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Download Article (PDF) DEMONSTRATIO MATHEMATICA Vol. X No 1 1977 Jan Romanski ON CONNECTIONS BETWEEN ALGEBRAIC AND KRIPKE SEMANTICS The paper deals with algebraic and Kripke semantics for intuitionictic theories. In [l] M.C. Fitting proved that in propositional case "both these semantics are equivalent (in some sence). The aim of this paper is to state the conditions under which one can construct an algebraic model "inside" the given Kripke structure and conversely. An algebraic model is a pair & = (B,h) where B is a nondegenerate pseudo-Boolean algebra and h is a mapping from the set FORM of all propositional formulas into B, such that for every formula X, Y following conditions hold: (AA) h(Xv Y)=h(X) v h(Y) (AC) h(Xa Y)=h(X)a h(Y) (AI) h(X—Y)=h(X) —-h(Y) (AN) h( -i X)=-ih(X). A Kripke model is a triple K=(S,^fN), where (S,^) is a partially ordered set and |= is a relation (of inheri- tance of information) between elements of S and formulas, such that for every formula X, Y and for- every, s, t in S the following conditions hold: (Ka) X is an atomic formula, s NX, sét => t NX (KA) s l=X v Y iff s NX or s NY (KC) s NX A Y iff s NX and s NY (Kl) s NX—Y iff s*St implies non(t NX) or tNY (KN) sNiX iff sét implies non(t NX). - 123 - 2 J. Roraanski Let M be any model (of A or K type). A formula X is satisfied in M, Mil—X, iff h(X)=1B or, respectively, for every s in S s I wish to underline here that the set of sentences satisfied in the given model is not a logic, because this set is not invariant under substitutions. It is a theory, only. Theorem 1 [1]. For every model A (or K) there exists a model K (resp. A) such that for every formula X All-X iff k||-X holds. Outline of the proof: Let K=(S,4, l=) be a Kripke model. Let B be a pseudo-Boolean algebra of all open sets in the order topology on S, i.e. aeB iff ass and for every s, t e. S s4t, sea =*tca. Let h(X)=jse£; sl= X}. One can show that for every formula X h(X) is an open set, A=(B ,h) is an algebraic model and, moreover, g||-X iff A||-X. Let A=(B,H) be an algebraic model. Let S be a set of all prime (i.e. disjunctive) filters on B ordered by inclusion and let s (=X iff h(X)es for every filter s in B. It may be shown that K=(S,£, is a Kripke model and for every formula X A||-X iff K||-X holds. The question iss can this construction be used with success in the first order case? First of all let us complete definitions. Language: There are no functional symbols. There is an infinite countable set V of individual constants. (•) and (E-) stands for universal and existential quantifiers. (x)X(x), (Ex)X(x) denotes sentences such that X is a formula with x as the only free variable. The set of all sentences is denoted by SENT. Algebraic model. In the pair A=(B,h) B is an to-comple- te pseudo-Boolean algebra and the mapping h satisfies (AA), (AC), (AI), (AIT) and - 124 - Algebraic and Kripke semantics 3 (AH) h((x)X(x))=f>(X(v)) vtV (AE) h((Ex)X(x)= (J h(X(v)). v eV A Kripke model is a quadruple K=(S,4 , l=,D)f where (S, ^ ) is a partially ordered set, for every s in S D(s) is a nonempty subset of V and SENT such that for every s, t in S and for every sentence X, Y the following conditions hold: (Kd) s => D(s)c D(t) (Ka^j) X is an atomic formula and sf=X=> every, constant of X is in D(s), (Ka2) X is an atomic formula, s (=X, s<t=>t (= X, (KA), (KC), (KI), (KN) as in propositional case, (KU) st=(x)X(x) iff s4t=)tMX(v) for every v in D(t), (Kbi) st=(Ex)X(x) iff there exists a v in D(s) such that s f=X(v). One can check here that our models are realizations of the language in the set of terms (because individual constants are only terms). Now let A=(B,h) he an algebraic model. It has turned out to be impossible to construct a Kripke model "inside" A. There are two reasons. First of all, evrry individual constant appears in every filter in B, e.g. in the sentence X(v) —X(v). Thus we have to define D(s)=V for every s in S. But it is a well known fact that in a Kripke model with the constant function D the sentence (*) (x)(P(x) vQ) — (x)P(x) v Q (x is not free in Q) is satisfied arid there are algebraic models in which (*) fails to hold. Moreover in the construction analogous to that of the propositional case we have to extend given filters to the filters preserving infinite joins and meets and such the extention is, generally, impossible. - 125 - 4 J. Romaiiski Theorem 2. Let A=(B,h) be an algebraic model with the algebra B finite. There exists a Kripke model K, such that for every sentence X A If—X iff K||-X holds. Moreover S is a finite set. The proof is obvious,'analogous to that of the theorem 1. Let K=(S,4 , f= ,D) be a Kripke model. Let B be the algebra of all open subsets of S and let h(X)={s; st=X} for every sentence X. It follows from (Ka2) that h(X) is always an open set. It can be proved that conditions (AA), (AC), (AI), (AN), (AE) hold. Now let V={aQ,a1,a2,.. let S be the set of all nonnegative integers with the natural order, let DinJ^a^a^ ,... ,aQJ, n=0,1,2,... Let P be the only (unuary) predicate symbol of the language and let nf=P(ak) iff k^n. The structure K=(S, , ,D) is a model for the sentence (x)P(x), but h(P(ak))={k,k+1,k+2,...} and h(P(av 0 h(P(a, ))=0. This example shows that condi- * akeV k tion (AU) cannot be proved. Theorem 3. Let K=(S,^,I=,D) be a Kripke model with the function D constant, i.e. D(s)=V for every s in S. There exists an algebraic model A with the property K ||-X iff AII—X for every -X SENT. Proof. It remains to show that the condition (AU) holds. In a Kripke model with the constant domain function the condition (KU) is equivalent to s |=(x)X(x) iff for every v in V s^X(v). Taking this into consideration we have h((x)X(x)={s;s t=(x)X(x)}={s;s *=X(v) for every v in V} = = n{s;sl=X(v)}= f| &(X(v)). vcV veV Moreover glbx iff s |=X for every s in S iff h(X)=S=1B iff AII—X which completes the proof. - 126 - Algebraic and Kripke semantics 5 REFERENCES [1] M.C. F i t t i n g: Intuitionistic logic model theory and forcing, Studies in Logic and the Foundations of Mathematics, Amsterdam 1969. INSTITUTE OP MATHEMATICS, UNIVERSITY OP WARSAW Received April 27, 1976. - 127 - .
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