part i. order, posets, lattices and residuated lattices in...

Introduction Introduction Introduction

PART I.Order, posets, lattices and residuated lattices in

logic

October 22, 2007

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PART I. Order, posets, lattices and residuated lattices in logic


Applied Logics

Classical Boolean logic is the logic of mathematics, whosehistorical challenger has been Intuitionistic Logic.

Recently, there has appeared applied logical systems such as• Girard’s Linear Logic,• Zadeh’s Fuzzy Logic,• Hajek’s GUHA Logic.In 1920’s Lukasiewicz introduced Many-valued Logic.All such logics are special cases of Hohle’s Monoidal Logic; we arenow going to study this logic from an algebraic point of view; eachabove mentioned logic has an algebraic counterpart which is aresiduated lattice.We are going to show that first order Monoidal Logic and all itspresented axiomatic extensions are complete logics.



Applied Logics

Classical Boolean logic is the logic of mathematics, whosehistorical challenger has been Intuitionistic Logic.Recently, there has appeared applied logical systems such as• Girard’s Linear Logic,

• Zadeh’s Fuzzy Logic,• Hajek’s GUHA Logic.In 1920’s Lukasiewicz introduced Many-valued Logic.All such logics are special cases of Hohle’s Monoidal Logic; we arenow going to study this logic from an algebraic point of view; eachabove mentioned logic has an algebraic counterpart which is aresiduated lattice.We are going to show that first order Monoidal Logic and all itspresented axiomatic extensions are complete logics.



Applied Logics

Classical Boolean logic is the logic of mathematics, whosehistorical challenger has been Intuitionistic Logic.Recently, there has appeared applied logical systems such as• Girard’s Linear Logic,• Zadeh’s Fuzzy Logic,

• Hajek’s GUHA Logic.In 1920’s Lukasiewicz introduced Many-valued Logic.All such logics are special cases of Hohle’s Monoidal Logic; we arenow going to study this logic from an algebraic point of view; eachabove mentioned logic has an algebraic counterpart which is aresiduated lattice.We are going to show that first order Monoidal Logic and all itspresented axiomatic extensions are complete logics.



Applied Logics

Classical Boolean logic is the logic of mathematics, whosehistorical challenger has been Intuitionistic Logic.Recently, there has appeared applied logical systems such as• Girard’s Linear Logic,• Zadeh’s Fuzzy Logic,• Hajek’s GUHA Logic.

In 1920’s Lukasiewicz introduced Many-valued Logic.All such logics are special cases of Hohle’s Monoidal Logic; we arenow going to study this logic from an algebraic point of view; eachabove mentioned logic has an algebraic counterpart which is aresiduated lattice.We are going to show that first order Monoidal Logic and all itspresented axiomatic extensions are complete logics.



Applied Logics

Classical Boolean logic is the logic of mathematics, whosehistorical challenger has been Intuitionistic Logic.Recently, there has appeared applied logical systems such as• Girard’s Linear Logic,• Zadeh’s Fuzzy Logic,• Hajek’s GUHA Logic.In 1920’s Lukasiewicz introduced Many-valued Logic.

All such logics are special cases of Hohle’s Monoidal Logic; we arenow going to study this logic from an algebraic point of view; eachabove mentioned logic has an algebraic counterpart which is aresiduated lattice.We are going to show that first order Monoidal Logic and all itspresented axiomatic extensions are complete logics.



Applied Logics

Classical Boolean logic is the logic of mathematics, whosehistorical challenger has been Intuitionistic Logic.Recently, there has appeared applied logical systems such as• Girard’s Linear Logic,• Zadeh’s Fuzzy Logic,• Hajek’s GUHA Logic.In 1920’s Lukasiewicz introduced Many-valued Logic.All such logics are special cases of Hohle’s Monoidal Logic; we arenow going to study this logic from an algebraic point of view; eachabove mentioned logic has an algebraic counterpart which is aresiduated lattice.

We are going to show that first order Monoidal Logic and all itspresented axiomatic extensions are complete logics.



Applied Logics

Classical Boolean logic is the logic of mathematics, whosehistorical challenger has been Intuitionistic Logic.Recently, there has appeared applied logical systems such as• Girard’s Linear Logic,• Zadeh’s Fuzzy Logic,• Hajek’s GUHA Logic.In 1920’s Lukasiewicz introduced Many-valued Logic.All such logics are special cases of Hohle’s Monoidal Logic; we arenow going to study this logic from an algebraic point of view; eachabove mentioned logic has an algebraic counterpart which is aresiduated lattice.We are going to show that first order Monoidal Logic and all itspresented axiomatic extensions are complete logics.



Order is a fundamental concept in logic. Indeed, already in twovalued logic we assume that the truth value true is in some sense’more’ or ’bigger’ than the truth value false. When talking aboutlogics were truth is by degrees we implicitely assume that thesedegrees are put to some order, the top and bottom elements beingtrue and false, respectively.

Moreover, it is natural to assume that,given two degrees of truth, if they are not comparable, they atleast have the greatest lower bound and the least upper bound. Inthis way we have entered the realm of lattices. Anotherfundamental assumpion is that logical connectives and and impliesare bound by the following: a sentence α and β has at most thesame degree of truth that a sentence γ if, and only if α has atmost the same degree of truth that a sentence β implies γ. Thenwe talk about residuated lattices.



Order is a fundamental concept in logic. Indeed, already in twovalued logic we assume that the truth value true is in some sense’more’ or ’bigger’ than the truth value false. When talking aboutlogics were truth is by degrees we implicitely assume that thesedegrees are put to some order, the top and bottom elements beingtrue and false, respectively. Moreover, it is natural to assume that,given two degrees of truth, if they are not comparable, they atleast have the greatest lower bound and the least upper bound. Inthis way we have entered the realm of lattices.

Anotherfundamental assumpion is that logical connectives and and impliesare bound by the following: a sentence α and β has at most thesame degree of truth that a sentence γ if, and only if α has atmost the same degree of truth that a sentence β implies γ. Thenwe talk about residuated lattices.



Order is a fundamental concept in logic. Indeed, already in twovalued logic we assume that the truth value true is in some sense’more’ or ’bigger’ than the truth value false. When talking aboutlogics were truth is by degrees we implicitely assume that thesedegrees are put to some order, the top and bottom elements beingtrue and false, respectively. Moreover, it is natural to assume that,given two degrees of truth, if they are not comparable, they atleast have the greatest lower bound and the least upper bound. Inthis way we have entered the realm of lattices. Anotherfundamental assumpion is that logical connectives and and impliesare bound by the following: a sentence α and β has at most thesame degree of truth that a sentence γ if, and only if α has atmost the same degree of truth that a sentence β implies γ. Thenwe talk about residuated lattices.



