Correspondence and Completeness forGeneralized Quanti�ers. 1Natasha Alechina Michiel van Lambalgennatasha@fwi.uva.nl michiell@fwi.uva.nlDepartment of Mathematics and Computer ScienceUniversity of AmsterdamPlantage Muidergracht 241018 TV Amsterdam1 IntroductionA generalized quanti�er Q as de�ned by Mostowski (1957) is a class of subsets of theuniverse, so that a model M satis�es Qx'(x; �d) if the set of elements fe : M j= '[e; �d]gis in Q. Examples are: the ordinary existential quanti�er (interpreted as the set of allnon-empty subsets of the universe); the quanti�er "there are exactly 2"; a �lter quanti�er(where the only requirement onQ is that it is a �lter), "there are uncountably many" (wherethe domain is uncountable, and Q contains all uncountable subsets), etc. An obviousproperty of these quanti�ers is extensionality: if two formulas ' and are satis�ed bythe same sets of elements, Qx' is satis�ed if and only if Qx is. Actually Mostowski alsorequired that the generalized quanti�er is invariant under permutations of the universe, thusrestricting attention to quanti�ers related to cardinality. Subsequently, other generalizedquanti�ers were considered which do not have the property of permutation invariance, suchas topological quanti�ers or measure quanti�ers. For an overview of the subject, one mayconsult the collection "Model-Theoretic Logics", edited by Barwise and Feferman (1985).The very title of this volume makes it plain that generalized quanti�ers have been studiedmostly from a model theoretic point of view. There are various sound reasons for this focus:many quanti�ers are not axiomatizable, so proof theory wouldn't make much sense2 andeven if there exists an axiomatization as in the case of the quanti�er "uncountably many",the failure of interpolation (see Keisler (1970)) would lead one to suspect that even if onecould give a proof theory, it would not enjoy such pleasant properties as for instance cutelimination.Nonetheless, in van Lambalgen (1991) an attempt was made to develop Gentzen stylecalculi for the �lter quanti�ers "almost all" (due to H. Friedman; cf. Steinhorn (1985)) and"co-countably many".1This research was supported by the Netherlands Organization for Scienti�c Research (NWO) undergrant PGS 22-262.2Although one can sometimes give a proof theory by altering the semantics; cf. M. Mostowski (1991).1

The guiding intuition behind this attempt is the idea that one can set up a proofsystem once one can manage dependency relations between variables bound by quanti�ers.This idea is of course not restricted to generalized quanti�ers: it can also be found inFine's natural deduction systems using arbitrary objects and a dependency relation (cf. his(1985)), or in various forms of the functional interpretation of the existential quanti�er (cf.Gabbay and de Queiroz (1991)).For example, if Q is a quanti�er which determines a non-trivial free �lter, we have8yQx(x 6= y) but not Qx8y(x 6= y); the failure of commutation seems to indicate that xin some way depends on y, as it would if we were considering the quanti�er combination8y9x.So we think of axioms for quanti�ers such as QxQy' ! QyQx' (characteristic for"almost all") or 8yQx' ^ Qy8x' ! Qx8y' (characteristic for "co-countably many") asimplicitely determining a dependency pattern between variables; one may now ask whetherthis pattern can be made explicit, e.g. whether the properties of the dependence relation are�rst order describable. For example, in the case of the existential quanti�er, the dependencecan be taken to be functional, and the required �rst order description is given by the axiomsfor Skolem functions. Indeed, as Fine shows, one can also give a graphical representationof these dependencies.Somewhat surprisingly, the axioms mentioned do indeed determine a �rst order conditionon dependence, even conditions which are true for a paradigmatic case of dependence,namely, linear dependence in vector spaces. It is the purpose of this paper to explore thisphenomenon, both its extent and its limits, in greater detail.As a �rst step, we introduce an analogue of the expansion of a language by Skolemfunctions. Consider a language L82 which extends a �rst-order language with equalityby introducing a unary generalized quanti�er 2x. We use this notation to emphasize ananalogy between generalized quanti�ers and modal operators, that will become apparentbelow. The dual of 2x is 3x' =df :2x:'. (In the examples above a �lter quanti�er wouldcorrespond to 2.)De�nition 1 Let R(x; �y) be a relation of inde�nite arity. The intuitive interpretationof R(x; �y) is "x depends on �y". The standard translation � : L82 �! L(R) is de�nedinductively as follows:P (x1; : : : ; xn)� = P (x1; : : : ; xn);(: )� = : �;( 1 ^ 2)� = �1 ^ �2;(8x )� = 8x �(2x (x; �y))� = 8x(R(x; �y)! �(x; �y)). 2In other words, in the formula 2x (x; �y) the bound variable x depends on �y in a waydetermined solely by 2x, not by (this is di�erent from Skolem functions).As in the case of Skolem functions, we would like to prove that every theory T has aconservative extension to a theory T 0 such that([) T 0 ` 2x'(x; y1; : : : ; yn) � 8x(R(x; y1; : : : ; yn)! '(x; y1; : : : ; yn)):A moment's re ection shows that this can be true only for theories which are consistentwith the minimal logic 2

De�nition 2 The minimal logic Lmin for L82 is the smallest class of formulas closed withrespect to classical �rst-order logic and the following axiom schemata:A1 ` 2x>;A2 ` 8x('! )^ 2x'! 2x , provided 2x and 2x' have the same free variables;A3 ` 2x' ^2x ! 2x('^ ), provided 2x and 2x' have the same free variables;A4 ` 2x'! 2y', provided y is free for x in '. 2In order to talk about axioms (that is, formulas) and not schemata, we introduce thesubstitution rule ` �(P (x1; : : : ; xn))` �('(x1; : : : ; xn));provided P (�x) and '(�x) have precisely the same free variables. This restriction is necessarydue to the fact that A2 entails only a restricted form of extensionality.For theories which are consistent with the minimal model we can indeed prove that aconservative extension T 0 with ([) exists. First we de�ne the required model expansion.De�nition 3 A relational model for L82 is a triple M = <D;R; V >, where D is a non-empty domain, V interprets predicate symbols, and R is a relation of inde�nite arity on D,called the dependence relation. The notion of a formula being satis�ed in a model under avariable assignment is standard; the clause for 2x reads as follows:M j=� 2x'(x; �y) , 8d 2 D(R(d; �(�y))!M j=� '(d; �y))A relational canonical model is a model where the accessibility relation is de�ned asfollows: R(x; �y) = ^'(x;�y)2L822x'(x; �y)! '(x; �y): 2The existence of T 0 satisfying ([) now follows fromTheorem 1 Every Lmin-consistent set of L82 formulas has a Henkin canonical relationalmodel.Proof can be found in van Benthem & Alechina (1993). 2Clearly, then, this way of making dependency patterns explicit will not work for allunary quanti�ers; for quanti�ers which are not (the dual of) a �lter quanti�er a translationmore complicated than * might be necessary (see, however, the discussion in section 6.2).But it will be seen below that this simple case already has a rich theory.Given the translation *, a quanti�er axiom corresponds to a schema in the languageL(R); the main question then becomes: when can this schema be replaced by a �rst-ordercondition on R?The reader will have observed that both the set-up and the main problem are very muchanalogous to familiar themes in modal logic: R plays the role of the accessibility relation,and what we ask for is a Sahlqvist theorem, i.e. a characterization of a class of formulas forwhich there is a �rst order correspondent. To be more speci�c we need a de�nition:3

