number-theory

48 3 The principal-type algorithmSubcase IVb: M = PQ and PT(P) is atomic, say PT(P) = b. Then the conclusionsof A and AQ have form, respectively,(19) u1:B1,...,uP:O" W1:W1,...,wr:Wr i-- P:b,(20) v1:4)1,...,vq:4)q, W1:X1,...,Wr:Xr H Q:T.Choose any variable cto the pair of sequencesVars(SP) U Vars(OQ) and apply the unification algorithm(21) (Wl,...,Wr,b), (Xl,...,Xr,t +c).Subsubcase IVbl: the pair (21) has no unifier. Then PQ is not typable. [See thejustification below]Subsubcase IVb2: the pair (21) has a unifier. Then the unification algorithm givesan m.g.u. u; apply 3D2.5 to ensure that(22) Dom(u) = Vars(W1,(23) Range(u) r1V = 0,where V is the same as in (6). Then apply a to Op and L. By the definitionof u,u(b) = u(T-'c) = u(T)--'u(c),and thus the conclusions of u(Ap) and u(AQ) areu1 :e1 , ... , uP :BP , w1 :W1 , ... , Wr :W,* F a P : T* -+C*,vl:4l,...,vq:4) , W1:X1,...,Wr:Xr Q:T ,where * denotes application of the substitution u, andWI - Xl > ,Wr = XrChoose APQ to be the deduction obtained by applying ruleu(OQ); its conclusion isto u(Ap) andul:01 ,...,uP:o , v1 :01 ,...,vq:4q, W1:XI,...,Wr:Xr i-a PQ:C*.Justification of Subcase M. For IVb1 we must prove that if PQ is typable thenandhave a unifier, and for IVb2 we must prove thatOPQ is principal for PQ.Justification of IVbl. If PQ is typable there is a 0 whose conclusion has the form(8). By 2B2, 0 must have been built by applying (--*E) to two deductions Al and A2with conclusions (9) and (10). But P and Q have principal deductions Ap and AQ,so(24) Al = a'1(AP), A2 = r2(AQ)