Math. Log. Quart. 54, No. 2, 176 – 186 (2008) / DOI 10.1002/malq.200710029

n-fold filters in BL-algebras

Masoud Haveshki∗ and Esfandiar Eslami∗∗

Department of Mathematics, Shahid Bahonar University of Kerman, Kerman, Iran

Received 17 July 2007, revised 27 July 2007, accepted 14 August 2007Published online 7 March 2008

Key words n-fold (positive) implicative basic logic, n-fold (positive) implicative BL-algebra, n-fold (posi-tive) implicative filter, Godel algebra, Boolean algebra.

MSC (2000) 06F35, 03G25

In this paper we introduce n-fold (positive) implicative basis logic and the related algebras called n-fold (posi-tive) implicative BL-algebras. Also we define n-fold (positive) implicative filters and we prove some relationsbetween these filters and construct quotient algebras via these filters.

c© 2008 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

BL-algebras were inverted by P. Hajek [3] in order to provide an algebraic proof of the completeness theoremof “basic logic” (BL, for short) arising from the continuous triangular norms, familiar in the fuzzy logic frame-work. The language of propositional Hajek basic logic [3] contains the binary connectives ◦ and ⇒ and the con-stant 0. Axioms of BL are:(A1) (ϕ ⇒ ψ) ⇒ ((ψ ⇒ ω) ⇒ (ϕ ⇒ ω)).(A2) (ϕ ◦ ψ) ⇒ ϕ.(A3) (ϕ ◦ ψ) ⇒ (ψ ◦ ϕ).(A4) (ϕ ◦ (ϕ ⇒ ψ)) ⇒ (ψ ◦ (ψ ⇒ ϕ)).(A5a) (ϕ ⇒ (ψ ⇒ ω)) ⇒ ((ϕ ◦ ψ) ⇒ ω).(A5b) ((ϕ ◦ ψ) ⇒ ω) ⇒ (ϕ ⇒ (ψ ⇒ ω)).(A6) ((ϕ ⇒ ψ) ⇒ ω) ⇒ (((ψ ⇒ ϕ) ⇒ ω) ⇒ ω).(A7) 0 ⇒ ω.

The above notion is generalized to an algebraic system in which the required conditions are fulfilled (Defini-tion 2.1). Filters in BL-algebras are also defined in [3]. We defined the notions of implicative and positive im-plicative filters in [5]. In Section 2 we give some definitions and theorems which are needed in the rest of thepaper. In Section 3 we define n-fold implicative basic logic and n-fold implicative BL-algebra and state a com-pleteness theorem for this logic. Also in a particular case, we prove that 1-fold implicative basic logic is Godellogic and intuitionistic logic. Continuously in Section 4 we define n-fold implicative filters. Jun and Ko introdu-ced n-fold grisly deductive systems and proved in [6] that if D is a deductive system of a BL-algebra A, then thefollowing conditions are equivalent:

(a) D is an n-fold grisly deductive system of A.(b) For all x, y ∈ A, if P (xn+1 \ y) ∈ D, then P (xn \ y) ∈ D.(c) For all x, y, z ∈ A, if P (xn, y \ z) ∈ D, then P (P (xn \ y) \ P (xn \ z)) ∈ D (see [5]).

However, n-fold implicative filter and n-fold grisly deductive system are the same. In this paper we prove theabove theorem in a different way and we show that (a) and (b) and (c) are equivalent to

(d) xn → x2n ∈ D for all x ∈ A.

∗ e-mail: Haveshki masoud@yahoo.com∗∗ Corresponding author: e-mail: eeslami@mail.uk.ac.ir

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Math. Log. Quart. 54, No. 2 (2008) / www.mlq-journal.org 177

We use this condition (d) in the proof of theorems, in Section 4. In fact we prove theorems of this section in a dif-ferent way. Also we show that F is a 1-fold implicative filter if and only if A/F is a Godel algebra. In Section 5we define n-fold positive implicative basic logic and state a completeness theorem for this logic and prove that,too. Also in a particular case we prove that 1-fold positive implicative basic logic is a Boolean logic. In Section 6we define n-fold positive implicative filters, after that we state the equivalent conditions for n-fold positive im-plicative BL-algebras and show that F is a positive implicative filter if and only if A/F is an n-fold positive im-plicative BL-algebra. In a special case we show that F is a 1-fold positive implicative filter if and only if A/F isa Boolean algebra.