De�nition 4 If A is a quanti�er axiom, a correspondent in the sense of completeness is a�rst order condition Ay on R with the following two properties:i any set of sentences consistent with A has a relational model where Ay holds;ii A is satis�ed on any relational model where Ay holds. 2Recall that a modal logic L is called �rst order complete if there is a set � of �rst ordersentences in the language fR;=g (where R is the accessibility relation) such that`L ' if and only if for every Kripke model M : if M j= � then M j= ';equivalently, `L ' if and only if ' is true on any frame which satis�es �. This notion canbe reformulated for the generalized quanti�ers as followsDe�nition 5 If L is a generalized quanti�er logic, then L is �rst order complete if there isa set � of �rst order sentences in the language L(R) so that for every quanti�er formula ',`L ' if and only if for every relational model M : if M j= � then M j= '. 2Note if L is �nitely axiomatizable then by compactness � can be taken to be �nite.The de�nition 4 can be reformulated as follows: A has a correspondent for completenessAy if Lmin +A is �rst order complete with respect to the class of models satisfying Ay.In modal logic one can even de�ne a stronger notion of completeness. A modal logic Lis canonical if it is �rst order complete and the corresponding set � of �rst order sentencesholds in the canonical model for L. For generalized quanti�ers, there is no unique canonicalmodel. However, we can approximate the modal concept as follows:De�nition 6 If L is a generalized quanti�er logic, then L is canonical if L is �rst ordercomplete and the corresponding set � of �rst order sentences holds in every !-saturatedcanonical relational model of L. 2This de�nition is motivated by the fact that in modal logic canonical models are !-saturated.The correspondence theory for generalized quanti�ers developed in van Benthem andAlechina (1993) is not quite adequate for our purposes, since it is based on the notion offrame correspondence:De�nition 7 If A is a quanti�er axiom, a frame correspondent is a �rst order condition A?on R such that <D;R> j= A? if and only if for any interpretation V , <D;R; V> j= A. 2It will be seen below that the two notions of having a correspondent are di�erent;for example, there are formulas which have a frame correspondent, but do not have acorrespondent for completeness. However, the following holds:Proposition 1 If A has a correspondent for completeness Ay and a frame correspondentA?, then `FOL Ay ! A?.Proof. Let <D;R> j= Ay. Then by the De�nition 4 (ii), for every interpretation V<D;R; V> j= A. By the De�nition 7, <D;R> j= A?. 2We restate the result on frame correspondence of van Benthem and Alechina (1993)here, and give an outline of its proof, because we shall need some concepts occurring in theproof in the sequel. 4

Theorem 2 (Correspondence part of the Sahlqvist theorem). All formulas of the "Sahlqvistform" '! , where1. ' is constructed from� atomic formulas, possibly pre�xed by 2x, 8;� formulas in which predicate letters occur only negativelyusing ^;_;3x; 92. in all predicate letters occur only positivelyhave a frame correspondent. 3We shall sketch the proof of the theorem here.A formula � is valid in a frame if the following second-order formula is: 8P1 : : :8Pn��,where P1; : : : ; Pn are all predicate letters in �. If this formula is equivalent to a �rst-orderformula without P1; : : : ; Pn, then � de�nes a �rst-order condition on frames. The proof ofthe theorem is based on the method of minimal substitutions, due to van Benthem (1983).After performing some syntactical transformations on � the task can be reduced to provingthat the following formula has a �rst-order equivalent:8P1 : : :8Pn8�u( j Rj ^ i 8�x(Ri ! Pi)^ k Pk ! )where Vj Rj corresponds to the translation of the truth conditions for 3-quanti�ers in theantecedent, Vi 8�x(Ri ! Pi) corresponds to the translations of occurrences of predicatesymbols preceded by 2-quanti�ers, Vk Pk corresponds to occurrences of predicate symbolsnot preceded by 2's, and is a positive formula.For example, assume that � = 3x(2yP (x; y) ^ S(x)) ! 2zS(z). Then its translationreads 8P8S8x(R(x)^ 8y(R(y; x)! P (x; y))^ S(x)! 8z(R(z)! S(z))):Here Vj Ri = R(x), Vj 8�x(Ri ! Pi) = 8y(R(y; x)! P (x; y)), and Vk Pk = S(x).Note that we quantify over all possible assignments to the predicate symbols, so a secondorder quanti�er 8Pni can be instantiated using any suitable set of n-tuples. That is what weare going to do. If Pi occurs in the antecedent in a subformula of the form 8�x(Ri ! Pi(�x)),the minimal substitution for this occurrence of Pi is precisely the set of those tuples forwhich Ri holds. (In the running example, P (u1; u2) := R(u2; u1) ^ u1 = x.) If Pi occursas a member of conjunction Vk Pk , then the minimal substitution for this occurrence is asingleton set: in the example, S(u) := u = x. Finally, the minimal substitution for Pi is thedisjunction of minimal substitutions for its occurrences. (Precise de�nitions and details aregiven in van Benthem and Alechina (1993)). Denote the result of the substitution �0. (Inthe example, �0 is8x(R(x)^ 8y(R(y; x)! R(y; x)^ x = x) ^ x = x! 8z(R(z)! z = x));which is equivalent to 8x(R(x)! 8z(R(z)! z = x)).)3This theorem can be made slightly more general by allowing an arbitrary long pre�x of 2-quanti�ersbefore the formula, constant formulas in the antecedent(formulas which contain only =, > and ? as predicatesymbols), and also stating that a conjunction of Sahlqvist formulas is a Sahlqvist formula.5

�0 follows from 8 �P�� as a substitutional instance. The converse also holds, as follows.Denote the substitutions for Pi as pi. Each pi is a �rst-order formula. The pi's werechosen so that the antecedent of �0 becomes trivially true except for the �rst-order partVjRj , �0 = 8�u( j Rj ^ i 8�x(Ri ! Ri) ^^(w = w)! (p1; : : : ; pn));that is �0 = 8�u(Vj Rj ! (p1; : : : ; pn)). The consequent (p1; : : : ; pn) is a �rst-orderformula. Assume that �0 is true and there is a valuation (an assignment to predicatesymbols) V which makes the antecedent of �� true:<D;R; V> j= j Rj ^ i 8�x(Ri ! Pi) ^ k Pk<D;R> j= j Rj ! (p1; : : : ; pn)):We want to show that under this valuation also the consequent of ��, (P1; : : : ; Pn), istrue. First of all, observe that (p1; : : : ; pn) is true. The substitutions pi which we used areminimal in the sense that always when the antecedent of �� is true under some assignmentof values to predicate letters, then this assignment should contain the tuples satisfying pi:f< d1; : : : ; dn >: pi(d1; : : : ; dn)g � V (Pi);that is, <D;R; V > j= 8 �d(pi( �d) ! Pi( �d)). (P1; : : : ; Pn) follows from (p1; : : : ; pn) bymonotonicity of . 2A disadvantage of the notion of frame correspondence is that there may be consistentquanti�er logics which are frame incomplete, i.e. logics L such that for no D;R: for all V ,<D; V;R> j= A. In particular this holds for logics where the quanti�er determines a free�lter, such as 'almost all' or 'co-countably many'. It will be seen below that this is mainlydue to the property of upwards monotonicity for the �lter quanti�er; unlike the modal case,where monotonicity is automatic given Kripke semantics, this property is highly non-trivialhere. Actually, inspection of the proofs below shows that this result can be strengthened:the troublesome property is extensionality, that is, 8x(' � ) ! (2x' � 2x ). This isinteresting in view of the fact that extensionality is generally taken to be the sine qua nonfor logicality of a quanti�er.2 Statement of the result and idea of the proofWe now formulate the completeness part of the Sahlqvist theorem, which describes a classof formulas ' de�ning �rst-order conditions on R so that for any logic L in the languageof L82 which has a canonical relational model, L with ' as an axiom is complete for theclass of models where R has the �rst order property corresponding to '. This class isstrictly smaller than the class of formulas described above. We shall call those formulasweak Sahlqvist formulas.Theorem 3 (Completeness part of the Sahlqvist theorem) All "weak Sahlqvist formulas" �of the form VQz1 : : :Qzn(A! B), where n � 0, each Q is either 8 or 2, and1. A is constructed from 6

