Correspondence and Completeness for Generalized Quanti ers. 1
Natasha Alechina Michiel van Lambalgen
[email protected]
[email protected]
Department of Mathematics and Computer Science University of Amsterdam Plantage Muidergracht 24 1018 TV Amsterdam
1 Introduction
A generalized quanti er Q as de ned by Mostowski (1957) is a class of subsets of the universe, so that a model M satis es Qx'(x; d) if the set of elements fe : M j= '[e; d]g is in Q. Examples are: the ordinary existential quanti er (interpreted as the set of all non-empty subsets of the universe); the quanti er "there are exactly 2"; a lter quanti er (where the only requirement on Q is that it is a lter), "there are uncountably many" (where the domain is uncountable, and Q contains all uncountable subsets), etc. An obvious property of these quanti ers is extensionality: if two formulas ' and are satis ed by the same sets of elements, Qx' is satis ed if and only if Qx is. Actually Mostowski also required that the generalized quanti er is invariant under permutations of the universe, thus restricting attention to quanti ers related to cardinality. Subsequently, other generalized quanti ers were considered which do not have the property of permutation invariance, such as topological quanti ers or measure quanti ers. For an overview of the subject, one may consult 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 studied mostly 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 and even 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 one could give a proof theory, it would not enjoy such pleasant properties as for instance cut elimination. Nonetheless, in van Lambalgen (1991) an attempt was made to develop Gentzen style calculi for the lter quanti ers "almost all" (due to H. Friedman; cf. Steinhorn (1985)) and "co-countably many". This research was supported by the Netherlands Organization for Scienti c Research (NWO) under grant PGS 22-262. 2 Although one can sometimes give a proof theory by altering the semantics; cf. M. Mostowski (1991). 1
1
The guiding intuition behind this attempt is the idea that one can set up a proof system 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 in Fine'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 have 8yQx(x 6= y) but not Qx8y(x 6= y); the failure of commutation seems to indicate that x in some way depends on y , as it would if we were considering the quanti er combination 8y9x. So we think of axioms for quanti ers such as QxQy' ! QyQx' (characteristic for "almost all") or 8yQx' ^ Qy 8x' ! Qx8y' (characteristic for "co-countably many") as implicitely determining a dependency pattern between variables; one may now ask whether this 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 dependence can be taken to be functional, and the required rst order description is given by the axioms for Skolem functions. Indeed, as Fine shows, one can also give a graphical representation of these dependencies. Somewhat surprisingly, the axioms mentioned do indeed determine a rst order condition on 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 this phenomenon, both its extent and its limits, in greater detail. As a rst step, we introduce an analogue of the expansion of a language by Skolem functions. Consider a language L82 which extends a rst-order language with equality by introducing a unary generalized quanti er 2x . We use this notation to emphasize an analogy between generalized quanti ers and modal operators, that will become apparent below. The dual of 2x is 3x' =df :2x :'. (In the examples above a lter quanti er would correspond to 2.) De nition 1 Let R(x; y) be a relation of inde nite arity. The intuitive interpretation of R(x; y) is "x depends on y". The standard translation : L82 ?! L(R) is de ned inductively as follows: P (x1 ; : : :; xn) = P (x1; : : :; xn); (: ) = : ; ( 1 ^ 2 ) = 1 ^ 2; (8x ) = 8x (2x (x; y)) = 8x(R(x; y) ! (x; y)). 2 In other words, in the formula 2x (x; y) the bound variable x depends on y in a way determined 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 a conservative 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 consistent with the minimal logic 2
De nition 2 The minimal logic Lmin for L82 is the smallest class of formulas closed with
respect 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 '. 2 In order to talk about axioms (that is, formulas) and not schemata, we introduce the substitution rule ` (P (x1; : : :; xn)) ` ('(x1; : : :; xn)); provided P (x) and '(x) have precisely the same free variables. This restriction is necessary due 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 a conservative 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 = , where D is a nonempty 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 a variable 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 as follows: ^ R(x; y) = 2x'(x; y) ! '(x; y): '(x;y)2L82
The existence of T 0 satisfying ([) now follows from
2
Theorem 1 Every Lmin -consistent set of L82 formulas has a Henkin canonical relational model.
Proof can be found in van Benthem & Alechina (1993).
2
Clearly, then, this way of making dependency patterns explicit will not work for all unary quanti ers; for quanti ers which are not (the dual of) a lter quanti er a translation more 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 language L(R); the main question then becomes: when can this schema be replaced by a rst-order condition on R? The reader will have observed that both the set-up and the main problem are very much analogous 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 for which 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. 2 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 (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 can be reformulated for the generalized quanti ers as follows De nition 5 If L is a generalized quanti er logic, then L is rst order complete if there is a 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= '. 2 Note 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 completeness y A 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 L is canonical if it is rst order complete and the corresponding set of rst order sentences holds in the canonical model for L. For generalized quanti ers, there is no unique canonical model. 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 order complete and the corresponding set of rst order sentences holds in every ! -saturated canonical relational model of L. 2 This 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 and Alechina (1993) is not quite adequate for our purposes, since it is based on the notion of frame correspondence: De nition 7 If A is a quanti er axiom, a frame correspondent is a rst order condition A? on R such that j= A? if and only if for any interpretation V , j= A. 2 It 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 a correspondent for completeness. However, the following holds: Proposition 1 If A has a correspondent for completeness Ay and a frame correspondent A?, then `FOL Ay ! A?. Proof. Let < D; R> j= Ay . Then by the De nition 4 (ii), for every interpretation V j= A. By the De nition 7, j= A? . 2 We 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 the proof in the sequel. 4
Theorem 2 (Correspondence part of the Sahlqvist theorem). All formulas of the "Sahlqvist form" ' ! , where 1. ' is constructed from
atomic formulas, possibly pre xed by 2x, 8; formulas in which predicate letters occur only negatively using ^; _; 3x; 9 2. in
all predicate letters occur only positively
have a frame correspondent.