Definition

Assume A is a non–void set and R a binary relation on A. R is apre–order (or quasi–order) on A if

R is reflexive if ∀x ∈ A, xRx and

R is transitive if xRy , yRz imlies xRz where x , y , z ∈ A.

A pre–order R is a partial order on A if

R is anti-symmetric: if xRy , yRx then x = y where x , y ∈ A.A partial order is denoted by ≤. A partial order is a totalorder on A if ∀x , y ∈ A, x ≤ y or y ≤ x .

A pre–order R is an equivalence on A if

R is symmetric: if xRy then yRx where x , y ∈ A.An equivalence relation on A is denoted by ∼.

Denote |x | = {y ∈ A|x ∼ y}, x ∈ A, the equivalence class definedby x , A/ ∼= {|x |; x ∈ A}, set of all equivalence classes.



Definition





R is anti-symmetric: if xRy , yRx then x = y where x , y ∈ A.

A partial order is denoted by ≤. A partial order is a totalorder on A if ∀x , y ∈ A, x ≤ y or y ≤ x .






Definition





R is anti-symmetric: if xRy , yRx then x = y where x , y ∈ A.A partial order is denoted by ≤.

A partial order is a totalorder on A if ∀x , y ∈ A, x ≤ y or y ≤ x .






Definition





R is anti-symmetric: if xRy , yRx then x = y where x , y ∈ A.A partial order is denoted by ≤. A partial order is a totalorder on A if ∀x , y ∈ A, x ≤ y or y ≤ x .






A poset A is called a chain if the partial order ≤ is a total order.

Lemma (1)

Let ∼ is an equivalence relation on A. Then for all x , y ∈ A

x ∈ |x |,x ∼ y iff |x | = |y | iff x ∈ |y |,if x ∼ y does not hold, then |x | ∩ |y | = ∅.

Proof. Exercise.

Let ∼ is an equivalence relation on A and f : Am 7→ A an m–aryoperation on A. If conditions a1 ∼ b1, · · · , am ∼ bm implyf (a1, · · · , am) ∼ f (b1, · · · , bm) then ∼ is a congruence withrespect to f . Then define g(|a1|, · · · , |am|) = |f (a1), · · · , f (am)|and we obtain m–ary operation on A/ ∼.




Lemma (1)


x ∈ |x |,

x ∼ y iff |x | = |y | iff x ∈ |y |,if x ∼ y does not hold, then |x | ∩ |y | = ∅.

Proof. Exercise.





Lemma (1)


x ∈ |x |,x ∼ y iff |x | = |y | iff x ∈ |y |,

if x ∼ y does not hold, then |x | ∩ |y | = ∅.

Proof. Exercise.





Lemma (1)



Proof. Exercise.





Lemma (1)



Proof. Exercise.

Let ∼ is an equivalence relation on A and f : Am 7→ A an m–aryoperation on A.

If conditions a1 ∼ b1, · · · , am ∼ bm implyf (a1, · · · , am) ∼ f (b1, · · · , bm) then ∼ is a congruence withrespect to f . Then define g(|a1|, · · · , |am|) = |f (a1), · · · , f (am)|and we obtain m–ary operation on A/ ∼.




Lemma (1)



Proof. Exercise.

Let ∼ is an equivalence relation on A and f : Am 7→ A an m–aryoperation on A. If conditions a1 ∼ b1, · · · , am ∼ bm implyf (a1, · · · , am) ∼ f (b1, · · · , bm) then ∼ is a congruence withrespect to f .

Then define g(|a1|, · · · , |am|) = |f (a1), · · · , f (am)|and we obtain m–ary operation on A/ ∼.




Lemma (1)



Proof. Exercise.




Lemma (2)

Let R be a pre–order on A. Define a binary operation E on A by

xEy iff xRy and yRx.

Then E is an equivalence relation on A. Moreover, for eachx , y ∈ A define on A/E a binary relation S by

|x |S |y | iff xRy.

Then S is an order relation on A/E. Proof. Exercise.

If there is a partial order ≤ on a set A, then A is called a poset.Assume A is a poset and x , y ∈ A. Any element z ∈ A such thatx , y ≤ z is called an upper bound of {x , y}. A lower bound of{x , y} is an element w ∈ A such that w ≤ x , y . An upper boundand a lower bound of any subset X ⊆ A are defined similarly. Theleast upper bound of {x , y} is an upper bound z such that z ≤ z0

for any other upper bound of {x , y}.



Lemma (2)



Then E is an equivalence relation on A.

Moreover, for eachx , y ∈ A define on A/E a binary relation S by

|x |S |y | iff xRy.






Lemma (2)




|x |S |y | iff xRy.

Then S is an order relation on A/E.

Proof. Exercise.





Lemma (2)




|x |S |y | iff xRy.






Lemma (2)




|x |S |y | iff xRy.


If there is a partial order ≤ on a set A, then A is called a poset.

Assume A is a poset and x , y ∈ A. Any element z ∈ A such thatx , y ≤ z is called an upper bound of {x , y}. A lower bound of{x , y} is an element w ∈ A such that w ≤ x , y . An upper boundand a lower bound of any subset X ⊆ A are defined similarly. Theleast upper bound of {x , y} is an upper bound z such that z ≤ z0




Lemma (2)




|x |S |y | iff xRy.


If there is a partial order ≤ on a set A, then A is called a poset.Assume A is a poset and x , y ∈ A. Any element z ∈ A such thatx , y ≤ z is called an upper bound of {x , y}.

A lower bound of{x , y} is an element w ∈ A such that w ≤ x , y . An upper boundand a lower bound of any subset X ⊆ A are defined similarly. Theleast upper bound of {x , y} is an upper bound z such that z ≤ z0




Lemma (2)




|x |S |y | iff xRy.


If there is a partial order ≤ on a set A, then A is called a poset.Assume A is a poset and x , y ∈ A. Any element z ∈ A such thatx , y ≤ z is called an upper bound of {x , y}. A lower bound of{x , y} is an element w ∈ A such that w ≤ x , y .

An upper boundand a lower bound of any subset X ⊆ A are defined similarly. Theleast upper bound of {x , y} is an upper bound z such that z ≤ z0




Lemma (2)




|x |S |y | iff xRy.


If there is a partial order ≤ on a set A, then A is called a poset.Assume A is a poset and x , y ∈ A. Any element z ∈ A such thatx , y ≤ z is called an upper bound of {x , y}. A lower bound of{x , y} is an element w ∈ A such that w ≤ x , y . An upper boundand a lower bound of any subset X ⊆ A are defined similarly.

