Interpolation in \Lukasiewicz logic and amalgamation of MV-algebras Daniele Mundici Dept. of...

Post on 16-Dec-2015

218 views 1 download

Tags:

Transcript of Interpolation in \Lukasiewicz logic and amalgamation of MV-algebras Daniele Mundici Dept. of...

Interpolation in \Lukasiewicz logic and amalgamation of MV-algebras

Daniele MundiciDept. of Mathematics “Ulisse Dini”

University of Florence, Florence, Italy

mundici@math.unifi.it

2-simplex

0-simplex

1-simplex

3-simplex

we all know what a simplex in Rn is

polyhedron P= finite union of simplexes Si in Rn

P need not be convex, nor connected

a polyhedron P = USi is said to be rational

if so are the vertices of every simplex Si

our main themes:

rational polyhedra and\Lukasiewicz logic

Chapter 1: Local Deduction

as a main ingredient of interpolation and amalgamation

\Lukasiewicz logic L∞

• FORMULAS are exactly the same as in boolean logic

• any VALUATION V evaluates formulas into the real unit interval [0,1] via the inductive rules:

• V(¬F) = 1–V(F)

• V(F —> G) = min(1, 1–V(F)+V(G))

• Therefore, every valuation V is uniquely determined by its values on the variables: V(X1),...,V(Xn)

• CONSEQUENCE RELATION: F |– G means that every valuation satisfying F also satisfies G

formulas yield functions f:[0,1]n—>[0,1] as boolean formulas yield f:0,1n—>0,1

• every formula F(X1,...,Xn) determines a map fF : [0,1]n —>[0,1] by

• fXi = the ith coordinate map

• f¬F = 1 – fF

• fF —> G = min(1, 1 – fF + fG)

definable functions of one variable

the ONESET fF-1(1) of fF

is the set of valuations satisfying the formula F

oneset(fF)=zeroset(¬fF)

for each formula F, its associated function fF is continuous, linear, and each linear piece has integer coefficients (for short, fF is a McNaughton function)

oneset of fF = Mod(F)

• by induction on the number of connectives in F, the oneset of fF is a rational polyhedron, and so is the oneset of f¬F and of fF —> G

EACH ZEROSET AND EACH ONESET IS A RATIONAL POLYHEDRON IN [0,1]n

(Local) Deduction Theorem

Theorem. For any two formulas A and B, the following conditions are equivalent:

1. Every valuation satisfying A also satisfies B

2. For some m=1,2,... the formulaA—>(A—>(A—>...—>(A—>(A—>B))...)) is a tautology

3. B is obtained from A and the tautologies via Modus Ponens

PROOF. 2—>3 easy; 3—>1 induction; 1—>2 is proved geometrically

assume oneset(fA) contained in oneset(fB)

1

1

let T be a triangulation of [0,1] such that the functions fA and fB both formulas A and B are linear over each interval of T

fA

fB

fA & fA

< fA

1

1

fA&A

fB

applying \Lukasiewicz conjunction to A, from the formula A&A we get obtain a minorant fA&A of fA, still with the same one set of fA

Recall definitionP&Q = ¬(P —> ¬Q)

fA & fA

& fA

< fA & fA < fA

1

1

fkA

fB

by iterated application of the \Lukasiewicz

conjunction we obtain a function

fkA= fA&fA&...&fA

with the same oneset of fA, and with the additional property

that fAk ≤ fB

for large k this will hold at every simplex of T

1

1

fA

fB in other words, we have the tautology Ak

—>B, which is the same as the desired tautology

A—>(A—>(A—>...—>(A—>(A—>B))...))

