algebraic semantics for the logic of multiple-source

Algebraic Semantics for the Logic ofMultiple-source Approximation Systems

Mohua Banerjee Md. Aquil Khan

Indian Institute of Technology Kanpur

ICLA 2009

Mohua Banerjee, Md. Aquil Khan Algebraic Semantics for the Logic of Multiple-source Approximation Systems

Rough Set Theory

Out break of some disease.

Totally unaffected class −→ classes 1 to 8.

Totally affected class −→ class 9.

Partially affected class −→ class 10.

Q. Is ‘A’, a class 8 student, affected?

A. No.

Q. Is ‘B’, a class 9 student, affected?

A. Yes.

Q. Is ‘C’, a class 10 student, affected?

A. Possibly, but not certainly.

Rough Set Theory

A. No.

A. Yes.

Rough Set Theory

A. No.

A. Yes.

Rough Set Theory

A. No.

A. Yes.

Rough Set Theory

A. No.

A. Yes.

Rough Set Theory

A. No.

A. Yes.

Rough Set Theory

A. No.

A. Yes.

Rough Set TheoryPawlak Approximation Space

Approximation space [Pawlak’82]

(U,R), where R is an equivalence relation on U.

XX B(X )

X B(X )

XX B(X )

Boundary element

U := Set of students;

aRb iff a and b are in the same class;

X := Set of affected students.

1 2 3 4 5

6 7 8 9 10

1 2 3 4 5

6 7 8 9 10

1 2 3 4 5

6 7 8 9 10X

1 2 3 4 5

6 7 8 9 10

1 2 3 4 5

6 7 8 9 10XX

1 2 3 4 5

6 7 8 9 10

Multiple-source Approximation Systems

Multiple-source Approximation Systems(MSAS) [Banerjee andKhan ’08]

F := (U, {Ri}i∈N), where

U is a non-empty set,

N an initial segment of the set of positive integers, and

each Ri , i ∈ N, is an equivalence relation on the domain U.

|N| is referred to as the cardinality of F and is denoted by |F|.

Language

a non-empty countable set Var of variables,

a (possibly empty) countable set Con of constants,

a non-empty countable set PV of propositional variables and

the propositional constants >,⊥.

Terms T := Var ∪ Con.

Wffs:=>|⊥|p|¬α|α ∧ β|〈t〉α|∀xαp ∈ PV , t ∈ T , x ∈ Var , and α, β are wffs.

F−→ Set of all wffs;

F−→ Set of all closed wffs;

Let Γ be a set of wffs of LMSAS .

Interpretation for Γ

M := (F,V , I ), where

F := (U, {Ri}i∈N) is a MSAS,

V : PV → P(U) and

I : Con(Γ) → N.

Assignment for an interpretation M is a mapv : Term(Γ) → N such that v(c) = I (c), for each c ∈ Con(Γ).

satisfiability

M, v ,w |= p, if and only if w ∈ V (p), for p ∈ PV .

M, v ,w |= 〈t〉α, if and only if there exists w ′ in U such thatwRv(t)w

′ and M, v ,w ′ |= α.

M, v ,w |= ∀xα, if and only if for every assignment v ′

x-equivalent to v , M, v ′,w |= α.

α is valid, denoted |= α, if and only if M, v ,w |= α, for everyinterpretation M := (µ,V , I ), assignment v for M and object win the domain of F.

LMSASRough Set Interpretation

Consider F := (U, {Ri}i∈N), M := (F,V , I ), and assignmentv ;

let V (α) := {w ∈ U : M, v ,w |= α}.

V (〈t〉α) = V (α)Rv(t);

V ([t]α) = V (α)Rv(t)

V (∀x [x ]α) =⋂

i V (α)Ri

V (∃x [x ]α) =⋃

i V (α)Ri

V (∀x〈x〉α) =⋂

i V (α)Ri;

V (∃x〈x〉α) =⋃

i V (α)Ri.

Axioms

1 All axioms of classical propositional logic.

2 ∀xα → α[t/x ], where α admits the term t for the variable x .

3 ∀x(α → β) → (α → ∀xβ), where the variable x is not free inα.

4 ∀x [t]α → [t]∀xα, where the term t and variable x aredifferent.

5 [t](α → β) → ([t]α → [t]β).

6 α → 〈t〉α.

7 α → [t]〈t〉α.

8 〈t〉〈t〉α → 〈t〉α.

Rules of inference

∀. α MP. α N. α∀xα α → β [t]α

Soundness and Completeness Theorem

` α if and only if |= α

Algebraic Semantics of LMSAS

Boolean Algebra with Operators(BAOs)

BAOs A := (A,∩,∼, 1, fi )i∈∆ where

(A,∩,∼, 1) is a Boolean algebra (BA) and

for each i ∈ ∆, fi is an unary operator satisfying1 f (1) = 1;2 f (a ∩ b) = fa ∩ fb.

