra, then St_r({0}) = {1};

2. If E is a good EQ-algebra, then St_r({0}) = {1} if and only if x → y, y → x ∈ St_r({0}) implies x = y for any x, y ∈ E.

3. St_r({0}) = A if and only if 0 ∈ St_l(A).

Proof

1. Let x ∈ St_r({0}). Then x → 0 = 0 and thus x = ¬¬x = (x → 0) → 0 = 0 → 0 = 1. This implies St_r({0} = {1}.

2. Let St_r({0}) = {1} and x → y, y → x ∈ St_r({0}) for any x, y ∈ E. Then x → y, y → x = 1 and so x = y. Conversely, let a ∈ St_r({0}). Since a → 1 = 1 ∈ St_r({0}) and 1 → a = a ∈ St_r({0}), by hypothesis we have a = 1 and therefore St_r({0}) = {1}.

3. St_r({0}) = A iff ¬x = x → 0 = 0 for all x ∈ A iff 0 ∈ St_l(A).□

Definition 3.9

A prefilter F of E is called a normal prefilter of E if z ∈ F and z → ((y → x) → x) ∈ F imply (x → y) → y ∈ F for all x, y, z ∈ E.

Example 3.10

Let E = {0, a, b, 1} with 0 < a < b < 1. Define operations ⊗ and ∼ as follows:

Then (E, ∧, ⊗, ∼, 1) is an EQ-algebra [5]. Routine calculation shows that {b, 1} is a normal prefilter of E.

Lemma 3.11

Let F be a prefilter of a good EQ-algebra E. Then F is a normal prefilter of E if and only if (y → x) → x ∈ F implies (x → y) → y ∈ F for all x, y ∈ E.

Proof

Let F be a normal prefilter of E and (y → x) → x ∈ F for any x, y ∈ E. As 1 → ((y → x) → x) = (y → x) → x ∈ F and 1 ∈ F, we have (x → y) → y ∈ F. Conversely, let z ∈ F and z → ((y → x) → x) ∈ F. Since F is a prefilter of E, then (y → x) → x ∈ F and by hypothesis (x → y) → y ∈ F.□

Theorem 3.12

Let E be a good EQ-algebra.

1. If F is a normal prefilter of E, then St_r({0}) ⊆ F;

2. If F = {1} is a normal prefilter of E, then St_r({0}) = {1};

3. St_r({0}) is a normal prefilter if and only if St_r({0}) is the intersection of all normal prefilters of E;

4. If St_r({0}) = E – {0}, then St_r({0}) is a normal prefilter of E.

Proof

(1) Let x ∈ St_r({0}). Then (x → 0) → 0 = 1 ∈ F. Since F is a normal prefilter of E, by Lemma 3.11 we have (0 → x) → x = x ∈ F.

(2) and (3) By (1).

(4) Since St_r({0}) = E – {0} and St_r({0}) is a prefilter of E, we can easily prove that (x → y) → y, (y → x) → x ∈ St_r({0}) for all x, y ∈ E.□

Definition 3.13

An EQ-algebra E is said to be simple if it has no non-trivial prefilter.

Example 3.14

Let E = {0, a, b, c, d, 1} with 0 < a < c < d < 1, 0 < b < c < d < 1. Define operations ⊗ and ∼ on E as follows:

Then one can easily verify that (E, ∧, ⊗, ∼, 1) is a simple EQ-algebra.

Theorem 3.15

Let E be an idempotent and residuated EQ-algebra. Then the following are equivalent:

1. E is simple;

2. St_r({a}) = {1} for any a ≠ 1 in E.

Proof

(1) ⇒ (2) Suppose that E is simple. If for some a ≠ 1 such that St_r({a}) ≠ {1}, then there exists x ≠ 1 such that x ∈ St_r({a}). So, we have x → a = a. By the simplicity of E, we get < x > = E and hence for some n ≥ 1, x →ⁿ a = 1 as a ∈ E. On the other hand, since E is idempotent and residuated, it follows from Proposition 2.5 that x →ⁿ a = x → (x →^n–1 a) = (x ⊗ x) →^n–1 a = x →^n–1 a = ⋯ = x → a. This shows a = 1, which is a contradiction. Therefore St_r({a}) = {1} for any a ≠ 1.