3
We 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-order formula without P1; : : :; Pn , then de nes a rst-order condition on frames. The proof of the 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 proving that the following formula has a rst-order equivalent:
^ ^ ^ 8P1 : : : 8Pn8u( Rj ^ 8x(Ri ! Pi ) ^ Pk ! ) j
i
k
V where j Rj Vcorresponds to the translation of the truth conditions for 3-quanti ers in the antecedent, i 8x(Ri ! Pi ) corresponds to the translations of occurrences of predicate V symbols preceded by 2-quanti ers, k Pk corresponds to occurrences of predicate symbols not preceded by 2's, and is a positive formula. For example, assume that = 3x(2y P (x; y ) ^ S (x)) ! 2z S (z ). Then its translation reads
8P 8S 8x(R(x) ^ 8y(R(y; x) ! P (x; y)) ^ S (x) ! 8z(R(z) ! S (z))): V V V Here j Ri = R(x), j 8x(Ri ! Pi ) = 8y (R(y; x) ! P (x; y )), and k Pk = S (x).
Note that we quantify over all possible assignments to the predicate symbols, so a second order quanti er 8Pin can be instantiated using any suitable set of n-tuples. That is what we are going to do. If Pi occurs in the antecedent in a subformula of the form 8x(Ri ! Pi (x)), the minimal substitution for this occurrence of Pi is precisely the set of those tuples for which Ri holds. (In the running V example, P (u1; u2) := R(u2; u1) ^ u1 = x.) If Pi occurs as a member of conjunction k Pk , then the minimal substitution for this occurrence is a singleton set: in the example, S (u) := u = x. Finally, the minimal substitution for Pi is the disjunction of minimal substitutions for its occurrences. (Precise de nitions and details are given in van Benthem and Alechina (1993)). Denote the result of the substitution 0 . (In the example, 0 is
8x(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)).) This theorem can be made slightly more general by allowing an arbitrary long pre x of 2-quanti ers before the formula, constant formulas in the antecedent(formulas which contain only =, > and ? as predicate symbols), and also stating that a conjunction of Sahlqvist formulas is a Sahlqvist formula. 3
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0 follows from 8P 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 were 0 chosen V R , so that the antecedent of becomes trivially true except for the rst-order part j j ^ ^ ^ 0 = 8u( Rj ^ 8x(Ri ! Ri) ^ (w = w) ! (p1 ; : : :; pn)); j i V 0 that is = 8u( j Rj ! (p1; : : :; pn )). The consequent (p1 ; : : :; pn) is a rst-order formula. Assume that 0 is true and there is a valuation (an assignment to predicate symbols) V which makes the antecedent of true: ^ ^ ^ j= Rj ^ 8x(Ri ! Pi) ^ Pk i
j
^ j= Rj ! (p1 ; : : :; pn)):
k
j
We want to show that under this valuation also the consequent of , (P1 ; : : :; Pn ), is true. First of all, observe that (p1; : : :; pn ) is true. The substitutions pi which we used are minimal in the sense that always when the antecedent of is true under some assignment of 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= 8d(pi(d) ! Pi (d)). (P1 ; : : :; Pn) follows from (p1 ; : : :; pn ) by
monotonicity of . 2 A disadvantage of the notion of frame correspondence is that there may be consistent quanti er logics which are frame incomplete, i.e. logics L such that for no D; R: for all V , 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 mainly due 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-trivial here. Actually, inspection of the proofs below shows that this result can be strengthened: the troublesome property is extensionality, that is, 8x(' ) ! (2x ' 2x ). This is interesting in view of the fact that extensionality is generally taken to be the sine qua non for logicality of a quanti er.
2 Statement of the result and idea of the proof We now formulate the completeness part of the Sahlqvist theorem, which describes a class of formulas ' de ning rst-order conditions on R so that for any logic L in the language of L82 which has a canonical relational model, L with ' as an axiom is complete for the class of models where R has the rst order property corresponding to '. This class is strictly smaller than the class of formulas described above. We shall call those formulas weak Sahlqvist formulas.