2 Preliminaries

Definition 2.1 [3, Definition 2.3.2] A BL-algebra is an algebra (A,∧,∨, ∗,→, 0, 1) with four binary opera-tions ∧,∨, ∗,→ and two constants 0, 1 such that:(BL1) (A,∧,∨, 0, 1) is a bounded lattice;(BL2) (A, ∗, 1) is a commutative monoid;(BL3) ∗ and → form an adjoint pair; i. e., c ≤ a → b if and only if a ∗ c ≤ b for all a, b, c ∈ A;(BL4) a ∧ b = a ∗ (a → b);(BL5) (a → b) ∧ (b → a) = 1.

We denote

xn = x ∗ · · · ∗ x︸ ︷︷ ︸

n times

if n > 0 and x0 = 1

for all x ∈ A. A BL-algebra is called a Godel algebra if a2 = a ∗ a = a for all a ∈ A.Lemma 2.2 [1, 2, 3] In each BL-algebra A, the following relations hold for all x, y, z ∈ A:

(1) x ∗ (x → y) ≤ y.(2) x ≤ (y → (x ∗ y)).(3) x ≤ y if and only if x → y = 1.(4) x → (y → z) = x ∗ y → z = y → (x → z).(5) If x ≤ y, then y → z ≤ x → z and z → x ≤ z → y.(6) y ≤ (y → x) → x.(7) y → x ≤ (z → y) → (z → x).(8) x → y ≤ (y → z) → (x → z).(9) x ∗ y → z = x → (y → z).Definition 2.3 [3] A filter of a BL-algebra A is a non-empty subset F of A such that for all a, b ∈ A,

(1) a, b ∈ F implies a ∗ b ∈ F ,(2) a ∈ F and a ≤ b imply b ∈ F .Definition 2.4 [7, p. 2] A non-empty subset D of a BL-algebra A is called a deductive system if

(a) 1 ∈ D;(b) if x, x → y ∈ D, then y ∈ D, for all x, y ∈ A.Theorem 2.5 [7, p. 3] A non-empty subset D of A is a deductive system if and only if D is a filter of A.Definition 2.6 [6, Definition 3.6] Let D be a non-empty subset of A. Then D is called a (k,m, n)-fold grisly

deductive system of A if it satisfies:(a) 1 ∈ D;(b) for all x, y, z ∈ A, if P (xk, y \ z) ∈ D and P (xm \ y) ∈ D, then P (xn \ z) ∈ D, where

P (a1, a2 \ a3) := (a1 → (a2 → a3)) and P (a1 \ a2) := (a1 → a2).

If k = m = n in (b), we say that D is an n-fold grisly deductive system of A.

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178 M. Haveshki and E. Eslami: n-fold filters in BL-algebras

Theorem 2.7 [3, Lemma 2.3.14] Let F be a filter of a BL-algebra A. Define:

x ≡F y if and only if x → y ∈ F and y → x ∈ F .

Then ≡F is a congruence relation on A.The set of all congruence classes is denoted by A/F , i. e., A/F := {[x] | x ∈ A}, where

[x] := {y ∈ A | x ≡F y}.Define ∗,→,∧,∨ on A/F as follows:

[x] ∗ [y] = [x ∗ y], [x] → [y] = [x → y], [x] ∧ [y] = [x ∧ y], [x] ∨ [y] = [x ∨ y].

Therefore (A/F, ∗,→,∧,∨, [1], [0]) is a BL-algebra which is called the quotient BL-algebra with respect to F .

3 n-fold implicative basic logic and n-fold implicative BL-algebra

From now on (A,∧,∨, ∗,→, 0, 1) or simply A is a BL-algebra and n is a natural number.

Definition 3.1 Axioms of an n-fold implicative basic logic are those of BL plus

ϕn+1 ≡ ϕn,

where

ϕn = ϕ ◦ · · · ◦ ϕ︸ ︷︷ ︸

n times

and where ϕ ≡ ψ denotes ϕ ⇒ ψ plus ψ ⇒ ϕ.

Definition 3.2 An n-fold implicative BL-algebra is a BL-algebra (A,∧,∨, ∗,→, 0, 1) satisfying xn+1 = xn,for all x ∈ A, where

xn = x ∗ · · · ∗ x︸ ︷︷ ︸

n times

.