(2) ⇒ (1) Suppose St_r({a}) = {1} for all a ≠ 1 in E. To prove that E is simple, it suffice to show that < y > = E for any y ≠ 1. Now if x ∈ E, x ≠ 1 such that y →ⁿ x ≠ 1, then by hypothesis y → (y →ⁿ x) ≠ y →ⁿ x. That is, y →ⁿ⁺¹ x ≠ y →ⁿ x which contradicts to the fact that E is idempotent and residuated. This indicates y →ⁿ x = 1 and hence x ∈ < y >. Therefore < y > = E for any y ≠ 1.□

Definition 3.16

Let A, B be two nonempty subsets of E. The right, left stabilizer of A with respect to B are defined by:

Example 3.17

Consider the EQ-algebra E in Example 3.2. It is not difficult to check that the right, left stabilizer of A with respect to B are St_r(A, B) = {b, d, 1}, St_l(A, B) = {0, a, d, 1} by taking A = {c, 1}, B = {d, 1}.

Proposition 3.18

Let A, B, A_i, B_i be nonempty subsets and F be a prefilter of E. We have:

1. If E is good and St_r(A, B) = E (St_l(A, B) = E), then A ⊆ B;

2. F ⊆ B iff St_r(F, B) = E;

3. If E is good, then F ⊆ B iff St_l(F, B) = E;

4. St_r(F, F) = E. Also, if E is good, then St_l(F, F) = E;

5. St_r(A) ⊆ St_r(A, F) and St_l(A) ⊆ St_l(A, F);

6. If A_i ⊆ B_i and A_j ⊆ B_j, then St_r(B_i, A_j) ⊆ St_r(A_i, B_j) and St_l(B_i, A_j) ⊆ St_l(A_i, B_j);

7. If E is separated, then St_r(A, {1}) = St_r(A) and St_l(A, {1}) = St_l(A);

8. St_r(A, ⋂ B_i) ⊆ ⋂ St_r(A, B_i) and St_l(A, ⋂ B_i) ⊆ ⋂ St_l(A, B_i).

Proof

1. Let St_r(A, B) = E (St_l(A, B) = E). Then for any x ∈ A, (x → x) → x = 1 → x ∈ B. Hence A ⊆ B.

2. Let F ⊆ B and a ∈ E. Then for any x ∈ F, x ≤ (a → x) → x. Since F is a prefilter of E, we have (a → x) → x ∈ F and so (a → x) → x ∈ B. It follows that St_r(F, B) = E.

3. Let F ⊆ B and a ∈ E. Since E is good, then for any x ∈ F, x ≤ (x → a) → a. Considering that F is a prefilter of E, we obtain that (x → a) → a ∈ F and thus (x → a) → a ∈ B. It follows that St_l(F, B) = E.

4. By (2) and (3).

5. Let a ∈ St_r(A). Then for all x ∈ A, a → x = x and hence (a → x) → x = 1 ∈ F. This implies a ∈ St_r(A, F), that is, St_r(A) ⊆ St_r(A, F). Similarly, St_l(A) ⊆ St_l(A, F).

6. Let a ∈ St_r(B_i, A_j). Then (a → x) → x ∈ A_j for all x ∈ B_i. Hence (a → x) → x ∈ B_j for all x ∈ A_i, which shows St_r(B_i, A_j) ⊆ St_r(A_i, B_j). Similarly, St_l(B_i, A_j) ⊆ St_l(A_i, B_j).

