Embedding logics into product logic. Abstract. We construct a faithful interpretation of Lukasiewicz's logic in the product logic (both

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1 1 Embedding logics into product logic Matthias Baaz Petr Hajek Jan Krajcek y David Svejda Abstract We construct a faithful interpretation of Lukasiewicz's logic in the product logic (both propositional and predicate). Using known facts it follows that the product predicate logic is not recursively axiomatizable. We prove a completeness theorem for the product logic extended by a unary connective 4 of Baaz [1]. We show that Godel's logic is a sublogic of this extended product logic. We also prove NP-completeness of the set of propositional formulas satisable in product logic (resp. in Godel's logic). 1 Introduction We shall be concerned with many-valued logics in this paper; in particular, in Lukasiewicz's logic L, Godel's logic G and product logic P. Our aim is to obtain information about complexity of these logics in terms of recursive theory (in the case of predicate logic) or in terms of computational complexity theory (in the case of propositional logic). Scarpellini [13] and Mundici [9] provide such information for Lukasiewicz's logic. Hence we shall concentrate on the other two logics. All three logics have the same language. The propositional connectives are 0; 1; &;!. The truth-values are drawn from the unit interval [0; 1]. The values of 0; 1 are always 0 and 1 respectively. The three logics we consider compute the value of & by a t-norm and the value of! by the corresponding residuum. In particular, in Lukasiewicz's logic x & L y = max(0; x + y? 1) and in Godel's logic and and in product logic and 1 if x y x! L y = 1? x + y otherwise x & G y = min(x; y) 1 if x y x! G y = y otherwise x & P y = x y 1 if x y x! P y = otherwise y x Partially supported by the grant No. A /96 of the Grant Agency of the Academy of Sciences of the Czech Republic and by the COST Action 15. y Partially supported by the US - Czechoslovak Science and Technology Program grant No , and by the grant No of the Grant Agency of the Academy of Sciences of the Czech Republic.

2 The negation is dened as : x = x! 0 Minimum (^) and maximum (_) can be dened in all three logics as x ^ y = x & (x! y) and x _ y = ((x! y)! y) ^ ((y! x)! x) respectively. An evaluation is a mapping of the set of propositional variables to the unit interval [0,1]. Every evaluation e can be extended using particular truth functions to the evaluation e of all formulas (e.g. the truth value of a formula '& in Lukasiewicz's logic will be e L ('& ) = e L (') & L e L ( ) etc.) using the truth tables of the particular logic. In the propositional case we ask about the existence of a complete axiomatization of the set of 1-tautologies. These are formulas that obtain the truth value 1 by each evaluation. In all three cases a complete axiomatization exists. On the other hand, if we introduce models of multivalued predicate calculi and ask again about a complete axiomatization of 1-tautologies we get the positive answer for the Godel's logic (see [14]), and negative (due to [13]) for the Lukasiewicz. In the following we will see how to embed Lukasiewicz's logic into product and use the fact that it is not recursively axiomatizable to prove the recursive non-axiomatizability of product predicate calculus. We then study an extension of product logic that contains G, and complexity of propositional satisability in P and G. We prove NP-completeness of the propositional satisability in P and G. Embedding Lukasiewicz's logic into product logic In this section we will show that Lukasiewicz's t-norm x & L y = max(0; x + y? 1) can be isomorphically transformed to (restricted) product on [a, 1] for arbitrary xed 0 < a < 1. This fact is also a direct consequence of the result of [8]. Then we will present a translation of formulas such that ' is 1-tautology of Lukasiewicz's logic if and only if its particular translation is 1-tautology of product logic. We apply this isomorphism also to models of corresponding predicate calculi and as a conclusion we get, via Scarpellini result that the set of 1-tautologies of Lukasiewicz's logic is not recursively enumerable, that the same holds for the set of 1-tautologies of product predicate calculus. Lemma 1 For each 0 < a < 1, [0,1] with Lukasiewicz's conjunction x & L y = max(0; x + y? 1) is isomorphic to [a; 1] with restricted product x& P ay = max(a; x y). Proof The isomorphism is f a (x) = a 1?x from which f?1 a (y) = 1? log a y. We can see that f a (x)& P af a (y) = max(a; f a (x) f a (y)) = max(a; a?x?y ), and f a (x& L y) = a 1?max(0;x+y?1). Both give a in the case when x + y 1 and a 1?(x+y?1) otherwise. Now, for each formula ' with propositional variables in fp 1 ; :::; p n g we dene a translation ' using one new propositional variable p 0. Let (for i f1; :::; ng): 0 p 0 p i p 0 _ p i ('& ) p 0 _ (' & ) ('! ) '! In particular (:') '! p 0. We will prove rst a technical lemma, that will demonstrate the correspondence between e L and e P.

