From Free Agebras to Proof Bounds - unimi.itusers.mat.unimi.it/users/ghilardi/allegati/nashville.pdf · 2013-07-21 · From Free Agebras to Proof Bounds N. Bezhanishvili1 and S. Ghilardi2

From Free Agebras to Proof Bounds

N. Bezhanishvili1 and S. Ghilardi2

1University of Utrecht

2Università degli Studi di Milano

TACL, Nashville, July 31, 2013

N. Bezhanishvili - S. Ghilardi (UU & UNIMI) From free algebras ... TACL 2013 1 / 47

20 Years in Free Modal Algebras Constructions

Step-by-step constructions of free modal algebras have longstandingtradition:

they can traced back to Fine normal forms (1975);Abramsky (1988, published much later in 2005);Ghilardi (1995; for Heyting case 1992, for S4 case 2010);coalgebraic literature (N. Bexhanishvili, Kurz, Kupke, Gehrke,Pattinson, Schröder, Venema, ... 2000–> );recent advances in a paper by Coumans-vanGool (JLC 2013).























Coumans-van Gool emphasis is on partial algebras; a lightreformulation of their point of view uses two-sorted structures.

These are called step-algebras and step-frames in (N.Bezhanishvili-Ghilardi-Jibladze, in press).

We review step-algebras and step-frames and then discuss ongoingwork concerning proof theoretic applications

We anticipate the kind of proof theoretic aspects we want toinvestigate.




















Logics and Inference Systems

A logic L is a set of formulae containing tautologies, Aristotle’s law andclosed under necessitation, modus ponens and uniform substitution.

An inference system Ax for L is a set of inference rules

φ1(x), . . . , φn(x)

ψ(x)(1)

satisfying the following conditions



A logic L is a set of formulae containing tautologies, Aristotle’s law andclosed under necessitation, modus ponens and uniform substitution.

An inference system Ax for L is a set of inference rules

φ1(x), . . . , φn(x)

ψ(x)(1)

satisfying the following conditions



all rules in Ax are derivable from L;from the rules in Ax all formulae in L are provable;all formulae occurring as premises or as conclusions of rules inAx have modal degree at most 1;if a propositional variable occurs in a rule r ∈ Ax , then it has anoccurrence in r which is located inside a modal operator.

Rules satisfying the last two conditions are called reduced.



















For every L one can find an Ax which is suitable for L (there are alsocanonical methods for that), but ...... not all equivalent inference systems are proof-theoretically equallygood ...... for some of them proof search may become intricated ...... we want at least the following: to prove a formula φ, only formulaeup to the modal degree of φ are needed.

Since the above property leads to decidability, it won’t be possible toget it in general, but we want to have some criteria to recognize goodsystems.



















As an example, take GL-system. It can be axiomatized by the singleaxiom �(�x → x)→ �x .An inference system for L consists of transitivity and Godel-Lob rules

�+y → x�y → �x

�x → xx

(2)

An alternative one (the good one!) consists of the Avron rule

�+x ∧�y → y�x → �y

(3)


Outline

1 Free Modal AlgebrasThe Local MethodAdding EquationsStep Correspondence

2 The Step Embedding TheoremAn ExampleSome Case StudiesConclusions


Outline




Free Modal Algebras




Free Modal Algebras The Local Method





One-Step Modal Algebras

Definition1 A one-step modal algebra is a quadruple (A0,A1, i0,♦0), where

A0,A1 are Boolean algebras, i0 : A0 → A1 is a Boolean morphism,and ♦0 : A0 → A1 is a semilattice morphism. The algebras A0,A1are called the source and the target Boolean algebras of theone-step modal algebra (A0,A1, i0,♦0).

2 A one-step extension of the one-step modal algebra (A0,A1, i0,♦0)is a one-step modal algebra (A1,A2, i1,♦1) satisfying i1♦0 = ♦1i0.




Definition1 A one-step modal algebra is a quadruple (A0,A1, i0,♦0), where

A0,A1 are Boolean algebras, i0 : A0 → A1 is a Boolean morphism,and ♦0 : A0 → A1 is a semilattice morphism. The algebras A0,A1are called the source and the target Boolean algebras of theone-step modal algebra (A0,A1, i0,♦0).

2 A one-step extension of the one-step modal algebra (A0,A1, i0,♦0)is a one-step modal algebra (A1,A2, i1,♦1) satisfying i1♦0 = ♦1i0.