Theorem 3.3 (Completeness) n-fold implicative basic logic is complete, that is, for each formula ϕ, the fol-lowing conditions are equivalent:

(1) n-fold implicative basic logic proves ϕ.(2) ϕ is an A-tautology for each linearly ordered n-fold implicative BL-algebra A.(3) ϕ is an A-tautology for each n-fold implicative BL-algebra A.

P r o o f. This follows from [3, Theorem 2.3.22] because an n-fold implicative basic logic is a schematic ex-tension of BL.

Corollary 3.4 A 1-fold implicative basic logic is a Godel logic.

P r o o f. This follows from [5, Theorems 3.5 and 3.7].

Corollary 3.5 A 1-fold implicative basic logic is an intuitionistic logic.

P r o o f. By [3, Lemma 4.2.7], this is clear.

Theorem 3.6 If A is an n-fold implicative BL-algebra, then A is an (n + 1)-fold implicative BL-algebra.

P r o o f. Since A is n-fold implicative BL-algebra, xn+1 = xn for all x ∈ A. But

xn+2 = xn+1 ∗ x = xn ∗ x = xn+1

and hence A is an (n + 1)-fold implicative BL-algebra.

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Math. Log. Quart. 54, No. 2 (2008) / www.mlq-journal.org 179

By the following example we show that the converse is not true.Example 3.7 Let B = {0, a, b, 1}. Define ∗ and → as follows:

∗ 0 a b 10 0 0 0 0a 0 0 0 a

b 0 0 a b

1 0 a b 1

, → 0 a b 10 1 1 1 1a b 1 1 1b a b 1 11 0 a b 1

.

Then (B,∧,∨, ∗,→, 0, 1) is a BL-algebra and it is clear that B is a 3-implicative BL-algebra but since b3 = b2,B is not a 2-implicative BL-algebra.

4 n-fold implicative filters

Definition 4.1 A non-empty subset F of a BL-algebra A is called an n-fold implicative filter of A if it satisfies:(1) 1 ∈ F ;(2) xn → (y → z) ∈ F , xn → y ∈ F imply xn → z ∈ F , for all x, y, z ∈ F .Theorem 4.2 Any n-fold implicative filter of A is a filter of A.

P r o o f. Let x, x → y ∈ F . Hence 1 → x ∈ F and 1 → (x → y) ∈ F . But 1 = 1n, thus y = 1 → y ∈ F ,that is, F is a filter of A.

The following example shows that the converse of Theorem 4.2 is not true.Example 4.3 Let B = {0, a, b, 1}. Define ∗ and → as follows:

∗ 0 a b 10 0 0 0 0a 0 0 a a

b 0 a b b

1 0 a b 1

, → 0 a b 10 1 1 1 1a a 1 1 1b 0 a 1 11 0 a b 1

.

Then (B,∧,∨, ∗,→, 0, 1) is a BL-algebra and it is clear that F = {b, 1} is a filter, while it is not a 1-implicativefilter since a → (a → 0) ∈ F and a → a ∈ F but a → 0 /∈ F .

Theorem 4.4 For any a ∈ A, A(a) := {x ∈ A | a ≤ x} is a filter if and only if a ≤ x whenever a ≤ y → xand a ≤ y for all x, y ∈ A.

P r o o f. Let A(a) be a filter and a ≤ y → x and a ≤ y, then y → x ∈ A(a) and y ∈ A(a). Since A(a) is a fil-ter, x ∈ A(a), that is, a ≤ x. Conversely, since a ≤ 1, 1 ∈ A(a). If x, x → y ∈ A(a), then a ≤ x and a ≤ x → y.By assumption a ≤ y and hence y ∈ A(a). Hence A(a) is a filter.

Theorem 4.5 Let a be an element of A. If A(a) is an n-fold implicative filter of A, then an+1 → (x → y) = 1,an+1 → x = 1 imply an+1 → y = 1, for all x, y ∈ A.

P r o o f. Let A(a) be an n-fold implicative filter and an+1 → (x → y) = 1, an+1 → x = 1. Since

a → (an → (x → y)) = an+1 → (x → y) = 1,

an → (x → y) ∈ A(a). Similarly an → x ∈ A(a). Since A(a) is an n-fold implicative filter, an → y ∈ A(a),that is, a ≤ an → y. Thus an+1 → y = 1.

Theorem 4.6 Let F be a filter of A. Then for all x, y, z ∈ A the following conditions are equivalent:(a) F is an n-fold implicative filter of A.(b) xn → x2n ∈ F for all x ∈ A.(c) xn+1 → y ∈ F implies xn → y ∈ F .(d) xn → (y → z) ∈ F implies (xn → y) → (xn → z) ∈ F .

