Tóth, K.

English

Tóth K.

University of Szeged

Completeness in non-simple and stable modal logics B y K . TÓTH In my work [1] I have defined the syntax and semantics of modal logics. Also, in-ference systems and completeness theorems for simple, non-stable logics have been included. Unfortunately, the methods used there cannot apply directly to non-simple and stable logics. In this paper I give a modification of the method and prove completeness theorems for the cases not covered in [1]. In fact, this paper is a continuation to [1], all non-common notions and notations are introduced there. § 1. Completeness in non-simple logics The notion of consistency is defined in [1]. DEFINITION. The set of formulae is complete if the following conditions are satisfied: _ (i) a is consistent; (ii) If si contains variables only from n(a), then either A£a. or (iii) Let sd contain variables only from 7r(a). If then there exists a variable ad-n(a) such that a is free for JC and s/[xfa]£<x; (iv) Let / be «-argument function symbol and let xx, ..., xn£n(a). There exists a variable a£_n (a) such that for all classical f o r m u l a ^ the fact f(xu ...,x„) is free for a in si implies that the two assertions si^a. and st{alf(xx, ..., x„)]€oc are equivalent. \ Theorem 1. If a is consistent, then there exists a complete set /? such that ctQfi. Proof. Parallel to the proof of Theorem 5 in [1] using the following Lemma. LEMMA. Let / be an «-argument function symbol, a a consistent set and A$7T(A). Moreover, let a ' = a U { j j / : .si is a classical formula, f ( x x , ..., xn) is free for a in and si[alf(x1, . . . , x„)]6a}. Then a ' is consistent. Proof. In contrary, let us suppose that there exist the formulae six, ...,sik, such that six, ...,sik£<x, ^ [ a / / ^ , ..., *„)], ... ,.3§,[al f ( x x , ...,x„)]€a 378 K. Tóth and | (si1A...AsfkA@1A...A@l). Applying R2 and A6.b we have h-Va ~(sixA...As!kA®xA...A381) h- ~(stfiA...AsfkA&1[a/f(x1,..., x„ ) ]A . . .Aa , [a l f ( x l t ...,xn)]) which is a contradiction. Definition of a complete system of formula-sets is just the same as in [l]-how-ever, item (iii) can be omitted by the remark above. The theorem remains valid for the new concept: Theorem 2. If a is a complete set of formulae, then there exists a complete system of sets M such that a The completeness result follows easily from this theorem. Theorem 3. Let a non-simple, non-stable modal logic be given. If si cannot be derived in this logic, then is satisfiable. Proof. By the previous theorems, there exist a complete set a and a complete system of sets M such that ~ si£ct, a£M. We assume, by the definition of a complete set, that is a function for which the following property holds: if a£M, / is «-argu-ment function symbol, x l 5 x„^n(a), then v(a,f(x1, ..., x„)) is a variable, such that for all classical formula 38, if f ( x . . . , x„) is free for v(a, f ( x x , . . . , x„)) in 38, then the two assertions and 3S[V(OL, f i x j , . . . , xn))/ f(XX, ...,x„)]£oc are equivalent. Let us introduce the notations: N= {j9: P£M and P+ * 0}; If p, y£M, then PRy {(P+ £ y and P+ * 0) or (p+ = 0 and y=P)); \P(P) I = *(P); fp^ix!, ...,xn) = v(P,f(x1, ...,x„)), where x , , ..., x„£n(p); rpm(xi> xn) r(*i> •••> xn)£P, where x l 5 ..., xnen(P). It is clear that (M, N, a, R, P) is a model. Let us extend the domain of v as follows: let v(P,x)=x, where x£it()3); and let v(P, /(T^ ..., T„)) = d(P, f(v(P, x^, ..., ...,V(P, T„))), where TX, . . . , T„ are terms containing variables from N(P) exclusively. The following assertions can be proved by (the usual) induction: Let k be an interpretation and x the corresponding valuation. (i) If then X(T,P)=V(P,T[Xi, ..., xjk(xO, ...,k(xm)]), where x l 5 . . . , x m are all variables occuring in r. (ii) If then p\=SS[k]o3S[Xi, ...,xjk(xi), ..., k(xm)]<ip; where x 1 ; . . . , x m are all variables occuring in 38. In particular, it follows that ~ si is valid in the model ( M , N, a, R, P). Properties K1—K3 can be proved just as in [1, Theorem 7]. 