The ‘universal’ one-step extension of a given (A0,A1, i0,♦0) is built byusing pushouts (A2 ' A1 +V (A0) V (A1)) as follows

V (A0) V (A1)

A1 A2i1

♦T1

V (i0)

♦T0

where we used the bijective correspondence given by Vietoris functor

♦ : C −→ D (in SemiL)

♦T : V (C) −→ D (in Bool)



One-Step Modal AlgebrasWe can use the above construction to built a chain; we start with afinite Boolean algebra B0 and build the one step modal algebrai0,♦0 : B0 ⇒ B0 + V (B0), where i0,♦T

0 are the coproduct injections.Then we go on with universal one-step extensions. We have a chain

B0 ⇒ B1 ⇒ B2 ⇒ · · · (4)

The upper morphisms ik are Boolean morphisms and the lowermorphisms ♦k are Semilattice morphisms. Since we have♦k+1 ◦ ik = ik+1 ◦ ♦k , we can define

♦ : lim→

Bk −→ lim→

Bk

This construction turns out to give the free modal algebra over thefinite Boolean algebra B0.





B0 ⇒ B1 ⇒ B2 ⇒ · · · (4)


♦ : lim→

Bk −→ lim→

Bk






B0 ⇒ B1 ⇒ B2 ⇒ · · · (4)


♦ : lim→

Bk −→ lim→

Bk



Free Modal Algebras Adding Equations





Adding equations

The point is now to add equations axiomatizing modal algebras for agiven logic. Following a suggestions by Coumans & Van Gool (inpress), we transform axioms into inferences of modal degree 1.

For instance, in case of K4, the axiom

�x → ��x

is transformed into the rule

�x → y�x → �y

(5)



Adding equations

The point is now to add equations axiomatizing modal algebras for agiven logic. Following a suggestions by Coumans & Van Gool (inpress), we transform axioms into inferences of modal degree 1.

For instance, in case of K4, the axiom

�x → ��x

is transformed into the rule

�x → y�x → �y

(5)



Adding equations

Equation (5) can be interpreted in a one-step modal algebra(A0,A1, i ,♦), where is says

∀x , y ∈ A0 (�x ≤ i(y)⇒ �x ≤ �y). (6)

Thus it is sufficient, whenever we build one-step extensions, to dividethe newly built Boolean algebra by the minimum congruenceleading (6) to hold.

All this is quite formal ... let’s have a better look !



Adding equations


∀x , y ∈ A0 (�x ≤ i(y)⇒ �x ≤ �y). (6)





Adding equations


∀x , y ∈ A0 (�x ≤ i(y)⇒ �x ≤ �y). (6)





Adding equations

There is a point to care indeed. In the quotient chain

B̃0 ⇒ B̃1 ⇒ B̃2 ⇒ · · · (7)

we are not guaranteed that the upper Boolean morphisms ik areinjective.Thus, we can claim that the colimit algebra is the free algebra (modulothe extra conditions), but we cannot claim that the quotient algebra B̃kis the algebra of formulae of modal complexity less or equal to k .To claim it, we need an extra argument showing injectivity of the ik .This amounts to show that provable formulae of a certain modalcomplexity have proofs of that complexity. The required argumentcannot be general, because it implies decidability of the logic!



Adding equations


B̃0 ⇒ B̃1 ⇒ B̃2 ⇒ · · · (7)




Adding equations


B̃0 ⇒ B̃1 ⇒ B̃2 ⇒ · · · (7)



Free Modal Algebras Step Correspondence





Applying Duality

To get a better understanding of the situation, it is nice to apply duality.Since we are interested in finite algebras, duality needs not topologicalmachinery and is easy.

Definition1 A one-step frame is a quadruple (W1,W0, f ,R), where W0,W1 are

sets, f : W1 −→W0 is a function and R ⊆W1 ×W0 is a relation.2 A one-step extension of the one-step frame (W1,W0, f ,R) is a

one-step frame (W2,W1,g,S) satisfying f ◦ S = g ◦ R.



Applying Duality

To get a better understanding of the situation, it is nice to apply duality.Since we are interested in finite algebras, duality needs not topologicalmachinery and is easy.

Definition1 A one-step frame is a quadruple (W1,W0, f ,R), where W0,W1 are

sets, f : W1 −→W0 is a function and R ⊆W1 ×W0 is a relation.2 A one-step extension of the one-step frame (W1,W0, f ,R) is a

one-step frame (W2,W1,g,S) satisfying f ◦ S = g ◦ R.