Theorem 3V(Completeness part of the Sahlqvist theorem) All "weak Sahlqvist formulas" of the form Qz1 : : :Qzn (A ! B ), where n 0, each Q is either 8 or 2, and 1. A is constructed from
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a. atomic formulas, possibly with a quanti er pre x Qx1 : : :Qxk , where each Q is a 2- 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 positively 3. every occurrence of a predicate letter has the same free variables 4 have a correspondent in the sense of completeness; moreover, this correspondent holds in every ! -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 not satisfy one of them, then it need not have a correspondent for completeness. The conditions which 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 antecedent and all occurrences of a predicate symbol have the same free variables. The necessity of those 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 S 2xP (x; y) ! 2x(P (x; y) _ S (x; z)) is valid in C , then the accessibility relation in C has the property R(x; yz) ! R(x; y). Proof. It is easy to see (since the consequent is monotone in S ) that the axiom above is equivalent to 2x P (x; y) ! 2x (P (x; y) _?(x; z)). We translate the validity conditions using second-order quanti ers which range only over de nable relations of C (this is the di erence
with the case of the frame correspondence). To emphasize this di erence we use quanti ers 8'. 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. Using this notation the validity condition of the axiom reads as follows:
8'[8x(R(x; y) ! '(x; y)) ! 8x(R(x; yz) ! '(x; y) _ ?(x; z))] This is equivalent to 8'[8x(R(x; y) ! '(x; y)) ! 8x(R(x; yz) ! '(x; y))], and, in turn, V to '(x;y) f8x(R(x; yz) ! '(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 the Intersection Lemma; in the given case the proof is obvious), we obtain ^ 8x(R(x; yz) ! f'(x; y)) : 8x(R(x; y) ! '(x; y))g) '(x;y)
Actually, the sucient condition is that all positive occurrences of a predicate letter in the antecedent have the same free variables, say, y, all negative occurrences of a predicate letter in the antecedent and all occurrences in the consequent have the same free variables, say, z, and y z. It can be shown that in this case the condition 3 can be forced to hold. But we assume that this holds from the very beginning to make the formulation of the theorem readable. 5 Henceforth, "canonical model" will always mean "canonical relational model". 4
7
V
But in C '(x;y) f'(x; y)) : 8x(R(x; y) ! '(x; y))g) R(x; y). Substituting R(x; y) instead of the in nite conjunction yields the rst-order equivalent 8x(R(x; yz) ! R(x; y)). It is easy to check that in every model where 8x(R(x; yz) ! R(x; y)) holds, the axiom is valid. 2 The general case is slightly more complicated because to obtain a correspondent we must sometimes move to a di erent canonical model, namely, to an ! -saturated canonical model. The existence of such model for every Lmin -consistent set of sentences is proved in section 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 same free variables, closed with respect to ^, then in an ! -saturated model
^
^ fB(') : ' 2 X g B( f' : ' 2 X g):
The proof of the Theorem 3 consists of the same three ingredients as those in the example: translation of the validity conditions of an axiom (eventually accompanied by some syntactic transformations), application of the Intersection Lemma, and making use of the fact that for some rst-order expression R with R as the only predicate symbol
R(x; y)
^
'(x;y)
f'(x; y) : 8x(R(x; y) ! '(x; y))g:
Of course, R is nothing else than a "minimal substitution", but of a special kind, which will be formally de ned below.
De nition 8 Let M be a canonical model, and A a conjunction of atomic formulas which
are pre xed by universal and 2-quanti ers, so that all occurrences of a predicate symbol have the same free variables. Every occurrence of a predicate symbol P in A is therefore of the form Q i xP (x; y), where Q i is the V quanti erVpre x of the ith occurrence. P has a good minimal substitution in A if M j= f'(x; y) : i Q i x'(x; y)g p, where p is a rst-order formula built using the predicates R, =, > and ? only. 2 For 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 form 2xP (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: V IfV M j= 9xR(x; y ), then M j= f'(y ) : 9xR(x; y ) ! '(y )g >(y ). If not, then M j= f'(y) : 9xR(x; y) ! '(y)g ?(y). This means that axioms with vacuous 2's in the
antecedent can have di erent correspondents in di erent models. Henceforth we assume that all quanti ers are non-vacuous. Formally, this corresponds to assuming that R is always non-empty, or that the axiom 3x>(x; y) holds in all canonical models. This is an innocuous assumption, because if R is empty, every L82-formula is equivalent 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 this is no loss of generality can be seen as follows. Let P occur in A with a purely universal pre x, 8x1 : : : 8xn P (x; y). Then this occurrence implies all other possible occurrences of P in A, and A is equivalent to a conjunction where 8x1 : : : 8xn P (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 2quanti ers, so that all occurrences of a predicate symbol in A have the same free variables. Then every atomic formula in A has a good minimal substitution, and this minimal 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 n
there is an element an in the domain of M such that '1(an ); : : :; 'n(an ) is true in M . If M is just an ordinary Henkin model, there does not necessarily exist an elementV a such that for every ' in X '(aV) is true in M . Among other things this implies that 3x f' : ' 2 X g is not equivalent to f3x' : ' 2 X g in M . But in the proof we do need that
^ ^ M j= 3x f' : ' 2 X g f3x' : ' 2 X g
(an analogue of Esakia's lemma). We therefore move from the original Henkin model to a mildly saturated extension.
Theorem 4 Every consistent set of L82 formulas has a model A which is !-saturated and canonical, that is
i 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 consistent set of formulas has a canonical model C where ^ RC (x; y) 2x'(x; y) ! '(x; y) '(x;y)2L82
By the truth de nition
C 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) such that if 2 L82 C j= , C j=
where is the standard translation of .
