Semantics for a Logic of Essence

KIT FINE

SEMANTICS FOR THE LOGIC OF ESSENCE Received on 28 April 1999; revised on 17 November 1999

ABSTRACT. This paper provides a possible worlds semantics for the system of the author’s previous paper ‘The Logic of Essence’. The basic idea behind the semantics is that a statement should be taken to be true in virtue of the nature of certain objects just in case it is true in any possible world compatible with the nature of those objects. It is shown that a slight variant of the original system is sound and complete under the proposed semantics. KEY WORDS: essence, modality, possible worlds, semantics

In a previous paper ‘The Logic of Essence’, I presented a system for the logic of essence. In this paper, I present a semantics for a variant of the system and prove it complete with respect to that semantics. The main primitive of the system is an indexed operator F . Where A is a sentence and F a predicate, F A may be read: it is true in virtue of the nature of (some or all) the F’s that A. A statement of the form F A is only taken to be true when each of the objects mentioned in A is involved in the nature of one of the F’s. Thus it is not taken to be true in virtue of the nature of the number 2 that Socrates = Socrates. The system also contains symbols for what I call ‘rigid predicates’. These are predicates that might be expressed in the form: x is either a1 or a2 or . . . , where a1 , a2 , . . . are rigid names. Thus rigid predicates are essentially lists. These predicates are required in order to give a reasonable formulation of the reduction theses for the operator F . The basic idea behind the semantics is that a statement should be taken to be true in virtue of the nature of certain objects if it is true in any world compatible with the nature of those objects. We shall make the simplifying assumption that each world is compatible with the nature of all and only those objects that it contains. Thus the condition for a statement to be true in virtue of the nature of certain objects is that it should be true in all those worlds that contain those objects. However, the presence of an object in a world is not taken to guarantee its existence but merely its possibility. Thus each world will be taken to embody its own ‘view’ of which objects are possible and which are not. The proof of completeness for the system is not straightforward. The difficulties essentially arise from the presence of rigid predicates. For in Journal of Philosophical Logic 29: 543–584, 2000. © 2000 Kluwer Academic Publishers. Printed in the Netherlands.

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constructing a canonical model, we will need to add new ‘witnessing’ predicates. But since we cannot quantify over these predicates, it is hard to maintain control on the consistency of the resulting construction. One novelty of the proof is a technique for eliminating the predicates through approximating definitions rather than through quantification. Another is a device for getting the canonical model into ‘equilibrium’ by splitting it into parts. The first two sections are devoted to the language of the logic (Section 1) and the system of proof (Section 2). The next section gives the semantics (Section 3). The remaining six sections develop the completeness proof. The first three (Sections 4–6) provide lemmas crucial to the construction of the canonical model. The next two sections (Sections 7, 8) show how to build up a ‘diagram’ of the model; and the last section (Section 9) shows how to obtain the model itself. The reader might find it helpful to have the previous paper ‘The Logic of Essence’ at hand (henceforth abbreviated to ‘LE’) and also to consult the papers ‘Essence and Modality’ and ‘Senses of Essence’ for further explanation of the notion of essence and for general philosophical motivation.1

1. R EVIEW OF THE L ANGUAGE

The vocabulary of the system consists of the following symbols: (i) denumerably many individual variables; (ii) denumerably many individual constants; (iii) denumerably many n-place pure predicate symbols, for n = 0, 1, 2, . . .; (iv) denumerably many 1-place rigid predicate symbols; (v) the 2-place identity predicate =, and the 2-place dependency predicate ≥; Q (vi) the logical constants ∼, ∨, and ; (vii) the essentialist operator symbol ; (viii) the abstraction operator λ; (ix) the bracketing devices ( , ), [ and ], and :. The language of LE did not contain constants, but they are useful for the semantics and for the proof of completeness. The ‘pure’ predicates are ones involving no reference to any object. We use ‘r-predicate symbol’ as shorthand for ‘rigid predicate symbol’. The formulas and predicates of the system are defined by the following simultaneous induction:

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(i) if F is an n-place predicate and t1 , . . . , tn are variables or constants, then Ft1 . . . tn is a formula; Q (ii) standard rules for ∼ and ∨ and for the quantifier . (iii) If A is a formula and F is a 1-place predicate, then [F : A] is a formula; (iv) Each n-place predicate symbol is an n-place predicate; (v) If A is a formula and x a variable, then λxA is a 1-place predicate. We adopt standard meta-linguistic terminology and conventions. In particular, we use: Q (i) > for x(x = x), where x is the first free variable under some standard ordering of the variables; (ii) W ⊥ for ∼>; (iii) 1≤i≤n Ai for ⊥ if n = 0, for A1 if n = 1, and for (. . . (A1 ∨ A2 ) ∨ · · · ∨ An ) if n > 1 (and similarly for the generalized conjunction V 1≤i≤n Ai ). We use F and G etc. for arbitrary 1-place predicates, P, Q, R and S etc. for arbitrary r-predicate symbols, and H for an arbitrary n-place predicate, n ≥ 0. We use boldface to indicate a list of symbols. Thus x indicates a list of n variables x1 , . . . , xn , n ≥ 0; and similarly for F and P. In the formula [F : B], the predicate F is called the delimiter. A predicate W is said to be rigid if it is either an r-predicate symbol or is of the form λx 1≤i≤nAi , n ≥ 0, where each formula Ai , i = 1, . . . , n, is either of the form Px or of the form x = y for some variable y distinct from x. We adopt the following further abbreviations: (I) Variants on . (i) F A for [F : A]; (ii) ♦[F : A] (and ♦F A) for ∼[F : ∼A]. (II) Dependency Notions (i) x ≤ y for y ≥ x; (ii) x ≤ F for 6y(Fy & x ≤ y), y the first variable distinct from x and not free inQ F; Q (iii) Cl(F) for x y(Fx & x ≥ y → Fy). Cl(F) states that F is closed under the relation of dependence. (III) Constituency Let E be a formula or predicate. Suppose that x1 , . . . , xm are the free variables of E in order of appearance and that P1 , . . . , Pn are the r-predicate symbols of E, likewise in order of appearance. We then use:

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(i) xηE for

W

1≤i≤m x

= xi ∨

W

1≤j ≤n Pj x.

The intuitive meaning of HxηE is that x occurs in the proposition (or relation) expressed by E. (IV) Predicational Notions W (i) V for λx>; (ii) for λx⊥; (iii) (y) for λx(x = y), x the first variable distinct from y; (iv) |E| for λx(xηE), x the first variable not free in E; (v) cF for λx(x ≤ F), xWthe first variable not free in F; (vi) (F1 , . . . , Fn ) for λx 1≤i≤nFi x, where n ≥ 0 and x is the first variable not to occur free in any of F1 , . . . , Fn ; (vii) F − G for λx(Fx & ∼Gx); Q (viii) F ⊂ G for x ∼(Fx & ∼Gx), where F and G are 1-place predicates and x is the first variable not to occur free in F or G; (ix) R ≈ S for (R ⊂ S & S ⊂ R). The reader should distinguish between the object-language ⊂ and the meta-language ⊆. In using (t), t a variable or constant, as part of a delimiter we will usually omit the brackets. Outermost brackets of the disjunctive predicate (F1 , . . . , Fn ) under (vi) may also be omitted. Thus, combining these two conventions, we may use x,y A to abbreviate λz((x)z∨(y)z)A.

2. R EVIEW OF THE S YSTEM

The axioms and rules of the system E5 are as follows: (I) Standard Classical Axioms and Rules: We may use any of the standard axiomatizations of classical first-order logic. However, we shall suppose that the sole rules of inference are Modus Ponens (MP) and Universal Generalization (Gen). (II) Modal Axioms and Rules: (i) (ii) (iii) (iv) (v)

F A → A; F (A → B) → (F A → F B); ∼ F A → F,|A| ∼ F A, F rigid; A \ |A| A; F ⊂ G → (F A → G A);

(III) Predicational Axioms and Rules: (i) λxAy ↔ A(y), where A(y) is the result of substituting y freely for free occurrences of x in A;

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(ii) Px → P Px; (iii) (Predicate Elimination) P ≈ F → A/A, as long as the predicate P does not occur in A or F. Q (iv) (Modal Predicate Q Elimination) A → (Q ⊂ P → P x(Q ≈ F → B)/A → P x(F ⊂ P → B), where Q is distinct from P and does not occur in F or B. (IV) Dependency Axioms: (i) cF A → F A; (ii) F A & xηA → x ≤ F; Rule (III)(iv) was not included in the original formulation of E5. It seems necessary for the proof of completeness. Suppose we add quantifiers for r-predicate symbols to the language and then add standard axioms and rules for these quantifiers to the system, plus the axiom: 6P: 6P(P ≈ F), for P not free in F. The rules (III)(iii) and (iv) can then be derived. For suppose `P ≈ F → A. By quantificational reasoning, `6P(P ≈ F) → A and so, byQ6P and Modus Ponens, `A. Now suppose `A → (Q ⊂ P → Q P x(Q ≈ F → B). Then, by quantificational reasoning, `A → Q(Q ⊂ P → Q P x(Q ≈ F → B). But then by modal reasoning (as in Theorem V.1 Q Q of LE), `A →  Q(Q ⊂ P → x(Q ≈ F → B), hence `A → P Q Q P Qx Q(Q ⊂ P → (Q ≈ F → B)), and hence: Q (a) `A → P x(6Q(Q ⊂ P & QQ ≈ F) → B). From axiom 6P, `P x6Q(Q ≈ F), and hence: (b) `P x(F Q ⊂ P → 6Q(Q ⊂ P & Q ≈ F)). But then from (a) and (b), `A → P x(F ⊂ P → B). The rules (iii) and (iv) appear to be the price we must pay in order to avoid quantification over r-predicates. However, I do not know whether either or both of the rules could be dropped without loss of theorems.

3. S EMANTICS FOR THE S YSTEM

A model M is a quadruple hW, I, , φi, where: (i) W (worlds) is a non-empty set; (ii) I (individuals) is a function taking each w ∈ W into a non-empty set Iw ; S (iii)  (dependence) is a reflexive and transitive relation on w∈W Iw with respect to which each world is closed (a ∈ Iw and a  b implies b ∈ Iw );

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(iv) φ (valuation) is a function that takes each constant a into an individual φ(a) of some Iw (w ∈ W ), each r-predicate symbol H into a subset φ(H) of some Iw , and each world w and pure n-place predicate symbol F into a set φ(F, w) of n-tuples of Iw . S Given any subset J of Iw , we let the closure c(J ) of J (in M) be {b : a  b for some a ∈ J }. Thus the subset J is closed if c(J ) ⊆ J . It is readily verified that c satisfies the standard closure conditions: J ⊆ c(J ); cc(J ) ⊆ c(J ); and c(J1 ) ∪ c(J2 ) = c(J1 ∪ J2 ). Let M be a model and E a sentence or closed predicate whose constants are a1 , . . . , am and whose r-predicate symbols are P1 , . . . , Pn . Then the objectual content [E]M of E (in M) is taken to be {φ(a1 ), . . . , φ(am )} ∪ φ(P1 ) ∪ · · · ∪ φ(Pn ); and we say that E is defined (in M) at w ∈ W if [E]M ⊆ Iw . The index M will usually be dropped from [E]M and from other model-indexed terms. S The model M is said to be full if for each a ∈ Iw there is a constant a for which φ(a) = a. Given a full model M, we now define w  A, the truth of the sentence A at w, and Hw , the extension of the predicate H at w. We shall presuppose that w  A is informally defined only when A is (in the technical sense) defined in M at w and that Hw is defined only when H is defined at w. The clauses are then as follows: (i) (a) w  Ha1 . . . an iff hφ(a1 ), . . . , φ(an )i ∈ Hw , where H is any nonlogical predicate; (i) (b) w  a = b iff φ(a) = φ(b); (i) (c) w  a ≥ b iff φ(a)  φ(b); (ii) w  ∼A iff not w  A; (iii) w  Q (A ∨ B) iff w  A or w  B; (iv) w  xA(x) iff w  A(a) whenever φ(a) ∈ Iw ; (v) w  F A iff (a) [A]M ⊆ c(Fw ) and (b) v  A whenever Iv ⊇ Fw ; (vi) Fw = φ(w, F); (vii) Pw = φ(P); (viii) λxA(x)w = {a ∈ Iw : w  A(a)}. Truth and extension for non-full models may be defined by extending the language and the model with a new constant for each individual of the model. We omit the obvious details. We cite, without proof, the following elementary results, whose use will often be tacit: LEMMA 1. Given any model M and world w in M, (i) [∼A] = [A], [(A ∨ B)] = [A] ∪ [B], [F A] = [F] ∪ [A], and [|A|] = [A];