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180 M. Haveshki and E. Eslami: n-fold filters in BL-algebras

P r o o f.(a) ⇒ (b): Let x ∈ A, hence by Lemma 2.2,

xn → (xn → x2n) = x2n → x2n = 1 ∈ F and xn → xn = 1 ∈ F.

Since F is an n-fold implicative filter, we get xn → x2n ∈ F .(b) ⇒ (a): Let x, y, z ∈ A be such that xn → (y → z) ∈ F , xn → y ∈ F . Since

(xn → (y → z)) ∗ (xn → y) ∗ xn ∗ xn = (xn ∧ (y → z)) ∗ (xn ∧ y) ≤ (y → z) ∗ y = y ∧ z ≤ z,

then (xn → (y → z)) ∗ (xn → y) ≤ x2n → z. Since xn → (y → z) ∈ F , xn → y ∈ F , we get

(xn → (y → z)) ∗ (xn → y) ∈ F

and so x2n → z ∈ F . By Lemma 2.2, xn → x2n ≤ (x2n → z) → (xn → z). On the other hand, x2n → z ∈ Fand xn → x2n ∈ F , then xn → z ∈ F . Hence F is an n-fold implicative filter of A.

(b) ⇒ (c): Since (b) holds, F is an n-fold implicative filter of A. On the other hand,

xn+1 → y = xn → (x → y) ∈ F and xn → x = 1 ∈ F,

hence xn → y ∈ F .(c) ⇒ (b): We have

xn+1 → (xn−1 → x2n) = x2n → x2n = 1 ∈ F,

hence by (c), xn → (xn−1 → x2n) ∈ F . But

xn+1 → (xn−2 → x2n) = x2n−1 → x2n = xn → (xn−1 → x2n) ∈ F,

that is, xn+1 → (xn−2 → x2n) ∈ F , and so xn → (xn−2 → x2n) ∈ F . By repeating the process n times weget xn → (x0 → x) = xn → (1 → x) = xn → x2n ∈ F .

(b) ⇒ (d): Let xn → (y → z) ∈ F , by Lemma 2.2,

xn → (y → z) ≤ xn → ((xn → y) → (xn → z))= xn → (xn → ((xn → y) → z))= x2n → ((xn → y) → z).

Hence x2n → ((xn → y) → z) ∈ F . By (b), we have xn → x2n ∈ F , now by Lemma 2.2,

x2n → ((xn → y) → z) ≤ (xn → x2n) → (xn → ((xn → y) → z)).

Then we get (xn → y) → (xn → z) = (xn → ((xn → y) → z)) ∈ F .(d) ⇒ (b): Since

xn → (xn → x2n) = x2n → x2n = 1 ∈ F,

by (d) we have xn → x2n = (xn → xn) → (xn → x2n) ∈ F .

Theorem 4.7 Let F be a filter of a BL-algebra A. If F is an n-fold implicative filter, then F is an (n+1)-foldimplicative filter.

P r o o f. Let x, y ∈ A be such that xn+2 → y ∈ F . By Lemma 2.2,

xn+1 → (x → y) = xn+2 → y

and since F is an n-fold implicative filter, by Theorem 4.6, xn → (x → y) ∈ F . Hence xn+1 → y ∈ F , that is,F is an (n + 1)-fold implicative filter.

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Math. Log. Quart. 54, No. 2 (2008) / www.mlq-journal.org 181

By the following example we show that the converse is not true.Example 4.8 In Example 3.7, {1} is a 3-fold implicative filter but since

b3 → 0 = 1 ∈ {1} and b2 → 0 = b = 1,

{1} is not a 2-fold implicative filter.Corollary 4.9 In an n-fold implicative BL-algebra, the concepts of filters and n-fold implicative filters coin-

cide.

P r o o f. It follows from Theorem 4.6 and the definition of an n-fold implicative BL-algebra.

Theorem 4.10 A is an n-fold implicative BL-algebra if and only if {1} is an n-fold implicative filter of A.

P r o o f. If A is an n-fold implicative BL-algebra, then xn+1 = xn for all x ∈ A and so

xn+2 = xn+1 ∗ x = xn ∗ x = xn+1 = xn.

