Craig Interpolation Theorem for L!1! (or L!! )

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1 Craig Interpolation Theorem for L!1! (or L!! ) Theorem We assume that L 1 and L 2 are vocabularies. Suppose =!, where is an L 1 -sentence and is an L 2 -sentence of L!!. Then there is an L 1 \ L 2 -sentence of L!! such that 1 =! 2 =! Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

2 Proposition Suppose depends only on R in the sense that M = () N = whenever M and N have the same domain and the same interpretation of R. Then = $ where no non-logical symbols except R occurs in. Proof. Let S 1,...,S n be the non-logical symbols in or in addition to R. Let 0 be the result of replacing each S i in by a new symbol Si 0. Now =! 0. By the Interpolation Theorem there is containing only R such that =! and =! 0. Hence = $. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

3 Theorem (Beth Definability Theorem) Suppose the predicate S depends only on R in the sense that (R, S) ^ (R, S 0 ) = 8x(S(x) $ S 0 (x)). Then there is (R, x) where S does not occur such that (R, S) = 8x(S(x) $ (R, x)). Proof. By assumption Let (R, c) be such that = ( (R, S) ^ S(c))! ( (R, S 0 )! S 0 (c)). = ( (R, S) ^ S(c))! (R, c) and = (R, c)! ( (R, S 0 )! S 0 (c)). Then (R, S) = 8x(S(x) $ (R, x)). Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

4 Interpolation in second order logic Remark Single sorted logic second order logic satisfies trivially the Interpolation Theorem: Suppose =!, where is an L 1 -sentence and is an L 2 -sentence of second order logic. Then there is an L 1 \ L 2 -sentence of second order logic such that =! and =!, obtained as follows. Let L 1 \ L 2 consist (only) of the relation symbols R 1,...,R n, a(r i )=m i. We can write as (R 1,...,R n ). Let : 9X m 1...9X mn (X m 1,...,X mn ). Many sorted second order logic does not satisfy Interpolation Theorem (see later). Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

5 Towards many sorted interpolation Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

6 Let Un( ) be all sorts s such that a variable of sort s occurs universally quantified in. Similarly Un(S) for a set S of formulas. Let Ex( ) be all sorts s such that a variable of sort s occurs existentially quantified in. Similarly Ex(S) for a set S of formulas. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

7 Many Sorted Interpolation Theorem for L!1! (or L!! ) Theorem We assume that L 1 and L 2 are relational. Suppose =!, where is an L 1 -sentence and is an L 2 -sentence of L!1!. Then there is an L 1 \ L 2 -sentence of L!1! such that 1 =! 2 =! 3 Un( ) Un( ) 4 Ex( ) Ex( ). If and are strict, then so is. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

8 Proof: Let us assume that the claim of the theorem is false and derive a contradiction. Since =!, the set {, } has no models. We construct a consistency property for {, }. Let L = L 1 \ L 2. Suppose C s = {c s n : n 2 N} is a set of new constant symbols for each sort s of L 1 [ L 2. Let C = S s Cs. Given a set S of sentences, let S 1 consists of all L 1 [ C -sentences in S with only finitely many constant from C, and let S 2 consists of all L 2 [ C -sentences in S with only finitely many constant from C. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

9 Definition Let us say that separates S 0 and S 00 if 1 S 0 =, 2 S 00 =, 3 Un 0 ( ) Un(S 0 ), 4 Ex 0 ( ) Un(S 00 ), where Un 0 ( ) consists of sorts s 2 Un( ) and sorts of constants c 2 C occurring in, and Ex 0 ( ) consists of sorts s 2 Ex( ) and sorts of constants c 2 C occurring in. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

10 Definition Let consist of finite sets S of sentences of L!1! such that S = S 1 [ S 2 and : (?) There is no L [ C -sentence that separates S 1 and S 2. In the strict case we demand that all the sentences are strict. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

11 is a consistency property Case 0. {, } 2. True by assumption. (We may assume that is consistent and that is not valid.) Case 1. Suppose S 2 and consider c = c, where, for example, c 2 L 1 [ C. We let S1 0 = S 1 [ {c = c} and S2 0 = S 2 [ {c = c}. Suppose (c 0,...,c m 1 ) separates S1 0 and S0 2. Then clearly also (c 0,...,c m 1 ) separates S 1 and S 2, a contradiction. Case 2. Suppose (c) 2 S 2, (c) atomic, and c = c 0 2 S. Clearly S [ { (c 0 )} 2. Case 3. Negation: trivial. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

12 Case 4. Consider W i2i ' i, where for example W i2i ' i 2 S 1. We claim that for some i 2 I the sets S 1 [ {' i } and S 2 satisfy (?). Otherwise there is for each i 2 I some i that separates S 1 [ {' i } and S 2. Let = W i2i i. Then separates S 1 and S 2 contrary to assumption. Case 5. Consider ' i where for example V i2i ' i 2 S 1. Let S1 0 = S 1 [ {' i } and S2 0 = S 2. If separates S1 0 and S0 2, then clearly also separates S 1 and S 2. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

