theorem ::: (p|^a)|^c = p|^b implies a divides b  contraposed
  for p be non trivial Nat holds
    not a divides b implies (p|^a)|^c <> p|^b
proof
  let p be non trivial Nat;
  assume
  A1: not a divides b;
  A2: p|^(a*c) = p|^b iff (p|^a)|^c = p|^b by NEWTON:9;
  p > 1 by Def0;
  hence thesis by A1,A2,PEPIN:30;
end;