By the similar way, x2n = xn for all x ∈ A, that is, xn → x2n = 1 ∈ {1}, for all x ∈ A. By condition (b) of The-orem 4.6, {1} is an n-fold implicative filter of A. Conversely, let the filter {1} of A be n-fold implicative. Since

xn → (xn → xn+1) = x2n → xn+1 = 1 ∈ {1} and xn → xn = 1 ∈ {1},

we get xn → xn+1 ∈ {1}, that is, xn = xn+1. Hence A is an n-fold implicative BL-algebra.

The following theorem was proved by Jun and Ko in [6], we prove that in a simpler way using condition (b)in Theorem 4.6:

Theorem 4.11 Let F and G be filters of A such that F ⊆ G. If F is an n-fold implicative filter, then G is alsoan n-fold implicative filter.

P r o o f. Let F be an n-fold implicative filter of A. Then by Theorem 4.6, xn → x2n ∈ F for all x ∈ A, andso xn → x2n ∈ G for all x ∈ A. Hence G is an n-fold implicative filter.

Theorem 4.12 In any BL-algebra A, the following conditions are equivalent:(a) A is an n-fold implicative BL-algebra.(b) Every filter of A is an n-fold implicative filter.(c) {1} is an n-fold implicative filter.(d) xn = x2n for all x ∈ A.

P r o o f.(a) ⇒ (b): It is clear by the definition of an n-fold implicative BL-algebra and Theorem 4.6.(b) ⇒ (c): This is clear.(c) ⇒ (a): By Theorem 4.10, this is clear.(a) ⇒ (d): Let A be an n-fold implicative BL-algebra, hence xn+1 = xn for all x ∈ A. We have

xn+2 = xn+1 ∗ x = xn ∗ x = xn+1 = xn.

By repeating the process n times, we get xn = x2n for all x ∈ A.(d) ⇒ (a): If xn = x2n for all x ∈ A, then xn → x2n = 1 ∈ {1} for all x ∈ A. By Theorem 4.6, {1} is an

n-fold implicative filter. Since (a) and (c) are equivalent, A is an n-fold implicative BL-algebra.

The following theorem which appeared also in [6] is proved using condition (b) in Theorem 4.6, too:Theorem 4.13 Let F be a filter of A. Then F is an n-fold implicative filter if and only if A/F is an n-fold

implicative BL-algebra.

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182 M. Haveshki and E. Eslami: n-fold filters in BL-algebras

P r o o f. Let F be an n-fold implicative filter, by Theorem 4.6, xn → x2n ∈ F for all x ∈ A. Then

[x]n → [x]2n = [xn → x2n] = [1]

and so [x]n ≤ [x]2n, that is, [x]n = [x]2n, for all x ∈ A. Hence by Theorem 4.13, A/F is an n-fold implicativeBL-algebra. Conversely, let A/F be an n-fold implicative BL-algebra, then [x]n = [x]2n, for all x ∈ A. Hence

[xn → x2n] = [x]n → [x]2n = [1],

that is, xn → x2n ∈ F for all x ∈ A. Therefore by Theorem 4.6, F is an n-fold implicative filter.

Corollary 4.14 Let F be a filter of A. Then F is a 1-fold implicative filter if and only if A/F is a Godel al-gebra.

P r o o f. By [5, Theorems 3.5 and 3.7], a 1-fold implicative BL-algebra is a Godel algebra. By Theorem 4.13,we have the desired result.

5 n-fold positive implicative basic logic and n-fold positiveimplicative BL-algebra

Definition 5.1 Axioms of an n-fold positive implicative basic logic are those of BL plus

(ϕn ⇒ 0) ⇒ ϕ ≡ ϕ.

Definition 5.2 An n-fold positive implicative BL-algebra is a BL-algebra (A,∧,∨, ∗,→, 0, 1) if it satisfies

(xn → 0) → x = x,

for all x ∈ A.Theorem 5.3 Every n-fold positive implicative BL-algebra is an (n+1)-fold positive implicative BL-algebra.

P r o o f. Let A be an n-fold positive implicative BL-algebra. Then

(xn → 0) → x = x

for all x ∈ A. Since xn+1 ≤ xn, then xn → 0 ≤ xn+1 → 0 and so

(xn+1 → 0) → x ≤ (xn → 0) → x = x.

But by Lemma 2.2, x ≤ (xn+1 → 0) → x. Hence x = (xn+1 → 0) → x, that is, A is an (n + 1)-fold positiveimplicative BL-algebra.