a. atomic formulas, possibly with a quanti�er pre�x Qx1 : : :Qxk, where each Q is a2- or 8-quanti�er;b. formulas in which atomic formulas occur only negatively,c. constant formulas (where the only predicate letters are >;? and =),using ^ and _,2. in B all predicate letters occur only positively3. every occurrence of a predicate letter has the same free variables 4have a correspondent in the sense of completeness; moreover, this correspondent holds inevery !-saturated canonical relational model of a logic in which � is provable.All conditions of the theorem can be shown to be necessary, namely, if a formula does notsatisfy one of them, then it need not have a correspondent for completeness. The conditionswhich are common with the Theorem 2, are shown to be necessary in van Benthem &Alechina (1993). The additional conditions are: no existential quanti�ers in the antecedentand all occurrences of a predicate symbol have the same free variables. The necessity ofthose is shown in section 6.The idea of the proof of the Theorem 3 (very similar to the one used in Sambin &Vaccaro (1989)) can be illustrated by the following example.Example 1 Let C be a canonical5 model. We show that if for every P and S2xP (x; �y)! 2x(P (x; �y)_ S(x; �z))is valid in C, then the accessibility relation in C has the property R(x; �y�z)! R(x; �y).Proof. It is easy to see (since the consequent is monotone in S) that the axiom above isequivalent to 2xP (x; �y)! 2x(P (x; �y)_?(x; �z)). We translate the validity conditions usingsecond-order quanti�ers which range only over de�nable relations of C (this is the di�erencewith the case of the frame correspondence). To emphasize this di�erence we use quanti�ers8'. Note that due to the restricted Substitution Rule, if P is an n-place predicate symbol,then formulas which can be substituted for P must have precisely n variable places. Usingthis notation the validity condition of the axiom reads as follows:8'[8x(R(x; �y)! '(x; �y))! 8x(R(x; �y�z)! '(x; �y)_ ?(x; �z))]This is equivalent to 8'[8x(R(x; �y) ! '(x; �y)) ! 8x(R(x; �y�z) ! '(x; �y))], and, in turn,to V'(x;�y)f8x(R(x; �y�z)! '(x; �y)) : 8x(R(x; �y)! '(x; �y))g. Moving the conjunction inside(the proof that this can be done for any positive logical function of ', is given in theIntersection Lemma; in the given case the proof is obvious), we obtain8x(R(x; �y�z)! ^'(x;�y)f'(x; �y)) : 8x(R(x; �y)! '(x; �y))g)4Actually, the su�cient condition is that all positive occurrences of a predicate letter in the antecedenthave the same free variables, say, �y, all negative occurrences of a predicate letter in the antecedent and alloccurrences in the consequent have the same free variables, say, �z, and �y � �z. It can be shown that in thiscase the condition 3 can be forced to hold. But we assume that this holds from the very beginning to makethe formulation of the theorem readable.5Henceforth, "canonical model" will always mean "canonical relational model".7

But in C V'(x;�y)f'(x; �y)) : 8x(R(x; �y)! '(x; �y))g) � R(x; �y). Substituting R(x; �y) insteadof the in�nite conjunction yields the �rst-order equivalent 8x(R(x; �y�z)! R(x; �y)).It is easy to check that in every model where 8x(R(x; �y�z)! R(x; �y)) holds, the axiomis valid. 2The general case is slightly more complicated because to obtain a correspondent wemust sometimes move to a di�erent canonical model, namely, to an !-saturated canonicalmodel. The existence of such model for every Lmin-consistent set of sentences is proved insection 3. In a canonical !-saturated model the following lemma holds:Intersection Lemma If B is a positive formula and X is a set of formulas with the samefree variables, closed with respect to ^, then in an !-saturated model^fB(') : ' 2 Xg � B(^f' : ' 2 Xg):The proof of the Theorem 3 consists of the same three ingredients as those in theexample: translation of the validity conditions of an axiom (eventually accompanied bysome syntactic transformations), application of the Intersection Lemma, and making use ofthe fact that for some �rst-order expression R with R as the only predicate symbolR(�x; �y) � ^'(�x;�y)f'(�x; �y) : 8�x(R(�x; �y)! '(�x; �y))g:Of course, R is nothing else than a "minimal substitution", but of a special kind, whichwill be formally de�ned below.De�nition 8 Let M be a canonical model, and A a conjunction of atomic formulas whichare pre�xed by universal and 2-quanti�ers, so that all occurrences of a predicate symbolhave the same free variables. Every occurrence of a predicate symbol P in A is therefore ofthe form �Qi�xP (�x; �y), where �Qi is the quanti�er pre�x of the ith occurrence. P has a goodminimal substitution in A if M j= Vf'(�x; �y) : Vi �Qi�x'(�x; �y)g � p, where p is a �rst-orderformula built using the predicates R, =, > and ? only. 2For example, we have seen that if the only occurrence of P in A is of the form 2xP (x; �y),then P has a good minimal substitution in A: for every canonical model M ,M j=^f'(x; �y) : 2x'(x; �y)g � R(x; �y):Before we formulate the Closure Lemma, we shall get rid of two degenerate cases.(a) Let P be preceded by a vacuous 2: e.g., the only occurrence of P in A is of the form2xP (y). Then every canonical model M satis�es^f'(y) : 2x'(y)g �^f'(y) : 8x(R(x; y)! '(y))g �^f'(y) : 9xR(x; y)! '(y)g:If M j= 9xR(x; y), then M j= Vf'(y) : 9xR(x; y) ! '(y)g � >(y). If not, then M j=Vf'(y) : 9xR(x; y) ! '(y)g � ?(y). This means that axioms with vacuous 2's in theantecedent can have di�erent correspondents in di�erent models.Henceforth we assume that all quanti�ers are non-vacuous. Formally, this correspondsto assuming that R is always non-empty, or that the axiom 3x>(x; �y) holds in all canonicalmodels. This is an innocuous assumption, because if R is empty, every L82-formula isequivalent to a �rst-order formula (2x' becomes equivalent to > and 3x' becomes ?).8

We shall also assume that every quanti�er pre�x in A contains at least one 2. That thisis no loss of generality can be seen as follows.Let P occur in A with a purely universal pre�x, 8x1 : : :8xnP (�x; �y). Then this occurrenceimplies all other possible occurrences of P in A, and A is equivalent to a conjunction where8x1 : : :8xnP (�x; �y) is the only occurrence of P .M j= ^f'(�x; �y) : 8x1 : : :8xn'(�x; �y)g � >(�x; �y);so P has a good minimal substitution in A.Now we can state the Closure Lemma which is proved in section 4:Closure Lemma. Let A be a conjunction of atomic formulas pre�xed by 8 and 2-quanti�ers, so that all occurrences of a predicate symbol in A have the same freevariables. Then every atomic formula in A has a good minimal substitution, and thisminimal substitution is the same as the one used to obtain the frame correspondent.3 !-saturated models and the Intersection Lemma.Let X = f'1(x); '2(x); : : :g be a �nitely realizable type in a model M , that is, for every nthere is an element an in the domain ofM such that '1(an); : : : ; 'n(an) is true in M . If Mis just an ordinary Henkin model, there does not necessarily exist an element a such thatfor every ' in X '(a) is true in M . Among other things this implies that 3xVf' : ' 2 Xgis not equivalent to Vf3x' : ' 2 Xg in M . But in the proof we do need thatM j= 3x^f' : ' 2 Xg �^f3x' : ' 2 Xg(an analogue of Esakia's lemma). We therefore move from the original Henkin model to amildly saturated extension.Theorem 4 Every consistent set of L82 formulas has a model A which is !-saturated andcanonical, that isi every �nitely realizable type which contains �nitely many parameters is realized;ii RA(d; �d) =df V'(x;�d)2L82 2x'(x; �d)! '(d; �d)Proof From the completeness proof for the minimal logic we know that every consistentset of formulas has a canonical model C whereRC(x; �y) � ^'(x;�y)2L822x'(x; �y)! '(x; �y)By the truth de�nitionC j= 2x'(x; �y), 8x(RC(x; �y)) C j= '(x; �y))Therefore there is a �rst-order model C� (with R = RC just an ordinary predicate) suchthat if 2 L82 C j= , C� j= �where � is the standard translation of . 9