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(ii) [E] = |E|w for any sentence or closed predicate E and any world w at which E is defined; (iii) for F a closed rigid predicate, [F] = Fw whenever F is defined at w; (iv) (cF)w = c(Fw ) whenever F is defined at w. A sentence A is said to be valid if for every model M and every world w of M at which A is defined, A is true at w in M; and a formula A(x1 , . . . , xn ) is said to be valid if each of its closed instance A(a1 , . . . , an ) is valid. The notions of consequence and of satisfiability are defined similarly. THEOREM 2 (Soundness). Each theorem of E5 is valid. Proof. We prove by induction that each closed instance of each theorem is valid: (I) Standard Classical Axioms and Rules. Straightforward (though the case of Modus Ponens needs to treated with some care). (II) Modal Axioms and Rules. (i) Suppose w  F A. By clause (v)(b) in the truth-definition, v  A whenever Iv ⊇ Fw . But clearly Iw ⊇ Fw ; and so w  A. (ii) Suppose w  F (A → B) and w  F A. Then [(A → B)] ⊆ c(Fw ), and both v  (A → B) and v  A whenever Iv ⊇ Fw . But then [B] ⊆ c(Fw ) by Lemma 1(i) and v  B whenever Iv ⊇ Fw ; and so w  F B. (iii) Suppose w  ∼F A, for F rigid. By Lemmas 1(i) and (iii), [A] ⊆ |A|w and [F] ⊆ Fw ; and so [∼F A] ⊆ c((F, |A|)w ). We now distinguish two cases. (a) [A] * c(Fw ). Take any v for which Iv ⊇ Fw . Then F is defined at v and so, by Lemma 1(iii), [A] * c(Fv ). But then v  ∼F A for any such v; and so w  F,|A| ∼F A. (b) [A] ⊆ c(Fw ). Then v  ∼A for some v for which Iv ⊇ Fw . Take now any u for which Iu ⊇ Fw ∪ |A|w . Then F is defined at u, with Fu = Fv , and so Iv ⊇ Fu . So given v  ∼A, u  ∼F A and hence w  F,|A| ∼F A. (iv) Suppose A is valid. Consider any model M and any world w of M at which A is defined. Then [A] ⊆ |A|w and, for any v for which Iv ⊇ |A|w = [A], A is defined and so v  A. But then w  |A| A for any such w and so |A| A is valid. (v) Suppose w  F ⊂ G and w  F A. Then [A] ⊆ c(Fw ) and v  A whenever Iv ⊇ Fw . Since Fw ⊆ Gw , c(Fw ) ⊆ c(Gw ); and so [A] ⊆ c(Gw ). Take now any v for which Iv ⊇ Gw . Then since Gw ⊇ Fw , Iv ⊇ Fw and v  A. Therefore w  G A. (III) Predicational Axioms and Rules. (i) Straightforward. (ii) Suppose w  Pa. Then φ(a) ∈ φ(P); and so [Pa] ⊆ Pw = φ(P) ⊆

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c(Pw ). Taken any v for which Iv ⊇ Pw . Then clearly v  Pa; and so w  P Pa. (iii) Suppose A is not valid. Then for some model M and w in M at which A is defined, not w  A. The domain Iw contains an individual a. Let M 0 be the model like M except that each constant in F but not in A refers to a, each r-predicate symbol P in F but not in A has the null set as its extension, and P has the extension of F at w as its extension. Then P ≈ F → A is defined at w in M 0 , w  P ≈ F in M 0 and not w  A in M 0 . Therefore P ≈ F → A is not valid. (iv) Suppose the conclusion of the rule is not Q valid. Then for some model M and w in M, w  A and w  ∼P x(F(x) ⊂ P → B(x)) (where the Q variables x in F and B are now displayed). We Q distinguish two cases. (a) [ x(F (x) ⊂ P → B(x))] * cPw . Then [ x(P ≈ F(x) → Q B(x)] * cPw ; and so w  ∼(A x(P ≈ F(x) → B(x)), → (P ⊂ P →  P Q and Q A → (Q ⊂ P → P x(Q ≈ F(x) →QB(x)) is not valid. (b) [ x(F(x) ⊂ P → B(x))] ⊆ cPw . Then v  ∼ x(F(x) ⊂ P → B(x)) for some v with Iv ⊇ Pw ; and so there are a1 , . . . , an ∈ Iv with respective names a1 , . . . , an for which v  F(a1 , . . . , an ) ⊂ P and v  ∼B(a1 , . . . , an ). Let J be F(a1 , . . . , an )v and let M 0 be the model whose only difference from M is that J is assigned as the extension of Q. Since v  F(a1 , . . . , an ) ⊂ P in M, w  Q ⊂ P in M 0 . By stipulation, v  Q ≈ F(a1 , . . . , an ) in M 0 ; and, since v Q  ∼B(a1 , . . . , an ) in M, v  0 0 ∼B(a1 , . . . , an ) in M . But then Q w  ∼P x(Q ≈ F(x) → B(x)) in M ; and so A → (Q ⊂ P → P x(Q ≈ F(x) → B(x)) is not valid. (IV) Dependency Axioms. (i) Suppose w  cF A. Then [A] ⊆ c((cF)w ) = cc(Fw ) = c(Fw ) by Lemma 1(iv). Take now any v for which Iv ⊇ Fw . Then from the closure condition and Lemma 1(iv) again, Iv = c(Iv ) ⊇ c(Fw ) = (cF)w . So v  A; and hence w  F A. (ii) Suppose w  F A&bηA. Then [A] ⊆ (cF)w = c(Fw ) and φ(b) ∈ [A]. But then, for some a ∈ Fw , a  b; and so w  b ≤ F. It is interesting to note that there is an interpretation of the formal semantics in which each domain Iw is taken to consist of the objects that exist at w. A modal statement F A is then true at a world w iff each object mentioned in A depends for its existence on something that F’s and A is true in each world that contains the objects that exist in F. The proof to follow will then establish that the system is complete under this modalexistential interpretation of F . However, it would be a serious mistake to suppose that the notion of essence can therefore be analyzed in terms of modality and existence. Two notions can have the same logic, even the same formal semantics, and yet be intuitively quite different. Indeed, under

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the proposed interpretation, each existent will essentially exist and yet this is not a result which, intuitively, we want. Or again, it is hard to see under what sense of necessity the existence of {2} would entail that 2 ∈ {2} and yet the existence of 2 not entail that 2 ∈ {2}. 2

4. S YNTACTIC P RELIMINARIES

We present some of the basic notions required for the completeness proof and the motivation for using them. Languages. It will be important to keep track of the languages of the different worlds in the canonical model. We take a language to be a set of constants and r-predicate symbols. The language L(A) of a formula A is the set of individual constants and r-predicate symbols that occur in A. Similarly for the language of a predicate or of a set of formulas. We shall sometimes loosely talk of a sentence or predicate being in a given language L when we mean that its own language is included in the language L. Original Predicates. We shall suppose that a certain subclass of original individual constants and predicate symbols is specified in advance. The given predicate symbols are to include all of the pure predicate symbols and all but infinitely many r-predicate symbols; and, likewise, the given constants are to include all but infinitely many constants of the language. It is from this subclass of symbols that we draw the consistent set of sentences that is to be proved satisfiable. The remaining constants and r-predicate symbols are placed on reserve, to be used as ‘witnesses’ in the completeness proof. A predicate symbol is said to be original if it is from the given subclass and to be supplemental or an s-predicate symbol otherwise. A formula, predicate, or set of formulas is said to be original if all of its predicate symbols are original and is said to be supplemental otherwise. Note that no restriction is placed on which individual constants can occur in an the original formula. Restricted Formulas. We shall impose a global restriction on which supplemental sentences can be used in the worlds of the canonical model. The presence of sentences satisfying this restriction is required to establish that the canonical model has the required properties. A basic restricted sentence is one of the form: (i) F ⊂ S; (ii) S ⊂ F; (iii) Sa,

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where S is an s-predicate symbol and F is either an s-predicate symbol or an original closed predicate. A restricted sentence is then either an original sentence, a basic restricted sentence, a truth-functional compound of basic restricted sentences and original sentences, or a sentence of the form: (iv) S A, for S an s-predicate symbol and A an original sentence. We say that a restricted sentence B is negatable if ∼B is also a restricted sentence and that otherwise it is unnegatable. It should be evident that the unnegatable restricted sentences are exactly those sentences of the form S A listed under (iv). A set of sentences is said to be restricted if each of its members is restricted. From henceforth we shall use A (and its variants) for arbitrary unrestricted formulas, B, C, and D for restricted formulas, and 1 and 0 for sets of restricted sentences unless there is some explicit indication to the contrary. Completeness Concepts. A set of sentences 1 is said to be: consistent if not 1 `⊥; a theory if B ∈ 1 whenever 1 ` B and B is in L(1); N-complete if B ∈ 1 or ∼B ∈ 1 for any negatable B in L(1); Q-complete if B(a) ∈ 1 for some individual constant whenever a sentence 6xB(x) ∈ 1; (v) standardly complete if both N-complete and Q-complete; (vi) P -complete if Q ≈ F for some s-predicate symbol Q whenever F is an original closed predicate of L(1); (vii) fully complete if standardly complete and P-complete. (i) (ii) (iii) (iv)

Note the restriction Q in definitions (ii) and (iii) to the language of the given set 1. Thus if xB(x) belongs to the theory 1 then B(a) will not belong to 1 unless a is in the language of 1; and an N-complete theory need only ‘decide’ a sentence when it is negatable and in the language of the theory. A set of sentences 1 is said to extend or be an extension of another 10 if 1 ⊇ 10 and it is a safe extension of 1 if L(1) contains no original r-predicates not already in L(10 ). 10 is said to subsume 1 if 10 ` A for every sentence A in 1; and 1 and 10 are (deductively) equivalent if each subsumes the other. Companionship. Worlds accessible from one another in the canonical model will be called companions. To be more precise, if P is an s-predicate symbol, then the set of sentences 0 is said to be a P -companion to the set of sentences 1 if P ∈ L(1) and if, for any arbitrary sentence A in L(1) for

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which 1 ` P A, 0 ∪ {A} is consistent. Note that the sentence A in this definition need not be restricted. If 0 is a P-companion to 1, then both 0 and 1 must be consistent. For the sentence > is in L(1) and 1 ` P >. So, if 0 is a P-companion to 1, 0 ∪ {>} is consistent and so 0 itself is consistent. On the other hand, if 1 is inconsistent, then ⊥ is in L(1) and 1 ` P ⊥. But 0 ∪ {⊥} is inconsistent and so 0 is not a P-companion to 1. 0 is said to be a maximal P-companion to 1 if it is a P-companion and if no proper extension of 0 in the language of 0 is a P-companion to 1; and 0 is a broad P-companion to 1 if it is a P-companion and if B ∈ 0 whenever 0 ∪ {A : 1 ` P A and A is in L(1)} ` B and B is in L(0). It should be clear that every maximal P-companion is broad and that every broad P-companion is a theory. It is essential in the definition of companionship that the sentence A be required to belong to the language L(1). For we should allow Q {6x∼Px, ♦P P xPx} to be consistent.QBut this will requireQ(mutual) Pcompanions 1 andQ0 with ∼Pa, ♦P P xPx ∈ 1 and P xPx ∈ 0. Thus even though xPx is inconsistent with ∼Pa, we still wish 1 to be a P-companion to 0. Definitions. These are the means by which predicates in the canonical model are witnessed. Witnesses, in the form of rigid predicates, are required in order to establish that the canonical model has the appropriate S5-structure. Formally speaking, a definition is a sentence of the form Q ≈ F, for Q an s-predicate symbol and F a closed predicate. Q is said to be its defieniendum and F its definiens, and a definition is said to be nonmodal when its definiens is nonmodal. Unless otherwise stated, it will be assumed that F is an original predicate (containing no s-predicate symbols). A system of definitions is a set of definitions no two of which share a common definiendum. A system of definitions is said to be of its definiendia and to be in terms of the predicate symbols that occur in its definientia. We say that an s-predicate symbol Q is defined in the set of sentences 1 if Q ≈ F ∈ 1 for some original predicate F and that otherwise it is undefined.

5. D EFINITIONAL R EPLACEMENT

We show how a set of restricted sentences may be approximated by means of a system of definitions. We then consider two significant ways in which this result is of use in establishing the consistency. It should be noted that the definitions in 1d below may contain s-predicate symbols on the right and hence need not be restricted.

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THEOREM 1 (Definitional Replacement). Suppose that 1 is a standardly complete theory. For S a set of s-predicate symbols from 1, let 1c be the set of sentences of 1 containing no s-predicate symbols from S and let 10 be a finite set of sentences of 1. There is then a system of nonmodal definitions 1d of the s-predicate symbols of S in terms of the predicate symbols of L(1) − S for which 1c ∪ 1d subsumes 10 . Proof. It suffices to prove this result for any finite subset 100 of 1 − 1c for which 1c ∪ 100 subsumes 10 . For if 1c ∪ 1d subsumes 100 , it subsumes 1c ∪ 100 ; and so, given that 1c ∪ 100 subsumes 10 , 1c ∪ 1d subsumes 10 , by the transitivity of subsumption. Thus for the purposes of the proof, any sentence of 10 may be replaced by a finite number of sentences from 1 that deductively imply it and any sentences of 10 that belong to 1c may be dropped. Let us now describe two ways in which this may be done. First, let C1 , . . . , Cn be the truth-functional constituents that occur in the truthfunctionally complex sentences B of 10 (a truth-functional constituent being any sentence that is not, in primitive notation, either a negation or a disjunction). Then we may drop the sentences B of 10 in favour of the subset 100 of the sentences ±C1 , ±C2 , . . . , ±Cn that belong to 1 (±Ci is either Ci or its negation). For suppose B ∈ 10 . By 1 N-complete, either Ci or ∼Ci belongs to 1, for i = 1, . . . , n (since each Ci is negatable). By truth-functional logic, 100 ` B or 100 ` ∼B. But if 1 ` ∼ B, 1 is inconsistent. Thus, in either case, 100 ` B; and hence 100 subsumes 10 . Second, suppose that 100 contains a sentence of the form ∼F ⊂ G. By 1 Q-complete, (Fa & ∼Ga) ∈ 1 for some individual constant a. Let 1000 be the result of replacing each such sentence ∼F ⊂ G in 1 with the corresponding sentence Fa & ∼Ga. Then clearly 1000 subsumes 100 . It therefore suffices to prove the result for 1000 − 1c , which, for convenience, we redesignate as 10 . It should be clear that 10 consists of sentences of the following six kinds: (i) (ii) (iii) (iv) (v) (vi)

R ⊂ S, for R, S ∈ S; S ⊂ F, for S ∈ S and F a predicate from L(1) − S; F ⊂ S, for S ∈ S and F a predicate from L(1) − S; Sa, for S ∈ S; ∼Sa, for S ∈ S; S B, for S ∈ S and B an original sentence from L(1) − S.