The following example shows that the converse of Theorem 5.3 is not true.Example 5.4 In Example 3.7, B is a 3-fold positive BL-algebra but since (b2 → 0) → b = b, B is not a 2-fold

positive implicative BL-algebra.Theorem 5.5 The n-fold positive implicative basic logic is complete, that is, for each formula ϕ, the following

conditions are equivalent:(1) The n-fold positive implicative basic logic proves ϕ.(2) ϕ is an A-tautology for each linearly ordered n-fold positive implicative BL-algebra A.(3) ϕ is an A-tautology for each n-fold positive implicative BL-algebra.

P r o o f. This follows from [3, Theorem 2.3.22], because n-fold positive implicative basic logic is a schematicextension of BL.

Corollary 5.6 1-fold positive implicative basic logic is Boolean logic.

P r o o f. This follows from [5, Theorem 3.17].

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6 n-fold positive implicative filters

Definition 6.1 A non-empty subset F of A is called an n-fold positive implicative filter if it satisfies:(1) 1 ∈ F ;(2) x → ((yn → z) → y) ∈ F and x ∈ F imply y ∈ F , for all x, y, z ∈ A.Theorem 6.2 Every n-fold positive implicative filter of A is a filter of A.

P r o o f. Let F be an n-fold positive implicative filter and x, y ∈ A be such that x, x → y ∈ F . But we havethat x → ((yn → 1) → y) = x → y and so x → ((yn → 1) → y) ∈ F . Since x ∈ F and F is an n-fold positiveimplicative filter, y ∈ F . Hence F is a filter.

Theorem 6.3 Let F be a filter of A. Then the following conditions are equivalent:(a) F is an n-fold positive implicative filter.(b) (xn → 0) → x ∈ F implies x ∈ F , for all x ∈ A.(c) (xn → y) → x ∈ F implies x ∈ F , for all x, y ∈ A.

P r o o f.(a) ⇒ (c): Let F be an n-fold positive implicative filter of A and (xn → y) → x ∈ F . Since

1 → ((xn → y) → x) = ((xn → y) → x),

we get 1 → ((xn → y) → x) ∈ F . We have 1 ∈ F , thus x ∈ F .(c) ⇒ (b): This is clear.(b) ⇒ (a): Let x → ((yn → z) → y) ∈ F and x ∈ F . Since F is a filter, (yn → z) → y ∈ F . Since 0 ≤ z,

by Lemma 2.2, yn → 0 ≤ yn → z and so (yn → z) → y ≤ (yn → 0) → y. Hence (yn → 0) → y ∈ F and soby assumption we get y ∈ F . Therefore F is an n-fold positive implicative filter.

Theorem 6.4 Let F be a filter of a BL-algebra A. If F is an n-fold positive implicative filter, then F is an(n + 1)-fold positive implicative filter.

P r o o f. Let F be an n-fold positive implicative filter and x ∈ A be such that (xn+1 → 0) → x ∈ F . Sincexn+1 ≤ xn, we have xn → 0 ≤ xn+1 → 0 and so (xn+1 → 0) → x ≤ (xn → 0) → x. Since F is a filter,(xn → 0) → x ∈ F . Since F is an n-fold positive implicative filter, we have x ∈ F and so F is an (n + 1)-foldpositive implicative filter.

By the following example we show that the converse is not true.Example 6.5 In Example 3.7, {1} is a 3-fold positive implicative filter, but (b2 → 0) → b = 1 and b = 1,

hence {1} is not a 2-fold positive implicative filter.Theorem 6.6 Every n-fold positive implicative filter is an n-fold implicative filter.

P r o o f. Let F be an n-fold positive implicative filter of A. By Theorem 6.2, F is a filter of A. Let x, y ∈ Abe such that xn+1 → y ∈ F . Then by Lemma 2.2,

(xn+1 → y)n → (xn → y) = (xn+1 → y)n−1(xn+1 → y) → (xn → y)= (xn+1 → y)n−1 → ((xn+1 → y) → (xn → y))= (xn+1 → y)n−1 → ((xn+1 → y) → (xn−1 → (x → y)))= (xn+1 → y)n−1 → (xn−1 → ((xn+1 → y) → (x → y)))= (xn+1 → y)n−1 → (xn−1 → ((x → (xn → y)) → (x → y)))≥ (xn+1 → y)n−1 → (xn−1 → (xn → y) → y)= (xn+1 → y)n−1 → ((xn → y) → (xn−1 → y))= (xn → y) → ((xn+1 → y)n−1 → (xn−1 → y)).

