
theorem PSU:
  for a,b be non zero Integer holds Parity (a+b) = ((Parity a)+(Parity b)) iff
  (Parity a = Parity b) & parity ((Oddity a) div 2) = parity ((Oddity b) div 2)
  proof
    let a,b be non zero Integer;
    thus Parity (a+b) = ((Parity a)+(Parity b)) implies
    (Parity a = Parity b) &
      parity ((Oddity a) div 2) = parity ((Oddity b) div 2)
    proof
      assume
      A1: Parity (a+b) = ((Parity a) + (Parity b)); then
      A2: Parity a = Parity b by LEQ; then
      2*Parity a = Parity ((Parity a)*(Oddity a) + (Parity b)*(Oddity b)) by A1
      .= Parity ((Parity a)*((Oddity a)+ (Oddity b))) by A2
      .= (Parity (Parity a))*(Parity((Oddity a) + (Oddity b))) by ILP
      .= (Parity a)*Parity ((Oddity a)+(Oddity b));
      hence thesis by A1,LEQ,PPD,XCMPLX_1:5;
    end;
    assume
A1: (Parity a = Parity b) &
    parity ((Oddity a) div 2) = parity ((Oddity b) div 2);
    Parity (a+b) = Parity ((Parity a)*(Oddity a)+(Parity b)*(Oddity b))
    .= Parity ((Parity a)*((Oddity a)+ (Oddity b))) by A1
    .= (Parity(Parity a))*(Parity ((Oddity a)+(Oddity b))) by ILP
    .= 2*(Parity a) by A1,PPD;
    hence thesis by A1;
  end;