Applying Duality

All constructions we used have duals: coproducts are turned intoproducts, pushouts into pullbacks, Vietoris functor is turned intocovariant power set functor, etc.Thus we can easily build, starting from a finite set B∗0 the dual of theBoolean algebras chain (4)

B∗0 ⇔ B∗1 ⇔ B∗2 ⇔ · · ·

and (in presence of further axioms) the dual of the quotient chain (7)

B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·

Our problem is turned into the problem of showing that the upperfunctions are surjective.



Applying Duality


B∗0 ⇔ B∗1 ⇔ B∗2 ⇔ · · ·


B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·




Applying Duality


B∗0 ⇔ B∗1 ⇔ B∗2 ⇔ · · ·


B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·




Applying Duality


B∗0 ⇔ B∗1 ⇔ B∗2 ⇔ · · ·


B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·




Step Correspondence Theory

To clarify our goal we need to understand better what it means for (thedual of) a one-step frame (W1,W0, f ,R) to validate an inference rulelike the transitivity rule we got for K4.This happens iff

∀x , y ⊆W0 (�Rx ⊆ f ∗(y)⇒ �Rx ⊆ �Ry). (8)

The idea is to apply the step version of correspondence theory in orderto eliminate the second order quantifiers via Ackermann rule.





∀x , y ⊆W0 (�Rx ⊆ f ∗(y)⇒ �Rx ⊆ �Ry). (8)






∀x , y ⊆W0 (�Rx ⊆ f ∗(y)⇒ �Rx ⊆ �Ry). (8)





Applying adjunction ∃f a f ∗, Ackermann rule, set-theoretic definition of⊆, definition of �R and Ackermann rule again we get:

∀x , y ⊆W0. �Rx ⊆ f ∗(y)⇒ �Rx ⊆ �Ry∀x , y ⊆W0. ∃f (�Rx) ⊆ y ⇒ �Rx ⊆ �Ry∀x ⊆W0. �Rx ⊆ �R ∃f (�Rx)

∀x ⊆W0∀w ∈W0. w ∈ �Rx ⇒ w ∈ �R ∃f (�Rx).

∀x ⊆W0 ∀w ∈W0. R(w) ⊆ x ⇒ w ∈ �R ∃f (�Rx)

∀w ∈W0. w ∈ �R ∃f (�RR(w)).



Step Correspondence TheoryLast condition reads as:

∀w ∀v (R(w , v)→ ∃w1 (f (w1) = v & R(w1) ⊆ R(w))) . (9)

The possibility of getting a first order condition is subject to sufficientsyntactic conditions very similar to those of Sahlqvist theorem.Using the above first order characterization it is possible to give adirect description of the dual quotient chain

B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·

and to prove that the upper fuctions are surjective: details are in ourpaper (N. Bezhanishvili & Ghilardi & Jibladze, in press).We do not insist on such details here, because we want to show thatthe whole machinery becomes much more manageable if we let finitemodel property (for global consequence relation) to enter in thepicture.




∀w ∀v (R(w , v)→ ∃w1 (f (w1) = v & R(w1) ⊆ R(w))) . (9)


B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·





∀w ∀v (R(w , v)→ ∃w1 (f (w1) = v & R(w1) ⊆ R(w))) . (9)


B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·





∀w ∀v (R(w , v)→ ∃w1 (f (w1) = v & R(w1) ⊆ R(w))) . (9)


B̃∗0 ⇔ B̃∗1 ⇔ B̃∗2 ⇔ · · ·



The Step Embedding Theorem






Given a logic L, for a finite set of formulae Γ and for a formula φ, wewrite

Γ `L φ (10)

for the global consequence relation: this means that φ has a proof fromaxioms and rules of L with premises Γ (uniform substitution appliesonly to formulae in L, not to formulae in Γ). For transitive systems,Γ `L φ is the same as �+(

∧Γ)→ φ ∈ L.

For axioms systems Ax , we have a similar notion

Γ Àx φ (11)

Notice that if Ax is an inference system for L, the relations (10)and (11) are equivalent.




DefinitionAn inference system Ax has the bounded proof property (bpp) iffwhenever Γ Àx φ holds, there is a proof in Ax of φ with premises Γ inwhich formulae not exceeding the modal degree of formulae in Γ, φoccur.

DefinitionA logic L has the finite model property (fmp) iff whenever Γ Àx φ doesnot hold, then there is a Kripke model based on a finite frame for Lwhere the Γ are everywhere true and φ is not.