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We shall use this fact to build the saturated model which we need, because one can apply the standard procedure of constructing an ! -saturated extension of C . (While extending a model for a generalized quanti er is much more dicult, see for example Hodges (1985).) Take an ! -saturated elementary extension A of C . It is clear that
C 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 in A . But A is still a rst-order model; to make an L82 model A out of it, we could take the interpretation of R in A to be the accessibility relation in A, i.e. stipulate
A 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 as ^ RA(x; y) = 8x(R(x; y) ! '(x; y)) ! '(x; y): ' (x;y):'(x;y)2L82
Note that the intersection is only over the formulas ' (x; y) such that '(x; y) 2 L82 . We are done if we can show that Lemma 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, by the 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)). Let R(x; y) hold in A. We want to show that A j= (x; y). It is enough to show that R(x; y) implies RA(x; y). If this is so, we obtain (x; y) from R(x; y) and the fact that A 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 Vprove that for all , R(x; y) ! (8x(R(x; y) ! (x; y)) ! (x; y)). Therefore R(x; y) ! (8x(R(x; y) ! (x; y)) ! (x; y)), which means that R(x; y) implies RA (x; y). 2 Comment. C is an elementary extension of A with respect to L82 formulas, but not necessarily 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 formulas with the same free variables, closed with respect to ^, then in an ! -saturated model ^ ^ fB(') : ' 2 X g B( f' : ' 2 X g) 10
Proof By induction on the complexity of B. B(') = ': trivial, because Vf' : ' 2 X g = Vf' : ' 2 X g. 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 X g = fB1(') : ' 2 X g ^ fB2(') : ' 2 X g V (by theVassociativity of ) Vand, by the inductive hypothesis, the latter is equivalent to B1 ( f' : ' 2 X g) ^ B2 ( f' : ' 2 X g). Let B = B1 _ B2. V V Assume M; e j= B1 ( f' : ' 2 XVg) _ B2 ( f' : ' 2 X g). Then one of the disjuncts isV true, for example M; e j= B1 ( f'V: ' 2 X g). By the inductive hypothesis, M; e j= fB1(') : ' 2 X g. Clearly, M; e j= fB1(') _ B2(') : ' 2 X g. For the other direction, assume M; e 6j= B1(Vf' : ' 2 X g) _VB2 (Vf' : ' 2 X g). V This means that M; e 6j= B1 ( f'V: ' 2 X g) and M; e 6j= B2 ( fV' : ' 2 X g). By the inductive hypothesis, M; e 6j= fB1 (') : ' 2 X g and M; e 6j= fB2 (') : ' 2 X g. 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).VNote that X is closed under ^, therefore '1 ^ '2 2 X . But this means that M; e 6j= fB1 (') _ B2 (') : ' 2 X g. Let BV= 8xB1. Vf8xB1(') : ' 2 X g = 8x VfB1(') : ' 2 X g (because V 8 distributes over ), and by the inductive hypothesis this is equivalent to 8xB1 ( f' : ' 2 X g). Let B = 2xB1. This case is analogous, but since 2 distributes only over conjunctions of formulas with the same free variables, it is important that all formulas in X (and therefore in fB1(') : ' 2 X g) have the same free variables. Let B = 3xB1. This is the only non-trivial part, and here we need the fact that the model is ! -saturated. First of all, it would be convenient if the set Y = fB1 (') : ' 2 X g 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 X g. For every formula in Y , there in Y 0 which implies it (due to the monotonicity of B1 ). Therefore V Y 0 !is aVformula On the otherVhand, Y 0 Y (because X is closed under ^). Therefore V Y ! V YY0.. Analogously V f3x : 2 Y g = f3x : 2 Y 0g. So, it is suces to prove that ^ ^ f3x : 2 Y 0g = 3x f : 2 Y 0g V V Since f V: 2 Y 0 g implies for any 2 Y 0 , 3x f : 2 Y 0g implies 3x and therefore f3x : 2 Y 0 g. This proves the implication right to left. For the other direction, assume that the model satis es 3x for every 2 Y 0 . Due to the de nition of Y 0 and monotonicity of B1 , this means that for every n the model
Monotonicity in L82 is restricted to the formulas with the same free variables, but this is always the case in this proof. 6
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 Vthe model is ! -saturated, thereV is an element which satis es the whole set: R(u; e) ^ f (u; e) : 2 Y 0 g. Then 3x f : 2 Y 0g is true. Let B = 9xB1 : the proof is analogous to the previous case. 2
4 Closure Lemma Let A be a conjunction as in the condition of the Closure Lemma. We also assume that all quanti ers are non-vacuous and that every quanti er pre x contains at least one 2quanti er (cf. the end of section 2). Let A0 be the subformula of A which contains all and only of the predicate V Qoccurrences 0 symbol P . We shall use both the L -form of A , namely x P ( x ; y ), and its standard 82 i i V translation i 8x(Ri ! P (x; y)), where i runs over the occurrences of P . In the sequel W we call the Ri R-conditions. The standard translation of A0 is thus equivalent to 8x( i Ri ! P (x; y)). P (x; y) has a good minimal substitution in A if
_ i
^ _ Ri = f'(x; y) : 8x( Ri ! '(x; y))g: i
Note that good minimal substitutions are the same expressions which were used in the proof of Theorem 2 as minimal substitutions for occurrences of predicate symbols in the scope of 2- and 8-quanti ers.
Example 2 The R-condition corresponding to 2x2y P (x; y) is R(x) ^ R(y; x). Example 3 Let A0 = 8x2y P (x; y) ^ 2x8yP (x; y), then A0 = 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 for P in A must be therefore R(y; x) _ R(x). We are going to prove the existence of good minimal substitutions for all non-vacuous quanti er pre xes containing at least one 2. From this and from the considerations at the end of the section 2 the Closure Lemma will follow. But rst we need several propositions.