Chapter 2: Interpolation

(as a main tool to amalgamation)

interpolation/amalgamation

• Craig interpolation theorem fails in \Lukasiewicz logic, because the tautology x¬x—>y¬y has no interpolant

• deductive interpolation is like Craig interpolation, with the |– symbol in place of the implication connective (more soon)

• over the last 25 years, several proofs have been given of deductive interpolation for \Lukasiewicz infinite-valued propositional logic

• deductive interpolation, together with local deduction, is a main tool to prove the amalgamation theorem for the algebras of \Lukasiewicz infinite-valued logic

amalgamation: many proofs• the first proof of amalgamation used the categorical

equivalence between MV-algebras and unital lattice-ordered groups (relying on Pierce's amalgamation theorem).

• in the early eighties I heard from Andrzej Wro\’nski during one of his visits to Florence, that the Krakow group had a proof of the amalgamation property for MV-algebras without negation (i.e., Komori’s C algebras)

• recent proofs, like the proof by Kihara and Ono, follow by applying to MV-algebras results in universal algebra

• I will present a simple geometric proof of the amalgamation theorem, using Deductive Interpolation

background literature

F. Montagna, Interpolation and Beth's property in propositional many-valued logics: A semantic investigation, Annals of Pure and Applied Logic, 141: 148-179, 2006. This is based on:

N.Galatos, H. Ono, Algebraization, parametrized local deduction theorem and interpolation for substructural logics over FL, Studia Logica, 83:279-308, 2006. For the proof of Theorem 5.8, the following is needed:

A. Wro\'nski, On a form of equational interpolation property, In: Foundations of Logic and Linguistic, G.Dorn, P. Weingartner, (Eds.), Salzburg, June 19, 1984, Plenum, NY, 1985, 23-29. For the proof of Theorem I on page 25, the following is needed:

P.D. Bacisch, Amalgamation properties and interpolation theorems for equational theories, Algebra Universalis, 5:45-55, 1975.

(Deductive) Interpolation

If F |– G then there is a formula J such that F |– J, J |– G, and each variable of J is a variable of both F and G

our proof will be entirely geometrical

rational polyhedra are preserved under projection

the projectionof a (rational) polyhedron onto a (rational) hyperplaneis a (rational) polyhedron

we record this fact as the PROJECTION LEMMA

rational polyhedra are preserved under perpendicular cylindrification

we record this fact as the CYLINDRIFICATION LEMMA

oneset of fF = Mod(F)

recall: THE ZEROSET (AND THE ONESET) OF ANY \LUKASIEWICZ FORMULA IS A RATIONAL POLYHEDRON IN [0,1]n

we now prove the converse: EACH RATIONAL POLYHEDRON IN [0,1]n IS THE ZEROSET OF SOME \LUKASIEWICZ FORMULA

rational half-spaces in [0,1]n

a rational line L in [0,1]2

H is one of the half-planes bounded by L in the square [0,1]2

PROBLEM:Does there exist a formula F such that the zeroset of fF coincides with H ?mx+ny+p=0, with m,n,p integers, m>0

H

ANSWER: Yes, by induction on |m|+|n|

then every rational polyhedron is a zeroset

this blue half-space is a zerosetthen so is this rational triangle(formulas can express intersections)

and this rational polyhedron(formulas can express unions)

ANY RATIONAL POLYHEDRON IN [0,1]n IS THE ZEROSET OF SOME fF

this was known to McNaughton (1951)

FOLKLORE LEMMA

Rational polyhedra contained in the n-cube [0,1]n coincide with zerosets (and also coincide with onesets) of definable maps, i.e., functions of the form fF where F ranges over formulas in n variables

we record the FOLKLORE LEMMA by writing: RATIONAL

POLYHEDRA=ONESETS=MODELSETS

Deductive interpolation

PROOF. We may write var(F) = X u Z var(G) = Y u Z, for X,Y,Z pairwise disjoint sets of variables

Mod(F) = fF-1(1) = P, which by the Folklore Lemma is

a rational polyhedron in [0,1]XuZ

by the Projection Lemma, the projection of P onto RZ is a rational polyhedron Q contained in [0,1]Z

Mod(G) = fG-1(1) = R, a rational polyhedron in [0,1]YuZ

Mod(F) = P

X

Y

Z

Mod(G)=R

the hypothesis F |— G states that, in the space RXuYuZ

Mod(F) is contained in Mod(G)

Q

Q

Mod(F) = P

We then obtain the first half of interpolation: F |— J

X

Y

Z

by the Folklore Lemma, there is a formula J(Z) such that Q=Mod(J)

regarding J as a formula in the variables X,Z, then Mod(J) is this blue rectangle!