3 Lemma Let e be an evaluation of propositional variables including p, with e(p 0 ) > 0. Let e 0, e 00 be evaluations such that e 0 (p i ) = max(a; e(p i )) and e 00 (p i ) = f a?1 (e 0 (p i )), where a = e(p 0 ). Then for every formula ' not containing p 0 f a (e 00 L (')) = e P(' ). Proof The proof goes by induction on the complexity of '. Atomic formulas. For ' 0 it is obvious, if ' p then e P (p ) = e(p 0 _ p) = max(a; e(p)) (max(a; e(p))) = max(a; e(p)) f a (e 00 L (p)) = f a(f?1 a Composed formulas. { ' & e P (( &) ) = e P (p 0 _ ( & )) = max(a; e P ( & )) = max(a; e P ( ) e P ( )) f a (e 00 L ( &)) = f a(max(0; e 00 L ( ) + e00 L ()? 1)), which by the induction assumption equals to f a (max(0; f a?1 (e P ( ))+f a?1 (e P ( ))+1)) = max(a; a log ep ( a ) a log ep ( a ) ) = max(a; e P ( ) e P ( )) { '! e P ((! ) ) = e P (! ), which is 1 in the case when e P ( ) e P ( ) and ep ( ) ep ( ) otherwise f a (e 00 L (! )) = f a(min(1; 1? e 00( ) + L e00 L ())), which is 1 in the case when e00( ) L e 00 L () and f a(1? f a?1 (e P ( )) + f a?1 (e P ( ))) = f a (1 + log a e P ( )? log a e P ( )) = otherwise. ep ( ) ep ( ) Theorem 1 Let ' y denote the formula ::p 0! '. For each formula ' not containing p 0, ' is 1-tautology of Lukasiewicz's logic if and only if ' y is a 1-tautology of product logic. Proof If ' is 1-tautology of Lukasiewicz's logic then for every evaluation e, e L (') = 1. This holds in particular for the derived evaluation e 00 and thus e P (' ) = 1 (the case when e(p 0 ) = 0 is obvious). Conversely if ' y is 1-tautology of product logic, then for every evaluation e, e P (::p 0! ' ) = 1, in particular for every e such that e(p 0 ) > 0, e P (' ) = 1. We can see, that for every evaluation e, there is an evaluation e and a constant 0 < a < 1 such that e(p) = f a?1 (max(a; e(p))) for every propositional variable p. Let e(p 0 ) = a, from e P (' ) = 1 follows e L (') = 1 and therefore ' is 1-tautology of Lukasiewicz's calculus. Altogether we proved that Lukasiewicz's logic has a faithful interpretation in product logic..1 Recursive non-axiomatizability of product predicate logic Now we investigate a language of predicate calculus consisting of predicates P 1 ; P ; ::: and constants c 1 ; c ; :::. Models are dened as usual (cf. [6, ]), i.e. structures M = hm; (r P ) P ; (m c ) c i where M 6= ;, for each n-ary predicate symbol P, realization of P is a n-ary fuzzy relation r P (function from M n to [0,1]). And for each constant c, m c M is the realization of this constant. In the following let p 0 Q(c 0 ) be a new closed atomic formula. Now we present a similar translation * for the predicate calculus. It treats propositional connectives as above. For n-ary 3