We shall use this fact to build the saturated model which we need, because one can applythe standard procedure of constructing an !-saturated extension of C�. (While extendinga model for a generalized quanti�er is much more di�cult, see for example Hodges (1985).)Take an !-saturated elementary extension A� of C�. It is clear thatC j= , C� j= �, A� j= �;for every sentence of L82.Every type �nitely realizable in C is �nitely realizable in C� and is therefore realized inA�. But A� is still a �rst-order model; to make an L82 model A out of it, we could takethe interpretation of R in A� to be the accessibility relation in A, i.e. stipulateA j= 2x'(x; �y), 8x(R(x; �y)) A j= '(x; �y)):However, it is not obvious that A is still canonical.Instead we de�ne the accessibility relation anew in A. A will be the expansion < A�; RA >of A�, where RA is de�ned on A� asRA(x; �y) = ^'�(x;�y):'(x;�y)2L82 8x(R(x; �y)! '�(x; �y))! '�(x; �y):Note that the intersection is only over the formulas '�(x; �y) such that '(x; �y) 2 L82.We are done if we can show thatLemma 1 A j= ', A� j= '� for all formulas ' 2 L82.Proof By induction on the complexity of '. The only non-trivial case is ' = 2x (x; �y).To prove the direction from right to left, assume that A� j= (2x (x; �y))�, that is, A� j=8x(R(x; �y) ! �(x; �y)). We want to prove A j= 2x (x; �y), that is A j= 8x(RA(x; �y) ! (x; �y)).Let RA(x; �y) hold in A. By the de�nition of RA, A� j= 8x(R(x; �y) ! �(x; �y)) ! �(x; �y). We know that A� j= 8x(R(x; �y) ! �(x; �y)). Therefore A� j= �(x; �y) and, bythe inductive hypothesis, A j= (x; �y).From left to right: let A j= 2x (x; �y), that is A j= 8x(RA(x; �y) ! (x; �y)). LetR(x; �y) hold in A�. We want to show that A� j= �(x; �y). It is enough to show thatR(x; �y) implies RA(x; �y). If this is so, we obtain (x; �y) from R(x; �y) and the fact thatA j= 8x(RA(x; �y) ! (x; �y)), and hence applying the inductive hypothesis we also get �(x; �y).Let R(x; �y). Take an arbitrary formula �� such that 8x(R(x; �y) ! ��(x; �y)). Then��(x; �y). This way we prove that for all ��, R(x; �y)! (8x(R(x; �y)! ��(x; �y))! ��(x; �y)).Therefore R(x; �y) ! V��(8x(R(x; �y) ! ��(x; �y)) ! ��(x; �y)), which means that R(x; �y)implies RA(x; �y). 2Comment. C is an elementary extension of A with respect to L82 formulas, but notnecessarily with respect to L(R) formulas if R is interpreted as RA.Now we are ready to prove that in A the Intersection Lemma holds.Lemma 2 (Intersection Lemma) If B is a positive formula and X is a set of formulaswith the same free variables, closed with respect to ^, then in an !-saturated model^fB(') : ' 2 Xg � B(^f' : ' 2 Xg)10

Proof By induction on the complexity of B.� B(') = ': trivial, because Vf' : ' 2 Xg = Vf' : ' 2 Xg.B can be a formula with principal sign ^;_; 8; 9;3x or 2x (since B is positive).Assume that for the components of B the claim holds.� Let B = B1 ^B2;^fB1(')^B2(') : ' 2 Xg =^fB1(') : ' 2 Xg ^^fB2(') : ' 2 Xg(by the associativity of V) and, by the inductive hypothesis, the latter is equivalentto B1(Vf' : ' 2 Xg)^B2(Vf' : ' 2 Xg).� Let B = B1 _B2.Assume M; �e j= B1(Vf' : ' 2 Xg) _ B2(Vf' : ' 2 Xg). Then one of the disjunctsis true, for example M; �e j= B1(Vf' : ' 2 Xg). By the inductive hypothesis, M; �e j=VfB1(') : ' 2 Xg. Clearly, M; �e j= VfB1(')_ B2(') : ' 2 Xg.For the other direction, assume M; �e 6j= B1(Vf' : ' 2 Xg) _ B2(Vf' : ' 2 Xg).This means that M; �e 6j= B1(Vf' : ' 2 Xg) and M; �e 6j= B2(Vf' : ' 2 Xg). Bythe inductive hypothesis, M; �e 6j= VfB1(') : ' 2 Xg and M; �e 6j= VfB2(') : ' 2 Xg.Therefore there are '1 and '2 in X such that M; �e 6j= B1('1) and M; �e 6j= B2('2).Since B1 is also positive and therefore monotone, 6 M; �e 6j= B1('1 ^ '2) and M; �e 6j=B2('1 ^ '2), thus M; �e 6j= B1('1 ^'2)_B2('1 ^ '2). Note that X is closed under ^,therefore '1 ^ '2 2 X . But this means that M; �e 6j= VfB1(')_ B2(') : ' 2 Xg.� Let B = 8xB1. Vf8xB1(') : ' 2 Xg = 8xVfB1(') : ' 2 Xg (because 8 distributesover V), and by the inductive hypothesis this is equivalent to 8xB1(Vf' : ' 2 Xg).� Let B = 2xB1. This case is analogous, but since 2 distributes only over conjunctionsof formulas with the same free variables, it is important that all formulas in X (andtherefore in fB1(') : ' 2 Xg) have the same free variables.� Let B = 3xB1. This is the only non-trivial part, and here we need the fact that themodel is !-saturated. First of all, it would be convenient if the set Y = fB1(') :' 2 Xg had the following property: for every n, M j= n ! 1 ^ : : : ^ n�1, i 2 Y . This is not so in general, but we can consider instead of Y the set Y 0 =fB1('1); B1('1 ^ '2); B1('1 ^ '2 ^ '3); : : : : 'i 2 Xg. For every formula in Y ,there is a formula in Y 0 which implies it (due to the monotonicity of B1). ThereforeVY 0 ! VY . On the other hand, Y 0 � Y (because X is closed under ^). ThereforeVY ! VY 0. Analogously Vf3x : 2 Y g = Vf3x : 2 Y 0g. So, it is su�ces toprove that ^f3x : 2 Y 0g = 3x^f : 2 Y 0gSince Vf : 2 Y 0g implies for any 2 Y 0, 3xVf : 2 Y 0g implies 3x andtherefore Vf3x : 2 Y 0g. This proves the implication right to left.For the other direction, assume that the model satis�es 3x for every 2 Y 0. Dueto the de�nition of Y 0 and monotonicity of B1, this means that for every n the model6Monotonicity in L82 is restricted to the formulas with the same free variables, but this is always thecase in this proof. 11

satis�es 3x( 1^ : : :^ n), i.e. that for every n there is an element x satisfying R(x; �e)and 1(x; �e); : : : ; n(x; �e). Since the model is !-saturated, there is an element whichsatis�es the whole set: R(u; �e)^Vf (u; �e) : 2 Y 0g. Then 3xVf : 2 Y 0g is true.� Let B = 9xB1: the proof is analogous to the previous case. 24 Closure LemmaLet A be a conjunction as in the condition of the Closure Lemma. We also assume thatall quanti�ers are non-vacuous and that every quanti�er pre�x contains at least one 2-quanti�er (cf. the end of section 2).Let A0 be the subformula of A which contains all and only occurrences of the predicatesymbol P . We shall use both the L82-form of A0, namely Vi �Qi�xP (�x; �y), and its standardtranslation Vi 8�x(Ri ! P (�x; �y)), where i runs over the occurrences of P . In the sequel wecall the Ri R-conditions. The standard translation of A0 is thus equivalent to 8�x(WiRi !P (�x; �y)). P (�x; �y) has a good minimal substitution in A if_i Ri =^f'(�x; �y) : 8�x(_i Ri ! '(�x; �y))g:Note that good minimal substitutions are the same expressions which were used in the proofof Theorem 2 as minimal substitutions for occurrences of predicate symbols in the scope of2- and 8-quanti�ers.Example 2 The R-condition corresponding to 2x2yP (x; y) is R(x)^R(y; x).Example 3 Let A0 = 8x2yP (x; y)^ 2x8yP (x; y), thenA0� = 8x8y(R(y; x)! P (x; y))^ 8x8y(R(x)! P (x; y))which is equivalent to 8x8y(R(y; x)_R(x)! P (x; y)). The good minimal substitution forP in A must be therefore R(y; x)_R(x).We are going to prove the existence of good minimal substitutions for all non-vacuousquanti�er pre�xes containing at least one 2. From this and from the considerations at theend of the section 2 the Closure Lemma will follow. But �rst we need several propositions.Proposition 2 Let R be an R-condition, such thatR(�x; �y) �^f'(�x; �y) : 8�x(R(�x; �y)! '(�x; �y))g;then R(�x; �y) �^f (x; �y�z) : 8�x8�z(R(�x; �y)! (�x; �y�z))g;and vice versa: if R(�x; �y) �^f (x; �y�z) : 8�x8�z(R(�x; �y)! (�x; �y�z))g;then R(�x; �y) �^f'(�x; �y) : 8�x(R(�x; �y)! '(�x; �y))g:12