7. Let a ∈ St_r(A, {1}). Then for all x ∈ A, (a → x) → x = 1 and so a → x ≤ x. On the other hand, x ≤ a → x. It follows that a → x = x for all x ∈ A. Further a ∈ St_r(A). Conversely, let a ∈ St_r(A). Then a → x = x for all x ∈ A. It implies that (a → x) → x = 1 and hence a ∈ St_r(A, {1}). Therefore St_r(A, {1}) = St_r(A). The other part is similar.

8. Let a ∈ St_r(A, ⋂ B_i). Then (a → x) → x ∈ ⋂ B_i for all x ∈ A. Hence (a → x) → x ∈ B_i for every i and all x ∈ A. This shows a ∈ ⋂ St_r(A, B_i). Similarly, St_l(A, ⋂ B_i) ⊆ ⋂ St_l(A, B_i).□

Theorem 3.19

Let A ⊆ E and F be a prefilter of a good EQ-algebra E. If F is a normal prefilter of E, then St_r(A, F) = St_l(A, F).

Proof

By Lemma 3.11.□

Theorem 3.20

Let F(x) = {y ∈ E : x ≤ y} be a normal prefilter of a good EQ-algebra E for all x ∈ E. Then St_l(A, B) = St_r(A, B) for all A, B ⊆ E,

Proof

Let F(x) be a normal prefilter of E for all x ∈ E. We have (b → a) → a ∈ F((a → b) → b) as (a → b) → b ∈ F((a → b) → b). It implies (b → a) → a ≤ (a → b) → b. Symmetrically, (a → b) → b ≤ (b → a) → a. Therefore (a → b) → b = (b → a) → a for all a, b ∈ E and so St_r(A, B) = St_l(A, B).□

Lemma 3.21

In any good EQ-algebra E, y → x = ((y → x) → x) → x and ¬x = ¬¬¬x for all x, y ∈ E.

Proof

Let x, y ∈ E. By Proposition 2.4, y → x ≤ ((y → x) → x) → x and y ≤ (y → x) → x. Hence by (p4) ((y → x) → x) → x ≤ y → x. Therefore y → x = ((y → x) → x) → x.□

Theorem 3.22

Let E be a good EQ-algebra and A, B be two prefilters of E. Then St_r(A, B) is a prefilter of E.

Proof

Let a, a → b ∈ St_r(A, B) for any a, b ∈ E. Then for all x ∈ A, (a → x) → x, ((a → b) → x) → x ∈ B. Since x ≤ a → x, we have a → x ∈ A, and hence ((a → b) → (a → x)) → (a → x) ∈ B. Again by (p5), b → x ≤ (a → b) → (a → x). It follows that ((a → b) → (a → x)) → (a → x) ≤ (b → x) → (a → x). This implies (b → x) → (a → x) ∈ B. Considering a → x = ((a → x) → x) → x from Lemma 3.21, we obtain that ((a → x) → x) → ((b → x) → x) = (b → x) → ((a → x) → x) → x) = (b → x) → (a → x). Therefore (b → x) → x ∈ B and so b ∈ St_r(A, B). That is, St_r(A, B) is a prefilter of E.□

Definition 3.23

A nonempty subset F of a bounded lattice-ordered EQ-algebra E is called a Boolean prefilter of E if F is a prefilter and x ∨ ¬x ∈ F for all x ∈ E.

Example 3.24

Let E be the EQ-algebra given in Example 3.2. Then one can check that F = {b, c, d, 1} is a Boolean prefilter of E.

Theorem 3.25

Let E be a bounded good lattice-ordered EQ-algebra. If A is a prefilter and B is a Boolean prefilter of E, then St_r(A, B) is a Boolean prefilter of E.