4 predicate symbol P dene (P (t 1 ; :::; t n )) = p 0 _ P (t 1 ; :::; t n ), and nally put ((8x)') = (8x)'. We prove rst a technical lemma similar to the propositional case. Observe the importance of the fact that our embedding f a is a continuous mapping. Lemma 3 Let ' be a closed formula not containing Q. Let x [0,1]. There is a model M of Lukasiewicz's predicate logic with jj'jj L M = x if and only if there is a model M0 of product predicate logic such that jj' jj P M 0 = f a(x). Proof Let M = hm; (r P ) P ; (m c ) c i be a model of Lukasiewicz's predicate logic such that jj'jj L M = x. Pick a > 0 and construct a model M 0 = f a (M) = hm; (r 0 P ) P ; (m c ) c i by modifying r 0 P = f a(r P ) and adding r 0 Q (c 0) = a. Now it is easy to show by induction on complexity of ' that jj' jj P M 0 = f a (x). Let us show this just for the case when ' (8x) (x). Then jj'jj L M = inf xfjj jj L M = xg = y, jj' jj P M 0 = inf xfjj jj P M 0 = xg = inf f a(x)fjj jj L M = xg = f a(inf x fjj jj L M = xg = f a(y) (f a (x) is continuous). The other implication can be proven similarly. Lemma 4 Let ' y be again ::p 0! '. Then ' is a 1-tautology of Lukasiewicz's predicate logic if and only if ' y is a 1-tautology of the product predicate logic. Proof Let ' be a 1-tautology of Lukasiewicz's predicate calculus. By the preceding lemma for every model M, if jj' jj P M = x then there is a model M0 such that jj'jj L M 0 = f?1 a (x) = 1. From this x = 1 and ' is 1-tautology of product predicate logic. Converse implication follows similarly from the above lemma. As we know (see [10]), the set of 1-tautologies of Lukasiewicz's predicate calculus is -complete, hence not a -set. From this fact and the above lemma we can conclude following theorem. Theorem The set of all 1-tautologies of the product predicate logic is not a -set (and hence not recursively enumerable).. On monadic predicate logics Ragaz [10] shows the following: (4.8.3) The set of all 1- tautologies of the monadic Lukasiewicz logic is a 1 set. (4.8.1) The set of all satisable closed formulas of this logic with at least four unary predicates is 1 -complete. (A closed formula is satisable if it has the value 1 for at least one model.) In [11] he formulates the problem whether the set of all 1-tautologies of the logic in question is decidable, and, if not, if it is 1 -complete. Thus we get immediately the following corollary for the monadic product logic: Theorem 3 The set of satisable closed formulas of the monadic product logic with at least four predicates is not a 1 -set (i.e. it is not recursively axiomatizable). Let us mention in passing some false statements in the literature. As Ragaz mentions in [11], Scarpellini's claim saying that Rutledge has shown decidability of the set of 1-tautologies of the monadic Lukasiewicz predicate logic is false (since Rutledge's system allows only quantication of a single variable). Gottwald claims ([] p. 3) that Rutledge [1] has shown the axiomatizability of the set of 1-tautologies of the monadic Lukasiewicz predicate calculus, and (p. 37) that Ragaz has shown its undecidability. It follows from the quotations above that both claims are false. 4

5 3 Embedding Godel's logic into extended product logic Let us now enrich, as in [1], the language of the Godel's logic by a new unary connective 4 such that e(4') = 1 if e(') = 1 and e(4') = 0 otherwise. By [1] the extension of the deductive system of the Godel's logic by the following axioms: (41) 4' _ :4' (4) 4('! )! (4'! 4 ) (43) 4'! ' (44) 4'! 44' (45) 4(' _ )! (4' _ 4 ) and by the deduction rule ' 4' is a complete axiomatization of such extension of Godel's logic. Similarly we can extend the deduction system of the product logic (see [5]) in the same way. In the following we will denote this extended deduction system by P4. We introduce the notion of a 4-product algebra as a structure A = ha; & P ;! P ; 4; 0; 1i such that ha; & P ;! P ; 0; 1i is a product algebra in the sense of [5] and it satises the following 4- axioms: (4i) 4a _ :4a = 1 (4ii) 4(a! P b) (4a! P 4b) (4iii) (4iv) 4a a 4a 44a (4v) 4(a _ b) (4a _ 4b) (4vi) 41 = 1 First we will prove some basic facts, that will be used in the proof of completeness. Lemma 5 a) a b implies 4a 4b b) 4a = 4a & P 4a c) 4(a & P b) = 4a & P 4b Proof a) If a b then a! P b = 1 holds in each product algebra, by (4vi) and (4ii) 1 = 4(a! P b) 4a! P 4b and hence using the operation of taking the residuum (u (v! P w) i (u& P v) w) 4a 4b. b) 1 = a! P a a, from a) 4(a! P a) 4a, again using (4ii) we obtain 4a! P 4a 4a which is the same as 4a 4a & P 4a. Conversely 4a 4a! P 4a & P 4a and from the fact that in each product algebra :4a 4a! P 4a & P 4a we can derive 1 = 4a _ :4a 4a! P 4a & P 4a and hence 4a 4a & P 4a. c) 4a; 4b 4(a & P b) and from b) we get 4(a & P b) = 4(a & P b) & P 4(a & P b), so together we can conclude 4(a & P b) 4a & P 4b. The converse inequality 4a & P 4b 4(a & P b) is obvious. Denition 1 A lter over a 4-product algebra A is a subset F A which is a lter in the sense of product algebras (that means if a F and b F then also a & P b F and if a F and a b then also b F ), satisfying the additional condition if a F then also 4a F. 5