∆ = Neach fk satisfies the following additional conditions:

(B1) fka ≤ fk fka;(B2) fka ≤ a;(B3) a ≤ fkgka, where gk :=∼ fk ∼.

Boolean Algebra with Operators(BAOs)

BAOs A := (A,∩,∼, 1, fi )i∈∆ where

(A,∩,∼, 1) is a Boolean algebra (BA) and

for each i ∈ ∆, fi is an unary operator satisfying1 f (1) = 1;2 f (a ∩ b) = fa ∩ fb.

∆ = Neach fk satisfies the following additional conditions:

(B1) fka ≤ fk fka;(B2) fka ≤ a;(B3) a ≤ fkgka, where gk :=∼ fk ∼.

Definition

Let A := (A,∩,∼, 1, fi )i∈N −→ be a BAOs.

Assignment θ : PV → A.

θ : F → A where

θ([ci ]α) = fi (θ(α)), i ∈ N,

θ(∀xα) =⋂{θ(α(cj/x)) : j ∈ N}, provided the g.l.b. exists.

For α ∈ F , θ(α) := θ(cl(α)).

A := (A,∩,∼, 1, fi )i∈N −→ be a BAOs satisfying (B1)-(B3).

θ(α) exists for all α ∈ F if and only if⋂{θ(α(cj/x)) : j ∈ N}

exists for all α with only free variable x .

fi (a ∩ b) = fia ∩ fib;fi (

⋂X ) 6=

⋂fiX ;

fk⋂j

θ(α(cj/x)) 6=⋂j

fk θ(α(cj/x));

(∀x [t]α → [t]∀xα, t and x are different)

fi (a ∩ b) = fia ∩ fib;

fi (⋂

X ) 6=⋂

fk⋂j

θ(α(cj/x)) 6=⋂j

fk θ(α(cj/x));

⋂X ) 6=

⋂fiX ;

fk⋂j

θ(α(cj/x)) 6=⋂j

fk θ(α(cj/x));

⋂X ) 6=

⋂fiX ;

fk⋂j

θ(α(cj/x)) 6=⋂j

fk θ(α(cj/x));

Realization

A BAOs A := (A,∩,∼, 1, fi )i∈N satisfying (B1)-(B3) such that forevery assignment θ : PV → A the following is satisfied:

(R1) θ(α) exists for all α ∈ F ,

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)), where α has

only one free variable x .

Definition

A BAOs A := (A,∩,∼, 1, fi )i∈∆ satisfying (B1)-(B3), is said to bea complete BAOs (CBAOs) if it satisfies the following propertiesfor all X ⊆ A:

(A1)⋂

X and⋃

X exists,

(A2) fk⋂

X =⋂

fkX , k ∈ ∆,

Every CBAOs is a realization but not conversely.

Consider the BAOs which is not complete but which has only afinite number of distinct operators fi .

Definition

(A1)⋂

X and⋃

X exists,

(A2) fk⋂

X =⋂

fkX , k ∈ ∆,

Definition

(A1)⋂

X and⋃

X exists,

(A2) fk⋂

X =⋂

fkX , k ∈ ∆,

Definition

Let A := (A,∩,∼, 1, fi )i∈N be a realization.

A α ≈ β if and only if for every assignmentθ : PV → A, θ(α) = θ(β).

C α if A α ≈ > for all A ∈ C (Class of all CBAOs).

Soundness

If ` α then C α

Completeness

If C α then ` α

If 6` α, then there exists a CBAOs A := (A,∩,∼, 1, fi )i∈N and anassignment θ : PV → A such that θ(α) 6= 1.

Soundness

If ` α then C α

Completeness

If C α then ` α

If 6` α, then there exists a CBAOs A := (A,∩,∼, 1, fi )i∈N and anassignment θ : PV → A such that θ(α) 6= 1.

Completeness Theorem

Lindenbaum Algebra

Equivalence relation ≡ on F :

α ≡ β if and only if ` α ↔ β.

F|≡ := {[α] : α ∈ F}.

Operations on F|≡:

[α] ∩ [β] := [α ∧ β];∼ [α] := [¬α];fi [α] := [[ci ]α].

L := (F|≡,∩,∼, 1, fi )i∈N, where 1 = [>] is a BAOs.

Question : Is L a realization?Yes

Lindenbaum Algebra

F|≡ := {[α] : α ∈ F}.

[α] ∩ [β] := [α ∧ β];∼ [α] := [¬α];fi [α] := [[ci ]α].

Question : Is L a realization?

Lindenbaum Algebra

F|≡ := {[α] : α ∈ F}.

[α] ∩ [β] := [α ∧ β];∼ [α] := [¬α];fi [α] := [[ci ]α].