Proposition 2 Let R be an R-condition, such that ^ R(x; y) f'(x; y) : 8x(R(x; y) ! '(x; y))g; then and vice versa: if then
^ R(x; y) f (x; yz) : 8x8z(R(x; y) ! (x; yz))g; ^ R(x; y) f (x; yz) : 8x8z(R(x; y) ! (x; yz))g; ^ R(x; y) f'(x; y) : 8x(R(x; y) ! '(x; y))g: 12
Proof. Assume R(x; y) Vf'(x; y) : 8x(R(x; y) ! '(x; y))g. For every '(x; y) holds: '(x; y) 8z('(x; y) ^ >(z)). Therefore ^ R(x; y) f8z('(x; y) ^ >(z)) : 8x8z(R(x; y) ! ('(x; y) ^ >(z))g: Since for every ', 8z('(x; y) ^ >(z )) '(x; y) ^ >(z ), ^ R(x; y) f'(x; y) ^ >(z) : 8x8z(R(x; y) ! ('(x; y) ^ >(z))g: V Now we prove that R(x; y) f (x; yz) : 8x8z(R(x; y) ! (x; yz))g. Trivially, ^ ^ f (x; yz) : 8x8z(R(x; y) ! (x; yz)g ! f'(x; y)^>(z) : 8x8z(R(x; y) ! ('(x; y)^>(z))g V and this implies thatV f (x; yz) : 8x8z(R(x; y) ! (x; yz))g ! R(x; y). V Since R(x; y) ! f (x; yz) : 8x8z(R(x; y) ! (x; yz))g, we have R(x; y) f (x; yz) : 8x8z(R(x; y) ! (x; yz))g. For the other direction of the proposition, let
^ R(x; y) f (x; yz) : 8x8z(R(x; y) ! (x; yz))g: V It is easy to check that R(x; y) f8z (x; y; z) : 8x(R(x; y) ! 8z (x; y; z))g, and then the
reasoning goes as above: the set of ''s with free variables x; y satisfying the same condition is larger than the set of 8z (x; y; z), therefore its conjunction implies the given one; on the other hand, the set of ''s satisfying the condition is implied by R, therefore the two sets are equivalent. 2 To 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; xy) Vf'(z; x; y) : 8z8x(R(z; xy) ! '(z; x; y))g. Proof.V The direction from left to right is trivial. For the converse direction, we have to prove f'(z; x; y) : 8z 8x(R(z; xy) ! '(z; x; y))g ! R(z; xy), in other words,
^ '
^
(8x2z '(z; x; y) ! '(z; x; y)) ! (2z (z; x; y) ! (z; x; y)):
V
V
It suces to derive 2z (z; e; y) ! (d; e; y) from ' 8x2z '(z; x; y) ! '(d; e; y). Take an arbitrary (z; x; y). We substitute this formula for in the tautology above:
8x2z (2z (z; x; y) ! (z; x; y)): V We assume that the conjunction ' 8x2z '(z; x; y) ! '(d; e; y) holds. As a special case we obtain
8x2z (2z (z; x; y) ! (z; x; y)) ! (2z (z; e; y) ! (d; e; y)): V Since this holds for every , we can derive 2 (z; e; y) ! (d; e; y). z
2
Proposition 4 If Qx1; : : :; Qxn contains at least one 2-quanti er, and all quanti ers are
non-vacuous, and the only occurrence of P (x; y) in A is of the form Qx1 : : :Qxn P (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, is 8(u)12z1 8(u)22z2 : : : 8(u)k 2z 8(u)k+1P (x; y); k
where k > 0 (that is, there is at least one 2 in the pre x), and uz = x. The standard translation of this formula is
8x(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 i^ =k ~ )i ; y); R R(zi; (uz i=1
where (uz ~ )i are the variables bound by the quanti ers preceding 2z , is a good minimal substitution for P . By propositions 2 and 3 (observe that zi (uz ~ )i x), i
^ R(zi ; (uz ~ )i ; y) f'(x; y) : 8x(R(zi; (uz ~ )i ; y) ! '(x; y))g Note that here we essentially use the fact that the zi occur in P (x; y), that is, that 2quanti ers are non-vacuous. V It is easy to see that R(z ; (uz ~ ) ; y) ! Vf'(x; y) : 8x(V R(z ; (uz ~ ) ; y) ! '(x; y))g. i
i
i
i
To prove the other direction, namely
^
i
i
^ ^ f'(x; y) : 8x( R(zi; (uz ~ )i ; y) ! '(x; y))g ! R(zi ; (uz ~ )i ; y); i
i
we argue as before:
^ f'(x; y) : 8x(R(zi; (uz ~ )i ; y) ! '(x; y))g f'(x; y) : 8x( R(zi; (uz ~ )i ; y) ! '(x; y))g i
therefore
^ ^ ^ f'(x; y) : 8x( R(zi; (uz ~ )i ; y) ! '(x; y))g ! f'(x; y) : 8x(R(zi ; (uz ~ )i ; y) ! '(x; y))g i
V
V
and this means that f'(x; y) : 8x( i R(zi ; (uz ~ )i ; y) ! '(x; y))g ! R(zi; (uz ~ )i ; y). Since this holds for every i,
^
^ ^ ~ )i ; y); ~ )i ; y) ! '(x; y))g ! R(zi ; (uz f'(x; y) : 8x( R(zi; (uz i
i
that is, R is a good minimal substitution.