DefinitionAn inference system Ax has the bounded proof property (bpp) iffwhenever Γ Àx φ holds, there is a proof in Ax of φ with premises Γ inwhich formulae not exceeding the modal degree of formulae in Γ, φoccur.

DefinitionA logic L has the finite model property (fmp) iff whenever Γ Àx φ doesnot hold, then there is a Kripke model based on a finite frame for Lwhere the Γ are everywhere true and φ is not.




DefinitionA one-step frame (W1,W0, f ,R) is conservative iff (i) f is surjective and(ii) for every w1,w2 ∈W1 we have that

f (w1) = f (w2) & R(w1) = R(w2) ⇒ w1 = w2. (12)

Dually, a one-step modal algebra (A0,A1, i ,♦) is conservative iff (i) i isinjective and (ii) the set

{i(a) | a ∈ A0} ∪ {♦a | a ∈ A0}

generate A1 as a Boolean algebra.




DefinitionA one-step frame (W1,W0, f ,R) is conservative iff (i) f is surjective and(ii) for every w1,w2 ∈W1 we have that

f (w1) = f (w2) & R(w1) = R(w2) ⇒ w1 = w2. (12)

Dually, a one-step modal algebra (A0,A1, i ,♦) is conservative iff (i) i isinjective and (ii) the set

{i(a) | a ∈ A0} ∪ {♦a | a ∈ A0}

generate A1 as a Boolean algebra.




TheoremLet Ax be an inference system for a logic L. Then every finiteconservative step-frame validating Ax is a p-morphic image of a finiteKripke frame for L iff Ax has the bpp and L has the fmp.

Kripke frames are here seen as step-frames with identical steptransition function f ; the notion of a p-morphism between step-framesis the obvious one.




A p-morphism between step frames F ′ = (W ′1,W

′0, f′,R′) and

F = (W1,W0, f ,R) is a pair of surjective mapsµ : W ′

1 −→W1, ν : W ′0 −→W0 such that

f ◦ µ = ν ◦ f ′ and R ◦ µ = ν ◦ R′ . (13)

W ′1 W1

W ′0 W0

ν

f

µ

f ′

W ′1 W1

W ′0 W0

ν

R

µ

R′


The Step Embedding Theorem An Example





An example

Using the above theorem, it is easy to prove fmp and bpp for simplelogics like K ,K 4,T ,S4, . . . . As an example, let us consider the densityaxiom ��x → �x (we add this axiom to K ).

First, we turn the axiom into a rule; there is a default naif method forthat giving

y → �x�y → �x .

(14)

Second, applying step correspondence, we get the following first-ordercharacterization for validation of (14) in step frames:

∀w∀v (wRv ⇒ ∃k (wRf (k) & kRv)) (15)



An example



y → �x�y → �x .

(14)


∀w∀v (wRv ⇒ ∃k (wRf (k) & kRv)) (15)



An example



y → �x�y → �x .

(14)


∀w∀v (wRv ⇒ ∃k (wRf (k) & kRv)) (15)



An example

Third, we fix a finite conservative step-frame S = (W1,W0, f ,R)satisfying (15); we must find a finite frame F = (V ,S) which dense inthe standard sense

∀w∀v (wSv ⇒ ∃k (kSv & wSk)). (16)

and a surjective map µ : V −→W1 such that R ◦ µ = f ◦ µ ◦ S.

The idea is to take V := W1 and µ := idW1 , so that we need to check

∀w∀v (wRv ⇔ ∃w ′ (wSw ′ & f (w ′) = v)). (17)



An example


∀w∀v (wSv ⇒ ∃k (kSv & wSk)). (16)



∀w∀v (wRv ⇔ ∃w ′ (wSw ′ & f (w ′) = v)). (17)



An example


∀w∀v (wSv ⇒ ∃k (kSv & wSk)). (16)



∀w∀v (wRv ⇔ ∃w ′ (wSw ′ & f (w ′) = v)). (17)



An example

Some ingenuity is needed in the general case to find the appropriate Sbut there are templates. Our template is

∀w∀w ′ (wSw ′ ⇔ ∃v (wRv & f (w ′) = v)). (18)

Thus, taking into consideration that f is also surjective (because S isconservative)

∀v ∃w f (w) = v , (19)

we need the validity of the implication

(19) & (15) & (18) ⇒ (17) & (16).



An example


∀w∀w ′ (wSw ′ ⇔ ∃v (wRv & f (w ′) = v)). (18)


∀v ∃w f (w) = v , (19)


(19) & (15) & (18) ⇒ (17) & (16).



