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From: AAAI-83 Proceedings. Copyright ©1983, AAAI (www.aaai.org). All rights reserved. A THEOREM-PROVER Philippe FOR A DECIDABLE BESNARD - Rene SUBSET QUINIOU - OF DEFAULT Patrice LOGIC QUINTON IRISA - INRIA Rennes Campus de Beaulieu 35042 RENNES Cedex FRANCE Abstract II DEFAULT Non-monotonic such logic notions evolution. has as However been so first-order on the of proved our that it logic deals which a set algorithm and 801 we study the and the can of A.I., complete world is an attempt is but it The formulae. in the field decidability problem problem, predefinite subset [Rei unvalid 801 have allows can resolution First and of all, we recall which III is devoted to the named by the the on an allow more of default main goal facilities to represent example, we know default principles attractive plausible that ostriches..) as the of of our default theorem-prover saturation default prover is complete We which generally always provide some knowledge. can a then derived knowledge by use A default of particular several the each logic can Part for of consistent which 1: T=( a proof of the a cannot be rules. (1I.T) where the theory theories A is a set theory. T called The use by building extensions, T. A , -6 propose a heart IV. This terminates. 27 are a conclusion first-order first-order that defaults denotes inference A extends the default is a couple contains So as “until a fly(x) is interpreted v “C ) A={ A :MB B , A :MC C This default theory , C :MD D ] resolution the be to suppose fly”. is. and default x can That theory default if it is consistent that and T a consistent of which the This of classical defaults Example of schemata inference. of defaults class, and infer assume...” of inferences. fly, of inference in default in part For fly” (except :Mfly(x) fly(x) CONS(G)). patterns problem also to 6: prerequisite contrary, though default default defaults is to “a bird This default (noted plausible The consistency constitutes presented and the Default of default the as: “if x is a bird apprehend a is x can the class, of a special saturation. generalisation use logic common-sense penguins, consequent 801 avoid logic. A ordinary logic of represented bird(x) a saturation. presentation the [Al of consistent theory. solved by determines of default is a crucial named focus solvable the inference be free-defaults, technique a To set free-defaults, some the present logic which context which technique then selected in a first-order nicely logic However, of heuristics. clauses, of a set of formulae, logic, Non-monotonic by use of supported of non-monotonic in the formulae be to this fact. variable decidable logic we feature always knowledge. to remedy realisations this a fundamental cannot to by he has the inferences. applied bird(x) Decision-making subset, plausible be as proposed free-defaults. A Definition I INTRODUCTION fields logic proof-procedure a default additional rule default have it always of first-order [Rei allow obtain to test the that in rewriting technique We we introduce Next, which on the formulae. subset up. a theorem-prover designed is complete for the selected such of a resolution first-order Reiter issue propose This First theory with a restriction with. was initially of and theorem-prover We generalisation saturation, consistency terminates decidable for default to take into account knowledge unexplored. formulae is based named the far theorem-prover is an attempt incomplete LOGIC has 2 extensions: E,=Th( A , -B v “C , B 1 E2=Th( A , “B v “C , C , D 1 Definition: A formula to a default Ao. Al. theory .. . . Ak U (1) T (2) Q has a default (A,T) iff there of finite subsets CONS(As) For I- proof exists with a finite of A, We have respect sequence such proved Theorem that (A.T). E c Q there the following result 1: for any extension exists an extension [ Bes E of E’ of 831: a default (f(A),T) theory such that E’ lbi<k T U CONS(Ai) Prerequisites(Ai-,I I- III SATURATION (3) Ak=O ( u CONS( (4) T u (2) expresses be true the to that let the the is satisfiable. prerequisite consequent In this of a default of this default must to be in extension. Reiter has shown i.e. that respect to (A,T) iff extension order completeness Q has there exists Q. However This logic, an us to which for (4) show the saturation by which will of (A.T) without Defaults may not is even a subset decision “fly(Max) The only T itself. such The true, a kind Next, generalisation in the of resolution we of present the definition of the saturation