Theleast upper bound of {x , y} is an upper bound z such that z ≤ z0




Lemma (2)




|x |S |y | iff xRy.






The greatest lower bound of {x , y} is defined similarly.

We denoteby x ∧ y and x ∨ y the greatest lower bound and the least upperbound of {x , y}, respectively. If x ∧ y and x ∨ y exist in A for allelements x , y ∈ A, then A is a lattice. A lattice A is (countable)complete if

∨{x |x ∈ X} and

∧{x |x ∈ X} exist in A for any

(countable) subset X ⊆ A. It is an exercise to prove the following

Lemma (3)

Let A be a poset. Then (whenever the equations exist in A),

x ∧ x = x, x ∨ x = x (idempotency)

x ∧ y = y ∧ x, x ∨ y = y ∨ x (commutativity)

(x ∧ y) ∧ z = x ∧ (y ∧ z), (x ∨ y) ∨ z = x ∨ (y ∨ z) (assoc.)

x ∧ (x ∨ y) = x ∨ (x ∧ y) = x (absoption)

x ≤ y iff x ∧ y = x iff x ∨ y = y (consistency)

∧ and ∨ are isotone.



The greatest lower bound of {x , y} is defined similarly. We denoteby x ∧ y and x ∨ y the greatest lower bound and the least upperbound of {x , y}, respectively.

If x ∧ y and x ∨ y exist in A for allelements x , y ∈ A, then A is a lattice. A lattice A is (countable)complete if

∨{x |x ∈ X} and



Lemma (3)










The greatest lower bound of {x , y} is defined similarly. We denoteby x ∧ y and x ∨ y the greatest lower bound and the least upperbound of {x , y}, respectively. If x ∧ y and x ∨ y exist in A for allelements x , y ∈ A, then A is a lattice.

A lattice A is (countable)complete if

∨{x |x ∈ X} and



Lemma (3)










The greatest lower bound of {x , y} is defined similarly. We denoteby x ∧ y and x ∨ y the greatest lower bound and the least upperbound of {x , y}, respectively. If x ∧ y and x ∨ y exist in A for allelements x , y ∈ A, then A is a lattice. A lattice A is (countable)complete if

∨{x |x ∈ X} and


(countable) subset X ⊆ A.

It is an exercise to prove the following

Lemma (3)










The greatest lower bound of {x , y} is defined similarly. We denoteby x ∧ y and x ∨ y the greatest lower bound and the least upperbound of {x , y}, respectively. If x ∧ y and x ∨ y exist in A for allelements x , y ∈ A, then A is a lattice. A lattice A is (countable)complete if

∨{x |x ∈ X} and



Lemma (3)










Lemma (4)

Let L be a lattice. Then for all elements x , y , z ∈ L,

x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z) (1)

if, and only if x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z) (2)

Proof. We show that (1) implies (2). The converse is similar.

x ∨ (y ∧ z) = x ∨ (z ∧ y) (commut.)

= [(x ∨ (z ∧ x)] ∨ (z ∧ y) (absorb.)

= (x ∨ [(z ∧ x) ∨ (z ∧ y)] (assoc.)

= (x ∨ [(z ∧ (x ∨ y)] (by (1))

= [x ∧ (x ∨ y)] ∨ [z ∧ (x ∨ y)] (absorb.)

= [(x ∨ y) ∧ x ] ∨ [(x ∨ y) ∧ z ] (commut.)

= (x ∨ y) ∧ (x ∨ z) (by (1)).



A lattice L such that equation (1) holds (thus (2) holds, too) iscalled a distributive lattice.

Definition (Boolean algebra)

Assume in a distributive lattice L, for all elements x ∈ L, there isan element x∗ ∈ L (called lattice complement of x) such that, forall y ∈ L holds

(x ∧ x∗) ∨ y = y and (x ∨ x∗) ∧ y = y .

Then the lattice L = 〈L,≤,∨,∧,∗ 〉 is a Boolean algebra.

Exercise. Prove that in a Boolean algebra (a) the latticecomplement x∗ of x ∈ L is unique, (b) for all x , y ∈ L,x ∧ x∗ = y ∧ y∗ and x ∨ x∗ = y ∨ y∗, (c) x ∧ x∗ is the leastelement of L (a therefore noted by 0), (d) x ∨ x∗ is the greatestelement of L (a therefore noted by 1).






(x ∧ x∗) ∨ y = y and (x ∨ x∗) ∧ y = y .


Exercise. Prove that in a Boolean algebra (a) the latticecomplement x∗ of x ∈ L is unique,

(b) for all x , y ∈ L,x ∧ x∗ = y ∧ y∗ and x ∨ x∗ = y ∨ y∗, (c) x ∧ x∗ is the leastelement of L (a therefore noted by 0), (d) x ∨ x∗ is the greatestelement of L (a therefore noted by 1).






(x ∧ x∗) ∨ y = y and (x ∨ x∗) ∧ y = y .


Exercise. Prove that in a Boolean algebra (a) the latticecomplement x∗ of x ∈ L is unique, (b) for all x , y ∈ L,x ∧ x∗ = y ∧ y∗ and x ∨ x∗ = y ∨ y∗,

(c) x ∧ x∗ is the leastelement of L (a therefore noted by 0), (d) x ∨ x∗ is the greatestelement of L (a therefore noted by 1).






(x ∧ x∗) ∨ y = y and (x ∨ x∗) ∧ y = y .


Exercise. Prove that in a Boolean algebra (a) the latticecomplement x∗ of x ∈ L is unique, (b) for all x , y ∈ L,x ∧ x∗ = y ∧ y∗ and x ∨ x∗ = y ∨ y∗, (c) x ∧ x∗ is the leastelement of L (a therefore noted by 0),

(d) x ∨ x∗ is the greatestelement of L (a therefore noted by 1).






(x ∧ x∗) ∨ y = y and (x ∨ x∗) ∧ y = y .


Exercise. Prove that in a Boolean algebra (a) the latticecomplement x∗ of x ∈ L is unique, (b) for all x , y ∈ L,x ∧ x∗ = y ∧ y∗ and x ∨ x∗ = y ∨ y∗, (c) x ∧ x∗ is the leastelement of L (a therefore noted by 0), (d) x ∨ x∗ is the greatestelement of L (a therefore noted by 1).



Not all lattices are distributive and a distributive lattice need not tobe a Boolean algebra.

A lattice L is called completely distributive if

x ∧∨i∈Γ

yi =∨i∈Γ

(x ∧ yi ), (3)

x ∨∧i∈Γ

yi =∧i∈Γ

(x ∨ yi ) (4)

hold for all elements x ∈ L and all subsets {yi |i ∈ Γ} ⊆ L. Ofcourse, only complete lattices can (but need not) be completelydistributive. Unlike equations (1) and (2), the equations (3) and(4) are not equivalent conditions. It is easy to find examples oflattices that do not contain the least element 0 nor the largestelement 1. In logic applications is, however, natural to assume thatwe have the elements 0 and 1 available. We are now ready todefine residuated lattices.