Proof

By Theorem 3.22 St_r(A, B) is a prefilter of E. Let a ∈ E. Then from Proposition 2.4, for all x ∈ A, a ∨ ¬a ≤ ((a → x) → x) ∨ ((¬a → x) → x) ≤ ((a → x) ∧ (¬a → x)) → x = ((a ∨ ¬a) → x) → x. Since B is a Boolean prefilter, we have a ∨ ¬a ∈ B. It follows that ((a ∨ ¬a) → x) → x ∈ B, which shows St_r(A, B) is a Boolean prefilter of E.□

Theorem 3.26

Let E be a good EQ-algebra. If A is a prefilter and B is an obstinate prefilter of E, then St_r(A, B) is an obstinate prefilter of E.

Proof

By Theorem 3.22 St_r(A, B) is a prefilter of E. Now let a, b ∉ St_r(A, B). Then (a → x) → x ∉ B and (b → x) → x ∉ B for all x ∈ A. Considering a ≤ (a → x) → x, b ≤ (b → x) → x, we have a, b ∉ B. Since B is an obstinate prefilter, then a → b, b → a ∈ B. Thus ((a → b) → b) → b = a → b ∈ B and ((b → a) → a) → a = b → a ∈ B for all a, b ∈ A. This shows that St_r(A, B) is an obstinate prefilter of E.□

Theorem 3.27

Let E be a good EQ-algebra. Then (𝓟𝓕(E), ⊓, ⊔, {1}, E) is a relative pseudo-complement lattice, where St_r(F, G) is the relative pseudo-complement of F with respect to G in 𝓟𝓕(E).

Proof

By Theorem 3.22 St_r(F, G) ∈ 𝓟𝓕(E). Now we prove St_r(F, G) ∩ F ⊆ G for any F, G ∈ 𝓟𝓕(E). Let a ∈ St_r(F, G) ∩ F. Then a ∈ F and (a → x) → x ∈ G for all x ∈ F. Taking x = a, we have (a → a) → a = 1 → a = a ∈ G. Suppose that M ∈ 𝓟𝓕(E) such that M ∩ F ⊆ G. Let b ∈ M. Since b, x ≤ (b → x) → x for all x ∈ F and F, M are prefilters of E, we obtain (b → x) → x ∈ F ∩ M for all x ∈ F, which implies (b → x) → x ∈ G and so b ∈ St_r(F, G). That is, M ⊆ St_r(F, G). Therefore St_r(F, G) is the relative pseudo-complement of F with respect to G in 𝓟𝓕(E).□

Corollary 3.28

Let E be a good EQ-algebra. Then (𝓟𝓕(E), ⊓, ⊔, {1}, E) is a pseudo-complement lattice, where St_r(F) is the pseudo-complement of F in 𝓟𝓕(E).

Definition 4.1

Let μ be a fuzzy set of E and R be a fuzzy congruence relation on E. Define the fuzzy right-stabilizer and left-stabilizer of μ with respect to R as follows: for x ∈ E

where → is the residuated implication with respect to a left-continuous t-norm.

Example 4.2

Consider the EQ-algebra E in Example 3.2. We define a fuzzy set μ by μ(0) = μ(a) = μ(b) = μ(c) = μ(d) = 0.4, μ(1) = 1. One can check that μ is a fuzzy filter of E and also the fuzzy congruence relation R is generated by μ. Then the fuzzy right-stabilizer and left-stabilizer of μ with respect to R are StRr (μ) = E; StRl(μ)(0)=StRl(μ)(a)=StRl(μ)(c)=StRl(μ)(d)=StRl(μ)(1),StRl(μ)(b)=0.4 , where the residuated implication → is taken as ffukasiewicz implication.

Proposition 4.3

Let R be a fuzzy congruence relation on E and μ, ν be fuzzy sets of E. We have:

1. If μ ⊆ ν, then StRr(ν)⊆StRr(μ)  and  StRl(ν)⊆StRl(μ);

2. StRr(μ∪ν)=StRr(μ)∩StRr(ν)  and  StRl(μ∪ν)=StRl(μ)∩StRl(ν);

3. StRr(μ∩ν)=StRr(μ)∪AnnR(ν)  and  StRl(μ∩ν)=StRl(μ)∪StRl(ν);

4. StRr(χ{1})=E  and  StRl(χ{1})=R(⋅¯,⋅), where χ_{1} is defined by χ_{1}(1) = 1, χ_{1}(otherwise) = 0;

5. If E is bounded, then StRr(χ{0})=R(¬⋅,0)  and  StRl(χ{0})=R(⋅,1), where χ_{0} is defined by χ_{0}(1) = 1, χ_{0}(otherwise) = 0.