Proof. Assume R(�x; �y) � Vf'(�x; �y) : 8�x(R(�x; �y) ! '(�x; �y))g. For every '(�x; �y) holds:'(�x; �y) � 8�z('(�x; �y) ^ >(�z)). ThereforeR(�x; �y) �^f8�z('(�x; �y)^ >(�z)) : 8�x8�z(R(�x; �y)! ('(�x; �y) ^ >(�z))g:Since for every ', 8�z('(�x; �y) ^ >(�z)) � '(�x; �y) ^ >(�z),R(�x; �y) �^f'(�x; �y) ^ >(�z) : 8�x8�z(R(�x; �y)! ('(�x; �y) ^ >(�z))g:Now we prove that R(�x; �y) � Vf (�x; �y�z) : 8�x8�z(R(�x; �y)! (�x; �y�z))g. Trivially,^f (�x; �y�z) : 8�x8�z(R(�x; �y)! (�x; �y�z)g !^f'(�x; �y)^>(�z) : 8�x8�z(R(�x; �y)! ('(�x; �y)^>(�z))gand this implies that Vf (�x; �y�z) : 8�x8�z(R(�x; �y)! (�x; �y�z))g ! R(�x; �y).SinceR(�x; �y)! Vf (�x; �y�z) : 8�x8�z(R(�x; �y)! (�x; �y�z))g, we haveR(�x; �y) � Vf (�x; �y�z) :8�x8�z(R(�x; �y)! (�x; �y�z))g.For the other direction of the proposition, letR(�x; �y) �^f (�x; �y�z) : 8�x8�z(R(�x; �y)! (�x; �y�z))g:It is easy to check thatR(�x; �y) � Vf8�z (�x; �y; �z) : 8�x(R(�x; �y)! 8�z (�x; �y; �z))g, and then thereasoning goes as above: the set of ''s with free variables �x; �y satisfying the same conditionis larger than the set of 8�z (�x; �y; �z), therefore its conjunction implies the given one; on theother hand, the set of ''s satisfying the condition is implied by R, therefore the two setsare equivalent. 2To prove the next proposition, we shall use the following tautology of the minimal logic:Lmin ` 2x(2x� ! �) (the proof is easy).Proposition 3 R(z; �x�y) � Vf'(z; �x; �y) : 8z8�x(R(z; �x�y)! '(z; �x; �y))g.Proof. The direction from left to right is trivial. For the converse direction, we have toprove Vf'(z; �x; �y) : 8z8�x(R(z; �x�y)! '(z; �x; �y))g ! R(z; �x�y), in other words,' (8�x2z'(z; �x; �y)! '(z; �x; �y))! (2z (z; �x; �y)! (z; �x; �y)):It su�ces to derive V 2z (z; �e; �y) ! (d; �e; �y) from V' 8�x2z'(z; �x; �y) ! '(d; �e; �y). Takean arbitrary (z; �x; �y). We substitute this formula for � in the tautology above:8�x2z(2z (z; �x; �y)! (z; �x; �y)):We assume that the conjunction V' 8�x2z'(z; �x; �y)! '(d; �e; �y) holds. As a special case weobtain 8�x2z(2z (z; �x; �y)! (z; �x; �y))! (2z (z; �e; �y)! (d; �e; �y)):Since this holds for every , we can derive V 2z (z; �e; �y)! (d; �e; �y). 2Proposition 4 If Qx1; : : : ; Qxn contains at least one 2-quanti�er, and all quanti�ers arenon-vacuous, and the only occurrence of P (�x; �y) in A is of the form Qx1 : : :QxnP (�x; �y),then P has a good minimal substitution in A.13

Proof. The general form of the pre�x described in the condition of this proposition, is8(�u)12z18(�u)22z2 : : :8(�u)k2zk8(�u)k+1P (�x; �y);where k > 0 (that is, there is at least one 2 in the pre�x), and �u�z = �x. The standardtranslation of this formula is8�x(R(z1; (�u)1; �y) ^ : : :^ R(zk; (�u)k; zk�1; (�u)k�1; : : : ; z1; (�u)1; �y)! P (�x; �y))(since there is at least one 2-quanti�er in the pre�x).We have to show that R � i=ki=1R(zi; ( ~uz)�i; �y);where ( ~uz)�i are the variables bound by the quanti�ers preceding 2zi , is a good minimalsubstitution for P .By propositions 2 and 3 (observe that zi( ~uz)�i � �x),R(zi; ( ~uz)�i; �y) �^f'(�x; �y) : 8�x(R(zi; ( ~uz)�i; �y)! '(�x; �y))gNote that here we essentially use the fact that the zi occur in P (�x; �y), that is, that 2-quanti�ers are non-vacuous.It is easy to see thatViR(zi; ( ~uz)�i; �y)! Vf'(�x; �y) : 8�x(ViR(zi; ( ~uz)�i; �y)! '(�x; �y))g.To prove the other direction, namely^f'(�x; �y) : 8�x( i R(zi; ( ~uz)�i; �y)! '(�x; �y))g ! i R(zi; ( ~uz)�i; �y);we argue as before:f'(�x; �y) : 8�x(R(zi; ( ~uz)�i; �y)! '(�x; �y))g � f'(�x; �y) : 8�x( i R(zi; ( ~uz)�i; �y)! '(�x; �y))gtherefore^f'(�x; �y) : 8�x( i R(zi; ( ~uz)�i; �y)! '(�x; �y))g !^f'(�x; �y) : 8�x(R(zi; ( ~uz)�i; �y)! '(�x; �y))gand this means that Vf'(�x; �y) : 8�x(ViR(zi; ( ~uz)�i; �y)! '(�x; �y))g ! R(zi; ( ~uz)�i; �y). Sincethis holds for every i,^f'(�x; �y) : 8�x( i R(zi; ( ~uz)�i; �y)! '(�x; �y))g ! i R(zi; ( ~uz)�i; �y);that is, R is a good minimal substitution. 2Proposition 5 A disjunction of good minimal substitutions is a good minimal substitution,i.e. if for every i, 1 � i � n, Ri(�zi; �y) = Vf'(�x; �y) : 8�x(Ri(�zi; �y)! '(�x; �y))g, �zi � �x, thenWiRi(�zi; �y) = Vf'(�x; �y) : 8�x(WiRi(�zi; �y)! '(�x; �y))g.Proof. Since for every Rif'(�x; �y) : 8�x(_i Ri(�zi; �y)! '(�x; �y))g � f'(�x; �y) : 8�x(Ri(�zi; �y)! '(�x; �y))g;14