Question : Is L a realization?Yes

(R1) θ(α) exists for all α ∈ F .

θ(pi ) = [αi ], where p1, p2, . . . is an enumerationof the set PV.β∗ is obtained from β by uniform replacement ofpi ’s by αi ’s.(α(cj/x))∗ = α∗(cj/x), j ∈ N.⋂j

[α(cj/x)] exists and is given by [∀xα], where

α has only x as free variable.θ(α) := [α∗].

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)).

∀x [t]α → [t]∀xα, t and x are different.

θ(pi ) = [αi ], where p1, p2, . . . is an enumerationof the set PV.

β∗ is obtained from β by uniform replacement ofpi ’s by αi ’s.(α(cj/x))∗ = α∗(cj/x), j ∈ N.⋂j

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)).

θ(pi ) = [αi ], where p1, p2, . . . is an enumerationof the set PV.β∗ is obtained from β by uniform replacement ofpi ’s by αi ’s.

(α(cj/x))∗ = α∗(cj/x), j ∈ N.⋂j

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)).

α has only x as free variable.

θ(α) := [α∗].

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)).

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)).

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)).

(R2) fk⋂j

θ(α(cj/x)) =⋂j

fk θ(α(cj/x)).

Question

Given a realization A := (A,∩,∼, 1, fi )i∈N, does there exist aCBAOs B := (B,∩,∼, gi )i∈N and a BAOs monomorphism

r : A → B which also preserves all of⋂j

θ(α(cj/x)) and⋃j

θ(α(cj/x))?

Complex Algebra

F := (U, {Ri}i∈∆) −→ frame.The complex algebra of F (notation F+) is the expansion of thepower set algebra P(U) with operators mRi

: 2U → 2U , i ∈ ∆,where

mRi(X ) := {x ∈ U : For all y such that xRiy , y ∈ X}.

If F is MSAS, then F+ is a CBAOs satisfying (B1)-(B3)

Complex Algebra

F := (U, {Ri}i∈∆) −→ frame.The complex algebra of F (notation F+) is the expansion of thepower set algebra P(U) with operators mRi

: 2U → 2U , i ∈ ∆,where

mRi(X ) := {x ∈ U : For all y such that xRiy , y ∈ X}.

If F is MSAS, then F+ is a CBAOs satisfying (B1)-(B3)

Definition

A := (A,∩,∼, 1, fi )i∈∆ −→ BAOs.A+ := (FP(A),Qfi )i∈∆, where

FP(A) is the set of all prime filters of the Boolean algebra(A,∩,∼, 1);

F Qfi G if and only if fia ∈ F implies a ∈ G .

If ∆ = N and each fi satisfies (B1)-(B3), then A+ is a MSAS.

Definition

A := (A,∩,∼, 1, fi )i∈∆ −→ BAOs.A+ := (FP(A),Qfi )i∈∆, where

FP(A) is the set of all prime filters of the Boolean algebra(A,∩,∼, 1);

If ∆ = N and each fi satisfies (B1)-(B3), then A+ is a MSAS.

Stone Representation Theorem

A := (A,∩,∼, 1) −→ Boolean algebra.r : A → 2FP(A) defined by

r(x) := {F ∈ FP(A) : x ∈ F}

is a monomorphism.

Jonsson-Tarski Theorem

A := (A,∩,∼, 1, fi )i∈∆ −→ BAOs.r : A → 2FP(A) defined by

r(x) := {F ∈ FP(A) : x ∈ F}

is a BAOs embedding of A into (A+)+

Stone Representation Theorem

A := (A,∩,∼, 1) −→ Boolean algebra.r : A → 2FP(A) defined by

r(x) := {F ∈ FP(A) : x ∈ F}

is a monomorphism.

Jonsson-Tarski Theorem

A := (A,∩,∼, 1, fi )i∈∆ −→ BAOs.r : A → 2FP(A) defined by

r(x) := {F ∈ FP(A) : x ∈ F}

is a BAOs embedding of A into (A+)+

Is Jonsson-Tarski Theorem enough?

Monomorphism r given in Stone representation theorem andJonsson-Tarski theorem does not always preserve infinite joinsand meets in A.

We want r to preserve⋂j

θ(α(cj/x)).

It is impossible to extend Stone Representation Theorem insuch a way that the mapping r preserves all existing infinitejoins and meets in A (Tanaka and Ono).

A := (A,∩,∼, 1) −→ Boolean algebra.Q := {Qn ⊆ A : n ∈ N}, where each Qn is non-empty.

Q-Filter [Tanaka and Ono’01]

A prime filter F of A is called Q − filter , if it satisfies the followingfor each n ∈ N.

1 If Qn ⊆ F and⋂

Qn exists then⋂

Qn ∈ F .

