Solution to Turunen’s open problems on n -fold filters

Abstract

In this paper, we shall give answers to show that two open problems, which is proposed by Turunen et al. in (Erratum to:) n-Fold implicative basic logic is Gödel logic [Soft Computing 16(2012): 177–181,183]. That is, each n-fold positive implicative filter is an n-fold Boolean filter and in an n-fold fantastic BL-algebra, n-fold implicative basic logic and n-fold positive implicative basic logic coincide.

Keywords

Non-classical logics BL-algebras

1 Introduction and preliminaries

A residuated lattice L = (L, ∧ , ∨ , ⊗ , → , 0, 1) is an algebra, where (L, ∧ , ∨ , 0, 1) is a bounded lattice with the least element 0 and the greatest element 1, (L, ⊗ , 1) is a commutative monoid and (⊗ , →) forms an adjoint pair, i.e. x ⊗ y ≤ z iff x ≤ y → z. When we refer to a commutative residuated lattice, we will omit the word commutative since we will deal here only with the commutative case. A residuated lattice is said to be a BL-algebra, if x ∧ y = x ⊗ (x → y) and (x → y) ∨ (y → x) = 1, for all x, y ∈ L. Moreover, we denote x → 0 as ¬x and $x^{n} = \underset{n times}{\underset{︸}{x \otimes x \otimes \dots \otimes x}}$ and x⁰ = 1 for all x ∈ L.

Lemma 1.1. [3] Let L be a BL-algebra. Then for all x, y, z, x₁, x₂ ∈ L it holds that

x ≤ y ⇔ x → y = 1.

(y → x) ⊗ (z → w) ≤ (x → z) → (y → w).

(x → y) ⊗ (z → w) ≤ (x ⊗ z) → (y ⊗ w).

x → (y ∨ z) ≥ (x → y) ∨ (x → z).

(x ∨ y) → y = x → y.

x ∨ y = [(x → y) → y] ∧ [(y → x) → x]

(x → y) ⊗ (y → z) ≤ x → z. (Transitivity of →)

x → (y → z) = y → (x → z) = (x ⊗ y) → z. (Exchange rule)

x₁ ≤ x₂ implies x₂ → y ≤ x₁ → y.(Antito-nicity of the first variable of →)

x₁ ≤ x₂ implies y → x₂ ≤ y → x₁. (Isotonicity of the second variable of →)

A filter of a BL-algebra L is a nonempty subset F ⊆ L satisfying the following conditions: F₁: x ∈ F and x ≤ y imply y ∈ F; F₂: x, y ∈ F implies x ⊗ y ∈ F. Alternatively, a filter can be defined by properties F₃: 1 ∈ F; F₄: x, x → y ∈ F implies y ∈ F.

The items F₁ to F₄ will be used frequently, so we don’t cite them every time.

In [1], Haveshki and Eslami introduced n-fold filters in BL-algebras. Particularly, they defined two types of n-fold filters, that is, n-fold implicative filters and n-fold positive implicative filters.

Definition 1.2. [1] Let F be a nonempty subset of a BL-algebra L. Then F is called an n-fold implicative filter, if 1 ∈ F, xⁿ → (y → z) ∈ F and xⁿ → y ∈ F imply xⁿ → z ∈ F, for all x, y, z ∈ L.

Definition 1.3. [1] Let F be a nonempty subset of a BL-algebra L. Then F is called an n-fold positive implicative filter, if 1 ∈ F, x → [(yⁿ → z) → y] ∈ F and x ∈ F imply y ∈ F, for all x, y, z ∈ L.

For n-fold (positive) implicative filters, they obtained the following characterizations:

Theorem 1.4. [1] Let F be a filter of a BL-algebra L. Then for all x, y ∈ L the following assertions are equivalent:

F is an n-fold implicative filter.

xⁿ → x²ⁿ ∈ F.

xⁿ⁺¹ → y ∈ F implies xⁿ → y ∈ F.

Theorem 1.5. [1] Let F be a filter of a BL-algebra L. Then for all x, y ∈ L the following assertions are equivalent:

F is an n-fold positive implicative filter.

¬xⁿ → x ∈ F implies x ∈ F.

(xⁿ → y) → x ∈ F implies x ∈ F.

In [4], Turunen et al. introduced another type of n-fold filters named n-fold Boolean filter.