An example


∀w∀w ′ (wSw ′ ⇔ ∃v (wRv & f (w ′) = v)). (18)


∀v ∃w f (w) = v , (19)


(19) & (15) & (18) ⇒ (17) & (16).



An example

The implication

(19) & (15) & (18) ⇒ (17) & (16).

can be shown to be valid manually or by a prover (SPASS solves theproblem in less than half a second; the superposition proof it findstakes 47 lines).

As a consequence of this we get altogether bpp, fmp and first orderdefinability (decidability and canonicity also follows by generalreasons).



An example

The implication

(19) & (15) & (18) ⇒ (17) & (16).

can be shown to be valid manually or by a prover (SPASS solves theproblem in less than half a second; the superposition proof it findstakes 47 lines).

As a consequence of this we get altogether bpp, fmp and first orderdefinability (decidability and canonicity also follows by generalreasons).



An example

Let’s summarize the three steps:- First: produce the inference rules (there are automatic methods,

not always they give the good rules).- Second: apply correspondence theory (this is automatic).- Third: produce p-morphic extensions to standard frames (not

automatic, but there are templates); provers can discharge thefinal proof obbligation.



An example






An example






An example





The Step Embedding Theorem Some Case Studies





Some case studies

We wonder to which extent the above mechanization of themetatheory can be pushed.We analyzed some more significant cases. The first case is GL systemaxiomatized by the single axiom �(�x → x)→ �x .

First Step can be driven so that to obtain a rule which is equivalent (forour purposes) to Avron’s rule

�+x ∧�y → y�x → �y



Some case studies



�+x ∧�y → y�x → �y



Some case studies



�+x ∧�y → y�x → �y



Some case studies

Second Step (via Ackermann rule applied to fixpoint logic) gives

∀w R(w) ⊆ µ(Y ,w)∃f (f ∗(R(w)) ∩�RR(w) ∩�RY ). (20)

In finite one-step frames this simplifies to

∀w (R(w) ⊆ {f (w ′) | R(w ′) ⊂ R(w)}) . (21)

Notice that ⊂ is strict inclusion, so the above condition is a ‘step’irreflexivity.



Some case studies

Second Step (via Ackermann rule applied to fixpoint logic) gives

∀w R(w) ⊆ µ(Y ,w)∃f (f ∗(R(w)) ∩�RR(w) ∩�RY ). (20)

In finite one-step frames this simplifies to

∀w (R(w) ⊆ {f (w ′) | R(w ′) ⊂ R(w)}) . (21)

Notice that ⊂ is strict inclusion, so the above condition is a ‘step’irreflexivity.



Some case studies

Third Step is not difficult, but is not fully automatic. We can use thetemplate S for transitive systems, but then the resulting Kripke frame isnot irreflexive, so one needs to take the dijoint union of the irreflexivesubframes satisfying (17).

It should be noticed that, if we do the same analysis for the systemaxiomatized by transitivity and Löb rule, we get a weaker conditionthan (20). Using the fact that the condition is too weak, it is possible toprove formally that bpp fails.



Some case studies

Third Step is not difficult, but is not fully automatic. We can use thetemplate S for transitive systems, but then the resulting Kripke frame isnot irreflexive, so one needs to take the dijoint union of the irreflexivesubframes satisfying (17).

It should be noticed that, if we do the same analysis for the systemaxiomatized by transitivity and Löb rule, we get a weaker conditionthan (20). Using the fact that the condition is too weak, it is possible toprove formally that bpp fails.



Some case studies

Our second case study is the system S4.3 axiomatized via S4reflexivity and transitivity axioms plus �(�x → y) ∨�(�y → x).

First step: the inference rule extracted automatically from the axiom isnot good (bpp fails). Instead, we use Goré infinitely many rules:

· · ·�y → xj ∨∨

j 6=i �xi · · ·�y →

∨ni=1 �xi

(22)

The rules are indexed by n and the n-th rule has n premises,according to the values j = 1, . . . ,n.



Some case studies

Our second case study is the system S4.3 axiomatized via S4reflexivity and transitivity axioms plus �(�x → y) ∨�(�y → x).

First step: the inference rule extracted automatically from the axiom isnot good (bpp fails). Instead, we use Goré infinitely many rules:

· · ·�y → xj ∨∨

j 6=i �xi · · ·�y →

∨ni=1 �xi

(22)

The rules are indexed by n and the n-th rule has n premises,according to the values j = 1, . . . ,n.