2
Proposition 5 A disjunction of good minimal is a good minimal substitution, Vf'(xsubstitutions i.e. if for every i , 1 i n , R ( z ; y ) = ; y ) : 8 x ( R ( z W R (z ; y) = Vf'(x; y) : 8x(W iRi(z ; y) ! '(x; y))g. i i; y) ! '(x; y))g, zi x, then i i i i i i Proof. Since for every Ri _ f'(x; y) : 8x( Ri(zi; y) ! '(x; y))g f'(x; y) : 8x(Ri(zi; y) ! '(x; y))g; i
14
^
^ _ f'(x; y) : 8x(Ri(zi; y) ! '(x; y))g ! f'(x; y) : 8x( Ri(zi; y) ! '(x; y))g; i V W Ri, RWi ! f'(x; y) : 8x( i Ri(zi; y) ! '(x; y))g, and this implies WthatR is,! for Vf'every ( x ; y ) : 8x( i Ri (zi ; y) ! '(x; y))g. i i Now we prove that the implication also in the other direction. From Proposition 2 Vf'(x; y) :holds follows that if R ( z ; y ) = 8 x ( R zi; y) ! '(x; y))g, zi x, then Ri(zi; y) = i i i( Vf (z ; y) : 8z (R (z ; y) ! (z ; y))g. i i i iV i W Now, assume that f'(x; y) : 8x( i Ri (zi ; y) ! '(x; y))g holds and none of the Ri(zi ; y) holds. Then as we have just seen, there are formulas 1 ; : : :; n , such that for every i, 8z (R (z ; y) ! (z ; y)) and : (z ; y). Take the disjunction of these formulas, W (z ; y). i
i i
i i
i i
i i i
The formula is also false. An equivalent Wformula with free variables x; y, namely, W resulting ( z ; y ) _? (x), belongs to the set f'(x; y) : 8x( i Ri (zi ; y) ! '(x; y))g; but this formula i i i is false, a contradiction. 2
Lemma 3 (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 free variables. Then every atomic formula in A has a good minimal substitution, and this minimal substitution is the same as the one used to obtain a frame correspondent.
Proof. The lemma follows from the four propositions proved above.
2
5 Syntactic transformations We now describe the syntactic transformations which reduce the task of nding a correspondent 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 form Qz1 : : :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 , itself corresponds to a rst-order condition on R (namely, a conjunction of y ). So, it suces to prove that every i corresponds to a rst-order condition on R. Step 2. Writing down the validity conditions of i , we obtain
i
8'1 : : : 8'nQz1 : : :Qzm(A ! B); where 8'j are the quanti ers over de nable relations corresponding to the predicate symbols in i . If Pj is an n-place predicate symbol, then 8'j can be instantiated on any formula with precisely n variable places. Assume that m > 0 (if not, we can move to the next step). Then translating the truth conditions for the quanti ers in rst-order logic, we obtain
8'1 : : : 8'n 8z1 : : : 8zm (? ! (A ! B)); where ? is a conjunction of R-conditions corresponding to the 2-quanti ers in the pre x (if there are such quanti ers). This is equivalent to
8z1 : : : 8zm(? ! 8'1 : : : 8'n (A ! B)): 15
W Step 3. A may contain disjunctions; we use the fact that 8'1 : : : 8'n ( i AVi ! B ) is equivV alent to 8'1 : : : 8'n i (Ai ! B ), and since 8 distributes over ^, to i 8'1 : : : 8'n (Ai ! B ). Substituting this in the formula obtained on the previous step and applying the
same reasoning (now with conjunction in the consequent), we obtain
^ i
8z1 : : : 8zn (? ! 8'1 : : : 8'n (Ai ! B)):
It now suces to prove that each conjunct corresponds to a rst-order condition on R 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 ! B 0 )); where ?0 is rst-order, B 0 is a positive rst-order formula, and A0 is a conjunction of formulas (Qx1 : : :Qxk 'i ), where all quanti ers are non-vacuous. Assume that there is only one predicate letter P in . Then the reasoning goes as follows: P occurs in A0 in subformulas of the form (Qx1 : : :Qxk P (x; y)), where the x are bound and the y free (rename the bound variables V if necessary). The condition V 0 0 (*) can be rewritten as 8z1 : : : 8zm (? ! fB ('(x; y)) : i (Q i x'(x; y))g): V V Applying the Intersection Lemma, 8z1 : : : 8zm (?0 ! B 0 ( f'(x; y) : i (Q i x'(x; y))g)), and by the Closure Lemma (P has a good minimal substitution in A0 , say p), this is equivalent to 8z1 : : : 8zm (?0 ! B 0 (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. Split A0 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 ! B 0 )), and this in turn to 8z1 : : : 8zm (?0 ! 8'1 : : : 8'n?1 (A1 ! 8'n(A2 ! B0 ))). And we apply the Intersection Lemma and the Closure Lemma to 8'n (A2 ! B 0 ). .. . This way all second-order quanti ers which bind predicate symbols occurring both in the antecedent and in the consequent can be eliminated. 16
If B contains predicate symbols which are not in the antecedent, these can be replaced by 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 the original one. Analogously, a predicate symbol occurring only in the antecedent can be 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 are free in B . Since B is positive, B (P (x)) can be equivalent to B 0 ^ P (x) or B 0 _ P (x).
A0 ^ P (x) ! B 0 _ P (x) obviously corresponds to a rst order condition, namely a trivial one, and
A0 ^ P (x) ! B 0 ^ P (x) is equivalent to A0 ^ P (x) ! B 0 , the case which we treated above. Let y be the result of applying steps 1 { 6 to . We proved that if is an axiom, then in a canonical ! -saturated model y holds. The converse holds due to the argument used in 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. 2 From the proof of the theorem follows that for weak Sahlqvist formulas the correspondents are unique and equal to the frame correspondents. For weak Sahlqvist formulas which do not have (generalized) existential quanti ers in the consequent, correspondents hold in all canonical models; the remaining weak Sahlqvist formulas have correspondents which hold in all ! -saturated canonical models.