6 Lemma 6 The unit interval with truth functions x & P y = x y, x! P y = 1 if x y and y x otherwise, and the truth-function 4 such that 4(1) = 1, 4(x) = 0 for x < 1 is a linearly ordered 4- product algebra. The algebra of classes of provably equivalent formulas in P4 is a 4-product algebra. Proof The proof is standard. Theorem 4 The deduction system P4 is a complete axiomatization of the set of 1-tautologies of logic P4. To prove this completeness theorem we inspect the proof of the completeness theorem from [5] for the product logic in the following series of statements. Sublemma 4.1 Let A be a 4-product algebra and let F be a lter. Dene the corresponding equivalence a F b i (a! P b) F and (b! P a) F. Then F is a congruence, and the quotient algebra A= F is a 4-product algebra, which is linearly ordered if and only if F is an ultralter. Proof We only have to verify that a F b implies 4a F 4b. Assume a F b. Then a! P b; b! P a F, hence 4(a! P b); 4(b! P a) F, consequently 4a! P 4b; 4b! P 4a F and we get 4a F 4b. The rest is as in [5]. Sublemma 4. Let A be a 4-product algebra and a A, a 6= 1. Then there is an ultralter F on A not containing a. Proof The construction is the same, only observe that if F is a lter not containing c! P d, then F 0 = fuj(9v F )(u v & P 4(c! P d))g is the smallest lter extending F and containing c! P d. In the following we will check that it's really the case (if u 1 ; u ; u F 0 then also u 1 & P u ; 4u F 0 ). Assume u 1 ; u F 0. Then u i v i & P 4(c! P d) for suitable v i and i = 1;, hence u 1 & P u v 1 & P v & P 4(c! P d) & P 4(c! P d) = v 1 & P v & P 4(c! P d) by lemma 5 b). But v 1 & P v F, and hence u 1 & P u F 0. And if u F 0 then u v & P 4(c! P d) for some v F. From this 4u 4(v& P 4(c! P d)) = 4v & P 44(c! P d) = 4v & P 4(c! P d) and hence 4u F 0. Now we start the construction of an ultralter not containing 1 6= a A. Start with F = f1g. Now if F is any lter not containing a and c; d are such that if neither (c! P d) nor (d! P c) belongs to F, create F 0, F 00 as F 0 = fuj(9v F )(u v & P 4(c! P d))g and the same for F 00 as the smallest lter extending F and containing (d! P c). Let us suppose that a is member of both F 0 and F 00. Then for some v F a (v & P 4(c! P d)) _ (v & P 4(d! P c)) v & P (4(c! P d) _ 4(d! P c)) v & P 4((c! P d) _ (d! P c)) = v & P 41 = v & P 1 = v, which means that a F, but this is a contradiction. Take for F that of F 0, F 00 which doesn't contain a and iterate. 6

