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 Th60:
  0 in a & 1 in b & c = b-exponent a implies 0 in a div^ exp(b, c) 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 0 in n; then
    n = 0 by ORDINAL3:8; then
    a = (0 qua Ordinal)+^d by A2,ORDINAL2:35 .= d by ORDINAL2:30; then
    exp(b, c) c= d by A1,Def10; then
    d in d by A2;
    hence contradiction;
  end;