Definition 1.6. [4] Let F be a filter of a BL-algebra L. Then F is called an n-fold Boolean filter, if x ∨ ¬ xⁿ ∈ F, for all x ∈ L.

They also proved that each n-fold Boolean filter is an n-fold positive implicative filter, and left the converse as an open problem. That is,

Problem 1.7. Is each n-fold positive implicative filter an n-fold Boolean filter?

In [5], Turunen et al. also pointed out that the inaccuracy of Proposition 5 [4] implies that the problem of the relation of n-fold implicative basic logic and n-fold positive implicative basic logic remains open, contrary to what is alleged in the Conclusion. That is,

Problem 1.8. Under what conditions an n-fold implicative basic logic is an n-fold positive implicative basic logic?

However, Lele introduced n-fold fantastic filters in BL-algebras which will be used in the sequel.

Definition 1.9. [2] Let F be a nonempty subset of a BL-algebra L. Then F is called an n-fold fantastic filter, if 1 ∈ F, z → (y → x) ∈ F and z ∈ F imply [(xⁿ → y) → y] → x ∈ F, for all x, y, z ∈ L.

2 Solution to Turunen’s open problems

2.1 Solution to problem 1

In order to solve problem 1, some alternative definitions of n-fold Boolean filter and positive implicative filter will be given in the following part.

Theorem 2.1. Let F be a filter of a BL-algebra L. Then for all x, y ∈ L the following assertions are equivalent:

F is an n-fold Boolean filter.

x ∨ (xⁿ → y) ∈ F.

Proof. Assume that F is an n-fold Boolean filter. That is, x ∨ ¬ xⁿ ∈ F. The inequality x ∨ ¬ xⁿ ≤ x ∨ (xⁿ → y) leads that x ∨ (xⁿ → y) ∈ F.

Conversely, it follows immediately by takingy = 0.□

Theorem 2.2. Let F be a filter of a BL-algebra L. Then for all x, y ∈ L the following assertions are equivalent:

F is an n-fold positive implicative filter.

(¬ xⁿ → x) → x ∈ F.

(¬ xⁿ → y) → [(y → x) → x] ∈ F.

[¬ (x ∨ y) ⁿ → y] → (x ∨ y) ∈ F.

Proof. (i)⇔(ii) Assume that F is an n-fold positive implicative filter. It follows from the inequality x ≤ (¬ xⁿ → x) → x and Lemma 1.1(ix) with y = 0 that ¬ [(¬ xⁿ → x) → x] ⁿ ≤ ¬ xⁿ. Applying Lemma 1.1(ix) with y = (¬ xⁿ → x) → x and the Exchange rule, we have 1 = (¬ xⁿ → x) → (¬ xⁿ → x) = ¬ xⁿ → [(¬ xⁿ → x) → x] ≤ ¬ [(¬ xⁿ → x) → x] ⁿ → [(¬ xⁿ → x) → x]. Thus ¬ [(¬ xⁿ → x) → x] ⁿ → [(¬ xⁿ → x) → x] ∈ F. By Theorem 1.5(ii), we get (¬ xⁿ → x) → x ∈ F.

Conversely, it is obvious.

(ii)⇔(iii) Assume that (ii) holds. The transitivity of → leads that (¬ xⁿ → y) ⊗ (y → x) ≤ ¬ xⁿ → x. Using antitonicity of the first variable of → and Exchange rule, it holds that (¬ xⁿ → x) → x ≤ [(¬ xⁿ → y) ⊗ (y → x)] → x = (¬ xⁿ → y) → [(y → x) → x], so (¬ xⁿ → y) → [(y → x) → x] ∈ F.

Conversely, it follows immediately by taking y = ¬ xⁿ.

(iii)⇔(iv) Assume that (iii) holds. Substituting x ∨ y for x in (iii), we get [¬ (x ∨ y) ⁿ → y] → (x ∨ y) ∈ F.

Conversely, by hybrid monotonicity of →, we have [¬ (x ∨ y) ⁿ → y] → (x ∨ y) ≤ (¬ xⁿ → y) → [(y → x) → x], so (¬ xⁿ → y) → [(y → x) → x] ∈ F.□

Using the technique in Theorem 1, we can characterize n-fold positive implicative filters as follows:

Theorem 2.3. Let F be a filter of a BL-algebra L. Then for all x, y, z ∈ L the following assertions are equivalent:

F is an n-fold positive implicative filter.