7 Sublemma 4.3 Each 4-product algebra is a subdirect product of linearly ordered 4-product algebras. The proof is the same as in [5]. Sublemma 4.4 If A is a 4-product algebra then 41 = 1. Moreover if A is linearly ordered, then a 6= 1 implies 4a = 0. Proof If a 6= 1, then 4a 6= 1 (since 4a a). Thus if A is linearly ordered from (4a_:4a) = 1 we get :4a = 1 and hence 4a = 0. Sublemma 4.5 If an identity =, in the language of 4-product algebras, is valid in the unit interval algebra, then it is valid in all linearly ordered 4-product algebras. Proof Use the proof from [5] observing that, by the preceding lemma, each linearly ordered product algebra extends uniquely a linearly ordered 4-product algebra; thus the representation using ordered Abelian groups works here too. Now we can complete the proof of the completeness theorem as follows: if ' is a 1-tautology of P4 then the identity ' = 1 is valid in the unit interval 4-product algebra and hence in all linearly ordered 4-product algebras, thus, due to sublemma 4.3 in all 4-product algebras, among others in the algebra of classes of provably equivalent formulas of P4, which means that ' is provable in P4. Theorem 5 Both Godel's logic and its 4-extension are sublogics of P4. Proof The theorem follows from the simple fact that the formula ('! ('& )) _ 4('! ) denes in P4 Godel's implication, in other words e P4 (('! ('& )) _ 4('! )) = e G ('! ) for every evaluation e. This can be easily veried. 4 Computational complexity of propositional satisability Let us dene, analogously to the Boolean case, the set SAT for multi-valued logics L, P, G as the set of all formulas for which there is an evaluation of propositional variables e such that e(') > 0. It is well-known that the set of satisable formulas in Boolean logic SAT Boole is NP-complete (see [3]). Mundici [9] proved that SAT L is also NP-complete. In this section we shall establish the same for SAT P. First we will investigate the following translation of formulas: Denition Let ' be a formula with atoms p 1 ; : : :; p n and let I fp 1 ; : : :; p n g. Dene a formula ' I by induction on the logical complexity of ' as follows: 7

8 1. If ' = p i and p i I then put ' I := 0. If p i = I, put ' I := p i.. If ' is a constant then ' I := '. 3. Let ' = 1 &. If one of i I is 0, put 'I := 0. If one of i I is 1, put ' I := 3?i. I Otherwise put ' I := 1& I. I 4. Let ' = 1!. If I 1 = 0, put ' I := 1. If I 1 = 1, put ' I := I. If I 1 6= 0 and I = 0, put ' I := 0. If I = 1, put 'I := 1. Otherwise put ' I := I 1! I. Lemma 7 For any ' and I, ' I is either a constant (0 or 1), or a formula not containing either 0 or 1. Proof By inspection of Denition. Now we present some statements that are valid for both product logic and Godel's logic. In the following let stand for P or G. Lemma 8 Let ' be a formula not containing the constant 0. Then there is an evaluation e of propositional variables in ' such that e(') = 1. In particular, ' SAT. Proof Assign to all atoms 1. Then the value of ' is 1 as well. Theorem 6 SAT NP Proof 1. Given a formula ' that is composed from propositional variables p 1 ; : : :; p n, guess a subset I fp 1 ; : : :; p n g.. Compute ' I. 3. Accept if ' I 6= 0. 8

9 If ' SAT and e(') > 0, then I := fp i j e(p i ) = 0g will not reduce ' to the constant 0 (as all rules in Denition are consistent with the truth-tables of ). Conversely, if ' I 6= 0 for some I then by lemmas 7, 8 e(') = 1 for the assignment and hence ' SAT. 1 e(p i ) = if pi = I 0 if p i I We have proved that both SAT P and SAT G are in NP (the result for Godel's logic was already known (folklore)). Moreover, if we modify a given formula ' by adding double negation (::) before every propositional variable in ' we obtain a formula ' :: which is in SAT if and only if formula ' is in SAT Boole (as ::a = 1 for a 6= 0 in both P and G). Hence SAT Boole reduces to SAT and we have the following corollary. Corollary 7 Both SAT P and SAT G are NP-complete. We deduce few more corollaries. Corollary 8 If ' SAT P then, in fact, ' has the value 1 for some assignment. Lukasiewisz's logic has a related property. Namely, the set of formulas for which there exists an assignment giving it the value 1 in L is in NP (by the proof of [9, Thm.4]). Now let us denote by TAUT the set of all 1- tautologies of the corresponding logic. Lemma 9 TAUT P co-np. Proof We will see that the complement is in NP - let ' 6 TAUT P. This means by denition that there is an evaluation e of propositional variables such that e P (') < 1. That exists if and only if there is an subset I of propositional variables in ' and a positive evaluation e 0 of ' I (e 0 (p i ) > 0 for every p i in ' I ) such that e 0 P ('I ) < 1. It is sucient to put e(p i ) = 0 i p i I. Due to the isomorphism from section (9I)(9e positive)(e P (' I ) < 1) if V and only if (9I)(9e positive)(e L (' I ) < 1). Finally this holds if and only if (9I)(9e positive)(e L ( p i ^ :(' I )) > 0). p i' I Thus ' 6 V TAUT P i pi' I p i ^ :(' I ) SAT L and hence TAUT P co-np. The co-np completeness of TAUT P ' :: TAUT P. follows from the fact that ' TAUT Boole if and only if Corollary 9 TAUT P is an co-np complete set. All three sets TAUT are conp-complete, and formal systems for any of the three logics is thus a non-deterministic acceptor of a conp-complete set. The length-of-proofs question for any of L, G, P is therefore related to the NP =? conp problem in the same way as it is for Boolean logic (see [4] for basic denitions and facts or [7] for a general background). Hence we propose to study the 9