Not all lattices are distributive and a distributive lattice need not tobe a Boolean algebra. A lattice L is called completely distributive if

x ∧∨i∈Γ

yi =∨i∈Γ

(x ∧ yi ), (3)

x ∨∧i∈Γ

yi =∧i∈Γ

(x ∨ yi ) (4)

hold for all elements x ∈ L and all subsets {yi |i ∈ Γ} ⊆ L.

Ofcourse, only complete lattices can (but need not) be completelydistributive. Unlike equations (1) and (2), the equations (3) and(4) are not equivalent conditions. It is easy to find examples oflattices that do not contain the least element 0 nor the largestelement 1. In logic applications is, however, natural to assume thatwe have the elements 0 and 1 available. We are now ready todefine residuated lattices.




x ∧∨i∈Γ

yi =∨i∈Γ

(x ∧ yi ), (3)

x ∨∧i∈Γ

yi =∧i∈Γ

(x ∨ yi ) (4)

hold for all elements x ∈ L and all subsets {yi |i ∈ Γ} ⊆ L. Ofcourse, only complete lattices can (but need not) be completelydistributive.

Unlike equations (1) and (2), the equations (3) and(4) are not equivalent conditions. It is easy to find examples oflattices that do not contain the least element 0 nor the largestelement 1. In logic applications is, however, natural to assume thatwe have the elements 0 and 1 available. We are now ready todefine residuated lattices.




x ∧∨i∈Γ

yi =∨i∈Γ

(x ∧ yi ), (3)

x ∨∧i∈Γ

yi =∧i∈Γ

(x ∨ yi ) (4)

hold for all elements x ∈ L and all subsets {yi |i ∈ Γ} ⊆ L. Ofcourse, only complete lattices can (but need not) be completelydistributive. Unlike equations (1) and (2), the equations (3) and(4) are not equivalent conditions.

It is easy to find examples oflattices that do not contain the least element 0 nor the largestelement 1. In logic applications is, however, natural to assume thatwe have the elements 0 and 1 available. We are now ready todefine residuated lattices.




x ∧∨i∈Γ

yi =∨i∈Γ

(x ∧ yi ), (3)

x ∨∧i∈Γ

yi =∧i∈Γ

(x ∨ yi ) (4)

hold for all elements x ∈ L and all subsets {yi |i ∈ Γ} ⊆ L. Ofcourse, only complete lattices can (but need not) be completelydistributive. Unlike equations (1) and (2), the equations (3) and(4) are not equivalent conditions. It is easy to find examples oflattices that do not contain the least element 0 nor the largestelement 1.

In logic applications is, however, natural to assume thatwe have the elements 0 and 1 available. We are now ready todefine residuated lattices.




x ∧∨i∈Γ

yi =∨i∈Γ

(x ∧ yi ), (3)

x ∨∧i∈Γ

yi =∧i∈Γ

(x ∨ yi ) (4)

hold for all elements x ∈ L and all subsets {yi |i ∈ Γ} ⊆ L. Ofcourse, only complete lattices can (but need not) be completelydistributive. Unlike equations (1) and (2), the equations (3) and(4) are not equivalent conditions. It is easy to find examples oflattices that do not contain the least element 0 nor the largestelement 1. In logic applications is, however, natural to assume thatwe have the elements 0 and 1 available.

We are now ready todefine residuated lattices.




x ∧∨i∈Γ

yi =∨i∈Γ

(x ∧ yi ), (3)

x ∨∧i∈Γ

yi =∧i∈Γ

(x ∨ yi ) (4)

hold for all elements x ∈ L and all subsets {yi |i ∈ Γ} ⊆ L. Ofcourse, only complete lattices can (but need not) be completelydistributive. Unlike equations (1) and (2), the equations (3) and(4) are not equivalent conditions. It is easy to find examples oflattices that do not contain the least element 0 nor the largestelement 1. In logic applications is, however, natural to assume thatwe have the elements 0 and 1 available. We are now ready todefine residuated lattices.



Definition (Residuated lattice)

Assume L = 〈L,≤,∧,∨, 0, 1〉 is a lattice with the least and thegreatest elements 0, 1, respectively.

Assume further that L has amonoidal structure, i.e. there is a binary operation � calledproduct on L such that � is associative, commutative, isotone andx � 1 = x holds for all x ∈ L. Moreover, assume there is anotherbinary operation → called residuum of � on L such that, for allx , y , z ∈ L holds a Galois connection

x � y ≤ z if, and only if x ≤ y → z . (5)

Then the structure L = 〈L,≤,∧,∨,�,→, 0, 1〉 is a residuatedlattice and the couple 〈�,→〉 is an adjoint couple.

Before giving examples of residuated lattices we present twoTheorems




Assume L = 〈L,≤,∧,∨, 0, 1〉 is a lattice with the least and thegreatest elements 0, 1, respectively. Assume further that L has amonoidal structure, i.e. there is a binary operation � calledproduct on L such that � is associative, commutative, isotone andx � 1 = x holds for all x ∈ L.

Moreover, assume there is anotherbinary operation → called residuum of � on L such that, for allx , y , z ∈ L holds a Galois connection







Assume L = 〈L,≤,∧,∨, 0, 1〉 is a lattice with the least and thegreatest elements 0, 1, respectively. Assume further that L has amonoidal structure, i.e. there is a binary operation � calledproduct on L such that � is associative, commutative, isotone andx � 1 = x holds for all x ∈ L. Moreover, assume there is anotherbinary operation → called residuum of � on L such that, for allx , y , z ∈ L holds a Galois connection






Theorem (1)

Assume 〈�,→〉 is an adjoint couple of a residuated lattice L. Then

x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)

whenever these joins exist in L.

Proof. Let x ∈ L, {yi |i ∈ Γ} ⊆ L and both sides of (6) exist in L.Since � is isotone, x � yi ≤ x �

∨i∈Γ yi for all i ∈ Γ, therefore∨

i∈Γ(x � yi ) ≤ x �∨

i∈Γ yi . Conversely, as x � yi ≤∨

i∈Γ(x � yi )holds for all i ∈ Γ, we obtain by the Galois connection (5) thatyi ≤ x → [

∨i∈Γ(x � yi )] holds for all i ∈ Γ, and therefore also∨

i∈Γ yi ≤ x → [∨

i∈Γ(x � yi )]. Again by (5) we reasonx �

∨i∈Γ yi ≤

∨i∈Γ(x � yi ), and therefore the claim follows.



Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)


Proof. Let x ∈ L, {yi |i ∈ Γ} ⊆ L and both sides of (6) exist in L.

Since � is isotone, x � yi ≤ x �∨

i∈Γ yi for all i ∈ Γ, therefore∨i∈Γ(x � yi ) ≤ x �

∨i∈Γ yi . Conversely, as x � yi ≤

∨i∈Γ(x � yi )

holds for all i ∈ Γ, we obtain by the Galois connection (5) thatyi ≤ x → [


i∈Γ yi ≤ x → [∨


∨i∈Γ yi ≤




Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)



∨i∈Γ yi for all i ∈ Γ,

therefore∨i∈Γ(x � yi ) ≤ x �

∨i∈Γ yi . Conversely, as x � yi ≤

∨i∈Γ(x � yi )

holds for all i ∈ Γ, we obtain by the Galois connection (5) thatyi ≤ x → [


i∈Γ yi ≤ x → [∨


∨i∈Γ yi ≤




Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)




i∈Γ(x � yi ) ≤ x �∨

i∈Γ yi .

Conversely, as x � yi ≤∨



i∈Γ yi ≤ x → [∨


∨i∈Γ yi ≤




Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)




i∈Γ(x � yi ) ≤ x �∨


i∈Γ(x � yi )holds for all i ∈ Γ,

we obtain by the Galois connection (5) thatyi ≤ x → [


i∈Γ yi ≤ x → [∨


∨i∈Γ yi ≤




Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)




i∈Γ(x � yi ) ≤ x �∨



∨i∈Γ(x � yi )] holds for all i ∈ Γ,

and therefore also∨i∈Γ yi ≤ x → [

∨i∈Γ(x � yi )]. Again by (5) we reason

x �∨

i∈Γ yi ≤∨

i∈Γ(x � yi ), and therefore the claim follows.



Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)




i∈Γ(x � yi ) ≤ x �∨




i∈Γ yi ≤ x → [∨

i∈Γ(x � yi )].

Again by (5) we reasonx �

∨i∈Γ yi ≤




Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)




i∈Γ(x � yi ) ≤ x �∨




i∈Γ yi ≤ x → [∨


∨i∈Γ yi ≤

∨i∈Γ(x � yi ),

and therefore the claim follows.



Theorem (1)


x �∨i∈Γ

yi =∨i∈Γ

(x � yi ) (6)




i∈Γ(x � yi ) ≤ x �∨




i∈Γ yi ≤ x → [∨


∨i∈Γ yi ≤




Theorem (2)

Assume condition (6) always holds in a lattice L with a monoidalstructure. Then L can be (uniquely) residuated via

x → y =∨{z ∈ L| x � z ≤ y}. (7)

Proof. If a� b ≤ c then a ≤∨{z ∈ L| b � z ≤ c} = b → c and

vice versa. Thus, the Galois connection (5) holds.

By condition (7), residuum → of a product � is unique, we maytalk about the residuum of �.

The above Theorems imply that if (6) holds, then we can definethe corresponding residuum by (7) and vice versa.



Theorem (2)


x → y =∨{z ∈ L| x � z ≤ y}. (7)


vice versa.

Thus, the Galois connection (5) holds.





Theorem (2)


x → y =∨{z ∈ L| x � z ≤ y}. (7)


vice versa. Thus, the Galois connection (5) holds.





As a consequence we have, for example, that any completelydistributive lattice is residuated.

Indeed, they have bottom and topelements 0, 1, the meet operation ∧ is associative, commutative,isotone and for all x ∈ L, x ∧ 1 = x . Moreover, completelydistributiveness is exatly equation (6). Such algebras are Heytingalgebras, the algebras corresponding to Intuitionistic Logic.Another familiar class of residuated lattices are Boolean algebras.There again � = ∧ and the residuum is given via x → y = x∗ ∨ y -to see all the details is an exercise. Boolean algebras are thealgebras of Classical Logic. The real unit interval [0, 1] is a latticewith bottom and top elements, the order is the order ≤ of reals,x ∧ y = min{x , y} and x ∨ y = max{x , y}.



As a consequence we have, for example, that any completelydistributive lattice is residuated. Indeed, they have bottom and topelements 0, 1, the meet operation ∧ is associative, commutative,isotone and for all x ∈ L, x ∧ 1 = x . Moreover, completelydistributiveness is exatly equation (6).

Such algebras are Heytingalgebras, the algebras corresponding to Intuitionistic Logic.Another familiar class of residuated lattices are Boolean algebras.There again � = ∧ and the residuum is given via x → y = x∗ ∨ y -to see all the details is an exercise. Boolean algebras are thealgebras of Classical Logic. The real unit interval [0, 1] is a latticewith bottom and top elements, the order is the order ≤ of reals,x ∧ y = min{x , y} and x ∨ y = max{x , y}.



As a consequence we have, for example, that any completelydistributive lattice is residuated. Indeed, they have bottom and topelements 0, 1, the meet operation ∧ is associative, commutative,isotone and for all x ∈ L, x ∧ 1 = x . Moreover, completelydistributiveness is exatly equation (6). Such algebras are Heytingalgebras, the algebras corresponding to Intuitionistic Logic.

Another familiar class of residuated lattices are Boolean algebras.There again � = ∧ and the residuum is given via x → y = x∗ ∨ y -to see all the details is an exercise. Boolean algebras are thealgebras of Classical Logic. The real unit interval [0, 1] is a latticewith bottom and top elements, the order is the order ≤ of reals,x ∧ y = min{x , y} and x ∨ y = max{x , y}.



As a consequence we have, for example, that any completelydistributive lattice is residuated. Indeed, they have bottom and topelements 0, 1, the meet operation ∧ is associative, commutative,isotone and for all x ∈ L, x ∧ 1 = x . Moreover, completelydistributiveness is exatly equation (6). Such algebras are Heytingalgebras, the algebras corresponding to Intuitionistic Logic.Another familiar class of residuated lattices are Boolean algebras.There again � = ∧ and the residuum is given via x → y = x∗ ∨ y

-to see all the details is an exercise. Boolean algebras are thealgebras of Classical Logic. The real unit interval [0, 1] is a latticewith bottom and top elements, the order is the order ≤ of reals,x ∧ y = min{x , y} and x ∨ y = max{x , y}.



As a consequence we have, for example, that any completelydistributive lattice is residuated. Indeed, they have bottom and topelements 0, 1, the meet operation ∧ is associative, commutative,isotone and for all x ∈ L, x ∧ 1 = x . Moreover, completelydistributiveness is exatly equation (6). Such algebras are Heytingalgebras, the algebras corresponding to Intuitionistic Logic.Another familiar class of residuated lattices are Boolean algebras.There again � = ∧ and the residuum is given via x → y = x∗ ∨ y -to see all the details is an exercise.