2 If⋃

Qn exists and belongs to F then Qn ∩ F 6= ∅.

FQ(A) −→ Set of all Q − filters.

Definition

A := (A,∩,∼, 1, fi )i∈∆ −→ BAOs.AQ+ := (FQ(A),Qfi )i∈∆, where

If ∆ = N and each fi satisfies (B1)-(B3), then AQ+ is aMSAS.

Definition

A := (A,∩,∼, 1, fi )i∈∆ −→ BAOs.AQ+ := (FQ(A),Qfi )i∈∆, where

If ∆ = N and each fi satisfies (B1)-(B3), then AQ+ is aMSAS.

Theorem

A := (A,∩,∼, 1) −→ Boolean algebra.Q := {Qn ⊆ A : n ∈ N}, where each Qn is non-empty.r : A → 2FQ(A) defined by

r(x) := {F ∈ FQ(A) : x ∈ F}

is a monomorphism which preserves all⋂

Qn and⋃

Qn if theyexist.

Extension of the above theorem to BAOs ?

Theorem

A := (A,∩,∼, 1) −→ Boolean algebra.Q := {Qn ⊆ A : n ∈ N}, where each Qn is non-empty.r : A → 2FQ(A) defined by

r(x) := {F ∈ FQ(A) : x ∈ F}

is a monomorphism which preserves all⋂

Qn and⋃

Qn if theyexist.

Extension of the above theorem to BAOs ?

{Xn}n∈N and {Yn}n∈N be an enumeration of the setsQ∗ := {Qm ∈ Q :

⋂Qm ∈ A} and

Q∗ := {Qm ∈ Q :⋃

Qm ∈ A}.Following conditions are satisfied for each i ∈ N:

(C1) for any n,⋂

fiXn exists and satisfies that⋂fiXn = fi

⋂Xn,

(C2) for any z ∈ A and n, there exists m such that{fi (z → x) : x ∈ Xn} = Xm, wherez → x :=∼ z ∪ x ,

(C3) for any z ∈ A and n, there exists m such that{fi (y → z) : y ∈ Yn} = Ym.

Theorem

Suppose (C1)-(C3) are satisfied. Then r : A → 2FQ(A) defined by

r(x) := {F ∈ FQ(A) : x ∈ F}

is a BAOs embedding of A into the complex algebra (AQ+)+

which also preserves all of⋂

Xn and⋃

Can we use the above result to obtain the following ?

“Given a realization A := (A,∩,∼, 1, fi )i∈N, there exist aCBAOs B := (B,∩,∼, gi )i∈N and a BAOs monomorphism

θ(α(cj/x))”

No, as the set Q := {Qθα}, where Qθ

α := θ(α(cj/x)) is notcountable.

“Given a realization A := (A,∩,∼, 1, fi )i∈N, there exist aCBAOs B := (B,∩,∼, gi )i∈N and a BAOs monomorphism

θ(α(cj/x))”

No, as the set Q := {Qθα}, where Qθ

α := θ(α(cj/x)) is notcountable.

“If 6` α, then there exists a CBAOs A := (A,∩,∼, 1, fi )i∈N andan assignment θ : PV → A such that θ(α) 6= 1”

Proposition

Consider the Lindenbaum algebra L and canonical assignmentθc : PV → F|≡ which maps p to [p]. Then there exists a CBAOsB := (B,∩,∼, 1, gi )i∈N and a BAOs monomorphism r : F|≡ → Bsuch that

α ) =⋂

r(Qθc

Proof.

Qθc:= {Qθc

α } is countable.

Take any enumeration of Q.

(C1)-(C3) are satisfied.

Proposition

Consider the Lindenbaum algebra L and canonical assignmentθc : PV → F|≡ which maps p to [p]. Then there exists a CBAOsB := (B,∩,∼, 1, gi )i∈N and a BAOs monomorphism r : F|≡ → Bsuch that

α ) =⋂

r(Qθc

Proof.

Qθc:= {Qθc

α } is countable.

Take any enumeration of Q.

(C1)-(C3) are satisfied.

If C α then ` α

Consider Lindenbaum algebra L.

Canonical assignment θc : PV → F|≡ where θc(p) = [p].

There exists a CBAOs B := (B,∩,∼, 1, gi )i∈N and a BAOsmonomorphism r : A → B as above.

Consider the assignment γ : PV → B such that γ(p) = r([p]).

γ(α) = r([α]).

If 6` α, then [α] 6= 1 and hence γ(α) = r([α]) 6= 1

If C α then ` α

γ(α) = r([α]).

If C α then ` α

γ(α) = r([α]).

If C α then ` α

γ(α) = r([α]).

If C α then ` α

γ(α) = r([α]).

If C α then ` α

γ(α) = r([α]).

If C α then ` α

γ(α) = r([α]).

Thank you

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