[(xⁿ → y) → x] → x ∈ F.

[(xⁿ → y) → z] → [(z → x) → x] ∈ F.

{ [(x ∨ z) ⁿ → y] → z } → (x ∨ z) ∈ F.

Proof. (i)⇔(ii) It follows from Theorem 2.2 (ii) that (¬ xⁿ → x) → x ∈ F. Using the hybrid monotonicity of →, ¬xⁿ ≤ xⁿ → y leads to (¬ xⁿ → x) → x ≤ [(xⁿ → y) → x] → x. Thus [(xⁿ → y) → x] → x ∈ F.

Conversely, it is obvious by taking y = 0.

(i)⇔(iii) and (i)⇔ (iv) can be proved in a similar way using Theorem 2.2(iii) and (iv).□

Now we turn to the relationships between n-fold Boolean filters and n-fold positive implicative filters. The following lemma will be necessary.

Lemma 2.4. Let L be a BL-algebra. Then for all x ∈ L, ¬ (x ∨ ¬ xⁿ) ⁿ → ¬ xⁿ = 1.

Proof. By Lemma 1.1 (i), (ii) and (iii), we have the following inequalities

$\begin{matrix} \neg {(x \lor \neg x^{n})}^{n} \to \neg x^{n} & \geq & x^{n} \to {(x \lor \neg x^{n})}^{n} \\ \geq & {[x \to (x \lor \neg x^{n})]}^{n} \\ = & 1 . \end{matrix}$

Hence, ¬ (x ∨ ¬ xⁿ) ⁿ → ¬ xⁿ = 1.□

Theorem 2.5. Let F be a filter of a BL-algebra L. If F is an n-fold positive implicative filter, then F is an n-fold Boolean filter.

Proof. Assume that F is an n-fold positive implicative filter. By Theorem 2.2(iv), we have [¬ (x ∨ y) ⁿ → y] → (x ∨ y) ∈ F. In particular, let y = ¬ xⁿ, by Lemma 2.4 it holds that [¬ (x ∨ ¬ xⁿ) ⁿ → ¬ xⁿ] → (x ∨ ¬ xⁿ) = x ∨ ¬ xⁿ ∈ F. Hence F is an n-fold Boolean filter.□

Corollary 2.6. Let F be a filter of a BL-algebra L. Then F is an n-fold positive implicative filter if and only if F is an n-fold Boolean filter.

2.2 Solution to problem 2

In order to solve the problem 2, we give some alternative definitions of the n-fold fantastic filter at first.

Lemma 2.7. Let L be a BL-algebra. Then (x ∨ y) ⁿ → y = xⁿ → y for all x, y ∈ L.

Proof. The case for n = 1 follows immediately from Lemma 1.1(v). Assume the identity holds for n = k. Now, we prove the case for n = k + 1. Using the Lemma 1.1(v) and the Exchange rule, it holds that $\begin{matrix} (x \lor y)^{k + 1} \to y & = [(x \lor y)^{k} \otimes (x \lor y)] \to y \\ = (x \lor y) \to [(x \lor y)^{k} \to y] \\ = (x \lor y) \to (x^{k} \to y) \\ = x^{k} \to [(x \lor y) \to y] \\ = x^{k} \to (x \to y) \\ = x^{k + 1} \to y \end{matrix}$ Thus the identity holds.□

Theorem 2.8. Let F be a filter of a BL-algebra L. Then for all x, y ∈ L the following assertions are equivalent:

F is an n-fold fantastic filter.

y → x ∈ F implies [(xⁿ → y) → y] → x ∈ F.

(y → x) → {[(xⁿ → y) → y] → x} ∈ F.

[(xⁿ → y) → y] → (x ∨ y) ∈ F.

Proof. (i)⇒(ii) It is trivial.

(ii)⇒(iii) Applying Exchange rule, it holds that y → [(y → x) → x] = (y → x) → (y → x) =1 ∈ F. By assumption, we have [(((y → x) → x) ⁿ → y) → y] → [(y → x) → x] ∈ F. The inequality [(((y → x) → x) ⁿ → y) → y] → [(y → x) → x] ≤ [(xⁿ → y) → y] → [(y → x) → x] leads to [(xⁿ → y) → y] → [(y → x) → x] ∈ F. Using Exchange rule again, then (y → x) → {[(xⁿ → y) → y] → x} ∈ F.