Boolean algebras are thealgebras of Classical Logic. The real unit interval [0, 1] is a latticewith bottom and top elements, the order is the order ≤ of reals,x ∧ y = min{x , y} and x ∨ y = max{x , y}.



As a consequence we have, for example, that any completelydistributive lattice is residuated. Indeed, they have bottom and topelements 0, 1, the meet operation ∧ is associative, commutative,isotone and for all x ∈ L, x ∧ 1 = x . Moreover, completelydistributiveness is exatly equation (6). Such algebras are Heytingalgebras, the algebras corresponding to Intuitionistic Logic.Another familiar class of residuated lattices are Boolean algebras.There again � = ∧ and the residuum is given via x → y = x∗ ∨ y -to see all the details is an exercise. Boolean algebras are thealgebras of Classical Logic.

The real unit interval [0, 1] is a latticewith bottom and top elements, the order is the order ≤ of reals,x ∧ y = min{x , y} and x ∨ y = max{x , y}.



As a consequence we have, for example, that any completelydistributive lattice is residuated. Indeed, they have bottom and topelements 0, 1, the meet operation ∧ is associative, commutative,isotone and for all x ∈ L, x ∧ 1 = x . Moreover, completelydistributiveness is exatly equation (6). Such algebras are Heytingalgebras, the algebras corresponding to Intuitionistic Logic.Another familiar class of residuated lattices are Boolean algebras.There again � = ∧ and the residuum is given via x → y = x∗ ∨ y -to see all the details is an exercise. Boolean algebras are thealgebras of Classical Logic. The real unit interval [0, 1] is a latticewith bottom and top elements, the order is the order ≤ of reals,x ∧ y = min{x , y} and x ∨ y = max{x , y}.



Examples of residuated structures on [0, 1]

Godel algebra (BL):

x � y = min{x , y}, x → y =

{1 if x ≤ yy otherwise

Product algebra (BL):

x � y = xy , x → y =

{1 if x ≤ y

y/x otherwiseLukasiewicz algebra (MV):x � y = max{0, x + y − 1}, x → y = min{1, 1− x + y}.A structure Lc (MTL) where 0 < c < 1:

x � y =

{0 if x + y ≤ cmin{x , y} elsewhere

,

x → y =

{1 if x ≤ y ,max{c − x , y} elsewhere.



Examples of residuated structures on [0, 1]Godel algebra (BL):

x � y = min{x , y}, x → y =



x � y = xy , x → y =

{1 if x ≤ y


x � y =


,

x → y =





x � y = min{x , y}, x → y =



x � y = xy , x → y =

{1 if x ≤ y

y/x otherwise

Lukasiewicz algebra (MV):x � y = max{0, x + y − 1}, x → y = min{1, 1− x + y}.A structure Lc (MTL) where 0 < c < 1:

x � y =


,

x → y =





x � y = min{x , y}, x → y =



x � y = xy , x → y =

{1 if x ≤ y

y/x otherwiseLukasiewicz algebra (MV):x � y = max{0, x + y − 1}, x → y = min{1, 1− x + y}.

A structure Lc (MTL) where 0 < c < 1:

x � y =


,

x → y =





x � y = min{x , y}, x → y =



x � y = xy , x → y =

{1 if x ≤ y


x � y =


,

x → y =




A structure LsD (SD):

x � y =

{0 if x , y ∈ [0, 1

2 ]min{x , y} elsewhere

,

x → y =

1 if x ≤ y ,12 if y < x ≤ 1

2 ,y if y < x , 1

2 < x .

Theorem (Basic properties of residuated lattices)

For all elements x , y , x1, x2, y1, y2 ∈ L the following hold

x = 1 → x , (8)

1 = x → x , (9)

x � y ≤ x , y , (10)



Theorem (Basic properties of residuated lattices, continuation 1)

x � y ≤ x ∧ y , (11)

y ≤ x → y , (12)

x � y ≤ x → y , (13)

x ≤ y iff 1 = x → y , (14)

1 = x → y = y → x iff x = y , (15)

x → 1 = 1, (16)

0 → x = 1, (17)

x → (y → x) = 1, (18)

(x → y) → [(y → z) → (x → z)] = 1, (19)




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21).

By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y)

≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y

≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z ,

and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z .

Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff

[(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff

[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff

[(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.

Thus, (21) holds.




(x → y) → [(z → x) → (z → y)] = 1, (20)

(x � y) → z = x → (y → z), (21)

x → (y → z) = y → (x → z), (22)

(x1 → y1) → {(y2 → x2) → [(y1 → y2) → (x1 → x2)]} = 1. (23)

Proof. We establish only (21). By Galois connection (5) we have[x → (y → z)]� (x � y) ≤ (y → z)� y ≤ z , and again by (5),[x → (y → z) ≤ (x � y) → z . Conversely,(x � y) → z ≤ x → (y → z) iff [(x � y) → z ]� x ≤ y → z iff[(x � y) → z ]� (x � y) ≤ z iff [(x � y) → z ] ≤ (x � y) → z , true.Thus, (21) holds.



When we later talk about logics, the residuum operation will playan important role when considering implication. For the needs ofnegation we introduce a complement (do not mix up with latticecomplement!) of an element x ∈ L by stipulating x∗ = x → 0.

Theorem (Properties of complement)

Let L be a residuated lattice and x , y ∈ L. Then

x � x∗ = 0, (24)

x ≤ x∗∗, (25)

1∗ = 0, (26)

0∗ = 1, (27)

x → y ≤ y∗ → x∗, (28)

x∗ = x∗∗∗. (29)



The proof is left as an exercise.

We introduce a derived operationbi–residuum by setting x ↔ y = (x → y) ∧ (y → x) on L.Bi–residuum will turn useful when we talk about fuzzy equivalencerelations. The proof of the following theorem is an exercise

Theorem (Properties of bi–residuum)

Let L be a residuated lattice. Then for all elements in L hold

x ↔ 1 = x , (30)

x = y iff x ↔ y = 1, (31)

x ↔ y = y ↔ x , (32)

(x ↔ y)� (y ↔ z) ≤ (x ↔ z) (33)

(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∧ x2) ↔ (y1 ∧ y2), (34)



The proof is left as an exercise. We introduce a derived operationbi–residuum by setting x ↔ y = (x → y) ∧ (y → x) on L.

