reserve
  a,b,c,d,e for Ordinal,
  m,n for Nat,
  f for Ordinal-Sequence,
  x for object;
reserve S,S1,S2 for Sequence;

theorem Th59:
  0 in a & 1 in b & c = b-exponent a implies a div^ exp(b, c) in b proof assume
A1: 0 in a & 1 in b & c = b-exponent a;
    set n = a div^ exp(b, c);
    exp(b,c) <> 0 by A1; then
    consider d such that
A2: a = n*^exp(b, c)+^d & d in exp(b, c) by ORDINAL3:def 6;
    assume not n in b; then
    b*^exp(b,c) c= n*^exp(b,c) by ORDINAL1:16,ORDINAL2:41; then
    exp(b, succ c) c= n*^exp(b,c) & n*^exp(b,c) c= a
    by A2,ORDINAL2:44,ORDINAL3:24; then
    exp(b, succ c) c= a; then
    succ c c= c & c in succ c by A1,Def10,ORDINAL1:6; then
    c in c;
    hence contradiction;
  end;