Note that the predicate F in (ii) and (iii) will either be an original predicate or an s-predicate symbol from L(1) − S. The set 1d is defined as follows. Given any predicate symbol S ∈ S, let F1 , . . . , Fm be all of the closed predicates F from L(10 ) − S for which 10 ` S ⊂ F, and let

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a1 , . . . , an be all of the a in L(10 ) for which 10 ` ∼Sa. We then let FS be the predicate λx(F1 x& · · · &Fm x&x 6= a1 & · · · &x 6= an ) and let 1d consist of the definitions S ≈ FS for each S ∈ S. Clearly, 1d is a system of definitions of the required sort. It therefore remains to show that 1c ∪ 1d subsumes 10 . Given a sentence B of 10 , let B∗ be the result of replacing each spredicate symbol S of S in B with its definiens FS . It then suffices to show: (∗) 10 ` B∗ for each sentence B ∈ 10 . For then 1 ` B∗ and so, given that B∗ contains no predicate symbols of S and that 1 is a theory, it follows that 1c ` B∗ and hence, using the definitions of 1d , that 1c ∪ 1d ` B. Let us first note that: (∗∗) 10 ` S ⊂ FS for each S ∈ S. For if Fi x is a conjunct within FS , then 10 ` S ⊂ Fi by definition; and if x 6=Qaj is a conjunct within FS , then 10 ` ∼Saj by definition and so 10 ` x(Sx → x 6= aj ). We now consider each sentence B in 10 of the kinds (i)–(vi) in turn: (i) B = R ⊂ S for R, S ∈ S. We wish to show that 10 ` B∗ , i.e. that 10 ` FR ⊂ FS . Suppose that Fi x is a conjunct within the predicate FS . Then 10 ` S ⊂ Fi . Given that 10 ` R ⊂ S, 10 ` R ⊂ Fi ; and so Fi x is also a conjunct within FR . But then (a) FR ⊂ Fi is a theorem. Now suppose that x 6= aj is a conjunct within FS . Then 10 ` ∼Saj . Given that 10 ` R ⊂ S, Q 10 ` ∼Raj ; and so x 6= aj is also a conjunct within FR . But then (b) x(FR x → x 6= aj ) is a theorem. From (a) and (b), FR ⊂ FS is a theorem and hence deducible from 10 . (ii) S ⊂ F, for S ∈ S and F a closed predicate from L(10 ) − S. Then Fx is a conjunct within FS ; and so FS ⊂ F is a theorem and hence deducible from 10 . (iii) F ⊂ S, for S ∈ S and F a closed predicate from L(10 ) − S. By (∗∗), 10 ` S ⊂ FS . So since F ⊂ S ∈ 10 , 10 ` F ⊂ FS , as required. (iv) Sa, for S ∈ S. If Fi x is a conjunct within the predicate FS , then 10 ` S ⊂ Fi and so 10 ` Fi a. If x 6= aj is a conjunct within FS , then 10 ` ∼Saj and so 10 ` a 6= aj . (v) ∼Sa, for S ∈ S. Then a = aj for some j and so ∼FS a is a theorem.

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(vi) S A, for S ∈ S and A an original sentence from L(1). Given that S A ∈ 10 , it follows by (∗∗) and subsumption that 10 ` FS A. The restricted sentences were so chosen as to be adequate for establishing the relevant properties of the canonical model and yet susceptible to elimination in this way. The first application of the theorem shows that, under suitable conditions, any restricted consequence of a theory can be represented as an consequence within the original language. COROLLARY 2 (Substitution). Let 10 be a fully complete theory and 1 a safe and consistent extension of 10 . Suppose 1 ` A = A(b1 , . . . , bk , S1 , . . . , Sl ), where b1 , . . . , bk are all of the distinct constants and S1 , . . . , Sl all of the distinct s-predicate symbols that occur in A but not in 10 , k, l ≥ 0. There are then constants a1 , . . . , ak and s-predicate symbols R1 , . . . , Rl from 10 for which 10 ` A(a1 , . . . , ak , R1 , . . . , Rl ). Proof. Given that 1 ` A, 1+ ` A for some standardly complete extension 1+ of 1; and so we may suppose, without loss of generality, that 1 is already standardly complete. We may also suppose, without loss of generality, that k, l > 0. Given that 1 ` A = A(b1 , . . . , bk , S1 , . . . , Sl ), there will be a finite subset 10 of 1 for which 10 ` A. Let the S of the theorem be the set of s-predicate symbols of 1. Thus 1c will be the set of original sentences of 1, those that contain no s-predicate symbols. By the theorem, there will be a system of nonmodal definitions of the s-predicate symbols of 1 (in terms of the original predicate symbols of 1 and hence, by safety, of 10 ) for which 1c ∪ 1d ` A. So there will be sentences C1 , . . . , Cm of 1c and sentences D1 , . . . , Dn of 1d such that C1 , . . . , Cm , D1 , . . . , Dn ` A. We may suppose that the D1 , . . . , Dn are of the form S1 ≈ F1 , . . . , S1 ≈ Fl for certain original predicates F1 , . . . , Fl , since any definition of the form S ≈ F, where S is not one of S1 , . . . , Sl may be eliminated by the Rule of Predicate Elimination; and by 1c a theory, we may suppose that there is a single sentence C from 1c for which: (a) C, S1 ≈ F1 , . . . , Sl ≈ Fl ` A. By substituting b1 for all those constants of C and of F1 , . . . , Fl that do not occur in A, we may also suppose that b1 , . . . , bk are the only constants occurring in C or any of the Fi , i = 1, . . . , n. Let us write C in the form C(b1 , . . . , bk ) and each Fi in the form Fi (b1 , . . . , bk ). Then 1 ` C0 = 6x1 . . . xk C(x1 . . . xk ). Since C0 contains only original r-predicates and no constants, it belongs to the language of 10 ; and since 1 is a theory, C0 ∈ 10 . Since 10 is also Q-complete,

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C(a1 , . . . , ak ) ∈ 10 for some constants a1 , . . . , ak ; and since 10 is also Pcomplete, there are s-predicate symbols Ri , i = 1, . . . , l, for which Ri ≈ Fi (a1 , . . . , ak ) ∈ 10 . Thus C(a1 , . . . , ak ), R1 ≈ F1 (a1 , . . . , ak ), . . . , Rl ≈ Fl (a1 , . . . , ak ) ∈ 10 . But substituting a1 , . . . , ak for b1 , . . . , bk and R1 , . . . , Rl for S1 , . . . , Sl in (a), we obtain: (b) C(a1 , . . . , ak ), R1 ≈ F1 (a1 , . . . , ak ), . . . , Rl ≈ Fl (a1 , . . . , ak ) ` A(a1 , . . . , ak , R1 , . . . , Rl ), and hence 10 ` A(a1 , . . . , ak , R1 , . . . , Rl ), as required.

2

The next result shows that companionship is preserved under the addition of restricted sentences. COROLLARY 3. Suppose that 0 is a P-companion to the fully complete theory 1 and that 1+ is a consistent and safe extension of 1 that contains no constants or s-predicate symbols from 0 that are not already in 1. Then 0 is also a P-companion to 1+ . Proof. Since 1 contains the r-predicate symbol P, so does 1+ . Suppose now that 0 is not a P-companion to 1+ . Then, for some sentence A of L(1+ ), 1+ ` P A and 0 ` ∼A. Write A in the form A(b1 , . . . , bk , S1 , . . . , Sl ), where b1 , . . . , bk are all of the distinct constants and S1 , . . . , Sl all of the distinct s-predicate symbols that occur in A but not in 1. By the previous corollary, there are constants a1 , . . . , ak and s-predicate symbols R1 , . . . , Rl of L(1) for which 1 ` P A(a1 , . . . , ak , R1 , . . . , Rl ). Since b1 , . . . , bk , S1 , . . . , Sl do not occur in 0, 0 ` ∼A(a1 , . . . , ak , R1 , . . . , Rl ) and so 0 is not a P-companion to 1. 2

6. C OMPANIONSHIP

We establish some results that will help us construct suitable companions. We first establish three general results on the existence of companions and then show how companions may be extended to comply with various partial criteria for completeness. We say that a constant a is P -bound in the set 1 if a, P ∈ L(1) and 1 ` Pa and that an r-predicate symbol Q is P -bound in 1 if Q, P ∈ L(1) and 1 ` Q ⊂ P. We say that the set of sentences 0 is P -bound in 1 if each constant and r-predicate symbol of 0 is P-bound in 1; and we let LP (1) = {a : a is P-bound in 1} ∪ {Q : Q is P-bound in 1}. We may give similar definitions of ∼P-bound but with ∼Pa and ∼(Q ⊂ P) taking the place of Pa and Q ⊂ P.

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Given a set of sentences 1, let ♦P 1 = {♦P (B1 & · · · &Bn) : B1 , . . . , Bn ∈ 1}. Then for P-bound sets 0, the criterion for companionship can be stated in terms of ♦P 0: LEMMA 1. Suppose that 0 is P-bound in 1. Then 0 is a P-companion to 1 as long as P ∈ L(1) and 1 ∪ ♦P 0 is consistent. Proof. Suppose P ∈ L(1) but 0 is not a P-companion to 1. Then for some sentence A of L(1), 1 ` P A but 0 ` ∼A. Therefore, for some sentences B1 , . . . , Bn of 0, B1 , . . . , Bn ` ∼A. By truth-functional logic, `A → ∼(B1 & · · · &Bn ); and so by Theorem 4(ii) of LE, `(P A&|B1 | ⊂ P& · · · &|Bn | ⊂ P) → P ∼(B1 & · · · &Bn ). Since 0 is P-bound by 1, 1 ` |B1 | ⊂ P& · · · &|Bn | ⊂ P. But then 1 ` P ∼(B1 & · · · &Bn ) and 1 ∪ {♦P (B1 & · · · &Bn )} is not consistent. 2 Given a sentence A and a set of constants a, let 6aA be 6x1 . . . 6xn A0 , where A0 is the result of replacing the distinct constants a1 , . . . , an of a in A by distinct free variables x1 , . . . , xn (the choice of the free variables x1 , . . . , xn in this definition does not matter); and given a set of sentences 1 and a set of constants a, let 6a1 = {6a(A1 & · · · &An ) : A1 , . . . , An ∈ 1}. Then companionship for 0 can be stated in terms of companionship for 6a0: LEMMA 2. Let a be the set of constants of 0 that do not appear in 1. Then 0 is a P-companion to 1 as long as 6a0 is a P-companion to 1. Proof. Suppose that 0 is not a P-companion to 1 and that P ∈ L(1). Then, for some sentence A of L(1), 1 ` P A and 0 ` ∼A. So, for some sentences B1 , . . . , Bn of 0, B1 , . . . , Bn ` ∼A. Since the constants of a do not appear in A, it follows by quantificational reasoning that 6a(B1 & · · · &Bn ) ` ∼A. But then 6a0 is not a P-companion to 1. 2 Putting together these two results, we obtain: LEMMA 3. Suppose that each r-predicate symbol of L(0) and each constant of L(0) ∩ L(1) is P-bound in 1. Let a be the set of constants of 0 that do not appear in 1. Then 0 is a P-companion to 1 as long as P ∈ L(1) and 1 ∪ ♦P 6a0 is consistent. Proof. Suppose that P ∈ L(1) and 1 ∪ ♦P 6a0 is consistent. Under the given conditions, 6a0 is P-bound in 1; and so by Lemma 1, 6a0 is a P-companion to 1. But a is the set of constants of 0 that do not appear in 1; and so, by Lemma 2, 0 is a P-companion to 1. 2 We now show how to achieve partial compliance with N-completeness:

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LEMMA 4. Suppose that 0 is a P-companion to 1 and let A be any sentence. Then either 0 ∪ {A} or 0 ∪ {∼A} is a P-companion to 1. Proof. Suppose otherwise. Then, for some sentence B in L(1), 1 ` P B and 0, A ` ∼B and, for some sentence C in L(1), 1 ` P C and 0, ∼A ` ∼C. Clearly, (B&C) is in L(1) and so, by Theorem IV.4(iv) of LE, 1 ` P (B&C). But 0 ` (∼B ∨ ∼C); and so 0 ∪ {B&C} is inconsistent and hence 0 is not a P-companion to 1. 2 We turn to the demands of Q-completeness. There are two cases, according as to whether or not the witnessing constant is required to be P-bound. In the second case, we also add Pa to the complementary set 1. This is done in order that we may later show that the companion relationship is recriprocated. It should be noted that the witnessed existential in the formulation of these lemmas is not required to be a restricted sentence. LEMMA 5. Suppose that 0 ∪ {6xA(x)} is a P-companion to 1 and that a is a constant not appearing in 1, 0 or A(x). Then 0 ∪ {A(a)} is a Pcompanion to 1. Proof. Suppose otherwise. Then, for some sentence A0 of L(1) not containing a, 1 ` P A and 0, A(a) ` ∼A0 . Since the constant a does not appear in 0, A(x) or A0 , it follows by quantificational reasoning that 0, 6xA(x) ` ∼A0 and so 0 ∪ {6xA(x)} is not a P-companion to 1 after all. 2 LEMMA 6. Suppose that 0 ∪ 6x(Px&B(x)) is a P-companion to 1 and that 1 ` 6xPx. Let a be a constant not appearing in 1, 0 or B(x). Then 0 ∪ {Pa, B(a)} is a P-companion to 1 ∪ {Pa}. Proof. Suppose otherwise. Then, for some sentence A(a) of L(1) ∪ {a} : 1, Pa ` P A(a) and 0, Pa, B(a) ` ∼A(a). So 1 ` Pa → P A(a); so since a does not appear in 1, 1 ` Px → P A(x); and hence 1 ` Q x(Px → P A(x)). Q Since 1 ` 6xPx, it follows by Corollary V.2(i) of LE that 1 ` P x(Px →QA(x)). Since 0, Pa, B(a) ` ∼A(a) and a does not appear in 0, Q Q0 ` x(Px → (A(x) → ∼B(x)). But then 0, Qx(Px → A(x)) ` x(Px → ∼B(x)). Therefore 0 ∪ {6x(Px&B(x)), x(Px → A(x))} is inconsistent and so 0 ∪ {6x(Px&B(x)) is not a P-companion to 1. 2 We consider next compliance with P-completeness. There are five cases in all. The first is for when the witnessing predicate Q is not required to be P-bound. The others are for when it is. Just as in the case of the constants, reciprocation will require that the witnessing predicate in these four cases be P-bound on both left and right. The cases divide into two pairs. The first

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of each pair deals with the special case in which application of definitional replacement is not required, and the second of each pair with the more general case in which it is. The first of the two pairs deals with the case in which witnessing is done on the ‘right’, in 0, and the second of the two pairs with the case in which it is done on the ‘left’, in 1. The significance of this distinction is that the left side may contain original r-predicates that do not appear on the right. LEMMA 7. Suppose that 0 is a P-companion to 1 and that Q is an spredicate symbol that does not appear in 1 or 0 or F. Then 0 ∪ {Q ≈ F} is a P-companion to 1. Proof. Suppose otherwise. Then, for some sentence A of L(1), 1 ` P A and 0, Q ≈ F ` ∼A. Since Q does not appear in 0, F or A, it follows by Predicate Elimination that 0 ` ∼A; and so 0 is not a P-companion to 1. 2 LEMMA 8. Suppose that 0 is a P-companion to 1, that each r-predicate symbol of L(0) and each constant of L(0) ∩ L(1) is P-bound in 1, and that 0 ` F ⊂ P for some original closed predicate F of L(0). Let Q be an r-predicate symbol that does not appear in 1, 0 or F. Then 0 ∪ {Q ≈ F} is a P-companion to 1 ∪ {Q ⊂ P}. Proof. Suppose that 0 ∪ {Q ≈ F} is not a P-companion to 1 ∪ {Q ⊂ P}. Let a be the set of constants of 0 that do not appear in 1. Then by Lemma 3, 1 ∪ {Q ⊂ P} ∪ ♦P 6a(0 ∪ {F ≈ Q}) is inconsistent. So for some finite subset 10 = {B1 , . . . , Bl } of 1 and some finite subset 0 0 = {C1 , . . . , Cm } of 0, 10 ∪ {Q ⊂ P} ∪ {♦P 6a(0 0 ∪ {Q ≈ F})} is inconsistent. Let a1 , . . . , am be the constants of a that appear in the members of 0 0 or in F, and write each member Ci of 0 0 in the form C(a1 , . . . , am ) and F in the form F(a1 , . . . , am ). Then {B1 , . . . , Bl } ∪ {Q ⊂ P} ∪ {♦P 6x1 . . . xm (C1 (x1 , . . . , xm )& · · · &Cn (x1 , . . . , xm )&Q ≈ F(x1 , . . . , xm )} is inconsistent. Let B = (B1 & · · · &Bl ) and C(x1 , . . . , xm ) = (C1 (x1 , . . . , xm )& · · · &Cn(x1 , . . . , xm )) and write the sequence of variables x1 , . . . , xm as x. Then:
Q `B → Q ⊂ P → P x(Q ≈ F(x) → ∼C(x) . So by the modal rule of predicate elimination:
Q `B → P x F(x) ⊂ P → ∼C(x) . Q Hence 1 ` P x(F(x) ⊂ P → ∼C(x)). But 0 ` F(a1 , . . . , am ) ⊂ P Q and 0 ` C(a1 , . . . , am ). So the set { x1 . . . xn (F(x1 , . . . , xm ) ⊂ P → ∼C(x1 , . . . , xm )), F(a1 , . . . , am ) ⊂ P, C(a1 , . . . , am )} is inconsistent and 0 is not a P-companion to 1 after all. 2

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We may relax the requirement that each r-predicate symbol of 0 be Pbound in 1: LEMMA 9. Suppose that the standardly complete theory 0 is a P-companion to 1, that each original r-predicate symbol of 0 appears in 1, that each constant and r-predicate symbol of L(1) ∩ L(0) is P-bound in 1, and that 0 ` F ⊂ P for some original closed predicate F of L(0). Let Q be an r-predicate symbol that does not appear in 1, 0 or F. Then 0 ∪{Q ≈ F} is a P-companion to 1 ∪ {Q ⊂ P}. Proof. Suppose that 0 ∪ {Q ≈ F} is not a P-companion to 1 ∪ {Q ⊂ P}. Then, for some sentence A of L(1) ∪ {Q}, 1, Q ⊂ P ` P A and 0, Q ≈ F ` ∼A. So, for some finite subset 0 0 of 0, 0 0 Q ≈ F ` ∼A. We now apply the theorem on definitional replacement (V.1) to 0 and 0 0 , with S the set of s-predicate symbols of 0 that do not appear in 1. Thus 0 c is the subset of sentences from 0 that contain only s-predicate symbols from 1; and, since every original r-predicate symbol in 0 appears in 1, every r-predicate symbol of 0 c appears in 1 and hence is P-bound in 1. By the theorem, there is a system of definitions 0 d of the s-predicate symbols of S in terms of the s-predicate symbols of L(0) − S for which 0 c ∪ 0 d subsumes 0 0 . Since 0 0 , Q ≈ F ` ∼A, 0 0 ` Q ≈ F → ∼A. So 0 c , 0 d ` Q ≈ F → ∼A; and so 0 c , 0 d , Q ≈ F ` ∼ A. Since F is an original predicate, none of the definiendia of 0 d appear in F and, by stipulation, none of them is Q. Moreover, none can appear in A, since A is in L(1). So by successive applications of Predicate Elimination, 0 c , Q ≈ F ` ∼A. We now apply the previous lemma to 0 c and 1. Every constant and rpredicate symbol of L(0) ∩ L(1) is P-bound in 1, and so every constant and r-predicate symbol of 0 c is P-bound in 1. Since 0 is a P-companion to 1, 0 c ∪ {F ⊂ P} is a P-companion to 1 and so, by the previous lemma, 0 c ∪ {Q ≈ F} is a P-companion to 1 ∪ {Q ⊂ P}. But this is contrary to the fact that 1, Q ⊂ P ` P A and 0 c , Q ≈ F ` ∼A. 2 Without the theorem on Definitional Replacement (and the restriction in the class of sentences upon which it depends), it is hard to see how this crucial result might be established. Let us now consider the pair of cases in which the predicate is witnessed on the left: LEMMA 10. Suppose that 0 is a P-companion to 1, that each r-predicate symbol of L(0) and each constant of L(0) ∩ L(1) is P-bound in 1, and that 1 ` F ⊂ P for some original closed predicate F of L(1). Let Q be an r-predicate symbol that does not appear in 1, 0 or F. Then 0 ∪ {Q ⊂ P} is a P-companion to 1 ∪ {Q ≈ F}.

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Proof. Suppose otherwise. Then, for some sentence A of L(1), 1 ∪ {Q ≈ F} ` P A and 0, {Q ⊂ P} ` ∼A. Let a be the set of constants of 0 that do not appear in 1. Then by Lemma 3, 1∪{Q ≈ F}∪♦P 6a(0 ∪{Q ⊂ P}) is inconsistent. So, for some finite subset 0 0 = {B1 , B2 , . . . , Bm } of 0, 1 ∪ {Q ≈ F} ∪ ♦P 6a(0 0 ∪ {Q ⊂ P}) is inconsistent. Let a1 , a2 , . . . , an be the distinct constants of a that appear in the members of 0 0 , and write each member Bi of 0 0 in the form Bi (a1 , . . . , an ). Then 1 ∪ {Q ≈ F} ∪ {♦P 6x1 , . . . , xn (B1 (x1 , . . . , xn )& · · · &Bm (x1 , . . . , xn )&(Q ⊂ P))} is inconsistent. Let B abbreviate 6x1 . . . xn (B1 (x1 , . . . , xn )& · · · &Bm(x1 , . . . , xn )). Then: (∗) 1, Q ≈ F ` P ((Q ⊂ P) → ∼B). But 1 ` F ⊂ P; so 1, Q ≈ F ` Q ⊂ P; so by Theorem VI.2(i) of LE, 1, Q ≈ F ` P (Q ⊂ P); and so, given (∗), 1, Q ≈ F ` P ∼B. Since Q does not occur in F, B or 1, it follows by Predicate Elimination that 1 ` P ∼B. But given that 0 ` (B1 (a1 , . . . , an )& · · · &Bm (a1 , . . . , an )), 0 ∪ {∼B} is inconsistent; and so 0 is not a P-companion to 1 after all. 2 LEMMA 11. Suppose that the standardly complete theory 0 is a P-companion to 1, that each original r-predicate symbol of 0 appears in 1, that each constant and r-predicate symbol in L(1) ∩ L(0) is P-bound in 1, and that 1 ` F ⊂ P for some original closed predicate F of L(0). Let Q be an r-predicate symbol that does not appear in 1, 0 or F. Then 0 ∪ {Q ⊂ P} is a P-companion to 1 ∪ {Q ≈ F}. Proof. Obtained from Lemma 10 analogously to the way Lemma 9 is obtained from Lemma 8. 2 We turn finally to the question of reciprocation. Say that P is closed in 1 if P ∈ L(1) and 1 ` Cl(P). LEMMA 12 (Reciprocation). Suppose that 0 is a broad P-companion to 1, that every constant and r-predicate symbol P-bound in 0 is P-bound in 1, that P is closed in either 1 or 0, and that P ∈ L(0). Then 1 is a P-companion to 0. Proof. Suppose that 1 is not a P-companion to 0. Then, for some sentence A in L(0), 0 ` P A and 1 ` ∼A. So by modal reasoning: (∗) 1 ` ♦P ∼A. Since 0 ` P A, 0 ` |A| ⊂ cP from axiom (IV)(ii). Now it is readily shown that ` Cl(P) ≡ (cP ⊂ P). So if 0 ` Cl(P), then 0 ` |A| ⊂ P. Suppose, on the other hand, that 1 ` Cl(P) and hence that 1 ` cP ⊂ P. From Theorem VI.2(i) of LE, it follows that 1 ` P (cP ⊂ P); and since

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0, cP ⊂ P ` |A| ⊂ P, it follows from the fact that 0 is a broad companion that 0 ` |A| ⊂ P. Thus, in either case, 0 ` |A| ⊂ P. But then each constant and r-predicate symbol of A is P-bound in 0 and hence P-bound in 1; and so: (∗∗) 1 ` |A| ⊂ P. But from (∗) and (∗∗), it follows by Theorem IV.6(i) of LE that 1 ` P ♦P ∼A. Since P ∈ L(1), ♦P A is in L(1); and so, given that 0 ` P A, 0 ∪ {♦P ∼A} is inconsistent and 0 is not a P-companion to 1. 2 Under the conditions of the lemma, 0 will be a P-companion to 1 and 1 to 0. In this case, we say that they are (reciprocal) P-companions.