Example 4 The characteristic axiom of the "for almost all" quanti er (the Fubini property): 2x 2y P (x; y; z) ! 2y 2x P (x; y; z), corresponds to the following condition on R: R(y; z) ^ R(x; yz) ! R(x; z) ^ R(y; xz). Proof. Rewriting the validity conditions of the axiom gives 8'(8x8y(R(x; z) ^ R(y; xz) ! '(x; y; z)) ! 8y8x(R(y; z) ^ R(x; yz) ! '(x; y; z)) which is equivalent to
^
f8y8x(R(y; z) ^ R(x; yz) ! '(x; y; z)) : 8x8y(R(x; z) ^ R(y; xz) ! '(x; y; z))g:
By the Intersection Lemma,
^ 8y8x(R(y; z) ^ R(x; yz) ! f'(x; y; z)) : 8x8y(R(x; z) ^ R(y; xz) ! '(x; y; z))g);
while by the Closure Lemma
^ f'(x; y; z)) : 8x8y(R(x; z) ^ R(y; xz) ! '(x; y; z))g R(x; z) ^ R(y; xz)
in every canonical model. Thus we obtain the correspondent 8y8x(R(y; z) ^ R(x; yz) ! R(x; z) ^ R(y; xz)): 17
2
Example 5 The characteristic axiom of the "co-countably many" quanti er (Keisler's axiom): 8x2y P (x; y; z) ^ 2x 8yP (x; y; z) ! 2y 8xP (x; y; z), corresponds to 8x8y(R(y; z) ! R(y; xz) _ R(x; z)): Proof. The axiom is valid i 8'(8x8y(R(y; xz) ! '(x; y; z)) ^ 8x8y(R(x; z) ! '(x; y; z)) ! 8x8y(R(y; z) ! '(x; y; z)); namely, 8'(8x8y (R(y; xz) _ R(x; z)) ! '(x; y; z)) ! 8x8y (R(y; z) ! '(x; y; z)). This can be rewritten as
^
f8x8y(R(y; z) ! '(x; y; z)) : 8x8y(R(y; xz) _ R(x; z)) ! '(x; y; z))g;
by the Intersection Lemma,
^ 8x8y(R(y; z) ! f'(x; y; z)) : 8x8y(R(y; xz) _ R(x; z)) ! '(x; y; z))g; and by the Closure Lemma, 8x8y (R(y; z) ! R(y; xz) _ R(x; z)).
2
6 Non-existence of correspondents for completeness
6.1 Restrictions imposed in the Sahlqvist theorem
In this section we show that not all formulas which have a frame correspondent also admit a correspondent for completeness. First we prove this for the truth de nition we are working with; later the result is generalized. Recall that the de nition of Sahlqvist formulas (cf. Theorem 2) allowed existential and 3-quanti ers in the antecedent, while for weak Sahlqvist this is not allowed. The following theorem 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 8x8y8z(R(x; y) ! R(x; yz)), we show that it does not have a correspondent for completeness. Assume that Ay exists. The following axioms are consistent with A: 3xx = x and 8y :3y x = y . Hence, together with A they have a model where Ay holds. From Proposition 1 follows that Ay implies 8x8y8z(R(x; y) ! R(x; yz)). Since the correspondents of the other two axioms (being their *-translations) hold in every model, we have a model where 8x8y8z(R(x; y) ! R(x; yz)), 9xR(x) and 8x:R(x; x) hold simultaneously. But this is impossible. 2 An immediate consequence of the theorem is that "2 over ^"-combination in the antecedent cannot be allowed, since the axiom considered above can be written equivalently as 2x(' ^ ) ! 2x': Also, we have Corollary 1 Extensionality, that is 8x(' ) ! (2x' 2x ) does not have a correspondent in the sense of completeness. 18
Proof. One can check that extensionality is consistent with 3xx = x and :3y x = y, and that extensionality implies 3x' ! 3x(' _ ). Hence if extensionality would have a correspondent for completeness, it would imply the frame correspondent of 3x' ! 3x(' _ ). The rest of the argument is the same as above.
2
Corollary 2 3x' ! 2x' does not have a correspondent in the sense of completeness. Proof. The proof is analogous; consistency can be shown by constructing a model V the where D is in nite, R(x; y) , i x = 6 yi, and for each predicate P n , V (P n) = Dn. 2 Corollary 3 9x(x =6 y ^ '(x; y)) ! 3x('(x; y) _ ) does not have a correspondent in the sense of completeness.
Proof. Analogous.