13 Case 6. Consider S 2 and 9x s (x s ) 2 S 1. Let c 0 2 C s be such that c does not occur in S. We claim that the sets S 1 [ { (c 0 )} and S 2 satisfy (?). Otherwise there is some (c 0,...,c m 1 ) that separates S 1 [ { (c 0 )} and S 2. Let 1 0 (c 1,...,c m 1 )=9x s (x s, c 1,...,c m 1 ). We show that 0 (c 1,...,c m 1 ) separates S 1 and S 2, a contradiction. Checking this: S 1 = 0 (c 1,...,c m 1 ): S 1 [ { (c 0 )} = (c 0,...,c m 1 ) by assumption S 1 = (c 0 )! (c 0,...,c m 1 ) S 1 = 8x s ( (x s )! (x s, c 1,...,c m 1 )) S 1 = 9x s (x s )!9x s (x s, c 1,...,c m 1 ) S 1 = 9x s (x s, c 1,...,c m 1 ) as S 1 = 9x s (x s ) S 1 = 0 (c 1,...,c m 1 ) 1 If c 0 does not occur in, then we take 0 =. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

14 Case 6. (Contd.) S 2 = 0 (c 0,...,c m 1 ): S 2 = (c 0,...,c m 1 ) S 2 = 8x s (x s, c 1,...,c m 1 ) S 2 = 9x s (x s, c 1,...,c m 1 ) S 2 = 0 (c 1,...,c m 1 ) Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

15 Case 6. (Contd.) Un 0 ( 0 (c 1,...,c m 1 )) Un(S 1 ): s 0 2 Un 0 ( 0 (c 1,...,c m 1 )) s 0 2 Un 0 ( (c 0,...,c m 1 )) s 0 2 Un(S 1 [ { (c 0 )}). If s 0 2 Un( (c 0 )), then s 0 2 Un(9x s (x s )), whence s 0 2 Un(S 1 ). Ex 0 ( 0 (c 1,...,c m 1 )) Un(S 2 ): s 0 2 Ex 0 ( 0 (c 1,...,c m 1 )) s 0 2 Ex 0 ( (c 0,...,c m 1 )) [ {s} s 0 2 Un(S 2 ) [ {s} = Un(S 2 ), since c 0 occurs in (c 0,...,c m 1 ), and hence c 0 2 Un(S 2 ) by the choice of. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

16 Case 6. (Contd.) Consider S 2 and 9x s (x s ) 2 S 2. Let c 0 2 C s be such that c 0 does not occur in S. We claim that the sets S 1 and S 2 [ { (c 0 )} satisfy (?). Otherwise there is some (c 0,...,c m 1 ) that separates S 1 and S 2 [ { (c 0 )}. Let 2 0 (c 1,...,c m 1 )=8x s (x s, c 1,...,c m 1 ). We show that 0 (c 1,...,c m 1 ) separates S 1 and S 2, a contradiction. Checking this: S 1 = 0 (c 1,...,c m 1 ): S 1 = (c 0,...,c m 1 ) S 1 = 8x s (x s, c 1,...,c m 1 ) S 1 = 0 (c 1,...,c m 1 ) 2 If c 0 does not occur in, we choose 0 =. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

17 S 2 = 0 (c 1,...,c m 1 ): S 2 [ { (c 0 )} = (c 0,...,c m 1 ) S 2 = (c 0 )! (c 0,...,c m 1 ) S 2 = 8x s ( (x s )! (x s, c 1,...,c m 1 )) S 2 = 9x s (x s )!9x s (x s, c 1,...,c m 1 ) S 2 = 9x s (x s, c 1,...,c m 1 ) S 2 = 0 (c 1,...,c m 1 ) Un 0 ( 0 (c 1,...,c m 1 )) Un(S 1 ): s 0 2 Un 0 ( 0 (c 1,...,c m 1 )) s 0 2 Un 0 ( (c 0,...,c m 1 )) 3 s 0 2 Un(S 1 ) by the choice of. 3 Remember that c 0 occurs in. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

18 Ex 0 ( 0 (c 1,...,c m 1 )) Un(S 2 ): Suppose s 0 2 Ex 0 ( 0 (c 1,...,c m 1 )). Then s 0 2 Ex 0 ( (c 0,...,c m 1 )) because the universal quantifier in front of 0 adds no new existentially quantified variables. By our assumption about, s 0 2 Un(S 2 [ { (c 0 )}). If s 0 2 Un( (c 0 )), then s 0 2 Un(S 2 ). So in any case s 0 2 Un(S 2 ). Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