7. T WO - WORLD E QUILIBRIUM

As the first step towards constructing the canonical model, we show how to get two worlds in equilibrium, each a companion to the other and each fully complete. For simplicity, we describe the construction in three steps. First, we get an initial companion to be standardly complete; then we get it to be fully complete; and finally, we get both it and its complementary world to be fully complete. Each step of the construction uses up infinitely many new constants and s-predicates and it is important always to leave behind infinitely many constants and s-predicates for subsequent use. This is a requirement that we shall not usually bother to state. For the purposes of the induction, we need to impose some further requirements on the initial companions. Say that a set of sentences 1 is P-determinate if (i) P is in L(1), (ii) for each constant a of L(1), either 1 ` Pa or 1 ` ∼Pa, and (iii) for each r-predicate symbol Q of L(1), either 1 ` Q ⊂ P or 1 ` ∼(Q ⊂ P). Say that 1 and 0 are (exactly) P-matching if (i) 1 and 0 are P-determinate and (ii) LP (1) = LP (0) = L(1) ∩ L(0). We say that 0 is a matching P-companion to 1 if 0 is a P-companion to 1 and, in addition, 0 and 1 exactly P-match. For later purposes, we shall also say that 1 almost P-matches 0 if 1 and 0 are Pdeterminate and LP (1) ⊇ LP (0) = L(1) ∩ L(0) and that 0 is an almost matching P-companion to 1 if 0 is a P-companion to 1 and 1 almost P-matches 0. We shall also say that 1 and 0 are an almost matching pair of P-companions if they are P-companions and either 1 almost P-matches 0 or 0 almost P-matches 1. If 1 and 0 are an almost matching pair of P-companions, we shall say that 1 is anomalous if LP (1) ) LP (0). Thus

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when the set 1 is anomalous it will contain some P-bound constants or predicate symbols that do not appear in 0. LEMMA 1. Suppose that 0 is a matchinq P-comnanion to 1. Then there are safe extensions 0 + of 0 and 1+ of 1 such that: (i) 0 ∗ is a maximal matching P-companion to 1+ ; and (ii) 0 + is standardly complete. Proof. Let a0 , a1 , a2 , . . . be an infinite sequence of distinct constants not occurring in 1 or 0 (taking care to leave behind infinitely many constants from the whole language). Let L+ = L(0) ∪ {a0 , a1 , a2 , . . .}; and let B0 , B1 , B2 , be a repeating enumeration of the sentences of L+ (i.e. one in which each sentence occurs infinitely often). We define the sets 1n and 0n , for n = 0, 1, 2, . . . , by induction: n = 0. 10 = 1 and 00 = 0. n = k + 1. We first define the sets 10n and 0n0 : (a) If Bk is not in L(0k ) or 0k ∪ {Bk } is not a P-companion to 1k , then: 10k+1 = 1k ,

and

0 0k+1 = 0k .

(b) If Bk is in L(0k ) and 0k ∪ {Bk } is a P-companion to 1k , then: 10k+1 = 1k ,

and

0 0k+1 = 0k ∪ {Bk }.

We now define 0n and 1n : (c) If Bk is not in 0k0 or is not of the form 6xC(x), then: 1k+1 = 10k+1 ,

and

0 0k+1 = 0k+1 . 0 (d) If Bk ∈ 0k0 it of the form 6xC(x) and if 0k+1 ∪ {∼Pa, C(a)} is a P0 companion to 1k+1 , for a the first of the constants a0 , a1 , a2 , . . . not to occur in 1k or 0k , then:

1k+1 = 10k+1 , 0k+1

and  0 = 0k+1 ∪ ∼Pa, C(a) .

0 (e) If Bk ∈ 0k is of the form 6xC(x) and if 0k+1 ∪ {∼Pa, C(a)} is not a 0 P-companion to 1k+1 , for a the first of the constants a0 , a1 , a2 , . . . not to occur in 1k or 0k , then:

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1k+1 = 10k+1 ∪ {Pa}  0 0k+1 = 0k+1 ∪ Pa, C(a) . We prove by induction on n that 1n and 0n are matching P-companions: n = 0. By supposition. n = k + 1. We show that each of the operations specified under (a)–(e) above preserve the property of being a matching P-companion: 0 (a) Trivially true by IH since 10k+1 = 1k and 0k+1 = 0k . 0 (b) By construction, 0k+1 = 0k ∪ {Bk } is a P-companion to 10k+1 = 1k . 0 0 0 0k+1 is P-determinate since 0k+1 ⊇ 0k , L(0k+1 ) = L(0k ) and 0k is P0 determinate by IH. Also LP (0k+1 ) = LP (0k ). For if a (or Q) is P-bound 0 in 0k+1 but not in 0k then ∼Pa(∼(Q ⊂ P)) ∈ 0k by 0k P-determinate, and 0 so 0k+1 is inconsistent, contrary to its being a P-companion to 10k+1 . (c) Trivial. (d) By construction, 0k+1 is a P-companion to 1k+1 = 10k+1 . Since ∼Pa ∈ 0k+1 , it follows by IH that 0k+1 is P-determinate. Since a ∈ / L(1k+1 ), L(1k+1 ) ∩ L(0k+1 ) = L(1k ) ∩ L(0k ); since ∼Pa ∈ 0k+1 , LP (0k+1 ) = LP (0k ), for if Pa ∈ 0k+1 then 0k+1 is inconsistent. (e) By IH, 0k0 is a P-companion to 10k . So by Lemma VI.4, either 0k0 ∪ {6xC(x), 6x(C(x)& ∼Px)} or 0k0 ∪{6xC(x), ∼6x(C(x)& ∼Px)} is a Pcompanion to 10k . In the former case, 0k0 ∪ {C(a), ∼Pa} is a P-companion to 10k by Lemma VI.5; and since, by hypothesis, this is not so, the latter option holds. So 0k0 ∪ {6x(C(x)&Px)} is a P-companion to 10k ; and so, by 0 Lemma VI.6, 0k+1 = 0k+1 ∪ {Pa, Ca} is a P-companion to 1k+1 = 10k+1 ∪ {Pa}. Since Pa ∈ 0k+1 and Pa ∈ 1k+1 , it follows by IH that both 1k+1 and 0k+1 are P-determinate. Clearly, L(1k+1 ) = L(1k ) ∪ {a}, LP (1k+1 ) = LP (1k ) ∪ {a}, L(0k+1 ) = L(0k ) ∪ {a}, and LP (0k+1 ) = LP (0k ) ∪ {a}; 0 and therefore 0k+1 and 10k+1 exactly match. + Let 1 = 10 ∪ 11 ∪ · · · and 0 + = 00 ∪ 01 ∪ · · ·. Given that 0n is a matching P-companion to 1n for n = 0, 1, . . . , it is readily verified that 0 + is a matching P-companion to 1+ and that 0 + and 1+ contain no new original r-predicates. It remains to verify the various completeness requirements: Maximality. Suppose that 0 + is not a maximal P-companion to 1+ . Then for some sentence B in L(0 + ), 0 + ∪ {B} is a P-companion to 1+ but B ∈ / 0 + . Since B is in L(0 + ), B is in L(0k ) for some k; and so B = Bm for some m ≥ k. But then the condition for (b) in the definition of 0m+1 is satisfied and so B = Bm ∈ 0m+1 ⊆ 0 + after all. N-completeness. This follows from maximality. For given that 0 + is a P-companion to 1+ , it follows by Lemma VI.4 that either 0 + ∪ {B} or 0 + ∪ {∼B} is a P-companion to 1+ for any negatable B in L(0 + ). But then by the maximality of 0 + , either B ∈ 0 + or ∼B ∈ 0 + .

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Q-completeness. Suppose that B = 6xC(x) ∈ 0 + . Then B ∈ L(0k ) for some k. So B = Bm for some m ≥ k. But then the condition for (c) or (e) in the definition of 0m+1 is satisfied and, in either case, C(a) ∈ 0m+1 ⊆ 0 + for some constant a. 2 We now attempt to make the companion fully complete. Say that a set of sentences 1 is P-restricted (or that P is restrictive on 1) if every original r-predicate in L(1) is P-bound in 1. Then: LEMMA 2. Suppose that 0 is a matching P-companion to 1 and that either 0 or 1 is P-restricted. Then there are safe extensions 1+ and 0 + of 1 and 0 which are such that (i) 0 + is a maximal matching P-companion to 1+ and (ii) 0 + is fully complete. Proof. Let a0 , a1 , a2 , . . . be an infinite sequence of distinct constants not occurring in 1 or 0 and Q0 , Q1 , Q2 an infinite sequence of distinct s-predicate symbols not occurring in 1 or 0 (taking care, as always, to leave behind infinitely many constants and s-predicate symbols from the language). Let L+ = L(0) ∪ {a0 , a1 , a2 , . . .} ∪ {Q0 , Q1 , Q2 . . .}; and let F0 , F1 , F2 , . . . be a repeating enumeration of the original closed predicates of L+ . We define the sets 1n and 0n , for n = 0, 1, 2, . . . , by induction: n = 0. 10 = 1 and 00 = 0. + n = k + 1, k ≥ 0. We first define the auxiliary sets 1+ k and 0k . To this end, we suppose that the set A of constants {a0 , a1 , a2 , . . .} is partitioned into infinitely many infinite subsets A0 , A1 , A2 , . . . . We then let 1+ k and + 0k be safe extensions of 1k and 0k which conform to the conditions (i)– (ii) of the previous lemma and for which L(0k+ ) = L(0k )∪A0 ∪A1 ∪· · ·∪ Ak if such extensions exist (otherwise we may, for the sake of definiteness, + set 1+ k = 1k and 0k = 0k ). We now define 1k+1 and 0k+1 : (a) if Fk is not in L(0k+ ) or if neither ∼(Fk ⊂ P) nor Fk ⊂ P belongs to 0k+ , then: 1k+1 = 1+ k,

and

0k+1 = 0k+ . (b) If ∼(Fk ⊂ P) ∈ 0k+ and if Q is the first of the s-predicate symbols in + Q0 , Q1 , Q2 , . . . not to appear in 1+ k or 0k , then: 1k+1 = 1+ k, 0k+1 = 0k+ ∪ {Q ≈ Fk }.

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(c) If (Fk ⊂ P) ∈ 0k+1 and if Q is the first of the s-predicate symbols in + Q0 , Q1 , Q2 , . . . not to appear in 1+ k or 0k , then: 1k+1 = 1+ k ∪ {Q ⊂ P},

and

0k+1 = 0k+ ∪ {Q ≈ Fk }. Suppose that 1 (resp. 0) is P-restricted. We show by induction on n that 0n is a matching P-companion to 1n , that 0n (resp. 1n ) is P-restricted, and that 0n and 1n are respective safe extensions of 0 and 1. n = 0. Trivial. n = k + 1. By IH, 0k is a matching P-companion to 1k ; and so, by Lemma 1, 0k+ will be a matching P-companion to 1+ k . By IH, 0k (resp. 1n ) is P-restricted. Since by Lemma 1, 0k+ (1+ ) is a safe extension of 0k k + + (1k ), 0k (1k ) is also P-restricted. Furthermore, since by IH, 0k and 1k are safe extensions of 0 and 1, 0k+ and 1+ k are also safe extensions of 0 and 1. Let us now consider each of the three cases under the definition of 1n and 0n : (a) Trivial. (b) 0k+ is a matching P-companion to 1+ k ; and so, by Lemma VI.7, + 0k+1 = 0k+ ∪ {Q ≈ Fk } is a P-companion to 1k+1 = 1+ k . Now 0k P+ + matches 1k . For since Q ∈ / L(1k+1 ), L(1k+1 ) ∩ L(0k+1 ) = L(1k ) ∩ L(0k+ ) and, since 0k+1 ` ∼(Q ⊂ P) and Q ∈ / L(1k+1 ), Q is not P-bound in either 0k+1 or 1k+1 . Therefore LP (0k+1 ) = LP (0k+ ) and LP (1k+1 ) = LP (1+ k ) and so, by IH, L(1k+1 ) ∩ L(0k+1 ) = LP (0k+1 ) = LP (1k+1 ). + Moreover, 1+ k and 0k are P-determinate and so, since ∼(Q ⊂ P) ∈ 0k+1 , 1k+1 and 0k+1 are P-determinate. The remaining conditions are readily verified. (c) By IH, 0k+ is a P-companion to 1+ k . The case now divides according + as to whether 1+ or 0 is P-restricted: k k (c0 ) 0k+ is P-restricted. It is easily seen that 0k+ and 1+ k conform to the conditions of Lemma VI.9 and so it follows from the lemma that 0k+1 = 0k+ ∪ {Q ≈ Fk } is a P-companion to 1k+1 = 1+ k ∪ {Q ⊂ P}. Since Q is P-bound in 1k+1 and also in 0k+1 (given Q ≈ F, F ⊂ P ∈ 0), 1k+1 and 0k+1 are P-matching. The remaining conditions are then readily verified. 0 (c00 ) 1+ k is P-restricted. As for case (c ) but using Lemma VI.11 in place of Lemma VI.9. Let 1+ = 10 ∪ 11 ∪ · · · and 0 + = 00 ∪ 01 ∪ · · ·. Given that 0n and 1n are matching P-companions for n = 0, 1, . . . , it is readily verified that 0 + and 1+ are matching P-companions; and likewise, given that 0n and 1n are safe extensions of 0 and 1, it is evident that the same is true of