^f'(�x; �y) : 8�x(Ri(�zi; �y)! '(�x; �y))g !^f'(�x; �y) : 8�x(_i Ri(�zi; �y)! '(�x; �y))g;that is, for every Ri, Ri ! Vf'(�x; �y) : 8�x(WiRi(�zi; �y) ! '(�x; �y))g, and this impliesWiRi ! Vf'(�x; �y) : 8�x(WiRi(�zi; �y)! '(�x; �y))g.Now we prove that the implication holds also in the other direction. From Proposition 2follows that if Ri(�zi; �y) = Vf'(�x; �y) : 8�x(Ri(�zi; �y) ! '(�x; �y))g, �zi � �x, then Ri(�zi; �y) =Vf (�zi; �y) : 8�zi(Ri(�zi; �y)! (�zi; �y))g.Now, assume thatVf'(�x; �y) : 8�x(WiRi(�zi; �y)! '(�x; �y))g holds and none of theRi(�zi; �y)holds. Then as we have just seen, there are formulas 1; : : : ; n, such that for every i,8�zi(Ri(�zi; �y)! i(�zi; �y)) and : i(�zi; �y). Take the disjunction of these formulas, Wi i(�zi; �y).The resulting formula is also false. An equivalent formula with free variables �x; �y, namely,Wi i(�zi; �y)_?(�x), belongs to the set f'(�x; �y) : 8�x(WiRi(�zi; �y)! '(�x; �y))g; but this formulais false, a contradiction. 2Lemma 3 (Closure Lemma.) Let A be a conjunction of atomic formulas pre�xed by 8 and2-quanti�ers, so that all occurrences of a predicate symbol in A have the same free vari-ables. Then every atomic formula in A has a good minimal substitution, and this minimalsubstitution is the same as the one used to obtain a frame correspondent.Proof. The lemma follows from the four propositions proved above. 25 Syntactic transformationsWe now describe the syntactic transformations which reduce the task of �nding a corre-spondent for an axiom � to simple applications of the Intersection and Closure Lemmas,thus �nishing the proof of the Theorem 3.Step 1. Assume that a formula � is of the form Vi �i, where each �i is of the formQz1 : : :Qzm(A ! B), A and B as in the Theorem 3. Then if every conjunct in �corresponds to a �rst-order condition on R in the canonical model, say �yi , � itselfcorresponds to a �rst-order condition on R (namely, a conjunction of �yi ). So, itsu�ces to prove that every �i corresponds to a �rst-order condition on R.Step 2. Writing down the validity conditions of �i, we obtain8'1 : : :8'nQz1 : : :Qzm(A! B);where 8'j are the quanti�ers over de�nable relations corresponding to the predicatesymbols in �i. If Pj is an n-place predicate symbol, then 8'j can be instantiated onany formula with precisely n variable places.Assume thatm > 0 (if not, we can move to the next step). Then translating the truthconditions for the quanti�ers in �rst-order logic, we obtain8'1 : : :8'n8z1 : : :8zm(�! (A� ! B�));where � is a conjunction of R-conditions corresponding to the 2-quanti�ers in thepre�x (if there are such quanti�ers). This is equivalent to8z1 : : :8zm(�! 8'1 : : :8'n(A� ! B�)):15

Step 3. Amay contain disjunctions; we use the fact that 8'1 : : :8'n(WiA�i ! B�) is equiv-alent to 8'1 : : :8'nVi(A�i ! B�), and since 8 distributes over ^, toVi 8'1 : : :8'n(A�i !B�). Substituting this in the formula obtained on the previous step and applying thesame reasoning (now with conjunction in the consequent), we obtaini 8z1 : : :8zn(�! 8'1 : : :8'n(A�i ! B�)):It now su�ces to prove that each conjunct corresponds to a �rst-order condition onR in an !-saturated canonical model.Step 4. Ai may contain constant formulas (without predicate symbols other than =, >and ?). We move those to �. Let A� = A0 ^ �, where � are constant formulas:(�! (� ^A0 ! B�)) � (� ^ �! (A0 ! B�))Let us denote � ^ � as �0. Note that �0 is still �rst-order.Step 5. A0 may contain negative formulas: those we move to the consequent using(� ^ � ! B�) � (� ! :� _ B�)Note that the consequent is still positive.Step 6. Finally, we have a formula(�) 8z1 : : :8zm(�0 ! 8'1 : : :8'n(A0 ! B0));where �0 is �rst-order, B0 is a positive �rst-order formula, and A0 is a conjunction offormulas (Qx1 : : :Qxk'i)�, where all quanti�ers are non-vacuous.Assume that there is only one predicate letter P in �. Then the reasoning goes asfollows: P occurs in A0 in subformulas of the form (Qx1 : : :QxkP (�x; �y))�, where the�x are bound and the �y free (rename the bound variables if necessary). The condition(*) can be rewritten as 8z1 : : :8zm(�0 ! VfB0('(�x; �y)) : Vi( �Qi�x'(�x; �y))�g):Applying the Intersection Lemma, 8z1 : : :8zm(�0 ! B0(Vf'(�x; �y) : Vi( �Qi�x'(�x; �y))�g)),and by the Closure Lemma (P has a good minimal substitution in A0, say p), this isequivalent to 8z1 : : :8zm(�0 ! B0(p)), which is a �rst-order statement in R.Now we consider the general case, when there is more than one predicate symbol.Then we eliminate the second-order quanti�ers one by one in the following way. SplitA0 in two parts, A1 and A2, so that A2 contains all and only occurrences of Pn:8z1 : : :8zm(�0 ! 8'1 : : :8'n(A0 ! B0))is equivalent to 8z1 : : :8zm(�0 ! 8'1 : : :8'n(A1 ^ A2 ! B0)), and this in turn to8z1 : : :8zm(�0 ! 8'1 : : :8'n�1(A1 ! 8'n(A2 ! B0))). And we apply the Intersec-tion Lemma and the Closure Lemma to 8'n(A2 ! B0)....This way all second-order quanti�ers which bind predicate symbols occurring both inthe antecedent and in the consequent can be eliminated.16

If B contains predicate symbols which are not in the antecedent, these can be replacedby a �xed contradiction having the same parameters as the original atomic formula;since B is positive, and therefore monotone, the resulting formula is equivalent to theoriginal one. Analogously, a predicate symbol occurring only in the antecedent canbe replaced by a tautology.Assume that A contains a predicate symbol which does not have a quanti�er pre�x,that is, A ! B can be written as A0 ^ P (�x) ! B(P (�x)). By assumption, the �x arefree in B. Since B is positive, B(P (�x)) can be equivalent to B0 ^ P (�x) or B0 _ P (�x).A0 ^ P (�x)! B0 _ P (�x)obviously corresponds to a �rst order condition, namely a trivial one, andA0 ^ P (�x)! B0 ^ P (�x)is equivalent to A0 ^ P (�x)! B0, the case which we treated above.Let �y be the result of applying steps 1 { 6 to �. We proved that if � is an axiom, thenin a canonical !-saturated model �y holds. The converse holds due to the argument usedin the proof of the Theorem 2: we used the same minimal substitutions as in that proof,hence a correspondent in the sense of completeness is a frame correspondent. 2From the proof of the theorem follows that for weak Sahlqvist formulas the correspon-dents are unique and equal to the frame correspondents. For weak Sahlqvist formulas whichdo not have (generalized) existential quanti�ers in the consequent, correspondents hold inall canonical models; the remaining weak Sahlqvist formulas have correspondents whichhold in all !-saturated canonical models.Example 4 The characteristic axiom of the "for almost all" quanti�er (the Fubini prop-erty): 2x2yP (x; y; �z) ! 2y2xP (x; y; �z), corresponds to the following condition on R:R(y; �z) ^R(x; y�z)! R(x; �z) ^R(y; x�z).Proof. Rewriting the validity conditions of the axiom gives8'(8x8y(R(x; �z) ^ R(y; x�z)! '(x; y; �z))! 8y8x(R(y; �z)^ R(x; y�z)! '(x; y; �z))which is equivalent to^f8y8x(R(y; �z) ^R(x; y�z)! '(x; y; �z)) : 8x8y(R(x; �z)^ R(y; x�z)! '(x; y; �z))g:By the Intersection Lemma,8y8x(R(y; �z)^ R(x; y�z)!^f'(x; y; �z)) : 8x8y(R(x; �z) ^R(y; x�z)! '(x; y; �z))g);while by the Closure Lemma^f'(x; y; �z)) : 8x8y(R(x; �z) ^R(y; x�z)! '(x; y; �z))g � R(x; �z) ^ R(y; x�z)in every canonical model. Thus we obtain the correspondent8y8x(R(y; �z) ^R(x; y�z)! R(x; �z) ^R(y; x�z)): 217