= Mod(J)

Mod(F) = P

X

Y

Z

Mod(G)=R

in the space RYuZ , Mod(J) is contained in Mod(G)

Q=Mod(J)

regarding J as a formula in Y,Z, then Mod(J) is this blue rectangle!

We then obtain the second half of interpolation: J |— G

Chapter 3: Amalgamation

of the algebras of \Lukasiewicz logic,i.e., Chang MV-algebras

MV-algebras (in Wajsberg’s version)directly from \Lukasiewicz axioms

A—>(B—>A)

(A—>B)—>((B—>C)—>(A—>C))

((A—>B)—>B)—> ((B—>A)—>A)

(¬A—>¬B)—>(B—>A)

the amalgamation property

Z

A B

we have

the usual setup

Z

A B

D

we want

henceforth, all blue maps are one-one

we have

the embedding of Z into A

let us focus attention on the embedding of Z into A

without loss of generality , Z is a subalgebra of A

thus the set A is the disjoint union of Z and some set X, A=Z U X

Z

A

extending maps to homomorphisms

the identity map z—>z uniquely extends to a homomorphism sZ of the free MV-algebra FREEZ onto Z

similarly, the identity map a—> a uniquely extends to a homomorphism sA of FREEA onto A

let ker sZ and ker sA denote the kernels of these maps

Z

FREEZ

sZ

Z

A

FREEZ

FREEXUZ

ker(sA)

ker(sZ)

sA

sZ

all blue arrows are inclusions

all red arrows are surjections

intuitively, this trivial Largeness Lemma states that ker(sZ) is as large as possible in ker(sA).

LEMMA ker(sZ) = ker(sA) ∩FREEZ

Z

A B

Z

A B

FREEZ

FREEXUZ FREEYUZ

ker(sA)ker(sB)

ker(sZ)

sA

sz

sA

Z

A B

FREEZ

FREEXUZ FREEYUZ

ker(sA)ker(sB)

ker(sZ)

sA

sz

I = the ideal generated by ker(sA) U ker(sB)

FREEXUYUZ

sA

Z

A B

FREEZ

FREEXUZ FREEYUZ

ker(sA)ker(sB)

ker(sZ)

sA

sz

I = the ideal generated by ker(sA) U ker(sB)

FREEXUYUZ

sA

D

Z

A B

FREEZ

FREEXUZ FREEYUZ

ker(sA)ker(sB)

ker(sZ)

sA

sz

i = the ideal generated by ker(sA) U ker(sB)

FREEXUYUZ

sA

D

µµ(x/ ker(sA)) = x/i

there remains to be proved that µ

is one-one

e/i = 0 means that e is an element of i. In other words, (theories ~ ideals) a, b |– e for some a in ker(sA) and b in ker(sB)

Let e be an element of FREEXUYUZ such that e/i =0. We must prove e/ker(sA) = 0

end of the proof of amalgamation

Chapter 4:

Further geometric developments on projective MV-

algebras

why should we insist in giving many proofs of MV-amalgamation and interpolation?

• because MV-algebras provide a benchmark for other structures of interest in algebraic logic

• because interpolation and amalgamation are deeply related to many fundamental logical-algebraic-geometric notions:

• quantifier elimination, cut elimination, joint consistency, joint embedding, unification, projectives,...

