theorem
  K is having_valuation implies
  for x being Element of ValuatRing v st x <> 0.K
  holds power(K).(x,n) <> 0.K
  proof
    assume
A1: K is having_valuation;
    let x be Element of ValuatRing v;
    reconsider y = x as Element of K by A1,Th51;
    power(K).(y,n) = y|^n;
    hence thesis by Th16;
  end;