379 K. Tóth ¡Completeness in non-simple and stable modal logics § 2. Completeness in stable logics Theorem 4. If a is complete, then there exist a complete set ß and a complete system of sets M such that <xQß, ßdM and for every y£M, n(ß) = n(y). Before proving this theorem we give the completeness result for stable logics. Theorem 5. Let a stable modal logic be given. If the formula si cannot be derived in this logic, then ~ s i is satisfiable. Proof. Very similar to the proof of Theorem 3 above or Theorem 7 of [1] pro-vided complete system of sets M, given by Theorem 4, is used in the construction. § 3. Proof of Theorem 4 We introduce the following notations: let a be a set of formulae. By i/*(a) we shall mean the set of all formulae which contain variables only from 7i(a). Let R be a two-argument relation. We define the relation Rn, n finite, by the following recurrence: R° is the identity relation and let i?"+1 be defined by ARn+1B if there exists C such that AR"C and CRB. Then, R = (J R", where R is the reflexive, transitive closure of R. n = 0 In the following we shall deal with certain ordered triplets (a, M, R). Without further mentioning we always suppose that the following conditions hold for <«, M, R): (i) M is a set of complete sets, a£M, R is a binary relation on M. (ii) For every ß£M,ocRß and if SQR and for all ß^M aSß, then R = S. (iii) If a£n(ß), then there exists y such that a£n(S) if and only if yRö. (iv) a) If ßRy then ß+f)>l/(y)Qy and b) Let ß£M, Qsitß. If there is a y£M, such that ßRy and si£\]/(y), then there also is a ydM with ßRy and sidy. Assertion 1. For arbitrary triplet ( t x , M , R ) there is no ß£M such that ßRa. If ßRS and yRö then ß=y. Proof, (a) Consider the triplet (a, M, S>, where S is defined by ßSy if and only if ßRy and y^a. By the second clause (ii) above, R—S. (b) Let us suppose that ßRS, yR8 and /My. Let S be defined as ßx Sß2 if and only if ßxRß2 and (ßi, ß2)^(ß, ö). Then conditions above will hold for (a, M, S), but SQR and S^R which contradicts the second condition (ii). DEFINITION, ( a , M , R ) is called n-th order triplet if for every ß £ M there is a k (O^k^n) such that a R k ß . (a, M, R) is totally w-th order triplet if it is an n-th order triplet and if 0^k<n, a R k ß , ()si£ß, then there exists ydM for which ßRy and si^y. It is clear, that for every m (m^ri) the fact (a, M, R) is an n-th order triplet implies that (a, M, R) is an m-th order triplet too. Similarly, if (a, M, R) is a zero order triplet, then M={a), R=0, thus (a, M, R) is totally zero order. Let (a, M, R) be arbitrary, ß£M. Let us set . MIß = {y: ßRy), Rjß = RC\(M/ßxM/ß). 380 K. Tóth Assertion 2. If <a, M, R) is an w-th order triplet and aRfi then (/?, M/P, R/P) is an (« —l)-th order triplet. DEFINITION. Let us define the operation L by the following items: if (a, M, R) is a 0-order triplet, then L{a, M, R)—a, if w>0 and (a, M, R) is an w-th order triplet, then let L(u, M, R)=a\J (J si is a conjunction of formulae from the aRP set L(P,Mfp,R/p)}. Theorem 6. Let (a, M, R) be an w-th order triplet. Then L(a, M, R) is con-sistent. Proof. We proceed by induction on n. If « = 0 then the assertion clearly holds. Let n > 0 , and assume the contrary, i.e. there exist sily si2, ... and a conjunction SS1 of formulae from M\px, RjP^), a conjunction SS2 of formulae from L(P2,MIP2,R/P2) etc., such that I - ~ Asi2A... A A O^aA.. . ) that is I st2\/...\Jn • We can assume that all /?,, Pj are distinct, for if not, then can apply h- • • - • ~ Hence we obtain a form in which all sets /?,-, Pj are distinct. Let x be a variable of such that It follows from Assertion 1 and condition (iii) that x does not occur in the formulae s/1,s#2, or SS2, .... Apply rule R2 for all variables not occuring in n(a): h - ~ ^ i V ~ ^ 2 V . . . V V x 1 1 V x 1 2 . . . • ~ ^ 1 V V x 2 1 Vx2 2 . . . • Since the fixed logic is stable we can repeatedly apply the axiom \j'xHsi— • \/xsi and obtain I- ~ j ^ V ~ si2\l...\l • V x u Vx12 . . . • Vx21 Vx2 2 . . . ~ 3S2\/... where all free variables are from n(a). Since bound variables can be substitued by suitable ones from n{a) we have H ~ j / i V ~ j* ,V. . . 'V • V * i i V * i t . . . ~ • V * n Vx 2 2 a complete, so this possible only when some disjunctive terms, e.g. • V*n ^x' i 2 . . . (For if then j^jCa which contradicts the completeness of a.) So V*ix Vxi2 i L(px, Mlplt R/Pi) and We concluded that L(Plt M/px , is consistent, wich proves the theorem. Theorem 7. If (a, M, R) is an n-th order triplet, aRy and P is a complete set such that L(tx, M, R)QP then P+{JL(y, M/y, R/y) is consistent. 