our default is [ Bos decidable is a predefinite clause are occur also - the 81 ] for clause of problem saturation saturation only already cited, This behaviours ([ Rei we present below of the Max in the default doesn’t fly the ({S] that while ,T) we expect generally is For to transformed birds all the type of are predicate, theory resolution variable clause symbols clauses. A if: that occur in the occuring in a positive literal one. for context an application of databases, since formulae are function--free - “bird(Max), contain particular symbols. is acceptable - a variable in a negative restriction which bird(x) :Mfly(x) 6= fly(x) and extension because free-defaults is predefinite function constant a default: ) It doesn’t or counter-intuitive to those of T={ prerequisite have In addition example be a -- all the variables 6 Defaults an introduce saturation. useful of the technique, resolved. 811). sets. be A Definition with clearlv , that consider the default proof extension problem leads for result a default (1) and membership semi-decidable. first the a formula contains the we first named proof-procedure. proofs, which chapter technique without variables the prodefinite example, the in Type,(x,) clauses all a...& function is symbol. defined in by a clauses. clause the C(x *.., Xn) 1’ Typen(xn) can variable predefinite => C(x,, be clause . . .. x,1. do. Consider :Mbird(x)=>fly(x) bird(x)=>fly(x) 6’= 6’ leads to becomes cannot default later the formula true, then the default be used theory permanent knowledge and (6’. contains & “fly(Max) 6’ instantiated is not by Max in the new expresses a true expressed in by then we he flies. 6’ opposite 6. he flies. to the situational By 6’ we always By 6 only We now when generalise know 6= f(6) P(x) :MR(x) R(x) is called I---> a free-default. f(6)- that x is a bird the previous transformation: :MP(x)=>R(x) P(x)=>R(x) or Theorem clauses The 2: a satisfiability set of a saturated variable of saturation one produced clause. subsumed by one A-ordering leftmost is parent The to produce, the tautologies to are and restrict previous test clauses. the The first. all most recently and ancestors for the literals permit to variable clauses equivalent to the means is clause. variable Due a which of their [ Kow 691 literal of variable empty logically clauses. gives is is of C. the of a set of predefinite saturation clauses of C either predefinite it contains of a set of predefinite resolvants inferences. by a clause iff by resolution, the variable of 2 clauses saturated of predefinite principle predefinite is subsumed saturation theorem. the C of resolvant is inconsistent produces, set prerequisite knowledge that E’ bird(Max) a set if each a tautology T U { bird(Max)}) if x is a bird know “bird(Max) without default The and If saturated extension unique a E’=Th(“fly(Max),bird(Max)=>fly(Max)) “bird(Max). Definition: ( ~ :M(x) bird(x)=>fly(x)) (x1 bird(x)=>fly(x) the not clauses produced. resolution the the number upon of Example F= ( -C 2: saturation of IV DEFAULT F = -C v E A The v D A is a free-default 1 =BvC C C = -B 3 = rtc 2, c3)= 4 default We 2 c Q C v D now prover only the consider = rtc cd)= D v E 1' 5 =AvC C default set. formula Let to Prerequisites(A’)=0 C BY SATURATION v D , A v C , -A I v E , B v C , -6 c PROOF theories A’ be be (A. T) a subset proved. where of A and Remark that thus (2’) T U CONS(A”) (3’) A” = 0 I- Prerequisites(6’) 6 c = -A 7 C It is worth-noting = rtc 8 c7)= = rtc has not cs)= E 1' 9 set C 6' C The default (cl, The empty .. .. cg } is saturated. been produced then the (§ II) are and (4) of the default and T U CONS(A’) by sets, clause these I =I B Saturation It is that later in the clause possible goal Let the we recursively the subsumed in sets the of example notion of predefinite clause 2. useless production. of F on E, produces defined a It is clauses To S(E,F) by E U (c} .c , F’ - (c} 1 the .r E: F’ if *or and r r is r literal process Example is not the resolvant A v B { } a clause upon a clause saturation S(E,F)=S( by the of E leftmost E. and A v B F={ } “A , ( S(0. ( A v B , -A ) A v B , -A , B Theorem 3: S(E,F) is finite iff Theorem 4: S(E,F) is saturated Corollary 5: S(0.C) B E and , ( ) ) , 0 1 1 6i, F are finite. equivalent to C. a if E is saturated. saturated logically T It then Then for proof During U ( -Qj This procedure over A’=( 61 ), the set of the successively S(0. T). T U CONS(62)), 62))... remark remains true, so: I= I SG(0.T). CONS(G1)) S(0, T U CONS(G2)) I= I S(S(0.T). CONS(B2)) T U CONS(G1. S(S(0, S(0. then, the every T U CONS(A’)) also CONEi are only contains or A’ subset contains the empty it. Due to this minimally the of A containing clause optimisation, inconsistent sets U T computed. By operating a global to compute consistent CONS(G2)) T U CONS( for it is useless 62)) T U CONS(GIH, example clause we never evaluates A’=0, for computes be example. (A’ ranges S(0. can it is useless CONS(A’). U The 62}... saturation proof of T U CONS(A’). T U CONStA’) of the T U CONS(6 1)) S(0. that of first. T U CONS( default be true). ( “Q]) computed and no addition, U S(0. If for and to S(0. resolution, T U CONS(G1. empty ) In of T U CONEi .. . . 