Bi–residuum will turn useful when we talk about fuzzy equivalencerelations. The proof of the following theorem is an exercise



x ↔ 1 = x , (30)

x = y iff x ↔ y = 1, (31)

x ↔ y = y ↔ x , (32)

(x ↔ y)� (y ↔ z) ≤ (x ↔ z) (33)

(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∧ x2) ↔ (y1 ∧ y2), (34)



The proof is left as an exercise. We introduce a derived operationbi–residuum by setting x ↔ y = (x → y) ∧ (y → x) on L.Bi–residuum will turn useful when we talk about fuzzy equivalencerelations.

The proof of the following theorem is an exercise



x ↔ 1 = x , (30)

x = y iff x ↔ y = 1, (31)

x ↔ y = y ↔ x , (32)

(x ↔ y)� (y ↔ z) ≤ (x ↔ z) (33)

(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∧ x2) ↔ (y1 ∧ y2), (34)



The proof is left as an exercise. We introduce a derived operationbi–residuum by setting x ↔ y = (x → y) ∧ (y → x) on L.Bi–residuum will turn useful when we talk about fuzzy equivalencerelations. The proof of the following theorem is an exercise



x ↔ 1 = x , (30)

x = y iff x ↔ y = 1, (31)

x ↔ y = y ↔ x , (32)

(x ↔ y)� (y ↔ z) ≤ (x ↔ z) (33)

(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∧ x2) ↔ (y1 ∧ y2), (34)



Theorem (Properties of bi–residuum, continuation)

(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∨ x2) ↔ (y1 ∨ y2), (35)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 � x2) ↔ (y1 � y2), (36)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 → x2) ↔ (y1 → y2). (37)

Theorem (Properties of a complete residuated lattice L)

Assume x ∈ L, {yi |i ∈ Γ} ⊆ L. Then

x →∧i∈Γ

yi =∧i∈Γ

(x → yi ), (38)∨i∈Γ

yi → x =∧i∈Γ

(yi → x) (39)




(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∨ x2) ↔ (y1 ∨ y2), (35)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 � x2) ↔ (y1 � y2), (36)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 → x2) ↔ (y1 → y2). (37)



x →∧i∈Γ

yi =∧i∈Γ

(x → yi ), (38)

∨i∈Γ

yi → x =∧i∈Γ

(yi → x) (39)




(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∨ x2) ↔ (y1 ∨ y2), (35)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 � x2) ↔ (y1 � y2), (36)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 → x2) ↔ (y1 → y2). (37)



x →∧i∈Γ

yi =∧i∈Γ

(x → yi ), (38)∨i∈Γ

yi → x =∧i∈Γ

(yi → x)

(39)




(x1 ↔ y1) ∧ (x2 ↔ y2) ≤ (x1 ∨ x2) ↔ (y1 ∨ y2), (35)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 � x2) ↔ (y1 � y2), (36)

(x1 ↔ y1)� (x2 ↔ y2) ≤ (x1 → x2) ↔ (y1 → y2). (37)



x →∧i∈Γ

yi =∧i∈Γ

(x → yi ), (38)∨i∈Γ

yi → x =∧i∈Γ

(yi → x) (39)



Theorem (Properties of a complete ... continuation)

∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)

∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi , (41)

The definitions of∨

and∧

as well as the Galois connection (5)are essential in the prooves (an exercise), for example to prove(40), we realize that (yi → x) ≤

∧i∈Γ yi → x holds for each index

i ∈ Γ (as → is antitone in the first variable) and therefore (40)holds. Particular instances of (39) and (40) are the following

(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.




∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi ,

(41)


and∧




(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.




∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi , (41)


and∧




(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.




∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi , (41)


and∧

as well as the Galois connection (5)are essential in the prooves (an exercise),

for example to prove(40), we realize that (yi → x) ≤



(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.




∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi , (41)


and∧

as well as the Galois connection (5)are essential in the prooves (an exercise), for example to prove(40), we realize that

(yi → x) ≤∧

i∈Γ yi → x holds for each indexi ∈ Γ (as → is antitone in the first variable) and therefore (40)holds. Particular instances of (39) and (40) are the following

(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.




∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi , (41)


and∧



i ∈ Γ (as → is antitone in the first variable)

and therefore (40)holds. Particular instances of (39) and (40) are the following

(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.




∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi , (41)


and∧



i ∈ Γ (as → is antitone in the first variable) and therefore (40)holds.

Particular instances of (39) and (40) are the following(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.




∨i∈Γ

(yi → x) ≤∧i∈Γ

yi → x , (40)∨i∈Γ

(x → yi ) ≤ x →∨i∈Γ

yi , (41)


and∧




(∨

i∈Γ yi )∗ =

∧i∈Γ y∗i ,

∨i∈Γ y∗i ≤ (

∧i∈Γ yi )

∗.



We finish this introduction by recalling some terminology.

Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.

∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧.

– Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.

∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.

∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L.

–Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.

∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.

∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



We finish this introduction by recalling some terminology. Aresiduated lattice L is∗ a (commutative) Girard monoid if the complement is involutive,i.e., x = x∗∗ for all x ∈ L. Complete Girard monoids – that is,Girard monoids whose lattice structure contains all meets and joins– are called Girard quantales.∗ a Heyting algebra if the product operation � coincides with theoperation ∧. – Involutive Heyting algebras are Boolean algebras.∗ called divisible if, for all x , y ∈ L, x ≤ y , there is z ∈ L such thatx = z � y . This condition is equivalent to x ∧ y = x � (x → y) forall x , y ∈ L.∗ called prelinear if (x → y)∨ (y → x) = 1 holds for all x , y ∈ L. –Another name is MTL–algebra.∗ a BL–algebra if it is prelinear and divisible.∗ an MV –algebra if it is a divisible Girard monoid or, equivalenty, aBL–algebra with an involutive complement.



∗ called semi–divisible if, for all x , y ∈ L,

(x∗ → y∗) → y∗ = (y∗ → x∗) → x∗. (42)

Equation (42) is equivalent to

(x∗ ∧ y∗)∗ = [x∗ � (x∗ → y∗)]∗ (43)

and, moreover, a subset MV (L) = {x∗| x ∈ L} (non–void as0, 1 ∈ MV (L) and called the MV–center of L) generates anMV –algebra, where the operations �MV and ∨MV are defined via

x∗ �MV y∗ = (x∗ � y∗)∗∗, x∗ ∨MV y∗ = (x∗ ∨ y∗)∗∗. (44)

The order ≤ and the operations →,∗ ,∧ are those on L.



Filters and deductive systems on residuated lattices

Definition

Let L be a residuated lattice. A non–void set F ⊆ L is a filter on Lif (i) x , y ∈ F implies x � y ∈ F and (ii) x ∈ F , x ≤ y impliesy ∈ F . Further a non–void set D ⊆ L is a deductive system on L if(a) 1 ∈ D and (b) x , x → y ∈ D implies y ∈ D.