381 K. Tóth ¡Completeness in non-simple and stable modal logics Proof. In contrary, let us assume that there are formulae 3SdP+ and a con-junction of elements from L(y, M/y, R/y) such that i.e. By R3, we obtain | - • ¿ i — • i.e. f - In accordance with our conditions, n@tdP and so which contradicts the completeness of p. DEFINITION. We say that (a, M, R) is a continuation of (P, N, S) if there exists a function / : N'-*M, such that f{fi)=a., if y^N then y g / ( y ) , and if yS5 then f(y)Kf(Sl Theorem 8. If (a, M, R) is an w-th order triplet and /J is a complete set for which L(a, M, R)^P, then there exists a totally n-th order triplet (¡3, N, S> which is a continuation of (a, M, R). Proof We proceed by induction on w. If « = 0 , the assertion follows. Let n>0 . If a + = 0, then for all sidtyid), Qsida, so QsidP. In particular if si is a negation of a tautology, then \ - ( ) s i t h u s for every ¡/(P), ()@dp, i.e. P+=V). It is impossible that pSy, by definition, hence (fi, {/?}, 0) is a totally «-th order triplet and this is a continuation of (a, M, R) — (a, {a}, 0). Let a + and so As we see Qs idP implies the consistency of P+U{jtf), hence we can assume that P+U {si}Q8^, 8^ is complete. By the pre-vious theorem, /?+U L(y, M/y, R/y) is consistent, too, provided txRy, thus there is a complete set 8y such that fi+UL(y, M/y, R/y)QSy. It is clear, that the new variables, introduced in these steps, may be chosen so that the sets n(8Si,)\n(P), ..., n(8y)\n(P), ... are pairwise disjoint. As (5^, {¿.a,}, 0) is an (n — l)-th order triplet, and since L (8^, {8^}, 0) ^ 8^ it follows that a totally (« — l.)-th order continuation (8^, M^, RJ) of (8^, {<5.̂ }, 0) exists. Since L(y, M/y, R/y)Q.8y by the induction hypothesis, it follows that there exist My, Ry such that (8y, My, Ry) is a totally («— l)-th order triplet and it is a continuation of (y, M/y, R/y). We may assume that the common variables of any two sets n ( \ ) M J ) , .. . , . . . , i ( U M 7 ) , ... are contanied in n(P). Let 8dN, provided 8=P , or if there is an si, such that QjsidP and 8dM^, or if there is a y, such that a Ry and 8£My. Let 82(LN and provided if 8X=P and there is an si such that QsidP and 82=8or if 8X=P and there is an y for which a Ry and 82=8y, or if there is an si, such that QsidP and 8XR^82, or if there is an y, such that a Ry and 81Ry82. It is obvious, that the conditions (i)—(iv) hold for (/?, N, S). Also, it is a totally «-th order triplet and is a continuation of (a, M, R). Now we can return to the proof of Theorem 4: Let (oc0, M0, R0) = (a, {a}, 0) i.e. a totally 0-order triplet. Let us suppose, that for some «, a totally «-th order triplet {a„, M„, R„) is defined. By Theorem 6, L(a„, M„, Rn) is consistent, and hence there is a complete set a„+ 1 , such that L(a„, M„, Rn)Q ot„+1. By Theorem 8, there exist M„+1, Rn+1 such that (a„+1 , M„+1, R„+1) is a totally (« + l)-th order triplet and it is a continuation of <a„, M„, Rn). Thus, there exists a function /„: M„—Afn+1 such that/„(«„) = a„+1 and p(LMn implies pQfn(p) and if pRny, then/„(^)i?n+1/n(y). 382 K. Tóth ¡Completeness in non-simple and stable modal logics Let /?= (J a„ and M=\ | J y„: yk£Mk and for all i^k, y i + 1 =/¡(7,)}. Since union n=0 ln=fc J of increasing complete sets is also complete we have that every element of M is complete. Let y£M, and ()stf£y. For y— (J y„, there exists an /, such that n=k 0 0 y, and hence there also exists <5, for which 7,7?, <5, and Let <5 = | J <5„. 0 0 0 0 y+= U 7 n + i ' U and si^b, thus M is a complete system of sets. n = l n = l Let a£n(li). For some k,a£n(ak), and if />&, then adn(ak),' too. If 5 then for some /, <5 = (J <5„. We may assume that k-=-.i and so Since n = i (a,, Mi, Ri) is a totally /-th order triplet, we have 7r (a,) Q 7t (<),), and thus a£n(S). 0 0 Let a£n(S) for some SdM. We may assume that S= (J Sk and a£n(dk). For n=k L(ak, Mk, Rk)Quk+1, a£n(ock+1) and hence a£n(fi). We gained, that for every 5£M, n(fi)=n(b) which completes the proof of Theorem 4 and also the completeness theorem. References [1] T6TH, K., Modal logics with function symbols, Acta Cybernet., v. 4, 1979, pp. 291—302. (Received July 13, 1979) 

Completeness in non-simple and stable modal logics

http://acta.bibl.u-szeged.hu/12298/1/cybernetica_004_fasc_004_377-382.pdf

Completeness in non-simple and stable modal logics

Abstract

Similar works

Full text

Available Versions

University of Szeged