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0 + and 1+ . It remains to show that 0 + is fully complete and a maximal P-companion. Since 0 + is the union 00 ∪ 01 ∪ 02 ∪ · · · of an increasing chain of sets 00 ⊆ 01 ⊆ 02 ⊆ · · · that are standardly complete, it is readily verified that 0 + itself is standardly complete. Likewise, given that each 0n , for n = 1, 2, 3, . . . is a maximal P-companion to 1n and hence to 1+ , it is readily verified that 0 + is a maximal P-companion to 1+ . It remains to establish P-completeness. Suppose that F is an original closed predicate in L(0 + ). Then either F ⊂ P or ∼(F ⊂ P) belongs to 0 + by N-completeness; and so either F ⊂ P or ∼(F ⊂ P) belongs to 0k for some k. Now F = Fm for some m ≥ k; and so the condition for either (b) or (c) in the definition of 0m+1 is satisfied. But in either case, Q ≈ Fm ∈ 0m+1 for some s-predicate symbol Q. We now get both sets to be fully complete: LEMMA 3. Suppose that 0 is a matching P-companion to 1, with P either closed in 1 or restrictive on 0. There are then safe extensions 1+ and 0 + of 1 and 0 which are maximal matching P-companions and fully complete. Proof. Clearly, by Lemma 2, we can suppose that 0 is a maximal Pcompanion to 1 and fully complete. We now define 1n and 0n by induction on n = 0, 1, . . .: n = 0. Set 10 = 1 and 00 = 0. n = 2k + 1, k ≥ 0. Let 12k+1 and 02k+1 be safe extensions of 12k and 02k which are such that 12k+1 it fully complete and a maximal Pcompanion to 02k+1 (if such sets do not exist, let 12k+1 = 12k and 02k+1 = 02k ). n = 2k, k > 0. Let 02k and 12k be safe extensions of 02k−1 and 12k−1 which are such 02k is fully complete and a maximal P-companion to 12k (if such sets do not exist, let 02k = 02k−1 and 12k = 12k−1 ). We now show by induction on n = 0, 1, 2, . . . that 0n and 1n are safe extensions of 0 and 1 and that 0n is a maximal matching and fully complete P-companion to 1n for even n and that 1n it a maximal matching and fully complete P-companion to 0n for odd n. n = 0. Trivial. n = 2k + 1. By IH, 02k and 12k are safe extensions of 0 and 1 and 02k is a fully complete maximal matching P-companion to 12k . We now apply the Reciprocation Lemma (VI.12) to 02k and 12k . Since 02k is a maximal matching P-companion to 12k , it is a broad P-companion to 12k ; since 02k and 12k exactly P-match, every constant and r-predicate symbol P-bound in 02k is P-bound in 12k and, also, P ∈ L(02k ); and since 1 or 0 is P-closed, 12k or 02k is P-closed. The conditions for the application of

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Lemma VI.12 are therefore satisfied and so we may conclude that 12k is a P-companion to 02k . We now apply Lemma 2. Since 02k is a matching P-companion to 12k , 12k is a matching P-companion to 02k . Since 1 or 0 is P-restricted and 12k and 02k are safe extensions of 1 or 0, either 12k or 02k is P-restricted. The conditions for the application of the lemma are therefore satisfied and so may conclude that there exist safe extensions 12k+1 and 02k+1 of 12k and 02k (and hence of 1 and 0) with 12k+1 a maximal matching and fully complete P-companion to 02k+1 . n = 2k. Similar to the previous case, but with left and right reversed. Let 1+ = 10 ∪ 11 ∪ · · · and 0 + = 00 ∪ 01 ∪ · · ·. Since 1+ is the union 11 ∪ 13 ∪ 15 ∪ · · · of an increasing chain 11 ⊆ 13 ⊆ 15 ⊆ · · · of fully complete sets, it is fully complete. Likewise, since 0 + is the union 00 ∪ 02 ∪ 04 ∪ · · · of an increasing chain 00 ⊆ 02 ⊆ 04 ⊆ · · · of fully complete sets, it is also fully complete. Furthermore, since 1n is a matching P-companion to 0n for n = 1, 3, 5, . . . , it is readily verified that 1+ is a matching P-companion to 0 + ; and, similarly, 0 + is a matching P-companion to 1+ . Finally, 1n and 0n are safe extensions of 1 and 0 and hence so are 1+ and 0 + . We conclude this section with two results that show how the initial conditions for the previous constructions can be met. LEMMA 4. Suppose that 1 is a consistent set of sentences that yields both ♦P B and |B| ⊂ P, for P ∈ L(1). Then 0 = {B} is a P-companion to 1. Proof. Suppose otherwise. Then there is a sentence A in L(1) such that 1 ` P A and B ` ∼A. So ` A → ∼B. But then by Theorem IV.4(vi) of LE, ` P A&(|B| ⊂ P) → P ∼B; and so 1 is inconsistent. 2 Let us say that P is effectively closed in 1 if whenever 1 ` P B for B in L(1) and a (resp. Q) belongs to B, then Pa (resp. Q ⊂ P) is deducible from 1. Clearly if P is closed in 1 then it is effectively closed in 1. However, the converse need not hold even when 1 is N-complete, since Cl(P) is not in general a restricted sentence. LEMMA 5. Suppose that P is effectively closed in 1, with 1 Pdeterminate, that 0 is a P-companion to 1, and that L(0) ∩ L(1) ⊆ LP (1) and L(0) − L(1) ⊆ L∼P (0). Let 0 + = 0 ∪ {B : 1 ` P B, B in L(1)}. Then 0 + is a matching P-companion to 1. Proof. It is readily seen that 0 + is a P-companion to 1. Let 1P = {B : 1 ` P B, B in L(1)}. To verify that 0 + and 1 exactly match, we first establish the following identities:

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(∗) LP (1) = L(1P ) = LP (0 + ). We merely consider the case of the constants of a language, since the proof for the case of r-predicates is similar. Note first that LP (1) is a subset of both L(1P ) and LP (0 + ). For suppose a ∈ LP (1). Then 1 ` Pa. But then 1 ` P Pa; and since a, P ∈ L(1), a belongs to both L(1P ) and LP (0 + ). Now suppose a ∈ L(1P ). Then 1 ` P B for some sentence B in L(1) that contains a. Since P is effectively closed in 1, 1 ` Pa; and so a ∈ LP (1). Thus LP (1) = L(1P ). Finally, let us establish that LP (0 + ) ⊆ L(1P ). Suppose a ∈ LP (0 + ). Then either a ∈ L(1P ) or a ∈ L(0). In the former case, there is nothing to prove. In the latter case, either a ∈ L(1) or a ∈ / L(1). In the former subcase, a ∈ LP (1) since L(0) ∩ L(1) ⊆ LP (1), and so a ∈ L(1P ) by the previously proved identity. In the latter case, a ∈ L∼P (0) since L(0) − L(1) ⊆ L∼P (0) = L. But then 0 + is inconsistent and cannot be a P-companion to 1. We now show that 0 + is P-determinate (again ignoring the case of predicate symbols). Suppose a ∈ L(0 + ). Then either a ∈ L(1P ) or a ∈ L(0) − L(1P ). In the former case, a ∈ LP (1) by (∗). But then 1 ` P Pa and so Pa ∈ 1P ⊆ 0 + . In the latter case, a ∈ L(0) − LP (0) by (∗) and, since L(0) ∩ L(1) ⊆ LP (1), a ∈ / L(1) and so, given L(0) − L(1) ⊆ L∼P (0), 0 ` ∼Pa. Furthermore, LP (0 + ) = LP (1) by (∗) and:
L(1) ∩ L(0 + ) = L(1) ∩ L(0) ∪ L(1P )

= L(1) ∩ L(0) ∪ LP (1) by (∗)

= L(1) ∩ L(0) ∪ L(1) ∩ LP (1) = LP (1) since L(1) ∩ L(0) ⊆ LP (1). And therefore 0 + and 1 exactly match.

2

We conclude with a result concerning a single set of sentences: LEMMA 6. Any consistent set of sentences 0 (containing only original predicate symbols and constants) has a safe and fully complete extension 0 + . Proof. The result can be proved directly using classical techniques to ensure standard completeness and the rule of predicate to enW elimination + sure P-completeness. Alternatively, we can add P ≈ to 1 , let 1 = ∅, and apply Lemmas 5 and 2. 2

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8. G LOBAL E QUILIBRIUM

We show how infinitely many worlds can be put into ‘equilibrium’ in such a way that all stated possibilities are realized. The worlds will be arranged in a tree with the arrows from one node to another representing the possibilities. Thus when a P-possibility at one node is realized by another node, there will be a ‘P-arrow’ in the tree linking the one node to the other. We must show how more and more arrows can be added to the tree without disturbing the equilibrium of the tree as a whole. We begin with some terminology concerning trees. We suppose given an infinite number of ‘world constants’ w0 , w1 , w2 , . . . . A diagram D is then a set consisting of: (i) pairs hw, Bi, for w a world constant and B a (restricted) sentence; (ii) triples hw, v, Pi, for w and v distinct world-constants and P an spredicate symbol. We shall suppose that for any given world-constants w and v there is at most one P for which hw, v, Pi ∈ D. Given a diagram D, we let: (i) (ii) (iii) (iv)

WD = {w : w is component of an ordered pair or triple in D}; Dw = {BS : hw, Bi ∈ D} for any w ∈ WD ; L(D) = {L(Dw) : w ∈ WD }; D = {hw, vi : hw, v, Pi ∈ D}.

When hw, v, Pi ∈ D, we shall say that w is linked to v (by P) in D. We call the pair hw, vi an arrow in D if hw, v, Pi ∈ D for some P and we call the pair hw, vi a link in D if either hw, vi or hv, wi is an arrow. Given a diagram D, let TD be hWD , D i. D is then said to be a tree diagram if TD is a tree. Suppose that w and v are two distinct nodes on a tree TD . Then by the path from w to v is meant the sequence of distinct nodes w = w0 , w1 , . . . , wn = v such that, for i = 0, 1, . . . , n − 1, hwi , wi+1 i is a link in the tree. Clearly, there is a unique path from any node of a tree TD to any other node. A subtree T0 of a tree T is a restriction of T in which paths are preserved (any path of T0 is a path of T). If T is the tree of a diagram D, then a subdiagram D0 of D may be associated with a subtree of T in the obvious way. We distinguish four types of extensions of a diagram. The tree diagram D+ is said to be an extension (simpliciter) of the tree diagram D if (i) + TD is a subtree of TD+ and (ii) for every w ∈ WD , Dw ⊆ D+ w ; D is + a conservative extension of D if, in addition, TD+ = TD ; D is a safe extension of D if it contains no original r-predicate symbols not already in D; and it is a normal extension of D if it is both safe and conservative.

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Let us now specify some of the desirable properties of diagrams. Their presence will help us in constructing a ‘canonical’ model from the diagram. Suppose that hw, v, Pi ∈ D. Then the arrow hw, vi in D is said to be regular if Dw and Dv are a maximal matching pair of fully complete P-companions with P closed in Dw and restrictive on Dv . The arrow is said to be almost regular if Dw and Dv are an almost matching pair of fully complete P-companions for which P is closed in Dw and Dv is Prestricted. If an arrow hw, vi of D is almost regular, then w (v) is said to be its anomalous node if Dw (resp. Dv ) is the anomalous member of the pair Dw , Dv . A tree diagram D itself is said to be regular if either it is has a single node w for which Dw is fully complete or it has more than one node and: (i) each arrow in the diagram is regular; and (ii) if w = w0 , w1 , . . . , wn = v is a path from w to v and a (or Q) ∈ L(Dw) ∩ L(Dv ), then a (resp. Q) ∈ Dwi for i = 1, . . . , n − 1. We call (ii) the path condition and if a (resp. Q) belongs to each member of the path p = w0 , w1 , . . . , wn , then we call p an a-path (resp. Q-path). The diagram is said to be almost regular under the same conditions but with (i) replaced by: (i0 ) each arrow in the diagram is almost regular and there is at most one anomalous node in the diagram. Let us explain how this terminology arises in the construction of the canonical model. Suppose that the sentence ♦P B ∈ Dw . Then we may wish to construct a world v that is linked to w by P and for which B ∈ Dv . By previous results we can ensure that Dw and Dv exactly match. But we may now have another sentence ♦Q C ∈ Dw and this may require us to construct another world u linked to w by Q. However, in getting Dw and Du to exactly match, we will lose the exact match between Dw and Dv . They will almost match, with w becoming an anomalous node. We shall show how these anomalies can be ironed out. But first we need a result on the behaviour of a world-language under extensions: To show that a diagram can retain regularity when we add new possibilities, we shall need to split the resulting diagram into two. Given a tree T, we say that hT1 , T2 , wi is a splitting of T if T1 and T2 are subtrees of T whose only node in common is w and which exhaust the nodes in T. Likewise, hD1 , D2 , wi is said to be a splitting of a tree diagram D if, for some T1 and T2 , hT1 , T2 , wi is a splitting of TD and D1 and D2 are the respective restrictions of D to T1 and T2 . We show how one half of a splitting can be regularized without harm to the other:

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LEMMA 1. Suppose that D is an almost regular tree-diagram and that hE, F, w0 i is a splitting of D in which F contains the sole anomolous node (if any) of D. Let F+ be a normal extension of F which contains no constants or s-predicate symbols that occur in E but not in F. Then the diagram D+ = E ∪ F+ is almost regular with w0 as its sole anomalous node. Proof. Let us verify the various conditions required for D+ to be almost regular: + First, take any w in WD . If w is in WF , then D+ w = Fw is fully complete since F+ is regular. If w is in WE − {w0 }, then D+ w = E w is fully complete since E is almost regular. Now take any hw, v, Pi ∈ D+ . Then either hw, v, Pi ∈ E or hw, v, Pi ∈ F. In either case, since E and F are both almost regular, P is closed in Dw and restrictive in Dv . Given that P is closed in Dw , it is closed in D+ w ⊇ Dw . + Given that Dv is P-restricted and that D contains no original r-predicate symbols that do not occur in D, D+ v is also P-restricted. Next, suppose hw, vi is either an arrow in F or an arrow of E for which neither w nor v is w0 . In the first case, hw, vi is a regular arrow of D+ since + + + F+ is regular and both D+ w = Fw and Dv = Fv ; and in the second case, + hw, vi is a regular arrow of D since E is almost regular, neither w nor v + is anomalous and both D+ w = Ew and Dv = Ev . In the remaining case, hw, v, Pi ∈ E with either w = w0 or v = w0 . Suppose that w = w0 (the case in which v = w0 is entirely similar). We must then show that hw, vi is an almost regular arrow in D+ with w = w0 the anomalous node. Since F contains the sole anomalous node of D, that node must be w = w0 . So, given that E is almost regular, Ew almost matches Ev . But + + + then the only way D+ w = Fw could fail to almost match Dv = Ev is if Fw contained a constant or r-predicate symbol s that belonged to Ev but not to Dw . If s is new to F+ then it cannot belong to E by supposition. If s is in F, then it is in Fu for some u. But there is a path from u to v that goes through w; and so, by the application of the path condition to D, s is in Dw after all. Since D is almost regular, 0 = Dv is a P-companion to 1 = Dw . Letting 1+ = D+ w , it is clear that the conditions for the application of + Corollary V.3 are satisfied; and so D+ v = Dv is a P-companion to Dw . + + Now letting 0 = Dv and 1 = Dw , it is clear that the conditions for the application of the Reciprocation Lemma (VI.12) will be satisfied; for 0 is a maximal P-companion to Dw , hence a maximal P-companion to 1, and so a broad P-companion to 1. Therefore, by the lemma, D+ w is a P-companion to D+ . v

