reserve G for Group,
  a,b for Element of G,
  m, n for Nat,
  p for Prime;

theorem Th2:
  for n,m being Nat holds n divides p |^ m implies
  ex r be Nat st n = p |^ r & r <= m
proof
  let n,m be Nat;
  assume
A1: n divides p |^ m;
  n > 0
  proof
    assume n <= 0;
    then n = 0;
    hence contradiction by A1,INT_2:3;
  end;
  then n in NatDivisors (p|^ m) by A1,MOEBIUS1:39;
  then n in {p|^ k where k is Element of NAT : k <= m} by NAT_5:19;
  then consider r be Element of NAT such that
A2: n = p|^ r & r <= m;
  take r;
  thus thesis by A2;
end;
