
theorem Th51:
  for f being Polynomial of F_Complex st deg(f) >= 1 for rho being
  Element of F_Complex st Re(rho) < 0 holds f is Hurwitz implies F*(f,rho) div
  rpoly(1,rho) is Hurwitz
proof
  let f be Polynomial of F_Complex;
  assume
A1: deg(f) >= 1;
  let rho be Element of F_Complex;
  assume
A2: Re(rho) < 0;
  reconsider ef = eval(f,rho), ef1 = eval(f*',rho) as Element of F_Complex;
  eval(ef1 * f - ef * (f*'),rho) = eval(ef1*f,rho) - eval(ef*(f*'),rho) by
POLYNOM4:21
    .= ef1 * eval(f,rho) - eval(ef*(f*'),rho) by POLYNOM5:30
    .= ef1 * eval(f,rho) - ef * eval((f*'),rho) by POLYNOM5:30
    .= 0.F_Complex by RLVECT_1:15;
  then rho is_a_root_of (ef1 * f - ef * (f*')) by POLYNOM5:def 7;
  then consider s being Polynomial of F_Complex such that
A3: ef1 * f - ef * (f*') = rpoly(1,rho) *' s by Th33;
  assume
A4: f is Hurwitz;
  then |.eval(f,rho).| < |.eval(f*',rho).| by A1,A2,Th49;
  then ef1 * f - ef * (f*') is Hurwitz by A1,A4,Th50;
  then
A5: s is Hurwitz by A3,Th41;
  -1 < deg rpoly(1,rho) by Th27;
  then
A6: deg 0_.(F_Complex) < deg rpoly(1,rho) by Th20;
  ef1 * f - ef * (f*') = s *' rpoly(1,rho) + 0_.(F_Complex) by A3,POLYNOM3:28;
  hence thesis by A5,A6,Def5;
end;