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Finally, we must show that D+ conforms to the path condition. The only difficulty concerns the case in which there is a path from v in E to u in F (where neither v nor u is w0 ) with the constant or r-predicate symbol + s in both D+ v = Ev and D u = Fu . But then s already occurs in F and so belongs to some Ft . But any path from v to t will go through w0 and so, by the application of the path condition to D, s belongs to Dw and hence to D+ 2 w. LEMMA 2. Any finite almost regular diagram D has a normal regular extension D+ . Proof. By induction on the number m ≥ 1 of nodes in D. If D is already regular, there is nothing to prove. So we may assume that D contains exactly one anomalous node w1 . m = 1. Trivial, since the notions of regular and almost regular coincide for diagrams with one node. m = 2. Let the two worlds be w and v and suppose that hw, v, Pi ∈ D with v not anomalous. We first apply Lemma VII.5. Clearly, P is closed in Dw . Since Dw is N-complete, it is P-determinate. Dv is a P-companion to Dw by supposition. L(0) ∩ L(1) ⊆ LP (1), since 1 almost matches. Also L(0) − L(1) ⊆ L∼P (1) since if a constant or r-predicate symbol is in L(0) − L(1), then it is not P-bound in 0 and so, by 0 N-complete, it is ∼P-bound in 0. Since the condition for its application are met, it follows from the lemma that (Dv )+ = Dv ∪ {B : Dw ` P B, B in L(Dw )} is a matching P-companion to Dw . So by Lemma VII.3, there are safe and fully complete extensions 0 + and 1+ of Dw and (Dv )+ which are maximal matching P-companions and fully complete. Let D+ be the conservative + + + + extension of D for which D+ w = 1 and Dv = 0 . Then clearly D is a regular tree-diagram. The proof in case v is anomalous is similar but with the role of w and v reversed. In this case, P may not be closed in Dv but it will be effectively closed. For suppose Dv ` P B for B in L(Dv) and that a is a constant of B. If Pa ∈ / Dv , then ∼Pa ∈ Dv by Dv N-complete. But since P is closed in Dw , Dw ` P Cl(P); and so Dv ∪ {Cl(P)} is inconsistent and Dv is not a P-companion to Dw after all. m > 2. It should be clear that there is a splitting hT1 , T2 , w0 i of TD in which neither T1 nor T2 consists of a single node. For if TD takes the form of a single branch, then we may split the tree at any node other than the root or the tip; whereas if TD forks at the node w and if v is one of the nodes immediately above w, then we may let T1 consist of all the nodes below w, w itself, and all of the nodes above v and let T2 consist of w and all of the remaining nodes not in T1 .

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Let E and F be the restrictions of the diagram D to the T1 and T2 ; and let us assume that the anomalous node w1 of D belongs to F (the proof for the case in which it belongs to E is similar, but with the roles of E and F reversed). We now define the diagrams En and Fn by induction on n = 0, 1, 2, . . .: n = 0. Set E0 = E and F0 = F. n = 2k + 1, k ≥ 0. We let E2k+1 = E2k and let F2k+1 be a regular extension of F2k whose new constants and s-predicate symbols do not belong to E2k and which contains no new original r-predicate symbols (if such an extension does not exist, we let F2k+1 be F2k ). n = 2k, k > 0. We let F2k = F2k−1 and let E2k be a regular extension of E2k−1 whose new constants and s-predicate symbols do not belong to F2k−1 and which contains no new original r-predicate symbols (if such an extension does not exist, we let E2k be E2k−1 ). Let Dn = En ∪ Fn for n = 0, 1, 2, . . . . Then we may show by a subsidiary induction on n that Dn is an almost regular diagram whose sole anomalous node when n > 0 is w0 and that Fn is regular for odd n and En regular for positive and even n: n = 0. Trivial. n = 2k + 1, k ≥ 0. By IH (for n), D2k is an almost regular diagram whose sole anomalous node belongs to T2 . Since F2k has fewer than m nodes, it follows by the main IH (for m) that F2k has a normal regular extension F+ 2k . Clearly, we may arrange for all of the new constants and spredicate symbols of F+ 2 not to occur in E2k . So by the construction, F2k+1 will be such an extension F+ 2k . But then, by Lemma 1, D2k+1 will be an almost regular diagram whose sole anomalous node is w0 . n = 2k, k > 1. The proof is the same, mutatis mutandis, as for the previous case. Let E+ = E0 ∪E1 ∪E2 ∪· · ·, F+ = F0 ∪F1 ∪F2 ∪· · ·, and D+ = E+ ∪F+ . Since each Dn , for n > 0 is a safe and almost regular extension of D whose sole anomalous is w0 , the same is true of D+ . But D+ is also regular. For E+ is the union of an increasing chain E2 ⊆ E4 ⊆ E6 ⊆ · · · of regular diagrams and so must itself be regular. Similarly for F+ . But then every arrow in D+ = E+ ∪ F+ is regular and so D+ is itself regular. 2 We must now show how all possibilities may be ‘witnessed’ in a diagram. Say that a diagram D is M-complete (‘M’ for possibility) if ♦cF , |B| ⊂ cF, cF ≈ P ∈ Dw implies that, for some v ∈ WD , hw, v, Pi ∈ D and B ∈ Dv . Note that no requirement is directly imposed on ♦P B since it will not, in general, be a restricted sentence. A diagram is said to be nice if it is regular and M-complete.

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THEOREM 3. Any regular finite tree-diagram D has a nice extension D+ . Proof. Let w0 , w1 , w2 , . . . , a0 , a1 , a2 , . . . and Q0 , Q1 , Q2 , . . . be respective enumerations of infinitely many world-constants, individual constants and s-predicate symbols that do not appear in D. Let W = {w0 , w1 , w2 , . . .}, A = {a0 , a1 , a2 , . . .} and Q = {Q0 , Q1 , Q2 , . . .} and, for L = L(D), let L+ = L ∪ A ∪ Q. Partition A and Q into infinitely many infinite sets A0 , A1 , A2 , . . . and Q0 , Q1 , Q2 , . . .; and let Ln = L ∪ A0 ∪ A1 ∪ · · · ∪ An ∪ Q0 ∪ Q1 ∪ · · · ∪ Qn for n = 0, 1, 2, . . . . Let M0 , M1 , M2 , . . . be a repeating enumeration of all quadruples of the form hw, B, F, Pi, where w is a world constant from WD ∪ W, B is an original sentence from L+ , F is an original closed predicate from L+ , and P is an s-predicate symbol from L+ . We define the diagrams D0 , D1 , D2 , . . . as follows: n = 0. D0 = D. n = k + 1. Suppose that Mk = hw, B, F, Pi. If it is not the case that ♦cF A, |B| ⊂ cF, cF ≈ P ∈ Dk,w , then let Dk+1 = Dk . If it is the case that ♦cF B, |B| ⊂ cF, cF ≈ P ∈ Dk,w , then let D+ k = Dk ∪ {hw, v, Pi, hv, Bi}, for v the first world-constant that does not appear in Dk , and let Dk+1 be a normal regular extension of D+ k in the language Lk+1 (if no such extension exists, we may let Dk+1 = D+ k ). We may show by induction on n = 0, 1, 2, . . . that Dk is a regular diagram: n = 0. Trivial. n = k + 1. Let 1 = Dk,w and 0 = {B}. By Lemma VII.4, 0 is a Pcompanion to 1. Now it is clear that the conditions for the application of Lemma VII.5 are satisfied; and so 0 + = 0 ∪ {B : 1 ` P B, B in L(1)} and 1 are matching P-companions. But then by Lemma VII.3, there are safe extensions 1∗ and 0 ∗ of 1 and 0 that are maximally matching Pcompanions and fully complete. Clearly, we can arrange that none of the new symbols of 1∗ or 0 ∗ occur in Dk . Let D+ k = Dk ∪ {hw, Ci : C ∈ ∗ ∗ 1 } ∪ {C : hv, Ci ∈ 0 } ∪ {hw, v, Pi}, for v a new world-constant. Then given by IH that Dk is regular, it is readily verified that D+ k is almost regular with w the sole anomalous node. So by Lemma 2, it has a regular extension containing {hw, v, Pi, hv, Bi}; and Dk+1 is therefore such a regular extension. Let D+ = D0 ∪ D1 ∪ D2 ∪ · · · . Then since D+ is the union of an increasing chain D0 ⊆ D1 ⊆ D2 ⊆ · · · of regular diagrams, it is itself a regular diagram. Suppose now that ♦cF B, |B| ⊂ cF, cF ≈ P ∈ Dw for some world-constant w of D+ . Then, for some k, ♦cF B, |B| ⊂ cF,

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cF ≈ P ∈ Dk,w . But, for some l > k, M1 = hw, B, F, Pi. So by the construction, there is a v for which hw, v, Pi, hv, Bi ∈ Dk+1 ⊆ D+ . 2 COROLLARY 4. Suppose that 1 is a consistent set of sentences with original individual constants and predicate symbols. There is then a nice diagram with root w0 for which Dw0 contains 1. Proof. By Theorem 3 and Lemma VII.6. 2

9. T HE C ANONICAL M ODEL

We construct a canonical model from a nice diagram in the obvious way and show that it verifies at each world the very sentences belonging to that world. We first give a breakdown of the results we shall require. The truth of -sentences is transferred along appropriate arrows: LEMMA 1. Suppose that D is a nice diagram and that hw, v, Pi ∈ D. Then: (i) Dw ` P B with B in L(Dw ) implies B ∈ Dv ; (ii) P B ∈ Dw implies P B ∈ Dv . Proof. (i) Suppose Dw ` P B with B in L(Dw ). Since Dw and Dv exactly P-match, B is in L(Dv ). But then given that Dv is a maximal Pcompanion to Dw , B ∈ Dv . (ii) Suppose P B ∈ Dw . Then P P B is in L(Dw ) (though not necessarily restricted) and, by Theorem IV.6(ii) of LE, Dw ` P P B. So given that P B is among the restricted sentences of L(Dw ), it follows by (i) that P B ∈ D v . 2 The r-predicate symbols behave as rigid predicates: LEMMA 2 (Predicate Rigidity). Let D be a nice diagram and P an rpredicate symbol. Then: (i) if Qa ∈ Dw and Q ∈ L(Dv ) then Qa ∈ Dv ; (ii) if Q ∈ L(Dw ) and a ∈ L(Dv ) whenever Qa ∈ Dw then Q ∈ L(Dv ). Proof. Let w = w0 , w1 , . . . , wn = v be the path from w to v in TD . The result, in both cases, is proved by induction on n: (i) n = 0. Trivial. n = k + 1. Suppose Qa ∈ Dw and Q ∈ L(Dv ). Since Q ∈ L(Dw ) ∩ L(Dv ), it follows by the path condition that Q ∈ L(Dwk ). So by

