theorem
  K is having_valuation & v.a = 1 implies normal-valuation(v) = v
  proof
    set f = normal-valuation(v);
    assume that
A1: K is having_valuation and
A2: v.a = 1;
    let a be Element of K;
    thus v.a = (f.a)*least-positive(rng v) by A1,Def10
    .= f.a*1 by A2,Th34
    .= f.a by XXREAL_3:81;
  end;