Some case studies

Second Step: correspondence theory applies to these rules.Interpreting the results in finite frames one gets

∀w ∀S ⊆ R(w) ∃v ∈ S ∃w ′ (f (w ′) = v & S ⊆ R(w ′) ⊆ R(w)). (23)

Third Step: the same method used in GL case shows that one-stepframes satisfying (23) are p-morphic images of Kripke frames for S4.3.This establishes bpp and fmp.



Some case studies

Second Step: correspondence theory applies to these rules.Interpreting the results in finite frames one gets

∀w ∀S ⊆ R(w) ∃v ∈ S ∃w ′ (f (w ′) = v & S ⊆ R(w ′) ⊆ R(w)). (23)

Third Step: the same method used in GL case shows that one-stepframes satisfying (23) are p-morphic images of Kripke frames for S4.3.This establishes bpp and fmp.



Some case studies

As a further case study let us consider S5.

First Step The following rule has been proposed in the literature:

�Γ⇒ y ,�∆

�Γ⇒ �y ,�∆ .(24)

In the resulting system, cuts cannot be completely eliminated, but canbe limited to subformulae of the sequent to be proved. This ‘analytic’cut-elimination property is sufficient to imply the bpp, and thus weshould be able to get the bpp directly by our methods.



Some case studies

As a further case study let us consider S5.

First Step The following rule has been proposed in the literature:

�Γ⇒ y ,�∆

�Γ⇒ �y ,�∆ .(24)

In the resulting system, cuts cannot be completely eliminated, but canbe limited to subformulae of the sequent to be proved. This ‘analytic’cut-elimination property is sufficient to imply the bpp, and thus weshould be able to get the bpp directly by our methods.



Some case studies

Second Step Correspondence theory gives

∀w∀v (wRv → ∃w̃ (f (w̃) = v & R(w) = R(w̃))). (25)

Third Step Step frames satisfying the above property are easily seento be p-morphic images of reflexive, transitive, symmetric Kripkeframes; this establishes bpp and fmp.



Some case studies

Second Step Correspondence theory gives

∀w∀v (wRv → ∃w̃ (f (w̃) = v & R(w) = R(w̃))). (25)

Third Step Step frames satisfying the above property are easily seento be p-morphic images of reflexive, transitive, symmetric Kripkeframes; this establishes bpp and fmp.



Some case studies

As a final case study consider the system obtained by adding to K theaxiom ��x ↔ �x . This is density+transitivity; we can join the rules wealready used for density and transitivity. This is not a good idea: bppfails!Instead, we use the following couple of rules suggested to us by G.Mints:

�+Γ→ α

�Γ→ �αΓ,�∆⇒ �α�Γ,�∆⇒ �α

(26)



Some case studies

As a final case study consider the system obtained by adding to K theaxiom ��x ↔ �x . This is density+transitivity; we can join the rules wealready used for density and transitivity. This is not a good idea: bppfails!Instead, we use the following couple of rules suggested to us by G.Mints:

�+Γ→ α

�Γ→ �αΓ,�∆⇒ �α�Γ,�∆⇒ �α

(26)



Some case studies

Second Step Correspondence theory gives, besides steptransitivity (9), the condition

∀w∀v (wRv → ∃w ′ (w ′Rv & {f (w ′)} ∪ R(w ′) ⊆ R(w))). (27)

Third Step Step frames satisfying the above property are easily seento be p-morphic images of transitive and dense Kripke frames; thisestablishes bpp and fmp.

Notice that the fact that (15)+ (9) do not imply (27) is a formalargument proving that bpp fails is we adopt the old rule (14) in atransitive context. Thus, at least in principle, model finders can beused as automatic supports for showing that bpp fails.



Some case studies







Some case studies






The Step Embedding Theorem Conclusions





Conclusions

step methods (via step embedding theorem and stepcorrespondence) seem to be quite effective in jointly proving bppand fmp;in simple cases the application of the methods is fully automatic(in a sense we are mechanizing the metatheory of modal logic!);in more complex cases some ingenuity is needed, still uniformarguments often work;the scalability of the methods are to be tested to morecomplicated logics arising in computer science applications.



Conclusions




Conclusions




Conclusions



Documents

From Free Agebras to Proof Bounds - unimi.itusers.mat.unimi.it/users/ghilardi/allegati/nashville.pdf · 2013-07-21 · From Free Agebras to Proof Bounds N. Bezhanishvili1 and S. Ghilardi2