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Automated Theorem Proving by Wolfgang Bibel (auth.)

By Wolfgang Bibel (auth.)

Since either the coments and the constitution of the booklet looked to be winning, in basic terms minor alterations have been made. specifically, a few fresh paintings in ATP has been included in order that the ebook keeps to mirror the cutting-edge within the box. the main major switch is within the caliber of the format together with the elimination of a couple of inaccuracies and typing mistakes. R. Caferra, E. Eder, F. van der Linden, and J. Muller have stuck vanous minor error. P. Haddawy and S.T. Pope have supplied many stilistic advancements of the English textual content. final no longer least, A. Bentrup and W. Fischer have produced the gorgeous format. The vast paintings of typesetting was once financally supported inside of ESPRIT professional­ ject 415. Munchen, September 1986 W. Bibel PREFACE one of the goals of mankind is the single facing the mechanization of human notion. because the global this day has develop into so complicated that people it appears fail to regulate it accurately with their highbrow presents, the conclusion of this dream may be appeared at the same time whatever like a need. nevertheless, the incredi­ ble advances in machine expertise allow it seem as a true possibility.

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F, - F In normal form, if F hAUT F then 1= F 46 I I. PROOF. We use notation as m c . 5. Let LOGIC KG denote that literal K occurs in clause p lj through F is of the form through F, it is obvious that path p THE CONNECTION METHOD IN PROPOSITIONAL {Kc} for some KG and some F is complementary if F is complemen- tary. In order to show the converse, assume that F is complementary. , 1 Lr } ~ p and p is in p. {N,1 N } , respectively. We want to show that p {M,lM}={U,lLr} and complemenHowever, if {N,1N}={L'f,lU}, thus is complementary also in this case.

K~m} ~ d after dele- d={Kt" ... ,K~m ,K2~11 , ... ,K~n} , for and KjJ, l~i~m , l~j~n , l~m~n . For two matrices F, F in normal form we say that F is obtained from F by subsumption or SUBS-reduction, in symbols F F=F' U{c,d} and F=F'lj{c} for some F' ,c, hUBS F , if they are of the form and d such that c subsumes o d. L. iff For any two matrices F m normal form, if F hUBS F then 1= F 1= F . PROOF. 7, any path p lj {KG }U {U } where p lj {KG} F F, is complementary. plj {KG }lj {Kd } , thus also p through F IS of the form is a path through F.

But we are now in the position to bring the two together. T. A matrix is complementary iff it is valid. PROOF. We prove this statement by induction on the size a(F) of the matrix, denoted by F. 1 of a this means that F is a literal or the empty matrix. In the case F = (/J the theorem trivially holds. tl there is a model in which F is false. Hence F is not valid. p2 there is exactly one path through F, viz. {F) , which obviously is not complementary. 1 this means that F={(k h P1), ... ,(kn ,Pn )} for n literals (ki,Pi ), n ~ 1 , or F ={{},L1, ...

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