Lemma

A ⊆ L is a filter on L iff A is a deductive system on L.

Proof. Let A be a filter on L. Then by (ii) 1 ∈ A. Letx , x → y ∈ A. Then by (i), (x → y)� x ∈ A. But(x → y)� x ≤ y so that by (ii) y ∈ A. Conversely, let A be a dson L. If x ∈ A, x ≤ y , then x → y = 1 ∈ A which by (b) impliesy ∈ A, consequently (ii) holds.




Definition

Let L be a residuated lattice. A non–void set F ⊆ L is a filter on Lif

(i) x , y ∈ F implies x � y ∈ F and (ii) x ∈ F , x ≤ y impliesy ∈ F . Further a non–void set D ⊆ L is a deductive system on L if(a) 1 ∈ D and (b) x , x → y ∈ D implies y ∈ D.

Lemma






Definition

Let L be a residuated lattice. A non–void set F ⊆ L is a filter on Lif (i) x , y ∈ F implies x � y ∈ F and (ii) x ∈ F , x ≤ y impliesy ∈ F .

Further a non–void set D ⊆ L is a deductive system on L if(a) 1 ∈ D and (b) x , x → y ∈ D implies y ∈ D.

Lemma






Definition

Let L be a residuated lattice. A non–void set F ⊆ L is a filter on Lif (i) x , y ∈ F implies x � y ∈ F and (ii) x ∈ F , x ≤ y impliesy ∈ F . Further a non–void set D ⊆ L is a deductive system on L if

(a) 1 ∈ D and (b) x , x → y ∈ D implies y ∈ D.

Lemma






Definition


Lemma






Definition


Lemma


Proof. Let A be a filter on L. Then by (ii) 1 ∈ A.

Letx , x → y ∈ A. Then by (i), (x → y)� x ∈ A. But(x → y)� x ≤ y so that by (ii) y ∈ A. Conversely, let A be a dson L. If x ∈ A, x ≤ y , then x → y = 1 ∈ A which by (b) impliesy ∈ A, consequently (ii) holds.




Definition


Lemma


Proof. Let A be a filter on L. Then by (ii) 1 ∈ A. Letx , x → y ∈ A. Then by (i), (x → y)� x ∈ A. But(x → y)� x ≤ y so that by (ii) y ∈ A.

Conversely, let A be a dson L. If x ∈ A, x ≤ y , then x → y = 1 ∈ A which by (b) impliesy ∈ A, consequently (ii) holds.




Definition


Lemma


Proof. Let A be a filter on L. Then by (ii) 1 ∈ A. Letx , x → y ∈ A. Then by (i), (x → y)� x ∈ A. But(x → y)� x ≤ y so that by (ii) y ∈ A. Conversely, let A be a dson L.

If x ∈ A, x ≤ y , then x → y = 1 ∈ A which by (b) impliesy ∈ A, consequently (ii) holds.




Definition


Lemma





To see that (i) holds too, let x , y ∈ A. Since (ii) holds andx ≤ y → (x � y), it follows that

y → (x � y) ∈ A, moreover, sincey ∈ A, also x � y ∈ A by (b). Hence (i) is valid. The proof iscomplete.

If X ⊆ L, a filter generated by X , denoted G (X ), is the smallestfilter containing X . It is an exercise to prove that G (∅) = {1} andif X 6= ∅ then

G (X ) = {y ∈ L| x1 � · · · � xn ≤ y for some x1, · · · , xn ∈ X}.

Further, if F is a filter on L and x ∈ L then

G (F , x) = {y ∈ L| f � xn ≤ y for some f ∈ F , n ∈ N}.

A filter F is proper if F 6= L. A proper filter is maximal if it is notstrictly contained in any proper filter on L.



To see that (i) holds too, let x , y ∈ A. Since (ii) holds andx ≤ y → (x � y), it follows that y → (x � y) ∈ A,

moreover, sincey ∈ A, also x � y ∈ A by (b). Hence (i) is valid. The proof iscomplete.








To see that (i) holds too, let x , y ∈ A. Since (ii) holds andx ≤ y → (x � y), it follows that y → (x � y) ∈ A, moreover, sincey ∈ A, also x � y ∈ A by (b).

Hence (i) is valid. The proof iscomplete.








To see that (i) holds too, let x , y ∈ A. Since (ii) holds andx ≤ y → (x � y), it follows that y → (x � y) ∈ A, moreover, sincey ∈ A, also x � y ∈ A by (b). Hence (i) is valid. The proof iscomplete.









If X ⊆ L, a filter generated by X , denoted G (X ), is the smallestfilter containing X .

It is an exercise to prove that G (∅) = {1} andif X 6= ∅ then












A filter F is proper if F 6= L.

A proper filter is maximal if it is notstrictly contained in any proper filter on L.



Lemma

A proper filter F ⊆ L is maximal if and only if for all x ∈ L \ F ,there is an n ∈ N such that (xn)∗ ∈ F .

Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F .

ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .

Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .

Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.

Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N .

Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).

Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal.

The proof is complete.



Lemma


Proof. Let F be a maximal filter and x ∈ L \ F . ThenG (F , x) = L, so that 0 = f � xn for some f ∈ F and some n ∈ N .Then, by residuation, f ≤ (xn)∗ which implies (xn)∗ ∈ F .Conversely, suppose that the condition holds and that F is proper.Let x /∈ F . Then x ∈ G (F , x), so xm ∈ G (F , x) for all m ∈ N . Onthe other hand there is an n ∈ N such that (xn)∗ ∈ F ⊆ G (F , x).Then, however, 0 = xn � (xn)∗ ∈ G (F , x), so that G (F , x) = Lwhich implies that F is maximal. The proof is complete.



ExercisesExercise 1. Assume A is the set of all human beings. Define a binary relation Ron A by xRy if person x understands person’s y language. Is R a quasi–orderon A?

Exercise 2. Prove Lemma 1.



Exercise 5. Prove that in a Boolean algebra (a) the lattice complement x∗ ofx ∈ L is unique, (b) for all x , y ∈ L, x ∧ x∗ = y ∧ y∗ and x ∨ x∗ = y ∨ y∗, (c)x ∧ x∗ is the least element of L, (d) x ∨ x∗ is the greatest element of L.

Exercise 6. Prove that the above mentioned structures (Godel, Product,Lukasiewicz, Lc and LsD) are residuated lattices.

Exercise 7. Prove equations (8) – (23).



Exercise 10. Prove equations (38), (39) and (41).


part i. order, posets, lattices and residuated lattices in...

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