(iii)⇒(iv) Taking x = x ∨ y in (iii). It follows from Lemma 2.7 that [(xⁿ → y) → y] → (x ∨ y) ∈ F.

(iv)⇒(i) It is obvious that 1 ∈ F. Assume that z, z → (y → x) ∈ F. Then y → x ∈ F. By Lemma 1.1(vi) and isotonicity of the second variable of →, it holds that [(xⁿ → y) → y] → (x ∨ y) ≤ [(xⁿ → y) → y] → [(y → x) → x], and hence [(xⁿ → y) → y] → [(y → x) → x] ∈ F. The Exchange rule leads (y → x) → {[(xⁿ → y) → y] → x} ∈ F. Hence, [(xⁿ → y) → y] → x ∈ F.□

Theorem 2.9. Each n-fold Boolean (positive implicative) filter is an n-fold fantastic filter, but the converse does not hold in general.

Proof. Assume that F is an n-fold Boolean filter. It follows from Theorem 2.2 (iv) that [¬ (x ∨ y) ⁿ → y] → (x ∨ y) ∈ F. By hybrid monotonicity of →, we get [¬ (x ∨ y) ⁿ → y] → (x ∨ y) ≤ [(xⁿ → y) → y] → (x ∨ y), and hence [(xⁿ → y) → y] → (x ∨ y) ∈ F. It follows from Theorem 2.8 (iv) that F is an n-fold fantastic filter.□

Example 2.10. Let L ={ 0, a, b, 1 }. Define ⊗ and → as follows:

⊗	a	b	1
0	0	0	0
a	0	0	a
b	0	a	b
1	a	b	1

→	0	a	b	1
0	1	1	1	1
a	b	1	1	1
b	a	b	1	1
1	0	a	b	1

It is trivial to verify that F = {1} is a 2-fold fantastic filter, but since b ∨ ¬ b² = b ∉ F, F is not a 2-fold Boolean filter.

Theorem 2.11. Let F be a filter of a BL-algebra L. If F is both an n-fold implicative and fantastic filter, then F is an n-fold Boolean filter.

Proof. Assume that F is both an n-fold implicative and fantastic filter. By Theorem 1.4 (ii), we have xⁿ → x²ⁿ ∈ F. It follows from Lemma 1.1 (ii) and Exchange rule that xⁿ → x²ⁿ ≤ ¬ x²ⁿ → ¬ xⁿ = (xⁿ → ¬ xⁿ) → ¬ xⁿ, and hence (xⁿ → ¬ xⁿ) → ¬ xⁿ ∈ F. Using Theorem 2.8 (iv), we get [(xⁿ → y) → y] → (x ∨ y) ∈ F. Taking y = ¬ xⁿ, then x ∨ ¬ xⁿ ∈ F. Hence F is an n-fold Boolean filter.□

Corollary 2.12. Let F be a filter of a BL-algebra L. Then F is an n-fold Boolean filter if and only if F is both an n-fold implicative and fantastic filter.

Theorem 2.8 (iii) or (iv) shows that the extension property (Theorem 6.12 [1]) holds for n-fold fantastic filters. Moreover, we have the following result:

Theorem 2.13. Let F be a filter of a BL-algebra L. Then for all x, y ∈ L the following assertion are equivalent:

{1} is an n-fold fantastic filter.

each filter is an n-fold fantastic filter.

L is an n-fold fantastic BL-algebra.

(y → x) → {[(xⁿ → y) → y] → x} =1.

[(xⁿ → y) → y] → (x ∨ y) =1.

Proof. The proof is similar to that in Theorem 4.12[1].□

Corollary 2.14. In n-fold fantastic BL-algebra, each n-fold implicative BL-algebra is an n-fold positive implicative BL-algebra.

3 Conclusions

It is well known that special filters play a vital role in investigating the structure of a logical system. In this paper, two problems were solved. It is proved that n-fold positive implicative filters are equivalent to n-fold Boolean filters; in an n-fold fantastic BL-algebra, each n-fold implicative BL-algebra is an n-fold positive implicative BL-algebra.

Future research will focus on defining other types of n-fold filters, and investigating the relationships among logic algebras and these filters.

Acknowledgments

This research was supported by the AMEP (DYSP) of Linyi University, the NSF of Shandong Province (Grant No. ZR2013FL006, ZR2014AL009).

References

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