We show that

(xn+1 → y)n → (xn → y) ≥ (xn → y)2 → ((xn+1 → y)n−2 → (xn−2 → y)).

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184 M. Haveshki and E. Eslami: n-fold filters in BL-algebras

For this consider

(xn → y)(xn+1 → y)n−1xn−1 = (xn → y)(xn+1 → y)n−2(xn+1 → y)xxn−2

= (xn → y)(xn+1 → y)n−2xn−2x(xn+1 → y).

Since xn+1 → y ≤ xn+1 → y = x → (xn → y), then x(xn+1 → y) ≤ xn → y, we get

(xn → y)(xn+1 → y)n−1xn−1 ≤ (xn → y)2(xn+1 → y)n−2xn−2.

Hence

((xn → y)2(xn+1 → y)n−2xn−2) → y ≤ ((xn → y)(xn+1 → y)n−1xn−1) → y

and so

(xn → y)2 → ((xn+1 → y)n−2 → (xn−2 → y))≤ (xn → y) → ((xn+1 → y)n−1 → (xn−1 → y)).

But we had

(xn+1 → y)n → (xn → y) ≥ (xn → y) → ((xn+1 → y)n−1 → (xn−1 → y)).

Then

(xn+1 → y)n → (xn → y) ≥ (xn → y)2 → ((xn+1 → y)n−2 → (xn−2 → y)).

Hence by repeating the process n times we get

(xn+1 → y)n → (xn → y) ≥ (xn → y)2 → ((xn+1 → y)n−2 → (xn−2 → y))≥ · · ·≥ (xn → y)n → ((xn+1 → y)0 → (x0 → y))= (xn → y)n → (1 → (1 → y))= (xn → y)n → y.

Hence ((xn → y)n → y) → ((xn+1 → y)n → (xn → y)) = 1 and so

(xn+1 → y)n → (((xn → y)n → y) → (xn → y)) = 1.

Since F is a filter and xn+1 → y ∈ F , we get (xn+1 → y)n ∈ F and so ((xn → y)n → y) → (xn → y) ∈ F .Since F is an n-fold positive implicative filter, by Theorem 6.3 we have xn → y ∈ F . Hence by Theorem 4.6,F is an n-fold implicative filter.

The following example shows that the converse of Theorem 6.6 is not true.Example 6.7 Let B = {0, a, b, c, 1}. Define ∗ and → as follows:

∗ 0 c a b 10 0 0 0 0 0c 0 c c c c

a 0 c a c a

b 0 c c b b

1 0 c a b 1

, → 0 c a b 10 1 1 1 1 1c 0 1 1 1 1a 0 b 1 b 1b 0 a a 1 11 0 c a b 1

.

Then (B,∧,∨, ∗,→, 0, 1) is a BL-algebra. It is clear that F = {b, 1} is a 2-fold implicative filter but it is not a2-fold positive implicative filter, since (a2 → 0) → a = 1 ∈ F and a /∈ F .

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Math. Log. Quart. 54, No. 2 (2008) / www.mlq-journal.org 185

Lemma 6.8 In the n-fold positive implicative BL-algebra, the notions of an n-fold positive implicative filterand of a filter coincide.

P r o o f. By definition of an n-fold positive implicative BL-algebra and Theorem 6.3, this is clear.

Theorem 6.9 Let A be a BL-algebra. Then the following conditions are equivalent:(a) A is an n-fold positive implicative BL-algebra.(b) Every filter of A is an n-fold positive implicative filter.(c) {1} is an n-fold positive implicative filter.

P r o o f.(a) ⇒ (b): By Lemma 6.8, this is clear.(b) ⇒ (c): This is clear.(c) ⇒ (a): Let {1} be an n-fold positive implicative filter of A. Consider x ∈ A and let

t = ((xn → 0) → x) → x.

Then by Lemma 2.2 we have

(tn → 0) → t = (tn → 0) → (((xn → 0) → x) → x)= ((xn → 0) → x) → ((tn → 0) → x)≥ (tn → 0) → (xn → 0)≥ xn → tn = 1.

