theorem
  K is having_valuation & 0 <= v.a implies normal-valuation(v).a <= v.a
  proof
    set f = normal-valuation(v);
    set l = least-positive(rng v);
    assume
A1: K is having_valuation;
    then
A2: v.a = f.a*l by Def10;
    assume 0 <= v.a;
    then
A3: 0 <= f.a by A1,Th41;
    0.qua ExtInt+1 <= l by Th10;
    then f.a*1 <= f.a*l by A3,Th3;
    hence thesis by A2,XXREAL_3:81;
  end;