Example 5 The characteristic axiom of the "co-countably many" quanti�er (Keisler's ax-iom): 8x2yP (x; y; �z) ^2x8yP (x; y; �z)! 2y8xP (x; y; �z), corresponds to8x8y(R(y; �z)! R(y; x�z) _R(x; �z)):Proof. The axiom is valid i�8'(8x8y(R(y; x�z)! '(x; y; �z))^ 8x8y(R(x; �z)! '(x; y; �z))! 8x8y(R(y; �z)! '(x; y; �z));namely, 8'(8x8y(R(y; x�z) _R(x; �z))! '(x; y; �z))! 8x8y(R(y; �z)! '(x; y; �z)). This canbe rewritten as^f8x8y(R(y; �z)! '(x; y; �z)) : 8x8y(R(y; x�z)_ R(x; �z))! '(x; y; �z))g;by the Intersection Lemma,8x8y(R(y; �z)!^f'(x; y; �z)) : 8x8y(R(y; x�z)_ R(x; �z))! '(x; y; �z))g;and by the Closure Lemma, 8x8y(R(y; �z)! R(y; x�z) _R(x; �z)). 26 Non-existence of correspondents for completeness6.1 Restrictions imposed in the Sahlqvist theoremIn this section we show that not all formulas which have a frame correspondent also admit acorrespondent for completeness. First we prove this for the truth de�nition we are workingwith; later the result is generalized.Recall that the de�nition of Sahlqvist formulas (cf. Theorem 2) allowed existential and3-quanti�ers in the antecedent, while for weak Sahlqvist this is not allowed. The followingtheorem gives the reason why.Theorem 5 3x'! 3x('_ ) does not have a correspondent in the sense of completeness.Proof. Let us call 3x' ! 3x(' _ ) A. Although A does have a frame correspondent,namely 8x8�y8�z(R(x; �y) ! R(x; �y�z)), we show that it does not have a correspondent forcompleteness. Assume that Ay exists. The following axioms are consistent with A: 3xx = xand 8y:3yx = y. Hence, together with A they have a model where Ay holds. FromProposition 1 follows thatAy implies 8x8�y8�z(R(x; �y)! R(x; �y�z)). Since the correspondentsof the other two axioms (being their *-translations) hold in every model, we have a modelwhere 8x8�y8�z(R(x; �y)! R(x; �y�z)), 9xR(x) and 8x:R(x; x) hold simultaneously. But thisis impossible. 2An immediate consequence of the theorem is that "2 over ^"-combination in the an-tecedent cannot be allowed, since the axiom considered above can be written equivalentlyas 2x(' ^ )! 2x':Also, we haveCorollary 1 Extensionality, that is 8x(' � ) ! (2x' � 2x ) does not have a corre-spondent in the sense of completeness. 18

Proof. One can check that extensionality is consistent with 3xx = x and :3yx = y,and that extensionality implies 3x' ! 3x(' _ ). Hence if extensionality would have acorrespondent for completeness, it would imply the frame correspondent of 3x'! 3x('_ ). The rest of the argument is the same as above. 2Corollary 2 3x'! 2x' does not have a correspondent in the sense of completeness.Proof. The proof is analogous; the consistency can be shown by constructing a modelwhere D is in�nite, R(x; �y), Vi x 6= yi, and for each predicate Pn, V (Pn) = Dn. 2Corollary 3 9x(x 6= y ^ '(x; y))! 3x('(x; y)_ ) does not have a correspondent in thesense of completeness.Proof. Analogous. 2The clause of the Theorem 3 forbidding occurrences of the same predicate letter withdi�erent free variables is also necessary. Suppose that there is a variable in the an-tecedent not occurring in the consequent, as in 2y(2xP (x; y) ! 2zP (x; z)), is equivalentto 3y(2xP (x; y) ^ >(x)) ! 2zP (x; z), and 8y(2xP (x; y) ! 2zP (x; z)) is equivalent to9y2xP (x; y)! 2zP (x; z):This shows that the class of weak Sahlqvist formulas is strictly smaller than the class ofall Sahlqvist formulas and that none of the conditions of the Theorem 3 can be dropped.Digression. Re ection on the proofs of Theorems 2 and 3 suggests a closer look atthe behaviour of singleton sets, which are used as minimal substitutions in the proof of theTheorem 2 (and in modal logic), but do not occur as good minimal substitutions in theproof of the Theorem 3. Note that singletons as minimal substitutions are used precisely inthe cases ruled out in the Theorem 3. The semantical correlate of "singletons as minimalsubstitutions" is distinguishability. A model is called distinguishable if every element isuniquely determined by the set of formulas in one free variable which are true for thiselement. For example, canonical models for modal logic are distinguishable. But in general,our models will not be distinguishable. Suppose an !-saturated canonical model satis�es2xx 6= y and extensionality for 2x, and d is the unique element satisfying Vf'(x) : '(d)g.We show that 8x(R(x) ! x 6= d). In a distinguishable model this would hold for everyelement, hence R would be empty.Suppose 9x(R(x)^x = d), then for all ' such that '(d), 9x(R(x)^'(x)). Extensionalityimplies for every formula 9x(R(x)^ (x))! 9x(R(x; d)^ (x; d)):Hence 9x(R(x; d)^'(x)), and by !-saturation 9x(R(x; d)^V'(x)). It follows that R(d; d),a contradiction. 26.2 Other truth de�nitionsLet us �rst of all compare the results of this paper with the results previously obtained formodal logic. Recall that a modal logic L is called �rst order complete if there is a set � of�rst order sentences in the language fR;=g such that`L ' if and only if for every Kripke model M : if M j= � then M j= ';19

Sahlqvist's theorem tells us that modal logics axiomatized by Sahlqvist formulas, such asK4, S4, S5, etc. are �rst order complete. When we transfer this concept of �rst ordercompleteness to generalized quanti�ers (cf. De�nition 5), we see that there is an impor-tant quanti�er logic axiomatized by Sahlqvist formulas which is not �rst order complete,namely the "almost all" quanti�er axiomatized by L1 { L4 below plus 3x' ! 3x(' _ )(which is equivalent to L1 { L5 given the minimal logic). We do have �rst order com-pleteness for logics axiomatized by weak Sahlqvist formulas, but since extensionality is notweak Sahlqvist, this result is not very interesting taken by itself. Indeed, we can obtaineven stronger negative results, as follows. The reader may have observed that for general-ized quanti�er logics a stronger form of �rst order completeness can be de�ned: the truthde�nition we considered above is universal, but nothing prevents us from considering morecomplex truth de�nitions. Might we not obtain a �rst order correspondent to extensionalitythis way? E.g. the following truth de�nition studied in Jervell (1975), Mijajlovic (1985)and Krynicki (1990) trivially yields extensionality:2x'(x) if and only if 9y8x(R(x; y)! '(x)),where R is a new binary predicate and y does not occur free in '. But in this case Keisler'saxiom for "co-countably many" 8x2y' ^2x8y'! 2y8x'corresponds to a schema, not to a �rst order condition as above.This is not accidental. For instance, for the quanti�er "almost all" it can be shown thatany truth de�nition, however complex, involving an accessibility relation R, will make atleast one axiom correspond to a schema.In this subsection, we shall consider even more general truth de�nitions involving arelation R(x; Y ), where Y is a �nite subset of the domain. Consideration of this relationis natural, because one might argue that dependence really is a relation between objectsand �nite sets of objects (compair algebraic or linear dependence). Such truth de�nitions,where quanti�ers over �nite sets are allowed, will be called weak second order. The truthde�nition that we employed up till now can be expressed in the new language as follows:2x'(x; y1; : : : ; yn) ,, 9Y [8x(R(x; Y )! '(x; y1; : : : ; yn))^y1 2 Y ^ : : :^yn 2 Y ^8z 2 Y (z = y1_ : : :_z = yn)]For any given weak second order de�nition, we may now ask whether quanti�er axioms havecorrespondents for completeness in the language fR;2;=g.Theorem 6 For any weak second order truth de�nition, the conjunction of the followingformulas (the Friedman axioms, see Steinhorn (1985)) will not have a correspondent forcompleteness:L1 3xx = xL2 2xx 6= yL3 2x' ^ 2x ! 2x(' ^ )L4 2x2y'! 2y2x' 20