Proof

1. Let μ ⊆ ν. For x ∈ E, it follows from (p4) that Stlr (ν)(x) = ∧_z∈E{ν(z) → R(z → x, x)} ≤ ∧_z∈E{μ(z) → R(z → x, x)} = Stlr (μ)(x). The proof of StRr(ν)⊆StRr(μ) is similar.

2. For x ∈ E, Stlr (μ ∪ ν)(x) = ∧_z∈E{μ(z) ∨ ν(z) → R(z → x, x)} = ∧_z∈E{(μ(z) → R(z → x, x)) ∧ (ν(z) → R(z → x, x))} = Stlr(μ)∩Stlr(ν). Similarly, StRr(μ∪ν)=StRr(μ)∩StRr(ν).

3. For x ∈ E, StRl (μ ∩ ν)(x) = ∧_z∈E{μ(z) ∧ ν(z) → R(z → x, x)} = ∧_z∈E{(μ(z) → R(z → x, x)) ∨ (ν(z) → R(z → x, x))} = StRl(μ)∪StRl(ν). Similarly, StRr(μ∩ν)=StRr(μ)∪StRr(ν).

4. For x ∈ E, StRr (χ_{1}) = E follows from StRr (χ_{1})(x) = ∧_z∈E{χ_{1}(z) → R(x → z, z)} = 1→ R(1, x → 1) = R(1, 1) = 1 and StRl (χ_{1}) = R(⋅̄, ⋅) follows from StRl (χ_{1})(x) = ∧_z∈E{χ_{1}(z) → R(z → x, x)} = 1 → R(1 → x, x) = R(x̄, x).

5. For x ∈ E, StRr (χ_{0}) = R(¬ ⋅, 0) follows from StRr (χ_{0})(x) = ∧_z∈E{χ_{0}(z) → R(x → z, z)} = 1 → R(x → 0, 0) = R(¬ x, 0) and StRl (χ_{0}) = R(⋅, 1) follows from StRl (χ_{0})(x) = ∧_z∈E{χ_{0}(z) → R(z → x, x)} = 1 → R(0 → x, 1) = R(1, x).□

Lemma 4.4

Let R be a fuzzy congruence relation on a good lattice-ordered EQ-algebra E. Then R(x, y) = R(x ↔ y, 1) for any x, y ∈ E, where x ↔ y = (x → y) ∧ (y → x).

Proof

It is clear that R(x ↔ y, 1) = R(x ↔ y, y ↔ y) ≥ R(x, y) ∧ R(y, y) = R(x, y). On the other hand, since x ⊗ (x → y) ≤ y, we have R(x ↔ y, 1) = R(x ↔ y, 1) ∧ R(x, x) = R(x ⊗ (x ↔ y), x ⊗ 1) = R(x ⊗ (x ↔ y), x) ∧ R(y, y) ≤ R(((x ⊗ (x ↔ y)) ∨ y), x ∨ y) = R(y, x ∨ y). By the similar way, R(x ↔ y, 1) ≤ R(x, x ∨ y). Hence R(x ↔ y, 1) ≤ R(x, x ∨ y) ∧ R(y, x ∨ y) ≤ R(x ∧ (x ∨ y), y ∧ (x ∨ y)). It follows that R(x, y) = R(x ↔ y, 1).□

Theorem 4.5

Suppose that E is a prelinear and good lattice-ordered EQ-algebra satisfying x ∧ (x → y) ≤ y for all x, y ∈ E, and μ is a fuzzy set of E and R is a fuzzy congruence relation on E. Then StRr (μ) is a fuzzy prefilter of E.