10 lengths of proofs in L, G and P with an ultimate goal to show that they cannot be polynomially bounded (in the lengths of formulas being proved) in any formal system. For Boolean case such lower bounds are known only for few formal systems, all weaker than the usual Hilbert-style system (cf. [7]). The lower bounds in the Boolean case seem to be possible because the systems operate with formulas of various restricted types only (e.g. clauses in resolution). It would be interesting should it be possible to prove a lower bound for a natural formal system for one of L, G or P, that operates with all formulas. References [1] Baaz, M. Innite-valued Godel Logics with 0-1-Projections and Relativizations. in: (Hajek, ed.) Godel'96 - Logical Foundations of Mathematics, Computer Science and Physics - Kurt Godel's legacy, Springer Verlag, (1996). [] Gottwald, S. Mehrwertige Logik, Berlin: Akademie-Verlag, (1988). [3] Cook, S A. The complexity of theorem proving procedures, in: Proc. 3 rd Annual ACM Symp. on Theory of Computing, (1971), pp ACM Press. [4] Cook, S. A., and Reckhow, A. R. The relative eciency of propositional proof systems, J. Symbolic Logic, 44(1), (1979), pp [5] Hajek, P., Esteva, F., and Godo, L. A complete many-valued logics with product conjunction, Archive for Mathematics Logic vol. 35, (1996), p [6] Hajek, P. Fuzzy logic and arithmetical hierarchy II., to appear in Studia Logica, (1997). [7] Krajcek, J. Bounded arithmetic, propositional logic and complexity theory. Encyclopedia of Mathematics and Its Applications, Vol. 60, Cambridge University Press, Cambridge - New York - Melbourne, (1995). [8] Mostert, P.S., Shields, A.,L. On the structure of semigroups on a compact manifold with boundary, Annals of Math., Vol. 65, (1957), p [9] Mundici, D. Satisability in many-valued sentential logic is NP-complete., Theor. Computer Sci., 5, (1987), pp [10] Ragaz, M. Arithmetische Klassikation der Formelnmengen der unendlichwertigen Logik Thesis, ETH Zurich, (1981). [11] Ragaz, M. Die Unentscheidbarkeit der einstelligen unendlichwertigen Pradikatenlogik, Arch. Math. Logik 3, (1983), pp [1] Rutledge, J.,D. A preliminary investigation of the innitely many-valued predicate calculus Thesis, Cornell University, (1959). [13] Scarpellini, B. Die Nichtaxiomatisierbarkeit des unendlichwertigen Pradikatenkalkuls von Lukasiewicz., J. Symb. Log. 7, (196), p [14] Takeuti, T., Titani, S.: Intuitionistic fuzzy logic and intuitionistic fuzzy set theory, Journ. Symb. Log. 49, (1984), pp

11 Addresses: Matthias Baaz: Institut fur Algebra und Diskrete Mathematik E118., Technische Universitat Wien, A-1040 Vienna, Austria. Petr Hajek and David Svejda: Institute of Computer Science, Academy of Sciences, Pod vodarenskou vez, 18 07, Prague, Czech Republic. fhajek, Jan Krajcek: Mathematical Institute and Institute of Computer Science, Academy of Sciences, Zitna 5, , Prague, Czech Republic 11