number-theory

8F Stretching, shrinking and completeness 135Define M* to be the result of replacing Bp+r in M by a copy of B,, (after changingbound variables in this copy to avoid clashes). Then M* has an argument-branchwith length d + r. (Its members areNo,...,Np+rlNpp+1,...,N°,where for i = 0,..., p + r each N` has the same position in M* as Ni had in M, andfor j = p + 1, ... , d we have N = Np) Hence by 8E4.1,Depth(M*) >d + r >d + 1.To complete the proof of (i) it only remains to show that M* is a genuine typedterm. This will be done by applying Lemma 5B2.1(ii) on replacement in typed terms.First, for i = 0, ... , d let F be the context that assigns to the initial abstractors ofNi the types they have in M. Since M has no bound-variable clashes the variablesin I'o,...,rd are all distinct, so(4) 1'o U ... U I'd is consistent.Also every variable free in B; is bound in one of No_., Ni because M is closed andBi is in Ni. Hence, by the definitions of typed term (5A1) and Con (5A4),(5) B1 E TT(FO U ... U ri), Con(B;) fo U ... U Fi.To apply 5B2.1(ii) it is enough to show that the set(6) Con(Bp) U Con(M) U FO U ... U Fp+ris consistent. (The abstractors in M whose scopes contain Bp+r are exactly the initialabstractors of N0....,Np+r.) But M is closed, so Con(M) = 0. And by (5),Thus (6) is consistent by (4).Con(Bp) s roU...Ufp s foU...UFp+r.8F3 Detailed Shrinking Lemma>- D(T) then(i) it has a memberM*Twith(cf. 8D3.) If Long(T) has a member Mt with depth(ii) it has a member NT withDepth(Mt) - IITII - 2 because T is composite since atomic types have no inhabitants.By 8E4.1, M has at least one argument-branch with length d; to reduce the depthof M we must shrink all these branches. Consider any such branch; just as in theproof of the stretching lemma 8F2 it has form(1) (No,...,Nd),