• let us briefly review what is known about finitely generated projective MV-algebras, i.e., retracts of FREEn for some n

• this is joint work with Leonardo Cabrer, to appear in Communications in Contemporary Math., and based on earlier joint work on Algebra Universalis 62 (2009) 63–74.

projectives are routinely characterized by duality

• Every n-generated projective MV-algebra A is finitely presented (essentially, Baker)

• A is finitely presented iff A=M(P) for some polyhedron P lying in some n-cube [0,1]n (Baker-Beynon duality)

• DEFINITION P is said to be a Z-retract if the MV-algebra M(P) is projective

• Problem: characterize Z-retracts, among all polyhedra

this property is not easy to handle;thus, we must find equivalent conditions for a polyhedron P to be retract of [0,1]n

a first property of Z-retracts:they are retracts of some cube [0,1]n

The elements of the fundamental group π1(P)

(introduced by Poincaré) of a connected polyhedron P are

the equivalence classes of the set of all paths with initial and

final points at a given basepoint p, under the

equivalence relation of homotopy. The fundamental

groups of homeomorphic spaces are isomorphic.

to check if P is a retract it suffices to check that all homotopy groups of P are trivial

equivalents for P to be a retract of [0,1]n

THEOREM. For any polyhedron P in [0,1]n the following conditions are equivalent:

(a) P is a retract of [0,1]n

(b) P is connected and all homotopy groups πi(P) are trivial

(c) P is contractible (can be continuously shrunk to a point).

Proof. (a)—>(b) by the functorial properties of the homotopy groups πi . The implications (b)—>(a) and (b)—>(c) follow from Whitehead theorem in algebraic topology. (c)—>(b) is trivial. QED

M(P) is not M(P) is not projectiveprojective

P is not a Z-P is not a Z-retract, because retract, because it is not simply it is not simply

connectedconnected

let P be this polyhedron

M(P) is not M(P) is not

projectiveprojective

a second property of Z-retracts:P must contain a vertex of [0,1]n

P is not a Z-retract, P is not a Z-retract, because it does not because it does not

contain any vertex of contain any vertex of the unit squarethe unit square

PROPOSITION If P is a Z-retract, then P has a triangulation Ω such that the affine hull of every maximal simplex in Ω contains some integer point of Rn

a third property: strong regularity

M(P) is not M(P) is not

projectiveprojective

this P is not a Z-this P is not a Z-retract: for, the affine retract: for, the affine hull of the hull of the vertical red vertical red

segment segment does not does not contain any integer contain any integer

pointpoint0

1

1

projectiveness: 3 necessary conditions

THEOREM (L.Cabrer, D.M., 2009) If A is a finitely generated projective MV-algebra, then up to isomorphism, A=M(P) for some rational polyhedron lying in [0,1]n such that

(i) P contains some vertex of [0,1]n,

(ii) P is contractible, and

(iii) P is strongly regular.

are these three conditions also sufficient for an MV-algebra A to be finitely generated projective ?

yes, when the maximal spectrum is one-dimensional

THEOREM (L.Cabrer, D.M.) Suppose the maximal spectrum of A is one-dimensional. Then A is n-generated projective if and only if A is isomorphic to M(P) for some contractible strongly regular rational polyhedron in [0,1]n containing a vertex of [0,1]n.

It is not known if these three conditions are sufficientin general.

They become sufficient if contractibility is strengthened to collapsibility

a sequence of collapses

a sequence of collapses

a sequence of collapses

a sequence of collapses

a sequence of collapses

a sequence of collapses

a sequence of collapses

a sequence of collapses

a sufficient condition for P to be a Z-retract, i.e., for M(P) to be projective

THEOREM (L.Cabrer, D.M., Communications in Contemporary Mathematics)

If P has a collapsible strongly regular triangulation containing a vertex of [0,1]n then M(P) is projective.

A is finitely presentedhomomorphismisomorphismindecomposableA is free n-generated A is n-generateddim(maxspec(A))=d A=M(P) is projective

A=M(P), P a polyhedronZ-mapZ-homeomorphismP is connectedP=unit cube [0,1]n P lies in [0,1]n dim(P)=dP is a Z-retract

algebra geometry

\Lukasiewicz logic and MV-algebras together are a rich source of geometric inspiration

thank youthank you