Proof

Let E be a prelinear and good EQ-algebra. Clearly, StRr (μ)(1) = ∧_{z ∈ E}{μ(z) → R((1 → z, z)} = 1 ≥ StRr (μ)(x). Now let x, y ∈ E. Since R(⋅, 1) is a fuzzy prefilter of E, from (3) of Proposition 2.4, 2.6, 2.10 and Lemma 2.16, 4.4 we have:

Therefore StRr (μ) is a fuzzy prefilter of E.□

Example 4.6

Let E = {0, a, b, c, d, e, f, 1} with 0 < a, b < d, a < c, d < f, c < e, f < 1. Define operations ⊗ and ∼ on E as follows:

Then (E, ∧, ⊗, ∼, 1) is a prelinear and good EQ-algebra. Also we obtain the implication → on E below:

It is easy to verify that x ∧ (x → y) ≤ y for all x, y ∈ E.

Definition 4.7

Let μ, ν be two fuzzy sets of E. Define the fuzzy right-stabilizer and the fuzzy left-stabilizer of μ with respect to ν by

where → is the residuated implication with respect to a left-continuous t-norm.

Example 4.8

Consider the EQ-algebra E in Example 3.2. Take two fuzzy sets μ, ν as follows: μ(0) = μ(a) = 0.4, μ(b) = 0.5, μ(c) = 0.8, μ(d) = μ(1) = 1; ν(0) = ν(a) = ν(b) = 0.2, ν(c) = ν(d) = ν(1) = 0.9. Then the fuzzy right-stabilizer and the left-stabilizer of μ with respect to ν are St^r(μ, ν)(0) = St^r(μ, ν)(a) = 0.8, St^r(μ, ν)(b) = St^r(μ, ν)(c) = 0.9, St^r(μ, ν)(d) = St^r(μ, ν)(1) = 0.7; St^l(μ, ν)(0) = St^l(μ, ν)(a) = 0.8, St^l(μ, ν)(b) = 0.2, St^l(μ, ν)(c) = St^l(μ, ν)(d) = St^r(μ, ν)(1) = 0.9, where the residuated implication → is taken as ffukasiewicz implication.

Proposition 4.9

Let μ, ν, μ_i, ν_i(i ∈ I) be fuzzy sets of E. Then we have:

1. St^r(χ_{1}, ν) = E if and only if ν(1) = 1;

2. If E is bounded, then St^l(χ_{0}, ν) = E if and only if ν(1) = 1;

3. If E is bounded, then St^r(χ_{0}, ν) = ν iff ν(¬ ¬ x) = ν(x);

4. If μ₁ ⊆ μ₂ and ν₁ ⊆ ν₂, then St^r(μ₂, ν₁) ⊆ St^r(μ₁, ν₂) and St^l(μ₂, ν₁) ⊆ St^l(μ₁, ν₂);

5. St^r(μ, ⋂_i∈Iν_i) = ⋂_i∈ISt^r(μ, ν_i) and St^l(μ, ⋂_i∈Iν_i) = ⋂_i∈ISt^l(μ, ν_i);

6. St^r(⋃_i∈Iμ_i, ν) = ⋂_i∈ISt^r(μ_i, ν) and St^l(⋃_i∈Iμ_i, ν) = ⋂_i∈ISt^l(μ_i, ν);

7. If E is bounded, then D(E) = StRr (χ_{0}, χ_{1}), where D(E) = {x ∈ E : ¬ x = 0}.

Proof

1. It follows from St^r(χ_{1}, ν)(x) = ∧_y∈E{χ_{1}(y)→ν((x → y) → y)} = ν(1).

2. It follows from St^l(χ_{0}, ν)(x) = ∧_y∈E{χ_{0}(y)→ν((y → x) → x)} = ν(1).

3. It follows from St^r(χ_{0}, ν)(x) = ∧_y∈E{χ_{0}(y)→ν((x → y) → y)} = ν(¬ ¬ x).

4. Since St^r(μ₂, ν₁)(x) = ∧_y∈E{μ₂(y)→ν₁((x → y) → y)} ≤ ∧_y∈E{μ₁(y)→ν₂((x → y) → y)} = St^r(μ₁, ν₂)(x), we get St^r(μ₂, ν₁)⊆ St^r(μ₁, ν₂). Similarly, St^l(μ₂, ν₁)⊆ St^l(μ₁, ν₂).