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IH, Qa ∈ Dwk . Suppose wk is linked by P to wk+1 = v in D. Then since Q ∈ L(Dwk ) ∩ L(Dwk+1 ), Q is P-bound in Dwk , i.e. Q ⊂ P ∈ Dwk . So from Qa ∈ Dwk it follows by subsumption that Dwk ` P Qa. But then Qa ∈ Dwk+1 by Lemma 1. (ii) n = 0. Trivial. n = k + 1. Suppose Q ∈ L(Dw ) and a ∈ L(Dv ) whenever Qa ∈ Dw . Then by the path condition, a ∈ L(Dwk ) whenever Qa ∈ Dw . So Q ∈ L(Dwk ) by IH and hence, by part (i), Qa ∈ Dwk iff Qa ∈ Dw . Suppose that wk is linked by P to wk+1 in D. Then Q ⊂ P ∈ Dwk . For otherwise, by Dwk N-complete, ∼(Q ⊂ P) ∈ Dwk ; and so, by Dwk Q-complete, Qa, ∼Pa ∈ Dwk for some constant a. Since Qa ∈ Dwk , Qa ∈ Dw and hence, by supposition, a ∈ L(Dv ). But then since a ∈ L(Dwk ) ∩ L(Dwk+1 ), it follows from the P-match between Dwk and Dwk+1 that a is P-bound in Dwk , contrary to the fact that ∼Pa ∈ Dwk . Therefore Q ⊂ P ∈ Dwk and so, again, by the P-match between Dwk and Dwk+1 , Q ∈ L(Dwk+1 ). Identity and dependence behave as they should: LEMMA 3. Let D be a nice diagram. Then: (i) a = b ∈ Dw and a ∈ L(Dv ) implies a = b ∈ Dv ; (ii) a ≥ b ∈ Dw and a ∈ L(Dv) implies a ≥ b ∈ Dv ; (iii) B(a) ∈ Dw and a = b ∈ Dv implies B(b) ∈ Dw . Proof. (i) Let w = w0 , w1 , . . . , wn = v be the path from w to v in TD . We prove that a = b ∈ Dwn by induction on n: n = 0. Trivial. n = k +1. By the Path Condition, a ∈ L(Dwk ). So by IH, a = b ∈ Dwk . Suppose that wk is linked to vk+1 by P in D. Since a ∈ L(Dwk )∩L(Dwk+1 ), it follows from the P-match between Dwk and Dwk+1 that Pa ∈ Dwk ∩Dwk+1 . By Theorem IV.7(i) of LE, ` a = b → a a = b; so by subsumption, Pa, a = b ` P a = b; and so, P a = b, Pb ∈ Dwk . But then a = b ∈ Dwk+1 by Lemma 1. (ii) Again, let w = w0 , w1 , . . . , wn = v be the path from w to v in TD . We prove that a ≥ b ∈ Dwm by induction on m. m = 0. Trivial. m = k + 1. By the Path Condition, a ∈ L(Dwk ). So by IH, a ≥ b ∈ Dwk . Suppose that wk is linked to wk+1 by P in D. Since a ∈ L(Dwk ) ∩ L(Dwk+1 ), it follows from the P-match between Dwk and Dwk+1 that Pa ∈ Dwk ∩ Dwk+1 . By Theorem IV.10(iii) of LE, ` a ≥ b → a a ≥ b; so by Subsumption and Chaining, Pa, a ≥ b ` P a ≥ b; and so, P a ≥ b ∈ Dwk . But then a ≥ b ∈ Dwk+1 by Lemma 1.

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(iii) Suppose B(a) ∈ Dw and a = b ∈ Dv . If a does not occur in B(a), there is nothing to prove. If it does, then a ∈ L(Dw ). But then a = b ∈ Dw by part (i); and so B(b) ∈ Dw by the logic of identity. 2 It is now easy to show that ‘=’ behaves in the manner of an equivalence relation on constants. Let KD be the set of constants that appear in D. Define the relation ≡ on KD by: a ≡ b if a = b ∈ Dw for some w in WD . LEMMA 4. ≡ is an equivalence relation. Proof. The verification of reflexivity and symmetry is straightforward. To establish transitivity, suppose that a = b ∈ Dw and b = c ∈ Dv . Then b = a ∈ Dw by the logic of identity. Since b ∈ L(Dw ), b = c ∈ Dw by Lemma 3(i); and so, again by the logic of identity, a = c ∈ Dw . 2 We can now define the canonical model for a nice diagram D. Given a ∈ KD , let a∗ = {b ∈ KD : b ≡ a}. The canonical model MD for D is then the quadruple hW, I, , φi, where: (i) (ii) (iii) (iv)

W = WD ; Iw = {a∗ : a ∈ L(Dw )};  = {ha∗ , b∗ i : a ≥ b ∈ Dw for some w ∈ W }; φ(a) = a∗ for each constant, φ(w, F) = {ha∗1 , . . . , a∗n i : Fa1 . . . an ∈ Dw } for each pure n-place predicate symbol F, and φ(P) = {a∗ : Pa ∈ Dw } for some w ∈ W } for each rigid predicate symbol P of D.

Note that the canonical model is full; each object has a name. LEMMA 5. For D nice, MD is indeed a model. Proof. 6x(x = x) ∈ Dw for any w. By Dw Q-complete, a = a ∈ Dw ; and so each Iw = {a∗ : a ∈ L(Dw )} is non-empty. It remains to show that each Iw is closed under . So suppose a∗ ∈ Iw and a∗  b∗ . Then a0 ≥ b0 ∈ Dv for some v ∈ WD , where a0 ≡ a and b0 ≡ b. So a0 ∈ L(Dw) by Lemma 3(i) and a0 ≥ b0 ∈ Dw by Lemma 3(ii). But then b0 ∈ L(Dw) and b∗ ∈ Iw , as required. 2 Sentences and predicates are defined at a world in the canonical model when they are in the language of that world: LEMMA 6 (Definability). Let D be a nice diagram. Then a sentence or closed predicate E is defined at w in the associated canonical MD iff E is in L(Dw ).

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Proof. E is defined at w ∈ WD iff φ(a) ∈ Iw and φ(P) ⊂ Iw for each constant a and r-predicate symbol P in E. E is in L(Dw ) iff a ∈ L(Dw ) and P ∈ L(Dw ) for each constant a and r-predicate symbol P in E. Therefore it suffices to show: (i) φ(a) ∈ Iw iff a ∈ L(Dw); and (ii) φ(P) ⊆ Iw iff P ∈ L(Dw). We consider each in turn: (i) φ(a) ∈ Iw iff a = b ∈ Dv for some v and some b ∈ L(Dw) iff a ∈ L(Dw ) by Lemma 3(i). (ii) Suppose that P ∈ L(Dv) for v ∈ WD . Then: φ(P) = {a∗ : Pa ∈ Du for some u ∈ WD } = {a∗ : Pa ∈ Dv } by Lemma 2(i). So: φ(P) ⊆ Iw

iff {a∗ : Pa ∈ Dv } ⊆ Iw iff {a : Pa ∈ Dv } ⊆ L(Dw ) by Lemma 3(i) iff P ∈ L(Dw ) by Lemma 2(ii).

2

THEOREM 7 (Truth). Let D be a nice diagram and MD = hW, I, ,φi its canonical model. Then for each w ∈ WD , sentence B and closed n-place predicate H: (a) w  B if B ∈ Dw , for any B in L(Dw ); and (b) Hw = {ha∗1 , . . . , a∗n i : Ha1 . . . an ∈ Dw }, for any H in L(Dw ). Proof. We should note that by Lemma 6 the stipulated side conditions under (a) and (b) are exactly those in which the sentence B and predicate H will be defined at w in MD . Normally, for the purposes of a proof by induction, we would replace (a) by the corresponding biconditional: (a)+ w  B iff B ∈ Dw , for any B in L(Dw). But this is only because B may figure in other more complex sentences. In the present case, unnegatable sentences of the form S C do not occur in more complex sentences; and so it suffices to establish (a)+ whenever B is negatable and (a) whenever B is unnegatable. The proof now proceeds by a simultaneous induction on B and G:

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(i) (a) B = Ha1 . . . an . w  Ha1 . . . an

iff hφ(a1 ), . . . , φ(an )i ∈ Hw iff ha∗1 , . . . , a∗n i ∈ Hw , by def. of φ iff Ha1 . . . an ∈ Dw by IH (b).

(i) (b) B = (a = b). wa=b

iff iff iff iff

φ(a) = φ(b) iff a∗ = b∗ a = b ∈ Dv for some v ∈ WD a = b ∈ Dw by Lemma 3(i) and the fact that a = b is in L(Dw ).

wa≥b

iff iff iff iff

φ(a)  φ(b) a∗  b∗ a ≥ b ∈ Dv for some v ∈ WD a ≥ b ∈ Dw by Lemma 3(ii) and the fact that a ≥ b is in L(Dw).

(i) (c)

(ii) B = ∼C. By Dw consistent and N-complete; (iii) B = Q (C ∨ D). Again, by Dw consistent and N-complete; (iv) B = xC(x). By Dw standardly complete. (v) B = F C. Suppose first that F C ∈ Dw . Since Dw is P-complete, F ≈ Q ∈ Dw for some s-predicate symbol Q (if F is already an s-predicate symbol, we may let Q = F). Let J = Fw . By IH, J = {a∗ : Fa ∈ Dw }; and, given F ≈ Q ∈ Dw , it readily follows that J = {a∗ : Qa ∈ Dw }. Now suppose J ⊆ Iv , i.e. {a∗ : Qa ∈ Dw } ⊆ {a∗ : a ∈ L(Dv )}. Then by Lemmas 2(i) and 3(i), {a : Qa ∈ Dw } ⊆ {a : a ∈ L(Dv )}; and so by Lemma 2(ii), Q ∈ L(Dv ). Let w = w0 , w1 , . . . , wn = v be a path from w to v. We now prove by induction on that Q C ∈ Dwm , m = 0, 1, . . . , n: m = 0. Trivial. m = k + 1. By IH, Q C ∈ Dwk . By the path condition, Q ∈ L(Dwk+1 ). Suppose that wk is linked to wk+1 by P in TD . Since Q ∈ L(Dwk ) ∩ L(Dwk+1 ), Q ⊂ P ∈ Dwk . So by Theorem IV.6(ii) of LE and subsumption, Dwk ` P Q C; and so by Lemma 1(ii), Q C ∈ Dwk+1 . Letting m = n in the above result, Q C ∈ Dv and so C ∈ Dv . But then by IH, v  C; and so w  F C.

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Suppose now that F C ∈ / Dw . Let us note that in this case F will be an original predicate and so, by Dw an N-complete theory, ♦F ∼C ∈ Dw . We distinguish two subcases. (1) |C| ⊂ cF ∈ / Dw . By Dw N-complete and the fact that C is an original sentence, ∼(|C| ⊂ cF) ∈ Dw . It now follows from IH (and an unproblematic application of (ii)–(iv)) that w  ∼(|C| ⊂ cF). But ` ∼(|C| ⊂ cF) → F C; and so, by Soundness, not w  F C. (2) |C| ⊂ cF ∈ Dw . By axiom (IV)(i), ♦cF ∼C ∈ Dw . By Dw Pcomplete, cF ≈ Q ∈ Dw for some s-predicate symbol Q. So by D Mcomplete, there is a v for which hw, v, Qi ∈ D and ∼ C ∈ Dv . By Dv consistent, C ∈ / Dv ; and by IH, not v  C. Let J = Fw . Then c(J ) = c(Fw ) = (cF)w  = a∗ : (cF)a ∈ Dw , from IH  = a∗ : Qa ∈ Dw , by cF ≈ Q ∈ Dw . Since Qa ∈ Dw implies Qa ∈ Dv , c(J ) ⊆ Iv and so J ⊆ c(J ) ⊆ Iv . Therefore, given not v  C, not w  F C. (vi) H a pure predicate symbol F. Then Fw = φ(w, F) by definition of Fw = {a∗ : Fa ∈ Dw } by definition of φ in MD . (vii) H a rigid predicate symbol P. Then Pw = φ(P) by definition of Pw = {a∗ : Pa ∈ Dv for some v ∈ WD } = {a∗ : Pa ∈ Dw } by Lemma 2(i) and the fact that P ∈ L(Dw ). (viii) H a λ-abstract λxA(x).  λxA(x)w = a∗ ∈ Iw : w  A(a) by definition of λxA(x)w  by IH = a∗ ∈ Iw : A(a) ∈ Dw  ∗ = a ∈ Iw : λxA(x)a ∈ Dw by Axiom III(i). 2 From Theorem 7 and Corollary VIII.4 we obtain our main result:

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THEOREM 8. Any consistent set of formulas of E5 has a model. This result can be extended to various stronger systems in an obvious way. Suppose, for example, that we add the domain axiom: Q Q (V) (ii) xPx → P xPx of LE. Then the resulting system can be proved complete for the condition that no domain of a model be properly included in any other domain. Or again, if an existence-predicate E is added to the language, then the axioms: Q (V) (i) xPx → ♦P Ex; and (≥ E) x ≥ y → (Ex → Ey), correspond to the conditions: (V) (i)∗ if a ∈ Iw then, for some v ∈ W, Iv ⊇ Iw and a ∈ φ(E, v); (≥ E)∗

b ∈ φ(w, E) whenever a ∈ φ(w, E) and a  b.

It would be of interest to extend the completeness results to the variants of E5 considered at the end of LE and also to systems with a different underlying modality. For some of these variants, the proofs of completeness are relatively straightforward;2 for others it is not. We should also note that there are other logics in which the modal operator may be naturally taken to be indexed to a group of items. We may talk, for example, of group knowledge or of group obligation or of what is provable by a group of proofs, i.e. within a system. It would be of interest to work out systems for these other cases and to develop a general account of systems of this sort.

N OTES 1 I should like to thank the referee and participants of conferences at Stanford (94) and

Uppsala (98) for helpful remarks on an earlier presensation of this material. 2 Let me note, in this connection, that Fabice Correia has established completeness for a propositional version of the system using a direct canonical model construction and Joshua Schechter has done the same for various systems with a weak underlying modality.

R EFERENCES Correia, F. (2000): A propositional logic of essence, to appear in J. Philos. Logic. Fine, K. (1994): Essence and modality, in J. Tomberlin (ed.), Philosophical Perspective 8, pp. 1–16, reprinted in Philosopher’s Annual, 1994.

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Fine, K. (1995): Senses of essence, in W. Sinnott-Armstrong (ed.), Modality, Morality and Belief, Cambridge Univ. Press, pp. 53–77. Fine, K. (1995): The logic of essence, J. Philos. Logic 24, 241–273.

Department of Philosophy 1879 Hall Princeton University Princeton, NJ 08544-10006, USA