19 Case 7. Consider S 2, (c 0 ), where c 0 2 C s and 8x s (x s ) 2 S 1. Exercise! Case 7. (Contd.) Consider (c 0 ), where c 0 2 C s and 8x s (x s ) 2 S 2. Exercise! Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

20 Appications of interpolation Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

21 Single sorted interpolation follows a fortiori. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

22 Many sorted interpolation gives preservation results. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

23 Theorem (Łoś-Tarski-Malitz) A formula is preserved by submodels if and only if it is logically equivalent to a universal formula. Even in L!1!. Definition Let EXT be the conjunction of 8x 1 9x 0 (x 0 = x 1 ) 8x 1 8y 1 (R 0 (x 1, y 1 ) $ R(x 1, y 1 )) This is true in ({M 0, M 1 }, R, R 0 ) iff (M 1, R 0 ) (M 0, R). Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

24 Theorem (Łoś-Tarski-Malitz) A formula is preserved by submodels if and only if it is logically equivalent to a universal formula. Even in L!1!. Definition Let EXT be the conjunction of 8x 1 9x 0 (x 0 = x 1 ) 8x 1 8y 1 (R 0 (x 1, y 1 ) $ R(x 1, y 1 )) This is true in ({M 0, M 1 }, R, R 0 ) iff (M 1, R 0 ) (M 0, R). Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

25 Proof. Let us assume the single sorted is written in sort 0 variables. Let 0 be the same written in sort 1 variables and with R replaced by R 0. Then EXT ^ 0 = EXT ^ 0 =, = The only common sort is 0, so has only sort 0 symbols. To see that is existential we note that if a sort 0 variable was universally quantified in, then it is universally quantified in EXT ^ 0, but there is no universally quantified sort 0 variable in EXT ^ 0. Thus is existential. Moreover, = $. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

26 Definition A flow formula is a formula of the vocabulary of Chu spaces made of atomic formulas and their negations in the lax 2-sorted logic of Chu spaces by means of ^, _, 9x 0, 8x 1. Definition A Chu transform is simple if its mappings g and h are identities. Lemma Flow formulas are preserved by simple Chu transforms. Proof. Easy induction on formulas. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

27 Theorem (van Benthem) A formula of 2-sorted first order logic of Chu spaces is preserved by simple Chu transforms if and only if it is equivalent to a flow-formula. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

28 Suppose, written in 0 and 1 is preserved by Chu transforms. Then is existential in 0: Let 0 be but in 2 and 3, and R 0 instead of R. Let EXT 1 : 8x 0 9x 2 x 0 = x 2 EXT 2 : 8x 0 8y 0 (R(x 0, y 0 ) $ R 0 (x 0, y 0 )) EXT 3 : 8x 3 9x 1 x 3 = x 1 This is true in ({M 0, M 1, M 2, M 3 }, R, R 0 ) iff id is a Chu transform M 0, M 1, R)! (M 2, M 3, R 0 ). We need 0 in 2 and 3. We want 2 to by existential only in 0, so on the left hand side we can have no universal 2. So we want ^ EXT 1 ^ EXT 2 ^ EXT 3 = ^ EXT 1 ^ EXT 2 ^ EXT 3 = 0, 0 = 0 Now 0 is in 2 and 3 and existential in 2. Clearly = from 0 by switching back to 0 and 1. $ if we get Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

29 So we have, existential in x 0 which is preserved by Chu transforms. We show that is universal in x 1. Let 0 be but R 0 instead of R. Let EXT 1 : 8x 0 9x 2 x 0 = x 2 EXT 2 : 8x 0 8y 0 (R(x 0, y 0 ) $ R 0 (x 0, y 0 )) EXT 3 : 8x 3 9x 1 x 3 = x 1 This is true in ({M 0, M 1, M 2, M 3 }, R, R 0 ) iff id is a Chu transform M 0, M 1, R)! (M 2, M 3, R 0 ). We need 0 in 0 and 1. We want 1 to by universal only in 0, so on the right hand side we can have no existential 1. So we want = (EXT 1 ^ EXT 2 ^ EXT 3 )! 0 = 0, 0 =(EXT 1 ^ EXT 2 ^ EXT 3 )! 0 Now 0 is in 0 and 1 and universal in 1. Moreover it is existential in 0, because is. Clearly = $ if we get from 0 by switching back to R. Jouko Väänänen (Helsinki and Amsterdam) Many-sorted logic January / 114

Introduction to Sets and Logic (MATH 1190)

Introduction to Sets and Logic (MATH 1190) Introduction to Sets Logic () Instructor: Email: shenlili@yorku.ca Department of Mathematics Statistics York University Sept 18, 2014 Outline 1 2 Tautologies Definition A tautology is a compound proposition