2
The clause of the Theorem 3 forbidding occurrences of the same predicate letter with di erent free variables is also necessary. Suppose that there is a variable in the antecedent not occurring in the consequent, as in 2y (2x P (x; y ) ! 2z P (x; z )), is equivalent to 3y (2xP (x; y ) ^ >(x)) ! 2z P (x; z ), and 8y (2x P (x; y ) ! 2z P (x; z )) is equivalent to 9y2xP (x; y) ! 2z P (x; z): This shows that the class of weak Sahlqvist formulas is strictly smaller than the class of all 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 at
the behaviour of singleton sets, which are used as minimal substitutions in the proof of the Theorem 2 (and in modal logic), but do not occur as good minimal substitutions in the proof of the Theorem 3. Note that singletons as minimal substitutions are used precisely in the cases ruled out in the Theorem 3. The semantical correlate of "singletons as minimal substitutions" is distinguishability. A model is called distinguishable if every element is uniquely determined by the set of formulas in one free variable which are true for this element. For example, canonical models for modal logic are distinguishable. But in general, our models will not be distinguishable. Suppose an ! -saturated canonicalVmodel satis es 2xx 6= y and extensionality for 2x, and d is the unique element satisfying f'(x) : '(d)g. We show that 8x(R(x) ! x 6= d). In a distinguishable model this would hold for every element, hence R would be empty. Suppose 9x(R(x) ^ x = d), then for all ' such that '(d), 9x(R(x) ^ '(x)). Extensionality implies for every formula
9x(R(x) ^ (x)) ! 9x(R(x; d) ^ (x; d)): V Hence 9x(R(x; d) ^ '(x)), and by ! -saturation 9x(R(x; d) ^ '(x)). It follows that R(d; d), 2
a contradiction.
6.2 Other truth de nitions
Let us rst of all compare the results of this paper with the results previously obtained for modal 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 as K4, S4, S5, etc. are rst order complete. When we transfer this concept of rst order completeness to generalized quanti ers (cf. De nition 5), we see that there is an important 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 completeness for logics axiomatized by weak Sahlqvist formulas, but since extensionality is not weak Sahlqvist, this result is not very interesting taken by itself. Indeed, we can obtain even stronger negative results, as follows. The reader may have observed that for generalized quanti er logics a stronger form of rst order completeness can be de ned: the truth de nition we considered above is universal, but nothing prevents us from considering more complex truth de nitions. Might we not obtain a rst order correspondent to extensionality this 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's axiom for "co-countably many"
8x2y ' ^ 2x8y' ! 2y 8x' 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 that any truth de nition, however complex, involving an accessibility relation R, will make at least one axiom correspond to a schema. In this subsection, we shall consider even more general truth de nitions involving a relation R(x; Y ), where Y is a nite subset of the domain. Consideration of this relation is natural, because one might argue that dependence really is a relation between objects and 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 truth de 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 have correspondents for completeness in the language fR; 2; =g.
Theorem 6 For any weak second order truth de nition, the conjunction of the following
formulas (the Friedman axioms, see Steinhorn (1985)) will not have a correspondent for completeness: L1 3xx = x
L2 2x x 6= y L3 2x ' ^ 2x ! 2x (' ^ ) L4 2x 2y ' ! 2y 2x '
20
L5 8x(' ! ) ! (2x ' ! 2x )
Proof. Choose a weak second order truth de nition and suppose that the conjunction of the Friedman axioms has a correspondent for completeness. Then the following statement would be true: (C) 9X 9D9R(X = D
^ '
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 must have 2x '(x; y) Qx'(x; y). The direction from right to left is trivial. For the converse it suces if Q is closed under countable intersections in L, for in that case V a 2x '(x; y) 8x(( Qx (x; y) ! (x; y)) ! '(x; y)) V Qx (x; y) ! Qx (x; y) Qx(V Qx (x; y) ! (x; y)) b whence (by L5) Qx'(x; y). To show that Q is closed under countable intersections in L, argue as follows: let Q be de ned on D (i.e. Q P (D) and D 2 Q), and letT(An )n2! be in L, such that An D and for all n, Qx(x 2 An ). We want to show Qx(x 2 n An ). De ne a relation T (x; y ) by
T (x; y) =df 8n(x 2 Acn ! 8m(y 2 Am ) _ 9m > n(y 2 Acm )) _ 8n(x 2 An ^ y 2 D) T ThenT 8xQyT (x; y ): for if x 2 n An , then y 2 D, and if for some n, x 62 An , then y 2 mn Am. Hence QxQyT (x; y)Tand by L4 QyQxT (x; y). T S c If y 2 Am , then fx : T (x; y )g = Tn An [ nm Acn ; and if y 2 m Am , then fx : T (x;Ty)g = D. It follows that m Am fy : QxT (x; y)g. If we have equality, also Qy(y 2 T m Am).SOtherwise, let m be suchTthat y 2 Acm and QxT (x; y). It follows that Qx(x 2 n An [ nm Acn ), whence Qx(x 2 n An ). 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 a canonical relational model, and that the Friedman axioms do not have existential quanti ers in the antecedent, hence their correspondents hold in every canonical model) shows that S satis es a 9xS (x) b :S (x; x) 21
c S (x; yz) ! S (x; y) d S (x; yz) ^ S (y; z) ! S (y; xz)
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; yz) ^ : (x; y)) that is (]) 8x(S (x; yz) ! (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; xz) ! x y ). Since 9z8x:P (x; z) implies (with the axiom of Dependent Choice, which holds in L) the existence of an in nite descending chain, we must have 8z9xP (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 following holds: 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)). 2
7 Conclusion As we write in the introduction, the aim of this paper is to study how implicit dependencies between variables, imposed by generalized quanti er axioms, can be made explicit. For this purpose, we de ne relational models for generalized quanti ers, where quanti ers are interpreted using a dependency relation between variables, and show that some axioms determine rst order conditions on this relation. The situation is analogous to the one in modal logic, where some modal axioms determine rst order properties of the accessibility relation (and some do not). We prove a Sahlqvist theorem for generalized quanti ers which 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 correspondent for completeness. The latter result is strengthened as follows. For any truth de 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 dependencies between variables can be made explicit, i.e. expressed in the language containing only 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.
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