5. Since St^r(μ, ⋂_i∈Iν_i) = ∧_y∈E{μ(y)→∧_i∈Iν_i((x → y) → y)} = ∧_y∈E ∧_i∈I(μ(y)→ν_i((x → y) → y)) = ∧_i∈I ∧_y∈E{μ(y)→ν_i((x → y) → y)} = ∧_i∈ISt^r(μ, ν_i)(x), then St^r(μ, ⋂_i∈Iν_i) = ⋂_i∈ISt^r(μ, ν_i).

6. Since St^l(⋃_i∈Iμ_i, ν) = ∧_y∈E{∨_i∈Iμ_i(y)→ν((y → x) → x)} = ∧_y∈E ∧_i∈I(μ_i(y)→ν((y → x) → x) = ∧_i∈I ∧_y∈E{μ_i(y)→ν((y → x) → x)} = ∧_i∈ISt^l(μ_i, ν)(x), then St^l(⋃_i∈Iμ_i, ν) = ⋂_i∈ISt^l(μ_i, ν).

7. For x ∈ D(E), StRr (χ_{0}, χ_{1}) = ∧_y∈E{χ_{0}(y)→χ_{1}((x → y) → y)} = χ_{1}((x → 0) → 0) = χ_{1}(1) = 1.□

Proposition 4.10

Let μ, ν be two fuzzy sets of a good EQ-algebra E. Then we have:

1. St^l(χ_{1}, ν) = ν;

2. If St^r(μ, ν) = E or St^l(μ, ν) = E, then μ ⊆ ν;

3. If ν is a fuzzy prefilter such that μ ⊆ ν, then St^l(μ, ν) = E. In particular, St^l(ν, ν) = E.

4. If μ is a fuzzy prefilter such that μ ⊆ ν, then St^l(μ, ν) = E;

5. If ν is a fuzzy prefilter, then ν ⊆ St^r(μ, ν).

Proof

1. Since St^l(χ_{1}, ν)(x) = ∧_y∈E{χ_{1}(y)→ν((y → x) → x)} = ν(x), we have St^∧ (χ_{1}, ν) = ν.

2. If St^r(μ, ν) = E, then for all x ∈ E, St^r(μ, ν)(x) = ∧_y∈E{μ(y)→ν((x → y) → y)} = 1. Hence μ(x)→ν((x → x) → x) = μ(x)→ν(x) = 1. This implies μ(x) ≤ ν(x) for all x ∈ E, that is, μ ⊆ ν. The other part is similar.

3. If μ ⊆ ν and ν is a fuzzy prefilter, then St^l(μ, ν)(x) = ∧_y∈E{μ(y)→ν((y → x) → x)} ≥ ∧_y∈E{μ(y)→ν(y)} = 1. This shows St^l(μ, ν) = E.

4. If μ ⊆ ν and μ is a fuzzy prefilter, then St^l(μ, ν)(x) = ∧_y∈E{μ(y)→ν((y → x) → x)} ≥ ∧_y∈E{μ((y → x) → x)→ν((y → x) → x)} = 1. This implies St^l(μ, ν) = E.

5. Since ν is a fuzzy prefilter, then St^r(μ, ν) = ∧_y∈E{μ(y)→ν((x → y) → y)} ≥ ∧_y∈Eν((x → y) → y)) ≥ ∧_y∈Eν(x) = ν(x). Hence ν ⊆ St^r(μ, ν).□