number-theory

3E The PT algorithm 47Let r - rl U r2. (This is a well-defined substitution since Ap and OQ were chosensuch that Vars(Ap) n Vars(OQ) = 0.) Then(11) Al = r(Ap), 02 = r(OQ),(12) Dom(r) = Vars(Op) U Vars(AQ).Now look at V1,...,Vr and a in (9). By comparing (9) and (1), since O1 - r(Op) wehave(13) (V1,...,Vr,a) = (r(W1),...,r(>Vr),r(P))And by comparing (10) and (12), since A2 =_ r(DQ),(14) (V1,...,Vr,a) = (r(XI),...,r(Xr),r(T))-Thus 41, ... , ivr, p) and (XI, ... , Xr, T) have a unifier, namely r.Justification of IVa2. We must prove that ApQ is principal for PQ, i.e. that anarbitrary deduction A of (8) is an instance of ApQ. For any such A, define Al, A2and r as above. By (6) and (12) we can express r as r =_ r' U r", where(15) r' - r V r" - r rpom(u).By (4), (13) and (14), r" is a unifier of(VP1,...,lPr,P),(XI,...,Xr,T).But u is an m.g.u. of this pair of sequences, so by the definition of m.g.u. (3D2) thereexists s such that r"(a) _ s(u(a)) for all a E Dom(u). We can clearly also assume(16) Dom(s) c Range(u).It follows that(17) r" =ext § o U.Since r - r' U r", we have by 3B4.1(ii),(18) r=extr'U(§ou).Now r', s, u satisfy conditions (i) and (ii) in the composition-extension lemma (3B6).In fact 3B6(i) holds because by (15) and (16) we haveDom(r') n (Dom(s) U Dom(u)) c V n (Range(u) U Dom(u))= 0and 3B6(ii) holds by (5) and (15). Hence by (18) and 3B6,Therefore by (11),Al - (r' U s)(u(Ap)),r =ext(G' U s) o U.A2 = (p U s)(u(AQ))by (5), (6),But A is a combination of Al and A2 by rule and ApQ is a similar combinationof u(Ap) and u(AQ). HenceA = (r' U s)(ApQ).