6n}. A). I= I only is linear preceding 1 B twice. unsatisfiable one S(0. } is (4) cannot A’={61, } ( A v B , -A, that belongs (because T U CONS(G1)). The -A =S( clause stopped with of C-Q), are A={61, S(0. saturation T U CONS( consistency by sets. =S( =( of the of S(0, discovered subsets subsumed of c with is called 3: E=( E F of because computed of a refutation A’={s2}, *either (1) T U CONEi U ( “Q)) if Q is a consequence Let E F is the T U CONSCA') empty search use (3) of S(0. given have By be joined of To solve elaboration is to test the clause and we saturate T U CONS(A’)), T U CONS(A’) happens S(E.F)=S( This So if the can (2) verified. T U CONS(A’) no clauses found, S(E.B)=E where the process clauses. the set noted way steps U (-Q}. SG(0, the then reach saturation. variable this during SO is useless. of qhese resolvant extended by process by c number of F be The saturation clause saturation order have E and a subsumed c to reduie by relaxing this In obvious produced is the by sets the immediatly proof. steps S(0, F is consistent, set that proof like saturation, some this, S(0. subsets a default proof T U CONS(A) of these is nothing U (-Q}). clauses but knowing are really computed. Note also S(0. I=I S(0, for that S(S(0. and saturation remains by sets extensions of U {“Q}, T U CONS(A)), T U CONS(A)) all V CONCLUSION T U CONS(A) can then the same provides a default {“Q}) We be computed for a kind every once query. of compilation have and of default The contains of the axioms theory. Is prover with is complete. main Theorem 7: the B Example prover always of a default proof terminates. further by saturation the T={ C v D, E] 6 =:M-C v “E 1 “C v “E The This default theory 6 -:M-D v “E 2- “D v “E ({ 6 1, 62, U CONS(G1, 6s)) E2=Th(T U CONS(G2, 6s)) S1=S(O. theory T)={ S2=SW0. T), =S1 U { S3=S(S(0, T), 4 =S2 U { S5=S(s(0. T), S6=SWS(0, =S2 CONS(G1)), , 0 U { [Al 801 prover: [Bes } U { “D 2 S1 U { v “E } [ Bos The B v “E ) computed can since answer a query increasing of the For S set: in Ss=SIS4,CONS(6s)) contains the of clauses to take a default scsl. {-D}>={ S(S2. ( -D})=S(S1. are empty clause. to answer any S(Si, advantage proof of C v D , E , “D u not “C the efficiently compilation prover is not a direction the subset deal about for of formulae with. non-monotonic vol. 13, logic 1980 for Note 198, a non-monotonic University G. & Siegel, au of de R. & Hayes, trees in automatic Intelligence 4, I, 1983 P. secours la non-monotonie of Aix-Marseille Kowalski, logic Rennes pp. II, 1981 P. J. theorem-proving 87-101, 1969 (“Q)) [Rei To with of the D is given v “E , “E Reiter, R. for default i Intelligence by: } 30 Reiter, interacting Proc. inclusion , 0 811 On } , D v “E 803 A logic reasoning vol. 13, pp. 81-132. 1980 Now we query. (“D}) t 691 Artificial Si’s. example, is to deal the by sets are CONS(6s)) Q we evaluate order on University Machine of to We clause P. Bossu, Semantic } clauses use these Besnard. 811 [Kow U{Bv-E} remaining issue La saturation 1) [ Rei =s3 is able Intelligence. 831 CONS(6s)I CONS(G2)), saturation provides as extending Special Technical CONS<6 CONS(G1)), T). This Artificial } B v “E S7=S(S(S(0, updating. Thesis, CONS(6s))= T), to axiom and variable once However, only REFERENCES } = S1 prover updating, subset =:MB v “E B v “E 3 to the , D v “E CONS(B2)) “E default upon realised. A proof-procedure v “E T), of the a subset consequents terminates. been theorem-prover CONS( “C S =S(S(S(B, is submitted C v D , E PROLOG. 6 3 } , T) has 2 extensions: El=Th(T default 6 the based It in investigations for of this in predefinite always and has adapted all and advantage querying phase theories and Our theorem-prover complete The 6: the a theorem-prover default be transformed implementing Theorem The free-defaults can form. proposed logic. IJCAI-81 R. & Criscuolo. G. defaults pp. 270-276, Vancouver, 1981