L5 8x('! )! (2x'! 2x )Proof. Choose a weak second order truth de�nition and suppose that the conjunction ofthe Friedman axioms has a correspondent for completeness. Then the following statementwould be true:(C) 9X9D9R(X = D<! ^R � D�X ^ 0R satis�es the correspondent for completenessfor the conjunction of the Friedman axioms0)The part in quotes is �0 since all second order quanti�ers range over X . Furthermore theformula X = D<! is �1, hence the entire statement is �1.From the Levy - Shoen�eld Absoluteness Lemma (cf. Jech (1978), p.120) it follows that(C) holds in the constructible universe L. We now show that this is a contradiction.Observe �rst that by using our truth de�nition in L, we may add a generalized quanti�erQ to the language f2;=g with standard interpretation; Q will satisfy the conjunction ofthe Friedman axioms for this language since R satis�es its correspondent for completeness.In L, de�ne a relation S by S(x; �y) =df ' Qx'(x; �y)! '(x; �y)and a new quanti�er 2 by 2x'(x; �y) =df 8x(S(x; �y) ! '(x; �y)). We show that we musthave 2x'(x; �y) � Qx'(x; �y). The direction from right to left is trivial. For the converse itsu�ces if Q is closed under countable intersections in L, for in that casea 2x'(x; �y) � 8x((V Qx (x; �y)! (x; �y))! '(x; �y))b V Qx (x; �y)! Qx (x; �y) � Qx(V Qx (x; �y)! (x; �y))whence (by L5) Qx'(x; �y).To show that Q is closed under countable intersections in L, argue as follows: let Q bede�ned on D (i.e. Q � P(D) and D 2 Q), and let (An)n2! be in L, such that An � D andfor all n, Qx(x 2 An). We want to show Qx(x 2 TnAn).De�ne a relation T (x; y) byT (x; y) =df 8n(x 2 Acn ! 8m(y 2 Am) _ 9m > n(y 2 Acm))_ 8n(x 2 An ^ y 2 D)Then 8xQyT (x; y): for if x 2 TnAn, then y 2 D, and if for some n, x 62 An, theny 2 Tm�n Am. Hence QxQyT (x; y) and by L4 QyQxT (x; y).If y 2 Acm, then fx : T (x; y)g= TnAn [Sn�mAcn; and if y 2 TmAm, thenfx : T (x; y)g = D. It follows that TmAm � fy : QxT (x; y)g. If we have equality, alsoQy(y 2 TmAm). Otherwise, let m be such that y 2 Acm and QxT (x; y). It follows thatQx(x 2 TnAn [ Sn�m Acn), whence Qx(x 2 TnAn).Hence 2x'(x; �y) � Qx'(x; �y), and it follows that 2 also satis�es the Friedman axioms.Applying the Sahlqvist algorithm to the Friedman axioms (note that we now have obtained acanonical relational model, and that the Friedman axioms do not have existential quanti�ersin the antecedent, hence their correspondents hold in every canonical model) shows that Ssatis�esa 9xS(x)b :S(x; x) 21

c S(x; �y�z)! S(x; �y)d S(x; y�z) ^ S(y; �z)! S(y; x�z)Furthermore, rewriting extensionality in terms of S shows that S satis�es, for every formula � which is a standard translation of a L82 formula 9x(S(x; �y)^ : �(x; �y))! 9x(S(x; �y�z) ^ : �(x; �y))that is (]) 8x(S(x; �y�z)! �(x; �y))! 8x(S(x; �y)! �(x; �y))We show that L does not have a de�nable well-ordering, which is a contradiction.De�ne P (x; �z) as S(x; �z)^ 8y(S(y; x�z)! x � y). Since 9�z8x:P (x; �z) implies (with theaxiom of Dependent Choice, which holds in L) the existence of an in�nite descending chain,we must have 8�z9xP (x; �z). Then we can choose x0 with S(x0) and 8y(S(y; x0)! x0 � y)and x1 with S(x1; x0) and 8y(S(y; x0x1)! x1 � y). Due to the property (]) the followingholds: 8y(S(y; x1)! x1 � y):S(x0)^S(x1; x0) implies together with d S(x0; x1), therefore x1 � x0. This yields x0 = x1,a contradiction with S(x1; x0) by b (cf. Theorem 1.5.3 in van Lambalgen (1994)). 27 ConclusionAs we write in the introduction, the aim of this paper is to study how implicit dependenciesbetween variables, imposed by generalized quanti�er axioms, can be made explicit.For this purpose, we de�ne relational models for generalized quanti�ers, where quan-ti�ers are interpreted using a dependency relation between variables, and show that someaxioms determine �rst order conditions on this relation. The situation is analogous to theone in modal logic, where some modal axioms determine �rst order properties of the accessi-bility relation (and some do not). We prove a Sahlqvist theorem for generalized quanti�erswhich de�nes a class of formulas having �rst order correspondents for completeness.However, some important formulas such as extensionality are proved not to have a cor-respondent for completeness. The latter result is strengthened as follows. For any truthde�nition which involves a dependency relation between objects and �nite sets of objects,the logic of the "almost all" quanti�er is not �rst order complete. In this sense, not all de-pendencies between variables can be made explicit, i.e. expressed in the language containingonly a dependency predicate. In a sequel to this paper (Alechina & van Lambalgen (1995))we shall prove that the property of having a correspondent for completeness is undecidable.8 ReferencesN. Alechina & M. van Lambalgen (1995). Generalized quanti�cation as substructurallogic. ILLC Report, Univ. of Amsterdam.J. Barwise & S. Feferman (1985). Model-theoretic logics. Springer Verlag.J.F.A.K. van Benthem & N. Alechina (1993). Modal quanti�cation over structured do-mains. ILLC Report ML-93-02, Univ. of Amsterdam. (To appear in M. de Rijke(ed.), Advances in intensional logic, Kluwer Academic Publishers 1995.)22

J.F.A.K. van Benthem (1983). Modal logic and classical logic. Bibliopolis, Napoli.J.F.A.K. van Benthem (1984). Modal correspondence theory. In: D. Gabbay and F.Guenthner (eds.), Handbook of Philosophical Logic, v.2, Reidel, 167-247.W. Hodges (1985). Building models by games. Cambridge Univ. Press.W. Hodges (1993). Model theory. Cambridge Univ. Press.K. Fine (1985). Natural deduction and arbitrary objects. J. Phil. Logic, 14(1), 57 { 107.T. Jech (1978). Set theory. Academic Press.H.R. Jervell (1975). Conservative end extensions and the quanti�er "the exist uncount-ably many". In S. Kanger (ed.), Proc. of the third Scandinavian logic symposium,North Holland, Amsterdam 63 { 80.H.J. Keisler (1970). Logic with the quanti�er "the exist uncountably many". Ann. Math.Logic, 1, 1 { 93.M. Krynicki (1990). Quanti�ers determined by partial orderings. Zeitschr. f. Math.Logik und Grund. d. Math. 36, 79 { 86.M. van Lambalgen (1990). The axiomatization of randomness. J. Symb. Logic, 55,1143{1167.M. van Lambalgen (1991). Natural deduction for generalized quanti�ers. In J. van derDoes & J. van Eijck (eds.), Generalized Quanti�er Theory and Applications, DutchNetwork for Language, Logic and Information, Amsterdam, 143 { 154. (To appear inCSLI Lecture Notes, Univ. of Chicago Press 1995.)M. van Lambalgen (1992). Independence, randomness and the axiom of choice. J. Symb.Logic 57, 1274{1304.M. van Lambalgen (1994). Independence structures in set theory. ILLC Report ML-94-04, Univ. of Amsterdam. To appear in Proceedings Logic Colloquium '93.Z. Mijajlovic (1985). On the de�nability of the quanti�er "there exist uncountablymany". Studia Logica 44, 257 { 264.A. Mostowski (1957). On a generalization of quanti�ers. Fundamenta Mathematicae 44,12 { 36.M. Mostowski (1991). Branched quanti�ers. Rozprawy Uniwersytetu Warszawskiego,Dzial Wydawnictw Filii UW w Bialymstoku.H. Sahlqvist (1975). Completeness and correspondence in the �rst and second ordersemantics for modal logic. In S. Kanger (ed.), Proc. of the third Scandinavian logicsymposium, North Holland, Amsterdam, 110 { 143.G.Sambin & V.Vaccaro (1989). A new proof of Sahlqvist's theorem on modal de�nabilityand completeness. J. Symb. Logic 54, 992{999.C.I. Steinhorn (1985). Borel structures and measure and category logics. In Barwise &Feferman (1985), 579 { 596. 23

R.J.G.B. de Queiroz & D.M. Gabbay (1991). The functional interpretation of the exis-tential quanti�er. Tech. Report, Department of Computing, Imperial College. Pre-sented at Logic Colloquium '91 , Uppsala, August 9{16 1991. Abstract J. Symb. Logic58(2), pp.753{754.

24