The last equality follows from x ≤ (xn → 0) → x = t. Then xn ≤ tn, that is, xn → tn = 1. Hence

(tn → 0) → t = 1 ∈ {1},and since {1} is an n-fold positive implicative filter, t = ((xn → 0) → x) → x = 1, that is, (xn → 0) → x ≤ x.On the other hand, by Lemma 2.2 we have (xn → 0) → x ≥ x. Hence we get (xn → 0) → x = x for all x ∈ A,that is, A is an n-fold positive implicative BL-algebra.

Theorem 6.10 Let F be a filter of A. Then A/F is an n-fold positive implicative BL-algebra if and only ifF is an n-fold positive implicative filter.

P r o o f. Let F be an n-fold positive implicative filter and x ∈ A be such that ([x]n → [0]) → [x] = 1. Then

[(xn → 0) → x] = ([x]n → [0]) → [x] = 1,

so (xn → 0) → x ∈ F . Since F is an n-fold positive implicative filter by Theorem 6.3, x ∈ F , hence [x] = [1]and so {[1]} is an n-fold positive implicative filter of A/F . Then by Theorem 6.9, A/F is an n-fold positive im-plicative BL-algebra. Conversely, let A/F be an n-fold positive implicative BL-algebra and let x ∈ A be suchthat (xn → 0) → x ∈ F . Then

[x] = ([x]n → [0]) → [x] = [(xn → 0) → x] = [1]

and so [x] = [1], that is, x ∈ F . It follows from Theorem 6.3 that F is an n-fold positive implicative filter.

Corollary 6.11 Let F be a filter of a BL-algebra A. Then A/F is a Boolean algebra if and only if F is a1-fold positive implicative filter.

P r o o f. By [5, Theorem 3.17], a 1-fold positive implicative BL-algebra is a Boolean algebra. The corollaryfollows from Theorem 6.10.

Theorem 6.12 Let F and G be filters of a BL-algebra A such that F ⊆ G. If F is an n-fold positive implica-tive filter, then G is an n-fold positive implicative filter.

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186 M. Haveshki and E. Eslami: n-fold filters in BL-algebras

P r o o f. Let x ∈ A be such that (xn → 0) → x ∈ G. Since F is an n-fold positive implicative filter, by The-orem 6.10, A/F is an n-fold positive implicative BL-algebra. Then

[(xn → 0) → x] = ([x]n → [0]) → [x] = [x]

and so ((xn → 0) → x) → x ∈ F ⊆ G. Since G is a filter and (xn → 0) → x ∈ G, x ∈ G. Hence by Theo-rem 6.3, G is an n-fold positive implicative filter.

7 Conclusion and future research

In this paper we introduced the notion of n-fold implicative and n-fold positive implicative basic logics as sche-matic extensions [3] of basic logic BL. The logic BL is known today as the mathematical fuzzy logic [4]. We usedthe results on schematic extensions and proved that both logics we introduced are complete. We also introducedthe algebraic counterpart of these logics as n-fold implicative BL-algebra and n-fold positive implicative BL-al-gebra. Therefore these two algebras are special cases of BL-algebras and correspondingly the logics are schematicextensions of mathematical fuzzy logic. We notice that 1-fold implicative basic logic is a Godel logic and 1-foldpositive implicative basic logic is a Boolean logic. Thus as we expect basic logic is an extension of Godel logicand Godel logic is an extension of mathematical logic. This is also true from algebraic point of view, since theLindenbaum algebras of mathematical logic, Godel logic, and basic logic are Boolean algebra, Godel algebra, andBL-algebra, respectively. As our future research work, we are investigating under what conditions an n-fold im-plicative basic logic is an n-fold positive implicative basic logic. In fact we use the role of special filters to showthis and some more beautiful results.

Acknowledgements This research has been supported in part by Fuzzy Sets and Its Application Center of Excellence,Shahid Bahonar University of Kerman, Iran.

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Aarhus. BRICS Report Science 2002.[3] P. Hajek, Metamathematics of Fuzzy Logic (Kluwer Academic Publishers, 1988).[4] P. Hajek, What is mathematical fuzzy logic? Fuzzy Sets and Systems 157, 597 – 603 (2006).[5] M. Haveshki, A. Borumand Saeid, and E. Eslami, Some types of filters in BL-algebras. Soft Computing 10, 657 – 664

(2006).[6] Y. B. Jun and J. M. Ko, Folding theory applied to BL-algebras. Central European J. Mathematics 2, 584 – 592 (2004).[7] E. Turunen, Boolean deductive system of BL-algebras. Arch. Math. Logic 40, 467 – 473 (2001).

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